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Unfortunately, there are two widely different definitions of the geometric distribution, with no clear consensus on which is to be used. Hence, the choice of definition is a matter of context and local convention. Fortunately, they are very similar. A series of Bernoulli trials is conducted until a success occurs, and a random variable
X
is defined as either
In either case, the geometric distribution is defined as the probability distribution of
X
Fortunately, these definitions are essentially equivalent, as they are simply shifted versions of each other. For this reason, the former is sometimes referred to as the shifted geometric distribution. In accordance with this convention, this article will use the latter definition for the geometric distribution; in particular,
X
represents the number of failures in the series of trials.
For example, consider rolling a fair die until a 1 is rolled. Rolling the die once is a Bernoulli trial, since there are exactly two possible outcomes (either a 1 is rolled or a 1 is not rolled), and their probabilities stay constant at
\frac{1}{6}
\frac{5}{6}
. The resulting number of times a 1 is not rolled is represented by the random variable
X
, and the geometric distribution is the probability distribution of
X
For a geometric distribution with probability
p
of success, the probability that exactly
k
failures occur before the first success is
(1-p)^kp.
\text{Pr}(X=k)
X
k
g(k;p)
, denoting the geometric distribution with parameters
k
p
This fact can also be observed from the above formula, as starting
k
from any particular value does not affect the relative probabilities of
X=k
. This is due to the fact that the successive probabilities form a geometric series, which also lends its name to the distribution.
The probability of success of a single trial is
\frac{1}{6}
, so the above formula can be used directly:
\begin{aligned} \text{Pr}(X=0) &= \bigg(\frac{5}{6}\bigg)^0\frac{1}{6} \approx .166\\ \text{Pr}(X=1) &= \bigg(\frac{5}{6}\bigg)^1\frac{1}{6} \approx .139\\ \text{Pr}(X=2) &= \bigg(\frac{5}{6}\bigg)^2\frac{1}{6} \approx .116\\ \text{Pr}(X=3) &= \bigg(\frac{5}{6}\bigg)^3\frac{1}{6} \approx .096\\ &\vdots \end{aligned}
This can also be represented pictorially, as in the following picture: The geometric distribution with
p=\frac{1}{6}
You are bored one day and decide to keep flipping an unfair coin until it lands on tails. It has a
60\%
chance of landing on heads. If you get tails on the
N^\text{th}
flip, the probability that
N
is an integer multiple of 3 can be expressed as
\frac{a}{b}
and
b
a+b.
The easiest to calculate is the mode, as it is simply equal to 0 in all cases, except for the trivial case
p=0
in which every value is a mode. This is due to the fact that
p>(1-p)^kp
p>0
The mean of a geometric distribution with parameter
p
\frac{1-p}{p}
\frac{1}{p}-1
The simplest proof involves calculating the mean for the shifted geometric distribution, and applying it to the normal geometric distribution. In the shifted geometric distribution, suppose that the expected number of trials is
E
. There is a probability
p
that only one trial is necessary, and a probability of
1-p
that an identical scenario is reached, in which case the expected number of trials is again
E
(this is a consequence of the fact that the distribution is memoryless). As such, the equation
E = p(1)+(1-p)(E+1) \implies E = (1-p)E+1
holds, so
E=\frac{1}{p}
As a result, the expected value of the number of failures before reaching a success is one less than the total number of trials, meaning that the expected number of failures is
\frac{1}{p}-1=\frac{1-p}{p}
_\square
Paddy is flipping a weighted coin, which displays heads with a probability of
\frac {1}{4}
. What is the expected number of coin flips he would need in order to get his first head?
Note that this makes intuitive sense: for example, if an event has a
\frac{1}{5}
probability of occurring per day, it is natural that to expect the event would occur in 5 days.
The variance of a geometric distribution with parameter
p
\frac{1-p}{p^2}
The geometric distribution has the interesting property of being memoryless. Let
X
be a geometrically distributed random variable, and
r
s
two positive real numbers. Then by this property
\text{P}(X>r+s | X>r) = {P}(X>s).
\begin{aligned} \text{Pr}(X=0)+\text{Pr}(X=1)+\text{Pr}(X=2) &=(0.7)^0(0.3)+(0.7)^1(0.3)+(0.7)^2(0.3)\\\\ &=0.657.\ _\square \end{aligned}
\begin{aligned} \text{Pr}(X=0)+\text{Pr}(X=1)+\text{Pr}(X=2)+\text{Pr}(X=3) &=(0.9)^0(0.1)+(0.9)^1(0.1)+(0.9)^2(0.1)+(0.9)^3(0.1) \\\\ &\approx 0.344.\ _\square \end{aligned} |
Spending and Tax Multipliers - Course Hero
Macroeconomics/Macroeconomics Models/Spending and Tax Multipliers
The spending multiplier is one of the key concepts of Keynesian economics. The spending multiplier is the factor by which gains in total output are greater than the change in spending that caused it. The spending multiplier is a formula for determining how much an increase in government spending will affect the economy. This concept is a way to describe how government intervention impacts the overall economy by its changes in spending. A fiscal injection magnifies the initial amount spent by the government in the economy due to the multiplier effect. Thus, for each dollar the government invests, the impact on the economy is multiplied. For instance, if the government invests $1 million on an infrastructure project, it can be seen as contributing more than $1 million to the economy. This creates a domino effect because additional workers must be hired for the project, who will receive incomes. In turn, their incomes are spent elsewhere, and that money spent becomes another person's income, and so on. The benefit of that project expenditure, then, is multiplied. The use of government spending adds to output because it contributes to the workers' disposable income, and the workers can either spend it or save it. The increase in disposable income raises consumption, which increases gross domestic product (GDP). According to the spending multiplier concept, the consumption, as a result of spending, will further increase future consumption and additionally increase GDP.
The numerical value of the multiplier is dependent on marginal propensity to consume (MPC) and marginal propensity to save (MPS). The basic equation for the spending multiplier is:
\text{Spending Multiplier}=\frac {1}{1-\text{MPC}}
MPC is the marginal propensity to consume. If a nation, on average, receives $20 and spends $16, the MPC would be 0.8. Then, following the equation, 1 is divided by
(1 - 0.8)
, or 1 is divided by 0.2. When 1 is divided by 0.2, the multiplier is 5. In this scenario, therefore, the economy benefited five times more than the monetary amount of the spending. The multiplier concept is helpful in many stimulus scenarios and is not limited to direct cash payments. A stimulus is the use of expansionary fiscal or monetary policy to help kickstart economic growth in a sluggish or depressed economy.
The multiplier effect implies that the government should invest money in the economy during an economic downturn. From the Keynesian viewpoint, when the economy is underperforming, government intervention is crucial to its recovery. Keynes argued that if the government curbs spending in an economic depression, which might be a natural choice when revenue is smaller, the economy could weaken further. The domino effect from the spending multiplier helps support the decision for a government to spend money, even during recessionary times. The multiplier concept influenced policy-making decisions in developed countries in the 20th century and continues to influence government policy today.
The tax multiplier is the effect on the economy from changes in tax policy. This multiplier is different than spending multipliers. Its effects are much smaller than spending multipliers. This is because when the government lowers taxes, it is not actually injecting new income into the economy as new spending would because consumers can choose to either spend or save. Conversely, the spending multiplier is a direct injection of spending. The tax multiplier is calculated as the negative MPC divided by the MPS, which can also be written as 1 minus the MPC.
\text{Tax Multiplier}=\frac{-\text{MPC}}{(1-\text{MPC})}
For example, if the government decides to increase expenditures and spend $10 million on a project, that money is injected in the economy. With an MPC of 0.8, the spending multiplier was shown to be 5. The total impact would be $50 million in the economy with a $10 million government spending project. If the total amount of tax dollars is reduced by $10 million, only the amount ($10 million) multiplied by the tax multiplier will be spent in the economy. Because a reduction in taxes leads to an increase in income, the relationship is negative. Thus, the tax multiplier is dependent on what amount a consumer will spend or save with the tax cut. In this example where
\text{MPC} = 0.8
, the tax multiplier is calculated as
-0.8/(1-0.8)
. Multiplying the value by the $10 million spent, the total injection from a tax cut of $10 million would be $40 million.
-\$10\text{,}000\times\left(\frac{-0.8}{1-0.8}\right)=-\$10\text{,}000\times-4=\$40\text{,}000
<Consumption Function>The Aggregate Expenditure Model |
NCERT Solutions Class 12 Physics Chapter 8 Electromagnetic Waves
The NCERT Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves provide detailed answers to textbook theory questions, numerical problems, worksheets and exercises. In Class 12 Physics, there are many complicated formulas and equations. In order to score good marks in the Class 12 term – II examination, it is important to solve the exercise questions provided at the end of each chapter using the NCERT Solutions for Class 12 Physics.
Most frequently, questions that are asked in CBSE Class 12 Physics term-wise exams directly appear from the NCERT textbook. Electromagnetism is one of the most frequently asked topics in the second term exam. Hence, students are suggested to refer to the NCERT Solutions for Class 12 Physics to attained a firm grip over the chapter as well as the subject. Now, download the NCERT Solutions for Class 12 Physics Chapter 8 from the link mentioned below.
Class 12 Physics NCERT Solutions Electromagnetic Waves Important Questions
Q 8.1) The Figure shows a capacitor made of two circular plates each of radius 12 cm and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of change of the potential difference between the plates.
The radius of each circular plate (r) is 12 cm or 0.12 m
The distance between the plates (d) is 5 cm or 0.05 m
The charging current (I) is 0.15 A
The permittivity of free space is
\varepsilon_{0} = 8.85\times 10^{-12}\; C^{2}N^{-1}m^{-2}
(a) The capacitance between the two plates can be calculated as follows:
C = \frac{\varepsilon _{0} A}{d}
A = Area of each plate =
\pi r^{2}
C = \frac{\varepsilon _{0} \pi r^{2}}{d}
\frac{8.85\times 10^{-12}\times \pi (0.12)^{2}}{0.05}
8.0032\times 10^{-12}\; F
= 80.032 pF
The charge on each plate is given by,
V is the potential difference across the plates
\frac{\mathrm{d} q}{\mathrm{d} t} = C \frac{\mathrm{d} V}{\mathrm{d} t}
\frac{\mathrm{d} q}{\mathrm{d} t}
= Current (I)
∴\frac{\mathrm{d} V}{\mathrm{d} t} = \frac{I}{C}
\frac{0.15}{80.032\times 10^{-12}} = 1.87\times 10^{9}\; V/s
Therefore, the change in the potential difference between the plates is
1.87\times 10^{9}\; V/s
Q 8.2) A parallel plate capacitor made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with an (angular) frequency of 300 rad s–1.
Capacitance of a parallel plate capacitor, C = 100 pF =
100\times 10^{-12}\; F
\omega = 300\; rad\;s^{-1}
(a) Rms value of conduction current, I =
\frac{V}{X_{c}}
X_{c}
= Capacitive reactance =
\frac{1}{\omega C}
∴ I = V\times \omega C
230\times 300\times 100\times 10^{-12}
6.9\times 10^{-6} A
6.9\; \mu A
Hence, the rms value of conduction current is
6.9\; \mu A
(b) Yes, conduction current is equivalent to displacement current.
B = \frac{\mu_{0}r}{2\pi R^{2}}I_{0}
\mu_{0}
4\pi \times 10^{-7}\; N\;A^{-2}
I_{0}
= Maximum value of current =
\sqrt{2}\; I
∴B = \frac{4\pi\times 10^{-7}\times 0.03\times \sqrt{2}\times 6.9\times 10^{-6}}{2\pi \times (0.06)^{2}}
1.63\times 10^{-11}\; T
Hence, the magnetic field at that point is
1.63\times 10^{-11}\; T
Q 8.3) What physical quantity is the same for X-rays of wavelength 10–10m, the red light of wavelength 6800 Å and radiowaves of wavelength 500m?
The speed of light (
3\times 10^{8}
m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.
Q 8.4) A plane electromagnetic wave travels in vacuum along the z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-y plane. They are mutually perpendicular.
Frequency of the wave, v = 30 MHz =
30\times 10^{6}\;s^{-1}
Speed of light in vacuum, C =
3\times 10^{8}
\lambda = \frac{c}{v}
\frac{3\times 10^{8}}{30\times 10^{6}}
Q 8.5) A radio can tune in to any station in the 7.5 MHz to 12 MHz bands. What is the corresponding wavelength band?
A radio can tune to minimum frequency,
v_{1} = 7.5\; MHz = 7.5\times 10^{6}\; Hz
Maximum frequency,
v_{2} = 12\; MHz = 12\times 10^{6}\; Hz
Speed of light, c =
3\times 10^{8}\; m/s
Corresponding wavelength for
v_{1}
\lambda_{1} = \frac{c}{v_{1}}
=\frac{3\times 10^{3}}{7.5\times 10^{6}} = 40\;m
v_{2}
\lambda_{2} = \frac{c}{v_{2}}
=\frac{3\times 10^{3}}{12\times 10^{6}} = 25\;m
Q 8.6) A charged particle oscillates about its mean equilibrium position with a frequency of 109 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Q 8.7) The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0=510 nT. What is the amplitude of the electric field part of the wave?
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
B_{0} = 510\; nT = 510\times 10^{-9}\; T
3\times 10^{8}\; m/s
Amplitude of electric field of an electromagnetic wave is given by the relation,
E = cB_{0} = 3\times 10^{8}\times 510\times 10^{-9} = 153\; N/C
Q 8.8) Suppose that the electric field amplitude of an electromagnetic wave is
E_{0} = 120\; N/C
and that its frequency is v = 50 MHz.(a) Determine
B_{0},\; \omega,\; k\;and\; \lambda
(b) Find expressions for E and B.
Electric field amplitude,
E_{0} = 120\; N/C
Frequency of source, v = 50 MHz =
50\times 10^{6}
3\times 10^{8}
B_{0} = \frac{E_{0}}{c}
\frac{120}{3\times 10^{8}}
40\times 10^{-8}\;=400\times 10^{-9} T = 400\; nT
Angular frequency of source is given by:
\omega =2\pi v=2\pi \times 50\times 10^{6}=3.14\times 10^{8}\,rads^{-1}
3.14\times 10^{8}
k = \frac{\omega }{c}
\frac{3.14\times 10^{8}}{3\times 10^{8}} = 1.05\; rad/m
Wavelength of wave is given by:
\lambda = \frac{c}{v}
\frac{3\times 10^{8}}{50\times 10^{6}}
\overline{E} = E_{0}\;sin(kx – \omega t)\;\widehat{j}
120\;sin[1.05x – 3.14\times 10^{8}t]\;\widehat{j}
\overline{B} = B_{0}\;sin(kx – \omega t)\;\widehat{k}
\overline{B} = (400 \times 10^{-9}) sin[1.05x – 3.14\times 10^{8}t]\;\widehat{k}
Q 8.9) The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for
different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
The energy of a photon is given as:
E = hv =
\frac{hc}{\lambda}
h = Planck’s constant =
6.6\times 10^{-34}\;Js
c = Speed of light =
3\times 10^{8}\;m/s
If the wavelength λ is in metre and the energy is in Joule, then by dividing E by 1.6 × 10-19 will convert the energy into eV.
E=\frac{hc}{\lambda \times 1.6\times 10^{-19}}\,eV
a) For Gamma rays, the wavelength ranges from 10-10 to 10-14 m, therefore the photon energy can be calculated as follows:
E=\frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-10}\times 1.6\times 10^{-19}}=12.4\times 10^3\approx 10^4\,eV
\lambda =10^{-10}\,m, energy = 10^{4}\,eV
\lambda =10^{-14}\,m, energy = 10^{8}\,eV
The energy for Gamma rays ranges from 104 to 108 eV.
b) The wavelength for X-rays ranges between 10-8 m to 10-13 m
For λ = 10-8,
Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{10^{-8}\times 1.6\times 10^{-19}}=124\approx 10^2\,eV
For λ = 10-13 m, energy = 107 eV
c) For ultraviolet radiation, the wavelength ranges from 4 × 10-7 m to 6 × 10-7 m.
For 4 × 10-7 m,
Energy = \frac{6.62\times 10^{-34}\times 3\times 10^8}{4\times 10^{-7}\times 1.6\times 10^{-19}}=3.1\approx 10^{10}\,eV
For 6 × 10-7 m, the energy is equal to 103 eV.
The energy of the ultraviolet radiation varies between 1010 to 103 eV.
d) For visible light, the wavelength ranges from 4 × 10-7 m to 7 × 10-7 m.
For 4 × 10-7, the energy is the same as above, that is 1010 eV
For 7 × 10-7 m, the energy is 100 eV
e) For infrared radiation, the wavelength ranges between 7 × 10-7 m to 7 × 10-14 m.
The energy for 7 × 10-7 m is 100 eV
The energy for 7 × 10-14 m is 10-3 eV
f) For microwaves, the wavelength ranges from 1 mm to 0.3 m.
For 1 mm, the energy is 10-3 eV.
For 0.3 m, the energy is 10-6 eV.
g) For radio waves, the wavelength ranges from 1 m to few km.
For 1 m, the energy is 10-6 eV.
The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source
Q 8.10) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1.
(c) Show that the average energy density of the E field equals the average energy density of the B field. [ c =
3\times 10^{8}\;m\;s^{-1}
Frequency of the electromagnetic wave, v =
2\times 10^{10}\;Hz
E_{0} = 48\;V\;m^{-1}
3\times 10^{8}\;m/s
\lambda = \frac{c}{v}
\frac{3\times 10^{8}}{2\times 10^{10}} = 0.015\; m
B_{0} = \frac{E_{0}}{c}
\frac{48}{3\times 10^{8}} = 1.6\times 10^{-7}\; T
U_{E} = \frac{1}{2}\; \epsilon _{0} \;E^{2}
U_{B} = \frac{1}{2\mu_{0}}B^{2}
\epsilon _{0}
\mu_{0}
E = cB …(1)
c = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}
E = \frac{1}{\sqrt{\epsilon_{0}\; \mu_{0}}}\; B
E^{2} = \frac{1}{\epsilon_{0}\; \mu_{0}}\; B^{2}
\epsilon_{0}\; E^{2} = \frac{B^{2}}{\mu_{0}}
\frac{1}{2}\; \epsilon_{0}\; E^{2} = \frac{1}{2}\; \frac{B^{2}}{\mu_{0}}
U_{E} = U_{B}
Q 8.11) Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 106 rad/s)t]} ˆi .
(a) The direction of motion is along the negative y-direction. i.e., along -j.
(b) The given equation is compared with the equation,
E = E0 cos (ky + ωt)
⇒ k = 1.8 rad/s
ω = 5.4 x 106 rad/s
λ = 2π/k = (2 x 3.14)/1.8 = 3.492 m
(c) Frequency, ν = ω/2π = 5.4 x 106/(2 x 3.14) = 0.859 x 106 Hz
(d) Amplitude of the magnetic field, B0 = E0/c
= 3.1/(3 x 108) = 1.03 x 10-8 T= 10.3 x 10-9 T= 10.3 nT
(e) Bz = B0 cos (ky + ωt)ˆk ={(10.3 nT) cos[(1.8 rad/m)y + (5.4 × 106 rad/s)t]} kˆ
Q 8. 12) About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) Average intensity of the visible radiation, I = P’/4πd2
Here, the power of the visible radiation, P’ = (5/100) x 100 = 5 W
At d = 1 m
I = P’/4πd2 = 5/(4 x 3.14 x 12) = 5/12.56 = 0.39 W/m2
(b) At d = 10 m
I = P’/4πd2 = 5/(4 x 3.14 x 102) = 5/1256 = 0.39 x 10-2 W/m2
Q 8.13) Use the formula λ m T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
We have the equation, λ m T = 0.29 cm K
⇒ T = (0.29/λ m )cm K
Here, T is the temperature
λ m is the maximum wavelength of the wave
For λ m = 10-4 cm
T = (0.29/10-4)cm K = 2900 K
For the visible light, λ m = 5 x 10-5 cm
T = (0.29/ 5 x 10-5 )cm K ≈ 6000 K
Note: a lower temperature will also produce wavelength but not with maximum intensity.
Q 8. 14) Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(e) 14.4 keV [energy of a particular transition in 57Fe nucleus
associated with a famous high-resolution spectroscopic method (Mössbauer spectroscopy)].
(a) Radio waves (short-wavelength end)
(b) Radio waves (short-wavelength end)
(d) Visible light (Yellow)
(e) X-rays (or soft γ-rays) region
Q 8.15) Answer the following questions:
(a) Ionosphere reflects waves in the shortwave bands.
(b) Television signals have high frequency and high energy. Therefore, it is not properly reflected by the ionosphere. Satellites are used to reflect the TV signals.
(c) Atmosphere absorbs X-rays, while visible and radio waves can penetrate it.
(d) Ozone layer absorbs the ultraviolet radiations from the sunlight and prevents it from reaching the surface of the earth and causing damage to life.
(e) If the atmosphere is not present, there would be no greenhouse effect. As a result, the temperature of the earth would decrease.
(f) The smoke clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’.
Chapter 8 Electromagnetic Waves of Class 12 Physics is categorized under the term – II CBSE Syllabus for 2021-22. Electromagnetism is a physical attraction that occurs in electrically charged particles. When a capacitor is charged using an external source, there can be a potential difference between two capacitive plates, we will show how to calculate that along with the displacement. This question will be solved using Kirchhoff’s rules. Students can make use of the NCERT Solutions for Class 12 to learn the correct methods of solving the exercise problems.
Concepts involved in Class 12 Physics Chapter 8 Electromagnetic Waves
We will determine the RMS value of the conduction current and we will be analysing the similarities between conduction current and displacement current. We will be analysing the similarities among the wavelengths of X-rays, red lights and radio waves. In this solution, you will be seeing questions on the wavelength of electromagnetic waves traveling in a vacuum.
Do you want to know what the frequency of electromagnetic waves produced by the oscillator is? Want to know about the electric field part of the harmonic electromagnetic wave in a vacuum? Check out the answers in the NCERT Solutions. We will be obtaining photo-energy of different parts of the electromagnetic spectrum and perceiving how to obtain different scales of photon energies of electromagnetic radiation.
We will be gaining knowledge on how to prove that the energy density of one field is equal to the average energy density of another field. We know that there are more fundamental forces such as weak and strong nuclear force and gravitational force. You will be finding questions on them in a different chapter. The questions mentioned in this chapter are very common in the second term exam and if prepared thoroughly, will definitely help you understand electromagnetism with ease.
BYJU’S provides class-wise NCERT Solutions, along with study materials, notes, books, assignments and sample papers prepared by top-notch subject experts of the country who have been involved in teaching the CBSE Syllabus for decades.
Do NCERT Solutions for Class 12 Physics Chapter 8 have answers for all the textbook questions?
The NCERT Solutions for Class 12 Physics Chapter 8 is available in PDF format designed by the subject experts. These solutions are completely based on the latest term – II CBSE Syllabus 2021-22 and cover all the important concepts for the exam. The textbook problems are solved in a stepwise manner as per the marks weightage in the second term exam. Both chapter wise and exercise wise PDF links are present in BYJU’S which can be accessed by the students to get their doubts clarified instantly.
How can we score full marks in NCERT Solutions for Class 12 Physics Chapter 8?
The NCERT Solutions for Class 12 Physics Chapter 8 are designed by experts at BYJU’S after conducting vast research on each concept. Every minute detail is explained in a comprehensive manner to help students score well in the class test as well as in term – II exams. It also helps students in doing their assignments given to them on time without any difficulty.
Are NCERT Solutions for Class 12 Physics Chapter 8 PDF enough to score well in the term – II exams?
NCERT Solutions for Class 12 Physics Chapter 8 are available in PDF format which can be downloaded and used by the students without any time constraints. The solutions are created by the highly experienced faculty at BYJU’S based on the latest term – II CBSE Syllabus and its guidelines. The exercise-wise solutions help students to gain an overall idea about the concepts which are important for the second term exams. Practising these questions on a regular basis will improve the time management and problem-solving abilities of students.
NCERT Grade 9 Maths Solutions NCERT Solutions For Class 9 Chemistry |
Divide array by vector along specified dimension - Simulink - MathWorks United Kingdom
Divide array by vector along specified dimension
The Array-Vector Divide block divides the values in the specified dimension of the N-dimensional input array A by the values in the input vector V.
The length of the input V must be the same as the length of the specified dimension of A. The Array-Vector Divide block divides each element of V by the corresponding element along that dimension of A.
Consider a 3-dimensional M-by-N-by-P input array A(i,j,k) and an N-by-1 input vector V. When the Divide along dimension parameter is set to 2, the output of the block Y(i,j,k) is
Y\left(i,j,k\right)=\frac{A\left(i,j,k\right)}{V\left(j\right)}
\begin{array}{l}1\le i\le M\\ 1\le j\le N\\ 1\le k\le P\end{array}
The output of the Array-Vector Divide block is the same size as the input array, A. This block accepts real and complex floating-point and fixed-point input arrays, and real floating-point and fixed-point input vectors.
The following diagram shows the data types used within the Array-Vector Divide block for fixed-point signals.
You can set the vector and output data types in the block dialog.
Divide along dimension
Specify the dimension along which to divide the input array A by the elements of vector V.
Click the Show data type assistant button to display the Data Type Assistant, which helps you set the Vector (V) data type parameter.
Array-Vector Add | Array-Vector Multiply | Array-Vector Subtract |
Computing Residual Income
Intrinsic Value w/Residual Income
Residual Income Method FAQs
There are many different methods of valuing a company or its stock. One could opt to use a relative valuation approach, comparing multiples and metrics of a firm in relation to other companies within its industry or sector. Another alternative would be valuing a firm based upon an absolute estimate, such as implementing discounted cash flow (DCF) modeling or the dividend discount method, in an attempt to place an intrinsic value on said firm.
One absolute valuation method which may not be so familiar to most, but is widely used by analysts, is the residual income method. In this article, we will introduce you to the underlying basics behind the residual income method and how it can be used to place an absolute value on a firm.
Residual income is the income a company generates after accounting for the cost of capital.
The residual income valuation formula is very similar to a multistage dividend discount model, substituting future dividend payments for future residual earnings.
Residual income models make use of data readily available from a firm's financial statements.
These models look at the economic profitability of a firm rather than just its accounting profitability.
An Introduction to Residual Income
When most hear the term residual income, they think of excess cash or disposable income. Although that definition is correct in the scope of personal finance, in terms of equity valuation residual income is the income generated by a firm after accounting for the true cost of its capital.
You might be asking, "but don't companies already account for their cost of capital in their interest expense?" Yes and no. Interest expense on the income statement only accounts for a firm's cost of its debt, ignoring its cost of equity, such as dividend payouts and other equity costs.
Looking at the cost of equity another way, think of it as the shareholders' opportunity cost, or the required rate of return. The residual income model attempts to adjust a firm's future earnings estimates to compensate for the equity cost and place a more accurate value on a firm. Although the return to equity holders is not a legal requirement, like the return to bondholders, in order to attract investors firms must compensate them for the investment risk exposure.
In calculating a firm's residual income, the key calculation is to determine its equity charge. Equity charge is simply a firm's total equity capital multiplied by the required rate of return of that equity, and can be estimated using the capital asset pricing model (CAPM).
Computing Residual Income and the Equity Charge
The formula below shows the equity charge equation:
Once we have calculated the equity charge, we only have to subtract it from the firm's net income to come up with its residual income. For example, if Company X reported earnings of $100,000 last year and financed its capital structure with $950,000 worth of equity at a required rate of return of 11%, its residual income would be:
Equity Charge - $950,000 x 0.11 = $104,500
As you can see from the above example, using the concept of residual income, although Company X is reporting a profit on its income statement, once its cost of equity is included in relation to its return to shareholders, it is actually economically unprofitable based on the given level of risk. This finding is the primary driver behind the use of the residual income method. A scenario where a company is profitable on an accounting basis, may still not be a profitable venture from a shareholder's perspective if it cannot generate residual income.
Given the opportunity cost of equity, a company can have positive net income but negative residual income.
Intrinsic Value With Residual Income
Now that we've found how to compute residual income, we must now use this information to formulate a true value estimate for a firm. Like other absolute valuation approaches, the concept of discounting future earnings is put to use in residual income modeling as well. The intrinsic, or fair value, of a company's stock using the residual income approach, can be broken down into its book value and the present values of its expected future residual incomes, as illustrated in the formula below.
\begin{aligned} &\text{V}_0 = BV_0 + \left \{ \frac {RI_1}{(1+r)^n} + \frac {RI_2}{(1+r)^{n+1}} + \cdots \right \}\\ &\textbf{where:}\\ &\textit{BV} = \text{Present book value}\\ &\textit{RI} = \text{Future residual income}\\ &\textit{r} = \text{Rate of return}\\ &\textit{n} = \text{Number of periods}\\ \end{aligned}
V0=BV0+{(1+r)nRI1+(1+r)n+1RI2+⋯}where:BV=Present book valueRI=Future residual incomer=Rate of returnn=Number of periods
As you may have noticed, the residual income valuation formula is very similar to a multistage dividend discount model, substituting future dividend payments for future residual earnings. Using the same basic principles as a dividend discount model to calculate future residual earnings, we can derive an intrinsic value for a firm's stock. In contrast to the DCF approach which uses the weighted average cost of capital for the discount rate, the appropriate rate for the residual income strategy is the cost of equity.
What Are the Pros of the Residual Income Method?
The residual income approach offers both positives and negatives when compared to the more often used dividend discount and discounted cash flows (DCF) methods. On the plus side, residual income models make use of data readily available from a firm's financial statements and can be used well with firms that do not pay dividends or do not generate positive free cash flow. Most importantly, as we discussed earlier, residual income models look at the economic profitability of a firm rather than just its accounting profitability.
What Are the Limitations of the Residual Income Method?
The biggest drawback of the residual income method is the fact that it relies so heavily on forward-looking estimates of a firm's financial statements, leaving forecasts vulnerable to psychological biases or historic misrepresentation of a firm's financial statements.
How Is a Firm's Residual Income Calculated?
Residual income is calculated as a company's net income less a charge for its cost of capital (known as the equity charge). The equity charge is computed from the value of equity capital multiplied by the cost of equity (often its required rate of return).
The residual income valuation approach is a viable and increasingly popular method of valuation and can be implemented rather easily by even novice investors. When used alongside the other popular valuation approaches, residual income valuation can give you a clearer estimate of the true intrinsic value of a firm maybe.
Bromwich, Michael, and Martin Walker. "Residual income past and future." Management Accounting Research, Vol. 9, No. 4, 1998, Pages 391-419. |
If f is continuous and integral 0 to 4 f(x)dx=10
If f is continuous and integral 0 to 4 f(x)dx=10, find integral 0 to 2
If f is continuous and integral 0 to 4
f\left(x\right)dx=10
, find integral 0 to 2
f\left(2x\right)dx
We will start working with the integral
{\int }_{0}^{2}f\left(2x\right)dx
2x=u
2dx=du
=\frac{1}{2}{\int }_{0}^{2}f\left(2x\right)\left(2dx\right)
Limits of integration will change from
{\int }_{0}^{2}
{\int }_{2×0}^{2×2}={\int }_{0}^{4}
\frac{1}{2}{\int }_{0}^{4}f\left(u\right)du
{\int }_{a}^{b}f\left(u\right)du={\int }_{a}^{b}f\left(x\right)dx
=\frac{1}{2}{\int }_{0}^{4}f\left(x\right)dx
{\int }_{0}^{4}f\left(x\right)dx=10
=\frac{1}{2}×10=5
{\int }_{0}^{2}f\left(2x\right)dx=5
\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0
x={t}^{2},y=2t,0\le t\le 5
\int 2{x}^{3}\mathrm{cos}\left({x}^{2}\right)dx
Provided a function that would be defined fo every natural number n as:
g\left(n\right)=\frac{2}{2}+\frac{1}{4}+\frac{1}{8}+\cdot \cdot \cdot +\frac{1}{{2}^{n-1}}+\frac{1}{{2}^{n}}
COnjecture a closed form form for g(n) (a simple expression devoid of signs or dots) while also using mathematical induction to prove that the conjecture is true.
Use a double integral to find the area of the region. The region inside the cardioid
r=1+\mathrm{cos}\theta
r=3\mathrm{cos}\theta
\int \frac{\sqrt{{x}^{2}-25}}{x}dx
Use geometry or symmetry, or both, to evaluate the double integral.
\int {\int }_{D}\sqrt{{R}^{2}-{x}^{2}-{y}^{2}}dA
D is the disk with center the origin and radius R. |
Equilibrium Price: Skateboards The demand for your factory-made sk
Antinazius 2021-11-15 Answered
Equilibrium Price: Skateboards
The demand for your factory-made skateboards, in weekly sales, is
q=-7p+50
If the selling price is $p. If you are selling them at that price, you can obtain
q=3p-30
Per week from the factory. At what price should you sell your skateboards so that there is neither a shortage nor a surplus? (Round your answer to the nearest cent.)
q=-7p+50
q=3p-30
To find: Price, p at which neither shortage nor surplus.
Analysis: at the equilibrium price,
\text{demand}=\text{supply}
-7p+50=3p-30
80=10p
8=p
p=8
Answer: the equilibrium price is $8.
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
Suppose X is a random variable such that
E\left(3X-7\right)=8
E\left(\frac{{X}^{2}}{2}\right)=19
What is Var
\left(70-2X\right)
\frac{dy}{dx}=x\sqrt{1-{y}^{2}}
Use a Numerical approach (i,e., a table of values) to approximate
\underset{x\to 0}{lim}\frac{{e}^{2}x-1}{x},
Provide support for your analysis (slow the table and explain your resoning).
learning features of expeirmental research in my research methods course
Why is randomization the best method for dealing with extraneous variables?
Plot each complex number. Then write the complex number in polar form. You may express the argument in degrees or radians.
1-i
-2\sqrt{3}+2i
-3-4i |
Distinct modes of collagen type I proteolysis by matrix metalloproteinase (MMP) 2 and membrane type I MMP during the migration of a tip endothelial cell: insights from a computational model — Physiome Model Repository
You are here: Home / Exposures / Karagiannis, Popel, 2006 / Distinct modes of collagen type I proteolysis by matrix metalloproteinase (MMP) 2 and membrane type I MMP during the migration of a tip endothelial cell: insights from a computational model
Component: Model
ddtM2=-kM2T2_onM2T2+kM2T2_offM2T2-kM2C1_onM2C1+kM2C1_off+kM2C1_catM2C1+kM2_actMT1T2M2proMT1ddtMT1=qMT1-kMT1_shedeffMT1MT1-kMT1T2_onMT1T2+kMT1T2_offMT1T2-kMT1C1_onMT1C1+kMT1C1_off+kMT1C1_catMT1C1-kMT1T2M2proMT1_onMT1MT1T2M2pro+kMT1T2M2proMT1_offMT1T2M2proMT1+kM2_actMT1T2M2proMT1ddtMT1_t=kMT1_shedeffMT1MT1ddtMT1C1=kMT1C1_onMT1C1-kMT1C1_off+kMT1C1_catMT1C1ddtMT1T2=kMT1T2_onMT1T2-kMT1T2_offMT1T2-kMT1T2M2pro_onMT1T2M2_p+kMT1T2M2pro_offMT1T2M2proddtMT1T2M2pro=kMT1T2M2pro_onMT1T2M2_p-kMT1T2M2pro_offMT1T2M2pro-kMT1T2M2proMT1_onMT1MT1T2M2pro+kMT1T2M2proMT1_offMT1T2M2proMT1ddtMT1T2M2proMT1=kMT1T2M2proMT1_onMT1MT1T2M2pro-kMT1T2M2proMT1_offMT1T2M2proMT1-kM2_actMT1T2M2proMT1ddtC1=-kMT1C1_onMT1C1+kMT1C1_offMT1C1-kM2C1_onM2C1+kM2C1_offM2C1ddtC1_D=kM2C1_catM2C1+kMT1C1_catMT1C1ddtMT1_cat=kMT1_shedeffMT1MT1ddtT2=-kM2T2_onM2T2+kM2T2_offM2T2+qT2-kMT1T2_onMT1T2+kMT1T2_offMT1T2ddtM2_p=qpro-kMT1T2M2pro_onMT1T2M2_p+kMT1T2M2pro_offMT1T2M2proddtM2T2=kM2T2_onM2T2-kM2T2_offM2T2-kM2T2_isoM2T2+kM2T2_negativeisoM2T2_starddtM2T2_star=kM2T2_isoM2T2-kM2T2_negativeisoM2T2_starddtM2C1=kM2C1_onM2C1-kM2C1_off+kM2C1_catM2C1ddtMT1T2_star=kM2_actMT1T2M2proMT1 |
1 Evaluating Negative Exponents
2 Completing Equations with Negative Exponents
Exponents tell you how many times any given number is multiplied by itself. For example, if you see
{\displaystyle 3^{3}}
, you know that you are going to multiply
{\displaystyle 3}
{\displaystyle 3}
times, which comes out to be
{\displaystyle 27}
. Negative exponents, on the other hand, tell you how many times you should divide by a number that is being multiplied by itself. Negative exponents can be written as
{\displaystyle 2^{-2},{\frac {(2^{-2})}{1}},{\frac {1}{(2^{2})}},}
{\displaystyle {\frac {1}{2x2}}}
. Negative exponents must become positive before an equation can be simplified. While it might seem tricky to get the hang of, calculating negative exponents is a simple process with constant rules.[1] X Research source
Evaluating Negative Exponents Download Article
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Get to know the basics of negative exponent expression. A negative exponent is usually written as a base number multiplied to the power of a negative number such as
{\displaystyle 3^{-3},5^{-2},}
{\displaystyle 7^{-4}}
. The larger number is known as a base number while the small number is the exponent, in this case the negative exponent. Exponents tell you how many times to multiply a number by itself. [2] X Research source
Both positive and negative exponents are also referred to as ‘powers’ or numbers that the base number is ‘raised to the power of’.
To solve an equation with a negative exponent, you must first make it positive.
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Convert negative exponents into fractions to simplify them. A negative exponent tells you that the base number is on the incorrect side of a fraction line. To simplify an expression with a negative exponent, you just flip the base number and exponent to the bottom of a fraction with a
{\displaystyle 1}
on top. Writing negative exponents as fractions will make it easier for you to understand how to work with them in an equation.[3] X Research source
To convert a negative exponent, create a fraction with the number 1 as the numerator (top number) and the base number as the denominator (bottom number).
Raise the base number to the power of the same exponent, but make it positive.
{\displaystyle 3^{-3},5^{-2},}
{\displaystyle 7^{-4}}
{\displaystyle {\frac {1}{(3^{3})}},{\frac {1}{(5^{2})}},}
{\displaystyle {\frac {1}{(7^{4})}}}
This process is known as the negative exponent rule.
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Simplify negative exponent expressions with unknown numbers. Once you understand the negative exponent rule, you can start to simplify more difficult exponent expressions. Things can get tricky at this stage since you will be working with unknown values such as ‘x’ or ‘y’, but luckily the rules to simplify such an equation never change.[4] X Research source
{\displaystyle 2x^{-1}}
{\displaystyle {\frac {2x^{-1}}{1}}}
which can then be simplified to
{\displaystyle {\frac {2}{({1x}^{1})}}}
{\displaystyle {\frac {2}{1x^{1}}}}
can then be simplified to
{\displaystyle {\frac {2}{x}}}
In this case, only ‘x’ became the denominator because it had the exponent.
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Understand how to solve for negative exponents in fraction form. Sometimes the exponent itself is a fraction. Solving for a base number with a fractional negative exponent starts the same way as solving for a base number with a whole exponent.[5] X Research source
To simplify a fractional negative exponent, you must first convert to a fraction.
If your starting base number is
{\displaystyle 16^{-1/2}}
, start by converting it to a fraction where the exponent becomes positive when the base number is switched to the denominator.
{\displaystyle 16^{-1/2}}
{\displaystyle {\frac {1}{16^{1/2}}}}
{\displaystyle {\frac {1}{16^{1/2}}}}
{\displaystyle {\frac {1}{\sqrt[{2}]{16}}}}
{\displaystyle {\frac {1}{\sqrt[{2}]{16}}}}
{\displaystyle {\frac {1}{4}}}
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Know the difference between negative bases and negative exponents. Negative bases have different rules than negative exponents when they are used in an equation. They do not need to be converted into fractions if the exponent is positive. Negative negative exponents must be converted into fractions to become positive.[6] X Research source
When an exponent is negative and a base number is positive, the expression must be converted into a fraction to make the exponent positive
{\displaystyle 6^{-2}={\frac {1}{6^{2}}}}
When an exponent is positive and a base number is negative, the base number will be multiplied by itself however many times the exponent shows us it should be.
{\displaystyle -5^{5}=-5*-5*-5*-5*-5=-3125.}
Use a calculator to complete exponent equations quickly. Calculators have specific functions for calculating exponents. Use the E, "^", or "e^x" button to raise any number to any power. Calculators make it easy to check your work and easily convert negative exponents. [7] X Research source
Remember to put negative exponent values in parentheses:
{\displaystyle 4E(-6)}
Solving exponential equations on a calculator will allow you to find answers more quickly without converting them into fractions.
Completing Equations with Negative Exponents Download Article
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Add exponents together if the multiplied base numbers are the same. If two identical base numbers are multiplied, you can add the negative exponents together. The base number will stay the same while the exponent will become a larger negative number.[8] X Research source
{\displaystyle 4^{-1/4}*4^{-1/4}}
{\displaystyle 4^{-1/2}}
You can further simplify
{\displaystyle 4^{-1/2}}
{\displaystyle {\frac {1}{4^{-1/2}}}}
{\displaystyle {\frac {1}{4^{-1/2}}}}
{\displaystyle {\frac {1}{\sqrt[{2}]{4}}}}
{\displaystyle {\frac {1}{2}}}
Subtract negative exponents if the divided base numbers are the same. Exponents with the same base number can be subtracted from one another. When you divide two base numbers with the same value and different exponents, you simply subtract the exponent values and keep the base number as it is. [9] X Research source
Because the exponent is negative, the subtraction will cancel out the second negative and make the exponent positive.
The exponents in
{\displaystyle {\frac {2^{-7}}{2^{-2}}}}
will subtract as
{\displaystyle (-7)-(-2)}
{\displaystyle (-7)+2}
The equation will simplify to
{\displaystyle 2^{-5}}
{\displaystyle {\frac {1}{2^{5}}}}
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Keep exponents the same when the base number is different. If two different base numbers with the same exponents are multiplied or divided, do not change the exponent value. When you multiply or divide numbers with different bases and the same negative exponents, the exponent number will not change. Multiply or divide the bases and keep the exponent the same.[10] X Research source
{\displaystyle 7^{-6}*8^{-6}}
{\displaystyle 56^{-6}}
{\displaystyle 5^{-1/6}*20^{-1/6}}
{\displaystyle 100^{-1/6}}
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Practice different equations to become a master of negative exponents. Once you understand the basics of working with negative exponents, it is a good idea to challenge yourself with different equations. The rules for negative exponents will never change. Once you learn the basic rules for negative exponents, your math homework will be a breeze.
{\displaystyle 16^{-1/4}+4^{-2}={\frac {1}{\sqrt[{4}]{16}}}+{\frac {1}{(4^{2})}}}
{\displaystyle {\frac {1}{\sqrt[{4}]{16}}}+{\frac {1}{(4^{2})}}={\frac {1}{2}}+{\frac {1}{16}}}
{\displaystyle {\frac {1}{2}}+{\frac {1}{16}}={\frac {8}{16}}+{\frac {1}{16}}}
{\displaystyle {\frac {8}{16}}+{\frac {1}{16}}={\frac {9}{16}}}
Categories: Mathematics | Exponents and Logarithms
Português:Calcular Expoentes Negativos |
The optional filter parameter, passed as the index to the Map or Map2 command, restricts the application of
\mathrm{with}\left(\mathrm{LinearAlgebra}\right):
A≔\mathrm{Matrix}\left([[1,2,3],[0,1,4]],\mathrm{shape}=\mathrm{triangular}[\mathrm{upper},\mathrm{unit}]\right)
\textcolor[rgb]{0,0,1}{A}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{4}\end{array}]
M≔\mathrm{Map}\left(x↦x+1,A\right)
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{4}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{5}\end{array}]
\mathrm{evalb}\left(\mathrm{addressof}\left(A\right)=\mathrm{addressof}\left(M\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
B≔〈〈1,2,3〉|〈4,5,6〉〉
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{4}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{5}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{6}\end{array}]
\mathrm{Map2}[\left(i,j\right)↦\mathrm{evalb}\left(i=1\right)]\left(\left(x,a\right)↦a\cdot x,3,B\right)
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{12}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{5}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{6}\end{array}]
\mathrm{Map}\left(x↦x+1,g\left(3,A\right)\right)
\textcolor[rgb]{0,0,1}{g}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{4}\\ \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{5}\end{array}]\right)
C≔\mathrm{Matrix}\left([[1,2],[3]],\mathrm{scan}=\mathrm{triangular}[\mathrm{upper}],\mathrm{shape}=\mathrm{symmetric}\right)
\textcolor[rgb]{0,0,1}{C}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{cc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\end{array}]
\mathrm{Map}\left(x↦x+1,C\right)
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{3}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{4}\end{array}]
[\begin{array}{cc}\textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{9}\\ \textcolor[rgb]{0,0,1}{9}& \textcolor[rgb]{0,0,1}{4}\end{array}] |
Test the condition for convergence of \sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)} and find the sum if
Test the condition for convergence of
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)\left(n+2\right)}
and find the sum if it exists.
Using Eulers
There is an alternate method and is as follows.
\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{\left(n-1\right)!}{\left(n+2\right)!}=\frac{12}{B}\left(n,3\right)
B\left(x,y\right)
is the Beta function. Using an integral form of the Beta function the summation becomes
S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)\left(n+2\right\}}
=\frac{1}{2}{\int }_{0}^{1}\left(\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1}\right)\left(1-x{\right)}^{2}dx
=\frac{1}{2}{\int }_{0}^{1}\frac{\left(1-x{\right)}^{2}}{1-x}dx=\frac{1}{2}{\int }_{0}^{1}\left(1-x\right)dx
=\frac{1}{4}
This leads to the known result
\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{4}
Alternatively, take
\frac{1}{1-x}=\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1}
and integrate three times with lower limit 0, giving
-\mathrm{log}\left(1-x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n}
x+\left(1-x\right)\mathrm{log}\left(1-x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n\left(n+1\right)}
\frac{3}{4}{x}^{2}-\frac{1}{2}x-\frac{1}{2}\left(1-x{\right)}^{2}\mathrm{log}\left(1-x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{{x}^{n}}{n\left(n+1\right)\left(n+2\right)}
\left(n+1\right)\left(n-2\right){a}_{n}=n\left({n}^{2}-n-1\right){a}_{n-1}-{\left(n-1\right)}^{3}{a}_{n-2}
{a}_{2}={a}_{3}=1
For each natural number n and each number
x\in \left[0,1\right]
{f}_{n}\left(x\right)=\frac{x}{nx+1}
Sum of the series:
\sum _{n=0}^{\mathrm{\infty }}\frac{1+n}{{3}^{n}}
Express the definite integral as an infinite series and find its value to within an error of at most
{10}^{-4}
{\int }_{0}^{1}\mathrm{cos}\left({x}^{2}\right)dx
\sum _{n=2}^{\mathrm{\infty }}\frac{{5}^{n}}{{12}^{n}}
Study the behaviour of
\sum _{n=1}^{\mathrm{\infty }}\frac{n+\mathrm{log}\left(n\right)}{{\left(n+\mathrm{cos}\left(n\right)\right)}^{3}}
\sum _{n=1}^{\mathrm{\infty }}\frac{\varphi \left(n\right)}{{7}^{n}+1} |
Assume that when adults with smartphones are randomly selected, 52% use them in meetings or classes. If 10 adult smartphone users are randomly selected, find the probability that fewer than 5 of them use their smartphones in meetings or classes.
52% of adults use their smartphones in meetings or classes. i.e.,
p=0.52
n=10
adult smartphone users are randomly selected, to find the probability that fewer than 5 of them use their smartphones in meetings or classes:
Let X denote the number of smartphone users who use their smartphones in meetings or classes and X follows Binomial distribution with number of trials
n=10
and probability of success
p=0.52
Probability mass function of Binomial variable is given by the formula:
P\left(X=x\right){=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}
Required probability is obtained as follows:
P\left(X<5\right)=P\left(X=0\right)+P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)
{=}^{10}{C}_{0}{\left(0.52\right)}^{0}{\left(1-0.52\right)}^{10-0}{+}^{10}{C}_{1}{\left(0.52\right)}^{1}{\left(1-0.52\right)}^{10-1}{+}^{10}{C}_{2}{\left(0.52\right)}^{2}{\left(1-0.52\right)}^{10-2}{+}^{10}{C}_{3}{\left(0.52\right)}^{3}{\left(1-0.52\right)}^{10-3}{+}^{10}{C}_{4}\left(0.52{\right)}^{4}\left(1-0.52{\right)}^{10-4}
=1×1×0.00065+10×0.52×0.0014+45×0.2704×0.0028+120×0.141×0.0059+210×0.73×0.012231
=0.0065+0.0728+0.341+0.99828+1.875
\approx 3.28773
Thus, the probability that fewer than 4 of them use their smartphones in meetings or classes is 3.28773
You start with $4 in your investment. Your money doubles every 6 years.How much money will your investment be worth in 54 years?
a. How many doublings will happen in 54 years? ___b. Your investment will be worth ___ in 54 years,
Which sequence matches the recursive formula?
{a}_{n}=2{a}_{n-1}+5
{a}_{1}=5
A) 5,10,15,20,...
B) 5,15,35,75,...
C) 5,15,25,35,...
D) 5,20, 35, 50,...
To calculate: The calcium present in 1 cup of milk and in 1 cup of cooked spinach if one day Shenika had 3 cups of milk and 1 cup of cooked spinach for a total of 1140 mg of calcium. The next day, she had 2 cups of milk and
1\frac{1}{2}
cups of cooked spinach for a total of 960 mg of calcium.
-3+\frac{2}{3}y-4-\frac{1}{3}y
f\left(x\right)=-9x+20
g\left(x\right)={x}^{2}+8
f\left(-5\right)\cdot g\left(-5\right)
To calculate: The solution of the equation
3{y}^{2}-4y=8-6y
12s-4t+7t-3s |
Knuth's Algorithm X - Wikipedia
Algorithm for exact cover problem
Algorithm X is an algorithm for solving the exact cover problem. It is a straightforward recursive, nondeterministic, depth-first, backtracking algorithm used by Donald Knuth to demonstrate an efficient implementation called DLX, which uses the dancing links technique.[1]
The exact cover problem is represented in Algorithm X by a matrix A consisting of 0s and 1s. The goal is to select a subset of the rows such that the digit 1 appears in each column exactly once.
Algorithm X functions as follows:
# If the matrix A has no columns, the current partial solution is a valid solution; terminate successfully.
Otherwise choose a column c (deterministically).
Choose a row r such that Ar, c = 1 (nondeterministically).
Include row r in the partial solution.
For each column j such that Ar, j = 1,
for each row i such that Ai, j = 1,
delete row i from matrix A.
delete column j from matrix A.
The nondeterministic choice of r means that the algorithm recurses over independent subalgorithms; each subalgorithm inherits the current matrix A, but reduces it with respect to a different row r. If column c is entirely zero, there are no subalgorithms and the process terminates unsuccessfully.
The subalgorithms form a search tree in a natural way, with the original problem at the root and with level k containing each subalgorithm that corresponds to k chosen rows. Backtracking is the process of traversing the tree in preorder, depth first.
Any systematic rule for choosing column c in this procedure will find all solutions, but some rules work much better than others. To reduce the number of iterations, Knuth suggests that the column-choosing algorithm select a column with the smallest number of 1s in it.
For example, consider the exact cover problem specified by the universe U = {1, 2, 3, 4, 5, 6, 7} and the collection of sets
{\displaystyle {\mathcal {S}}}
= {A, B, C, D, E, F}, where:
E = {2, 3, 6, 7}; and
F = {2, 7}.
This problem is represented by the matrix:
Algorithm X with Knuth's suggested heuristic for selecting columns solves this problem as follows:
Step 1—The matrix is not empty, so the algorithm proceeds.
Step 2—The lowest number of 1s in any column is two. Column 1 is the first column with two 1s and thus is selected (deterministically):
Step 3—Rows A and B each have a 1 in column 1 and thus are selected (nondeterministically).
The algorithm moves to the first branch at level 1…
Level 1: Select Row A
Step 4—Row A is included in the partial solution.
Step 5—Row A has a 1 in columns 1, 4, and 7:
Column 1 has a 1 in rows A and B; column 4 has a 1 in rows A, B, and C; and column 7 has a 1 in rows A, C, E, and F. Thus, rows A, B, C, E, and F are to be removed and columns 1, 4 and 7 are to be removed:
Row D remains and columns 2, 3, 5, and 6 remain:
Step 2—The lowest number of 1s in any column is zero and column 2 is the first column with zero 1s:
Thus this branch of the algorithm terminates unsuccessfully.
The algorithm moves to the next branch at level 1…
Level 1: Select Row B
Step 4—Row B is included in the partial solution.
Row B has a 1 in columns 1 and 4:
Column 1 has a 1 in rows A and B; and column 4 has a 1 in rows A, B, and C. Thus, rows A, B, and C are to be removed and columns 1 and 4 are to be removed:
Rows D, E, and F remain and columns 2, 3, 5, 6, and 7 remain:
Step 2—The lowest number of 1s in any column is one. Column 5 is the first column with one 1 and thus is selected (deterministically):
Step 3—Row D has a 1 in column 5 and thus is selected (nondeterministically).
Level 2: Select Row D
Step 4—Row D is included in the partial solution.
Step 5—Row D has a 1 in columns 3, 5, and 6:
Column 3 has a 1 in rows D and E; column 5 has a 1 in row D; and column 6 has a 1 in rows D and E. Thus, rows D and E are to be removed and columns 3, 5, and 6 are to be removed:
Row F remains and columns 2 and 7 remain:
Step 2—The lowest number of 1s in any column is one. Column 2 is the first column with one 1 and thus is selected (deterministically).
Row F has a 1 in column 2 and thus is selected (nondeterministically).
Level 3: Select Row F
Step 4—Row F is included in the partial solution.
Row F has a 1 in columns 2 and 7:
Column 2 has a 1 in row F; and column 7 has a 1 in row F. Thus, row F is to be removed and columns 2 and 7 are to be removed:
Step 1—The matrix is empty, thus this branch of the algorithm terminates successfully.
As rows B, D, and F are selected, the final solution is:
In other words, the subcollection {B, D, F} is an exact cover, since every element is contained in exactly one of the sets B = {1, 4}, D = {3, 5, 6}, or F = {2, 7}.
There are no more selected rows at level 3, thus the algorithm moves to the next branch at level 2…
There are no branches at level 0, thus the algorithm terminates.
In summary, the algorithm determines there is only one exact cover:
{\displaystyle {\mathcal {S}}^{*}}
= {B, D, F}.
Donald Knuth's main purpose in describing Algorithm X was to demonstrate the utility of dancing links. Knuth showed that Algorithm X can be implemented efficiently on a computer using dancing links in a process Knuth calls "DLX". DLX uses the matrix representation of the exact cover problem, implemented as doubly linked lists of the 1s of the matrix: each 1 element has a link to the next 1 above, below, to the left, and to the right of itself. (Technically, because the lists are circular, this forms a torus). Because exact cover problems tend to be sparse, this representation is usually much more efficient in both size and processing time required. DLX then uses dancing links to quickly select permutations of rows as possible solutions and to efficiently backtrack (undo) mistaken guesses.[1]
^ a b Knuth, Donald (2000). "Dancing links". arXiv:cs/0011047.
Knuth, Donald E. (2000), "Dancing links", in Davies, Jim; Roscoe, Bill; Woodcock, Jim (eds.), Millennial Perspectives in Computer Science: Proceedings of the 1999 Oxford-Microsoft Symposium in Honour of Sir Tony Hoare, Palgrave, pp. 187–214, arXiv:cs/0011047, Bibcode:2000cs.......11047K, ISBN 978-0-333-92230-9 .
Knuth's paper - PDF file (also arXiv:cs/0011047)
Knuth's Paper describing the Dancing Links optimization - Gzip'd postscript file.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Knuth%27s_Algorithm_X&oldid=1064914684" |
Complete Graphs Practice Problems Online | Brilliant
How many distinct complete graphs are there with at least 5 and less than 10 nodes?
Note: a graph is considered complete if for any two nodes on the graph, there is an edge between them.
K_4
be the complete graph with 4 nodes. What is the biggest degree of any node on this graph?
Note: the degree of a node is the number of edges that connect to it.
K_5
be the complete graph with 5 nodes, how many edges does it have?
How many complete graphs with at least 1 edge and at most 50 edges are there?
Define the distance between two distinct nodes to be the fewest number of edges in a path between the two nodes. What is the largest possible distance between two distinct nodes on the complete graph with 7 nodes? |
An electron is released from rest at the negative plate
An electron is released from rest at the negative plate of a parallel plate capa
protisvitfc 2021-11-16 Answered
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
\sigma =1.8×{10}^{-7}\mathrm{C}/{\mathrm{m}}^{2}
and the plates are separated by a distance of
1.5×{10}^{-2}\mathrm{m}.
How fast is the electron moving just before it reaches the positive plate?
The Electric field due to capacitor having surface charge density
\sigma
E=\frac{\sigma }{{ϵ}_{0}}
The acceleration of electron is given by
\stackrel{\to }{a}=\frac{\stackrel{\to }{F}}{m}
\stackrel{\to }{a}=\frac{q\stackrel{\to }{E}}{m}
We have to find the speed of electron as it reaches the plate
\stackrel{\to }{a}=\frac{q\sigma }{m{ϵ}_{0}}
\stackrel{\to }{a}=\frac{\left(1.6×{10}^{-19}C\right)\left(1.8×{10}^{-7}\frac{C}{{m}^{2}}\right)}{\left(9.11×{10}^{-31}kg\right)\left(8.85×{10}^{-12}N\cdot \frac{{m}^{2}}{{C}^{2}}\right)}
\stackrel{\to }{a}=3.572×{10}^{15}\frac{m}{{s}^{2}}
The velocity can be found from the equation of kinematics
{v}^{2}={v}_{0}^{2}+2\left(\stackrel{\to }{a}\right)s
{v}^{2}=0+2\left(3.572×{10}^{15}\frac{m}{{s}^{2}}\right)\left(1.5×{10}^{-2}m\right)
{v}^{2}=1.0716×{10}^{14}\frac{{m}^{2}}{{s}^{2}}
v=\sqrt{1.0716×{10}^{14}\frac{{m}^{2}}{{s}^{2}}}
\begin{array}{|c|}\hline v=1.035×{10}^{7}m/s\\ \hline\end{array}
v=1.035×{10}^{7}\frac{m}{s}
{F}_{A}
{F}_{B}
{F}_{A}=4200
{F}_{B}
{F}_{A}+{F}_{B}
\stackrel{\to }{B}
in acertain region is 0.128 ,and its direction is that of the z-axis in the figure.
What is the magnetic flux across the surface abcd in the figure?
What is the magnetic flux across the surface befc ?
Part C What is the magnetic flux across the surface aefd?
What is the net flux through all five surfaces that enclose the shaded volume?
Find the coordinate vector of p relative to the basis
S=\left\{P1,P2,p3\right\}\text{ }\text{ for }\text{ }P2.\left(b\right)p=2-x+{x}^{2};{p}_{1}=1+x,{p}_{2}=1+{x}^{2},{p}_{3}=x+{x}^{2}
Process to show that
\sqrt{2}+\sqrt[3]{3}
How can I prove that the sum
\sqrt{2}+\sqrt[3]{3}
is an irrational number ?? |
-1,0,1
\mathrm{ShortLexOrder}\left(u,v\right)
\mathrm{ShortLexOrder}\left(\mathrm{cat}\left(u,w\right),\mathrm{cat}\left(v,w\right)\right)
\mathrm{ShortLexOrder}\left(u,v\right)
\mathrm{ShortLexOrder}\left(\mathrm{cat}\left(w,u\right),\mathrm{cat}\left(w,v\right)\right)
\mathrm{ShortRevLexOrder}\left(u,v\right)
\mathrm{ShortRevLexOrder}\left(\mathrm{cat}\left(u,w\right),\mathrm{cat}\left(v,w\right)\right)
\mathrm{ShortRevLexOrder}\left(u,v\right)
\mathrm{ShortRevLexOrder}\left(\mathrm{cat}\left(w,u\right),\mathrm{cat}\left(w,v\right)\right)
The left and right recursive path orders, manifest in the LeftRecursivePathOrder(s1, s2) and RightRecursivePathOrder(s1, s2) commands, are defined for two strings
t
s
t
{s}_{-1}={t}_{-1}
{s}_{1..-2}
{t}_{1..-2}
{t}_{-1}<{s}_{-1}
{s}_{1..-2}
t
{s}_{-1}<{t}_{-1}
s
{t}_{1..-2}
s
t
{s}_{1}={t}_{1}
{s}_{2..-1}
{t}_{2..-1}
{t}_{1}<{s}_{1}
{s}_{2..-1}
t
{s}_{1}<{t}_{1}
s
{t}_{2..-1}
8
\mathrm{with}\left(\mathrm{StringTools}\right):
\mathrm{LexOrder}\left("abc","abd"\right)
\textcolor[rgb]{0,0,1}{-1}
\mathrm{LexOrder}\left("abd","abcd"\right)
\textcolor[rgb]{0,0,1}{1}
\mathrm{ShortLexOrder}\left("abd","abcd"\right)
\textcolor[rgb]{0,0,1}{-1}
\mathrm{RevLexOrder}\left("bcd","abd"\right)
\textcolor[rgb]{0,0,1}{1}
\mathrm{RevLexOrder}\left("bcd","bd"\right)
\textcolor[rgb]{0,0,1}{1}
\mathrm{ShortRevLexOrder}\left("aba","abc"\right)
\textcolor[rgb]{0,0,1}{-1}
\mathrm{ShortRevLexOrder}\left("abc","abc"\right)
\textcolor[rgb]{0,0,1}{0}
\mathrm{LeftRecursivePathOrder}\left("abc","abcc"\right)
\textcolor[rgb]{0,0,1}{-1}
\mathrm{RightRecursivePathOrder}\left("abc","acc"\right)
\textcolor[rgb]{0,0,1}{-1} |
A manufacturing machine has a 3% defect rate. If 4 items
A manufacturing machine has a 3% defect rate.
If 4 items are chosen at random, what is the probability that at least one will have a defect?
sweererlirumeX
Here, X denotes that the item is defective, which follows binomial distribution with
n=4ZKS\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}PSKp=0.03
The formula to find the binomial probability is,
P\left(X=x\right){=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{\left(n-x\right)}
Probability that at least one will have a defect:
The probability that at least one will have a defect is,
P\left(X\ge 1\right)=1-P\left(X=0\right)
=1{-}^{4}{C}_{0}{\left(0.03\right)}^{0}{\left(1-0.03\right)}^{\left(4-0\right)}
=1-0.8853
=0.1147
The probability that at least one will have a defect is 0.1147.
Eighty percent of households say they would feel secure if they had 50,000 in savings. You randomly select 8 households and ask them if they would feel secure if they had $50,000 in savings. Find the probability that the number that say they would feel secure is exactly five, (b) more than five, and (c) at most five
c.) Find the probability that the number that say they would feel secure is at most five.
P\left(<5\right)=
(round to three decimal places as needed).
A sample of 5 parts is drawn without replacement from a total population of 13 parts. Determine the probability of getting exactly 3 defective parts. The population is known to have 6 defective parts.
Assume a binomial probability distribution has p = 0.80 and n = 400.
a) what is the mean and standard deviation
b) is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain.
c) what is the probability of 300 to 310 successes? Use the normal approximation of the binomial distribution to answer this question. (Round your answer to four decimal places.)
The probability that a patient recovers from a stomach disease is 0.8. Suppose 20 people are known to have contracted this disease. What is the probability that exactly 14 recover?
1) State the formula for the Binomial Probability Distribution, also state the domain. 2) Tell what the requirement for the Binomial experiment are. 3) List both formulas for calculating the mean of a Binomial Probability Distribution. 4) List the formulas for the standard deviation of a Binomial Probability Distribution. |
How far from the center of the earth is the
How far from the center of the earth is the center of mass of the earth + moon s
alka8q7 2021-11-18 Answered
How far from the center of the earth is the center of mass of the earth + moon system? Data for the earth and moon can be found inside the back cover of the book.
The center of mass of a system consists of a set of obect numbered i=1,2,3... where i has mass
{m}_{i}
and is located at position
\left({x}_{i},{y}_{i}\right)
{x}_{cm}=\frac{1}{M}\sum _{i}{m}_{i}{x}_{i}
{y}_{cm}=\frac{1}{m}\sum _{i}{m}_{i}{y}_{i}
From the back cover of the book, the mass of earth is
{m}_{e}=5.98×{10}^{24}\text{ }kg
, the mass of Moon is
{m}_{m}=7.36×{10}^{22}\text{ }kg
, and the distnace between them is
R=3.84×{10}^{8}
We can conisder Earth and Moon to be along a straight line, and we ca take this straight line to be the x-axis.
Also we take the origin to be at the center of Earth, so Earths position is
{x}_{e}=0
and Moon's position is
{x}_{m}=R
Since our system consists of Earrth and Moon only, we can write equation (1) for our system as follows:
{x}_{c}m=\frac{{m}_{e}{x}_{e}+{m}_{m}{x}_{m}}{{m}_{e}+{m}_{m}}
Now , we enter the given values, so we get:
{x}_{cm}=\frac{0+\left(7.36×{10}^{22}\text{ }kg\right)\left(3.84×{10}^{8}\text{ }m\right)}{7.36×{10}^{22}+5.98×{10}^{24}\text{ }kg}=4.67×{10}^{6}m
{x}_{cm}=4.67×{10}^{6}
Therefore, the center of mass of the Earth-Moon system is at
4.67×{10}^{6}
m from the center of Earth which is less than Earth's radius
6.37×1{-}^{6}
An aeroplane starting from airport A flies 300kms East then 250 km at 30°West of North and then 150 km north to arrive at airpot B
Calculate vector position of each displacement in unit vector notation
Use differentiation to find a power series representation for
f\left(x\right)=\frac{1}{{\left(1+x\right)}^{2}}
, What is the radius of convergence?
Find the solution of the following Second Order Differential Equation
{y}^{″}-9y=0,y\left(0\right)=2,{y}^{\prime }\left(0\right)=0 |
Ryan's sister, Jollie, gets her height measured once a year. Here is the growth data that her mother has written down for the last six years:
2.5
3
2.25
4
1.75
3.5
How much has Jollie grown in the last six years?
Each of these measurements tells you how much Jollie has grown in one year.
How can you use these measurements to find out how much she has grown in the past six years?
Add the measurements together to find the total amount she has grown.
2.5
+
3
+
2.25
+
4
+
1.75
+
3.5
=
17
17
Find the mean, median, and mode of her growth over the last six years.
If you are having trouble starting this problem, refer to problems 1-110 and 1-112.
Don't remember what the mean, median, and mode are? Here's a quick review: The mean here is the number of inches Jollie would grow in a year if she grew the same amount each year. The median is the middle number if the numbers are put in order from least to greatest. The mode is the number that appears most often.
Jollie wants to convince her basketball coach that she is going to grow a lot this year. Which measure of central tendency should she use to try to convince the coach? Explain your choice.
Conner thinks that Jollie should use the median. Do you agree with him?
Would it really help Jollie to use a number that happened to fall in the middle of the data? |
An event has a probability p = \frac{5}{8}. Find
Karen Simpson 2021-12-12 Answered
An event has a probability
p=\frac{5}{8}
. Find the complete binomial distribution for
n=6
p=\frac{5}{8}=0.625
n=6
The binomial distribution is given by
P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right){p}^{x}{q}^{n-x}
P\left(X=x\right)=\left(\begin{array}{c}6\\ x\end{array}\right)\left(0.625{\right)}^{x}\left(1-0.625{\right)}^{6-x}
The probability distribution table is given as under
\begin{array}{|cc|}\hline x& p\left(x\right)\\ 0& 0.002781\\ 1& 0.027809\\ 2& 0.115871\\ 3& 0.257492\\ 4& 0.321865\\ 5& 0.214577\\ 6& 0.059605\\ \hline\end{array}
P(0) Probability of exactly 0 successes
If using a calculator, you can enter
\text{trials}=6,\text{ }p=0.625,
X=0
into a binomial probability distribution function (PDF). If doing this by hand, apply the binomial probability formula:
P\left(X\right)=\left(\begin{array}{c}n\\ x\end{array}\right)×{p}^{x}×\left(1-p{\right)}^{n-x}
The binomial coefficient,
\left(\begin{array}{c}n\\ x\end{array}\right)
\left(\begin{array}{c}n\\ x\end{array}\right)=\frac{n!}{X!\left(n-X\right)!}
The full binomial probability formula with the binomial coefficient is
P\left(X\right)=\frac{n!}{X!\left(n-X\right)!}×{p}^{x}×{\left(1-p\right)}^{n-x}
Where n is the number of trials, p is the probability if success on a single trial, and X is the number of successes. Substituting in values for this problem,
n=6,\text{ }p=0.625,
X=0
P\left(0\right)=\frac{6!}{0!\left(6-0\right)!}×{0.625}^{0}×{\left(1-0.625\right)}^{6-0}
Evaluting the expression, we have
P\left(0\right)=0.0027809143066406
If we apply the binomial probability formula, or a calculator's binomial probability distribution (PDF) function, to all possible values of X for 6 trials, we can construct a complete binomial distribution table. The sum of the probabilities in this table will always be 1. The complete binomial distribution table for this problem, with
p=0.625
and 6 trials is:
P\left(0\right)=0.0027809143066406
P\left(1\right)=0.027809143066406
P\left(2\right)=0.11587142944336
P\left(3\right)=0.25749206542969
P\left(4\right)=0.32186508178711
P\left(5\right)=0.21457672119141
P\left(6\right)=0.059604644775391
Five out of 25 students at AUA Applied Stat class are Lazio fans. Suppose that the number of Lazio fans among Applied Stat class follows binomial distribution. Find the probability that if you randomly chose five students none of them will be Lazio fan (round to 3 decimal places).
To find: The condition, mean and standard deviation for a binomial random variable approximation to normal.
To define: The continuity correction.
To explain: The procedure for applying the continuity correction method for finding the probability if using the normal approximation to the binomial distribution is used for finding the probability.
Out of 14 couples 13 of them had baby girls by using the preliminary test of the MicroSoft method of gender selection.
Calculate probability within both binomially distributed and normally distributed random variables.
A Finite final consists of 50 multiple choice questions. If each question has 5 choices and 1 right answer, find the probability that a student gets an A (i.e. 45 or better) by purely guessing on each question. |
Reference Values for Umbilical Cord Blood Gases of Newborns Delivered in Enshi Tujia and Miao Autonomous Prefecture
Department of Obstetrics, Affiliated Hospital of Hubei University for Nationalities, Enshi, China.
Objective: To define the normal ranges of umbilical cord blood values in Enshi Tujia and Miao Autonomous Prefecture. Methods: 1266 normal newborns were screened as umbilical cord blood gas samples. Normal neonates are full-term live pregnancies, single births, weight suitable for gestational age and heavier than the gestational age of the newborn without complications. Re-sults: The umbilical cord blood collected from 1266 newborns was analyzed in this study. The calculated reference range of the umbilical arterial PH was 7.16 - 7.39, of SBE was ?8.25 - 1.67 mmol/L, of lactic acid was 1.4 - 7.5 mmol/L, and of HCO3 was 15.60 - 30.70 mEq/L. Conclusions: This study confirmed the normal reference value of umbilical cord blood gas analysis in Enshi Tujia Miao Autonomous Prefecture.
Ethnic Minority, Umbilical Cord Blood, Blood Gas Analysis, Normal Referencevalue
Wu, H. , Yu, Q. , Wang, W. and Li, L. (2018) Reference Values for Umbilical Cord Blood Gases of Newborns Delivered in Enshi Tujia and Miao Autonomous Prefecture. Open Access Library Journal, 5, 1-6. doi: 10.4236/oalib.1105026.
Umbilical cord blood gas values are commonly indicators to evaluate neonatal birth status at home and abroad. It is an index with high specificity, which can effectively evaluate acid-base status and substance metabolism [1] . The neonatal cord blood gas results were used to evaluate whether the neonates had inhibition or not, to determine the cause of inhibition and to race against time for rescue. Umbilical cord blood gas values may also assiste valuations of an infant and indicate the occurrence of an acute, and intrapartum hypoxic event [2] . Similarly, Revathy Natesan S [3] reported the analysis of umbilical cord blood gas in 2212 neonates. The results showed that lactic acid and pH value were significantly correlated with the adverse outcomes of neonatal hypoxic-ischemic encephalopathy. However, many studies groups have reported that it is particularly susceptible to vary in the parameters of umbilical cord blood gas values, such as fetal age, production mode, maternal acid base balance, fetal hemoglobin concentration and so on. The normal values of the parameters still lack a unified reference range. Kattiya Manomayangkul et al. [4] reported that the so-called “normal” umbilical cord blood gas analysis may also vary depending on race and mode of work.
Enshi Tujia and Miao Autonomous Prefecture are the only minority Autonomous Prefecture in Hubei, China. It has special mountainous terrain, rich Selenium Resources and diet culture. There are a large number of births per year. At the end of 2016, the total population of Enshi was 4,040,100; the annual birth rate was 10.38 per thousand. In 2017, there were 42 thousand and 300 new born population [5] . But there is no statistical report on the normal reference range of umbilical cord blood gas analysis in Enshi. It will be a defect in evaluating the newborn birth situation, and predicting adverse outcomes and prognosis in Enshi Prefecture. This article will make a discussion.
1) Participants: In 2015-2018 years, a total of full-term pregnancies, single births, suitable gestational age or greater than gestational age in six hospitals in Enshi. The study has been approved by the hospital ethics committee.
2) Excluded objects: 1 minutes or 5 minutes of Apgar score of less than 7, anyone with abnormal symptoms of respiratory symptoms, 1 or more than 1 organ damage, with congenital heart and respiratory disease.
3) Personnel and equipment: The Apgar score was assessed by a trained obstetrician, midwife and newborn pediatrician in delivery room and operation room. Umbilical cord blood analyzer equipment model: Denmark, RADO ABL 90 FLEX.
4) Collection method: The umbilical cord about 10 cm in length was clamped with two sterile hemostatic forceps near the side of the fetus immediately after delivery without establishing spontaneous breathing. The umbilical cord was cut off at the outside of the clamp. Then the umbilical artery or venous blood was punctured with a heparinized syringe. The blood was sealed and sent to the obstetric Department of obstetrics. The blood gas analysis results obtained from the blood gas analyzer were recorded. The values of PH,
{\text{HCO}}_{3}^{-}
, cLac and SBE were detected.
5) Statistical methods: SPSS 17.0 software was used for statistical analysis. Apgar scores of 1 min and 5 min (>8) were used for single fetus, full-term, suitable weight and larger than gestational age, which included 1266 cases. The normal range of umbilical artery blood gas parameters is 1.96 standard deviations. Normal distribution method was used for data obeying normal distribution, and percentage method was used for data of skewed distribution to calculate the reference range. P < 0.05 was statistically significant.
One thousand six hundred and one term newborns were assessed in this study. 335 cases were excluded from inclusion criteria after birth. Data from the remaining 1266 newborns were then analyzed. Statistics of inclusion criteria are shown in Figures 1-4. The resulting reference ranges of the arterial umbilical cord blood of the newborns: The PH value was 7.16 - 7.39, the SBE value was −8.25 - 1.67 mmol/L, the lactic acid value was 1.4 - 7.5 mmol/L, and the HCO3 value was 15.60 - 30.70 mEq/L (Figures 1-4).
The normal value obtained in the study is slightly lower than the traditional normal range [6] . The analysis may be due to different inclusion criteria or different progress in labor. As long as the blood gas changes are not serious enough to produce organ function or organic damage, they should be considered physiological. Umbilical cord blood gas has been recognized as the most objective and reliable basis for evaluating fetal and neonatal oxygenation and acid-base status. Establishing a normal reference range of umbilical cord blood gas analysis value of neonates in Enshi Prefecture can improve the diagnosis rate of neonatal asphyxia,
Figure 1. Frequency distribution of umbilical artery PH value in 1266 vigorous human newborn infantsnormal distribution.
Figure 2. Frequency distribution of umbilical artery SBE value in 535 vigorous human newborn infantsnormal distribution.
Figure 3. Frequency distribution of umbilical artery Lac value in 562 vigorous human newborn infantsnormal distribution.
Figure 4. Frequency distribution of umbilical artery
{\text{HCO}}_{3}^{-}
value in 1236 vigorous human newborn infantsnormal distribution.
reduce the risk rate, improve the specificity of detection, and help to distinguish the causes of neonatal depression. Guiding neonatal further treatment, evaluating neonatal prognosis and even using umbilical cord blood gas analysis results, and avoiding medical disputes have important effects. Therefore, it is important to establish a normal reference range for umbilical cord blood gas analysis to improve the diagnosis of neonatal asphyxia, guide further rescue, and reduce the risk of adverse outcomes of neonates.
[1] Neonatal Umbilical Artery Blood Gas Index Research Group (2010) Multi Center Clinical Study of Umbilical Artery Blood Gas Index in Diagnosis of Neonatal Asphyxia. Chinese Journal of Pediatrics, 48, 668-673.
[2] Manomayangkul, K., Arunota, S., et al. (2016) Reference Values for Umbilical Cord Blood Gases of Newborns Delivered by Elective Cesarean Section. Journal of the Medical Association of Thailand, 99, 611-617.
[3] Revathy Natesan, S. (2016) Routine Measurements of Cord Arterial Blood Lactate Levels in Infants Delivering at Term and Prediction of Neonatal Outcome. The Medical Journal of Malaysia, 71, 131-133.
[4] Manomayangkul, K., Siriussawakul, A., Nimmannit, A., et al. (2016) Reference Values for Umbilical Cord Blood Gases of Newborns Delivered by Elective Cesarean Section. Journal of the Medical Association of Thailand, 99, 611-617.
[5] The People’s Government of Enshi Tujia and Miao Autonomous Prefecture, May 5, 2017. Enshi National Economic and Social Devel-opment Bulletin.
http://www.enshi.gov.cn/zzf/zc/
[6] Arikan, G.M., Scholz, H.S., Petru, E., et al. (2000) Cord Blood Oxygen Saturation in Vigorous Infantsat Birth: What Is Normal? BJOG, 107, 987-994. |
DirectionalDerivativeTutor - Maple Help
Home : Support : Online Help : Education : Student Packages : Multivariate Calculus : Interactive : DirectionalDerivativeTutor
Student[MultivariateCalculus][DirectionalDerivativeTutor] - plot or animate the directional derivative
DirectionalDerivativeTutor()
DirectionalDerivativeTutor(f(x,y), [x,y]=[a,b], [c,d], x=xmin..xmax, y=ymin..ymax, z=zmin..zmax)
(optional) point at which the gradient is evaluated
(optional) direction Vector
The DirectionalDerivativeTutor command launches a tutor interface that computes, plots, and animates the directional derivative of a function.
If f(x,y), [x,y]=[a,b], and [c,d] are not specified, DirectionalDerivativeTutor uses defaults.
The DirectionalDerivative command offers equivalent capabilities to DirectionalDerivativeTutor where interaction takes place in the worksheet interface. See the Student[MultivariateCalculus][DirectionalDerivative] help page.
When the DirectionalDerivativeTutor is running, interaction with the worksheet is not possible.
\mathrm{with}\left(\mathrm{Student}[\mathrm{MultivariateCalculus}]\right):
\mathrm{DirectionalDerivativeTutor}\left(\right)
\mathrm{DirectionalDerivativeTutor}\left({x}^{2}+{y}^{2},[x,y]=[1,2],[2,3]\right)
\mathrm{DirectionalDerivativeTutor}\left({x}^{2}+{y}^{2},[x,y]=[1,2],[2,3],x=-1..5,y=0..6,z=0..16\right)
Student, Student[MultivariateCalculus], Student[MultivariateCalculus][DirectionalDerivative] |
Consider a binomial experiment with n = 20\ and\ p
Consider a binomial experiment with
n=20\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p=.70
1. Compute f (12).
P\left(x\ge 16\right)
Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.”
(1) Compute f (12).
The value of the function f (12) is obtained below:
Let X denotes the random variable which follows binomial distribution with the probability of success 0.70 with the number trails 20.
That is, \(\displaystyle{n}={12},{p}={0.70},{q}={0.30}{\left(={1}-{0.70}\right)}\)
The probability distribution is given by,
\[P(X=x)=(\begin{array}{c}n\\ x\end{array})p^{x}(1-p)^{n-x};\ here\ x=0,1,2,...,n\ for\ 0 \le p \le 1\]
\(\displaystyle{f{{\left({12}\right)}}}={P}{\left({X}\le{12}\right)}\)
Use Excel to obtain the probability value for x equals 12.
4. Enter the number of success as 12.
5. Enter the Trails as 20
7. Enter the cumulative as True.
From the Excel output, the P-value is 0.2277
Thus, the value of the function f (12) is 0.2277.
(3) Compute the probability \(\displaystyle{P}{\left({x}\geq{16}\right)}\).
The probability \(\displaystyle{P}{\left({x}\geq{16}\right)}\) is obtained below as follows:
\(\displaystyle{P}{\left({x}\geq{16}\right)}={1}-{P}{\left({x}{ < }{16}\right)}\)
\(\displaystyle={1}-{P}{\left({x}\le{15}\right)}\)
\(\displaystyle={1}-{0.7625}\)
\(\displaystyle={0.2375}\)
The probability \(\displaystyle{P}{\left({x}\geq{16}\right)}\ {i}{s}\ {0.2375}\).
p=0.70
n=20
Formula binomial probability:
f\left(k\right)=\left(\begin{array}{c}n\\ k\end{array}\right)×{p}^{k}×\left(1-p{\right)}^{n-k}
f\left(12\right)=\left(\begin{array}{c}20\\ 12\end{array}\right)×{0.70}^{12}×\left(1-0.70{\right)}^{20-12}=0.114397
f\left(16\right)=\left(\begin{array}{c}20\\ 16\end{array}\right)×{0.70}^{16}×\left(1-0.70{\right)}^{20-16}=0.130421
c) Add the corresponding probabilities:
P\left(X\ge 16\right)=f\left(16\right)+f\left(17\right)+f\left(18\right)+f\left(19\right)+f\left(20\right)=0.2375
d) Use the complement rule for probabilities:
P\left(X\le 15\right)=1-P\left(X\ge 16\right)=1-0.2375=0.7625
e) The mean of a binomial distribution is the sample size n and the probability p:
\mu =np=20×0.70=14
f) The standard deviation of a binomial distribution is the square root of the product of the sample size n and the probabilities p and q. The variance is the square of the standard deviation.
{\sigma }^{2}=npq=np\left(1-p\right)=20\left(0.70\right)\left(1-0.70\right)=4.2
\sigma =\sqrt{npq}=\sqrt{np\left(1-p\right)}=\sqrt{20\left(0.70\right)\left(1-0.70\right)}\approx 2..0494
X\sim \text{Binomial}\left(n=20,\text{ }p=0.7\right)
F\left(x\right)\therefore P\left(x\right){=}^{n}{C}_{x}{P}^{x}\left(1-P{\right)}^{n-x}
=\frac{n!}{\left(n-x\right)!x!}{P}^{x}\left(1-P{\right)}^{n-x}
\therefore F\left(13\right){=}^{20}{C}_{13}\left(0.7{\right)}^{13}\left(1-0.7{\right)}^{20-13}
=\frac{20!}{7!13!}\left(0.7{\right)}^{13}\left(1-0.7{\right)}^{7}
F\left(13\right)=0.1642
F\left(16\right)=P\left(x=16\right){=}^{20}{C}_{16}\left(0.7{\right)}^{16}\left(1-0.7{\right)}^{20-16}
=\frac{20!}{16!4!}\left(0.7{\right)}^{16}\left(1-0.7{\right)}^{4}
F\left(16\right)=0.1304
P\left(X\ge 16\right)=1-P\left(x<16\right)
=1-P\left(X\le 15\right)
=1-0.7643(Using Binomial Table)
P\left(X\ge 16\right)=0.2357
P\left(X\le 15\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)
+P\left(x=4\right)+P\left(x=5\right)+P\left(x=6\right)+P\left(x=7\right)+P\left(x=8\right)
+P\left(x=9\right)+P\left(x=10\right)+P\left(x=11\right)+P\left(x=12\right)
+P\left(x=13\right)+P\left(x=14\right)+P\left(x=15\right)
=0+0+0+0
+0+0+0.0002+0.0010+0.0039
+0.120+0.308+0.0554+0.1144
+0.1643+0.1916+0.1789
P\left(X\le 15\right)=0.7643
F\left(x\right)=np
\because X\sim B\left(np\right)
=20×0.7
F\left(x\right)=14
var\left(x\right)=npq
\because q=1-p
=20×0.7\left(1-0.7\right)
var\left(x\right)4.2
\sigma =\sqrt{var\left(x\right)}=2.04
On each of three consecutive days the National Weather Service announces that there is a 50-50 chance of rain. Assuming that the National Weather Service is correct, what is the probability that it rains on at most one of the three days? Justify your answer. (Hint: Represent the outcome that it rains on day 1 and doesn’t rain on days 2 and 3 as RNN.)
I have a large jar of red and green M&M’s with 55% red M&M’s. I plan to pick 7 M&M’s.
1. I want to find the probability that I get 2 red M&M’s out of 7.
a) How many different ways can you get 2 red M&Ms out of 7? Use the formula for the binomial coefficient to find that number.
b) Use your answer from (d) and the binomial formula to calculate the probability that you get 2 red M&M’s out of 7.
c) Use your calculator to find the probability that I get 3 red candies out of the 7 picked
P\left(x\right)
n=3,\text{ }x=0,\text{ }p=0.8
Garam Doe’s daily chores require making 10 round trips by car between two towns. Once through with all ten trips, Mr. Doe can take the rest of the day off, a good enough motivation to drive above the speed limit. Experience shows that there is a 44% chance of getting a speeding fine on any round trip. What is the probability that the day will end with at most one speeding tickets (one or less)
During a visit to a local zoom, students either roam around the zoo solo or with thier friends. It is seen that 20% of the students roam around with thier freinds during a visit to a local zoo
In a sample of 6 studentsfind the prob that alteast 2 will roam around with thier friends in the zoo
use binomial prob function in excel to solve the question |
Match the binomial probability P(x<42) with the correct statement. a) P(there
Match the binomial probability P(x<42) with the correct statement.a) P(there
Match the binomial probability
P\left(x<42\right)
with the correct statement.
a) P(there are fewer than 42 successes)
b) P(there are at most 42 successes)
c) P(there are at least 42 successes)
d) P(there are more than 42 successes)
binomial probability :
P\left(x<42\right)
A. P(there are fewer than 42 successes)
n=20\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p=.70
P\left(x\ge 16\right)
a) Compute the number of vehicles expected to be hybrid.
b) Compute the probability that five of the sales were hybrid vehicles using Poisson distribution.
c) Compute the probability that five of the sales were hybrid vehicles using binomial distribution.
42% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorite nut. Find the probability that the number who say cashews are their favorite nut is (a) exactlythree, (b) at least four, and (c) at most two. If convenient, use technology to find the probabilities.
P\left(3\right)=
P\left(x\ge 4\right)=
P\left(x\le 2\right)=
Simplify, please:
\frac{12!}{8!4!} |
Cellular Automata/Excitable media - Wikibooks, open books for an open world
Cellular Automata/Excitable media
1 An introduction to excitable media
2 Discretization of differential equations using the explicit FTCS method
2.1 Single PDE
2.1.1 One-dimensional problem
2.2 System of PDE
3 Other PDE discretization methods
4 Modeling with cellular automata
4.1 Greenberg-Hastings Model
An introduction to excitable media[edit | edit source]
Excitable media are nonlinear dynamic systems known for exhibiting complex behavior that can be observed as pattern formation. They are usually defined by a reaction-diffusion differential equation.
{\displaystyle u_{t}(\mathbf {r} ,t)=D\nabla ^{2}u(\mathbf {r} ,t)+f(u(\mathbf {r} ,t))}
The diffusion part provides stability and propagation of information, the reactive part provides interesting local dynamics.
A common example of excitable media are prey-predator systems. Such systems are described by a system of differential equations, one function for each of the observed protagonists.
{\displaystyle u_{t}(\mathbf {r} ,t)=D_{u}\nabla ^{2}u(\mathbf {r} ,t)+f(u,v)}
{\displaystyle v_{t}(\mathbf {r} ,t)=D_{v}\nabla ^{2}v(\mathbf {r} ,t)+g(u,v)}
We will discuss two different approaches to modeling excitable media. Discretization of differential equations and modeling with cellular automata.
There are different ways to define boundary conditions for the reaction-diffusion equation.
The value of the function at the boundary is given explicitly
{\displaystyle u(\mathbf {r_{0}} ,t)}
Cyclic boundaries
If the initial condition
{\displaystyle u(x,0)}
is supposed to be periodic in space, cyclic boundary conditions can be used.
Zero-flux boundary conditions
If zero-flux is expected at the boundary than the component of the functions first derivative normal to the boundary is zero
{\displaystyle {\vec {n}}\cdot \nabla {u}=0}
at the boundary. This can be achieved by reflecting function values from the inside over the boundary to the outside.
Discretization of differential equations using the explicit FTCS method[edit | edit source]
The traditional method to simulate excitable media is discretization and numerical computation of the governing PDE. First the FTCS (forward-time centered-space method) discretization method is presented. Explicit methods are the simplest and the equations are similar to a cellular automaton, but are inadequate because of stability and convergence problems.
Single PDE[edit | edit source]
We will first observe a single PDE describing a single function.
{\displaystyle u_{t}(\mathbf {r} ,t)=D\nabla ^{2}u(\mathbf {r} ,t)+f(u(\mathbf {r} ,t))}
One-dimensional problem[edit | edit source]
In the one dimensional case the space vector becomes a single variable
{\displaystyle \mathbf {r} =x}
. The nabla operator becomes
{\displaystyle \nabla ^{2}={\frac {\partial ^{2}}{\partial {x^{2}}}}}
{\displaystyle {\frac {\partial u(x,t)}{\partial t}}=D{\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}+f(u(x,t))}
The partial differential equation is discretized.
{\displaystyle {\frac {u(x,t+\Delta {t})-u(x,t)}{\Delta {t}}}=}
{\displaystyle \quad D{\frac {u(x+\Delta {x},t)-2u(x,t)+u(x-\Delta {x},t)}{\Delta {x}^{2}}}}
{\displaystyle \quad +f(u(x,t))}
Forward-time centered-space method
Each finite element at time
{\displaystyle t+\Delta {t}}
is calculated from three neighboring elements at time
{\displaystyle t}
(see figure at the right).
{\displaystyle u(x,t+\Delta {t})=u(x,t)+\,}
{\displaystyle \quad +d(u(x+\Delta x,t)-2u(x,t)+u(x-\Delta x,t))}
{\displaystyle \quad +\Delta {t}f(u(x,t))}
where the diffusion number is
{\displaystyle d=D{\frac {\Delta {t}}{\Delta {x}^{2}}}}
The FTCS method is stable if
{\displaystyle d\leq 1/2}
For periodic boundaries present values at the left boundary
{\displaystyle x=0}
can be used to compute the future values at the right boundary
{\displaystyle x=L_{x}}
and the other way round.
{\displaystyle u(0,t+\Delta {t})=u(0,t)+d(u(0+\Delta x,t)-2u(0,t)+u(L_{x},t))+\Delta {t}f(u(0,t))\,}
{\displaystyle u(L_{x},t+\Delta {t})=u(L_{x},t)+d(u(0,t)-2u(L_{x},t)+u(L_{x}-\Delta x,t))+\Delta {t}f(u(L_{x},t))\,}
If there is zero-flux
{\displaystyle \nabla u(0,t)=0}
at the boundaries than values outside the boundary are reflections of values inside
{\displaystyle u(0+\Delta x,t)=u(0-\Delta x,t)}
{\displaystyle u(0,t+\Delta {t})=u(0,t)+2d(u(0+\Delta x,t)-u(0,t))+\Delta {t}f(u(0,t))\,}
{\displaystyle u(L_{x},t+\Delta {t})=u(L_{x},t)+2d(u(L_{x},t)+u(L_{x}-\Delta x,t))+\Delta {t}f(u(L_{x},t))\,}
Two-dimensional problem[edit | edit source]
In the two-dimensional case, the space vector becomes a variable pair
{\displaystyle \mathbf {r} =[x,y]}
{\displaystyle \nabla ^{2}={\frac {\partial ^{2}}{\partial {x^{2}}}}+{\frac {\partial ^{2}}{\partial {y^{2}}}}}
{\displaystyle {\frac {\partial u(x,y,t)}{\partial t}}=D\left({\frac {\partial ^{2}u(x,y,t)}{\partial x^{2}}}+{\frac {\partial ^{2}u(x,y,t)}{\partial y^{2}}}\right)+f(u(x,y,t))}
The partial differential equation is discretized using the forward-time centered-space method.
{\displaystyle {\frac {u(x,y,t+\Delta {t})-u(x,y,t)}{\Delta {t}}}=}
{\displaystyle \quad =D{\frac {u(x+\Delta {x},y,t)-2u(x,y,t)+u(x-\Delta {x},y,t)}{\Delta {x}^{2}}}+}
{\displaystyle \quad +D{\frac {u(x,y+\Delta {y},t)-2u(x,y,t)+u(x,y-\Delta {y},t)}{\Delta {y}^{2}}}+}
{\displaystyle \quad +f(u(x,y,t))}
{\displaystyle t+\Delta {t}}
is calculated from five neighboring elements at time
{\displaystyle t}
{\displaystyle u(x,y,t+\Delta {t})=u(x,t)+\,}
{\displaystyle \quad +d_{x}(u(x+\Delta {x},y,t)-2u(x,y,t)+u(x-\Delta {x},y,t))+}
{\displaystyle \quad +d_{y}(u(x,y+\Delta {y},t)-2u(x,y,t)+u(x,y-\Delta {y},t))+}
{\displaystyle \quad +\Delta {t}f(u(x,y,t))}
where the diffusion numbers are
{\displaystyle d_{x}=D{\frac {\Delta {t}}{\Delta {x}^{2}}}\quad d_{y}=D{\frac {\Delta {t}}{\Delta {y}^{2}}}}
{\displaystyle d_{x}\leq 1/2}
{\displaystyle d_{y}\leq 1/2}
The same ideas as in the one dimensional case can be used for two dimensions.
System of PDE[edit | edit source]
A system of PDE describes two functions that interact with each other (prey-predator).
{\displaystyle u_{t}(\mathbf {r} ,t)=D_{u}\nabla ^{2}u(\mathbf {r} ,t)+f(u,v)}
{\displaystyle v_{t}(\mathbf {r} ,t)=D_{v}\nabla ^{2}v(\mathbf {r} ,t)+g(u,v)}
The interaction is local, which means, the dispersion part can be computed separately for each equation, and than the reaction part is added to the result.
Other PDE discretization methods[edit | edit source]
Modeling with cellular automata[edit | edit source]
Greenberg-Hastings Model[edit | edit source]
Joe D. Hoffman, Numerical Methods for Engineers and Scientists
Toffoli T., Margolus N., Cellular Automata Machines: A New Environment for Modeling, The MIT Press (1987), Cambridge, Massachusetts
Toffoli T., Cellular automata as an alternative to Differential equations, in Modeling Physics, Physica 10D, (1984)
http://www.jweimar.de/paper-abstracts.html
Robert Fisch, Janko Gravner, David Griffeath, Threshold-Range Scaling of Excitable Cellular Automata
Robert Fisch, Janko Gravner, David Griffeath, Metastability in the Greenberg-Hastings Model
Marcus R. Garvie Finite difference schemes for reaction-diffusion equations modeling predator-prey interactions in MATLAB
Stephen Wolfram, A New Kind of Science, Wolfram Media, (2002)
Cellular automata interesting collection of books
http://www.schatten.info/info/ca/ca.html#CAandDG
http://psoup.math.wisc.edu/kitchen.html Primordial soup kitchen]
Dynamic Formation of a 3-strand Spiral
Belousov-Zhabotinsky oscillating chemical reaction
Belousov-Zhabotinsky oscillating chemical reaction continued
Living Spirals: Dictyostelium discoideum
Zoologisches Institut München -
CA modeling of the wave equation
Lattice-Gas Cellular Automata
Retrieved from "https://en.wikibooks.org/w/index.php?title=Cellular_Automata/Excitable_media&oldid=3229424" |
Data type that associates keys with values
"Dictionary (data structure)" redirects here. Not to be confused with data dictionary.
In computer science, an associative array, map, symbol table, or dictionary is an abstract data type that stores a collection of (key, value) pairs, such that each possible key appears at most once in the collection. In mathematical terms an associative array is a function with finite domain.[1] It supports 'lookup', 'remove', and 'insert' operations.
The dictionary problem is the classic problem of designing efficient data structures that implement associative arrays.[2] The two major solutions to the dictionary problem are hash tables and search trees.[3][4][5][6] In some cases it is also possible to solve the problem using directly addressed arrays, binary search trees, or other more specialized structures.
The name does not come from the associative property known in mathematics. Rather, it arises from the fact that we associate values with keys. It is not to be confused with associative processors.
The operations that are usually defined for an associative array are:[3][4][8]
Insert or put: add a new
{\displaystyle (key,value)}
pair to the collection, mapping the key to its new value. Any existing mapping is overwritten. The arguments to this operation are the key and the value.
{\displaystyle (key,value)}
Lookup, find, or get: find the value (if any) that is bound to a given key. The argument to this operation is the key, and the value is returned from the operation. If no value is found, some associative array implementations raise an exception, while others return a default value (zero, null, specific value passed to the constructor, ...).
The operations of the associative array should satisfy various properties:[8]
lookup(k, insert(j, v, D)) = if k == j then v else lookup(k, D)
lookup(k, new()) = fail, where fail is an exception or default value
remove(k, insert(j, v, D)) = if k == j then remove(k, D) else insert(j, v, remove(k, D))
remove(k, new()) = new()
where k and j are keys, v is a value, D is an associative array, and new() creates a new, empty associative array.
For dictionaries with very small numbers of mappings, it may make sense to implement the dictionary using an association list, a linked list of mappings. With this implementation, the time to perform the basic dictionary operations is linear in the total number of mappings; however, it is easy to implement and the constant factors in its running time are small.[3][10]
Hash table implementations[edit]
Hash tables need to be able to handle collisions: when the hash function maps two different keys to the same bucket of the array. The two most widespread approaches to this problem are separate chaining and open addressing.[3][4][5][11] In separate chaining, the array does not store the value itself but stores a pointer to another container, usually an association list, that stores all of the values matching the hash. On the other hand, in open addressing, if a hash collision is found, then the table seeks an empty spot in an array to store the value in a deterministic manner, usually by looking at the next immediate position in the array.
Tree implementations[edit]
Self-balancing binary search trees[edit]
Other trees[edit]
Lookup or Removal
O(1) O(n) O(1) O(n) No
O(log n) O(log n) O(log n) O(log n) Yes
O(log n) O(n) O(log n) O(n) Yes
O(n) O(n) O(1) O(1) No
Ordered dictionary[edit]
In Smalltalk, Objective-C, .NET,[21] Python, REALbasic, Swift, VBA and Delphi[22] they are called dictionaries; in Perl, Ruby and Seed7 they are called hashes; in C++, Java, Go, Clojure, Scala, OCaml, Haskell they are called maps (see map (C++), unordered_map (C++), and Map); in Common Lisp and Windows PowerShell, they are called hash tables (since both typically use this implementation); in Maple and Lua, they are called tables. In PHP, all arrays can be associative, except that the keys are limited to integers and strings. In JavaScript (see also JSON), all objects behave as associative arrays with string-valued keys, while the Map and WeakMap types take arbitrary objects as keys. In Lua, they are used as the primitive building block for all data structures. In Visual FoxPro, they are called Collections. The D language also has support for associative arrays.[23]
Permanent storage[edit]
^ Collins, Graham; Syme, Donald (1995). "A theory of finite maps". Higher Order Logic Theorem Proving and Its Applications. Lecture Notes in Computer Science. 971: 122–137. doi:10.1007/3-540-60275-5_61. ISBN 978-3-540-60275-0.
^ a b c d e Goodrich, Michael T.; Tamassia, Roberto (2006), "9.1 The Map Abstract Data Type", Data Structures & Algorithms in Java (4th ed.), Wiley, pp. 368–371
^ a b c d Mehlhorn, Kurt; Sanders, Peter (2008), "4 Hash Tables and Associative Arrays", Algorithms and Data Structures: The Basic Toolbox (PDF), Springer, pp. 81–98
^ a b Black, Paul E.; Stewart, Rob (2 November 2020). "dictionary". Dictionary of Algorithms and Data Structures. Retrieved 26 January 2022.
^ Knuth, Donald (1998). The Art of Computer Programming. Vol. 3: Sorting and Searching (2nd ed.). Addison-Wesley. pp. 513–558. ISBN 0-201-89685-0.
Look up associative array in Wiktionary, the free dictionary.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Associative_array&oldid=1086435023" |
Abstract: This study was conducted in Melka Wakena catchment; south eastern Ethiopia to assess land use/cover change (LULCC) and topographic elevation effect on selected soil quality/fertility parameters. 144 soil samples collected from 0 - 30 cm depth under three land cover types across three elevation gradients were analysed for selected soil quality/fertility parameters. Data were statistically analyzed using analysis of variance (ANOVA) and mean comparisons were made using Least Significant Difference (LSD). The soil properties examined generally showed significant variations with respect to land-use/land cover changes and elevation. Soil particles, soil organic carbon, total N, pH, available phosphorus, potassium and calcium content significantly decreased as forestland is converted into cropland/grassland. Heaviest soil deterioration was recorded in soils under cropland and followed by grassland soils. The conversion of natural forest to different land uses without proper soil conservation and management practices resulted in the overall decline of soil fertility quality. Thus, integrated land resource management approach is indispensable for sustaining agricultural productivity and the environmental health of the Melka Waken a catchment.
Keywords: Land Use/Cover, Resource Management, Soil Quality, Spatial Variation, Topographic Elevation, Soil Quality Parameters
C{h}_{Gl,Gl}=L{u}_{Cl}{{}_{,}}_{\text{}or\text{}Gl}-L{u}_{Fl}\times 100
Cite this paper: Hayicho, H. , Alemu, M. and Kedir, H. (2019) Assessing the Effects of Land-Use and Land Cover Change and Topography on Soil Fertility in Melka Wakena Catchment of Sub-Upper Wabe-Shebelle Watershed, South Eastern Ethiopia. Journal of Environmental Protection, 10, 672-693. doi: 10.4236/jep.2019.105040.
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http://www.soil-journal.net/2/63/2016/doi:10.5194/soil-2-63-2016
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[49] Abreha, K., Heluf, G., Tekalign, M. and Kibebew, K. (2012) Impact of Altitude and Land Use Type on Some Physical and Chemical Properties of Acidic Soils in Tsegede Highlands, Northern Ethiopia. Open Journal of Soil Science, 43, 223.
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CDS 110b: Random Processes - Murray Wiki
Revision as of 01:49, 30 January 2006 by Murray (talk | contribs) (→Frequently Asked Questions)
This lecture presents an introduction to random processes.
Quick review of random variables
Lecture Notes on Stochastic Systems
Reading: Friedland, Chapter 10
HW 4 - due 1 Feb
Hoel, Port and Stone, Introduction to Probability Theory - this is a good reference for basic definitions of random variables
Apostol II, Chapter 14 - another reference for basic definitions in probability and random variables
Q: Can you explain the jump from pdfs to correlations in more detail?
The probability density function (pdf),
{\displaystyle p(x;t)}
tells us how the value of a random process is distributed at a particular time:
{\displaystyle P(a\leq x(t)\leq b)=\int _{a}^{b}p(x;t)dx.}
You can interpret this by thinking of
{\displaystyle x(t)}
as a separate random variable for each time
{\displaystyle t}
The correlation for a random process tells us how the value of a random process at one time,
{\displaystyle t_{1}}
is related to the value at a different time
{\displaystyle t_{2}}
. This relationship is probabalistic, so it is also described in terms of a distribution. In particular, we use the joint probability density function,
{\displaystyle p(x_{1},x_{2};t_{1},t_{2})}
to characterize this:
{\displaystyle P(a_{1}\leq x_{1}(t_{1})\leq b_{1},a_{2}\leq x_{2}(t_{2})\leq b_{2})=\int _{a_{1}}^{b_{1}}\int _{a_{2}}^{b_{2}}p(x_{1},x_{2};t_{1},t_{2})dx_{1}dx_{2}}
Given any random process,
{\displaystyle p(x_{1},x_{2};t_{1},t_{2})}
descibes (as a density) how the value of the random variable at time
{\displaystyle t_{1}}
is related (or "correlated") with the value at time
{\displaystyle t_{2}}
. We can thus describe a random process according to its joint probability density function.
In practice, we don't usually describe random processes in terms of their pdfs and joint pdfs. It is usually easier to describe them in terms of their statistics (mean, variance, etc). In particular, we almost never describe the correlation in terms of joint pdfs, but instead use the correlation function:
{\displaystyle \rho (t,\tau )=E\{x(t)x(\tau )\}=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }x_{1}x_{2}p(x_{1},x_{2};t,\tau )dx_{1}dx_{2}}
The utility of this particular function is seen primarily through its application: if we know the correlation for one random process and we "filter" that random process through a linear system, we can compute the correlation for the corresponding output process (we'll see this in the next lecture).
Retrieved from "https://murray.cds.caltech.edu/index.php?title=CDS_110b:_Random_Processes&oldid=2979" |
The Laplace transform L {e^(-t^2)} exists, but without finding it
The Laplace transform L {e^(-t^2)} exists, but without finding it solve the initial-value problem y''+9y=3e^(-t^2), y(0)=0,y'(0)=0
\left\{{e}^{-{t}^{2}}\right\}
y{}^{″}+9y=3{e}^{-{t}^{2}},y\left(0\right)=0,{y}^{\prime }\left(0\right)=0
Apply the Laplace transform to both sides of the equation:
L\left(y{}^{″}+9y\right)=L\left(3{e}^{-{t}^{2}}\right)
⇒{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+9Y\left(s\right)=L\left(3{e}^{-{t}^{2}}\right)
⇒\left({s}^{2}+9\right)Y\left(s\right)=L\left(3{e}^{-{t}^{2}}\right)
⇒Y\left(s\right)=\frac{L\left(3{e}^{-{t}^{2}}\right)}{{s}^{2}+9}=\frac{1}{3}L\left(3{e}^{-{t}^{2}}\right)\frac{3}{{s}^{2}+9}
⇒Y\left(s\right)=L\left({e}^{-{t}^{2}}\right)xL\left(\mathrm{sin}\left(3t\right)\right)
Using the property of convolution:
f\left(t\right)\cdot g\left(t\right)←L\to F\left(s\right)G\left(s\right)
⇒y\left(t\right)={e}^{-{t}^{2}}\cdot \mathrm{sin}\left(3t\right)
⇒y\left(t\right)={\int }_{0}^{t}{e}^{-{\tau }^{2}}\sim \left(3\left(t-\tau \right)\right)d\tau
y\left(t\right)={\int }_{0}^{t}{e}^{-{\tau }^{2}}\sim \left(3\left(t-\tau \right)\right)d\tau
{y}^{″}+7{y}^{\prime }=0,y\left(0\right)=7,{y}^{\prime }\left(0\right)=2
find the Laplace transform of the given function.
f\left(t\right)=\left\{\begin{array}{ll}t,& 0\le t<1\\ 2-t,& 1\le t<2\\ 0,& 2\le t<\mathrm{\infty }\end{array}
F\left(s\right)=\frac{3}{\left({s}^{2}\right)}
{y}^{\prime }={\mathrm{cos}}^{2}x\mathrm{cos}y
\frac{dy}{dt}=ry\mathrm{ln}\left(\frac{K}{y}\right)
r=0.73
k=33.800kg,
\frac{{y}_{0}}{k}=0.27
Use the Gompertz model to find the predicted value of y(3)
\left(3x-2y+1\right)dx+\left(3x-2y+3\right)dy=0
Reverse Laplace transform of
\frac{3s-15}{2{s}^{2}-4s+10} |
Sensors | Free Full-Text | Simultaneous Measurements of Dose and Microdosimetric Spectra in a Clinical Proton Beam Using a scCVD Diamond Membrane Microdosimeter
Leveraging Deep Learning for Visual Odometry Using Optical Flow
Application of Measurement Sensors and Navigation Devices in Experimental Research of the Computer System for the Control of an Unmanned Ship Model
Loto, O.
Zahradnik, I.
Leite, A. Maia
Simultaneous Measurements of Dose and Microdosimetric Spectra in a Clinical Proton Beam Using a scCVD Diamond Membrane Microdosimeter
Oluwasayo Loto
Izabella Zahradnik
Amelia Maia Leite
Institut Curie, Radiation Oncology Department, PSL Research University, Proton Therapy Centre, Centre Universitaire, 91898 Orsay, France
Institut Curie, PSL Research University, University Paris Saclay, LITO, Inserm, 91898 Orsay, France
Current address: Diamond Sensors Laboratory, Centre Digiteo, CEA LIST, 91191 Gif Sur Yvette, France.
Academic Editor: Han Haitjema
A single crystal chemical vapor deposition (scCVD) diamond membrane-based microdosimetric system was used to perform simultaneous measurements of dose profile and microdosimetric spectra with the Y1 proton passive scattering beamline of the Center of Proton Therapy, Institute Curie in Orsay, France. To qualify the performance of the set-up in clinical conditions of hadrontherapy, the dose, dose rate and energy loss pulse-height spectra in a diamond microdosimeter were recorded at multiple points along depth of a water-equivalent plastic phantom. The dose-mean lineal energy (
{\overline{y}}_{D}
) values were computed from experimental data and compared to silicon on insulator (SOI) microdosimeter literature results. In addition, the measured dose profile, pulse height spectra and
{\overline{y}}_{D}
values were benchmarked with a numerical simulation using TOPAS and Geant4 toolkits. These first clinical tests of a novel system confirm that diamond is a promising candidate for a tissue equivalent, radiation hard, high spatial resolution microdosimeter in beam quality assurance of proton therapy. View Full-Text
Keywords: diamond; proton therapy; microdosimetry; radiation detectors; dosimeters; sensors diamond; proton therapy; microdosimetry; radiation detectors; dosimeters; sensors
Loto, O.; Zahradnik, I.; Leite, A.M.; De Marzi, L.; Tromson, D.; Pomorski, M. Simultaneous Measurements of Dose and Microdosimetric Spectra in a Clinical Proton Beam Using a scCVD Diamond Membrane Microdosimeter. Sensors 2021, 21, 1314. https://doi.org/10.3390/s21041314
Loto O, Zahradnik I, Leite AM, De Marzi L, Tromson D, Pomorski M. Simultaneous Measurements of Dose and Microdosimetric Spectra in a Clinical Proton Beam Using a scCVD Diamond Membrane Microdosimeter. Sensors. 2021; 21(4):1314. https://doi.org/10.3390/s21041314
Loto, Oluwasayo, Izabella Zahradnik, Amelia M. Leite, Ludovic De Marzi, Dominique Tromson, and Michal Pomorski. 2021. "Simultaneous Measurements of Dose and Microdosimetric Spectra in a Clinical Proton Beam Using a scCVD Diamond Membrane Microdosimeter" Sensors 21, no. 4: 1314. https://doi.org/10.3390/s21041314 |
Does \lim_{(x,y)\to(3,-1)}\frac{(x-3)(y+1)}{(x-3)^2+(y+1)^2} exist?
\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}
\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}
The value of the limit
\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}
do not exist as the denominator becomes 0.
\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}
as shown below
\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}=\frac{\left(0-3\right)\left(0+1\right)}{{\left(0-3\right)}^{2}+{\left(0+1\right)}^{2}}
=\frac{\left(-3\right)\left(1\right)}{9+1}
=-\frac{3}{10}
Therefore, the limits when x=0 and y=0
\underset{\left(x,y\right)\to \left(3,-1\right)}{lim}\frac{\left(x-3\right)\left(y+1\right)}{{\left(x-3\right)}^{2}+{\left(y+1\right)}^{2}}=-\frac{3}{10}
Determine which of the following limits exist, and find the limits which do exists.
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{3}+{y}^{3}}{{x}^{2}+{y}^{2}}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}+4x{y}^{2}+4{y}^{4}}{{x}^{2}+4{y}^{4}}
\underset{\left(x,y\right)\to \left(0,1\right)}{lim}\frac{\mathrm{arcsin}xy}{1-xy}
\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}
A tank initially contains 120 L of pure water. A mixture containing a concentration of γ g/L of salt enters the tank at a rate of 2 L/min, and the well-stirred mixture leaves the tank at the same rate. Find an expression in terms of γ for the amount of salt in the tank at any time t. Also find the limiting amount of salt in the tank as
t\to \mathrm{\infty }
\underset{x\to 0}{lim}\left\{\frac{2x-\mathrm{sin}\left\{x\right\}}{3x+\mathrm{sin}\left\{x\right\}}\right\}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{|x|}^{\frac{3}{2}}{y}^{2}}{{x}^{4}+{y}^{2}}\to 0
Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are
\mathrm{\infty }
-\mathrm{\infty }
, and describe the corresponding vertical asymptote.
\underset{x\to {0}^{±}}{lim}\frac{\mathrm{sin}x}{|x|} |
Selective Precipitation - Course Hero
General Chemistry/Equilibria of Other Reaction Classes/Selective Precipitation
In addition to providing the molar solubility of a compound, Ksp can be used to calculate whether a compound will form a precipitate at a given concentration. Also used in this calculation is the ion product (Qsp), a constant analogous to the solubility product constant but calculated from initial concentrations or other ion concentrations when the reaction is not at equilibrium.
{{{\rm{M}}_m}{{\rm{A}}_a}(s)}\rightleftharpoons{m{{\rm{M}}^{a+}}({aq})+a{{\rm{A}}^{m-}}(aq)}
Q_{\rm{sp}}=\lbrack{{\rm{M}}^{a+}\rbrack}^m\lbrack{{\rm{A}}^{m-}\rbrack}^a
Q_{\rm{sp}}=K_{\rm{sp}}
, the ion concentrations are equal to their equilibrium concentrations. This means the solution is exactly at its saturation point and no more precipitate will form. A precipitate should also not form if
Q_{\rm{sp}} \lt K_{\rm{sp}}
because the solution is unsaturated. If
Q_{\rm{sp}}\gt K_{\rm{sp}}
, the ion concentrations are greater than the molar solubility, and the solution is supersaturated. The equilibrium is shifted to the left, and the solution should form a precipitate.
If a solution contains two or more ions of the same charge that can all be precipitated out of solution by the same reagent, the ions can be separated by that reagent as long as the solubilities of the precipitates are different enough. Selective precipitation is a separation technique for compounds that share a common ion but have different solubilities and are precipitated out of solution.
Take, for example, a solution containing I– and Cl–, both of which form compounds with Ag+:
{{\rm{AgI}}(s)\rightleftharpoons{\rm{Ag}^ +}(aq)+{\rm{I}^-}(aq),\hspace{10pt}K_{\rm{sp}}=8.52\times10^{-17}}
{\rm{AgCl}}(s)\rightleftharpoons{\rm{Ag}^+}(aq)+{\rm{Cl}^-}(aq),\hspace {10pt}K_{\rm{sp}}=1.77\times10^{-10}
If Ag+ is slowly added to the solution, which precipitate will form first, AgI or AgCl? A precipitate will start to form once
Q_{\rm{sp}}\gt{K_{\rm{sp}}}
. In this case, Qsp is calculated from the concentration of ions before Ag+ is added. As Ag+ is added, the concentration of Ag+ increases, and therefore so does the value of Qsp. Since the Ksp for AgI is seven orders of magnitude less than that of AgCl, it follows that Qsp will be larger than Ksp for AgI while it is still less than Ksp for AgCl. Thus, AgI will form a precipitate before AgCl does.
If Ag+ is added to a solution of I- and Cl-, AgI forms a precipitate first because it has a lower molar solubility than AgCl.
The concentration of each species can be calculated using Qsp and Ksp. For example, given a solution of 0.1 M I– and 0.1 M Cl–, calculate [Ag+] when AgI first begins to precipitate:
K_{\rm{sp}}=\lbrack\rm{Ag}^+\rbrack\lbrack\rm{I}^-\rbrack=\lbrack\rm{Ag}^+\rbrack\lbrack 0.1\;\rm{M}\rbrack=8.52\times 10^{-17}
\lbrack\rm{Ag}^+\rbrack=8.52\times{10^{-16}}\;\rm{M}
, AgI begins to form a precipitate. Next, the concentration of I– remaining in solution when AgCl begins to precipitate can be calculated in order to reveal how successful the selective precipitation is at separating the compounds AgCl and AgI. To calculate [I–] when AgCl first starts to precipitate, first use the Ksp for AgCl to find [Ag+] at that point:
K_{\rm{sp}}=\lbrack\rm{Ag}^+\rbrack\lbrack\rm{Cl}^-\rbrack=\lbrack\rm{Ag}^+\rbrack\lbrack{0.1}\;\rm{ M}\rbrack=1.77\times 10^{-10}
\lbrack\rm{Ag}^+\rbrack=1.77\times{10^{-9}}\;\rm{ M}
when AgCl first starts to form. Finally, calculate [I–] when
\lbrack\rm{Ag}^+\rbrack=1.77\times{10^{-9}}\;\rm{ M}
to find the amount of I– left in solution:
\begin{aligned}K_{\rm{sp}}&=\lbrack\rm{Ag}^+\rbrack\lbrack\rm{I}^-\rbrack=(1.77\times{10}^{-9})\lbrack\rm{I}^-\rbrack=8.52\times{10^{-17}}\\{\lbrack}\rm{I}^-\rbrack&=\frac{8.52\times{10}^{-17}}{1.77\times{10}^{-9}}=4.81\times{10^{-8}}\;\rm{ M}\end{aligned}
By the time AgCl begins to precipitate, the concentration of I– has fallen from
1.0\times{10^{-1}}\;\rm{M}
4.81\times {10^{-8}}\;\rm{M}
, a difference of seven orders of magnitude. By this point, the I– is almost entirely removed from the solution in the form of solid AgI, and the separation of I– and Cl– is successful.
<Shifting the Solubility Equilibrium>Lewis Acids and Bases |
Design band-limited fractional delay FIR filter - MATLAB designFracDelayFIR - MathWorks 한êµ
h1 = 1×8
h2 = 1×32
h = 1×22
y\left[n\right]={h}_{fd}\left[n\right]âx\left[n\right]
\begin{array}{l}{h}_{fd}\left[n\right]=\mathrm{sinc}\left(nâfd\right),\\ \stackrel{^}{x}\left(t\right)=\underset{k}{â}x\left[k\right]\mathrm{sinc}\left(tâk\right)\\ â\stackrel{^}{x}\left(t+fd\right)=\underset{k}{â}x\left[k\right]\mathrm{sinc}\left(t+fdâk\right)\end{array}
{H}_{fd}\left(\mathrm{Ï}\right)={e}^{âj\mathrm{Ï}fd}
\stackrel{^}{x}\left(n+fd\right)ây\left[n\right]=\left(hâx\right)\left[n\right],\text{â}where\text{â}h\left[m\right]=\mathrm{sinc}\left(m+fd\right)·{K}_{N,\mathrm{β}}\left[m\right]
{K}_{N,\mathrm{β}}\left[m\right]
is a Kaiser window of length N and has a shape parameter β. The Kaiser window is designed to optimize the FIR frequency response, maximizing the combined bandwidths of both gain response and group delay response.
H\left(\mathrm{Ï}\right)={e}^{âj\mathrm{Ï}\left(fd+{i}_{0}\right)}
Given an FIR frequency response H(ω), the gain bandwidth is the largest interval [0 Ba] over which the gain response |H(ω)| is close to 1 up to a given tolerance value, tol.
{B}_{a}=\underset{\mathrm{Ï}}{\mathrm{max}}\left\{||H\left(\mathrm{ν}\right)|â1|<tol\text{â}â\text{â}0â¤\mathrm{ν}â¤\mathrm{Ï}\right\}
{B}_{g}=\underset{\mathrm{Ï}}{\mathrm{max}}\left\{|G\left(\mathrm{ν}\right)âfdâ{i}_{0}|<tol\text{â}â\text{â}0â¤\mathrm{ν}â¤\mathrm{Ï}\right\}
{B}_{c}=\mathrm{min}\left({B}_{a},{B}_{g}\right) |
Prove that discrete math the following statement (if true) or
Prove that discrete math the following statement (if true) or provide a counterexample (if false): For all n \geq 4, 2^n - 1 is not a prime number.
Prove that discrete math the following statement (if true) or provide a counterexample (if false): For all
n\ge 4,{2}^{n}-1
n=5 Then
{2}^{n}-1={2}^{5}-1=31
31 is a prime. So statement is false for n=5. As
5\ge 4
\cap \cup
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
p\left(n\right)-p\left(n-1\right)
1\right)=p\left(n\right)-p\left(n-1\right)
\left\{p\left(n\right){\right\}}_{n\in \mathbb{N}}
\prod _{i=1}^{\mathrm{\infty }}\frac{1}{1-{x}^{i}}
\left\{p\left(n-1\right){\right\}}_{n\in \mathbb{N}}
\prod _{i=1}^{\mathrm{\infty }}\frac{x}{1-{x}^{i}}
\prod _{i=1}^{\mathrm{\infty }}\frac{1}{1-{x}^{i}}-\prod _{i=1}^{\mathrm{\infty }}\frac{x}{1-{x}^{i}}=\left(1-
\prod _{i=1}^{\mathrm{\infty }}\frac{1}{1-{x}^{i}}
Suppose that A is the set of sophomores at your school and B is the set of students in discrete mathematics at your school. Express each of these sets in terms of A and B. a) the set of sophomores taking discrete mathematics in your school b) the set of sophomores at your school who are not taking discrete mathematics c) the set of students at your school who either are sophomores or are taking discrete mathematics |
ECDSA Private Keys Study of Security ()
National Technical University of Athens, Athens, Greece.
Kontogiannis, P. and Varvarigou, T. (2019) ECDSA Private Keys Study of Security. Open Access Library Journal, 6, 1-20. doi: 10.4236/oalib.1105423.
{F}_{p}
x,y
x\ne y
H\left(x\right)=H\left(y\right)
H\left(x\right)=y
{x}^{\prime }
{x}^{\prime }\ne x
H\left(x\right)=H\left({x}^{\prime }\right)
F\left(i\parallel z\right)
{y}^{2}={x}^{3}+Ax+B,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}A,B\in ℤ
\Delta =4\cdot {A}^{3}+27\cdot {B}^{2}\ne 0
E=\left\{\left(x,y\right):{y}^{2}={x}^{3}+Ax+B\right\}\cup \left\{O\right\}
{P}_{1}=\left({x}_{1},{y}_{1}\right)
{P}_{2}=\left({x}_{2},{y}_{2}\right)
E:{y}^{2}={x}^{3}+Ax+B
{P}_{1}\ne {P}_{2}
{x}_{1}={x}_{2}
{P}_{1}+{P}_{2}=0
{P}_{1}={P}_{2}
{y}_{1}=0
{P}_{1}+{P}_{2}=2{P}_{1}=0
{P}_{1}\ne {P}_{2}
{x}_{1}\ne {x}_{2}
\left\{\begin{array}{l}\lambda =\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\ \beta =-\lambda {x}_{1}+{y}_{1}=\frac{{y}_{1}{x}_{2}-{y}_{2}{x}_{1}}{{x}_{2}-{x}_{1}}\end{array}
⇒
{P}_{1}={P}_{2}
{y}_{1}\ne 0
\left\{\begin{array}{l}\lambda =\frac{3{x}_{1}^{2}+A}{2{y}_{1}}\\ \beta =-\lambda {x}_{1}+{y}_{1}=\frac{-{x}^{3}+Ax+2B}{2y}\end{array}
⇒
{P}_{1}+{P}_{2}=\left({\lambda }^{2}-{x}_{1}-{x}_{2},-{\lambda }^{3}+\lambda \left({x}_{1}+{x}_{2}\right)-\beta \right)
{y}^{2}={x}^{3}-x-1
{F}_{p}=\left\{0,1,2,\cdots ,p-1\right\}
{F}_{29}
\left\{0,1,2,3,\cdots ,28\right\}
{F}_{{p}^{n}}
{F}_{p}
B=\left\{{b}_{1},{b}_{2},\cdots ,{b}_{n}\right\}
a\in {F}_{{p}^{n}}
\alpha
a={a}_{1}\cdot {b}_{1}+{a}_{2}\cdot {b}_{2}+\cdots +{a}_{n}\cdot {b}_{n}
\left({a}_{1},{a}_{2},\cdots ,{a}_{n}\right)
{F}_{p}
{p}^{n}
{F}_{{p}^{n}}
{F}_{q}
q={p}^{n}
{F}_{{q}^{m}}
{p}^{m}
{F}_{{q}^{m}}
{F}_{q}
{p}^{m}
E:{y}^{2}+{a}_{1}\cdot xy+{a}_{3}\cdot y={x}^{3}+{a}_{2}\cdot {x}^{2}+{a}_{4}\cdot x+{a}_{6}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}\left(x,y\right)\in F.
{\alpha }_{1},{\alpha }_{2},{\alpha }_{3},{\alpha }_{4},{\alpha }_{6}\in F
\Delta \ne 0
\Delta
{E}_{1},{E}_{2}
\begin{array}{l}{E}_{1}:{y}^{2}+{a}_{1}\cdot xy+{a}_{3}\cdot y={x}^{3}+{a}_{2}\cdot {x}^{2}+{a}_{4}\cdot x+{a}_{6}\\ {E}_{2}:{y}^{2}+{{a}^{\prime }}_{1}\cdot xy+{{a}^{\prime }}_{3}\cdot y={x}^{3}+{{a}^{\prime }}_{2}\cdot {x}^{2}+{{a}^{\prime }}_{4}\cdot x+{{a}^{\prime }}_{6}\end{array}
u,r,s,t\in F
u\ne 0
\left(x,y\right)\to \left({u}^{2}\cdot x+r,{u}^{3}\cdot y+{u}^{2}\cdot s\cdot x+t\right)
{E}_{1}
{E}_{2}
\left(x,y\right)\to \left(\frac{x-3{a}_{1}^{2}-12{a}_{2}}{36},\frac{y-3{a}_{1}\cdot x}{216}-\frac{{a}_{1}^{3}+4{a}_{1}\cdot {a}_{2}-12{a}_{3}}{24}\right)
{F}_{p}
{y}^{2}={x}^{3}+a\cdot x+b\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}4\cdot {a}^{3}+27\cdot {b}^{2}\left(\mathrm{mod}p\right)\ne 0
{F}_{q}
#E\left({F}_{{q}^{n}}\right)
|#E\left({F}_{{q}^{n}}\right)-1-{q}^{n}|\le 2\cdot {q}^{\frac{n}{2}},\forall n\ge 1.
|#E\left({F}_{q}\right)-1-q|\le 2\cdot \sqrt{q},\forall n\ge 1
{F}_{p}
B\cdot {y}^{2}={x}^{3}+A\cdot {x}^{2}+x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}B\cdot \left({A}^{2}-4\right)\left(\mathrm{mod}p\right)\ne 0
{F}_{p}
{x}^{2}+{y}^{2}=1+d\cdot {x}^{2}\cdot {y}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}d\cdot \left(1-d\right)\left(\mathrm{mod}p\right)\ne 0
\left(x,y\right)\to \left(B\cdot u-\frac{A}{3},B\cdot v\right)
{v}^{2}={u}^{3}+a\cdot u+b,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}a=\frac{3-{A}^{2}}{3\cdot {B}^{2}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}b=\frac{2\cdot {A}^{3}-9\cdot A}{27\cdot {B}^{3}}
\left(x,y\right)\to \left(u/v,\left(u-1\right)/\left(u+1\right)\right)
B\cdot {v}^{2}={u}^{3}+A\cdot {u}^{2}+u,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}A=\frac{2\cdot \left(1+d\right)}{1-d}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{\hspace{0.17em}}B=\frac{4}{1-d}
{F}_{p},p>3
\left(x,y\right)
x,y\in {F}_{p}
{F}_{p}
{F}_{p}
\begin{array}{l}E:{y}^{2}={x}^{3}+A\cdot x+B,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}A,B\in {F}_{p}\\ \text{with}\text{\hspace{0.17em}}\text{restriction}\text{\hspace{0.17em}}4\cdot {A}^{3}+27\cdot {B}^{2}\left(\mathrm{mod}p\right)\ne 0\end{array}
E\left({F}_{p}\right)
x,x\in ℕ
Q=x\cdot P
O\left(n\right)
n=\underset{i}{\prod }{p}_{i}^{{e}_{i}}
O\left({\sum }_{i}{e}_{i}\left(\mathrm{log}n+\sqrt{{p}_{i}}\right)\right)
{F}_{p}
#E\left({F}_{p}\right)=N
N={p}_{1}{p}_{2}\cdots {p}_{n}
{n}_{i}\equiv {p}_{i}
T=\left(p,a,b,G,n,h\right)
p={2}^{256}-{2}^{32}-{2}^{9}-{2}^{8}-{2}^{7}-{2}^{6}-{2}^{4}-1
,is the size of the Galois field
{F}_{p}
h=1.
n\cdot P\left(\mathrm{mod}p\right)=0
E\left({F}_{p}\right)
h=\frac{N=#E\left({F}_{p}\right)}{n}
h\in ℕ
n\cdot \left(h\cdot P\right)=n\cdot G=0
G=h\cdot P.
{x}_{1}\cdot P,{x}_{2}\cdot P,{x}_{3}\cdot P,\cdots
{x}_{1},{x}_{2},{x}_{3},\cdots
x\cdot P=Q
O\left(p\right)
Q=x\cdot P
Q-a\cdot m\cdot P=b\cdot P
x,x\in ℤ
a,m,b\in ℤ
x=a\cdot m+b
P,Q
E\left({F}_{q}\right)
\left({x}_{1},{x}_{2},{x}_{3},\cdots \right)
\begin{array}{l}\text{Vector}\text{\hspace{0.17em}}1:{x}_{1}\cdot P,{x}_{2}\cdot P,{x}_{3}\cdot P,\cdots \\ \text{Vector}\text{\hspace{0.17em}}2:Q-{x}_{1}\cdot P,Q-{x}_{2}\cdot P,Q-{x}_{2}\cdot P,\cdots \end{array}
{x}_{i}\cdot P=Q-{x}_{j}\cdot P,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{where}\text{\hspace{0.17em}}i,j=1,2,3,\cdots
O\left(\sqrt{q}\right)
\left(a,A\right)
\left(b,B\right)
a\cdot P+b\cdot Q=A\cdot P+B\cdot Q,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}a,b,A,B\in ℤ
x=\left(a-A\right)\cdot {\left(B-b\right)}^{-1}\mathrm{mod}n
X\in 〈P〉
\left(c,d\right)
X=c\cdot P+d\cdot Q
f:〈P〉\to 〈P〉
X=f\left(X\right)
\stackrel{¯}{c},\stackrel{¯}{d}\in \left[0,n-1\right]
\stackrel{¯}{X}=\stackrel{¯}{c}\cdot P+\stackrel{¯}{d}\cdot Q
〈P〉
\left\{{S}_{1},{S}_{2},\cdots ,{S}_{L}\right\}
X=c\cdot P+d\cdot Q
f\left(X\right)=\stackrel{¯}{X}=\stackrel{¯}{c}\cdot P+\stackrel{¯}{d}\cdot Q
\stackrel{¯}{c}=c+{a}_{j}\mathrm{mod}n
\stackrel{¯}{d}=d+{b}_{j}\mathrm{mod}n
{X}_{0}\in 〈P〉
{\left\{{X}_{i}\right\}}_{i\ge 0}
{X}_{i}=f\left({X}_{i-1}\right)
i\ge 1
〈P〉
{X}_{w}={X}_{w+s}
s\ge 1
{X}_{i}={X}_{i-s},\forall i\ge w+s
\sqrt{\text{π}\cdot n/2}
t~\sqrt{\text{π}\cdot n/8}
s~\sqrt{\text{π}\cdot n/8}
\left({X}_{i},{X}_{2i}\right)
i=1,2,3,\cdots
{X}_{i}={X}_{2i}
{X}_{i},{X}_{j}⇒{X}_{i}={X}_{j}
i\ne j
k\in \left[w,w+s\right]
1.0308\cdot \sqrt{n}
{F}_{50101}
{F}_{100153}
{y}^{2}={x}^{3}+a\cdot {x}^{2}+b\cdot x+c
{F}_{p}
E\left({F}_{p}\right)
Q=d\cdot P
{F}_{100153}
E:{y}^{2}={x}^{3}+2065150\cdot {x}^{2}+x\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{with}\text{\hspace{0.17em}}x\in ℝ
P\left(\text{22},\text{72669}\right)
\left(Q,d\right)
{y}^{2}={x}^{3}-28486\cdot x+1926947
{y}^{2}\cong {x}^{3}-3\cdot x+4.621\cdot {10}^{76},x\in ℝ
P\left(4,21591\right)
P\left(4,28510\right)
P\left(7,24312\right)
P\left(16,2549\right)
[1] Secp256k1.
[2] Lange, T. and Bernstein, D.J. (2013) SafeCurves: Choosing Safe Curves for Elliptic-Curve Cryptography.
[3] Entropy (Information Theory).
[4] Narayanan, A., Bonneau, J., Felten, E., Miller, A. and Goldfender, S. (2015) Bitcoin and Cryptocurrency Technologies. Princeton University, Princeton, NJ.
[5] Johnson, D., Menezes, A. and Vanstone, S. (2001) The Elliptic Curve Digital Signature Algorithm (ECDSA). Department of Combinatorics & Optimization, University of Waterloo.
http://www.cs.miami.edu/home/burt/lea
rning/Csc609.142/ecdsa-cert.pdf
[6] Silverman, J.H. (2006) An Introduction to the Theory of Elliptic Curves. Brown University and NTRU Cryptosystems, Inc.
https://www.math.brown.edu/~jhs/Prese
ntations/WyomingEllipticCurve.pdf
[7] Benvenuto, C.J. (2012) Galois Field in Cryptography.
https://sites.math.washington.edu/~mo
rrow/336_12/papers/juan.pdf
[8] Edward Curve.
[9] Hasse’s Theorem on Elliptic Curves.
https://en.wikipedia.org/wiki/Hasse%
27s_theorem_on_elliptic_curves
[10] Rational Point.
[11] Rykwalder, E. (2014) The Math Behind Bitcoin.
[12] Corbellini, A. (2015) Elliptic Curve Cryptography: A Gentle Introduction.
http://andrea.corbellini.name/2015/05/17/
elliptic-curve-cryptography-a-gentle-introduction/
[13] Kha-lique, A., Sood, S. and Singh, K. (2010) Implementation of Elliptic Curve Digital Signature Algorithm. Internatuonal Journal of Computer Applications, 2, No. 2.
http://www.ijcaonline.org/volume2/
number2/pxc387876.pdf
[14] Corbellini, A. (2015) Elliptic Curve Cryptography: Breaking Security and a Comparison with RSA.
elliptic-curve-cryptography-breaking-security-
and-a-comparison-with-rsa/
[15] Qubd (2017)
https://github.com/qubd/mini_ecdsa/blob
/master/mini_ecdsa.py |
Two employees of Frontier Fence Company can install
100
feet of fence in three days. At the same rate, how many employees are needed to install
150
feet of fence in one day?
Since the employees work at the same rate, we can use the given work rate for two employees to find how many employees would be needed to install
150
feet of fence in one day.
Write a proportional equation for this situation.
\frac{100}{3} x = 2 \cdot 150
9 |
About Alexandru Froda: Romanian mathematician (1894 - 1973) | Biography, Facts, Career, Life
peoplepill id: alexandru-froda
Intro Romanian mathematician
Death 7 October 1973, Bucharest, Principality of Wallachia (aged 79 years)
University of Paris (1896-1968) doctorate
Alexandru Froda (July 16, 1894 in Bucharest, Romania – October 7, 1973 in Bucharest, Romania) was a well-known Romanian mathematician with important contributions in the field of mathematical analysis, algebra, number theory and rational mechanics. In his 1929 thesis he proved what is now known as Froda's theorem.
Alexandru Froda was born in Bucharest in 1894. In 1927 he graduated from the University of Sciences (now the Faculty of Mathematics from the University of Bucharest). He received his Ph.D. from the University of Paris and from University of Bucharest in 1929. He was elected president of the Romanian Mathematical Society in 1946. In 1948 he became professor at the Faculty of Mathematics and Physics at the University of Bucharest.
Froda's major contribution was in the field of mathematical analysis. His first important result was concerned with the set of discontinuities of a real-valued function of a real variable. In this theorem Froda proves that the set of simple discontinuities of a real-valued function of a real variable is at most countable.
In a paper from 1936 he proved a necessary and sufficient condition for a function to be measurable. In the theory of algebraic equations, Froda proved a method of solving algebraic equations having complex coefficients.
In 1929 Dimitrie Pompeiu conjectured that any continuous function of two real variables defined on the entire plane is constant if the integral over any circle in the plane is constant. In the same year Froda proved that, in the case that the conjecture is true, the condition that the function is defined on the whole plane is indispensable. Later it was shown that the conjecture is not true in general.
In 1907 Pompeiu constructed an example of a continuous function with a nonzero derivative which has a zero in every interval. Using this result Froda finds a new way of looking at an older problem posed by Mikhail Lavrentyev in 1925, namely whether there is a function of two real variables such that the ordinary differential equation
{\displaystyle dy=f(x,y)dx}
has at least two solutions passing through every point in the plane.
In the theory of numbers, beside rational triangles he also proved several conditions for a real number, which is the limit of a rational convergent sequence, to be irrational, extending a previous result of Viggo Brun from 1910.
In 1937 Froda independently noticed and proved the case
{\displaystyle n=1}
of the Borsuk–Ulam theorem.
https://id.loc.gov/authorities/names/ns2015000629
https://www.worldcat.org/identities/lccn-ns2015000629
Romanian musician and actress
Romanian female speed skater
Virgil-Daniel Gruia
Romanian politician from terpezița, dolj county |
Image-Based Vehicle Speed Estimation
Image-Based Vehicle Speed Estimation*
Md. Golam Moazzam1, Mohammad Reduanul Haque2, Mohammad Shorif Uddin1
2Department of Computer Science and Engineering, Daffodil International University, Dhaka, Bangladesh
Vehicle speed is an important parameter that finds tremendous application in traffic control identifying over speed vehicles with a view to reducing accidents. Many methods, such as using RADAR and LIDAR sensors have been proposed. However, these are expensive, and their accuracy is not quite satisfactory. In this paper, a video-based vehicle speed determination method is presented. The method shows satisfactory performance on standard data sets and gives that error rate of velocity estimation is within 10%.
Vehicle Detection, Speed Calculation, Background Subtraction, Vehicle Tracking
Traffic accidents are very dangerous as these result in injury and death of passengers and pedestrians in addition to the damage of vehicles and roads. Bangladesh is one of the top countries in the world where road accident rate is very high. A traffic accident is a great pain and loss of a nation. Though concerned authorities have taken many initiatives to minimize the road accident rate and increase the road safety, still every year thousands of people get killed and injured due to road accidents. According to statistics of road accidents and casualties from Bangladesh Police, in the year of 2016, 2566 dangerous accidents occurred and that caused 2463 deaths and 2134 injuries [1] . In fact, the actual number of accidents is very high as many accidents are not reported to concerned authorities. The speed of vehicle is considered as one of the main factors for road accidents, and, also it is an important traffic parameter, so detection of speed of a vehicle [2] - [7] is very significant for more smooth traffic management. Various methods based on RADAR (Radio Detection and Ranging) or LIDAR (Laser Infrared Detection and Ranging) or camera have been developed, but none of these techniques are perfect.
In this paper, an imaged-based vehicle speed estimation method is developed using physics-based velocity theory. It consists of a video camera that is placed a fixed location for capturing images and then a computation system works on the images to calculate the speed. Several video-based techniques are developed for detecting moving objects, such as temporal differentiation, optical flow, background subtraction, etc. Among these, background subtraction technique is simpler than other techniques. In this technique, the absolute difference of the background frame and the current frame is taken. Here, a hybrid technique is used that consists of an adaptive background subtraction technique and a three-frame differentiation method.
The method consists of five major modules such as image acquisition and enhancement, segmentation, centroid calculation, shadow removal and speed calculation.
The rest of the paper is organized as follows. Section II describes the theory and method along with implementation for vehicle speed estimation. In Section III, experimental result and discussions are given. Finally, conclusions are drawn in Section IV.
A flow diagram of the vehicle speed detection method is shown in Figure 1. A brief explanation of this algorithm is given below:
It consists of three major blocks. The first block is video image capturing and background subtraction. It consists of a stationary video camera that is placed a fixed location for capturing images. The primary task of the vehicle speed estimation method is to detect the moving vehicle from the video. A three-frame differencing technique is applied to know the motion pixels. So, the second block is the extraction of vehicle by using an adaptive background subtraction method. Here, the stationary pixels are background pixel and the moving pixels are foreground, as shown in Figure 2. This is accomplished through image enhancement (noise reduction), vehicle centroid and area calculation for getting the vehicle bounding box.
Let the position of a captured pixel is (x,y) at time t = n and its intensity is In(x,y). Using three frame differencing technique, an object is moving if the position of a pixel in the current image (In) and its position in the consecutive previous frames (In-1) and (In-2). Mathematically, motion is detected if the conditions of Equation (1) hold.
\left({I}_{n}\left(x,y\right)-{I}_{n-1}\left(x,y\right)>T{h}_{n}\left(x,y\right)\right)
\left({I}_{n}\left(x,y\right)-{I}_{n-2}\left(x,y\right)>Th\left(x,y\right)\right)
T{h}_{n}\left(x,y\right)
is the threshold value at pixel position (x,y).
Then the background subtraction image is achieved by the subtraction of the background Bn(x,y) frame from the current frame In(x,y) through Equation (2).
Figure 1. Flow diagram of the vehicle speed detection method.
Figure 2. Moving object extraction through background subtraction.
S{I}_{n}\left(x,y\right)={I}_{n}\left(x,y\right)-{B}_{n}\left(x,y\right)
S{I}_{n}\left(x,y\right)
be the subtracted image.
From the background subtracted image the noise pixels are removed based on outlier and the shadow pixels are removed based on the intensity of pixels, as the intensity of a shadow pixel is lower than the intensity of an object pixel. After that object tracking is done through segmentation by connectivity of pixels and labeling of objects. Each labelled object is bounded through a rectangle. Then the area of each labelled object is calculated. Tracking of each object (vehicle) is recorded when it enters the scene (at frame
S{I}_{0}
) and when it leaves the scene (at frame SIn). Centroid of vehicle in the respective frame can be easily determined from the labelled object using the x and y coordinates as
\left({x}_{c},{y}_{c}\right)=\left(\frac{\left({x}_{1}+{x}_{2}\right)}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)
Table 1. Vehicle speed measurement result.
\left({x}_{c},{y}_{c}\right)
The third and the final block is the calculation of the speed of the vehicle. Speed can be easily calculated by counting the frame numbers needed to enter and leave the labelled object and the distance it covers. The Euclidean distance of the centroids of nth and (n − 1)th frames gives the distance traveled by the respective object (vehicle). The frame rate of the video is then multiplied by the total number of frames. From this total time and distance, speed is measured and mapped in real time through Equation (5).
\text{Distance}=\sqrt{{\left({x}_{n-1}-{x}_{n}\right)}^{2}+{\left({y}_{n-1}-{y}_{n}\right)}^{2}}
\left(\left({x}_{n},{y}_{n}\right),\left({x}_{n-1},{x}_{n-1}\right)\right)
is the coordinates of the centroid pixel in nth frame and (n − 1)th frame, respectively.
\text{Speed}=\frac{\alpha \times \text{Distance}}{\left(\text{Frame}\left(n\right)-\text{Frame}\left(n-1\right)\right)\times \text{FrameRate}}
where, α is the calibration coefficient that maps image to object motion and can be calculated as
α = real height of the vehicle/image height of the vehicle.
For our experiment, we have used QMUL Dataset [8] for our experimentation. This is a traffic dataset whose video length is 1 hour and has 90,000 frames. The size of each frame is 360 × 288 pixels and the frame rate is 25 Hz.
Table 1 shows the experimental result. This table confirms that the speed of the vehicle calculated by our system is similar to the real speed of the vehicle, as the error rate is within 10%.
An image-based vehicle speed estimation system is presented in this paper, which is a good alternative to the traditional RADAR or LIDAR-based system. We have done experimentation through standard dataset and estimated the real speed of a vehicle. From the experimental findings, it is confirmed that the system works well with good accuracy as the error rate is within 10% limit. In the future, we will work for accuracy improvement as well as implementation of the system in real-life environment.
Moazzam, Md.G., Haque, M.R. and Uddin, M.S. (2019) Image-Based Vehicle Speed Estimation. Journal of Computer and Communications, 7, 1-5. https://doi.org/10.4236/jcc.2019.76001
1. Sharif Hossen, Md. (2019) Analysis of Road Accidents in Bangladesh. American Journal of Transportation and Logistics (AJTL), 2, 1-11.
2. Cathey, F.W. and Dailey, D.J. (2005) A Novel Technique to Dynamically Measure Vehicle Speed Using Uncalibrated Roadway Cameras. IEEE Intelligent Vehicles Symposium, Las Vegas, 6-8 June 2005, 777-782. https://doi.org/10.1109/ivs.2005.1505199
3. Douxchamps, D., Macq, B. and Chihara, K. (2006) High Accuracy Traffic Monitoring Using Road-Side Line Scan Cameras. IEEE Intelligent Transportation Systems Conference (ITLS2006), Toronto, September 2006, 875-878. https://doi.org/10.1109/itsc.2006.1706854
4. Ibrahim, O., ElGendy, H. and ElShafee, A.M. (2011) Towards Speed Detection Camera System for a RADAR Alternative. 2011 11th International Conference on ITS Telecommunications, St. Petersburg, 23-25 August 2011, 627-632. https://doi.org/10.1109/itst.2011.6060131
5. Dedeo˘glu, Y. (2004) Moving Object Detection, Tracking and Classification for Smart Video Surveillance. Master Thesis, Department of Computer Engineering, Bilkent University, Turkish.
6. Collins, R.T., Lipton, A.J., Kanade, T., Fujiyoshi, H., Duggins, D., Tsin, Y., Tolliver, D., Enomoto, N., Hasegawa, O., Burt, P. and Wixson, L. (2000) A System for Video Surveillance and Monitoring. The RoboticsInstitute, Carnegie Mellon University, Princeton, NJ.
7. Adnan, M.A. and Zainuddin, N.I. (2013) Vehicle Speed Measurement Technique Using Various Speed Detection Instrumentation. 2013 IEEE Business Engineering and Industrial Applications Colloquium (BEIAC), Langkawi, 7-9 April 2013, 668-672. https://doi.org/10.1109/beiac.2013.6560214
8. QMUL Junction Dataset (2016). http://personal.ie.cuhk.edu.hk/~ccloy/downloads_qmul_junction.html
*An earlier version of this paper is published in the Proc. of International Workshop on Computational Intelligence (IWCI), 12-13 December 2016, Dhaka, Bangladesh. |
Julian was calculating the area of a rectangular garden that was
7
23
feet long. He wrote down the product
7·23
, when he remembered that he could use the Distributive Property to make this calculation easier. He rewrote his product as
7(20+3)=7(20)+7(3)
Finish Julian's calculation. Did the Distributive Property make his calculation easier? Why or why not?
Remember your multiplication rules and the order of operations! Terms are separated by addition symbols (
+
) and you should simplify those terms first, before adding.
140+21=161
. The area of the garden is
161
squared feet. Do you think the Distributive Property made this problem easier to solve?
Now use the Distributive Property to calculate each of the following products.
4(63)
11(71)
Having trouble using the Distributive Property? This property states that when we multiply a sum by a number, we need to multiply each part of the sum by that number. Remember, it often useful to separate the number by place value.
In part (i), we can separate the sum (
63
) by place value (tens and ones). For example,
4(63)=4(60+3)
. Now fill in the expression:
4(60)\ + |
The function f(x)=2x^{3}-33x^{2}+168x+9 has one local minimum and one local
f\left(x\right)=2{x}^{3}-33{x}^{2}+168x+9
x=?
x=?
To estimate (approximate calculation) the local maximum and local minimum of the given function
y=f\left(x\right)
Now we know the critical values are at
x=4
x=7
. For further analysis graph the function between
x=3
x=8
(in the neighbourhood of the critical points)
f\left(x\right)=2{x}^{3}-33{x}^{2}+168x+9
The local max and min occur when
{f}^{\prime }\left(x\right)=6{x}^{2}-66x+168
=6\left({x}^{2}-11x+28\right)
=6\left(x-4\right)\left(x-7\right)=0
x=4,\text{ }x=7
Estimates: local minimum at
x=7
with value 255, local maximum at
x=4
with value 280
P\left(x\right)=-12{x}^{2}+2136x-41000
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
Which statement best describes the zeros of the function
h\left(x\right)=\left(x-6\right)\left({x}^{2}+8x+16\right)
A. The function has two distinct real zeros
B. The function has three distinct real zeros.
C. The function has one real and two complex zeros.
D. The function has three complex zeros.
Evaluate the algebraic expressions for the given values of the variables.
{\left(x-y\right)}^{2},x=5\text{ }\text{ and }\text{ }y=-3
{y}^{\prime }-xy+y=2y
Engineering statistics, i need solutions in 15 minutes please. MCQ/Engineering company has a task of checking compressive strength for 100 concrete cubes. The results revealed that 85 cubes passed the compressive strength test successfully and 15 cubes failed in the test. If 10 cubes are selected at random to be inspected by the company, determine the probability that the 8 cubes will pass the test and 2 cubes will fail in the test by using the Combinatorial Analysis. A-0.6553 B-0.2919 C-0.3415 D-0.4522 E-0.1156
1) A group of preschooters has 63 boys and 27 girl. What is the ratio of boys to all children? |
Waves/Geometrical Problems - Wikibooks, open books for an open world
Waves/Geometrical Problems
Figure 3.14: Refraction through multiple parallel
layers with different refractive indices.
Figure 3.15: Refraction through a
{\displaystyle 45^{\circ }}
{\displaystyle 45^{\circ }}
{\displaystyle 90^{\circ }}
prism.
Figure 3.16: Focusing of parallel rays by a parabolic mirror.
Figure 3.17: Refraction through a wedge-shaped prism.
The index of refraction varies as shown in figure 3.14:
{\displaystyle \theta _{1}}
, use Snell's law to find
{\displaystyle \theta _{2}}
{\displaystyle \theta _{2}}
{\displaystyle \theta _{3}}
From the above results, find
{\displaystyle \theta _{3}}
{\displaystyle \theta _{1}}
. Do
{\displaystyle n_{2}}
{\displaystyle \theta _{2}}
{\displaystyle 45^{\circ }}
{\displaystyle 45^{\circ }}
{\displaystyle 90^{\circ }}
prism is used to totally reflect light through
{\displaystyle 90^{\circ }}
as shown in figure 3.15. What is the minimum index of refraction of the prism needed for this to work?
Show graphically which way the wave vector must point inside the calcite crystal of figure 3.3 for a light ray to be horizontally oriented.
The human eye is a lens which focuses images on a screen called the retina. Suppose that the normal focal length of this lens is
{\displaystyle 4{\mbox{ cm}}}
and that this focuses images from far away objects on the retina. Let us assume that the eye is able to focus on nearby objects by changing the shape of the lens, and thus its focal length. If an object is
{\displaystyle 20{\mbox{ cm}}}
from the eye, what must the altered focal length of the eye be in order for the image of this object to be in focus on the retina?
Show that a concave parabolic mirror focuses incoming rays which are parallel to the optical axis of the mirror precisely at a focal point on the optical axis, as illustrated in figure 3.16. Hint: Since rays following different paths all move from the distant source to the focal point of the mirror, Fermat's principle implies that all of these rays take the same time to do so (why is this?), and therefore all traverse the same distance.
Use Fermat's principle to explain qualitatively why a ray of light follows the solid rather than the dashed line through the wedge of glass shown in figure 3.17.
Test your knowledge of Fermat's principle by finding the value of
{\displaystyle y}
{\displaystyle t}
is a minimum in equation (3.14). Use this to derive Snell's law.
Retrieved from "https://en.wikibooks.org/w/index.php?title=Waves/Geometrical_Problems&oldid=3664625" |
Revision as of 09:03, 28 July 2020 by Bot (talk | contribs) (Created page with "{{Publication | published = true | date = 2020-07-21 | authors = Nicolas Boutry, Rocio Gonzalez-Diaz, Maria-Jose Jimenez, Eduardo Paluzo-Hildago | title = Euler Well-Composedn...")
In this paper, we define a new flavour of well-composednesscalled Euler well-composedness, in the general setting of regular cell complexes: A regular cell complex is Euler well-composed if the Euler characteristic of the link of each boundary vertex is
{\displaystyle 1}
{\displaystyle I}
{\displaystyle {\big (}K(I),K({\bar {I}}){\big )}}
{\displaystyle K(I)}
{\displaystyle K({\bar {I}})}
{\displaystyle I}
{\displaystyle {\bar {I}}}
). Thena cell decomposition of a picture
{\displaystyle I}
{\displaystyle K(I)}
{\displaystyle K({\bar {I}})}
are Euler well-composed. We prove in this paper that, first, self-dual Euler well-composedness is equivalent to digital well-composedness in dimension 2 and 3, and second, in dimension 4, self-dual Euler well-composedness implies digital well-composednessthough the converse is not true. |
Use the binomial probability formula to find P(x) n= 16,
Use the binomial probability formula to find P(x) n= 16, x=3, p- 1/5
Use the binomial probability formula to find
P\left(x\right)
n=16,\text{ }x=3,\text{ }p-\frac{1}{5}
Answer is on photo
What is the probability of 18 to 22 succeses? p=.20 and n=100
P\left(x\right)
n=4,\text{ }x=2,\text{ }p=0.4
80% A poll is given, showing are in favor of a new building project. 8 people are chosen at random, what is the probability that exactly 3 of them favor the new building project?
Sue and Ann are taking the same English class but they do not study together, so whether one passes will be Independent of whether the other passes. In other words, "Sue passes and "Ann passes" are assumed to be independent events. The probebiity that Sue passes English is 0.8 and the probablity that Ann passes English is 0.75 What is the probability neither girl passes English?
Determine whether you can use the normal distribution to approximate the binomial distribution. If you can, use the normal distribution to approximate the indicated probabilities and sketch their graphs. If you cannot, explain why and use the binomial distribution to find the indicated probabilities.
A survey of adults in a region found that 33% name professional football as their favorite sport. You randomly select 11 adults in the region and ask them to name their favorite sport. Complete parts (a) through (d) below
(a) Find the probability that the number who name professional football as their favorite sport is exactly 8. |
In the first 4 papers each of hundred marks Ravi got 90, 75, 73, 85 marks If he wants an - Maths - Linear Inequalities - 6856945 | Meritnation.com
Let the marks obtained by him in 5th paper =x
Average of marks =
\frac{90+75+73+85+x}{5}
he wants an average of greater than or equal to 75 marks and less than80 marks. |
Circuit Behavior: Level 4-5 Challenges Practice Problems Online | Brilliant
8 resistors (orange color) are connected to form a regular octagon. 8 more resistors (blue color) connect the vertices of the octagon to its center. All the 16 resistors are of resistance
\SI{420}{\ohm}
If the connecting wires have negligible resistance, calculate the equivalent resistance (in ohms, rounded to the nearest integer) between the terminals
A
B
You are given a disk of thickness
h
with inner and outer radii
r_1
r_2
, respectively. If the resistivity of the disk varies as
\rho = \rho_0 \left|\sec \varphi\right|
\varphi
is the polar angle, find the resistance between the points
A
B
The inner and outer rims are metal rings with zero resistance.
\dfrac {r_2}{r_1} = e^2 \approx 7.389
\rho_0 = \SI{10}{\ohm \meter}
h= \SI{3}{\centi \meter}
In the circuit above, wire
AB
40\text{ cm}
and resistance per unit length
\SI[per-mode=symbol]{0.5}{\ohm\per\centi\meter}
. The voltmeter is ideal.
If we want to make the reading in the voltmeter vary with time as
V(t) = 2 \sin(\pi t) \ \si{\volt},
then what should be the velocity of the contact (the arrow-tipped end of the wire above) as a function of time?
If the velocity can be expressed as
A\sin(\omega t+\phi) \text{ cm}\,\text{s}^{-1},
0<\phi<\pi
, then enter the value of
\dfrac A{\omega- \phi}
by Gauri shankar Mishra
A useless wire having a total resistance of
48 \space \Omega
is cut into 48 equal pieces. Then, a regular Deltoidal Icositetrahedron as shown below.
If the equivalent resistance between two opposite points, where four edges meet together is
R \space \Omega
, then enter your answer as the value of
100R
This question is part of the set Platonic Electricity.
by Abhay Tiwari
The DC circuit above consists of two voltage sources
V_1
V_2
with internal resistances
R_1 = 1 \Omega
R_2 = 2 \Omega
respectively. There is a load
R_L = 3 \Omega
connected in parallel with the sources.
V_1
V_2
are variable quantities.
P_1
P_2
P_L
be the amounts of power in watts dissipated by
R_1
R_2
R_L
P_L = 10
watts, determine the minimum possible value of
P_1 + P_2 + P_L |
An insurance company found that 25% of all insurance policies
An insurance company found that 25% of all insurance policies are terminated before their maturity date. Assume that 10 polices are randomly selected from the company’s policy database. Assume a Binomial experiment.
What is the probability that at most eight policies are not terminated before maturity?
An insurance company found that all insurance policies terminated before their maturity date.
The Probability of insurance terminated before the maturity date is 25%
q=0.25
Then, the Probability of insurance not terminated before the maturity date is,
p=1-0.25=0.75
n=10
policies are randomly selected from the company’s policy database.
Let success be the insurance not terminated before the maturity date the required calculation can be defined by the formula,
P\left(X=x\right)=n{C}_{x}{p}^{x}{q}^{n-x}
n number o samples
x number of success
q probability of failure
to find the required probability:
P\left(X\le 8\right)=\sum _{x=0}^{8}10{C}_{x}{\left(0.75\right)}^{x}{\left(0.25\right)}^{10-x}
=1-\sum _{x=9}^{10}10{C}_{x}{\left(0.75\right)}^{x}{\left(0.25\right)}^{10-x}
=1-\left[10{C}_{9}{\left(0.75\right)}^{9}{\left(0.25\right)}^{10-9}+10{C}_{10}{\left(0.25\right)}^{10-10}\right]
=1-\left[0.1877+0.0563\right]
=1-\left[0.2440\right]
=0.7560
Therefore the probability that at most eight policies are not terminated before maturity is 0.7560 or 75.60%
p=0.25,q=1-p=1-0.25=0.75,n=10
Let X be the number of policies terminated before maturity-
Probability that at most 8 policies and not terminated before maturity-
P\left(X\le 8\right)=1-P\left(X>8\right)=1-\left[P\left(X=9\right)+P\left(X=10\right)\right]
=1-\left[{}^{\left\{10\right\}}{C}_{9}{\left(0.25\right)}^{9}{\left(0.75\right)}^{1}{+}^{10}{C}_{10}{\left(0.25\right)}^{10}\right]
PSK=1-(0.0000286+0.000009536)
=1-0.00002955=0.999997
An insurance company found that 25% of all insurance policies are terminated before their maturity date
⇒
Probability of policy termination
p=\frac{25}{100}=\frac{1}{4}
Probability of policy not terminating
=1-1/4=3/4
15 policies are randomly selected
⇒n=15
p\left(x\right){=}^{n}{C}_{x}{p}^{x}{q}^{n-x}
probability that more than 8 but less than 11 policies are terminated before maturing
⇒x=9,10
⇒probability=p\left(9\right)+p\left(10\right)
{=}^{15}{C}_{9}\left(\frac{1}{4}{\right)}^{9}\left(\frac{3}{4}{\right)}^{6}{+}^{15}{C}_{10}\left(\frac{1}{4}{\right)}^{10}\left(\frac{3}{4}{\right)}^{5}
=\left({3}^{5}/{4}^{15}\right)\left(5005\cdot 3+3003\right)
=\left({3}^{5}/{4}^{15}\right)\left(18018\right)
= 0.41% is the probability that more than 8 but less than 11 policies are terminated before maturing
How do you use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial for
n=5,\text{ }x=2,\text{ }p=0.25
Harry Ohme is in change of the electronics section of a large store. He has noticed that the probability that a customer who is just browsing will buy something is 0.3. Suppose that 15 customers browse in the electronics section each hour. What is the probability that no browsing customers will buy anithing during a specified hour?
If 40% of all commuters ride to wirk in carpools, find the probability that if 8 workers are selected, five will ride in carpools. Define your random variance X and determine X has a binomial or a geometric distribution.
Assume that a procedure yields a binomial distribution with
n=33
trials and a probability of success of
p=0.10
Use a binomial probability table to find the probability that the number of successes x is exactly 22. Using a binomial probability table
Sixty-eight percent of adults would still consider a car brand despite product/safety recalls. You randomly select 20 adults. Find the probability that the number of adults that would still consider a car brand despite product/safety recalls is at most one.
Distribution type: ? Probability: ?
You took a test with 40 multiple choice questions. Each question had 5 choices. What is the probability that you get at least 28 questions correct if you guessed in all of them. |
How do you find \frac{d^2y}{dx^2} by implicit differentiation where x^2y+xy^2=3x
\frac{{d}^{2}y}{{dx}^{2}}
by implicit differentiation where
{x}^{2}y+x{y}^{2}=3x
{x}^{2}y+x{y}^{2}=3x
with respect to x we get
2xy+{x}^{2}{y}^{\prime }+{y}^{2}+2xy{y}^{\prime }=3
{y}^{\prime }=\frac{3-{y}^{2}-2xy}{{x}^{2}+2xy}
for the second derivative we obtain
y{}^{″}=\frac{\left(-2y{y}^{\prime }-2y-2x{y}^{\prime }\right)\left(x+2xy\right)-\left(3-{y}^{2}-2xy\right)\left(2x+2y+2x{y}^{\prime }\right)}{{\left({x}^{2}+2xy\right)}^{2}}
Now plug the result for y' in this equation!
\frac{dw}{dt}
\frac{dw}{dt}
w=x\mathrm{sin}y,\text{ }x={e}^{t},\text{ }y=\pi -t
t=0
Value of differential equation at non existent points in the original function.
{y}^{2}+y={x}^{3}
Find a differential equation for
w\left(t\right)={\varphi }_{t}\left(z\right)=\frac{z+\mathrm{tan}t}{1-z\left(\mathrm{tan}t\right)}
Usually the question is the other way around, but (D. H. Sattinger and O. L. Weaver, 1986 Lie Groups and Algebras with Applications to Physics, Geometry, and Mechanics) pose it like this.
Earlier we showed that
{\varphi }_{t+s}={\varphi }_{t}\circ {\varphi }_{s}
Since tan is a periodic function, I thing that the differential equation they are looking for might be of order 2.
My efforts so far consists of calculating w'(t), and trying to identify w in there, but things get messy.
{f}^{{}^{\prime }}\left(x\right)=\gamma \frac{f\left(x\right)+{f}^{2}\left(x\right)}{\mathrm{log}\left(\frac{f\left(x\right)}{1+f\left(x\right)}\right)}
x\ge \text{ }\text{and}\text{ }f\left(0\right)\ge 0
f\left(x\right)=\frac{1}{-1+\mathrm{exp}\sqrt{2\gamma x+{\mathrm{log}\left(\frac{f\left(0\right)+1}{f\left(0\right)}\right)}^{2}}}
{\int }_{0}^{f\left(x\right)}\frac{\mathrm{log}\left(\frac{f\left(z\right)}{1+f\left(z\right)}\right)}{f\left(z\right)+{f}^{2}\left(z\right)}df={\int }_{0}^{x}\gamma dz
Finding solution of ODE using Laplace transform
y\left(0\right)=0
{y}^{\prime }+2y=f\left(x\right)
f\left(x\right)=0\text{ }\text{if}\text{ }x>1\text{ }\text{and}\text{ }f\left(x\right)=1\text{ }\text{if}\text{ }0\le x\le 1
I applied Laplace transform on both sides which gets
sY\left(s\right)+2Y\left(s\right)=\frac{1-{e}^{-s}}{s}⇒Y\left(s\right)=\frac{1-{e}^{-s}}{s\left(s+2\right)}=\frac{1}{s\left(s+2\right)}-\frac{{e}^{-s}}{s\left(s+2\right)}
Now I know inverse Laplace transform of
\frac{1}{s\left(s+2\right)}
but how to find inverse Laplace transform of second term? Or can we solve above ODE using different method?
How many solutions does the ODE have?
\left\{\begin{array}{l}{y}^{\prime }-{a}^{2}\left({y}^{\prime }{\right)}^{3}-\frac{\mathrm{sin}\left(x\right)}{x+y}=0\\ y\left(0\right)=1\end{array}
Write how many solutions does the system have for
a=0
a\ne 0
I have a question that asks me to show that the functional
I\left(x\right)={\int }_{\left\{{x}_{1}\right\}}^{{x}_{2}}\left({x}^{2}+3{y}^{2}\right){y}^{\prime }+2xydx
has an extremal that is independent of the choice of y that joins two arbitrary points
\left({x}_{1},{y}_{1}\right),\left({x}_{2},{y}_{2}\right)
So I tried solving for the extremal, which give me the euler lagrange equation:
6y{y}^{\prime }+2x=2x+6y{y}^{\prime }
That's not an error it seems that
\frac{d}{dx}\left(\frac{\partial F}{\partial {y}^{\prime }}\right)=\frac{\partial F}{\partial y}
Ok then no wonder that the path doesn't matter. But now I am trying to compute the value of the functional for the two points, I tried a straight line between the 2, but this is giving me a messy expression, it's computable but ugly:
J\left(y\right)={\int }_{\left\{{x}_{1}\right\}}^{{x}_{2}}s\left({x}^{2}+3{\left(sx+k\right)}^{2}\right)+2x\left(sx+k\right)dx
(where s and k are the linear coefficients of the line connecting the two points). |
Decode convolutional code using a posteriori probability (APP) method - Simulink - MathWorks 한êµ
Specifying Details of the Algorithm
Decode convolutional code using a posteriori probability (APP) method
The APP Decoder block performs a posteriori probability (APP) decoding of a convolutional code.
The input L(u) represents the sequence of log-likelihoods of encoder input bits, while the input L(c) represents the sequence of log-likelihoods of code bits. The outputs L(u) and L(c) are updated versions of these sequences, based on information about the encoder.
If the convolutional code uses an alphabet of 2n possible symbols, this block's L(c) vectors have length Q*n for some positive integer Q. Similarly, if the decoded data uses an alphabet of 2k possible output symbols, then this block's L(u) vectors have length Q*k.
This block accepts a column vector input signal with any positive integer for Q.
If you only need the input L(c) and output L(u), you can attach a Simulink Ground (Simulink) block to the input L(u) and a Simulink® Terminator (Simulink) block to the output L(c).
This block accepts single and double data types. Both inputs, however, must be of the same type. The output data type is the same as the input data type.
To define the convolutional encoder that produced the coded input, use the Trellis structure parameter. This parameter is a MATLAB® structure whose format is described in Trellis Description of a Convolutional Code. You can use this parameter field in two ways:
If you have a variable in the MATLAB workspace that contains the trellis structure, enter its name as the Trellis structure parameter. This way is preferable because it causes Simulink to spend less time updating the diagram at the beginning of each simulation, compared to the usage described next.
If you want to specify the encoder using its constraint length, generator polynomials, and possibly feedback connection polynomials, use a poly2trellis command within the Trellis structure field. For example, to use an encoder with a constraint length of 7, code generator polynomials of 171 and 133 (in octal numbers), and a feedback connection of 171 (in octal), set the Trellis structure parameter to
To indicate how the encoder treats the trellis at the beginning and end of each frame, set the Termination method parameter to either Truncated or Terminated. The Truncated option indicates that the encoder resets to the all-zeros state at the beginning of each frame. The Terminated option indicates that the encoder forces the trellis to end each frame in the all-zeros state. If you use the Convolutional Encoder block with the Operation mode parameter set to Truncated (reset every frame), use the Truncated option in this block. If you use the Convolutional Encoder block with the Operation mode parameter set to Terminate trellis by appending bits, use the Terminated option in this block.
You can control part of the decoding algorithm using the Algorithm parameter. The True APP option implements a posteriori probability decoding as per equations 20–23 in section V of [1]. To gain speed, both the Max* and Max options approximate expressions like
\mathrm{log}\underset{i}{â}\mathrm{exp}\left({a}_{i}\right)
by other quantities. The Max option uses max(ai) as the approximation, while the Max* option uses max(ai) plus a correction term given by
\mathrm{ln}\left(1+\mathrm{exp}\left(â|{a}_{iâ1}â{a}_{i}|\right)\right)
The Max* option enables the Scaling bits parameter in the dialog box. This parameter is the number of bits by which the block scales the data it processes internally (multiplies the input by (2^numScalingBits) and divides the pre-output by the same factor). Use this parameter to avoid losing precision during the computations.
Either Truncated or Terminated. This parameter indicates how the convolutional encoder treats the trellis at the beginning and end of frames.
Either True APP, Max*, or Max.
Number of scaling bits
An integer between 0 and 8 that indicates by how many bits the decoder scales data in order to avoid losing precision. This field is active only when Algorithm is set to Max*.
Disable L(c) output port
Select this check box to disable the secondary block output, L(c).
[1] Benedetto, S., G. Montorsi, D. Divsalar, and F. Pollara, “A Soft-Input Soft-Output Maximum A Posterior (MAP) Module to Decode Parallel and Serial Concatenated Codes,†JPL TDA Progress Report, Vol. 42-127, November 1996.
[2] Benedetto, Sergio and Guido Montorsi, “Performance of Continuous and Blockwise Decoded Turbo Codes.†IEEE Communications Letters, Vol. 1, May 1997, 77–79.
[3] Viterbi, Andrew J., “An Intuitive Justification and a Simplified Implementation of the MAP Decoder for Convolutional Codes,†IEEE Journal on Selected Areas in Communications, Vol. 16, February 1998, 260–264.
Viterbi Decoder | Convolutional Encoder |
Compiler optimizations - CS Notes Compiler optimizations | CS Notes
Linear IRs
Stack-machine code
Graphical IRs
Data-dependence graphs
Optimizing compilers perform optimizations to improve a program’s resource utilization. Generally the resource being optimized for is CPU time, but specialist compilers exist that optimize for other resources (e.g. code size, memory usage, disk accesses, etc.).
Optimization involves many subproblems that are computationally intractable. Therefore, heuristics are often used during optimization.
Typically, optimizations are run on an IR (Intermediate Representation) before the IR is passed to the code generator [1, P. 505].
IR (Intermediate Representation) is a language between the source code and the target language. It provides a layer of abstraction that:
Contains more details than the source
Contains less details than the target
An IR is designed to make processing of a program easier (e.g. optimization and translation). Some compilers translate through a series of intermediate languages during the compilation pipeline.
Linear IRs consist of sequentially-executed instructions. Linear IRs often resemble assembly code for an abstract machine.
Example LLVM IR:
3AC (Three-Address Code) is a linear IR. In 3AC there is at most one operator on the right side of an instruction.
A source-language expression like x - y * z might be translated into a sequence of 3AC instructions:
t1 = y * z
t2 = x - t1
3AC includes addresses and instructions.
A 3AC address can be one of:
A name (source-code names)
A compiler-generated temporary
Operators are either constants or registers.
Some common 3AC instruction forms:
Assignment (binary: x = y op z, unary: x = op y)
Copy instructions (x = y)
Unconditional jump (goto L, where L is a 3AC instruction with label L)
Conditional jump (if x goto L)
Conditional jump with relational operator (if x relop y goto L, relational operators: <, ==, etc.)
Indexed copy instructions (x = y[i], y[i] = x)
Address and pointer assignments (x = &y, x = *y, *x = y)
Procedure calls (implemented using param x for parameter)
3AC instructions can be represented as quadruples <op, arg_1, arg_2, result> [1, P. 366].
Stack-machine code is a linear IR that models the behavior of a stack machine. It is a form of one-address code.
Most operations take their operands from the stack. For example, a multiply instruction would remove the top 2 instructions from the stack, multiply them, and push the result back to the stack.
Example stack machine code:
Many interpreters execute stack-machine bytecode on a virtual stack machine (e.g. CPython, JVM).
Graphical IRs represent source code as a graph.
Graphical IRs can be trees:
Parse trees—graphical representation of derivation
ASTs—derivation with extraneous nodes removed
DAGs—ASTs where nodes can have multiple parents and identical subtrees are reused
A control-flow graph represents the possible paths through basic blocks, where a basic block is a sequence of operations that always execute together (unless an operation raises an exception).
Each node represents a basic code block. A directed edge
(B_i, B_j)
represents a possible transfer of control from
B_i
B_j
[2, P. 231].
Control-flow graphs are often used with another IR where the control-flow graph represents the relationships between blocks, and the operations within blocks are represented with another IR (e.g. a linear IR) [2, Pp. 231-2].
In compilers, a data-dependence graph represents the dependencies between individual instructions.
A data-dependence graph node
n
represents an operation. An edge
(n_i, n_j)
represents a definition value
n_j
n_i
that uses the value [2, P. 233].
Data dependence graphs are often supplementary data structures built from the definitive IR and discarded after use. They are used for instruction scheduling [2, P. 234].
The edges in a graph represent hard constraints (an operation
n_j
cannot run before operation
n_i
), the graph creates a partial order, where there are often many sequences that preserve the data dependencies of the graph. This property is exploited by out-of-order processors in order to schedule instruction efficiently [2, P. 233].
SSA (Static single-assignment form) is a property of an IR which requires that each variable is assigned exactly once and that every variable is defined before its use.
The process for transforming ordinary code into SSA involves replacing the target of each assignment with a new variable, and replacing each use of a variable with the version of the variable reaching that point.
Since control flow can’t be predicted in advance, there can be cases where a variable might refer to multiple versions. In this case, a variadic
Φ
(Phi) function is used. You can read
Φ(o_1, o_2, ..., o_n)
as “one of either
o_1
o_2
, …, or
o_n
A basic block is a sequence of code with no branches within itself, except to the entry, and no branches out, except at the exit.
A control-flow graph is a directed graph where the nodes are basic blocks and an edge
(B_1, B_2)
exists if execution can pass from the last instruction in
B_1
to the first instruction in
B_2
The body of a method (or procedure) can always be represented as a control-flow graph. There is one initial node, and all return nodes are terminal.
For languages like C, there are three optimization levels:
Local optimizations—applied to a basic block.
Global optimizations—applied to a single function, optimized across all basic blocks of that function.
Inter-procedural optimizations—applied across method boundaries.
Commonly, applying optimizations enables new optimizations (e.g. running a copy propagation optimization enables dead code elimination). Optimizing compilers repeat optimizations until no new optimizations are found (or a limit is reached).
Local optimizations only consider information local to a single basic block.
Common subexpression elimination is where identical expressions are replaced with a single variable holding the computed value. This optimization is easy to do when the IR is in SSA.
Copy propagation is where occurrences of direct assignments are replaced with their values. e.g. if y = x then z = 3 + y can become z = 3 + x. Copy propagation can enable dead code elimination and constant folding.
Similar to copy propagation, constant propagation is the process of substituting values of known constants in expressions. e.g. if x = 14 then y = 3 + x becomes y = 3 + 14.
Dead code elimination is the process of removing dead code, where dead code includes unreachable code and dead variables (variables written to but never read).
Constant folding is when the compiler reorganizes and evaluates constant expressions at runtime.
Example: x := 2 + 2 becomes x := 4.
Peephole optimization is an optimization technique applied to a short sequence of (usually contiguous) target language instructions. The sequence is known as the peephole.
The optimizer replaces the sequence with another sequence that produces the equivalent result but is faster.
Peephole optimizations are generally written as replacement rules:
Dataflow analysis is a variety of techniques that derive information about the flow of data along program execution paths. This enables global optimizations, e.g. global constant propagation [1, P. 597].
In dataflow analysis, a program point is a point in the program that is either before an instruction (the input state of an instruction) or after an instruction (the output state of an instruction). Dataflow analysis must consider all possible paths through program points that a program can take [1, P. 597].
Although Dataflow analysis can be run on program points, it can also be run on the boundaries of basic blocks (requiring less computation).
There are two main forms of dataflow analysis:
Forward flow analysis
Backward flow analysis
In forward flow analysis, the exit state of a program point is a function of the program point’s entry state. In backward flow analysis, the entry state of a program point is a function of the program point’s exit state.
In forward flow analysis you initialize an entry point before running the analysis. In backward flow analysis, you initialize exit points.
In forward flow analysis, the value of a block’s exit (
\text{out}_b
\text{transfer}\_b
is an output function of the block
b
(a transfer function).
And the value of a block’s entry (
\text{in}_b
Where the join operation combines the exit states of the predecessors of
b
, yielding the entry state of
b
Each data flow analysis has its own transfer function and join operation.
Backward flow analysis is the inverse:
Reaching definition analysis is a forward flow analysis that statically determines which definitions may reach a certain point.
In the following example, d2 is a reaching definition for d3 but d1 is not:
Reaching definition analysis can be defined as:
gen_b
is the set of all definitions introduced by
b
kill_b
is the set of all definitions that are overwritten by
b
Liveness analysis is a backward dataflow analysis used to calculate whether variables are live at each point in the program.
Liveness analysis can be used during register allocation to determine which registers should be favored [1, Pp. 608-9].
A variable is live at statement
s
There exists a statement
s'
that accesses
x
There is a path from
s
s'
The path has no intervening assignment to
x
A dead variable is one that is not live. A statement x = ... is dead code if x is dead, and can therefore be deleted.
Liveness analysis can be defined as:
def_b
is the set of variables defined in
b
prior to any use of that variable in
b
use_b
is the set of variables whose values may be used in
b
prior to any definition of the variable |
Tune Feedback Loops using looptune - MATLAB & Simulink - MathWorks Nordic
Specify Tunable Elements
Build Tunable Model of Feedback Loop
Tune Controller Parameters
This example shows the basic workflow of tuning feedback loops with the looptune command. looptune is similar to systune and meant to facilitate loop shaping design by automatically generating the tuning requirements.
This example uses a simple engine speed control application as shown in the following figure. The control system consists of a single PID loop and the PID controller gains must be tuned to adequately respond to step changes in the desired speed. Specifically, you want the response to settle in less than 5 seconds with little or no overshoot.
Use the following fourth-order model of the engine dynamics.
load rctExamples Engine
bode(Engine)
You need to tune the four PID gains to achieve the desired performance. Use a tunablePID object to parameterize the PID controller.
PID0 = tunablePID('SpeedController','pid')
PID0 =
Tunable continuous-time PID controller "SpeedController" with formula:
Type "pid(PID0)" to see the current value and "get(PID0)" to see all properties.
looptune tunes the generic SISO or MIMO feedback loop shown in the following figure. This feedback loop models the interaction between the plant and the controller. Note that this is a positive feedback interconnection.
For the speed control loop, the plant
G
is the engine model and the controller
C
consists of a PID controller and prefilter
F
To use looptune, create models for
G
C
. Assign names to the inputs and outputs of each model to specify the feedback paths between plant and controller. Note that the controller
C
has two inputs: the speed reference r and the speed measurement speed.
F = tf(10,[1 10]); % prefilter
G = Engine;
G.InputName = 'throttle';
G.OutputName = 'speed';
C0 = PID0 * [F , -1];
C0.InputName = {'r','speed'};
C0.OutputName = 'throttle';
Here, C0 is a generalized state-space model (genss) that depends on the tunable PID block PID0.
You can now use looptune to tune the PID gains subject to a simple control bandwidth requirement. To achieve the 5-second settling time, the gain crossover frequency of the open-loop response should be approximately 1 rad/s. Given this basic requirement, looptune automatically shapes the open-loop response to provide integral action, high-frequency roll-off, and adequate stability margins. Note that you could specify additional requirements to further constrain the design. For an example, see Decoupling Controller for a Distillation Column.
[~,C,~,Info] = looptune(G,C0,wc);
The final value is less than 1, indicating that the desired bandwidth was achieved with adequate roll-off and stability margins. looptune returns the tuned controller C. Use getBlockValue to retrieve the tuned value of the PID block.
PIDT = getBlockValue(C,'SpeedController')
PIDT =
with Kp = 0.000855, Ki = 0.00269, Kd = -7.83e-05, Tf = 0.877
Name: SpeedController
Use loopview to validate the design and visualize the loop shaping requirements implicitly enforced by looptune.
Next, plot the closed-loop response to a step command in engine speed. The tuned response satisfies the requirements.
T = connect(G,C,'r','speed'); % closed-loop transfer function from r to speed |
Transform IIR lowpass filter to IIR bandstop filter - MATLAB iirlp2bs
iirlp2bs
Transform Lowpass Filter to Bandstop Filter
IIR Lowpass Filter to IIR Bandstop Filter Transformation
Transform IIR lowpass filter to IIR bandstop filter
[num,den,allpassNum,allpassDen] = iirlp2bs(b,a,wo,wt)
[num,den,allpassNum,allpassDen] = iirlp2bs(b,a,wo,wt) transforms an IIR lowpass filter to an IIR bandstop filter.
The lowpass filter is specified using the numerator and denominator coefficients b and a respectively. The function returns the numerator and denominator coefficients of the transformed bandstop digital filter. The function also returns the numerator and denominator coefficients of the allpass mapping filter, allpassNum and allpassDen respectively.
For more details on the transformation, see IIR Lowpass Filter to IIR Bandstop Filter Transformation.
Transform a lowpass IIR filter to a bandstop filter using the iirlp2bs function.
Transform Filter Using iirlp2bs
Transform the prototype lowpass filter into a bandstop filter by placing the cutoff frequencies of the prototype filter at 0.25π and 0.75π.
[num,den] = iirlp2bs(b,a,0.5,[0.25 0.75]);
hfvt = fvtool(b,a,num,den);
legend(hfvt,"Prototype Filter (TF Form)",...
"Transformed Bandstop Filter")
[num2,den2] = iirlp2bs(ss(:,1:3),ss(:,4:6),0.5,[0.25 0.75]);
H\left(z\right)=\frac{B\left(z\right)}{A\left(z\right)}=\frac{{b}_{0}+{b}_{1}{z}^{-1}+\cdots +{b}_{n}{z}^{-n}}{{a}_{0}+{a}_{1}{z}^{-1}+\cdots +{a}_{n}{z}^{-n}},
b=\left[\begin{array}{ccccc}{b}_{01}& {b}_{11}& {b}_{21}& ...& {b}_{Q1}\\ {b}_{02}& {b}_{12}& {b}_{22}& ...& {b}_{Q2}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {b}_{0P}& {b}_{1P}& {b}_{2P}& \cdots & {b}_{QP}\end{array}\right]
H\left(z\right)=\prod _{k=1}^{P}{H}_{k}\left(z\right)=\prod _{k=1}^{P}\frac{{b}_{0k}+{b}_{1k}{z}^{-1}+{b}_{2k}{z}^{-2}+\cdots +{b}_{Qk}{z}^{-Q}}{{a}_{0k}+{a}_{1k}{z}^{-1}+{a}_{2k}{z}^{-2}+\cdots +{a}_{Qk}{z}^{-Q}},
H\left(z\right)=\frac{B\left(z\right)}{A\left(z\right)}=\frac{{b}_{0}+{b}_{1}{z}^{-1}+\cdots +{b}_{n}{z}^{-n}}{{a}_{0}+{a}_{1}{z}^{-1}+\cdots +{a}_{n}{z}^{-n}},
a=\left[\begin{array}{ccccc}{a}_{01}& {a}_{11}& {a}_{21}& \cdots & {a}_{Q1}\\ {a}_{02}& {a}_{12}& {a}_{22}& \cdots & {a}_{Q2}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {a}_{0P}& {a}_{1P}& {a}_{2P}& \cdots & {a}_{QP}\end{array}\right]
H\left(z\right)=\prod _{k=1}^{P}{H}_{k}\left(z\right)=\prod _{k=1}^{P}\frac{{b}_{0k}+{b}_{1k}{z}^{-1}+{b}_{2k}{z}^{-2}+\cdots +{b}_{Qk}{z}^{-Q}}{{a}_{0k}+{a}_{1k}{z}^{-1}+{a}_{2k}{z}^{-2}+\cdots +{a}_{Qk}{z}^{-Q}},
num — Numerator coefficients of transformed bandstop filter
Numerator coefficients of the transformed bandstop filter, returned as one of the following:
den — Denominator coefficients of transformed bandstop filter
Denominator coefficients of the transformed bandstop filter, returned as one of the following:
IIR lowpass filter to IIR bandstop filter transformation effectively places one feature of the original filter, located at frequency −wo, at the required target frequency location, wt1, and the second feature, originally at wo, at the new location, wt2. Choice of the feature subject to the lowpass to bandstop transformation is not restricted only to the cutoff frequency of an original lowpass filter. You can choose to transform any feature of the original filter like stopband edge, DC, deep minimum in the stopband, or others. It is assumed that wt2 is greater than wt1. Frequencies must be normalized to be between 0 and 1, with 1 corresponding to half the sample rate.
This transformation implements the "Nyquist Mobility," which means that the DC feature stays at DC, but the Nyquist feature moves to a location dependent on the selection of wo and wts.
Relative positions of other features of the original filter change in the target filter. This means that it is possible to select two features of the original filter, F1 and F2, with F1 preceding F2. After the transformation feature F2 will precede F1 in the target filter. In addition, the distance between F1 and F2 will not be the same before and after the transformation.
For more details on the lowpass to bandstop frequency transformation, see Digital Frequency Transformations.
[2] Nowrouzian, B., and L.T. Bruton. “Closed-Form Solutions for Discrete-Time Elliptic Transfer Functions.” In [1992] Proceedings of the 35th Midwest Symposium on Circuits and Systems , 784–87. Washington, DC, USA: IEEE, 1992. https://doi.org/10.1109/MWSCAS.1992.271206.
iirftransf | allpasslp2bs | zpklp2bs |
Evaluate the line integral, where C is the given curve C (\frac{x}{y})ds
\left(\frac{x}{y}\right)ds,C:x={t}^{3},y={t}^{4},1\le t\le 2
x={t}^{2},y=2t,0\le t\le 5
\left({x}^{2}+2xy-4{y}^{2}\right)dx-\left({x}^{2}-8xy-4{y}^{2}\right)dy=0
Math:Kimberly took her 6 nieces and nephews to a hockey game. She wants to buy them snacks. How much can she spend on snacks for each child if Kimberly wants to spend less than $33 in total
\int \frac{{e}^{t}}{\sqrt{{e}^{2t}+4}}dt
\int \left({e}^{x}-{e}^{-x}\right){\left({e}^{x}+{e}^{-x}\right)}^{3}dx
\int {\int }_{S}\left({x}^{2}z+{y}^{2}z\right)ds
{x}^{2}+{y}^{2}+{z}^{2}=9,z\ge 0
{\int }_{-1}^{1}dx{\int }_{1-{x}^{2}}^{2-{x}^{2}}f\left(x,y\right)dy |
Create univariate autoregressive integrated moving average (ARIMA) model - MATLAB - MathWorks France
1-0.5{L}^{1}+0.1{L}^{4}
1-{\varphi }_{1}{L}^{1}-{\varphi }_{4}{L}^{4}.
1+{\theta }_{1}{L}^{1}+{\theta }_{2}{L}^{2}+{\theta }_{3}{L}^{3}.
1-{\Phi }_{4}{L}^{4}-{\Phi }_{8}{L}^{8}.
1+{\Theta }_{4}{L}^{4}.
\varphi \left(L\right)=1-0.5L+0.1{L}^{2},
\Phi \left(L\right)=1-0.5{L}^{4}+0.1{L}^{8},
{y}_{t}=c+{\epsilon }_{t}
\mathit{c}
{\epsilon }_{\mathit{t}}\text{\hspace{0.17em}}
{\sigma }^{2}
\left(1+0.5{L}^{2}\right)\left(1-L\right){y}_{t}=3.1+\left(1-0.2L\right){\epsilon }_{t},
{\epsilon }_{\mathit{t}}
\Delta {y}_{t}=3.1-0.5\Delta {y}_{t-2}+{\epsilon }_{t}-0.2{\epsilon }_{t-1}.
{y}_{t}=1+\varphi {y}_{t-1}+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
\varphi
\left(1-{\varphi }_{1}L-{\varphi }_{2}{L}^{2}-{\varphi }_{3}{L}^{3}\right)\left(1-L\right){y}_{t}=\left(1+{\theta }_{1}L+{\theta }_{2}{L}^{2}\right){\epsilon }_{t}
{\epsilon }_{\mathit{t}}
{\sigma }^{2}
p
D
{y}_{t}={\epsilon }_{t}+{\theta }_{1}{\epsilon }_{t-1}+{\theta }_{12}{\epsilon }_{t-12},
{\epsilon }_{\mathit{t}}
{\sigma }^{2}
\left(0,1,1\right)×{\left(0,1,1\right)}_{12}
\left(1-L\right)\left(1-{L}^{12}\right){y}_{t}=\left(1+{\theta }_{1}L\right)\left(1+{\theta }_{12}{L}^{12}\right){\epsilon }_{t},
{\epsilon }_{\mathit{t}}
{\sigma }^{2}
{y}_{t}=0.05+0.6{y}_{t-1}+0.2{y}_{t-2}-0.1{y}_{t-3}+{\epsilon }_{t},
{\epsilon }_{t}
\mathit{t}
{y}_{t}=0.6{y}_{t-1}+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
\mathit{t}
\begin{array}{l}{y}_{t}=c+\varphi {y}_{t-1}+{\epsilon }_{t}+\theta {\epsilon }_{t-1}.\\ {\epsilon }_{t}={\sigma }_{t}{z}_{t}.\\ {\sigma }_{t}^{2}=\kappa +\gamma {\sigma }_{t-1}^{2}.\\ {z}_{t}\sim N\left(0,1\right).\end{array}
{y}_{t}={y}_{t-1}+{\epsilon }_{t},
{\epsilon }_{\mathit{t}}
{L}^{i}{y}_{t}={y}_{t-i}.
{y}_{t}=c+{x}_{t}\beta +{a}_{1}{y}_{t-1}+\dots +{a}_{w}{y}_{t-w}+{\epsilon }_{t}+{b}_{1}{\epsilon }_{t-1}+\dots +{b}_{v}{\epsilon }_{t-v}.
a\left(L\right){y}_{t}=c+{x}_{t}\beta +b\left(L\right){\epsilon }_{t}.
\varphi \left(L\right){\left(1-L\right)}^{D}\Phi \left(L\right){\left(1-{L}^{s}\right)}^{{D}_{s}}{y}_{t}=c+{x}_{t}\beta +\theta \left(L\right)\Theta \left(L\right){\epsilon }_{t}.
\varphi \left(L\right)
\varphi \left(L\right)=1-\varphi L-{\varphi }_{2}{L}^{2}-...-{\varphi }_{p}{L}^{p},
\Phi \left(L\right)
\Phi \left(L\right)=1-{\Phi }_{{p}_{1}}{L}^{{p}_{1}}-{\Phi }_{{p}_{2}}{L}^{{p}_{2}}-...-{\Phi }_{{p}_{s}}{L}^{{p}_{s}},
\theta \left(L\right)
\theta \left(L\right)=1+\theta L+{\theta }_{2}{L}^{2}+...+{\theta }_{q}{L}^{q},
\Theta \left(L\right)
\Theta \left(L\right)=1+{\Theta }_{{q}_{1}}{L}^{{q}_{1}}+{\Theta }_{{q}_{2}}{L}^{{q}_{2}}+...+{\Theta }_{{q}_{s}}{L}^{{q}_{s}},
q<\infty
E\left({y}_{t}\right)=\theta \left(L\right)0=0.
Var\left({y}_{t}\right)={\sigma }^{2}\sum _{i=1}^{q}{\theta }_{i}^{2}.
Cov\left({y}_{t},{y}_{t-s}\right)=\left\{\begin{array}{l}{\sigma }^{2}\left({\theta }_{s}+{\theta }_{1}{\theta }_{s-1}+{\theta }_{2}{\theta }_{s-2}+...+{\theta }_{q}{\theta }_{s-q}\right)\text{ if }s\ge q\\ 0\text{ otherwise}.\end{array}
\left\{{y}_{t};t=1,...,T\right\} |
(→Brief Description of the Study Test Case)
*Turbulence kinetic energy
** ''x – y'' cut of the field of time-averaged two-dimensional turbulent kinetic energy <math>\text{TKE}=\frac{1}{2}\left(\langle u'u'\rangle + \langle v'v'\rangle\right)/U_0^2</math>;
**2D TKE distribution along* ''y'' = 0;
**2D TKE distribution along ''y'' = 0;
**2D TKE distribution along* ''x'' = 1.5 ''D'' (in the gap between the cylinders);
**2D TKE distribution along ''x'' = 1.5 ''D'' (in the gap between the cylinders);
**2D TKE distribution along* ''x'' = 4.45 ''D'' (0.75 ''D'' downstream of the centre of the rear cylinder).
**2D TKE distribution along ''x'' = 4.45 ''D'' (0.75 ''D'' downstream of the centre of the rear cylinder).
All these and some other data are available on the
[https://info.aiaa.org/tac/ASG/FDTC/DG/Forms/AllItems.aspx?RootFolder=https%3A%2F%2Finfo.aiaa.org%2Ftac%2FASG%2FFDTC%2FDG%2FBECAN_files_%2FBANCII_category2&FolderCTID=0x0120007FEBD4B8002BD94694D9BBB199BB01DA web site of the BANC-I Workshop].
{|align="center" border="1" cellpadding="5"
!Windows!!Unix
|colspan="2" align="center"|[[Media:UFR2-12_readme.txt|readme.txt]]
|colspan="2" align="center"|[[Media:UFR2-12_Tandem_Cylinder_Problem_Statement_v1.08.pdf|problem_statement.pdf]]
|colspan="2" align="center"|[[Media:UFR2-12_problem2_data_and_guidelines.pdf|problem_data_and_guidelines.pdf]]
|[[Media:UFR2-12_data.zip|data.zip]]||[[Media:UFR2-12_data.tgz|data.tgz]]
|[[Media:UFR2-12_figures.zip|figures.zip]]||[[Media:UFR2-12_figures.tgz|figures.tgz]]
<!--on the
[https://info.aiaa.org/tac/ASG/FDTC/DG/Forms/AllItems.aspx?RootFolder=https%3A%2F%2Finfo.aiaa.org%2Ftac%2FASG%2FFDTC%2FDG%2FBECAN_files_%2FBANCII_category2&FolderCTID=0x0120007FEBD4B8002BD94694D9BBB199BB01DA web site of the BANC-I Workshop].-->
== Test Case Experiments ==
A detailed description of the experimental facility and measurement techniques is given in the original publications
[[UFR_2-12_References#2|[2-4]]] and available on the
[[UFR_2-12_References#2|[2-4]]]. <!--and available on the
So here we present only concise information about these aspects of the test case.
Finally, '''TUB''' applied their in-hose multi-block structured code ELAN in the framework of the ''incompressible'' flow assumption.
The pressure velocity coupling is based on the SIMPLE algorithm.
For the convective terms a hybrid approach [[UFR_2-12_References#15|[15]]]
with blending of 2<sup>nd</sup>-order central and upwind-biased TVD schemes was used.
The time integration was similar to that of NTS.
{\displaystyle {\theta }}
{\displaystyle U_{0}D/\nu }
{\displaystyle L/D}
{\displaystyle L_{z}/D}
{\displaystyle D}
{\displaystyle U_{0}}
{\displaystyle K}
{\displaystyle C_{p}=\langle (p-p_{0})\rangle /(1/2\rho _{0}U_{0}^{2})}
{\displaystyle {\left.\langle u\rangle /U_{0}\right.}}
{\displaystyle \theta }
{\displaystyle \theta }
{\displaystyle {\text{TKE}}={\frac {1}{2}}\left(\langle u'u'\rangle +\langle v'v'\rangle \right)/U_{0}^{2}} |
CTIC Financial Ratios - FinancialModelingPrep
\dfrac{Current Assets}{Current Liabilities}
\dfrac{Cash and Cash Equivalents + Short Term Investments + Account Receivables}{Current Liabilities}
\dfrac{Cash and Cash Equivalents}{Current Liabilities}
\dfrac{(Account Receivable (start) + Account Receivable (end))/2}{Revenue/365}
\dfrac{(Inventories (start) + Inventories (end))/2}{COGS/365}
\dfrac{DSO + DIO}{}
\dfrac{(Accounts Payable (start) + Accounts Payable (end))/2}{COGS/365}
3,183.90 DPO tells you how many days the company takes to pay its suppliers.
\dfrac{DSO + DIO − DPO}{}
-2,765.05 The cash conversion cycle (CCC = DSO + DIO – DPO) measures the number of days a company's cash is tied up in the production and sales process of its operations and the benefit it derives from payment terms from its creditors. The shorter this cycle, the more liquid the company's working capital position is. The CCC is also known as the "cash" or "operating" cycle.
\dfrac{Gross Profit}{Revenue}
\dfrac{Operating Income}{Revenue}
\dfrac{Income Before Tax}{Revenue}
\dfrac{Net Income}{Revenue}
\dfrac{Provision For Income Taxes}{Income Before Tax}
\dfrac{Net Income}{Average Total Assets}
\dfrac{Net Income}{Average Total Equity}
\dfrac{EBIT}{Average Total Asset − Average Current Liabilities}
\dfrac{Net Income}{EBT}
\dfrac{EBT}{EBIT}
\dfrac{EBIT}{Revenue}
\dfrac{Total Liabilities}{Total Assets}
\dfrac{Total Debt}{Total Equity}
\dfrac{Long−Term Debt}{Long−Term Debt + Shareholders Equity}
\dfrac{Total Debt}{Total Debt + Shareholders Equity}
\dfrac{EBIT}{Interest Expense}
\dfrac{Operating Cash Flows}{Total Debt}
\dfrac{Total Assets}{Total Equity}
\dfrac{Revenue}{NetPPE}
\dfrac{Revenue}{Total Average Assets}
\dfrac{Operating Cash Flow}{Revenue}
\dfrac{Free Cash Flow}{Operating Cash Flow}
\dfrac{Operating Cash Flow}{Total Debt}
\dfrac{Operating Cash Flow}{Short-Term Debt}
\dfrac{Operating Cash Flow}{Capital Expenditure}
\dfrac{Operating Cash Flow}{Dividend Paid + Capital Expenditure}
\dfrac{DPS (Dividend per Share)}{EPS (Net Income per Share Number}
\dfrac{Stock Price per Share}{Equity per Share}
-19.45 The price-to-book value ratio, expressed as a multiple (i.e. how many times a company's stock is trading per share compared to the company's book value per share), is an indication of how much shareholders are paying for the net assets of a company.
\dfrac{Stock Price per Share}{Operating Cash Flow per Share}
\dfrac{Stock Price per Share}{EPS}
\dfrac{Price Earnings Ratio}{Expected Revenue Growth}
\dfrac{Stock Price per Share}{Revenue per Share}
\dfrac{Dividend per Share}{Stock Price per Share}
\dfrac{Entreprise Value}{EBITDA}
\dfrac{Stock Price per Share}{Intrinsic Value}
-19.45 Helps investors determine whether a stock is trading at, below, or above its fair value estimate,A price/fair value ratio below 1 suggests the stock is trading at a discount to its fair value, while a ratio above 1 suggests it is trading at a premium to its fair value. |
The Research of Magneto-Rheological Fluid Yield Stress Model
The Research of Magneto-Rheological Fluid Yield Stress Model ()
Meng Ji, Yiping Luo*
Magnetorheological Fluid (MRF), as an advanced and smart material which was controlled by magnetic field, was a kind of stable suspension in which magnetic particle dissolved in base fluid. The yield stress, one of main performance parameters of MRF, was the demarcation point between liquid and solid. At present, the yield stress calculation model did not have a uniform standard. The research on yield stress model was significant to the research on MRF. First, the research was based on the MRF characteristic and the research status of MRF sheer yield stress; second the classic dipole model, local field dipole model, polarized pellet model, continuous models on the average had been calculated and compared. The classic dipole model and local field dipole model had a well ability to describe the yield stress of MRF.
MRF, Classic Dipole Model, Local Field Dipole Model, Polarized Pellet Model, Continuous Models
Ji, M. and Luo, Y. (2018) The Research of Magneto-Rheological Fluid Yield Stress Model. Open Access Library Journal, 5, 1-7. doi: 10.4236/oalib.1104643.
Magnetorheological fluid (MRF), as a new advanced material, was a type of stable suspension liquid which was composed of magnetizable particle and base solution. The MRF had a reversible ability that it could converse fluid into semisolid in millisecond in a changeable and stable magnetic field. The apparent viscosity increased 105 ~ 106 times and the maximum yield stress increased to 50 ~ 100 kPa. Using MRF could solve the problem that traditional machine cannot solve. This excellent ability is given a vast prospect of MRF application. In automobile, mechanical precision polishing and other aspects had a large market [1] [2] .
Magnetorheological fluid consists of magnetic particles, the base fluid and additives respectively. Magnetic particles were the main part of the performance, the base fluid was as carrier to carry magnetic particles and additive is to improve performance [3] [4] .
1) Magnetic particles
There exists a main tendency that the magnetic particles were hydroxyl iron powder, iron-cobalt alloy and iron-nickel alloy in nowadays research. Iron-cobalt alloy and iron-nickel alloy had a batter performance and were used in research application, but the high price was not ignored. The cost of hydroxyl iron powder was much low and it had an easier preparation technology. So hydroxyl iron powder was the best choice in MRF preparation technology and the performance is satisfied in the meantime. In order to guarantee MRF having a high yield stress, the magnetic particle had following characters:
① High permeability. The ability of magnetic energy level was depended on the permeability of magnetic-particle; a high permeability is a guarantee of yield stress.
② Low coercivity. Low coercive force for MRF can improve demagnetization and transfer, and transfer the reversible process.
③ Magnetic particle size. Results indicate that: if magnetic particles diameter in a suitable range increased, the yield stress will increase. But beyond the suitable scope, relationship could no longer exist. So it was also important to select a suitable diameter size, usually 1 - 10 microns.
2) Base fluid
Base fluid, as a carrier, had two kinds of magnetic carrier fluid: the nonmagnetic carrier fluid and magnetic carrier fluid which had been widely applied. The nonmagnetic carrier fluid was embodied in Bingham plastic fluids and Newtonian fluid respectively in magnetic field or without magnetic field. At present the common base products included silicone oil, water, and synthetic oil. Silicone oil was the most widely used. It was colorless, tasteless, high stability and low price. Base fluid usually had the following features:
① Low viscosity. Fluid of low viscosity is the important guarantee of MRF viscosity low zero field.
② Good stability, corrosion resistance. The stability of the base fluid directly affected the stability of magnetorheological fluid. Corrosion resistance was also in the bad environment to ensure safety work.
③ High density. Base fluid density is high, which can effectively prevent subsidence problems and improve the comprehensive performance.
Additive can improve the performance of MRF, such as:
① Reduce the settle-ability.
② The lubrication effect. Additive can prevent the sticky between solid particle and maintain the liquid homogeneity.
③ Improve magnetic susceptibility. Improve rheological property and enhance the polarization performance of magnetic particle.
The research status of MRF yield stress testing device
Ginder [5] research result based on utilizing finite element method indicated that maximum yield stress is proportional to the magnetic saturation magnetization intensity of particle. Weiss [6] indicated that yield stress decrease with temperature increase.
Felt [7] finished the research about the affect of volume fraction and particle size to MRF yield stress and indicated that volume fraction and particle size is proportional to MRF yield stress. Jiang [8] indicated that static yield stress is proportional to particle size.
2. MRF Yield Stress Calculation Model
2.1. The Classic Dipole Model
In Figure 1(a), R and
{\mu }_{1}
was particle radius and permeability,
{\mu }_{2}
was permeability of basic liquid, r was straight-line distance, φ was magnetic potential. When the magnetic field was
{H}_{0}
, particle inner and outer magnetic potential had a following relation:
{\nabla }^{2}{\phi }_{1}=0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(r<R\right)
{\nabla }^{2}{\phi }_{2}=0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(r>R\right)
Function (2-1) formed Laplace’s equations. boundary conditions were:
r=0,\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }{\phi }_{1}=0
r=\infty ,\text{ }\text{ }\text{ }\text{ }{\phi }_{2}=-{H}_{0}r\mathrm{cos}\theta
r=R,\text{ }\text{ }\text{ }\text{ }{\phi }_{1}={\phi }_{2},\text{ }\text{ }\text{ }{\mu }_{1}\frac{\partial {\phi }_{1}}{\partial r}={\mu }_{2}\frac{\partial {\phi }_{2}}{\partial r}
Solution of equations:
{\phi }_{1}=-\frac{3{\mu }_{2}}{{\mu }_{1}+2{\mu }_{2}}{H}_{0}r\mathrm{cos}\theta
{\phi }_{2}=\left(-r+\frac{{\mu }_{2}-{\mu }_{1}}{{\mu }_{1}+2{\mu }_{2}}\frac{{R}^{3}}{{r}^{2}}\right){H}_{0}\mathrm{cos}\theta
Figure 1. Magnetic field distribution of classic dipole model. (a) Single particle model diagram under magnetic field; (b) Magnetic field map of the dipole model.
The relation between magnetic field and magnetic potential was
H=-\nabla \phi
, the expression of particle inner and outer magnetic field:
{H}_{1}=\frac{3{\mu }_{2}}{{\mu }_{1}+2{\mu }_{2}}{H}_{0}\left(\mathrm{cos}\theta {e}_{r}-\mathrm{sin}\theta {e}_{\theta }\right)
{H}_{2}=\left(1-2\frac{{\mu }_{2}-{\mu }_{1}}{{\mu }_{1}+2{\mu }_{2}}\frac{{R}^{3}}{{r}^{3}}\right){H}_{0}\mathrm{cos}\theta {e}_{r}-\left(1+2\frac{{\mu }_{2}-{\mu }_{1}}{{\mu }_{1}+2{\mu }_{2}}\frac{{R}^{3}}{{r}^{3}}\right){H}_{0}\mathrm{sin}\theta {e}_{\theta }
{e}_{z}={e}_{r}\mathrm{cos}\theta -{e}_{\theta }\mathrm{sin}\theta
, we assumed
{k}_{1}=\frac{3{\mu }_{2}}{{\mu }_{1}+2{\mu }_{2}}
{k}_{2}=\frac{{\mu }_{2}-{\mu }_{1}}{{\mu }_{1}+2{\mu }_{2}}
, the Equation (2-4) simplify:
{H}_{1}={k}_{1}{H}_{0}{e}_{z}
{H}_{2}={H}_{0}{e}_{z}-\left(2\mathrm{cos}\theta {e}_{r}+\mathrm{sin}\theta {e}_{\theta }\right){k}_{2}{H}_{0}\frac{{R}^{3}}{{r}^{3}}
According to Equation (2-5), the inner magnetic field of particle is uniform magnetic field, particle outer magnetic field H2 is the first part,
{H}_{0}{e}_{z}
is a outer uniform magnetic field, the second part is addition magnetic field of MRF magnetization, The classic dipole model inside and outside the magnetic field distribution was shown in Figure 1(b).
m=\frac{4\text{π}{\mu }_{2}\left({\mu }_{1}-{\mu }_{2}\right){R}^{3}}{{\mu }_{1}+2{\mu }_{2}}{H}_{0}=-4\text{π}{\mu }_{2}{k}_{2}{R}^{3}{H}_{0}
\begin{array}{c}{H}_{2}={H}_{0}{e}_{z}+\frac{1}{4\text{π}{\mu }_{2}{r}^{3}}\left(2m\mathrm{cos}\theta {e}_{r}+m\mathrm{sin}\theta {e}_{\theta }\right)\\ ={H}_{0}+\frac{1}{4\text{π}{\mu }_{2}}\left(\frac{3\left(m\cdot r\right)r}{{r}^{5}}-\frac{m}{{r}^{3}}\right)\end{array}
The Equation (2-6) indicated that single particle crease a magnetic field in uniform magnetic field was equal to a magnetic dipole which dipole moment was m crease a magnetic field.
2.2. Local Dipole Model
Figure 2 was shown about local dipole model. R is particle radius, the magnetic dipole moment mi was in centre.
There exists a chaining particle i in uniform magnetic field and its magnetic dipole moment was that:
{{m}^{\prime }}_{i}=-3{\mu }_{2}{k}_{2}V{H}_{loc}
where, V was particle volume;
{H}_{loc}
was local field of particle i; H was the magnetic field creased by other particles in chain.
The magnetic field creased by n magnetized particles was that:
\begin{array}{c}{H}_{p}={H}_{1}+{H}_{2}+\cdots +{H}_{t-1}+{H}_{t}+\cdots +{H}_{n}\\ =\underset{q=1}{\overset{n-1}{\sum }}\frac{1}{4\text{π}{\mu }_{2}}\left(\frac{3\left({{m}^{\prime }}_{1}\cdot {r}_{q}\right){r}_{q}}{{r}_{q}^{5}}-\frac{{{m}^{\prime }}_{1}}{{r}_{q}^{3}}\right)\end{array}
Figure 2. Local field dipole model.
If n showed odd number, the magnetic field creased by centre particle i and alone H0 direction was shown that:
{H}_{p0}=\frac{4{{m}^{\prime }}_{1}}{4\text{π}{\mu }_{2}}\underset{q=1}{\overset{\left(n-1\right)/2}{\sum }}\frac{1}{{r}_{q}^{3}}=\frac{{{m}^{\prime }}_{1}}{\text{π}{\mu }_{2}{d}^{3}}\underset{q=1}{\overset{\left(n-1\right)/2}{\sum }}\frac{1}{{q}^{3}}
\begin{array}{c}{{m}^{\prime }}_{1}=-3{\mu }_{2}{k}_{2}V\left({H}_{0}+{H}_{p0}\right)\\ =-3{\mu }_{2}{k}_{2}V\left({H}_{0}+\frac{{{m}^{\prime }}_{1}}{\text{π}{\mu }_{2}{d}^{3}}\underset{q=1}{\overset{\left(n-1\right)/2}{\sum }}\frac{1}{{q}^{3}}\right)\end{array}
The expression about magnetic dipole moment
{{m}^{\prime }}_{1}
was that:
{{m}^{\prime }}_{1}=\frac{-4\text{π}{\mu }_{2}{R}^{3}{k}_{2}{H}_{0}}{1+4{k}_{2}{\left(\frac{R}{r}\right)}^{3}\underset{q=1}{\overset{\left(n-1\right)/2}{\sum }}\frac{1}{{q}^{3}}}
Based on the analysis to (2-11), the value of R had a stable value, the value of
{\left(R/d\right)}^{3}
depend on d, when the particle space d was large, the value of
{\left(R/d\right)}^{3}
was small, compared to classic dipole model, the moment of dipole
{{m}^{\prime }}_{1}
of local dipole model was very similar with the moment of dipole
m=-4\text{π}{\mu }_{2}{k}_{2}{R}^{3}{H}_{0}
of local dipole model.
2.3. Polarized Pellet Model
As regards polarized pellet model, there exist an assumption that magnetic particle only form isolated chain. The force between particles was utilized magnetic charge to calculate. Lemaire [9] calculated the function based on electric polarization ball:
{\tau }_{s}=\frac{2\varphi }{3\text{π}{a}^{2}}{\left\{3\mu {a}^{2}{H}^{2}f{\left(\frac{{\mu }_{i}-\mu }{{\mu }_{i}+2\mu }\right)}^{2}\right\}}^{m}
\mu ,{\mu }_{i}
was magnetic particle and MRF tape elongation respectively; a was radius of polarization sphere;
\varphi
was magnetic particle and MRF volume fraction ratio respectively; H was applied uniform magnetic field; f was polarzation coefficient (Figure 3).
Figure 3. Polarized pellet model.
Lemaire [9] calculated the MRF yield stress based on Equation (2-12), theoretical value tower above experimental value 60%. So, polarized pellet model had an obvious error.
2.4. Mean Continuous Field Model
Rosenweig [10] provided a mean continuous field model. He assumed that MRF was a kind of solid with isotropous magnetic susceptibility and yield stress (Figure 4).
Based on Maxwell tensor of stress and asymmetric stress effect, the function stress and strain was that:
\tau =\frac{1}{2}{\mu }_{0}{M}_{x}{H}_{0}=\frac{{\mu }_{0}{H}_{0}^{2}}{2}\left({\chi }_{\parallel }-{\chi }_{\perp }\right)\mathrm{sin}\alpha \mathrm{cos}\alpha
{\chi }_{\parallel },{\chi }_{\perp }
was parallel, vertical cylinder structure magnetic susceptibility;
\alpha
was the angle cylinder axis and magnetic field after yield.
Lemaire [11] calculated based on virtual work principle and mean field model. the expression was shown as following:
\tau =\frac{\partial W\left(r\right)}{\partial r}=-\frac{{\mu }_{0}{H}_{0}^{2}}{2}\frac{\partial {\chi }_{yy}}{\partial r}=\frac{{\mu }_{0}{H}_{0}^{2}}{2}\left({\chi }_{\parallel }-{\chi }_{\perp }\right)2\mathrm{sin}\alpha {\mathrm{cos}}^{3}\alpha
The function (2-14) had a redundant
2{\mathrm{cos}}^{2}\alpha
compared with function (2-13) because the calculation method may not be suitable for anisotropic material. The model ignored the magnetic concentrate effect between particles, so theoretical value will be lower than the test value.
According to the 4 models analyzed above, we can make a conclusion that: ① The classic dipole model had a concise and convenient expression, but it was only used when particle space was far enough. If particle space was near, the error could be out of control. ② Compared to the classic dipole model, local dipole model had a better accuracy in computation because of effect of other magnetic particle. It had a defeat that when the particle space was very near, the
Figure 4. Mean continuous field model.
accuracy decreased because the moment of dipole had an influence by other particle and decentralization effect. ③ Polarized pellet model could not consider the character of magnetic particle non-linearity and the error is obvious. ④ Mean continuous field model ignored the magnetic concentrate effect and the calculated value is lower than test value. Based on the four yield stress calculation model, we could make a reasonable hypothesis to analyze the major influence factor. A suitable assumption was a key to calculate the MRF yield stress.
[1] Bossis, G. and Volkova, O. (2003) Magneto-Rheology: Fluids, Structures and Rheology. Lecture Notes in Physics, 594, 202-230.
[2] Richter, L., Zipser, L. and Lange, U. (2001) Properties of Magneto-Rheological Fluids. Sensors and Materials, 13, 385-397.
[3] Carlson, J.D. (2002) What Makes a Good MR Fluid. Journal of Intelligent Material Systems and Structure, 13, 431-435.
[4] Zhu, X.C., Jing, X.J. and Li, C. (2012) Magnetorheological Fluid Dampers: A Review on Structure Design and Analysis. Journal of Intelligent Material Systems and Structures, 23, 839-873.
[5] Ginder, J.M. and Davis, L.C. (1994) Shear Stresses in Magnetorheological Fluids: Role of Magnetic Saturation. Applied Physics Letters, 65, 3410-3412.
[6] Weiss, K.D. and Duclos, T.G. (1994) Controllable Fluids: The Temperature De-pendence of Post-Yield Properties. World Scientific, Feldkirch, 20-23.
[7] Felt, D.W., Hagenbuchle, M. and Liu, J. (1996) Rheology of a Magnetorheological Fluid. World Scientific, Singapore, 738-746.
[8] Jiang, F.Q., Wang, Z.W. and Wu, J.Y. (1998) Magnetorheological Materials and Their Application in Shock Absorbers. World Scientific, Yonezawa, 494-500.
[9] Lemire, E. and Bossis, G. (1991) Yield Stress and Wall Effects in Magnetic Colloidal Suspensions. Journal of Physics D: Applied Physics, 24, 1473-1477.
[10] Rosensweig, R.E. (1995) On Magnetorheology and Electrorheology as Stasus of Unsymmetric Stress. Rheologica Acta, 39, 179-192.
[11] Lemiare, E. and Bossis, G. (1996) Deformation and Rupture Mechanisms of ER and MR Fluids. World Scientific, 11, 143-148. |
Scooter and Kayla are building a chicken coop in their back yard. The coop will be in the shape of a right triangle. One of the sides will be the wall of the garage. They have
11
feet of fencing, and one leg will be
4
If the garage forms the hypotenuse of the triangle, how long will the other leg be? How long will the hypotenuse be?
If the garage forms the hypotenuse, then the fencing will make up the legs of the triangle.
11
feet of fencing, and one side of the triangle is
4
feet long, what is the length of the other leg?
The other leg will be
7
feet long. To find the length of the hypotenuse use the Pythagorean theorem.
4^2 + 7^2 = x^2
If the garage wall is one leg of the triangle, how long will the hypotenuse be? How long will the leg along the garage be?
If the garage wall is a leg of the triangle instead of the hypotenuse, then its value cannot be greater than
7
The side with a value of
7
is, therefore, the new hypotenuse.
The leg along the garage will be
5.74
What is the area of each chicken coop in parts (a) and (b)?
The new Pythagorean equation is:
4^2 + x^2 = 7^2
To find each area, multiply the two legs (since they are perpendicular to each other and therefore form the base and the height) and divide that product by
2
The area for (a) is
14
^{2}
What is the area for (b)? |
Algebraic structure - Wikipedia
2 Common axioms
2.1 Equational axioms
2.2 Existential axioms
2.3 Non-equational axioms
3 Common algebraic structures
3.1 One set with operations
3.2 Two sets with operations
4 Hybrid structures
7 Different meanings of "structure"
Addition and multiplication are prototypical examples of operations that combine two elements of a set to produce a third element of the same set. These operations obey several algebraic laws. For example, a + (b + c) = (a + b) + c and a(bc) = (ab)c are associative laws, and a + b = b + a and ab = ba are commutative laws. Many systems studied by mathematicians have operations that obey some, but not necessarily all, of the laws of ordinary arithmetic. For example, the possible moves of an object in three-dimensional space can be combined by performing a first move of the object, and then a second move from its new position. Such moves, formally called rigid motions, obey the associative law, but fail to satisfy the commutative law.
Sets with one or more operations that obey specific laws are called algebraic structures. When a new problem involves the same laws as such an algebraic structure, all the results that have been proved using only the laws of the structure can be directly applied to the new problem.
In full generality, algebraic structures may involve an arbitrary collection of operations, including operations that combine more than two elements (higher arity operations) and operations that take only one argument (unary operations). The examples listed below are by no means a complete list, but include the most common structures that are important enough to receive a name.
Common axioms[edit]
Equational axioms[edit]
An axiom of an algebraic structure often has the form of an identity, that is, an equation such that the two sides of the equals sign are expressions that involve operations of the algebraic structure and variables. If the variables in the identity are replaced by arbitrary elements of the algebraic structure, the equality must remain true. Here are some common examples.
{\displaystyle *}
{\displaystyle x*y=y*x}
for every x and y in the algebraic structure.
{\displaystyle *}
is associative if
{\displaystyle (x*y)*z=x*(y*z)}
for every x, y and z in the algebraic structure.
{\displaystyle *}
is left distributive with respect to another operation
{\displaystyle +}
{\displaystyle x*(y+z)=(x*y)+(x*z)}
for every x, y and z in the algebraic structure (the second operation is denoted here as +, because the second operation is addition in many common examples).
{\displaystyle *}
is right distributive with respect to another operation
{\displaystyle +}
{\displaystyle (y+z)*x=(y*x)+(z*x)}
{\displaystyle *}
is distributive with respect to another operation
{\displaystyle +}
if it is both left distributive and right distributive. If the operation
{\displaystyle *}
is commutative, left and right distributivity are both equivalent to distributivity.
Existential axioms[edit]
Some common axioms contain an existential clause. In general, such a clause can be avoided by introducing further operations, and replacing the existential clause by an identity involving the new operation. More precisely, let us consider an axiom of the form "for all X there is y such that
{\displaystyle f(X,y)=g(X,y)}
", where X is a k-tuple of variables. Choosing a specific value of y for each value of X defines a function
{\displaystyle \varphi :X\mapsto y,}
which can be viewed as an operation of arity k, and the axiom becomes the identity
{\displaystyle f(X,\varphi (X))=g(X,\varphi (X)).}
The introduction of such auxiliary operation complicates slightly the statement of an axiom, but has some advantages. Given a specific algebraic structure, the proof that an existential axiom is satisfied consists generally of the definition of the auxiliary function, completed with straightforward verifications. Also, when computing in an algebraic structure, one generally uses explicitly the auxiliary operations. For example, in the case of numbers, the additive inverse is provided by the unary minus operation
{\displaystyle x\mapsto -x.}
Also, in universal algebra, a variety is a class of algebraic structures that share the same operations, and the same axioms, with the condition that all axioms are identities. What precedes shows that existential axioms of the above form are accepted in the definition of a variety.
Here are some of the most common existential axioms.
{\displaystyle *}
has an identity element if there is an element e such that
{\displaystyle x*e=x\quad {\text{and}}\quad e*x=x}
for all x in the structure. Here, the auxiliary operation is the operation of arity zero that has e as its result.
Given a binary operation
{\displaystyle *}
that has an identity element e, an element x is invertible if it has an inverse element, that is, if there exists an element
{\displaystyle \operatorname {inv} (x)}
{\displaystyle \operatorname {inv} (x)*x=e\quad {\text{and}}\quad x*\operatorname {inv} (x)=e.}
For example, a group is an algebraic structure with a binary operation that is associative, has an identity element, and for which all elements are invertible.
Non-equational axioms[edit]
The axioms of an algebraic structure can be any first-order formula, that is a formula involving logical connectives (such as "and", "or" and "not"), and logical quantifiers (
{\displaystyle \forall ,\exists }
) that apply to elements (not to subsets) of the structure.
Such a typical axiom is inversion in fields. This axiom cannot be reduced to axioms of preceding types. (it follows that fields do not form a variety in the sense of universal algebra.) It can be stated: "Every nonzero element of a field is invertible;" or, equivalently: the structure has a unary operation inv such that
{\displaystyle \forall x,\quad x=0\quad {\text{or}}\quad x\cdot \operatorname {inv} (x)=1.}
The operation inv can be viewed either as a partial operation that is not defined for x = 0; or as an ordinary function whose value at 0 is arbitrary and must not be used.
Common algebraic structures[edit]
One set with operations[edit]
Monoid: a semigroup with identity element.
Semiring: a ringoid such that S is a monoid under each operation. Addition is typically assumed to be commutative and associative, and the monoid product is assumed to distribute over the addition on both sides, and the additive identity 0 is an absorbing element in the sense that 0 x = 0 for all x.
Commutative ring: a ring in which the multiplication operation is commutative.
Field: a commutative ring which contains a multiplicative inverse for every nonzero element.
Heyting algebra: a bounded distributive lattice with an added binary operation, relative pseudo-complement, denoted by infix →, and governed by the axioms:
x → x = 1
x (x → y) = x y
y (x → y) = y
x → (y z) = (x → y) (x → z)
Two sets with operations[edit]
Group with operators: a group G with a set Ω and a binary operation Ω × G → G satisfying certain axioms.
Module: an abelian group M and a ring R acting as operators on M. The members of R are sometimes called scalars, and the binary operation of scalar multiplication is a function R × M → M, which satisfies several axioms. Counting the ring operations these systems have at least three operations.
Quadratic space: a vector space V over a field F with a quadratic form on V taking values in F.
Algebra-like structures: composite system defined over two sets, a ring R and an R-module M equipped with an operation called multiplication. This can be viewed as a system with five binary operations: two operations on R, two on M and one involving both R and M.
Inner product space: an F vector space V with a definite bilinear form V × V → F.
Clifford algebra: an associative
{\displaystyle \mathbb {Z} _{2}}
-graded algebra additionally equipped with an exterior product from which several possible inner products may be derived. Exterior algebras and geometric algebras are special cases of this construction.
Hybrid structures[edit]
Algebraic structures are defined through different configurations of axioms. Universal algebra abstractly studies such objects. One major dichotomy is between structures that are axiomatized entirely by identities and structures that are not. If all axioms defining a class of algebras are identities, then this class is a variety (not to be confused with algebraic varieties of algebraic geometry).
Identities are equations formulated using only the operations the structure allows, and variables that are tacitly universally quantified over the relevant universe. Identities contain no connectives, existentially quantified variables, or relations of any kind other than the allowed operations. The study of varieties is an important part of universal algebra. An algebraic structure in a variety may be understood as the quotient algebra of term algebra (also called "absolutely free algebra") divided by the equivalence relations generated by a set of identities. So, a collection of functions with given signatures generate a free algebra, the term algebra T. Given a set of equational identities (the axioms), one may consider their symmetric, transitive closure E. The quotient algebra T/E is then the algebraic structure or variety. Thus, for example, groups have a signature containing two operators: the multiplication operator m, taking two arguments, and the inverse operator i, taking one argument, and the identity element e, a constant, which may be considered an operator that takes zero arguments. Given a (countable) set of variables x, y, z, etc. the term algebra is the collection of all possible terms involving m, i, e and the variables; so for example, m(i(x), m(x, m(y,e))) would be an element of the term algebra. One of the axioms defining a group is the identity m(x, i(x)) = e; another is m(x,e) = x. The axioms can be represented as trees. These equations induce equivalence classes on the free algebra; the quotient algebra then has the algebraic structure of a group.
Some structures do not form varieties, because either:
Structures whose axioms unavoidably include nonidentities are among the most important ones in mathematics, e.g., fields and division rings. Structures with nonidentities present challenges varieties do not. For example, the direct product of two fields is not a field, because
{\displaystyle (1,0)\cdot (0,1)=(0,0)}
, but fields do not have zero divisors.
Different meanings of "structure"[edit]
In a slight abuse of notation, the word "structure" can also refer to just the operations on a structure, instead of the underlying set itself. For example, the sentence, "We have defined a ring structure on the set
{\displaystyle A}
," means that we have defined ring operations on the set
{\displaystyle A}
. For another example, the group
{\displaystyle (\mathbb {Z} ,+)}
can be seen as a set
{\displaystyle \mathbb {Z} }
that is equipped with an algebraic structure, namely the operation
{\displaystyle +}
^ Jonathan D. H. Smith (15 November 2006). An Introduction to Quasigroups and Their Representations. Chapman & Hall. ISBN 9781420010633. Retrieved 2012-08-02.
^ Ringoids and lattices can be clearly distinguished despite both having two defining binary operations. In the case of ringoids, the two operations are linked by the distributive law; in the case of lattices, they are linked by the absorption law. Ringoids also tend to have numerical models, while lattices tend to have set-theoretic models.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Algebraic_structure&oldid=1087418056" |
What is an acute angle? Learn the definition of acute angle and how to find acute angles in real life. Review some acute angle examples and acute angle shapes.
Acute Angle | Definition, Degree, & Examples
An acute angle is an angle that measures between
0°
90°
\frac{\pi }{2}
(in radians). Acute angles are always less than
90°
The word acute comes from Latin, acutus, meaning sharp or pointed. Anytime you see a pointy angle, you have an acute angle.
Facts About Acute Angles
Acute Angle Shapes
Make An Acute Angle
An acute angle is one of several angles you will encounter in geometry. Acute angles can be any degree greater than
0°
90°
Here is a tricky example:
Though the acute angle, measuring
89°
, is only
1°
off the right angle, it is still acute. Mathematics prefers precision, so even an
89.9°
angle cannot be called a right angle; it is an acute angle.
One acute angle will always measure between
0°
90°
Two acute angles can sum to be either greater than, less than, or equal to a right angle.
Two acute angles can be complementary angles (adding to
90°
Two acute angles alone cannot sum to make a straight angle (
180°
If acute angles are angles that measure between
0°
90°
, what is an angle called if it is
90°
exactly? And what if an angle measures more than
90°
A right angle is an angle whose measure is exactly
90°
\frac{\pi }{2}
90°
Here are examples of what acute, obtuse, and right angles can look like:
In addition to these angles, there are more angles to know, such as reflex angle, straight angle, and full angles. Learn more about the different types of angles you will encounter in geometry.
Acute angles are constantly popping up in geometry. You can find acute angles as exterior angles in shapes with five or more vertices like pentagons and octagons.
You can also find acute angles as interior angles in rhomboids and triangles.
Consider the isosceles and equilateral triangles, which depend on two and three acute angles for their construction.
How Many Acute Angles Are In An Acute Triangle?
In an acute triangle, all angles are acute. To be an acute triangle, it must have
3
acute angles.
A good example of an acute triangle is an equilateral triangle. The interior angles of all triangles must sum to
180°
, so in an equilateral triangle, where all three angles have the same measure, we know that each interior angle is an acute
60°
Examples of acute angles in real life are all around you. Acute angle things are sharp; they come to a sharp point. The sharp edge of a knife is an acute angle. A sharpened wood pencil has an acute end, and so does a pair of scissors:
You use angles in everyday life. Even the words you see in print are full of acute angles. The letters
A, K, M, N, R, V, W, X, Y, Z
are all formed with acute angles.
An example of an acute angle in the house are the hands of an analog clock, but only at four different hours (10, 11, 1, and 2 o'clock).
At 12 o'clock, the hands make a zero degree angle.
Acute angles are everywhere around you.
How To Make An Acute Angle
The easiest way to make or draw an acute angle is to write a capital letter A. This will create three acute angles.
The definition of an acute angle
The degree of acute angles
What shapes that are made with acute angles
Several examples of acute angles |
Electronics/Op-Amps - Wikibooks, open books for an open world
Electronics/Op-Amps
Electronics | Foreword | Basic Electronics | Complex Electronics | Electricity | Machines | History of Electronics | Appendix | edit
1 Op-Amp (operational amplifier)
2 741 Op-amp
3 Ideal Op-amps
4 Real Op-amps
4.1 DC Behaviour
4.2 AC Behaviour
5 Op-amp Configurations
7 Quick Design Process
8 Fully Differential Amplifier
8.2 Fully Differential Amplifier versus Instrumentation Amplifier
8.3 Applying a Quick Design Procedure
Op-Amp (operational amplifier)[edit | edit source]
Op-amp stands for operational amplifier. It is available in IC (Integrated Circuit) chip. It is an electronic circuit of many electronic components already connected and packaged inside a chip of many pins for external connection. Originally, op-amps were so named because they were used to model the basic mathematical operations of addition, subtraction, integration, differentiation, etc. in electronic analog computers. In this sense a true operational amplifier is an ideal circuit element.
741 Op-amp[edit | edit source]
The 741 op-amp has a symbol as shown.
Its terminals are:
V+: noninverting input
The 741 op-amp can be thought as Universal Voltage Difference Amplifier. The main function of the 741 op-amp is to amplify the difference of two input voltages which can be expressed mathematically like below
{\displaystyle V_{o}=A(V_{+}-V_{-})}
{\displaystyle V_{o}=A(V_{+}-V_{-})}
. The op-amp acts as a voltage difference amplifier.
It can be shown that an op-amp can function as a noninverting voltage amplifier or inverting voltage amplifier.
{\displaystyle V_{o}=AV_{+}}
{\displaystyle V_{-}=0}
. The op-amp acts as a noninverting voltage amplifier.
{\displaystyle V_{o}=-AV_{-}}
{\displaystyle V_{+}=0}
. The op-amp acts as an inverting voltage amplifier.
Also, it can be shown that an op-amp can function as a voltage comparator
{\displaystyle V_{o}=0}
{\displaystyle V_{-}=V_{+}}
. No difference between the two input voltages
Vo = VS+ .
{\displaystyle V_{+}>V_{-}}
. Non Inverting input voltage is greater than the Inverting input voltage
Vo = VS- .
{\displaystyle V_{-}>V_{+}}
. Inverting input voltage is greater than the Non Inverting input voltage.
Ideal Op-amps[edit | edit source]
The output voltage is the difference between the + and - inputs multiplied by the open-loop gain
Vout = (V+ − V−) * Avo.
The ideal op-amp has
an infinite open-loop gain,
infinite input impedances and
The input impedance is the impedance seen between the noninverting and inverting inputs.
The model for the op-amp is shown in Figure 1. Where V+ − V− is equal to vd; Rin is the input impedance; Rout is the output impedance; Avo is the open loop gain; and Rs is the source impedance.
Figure 1: The model of the op-amp.
Using voltage divider rule, we can determine the voltage vd.
{\displaystyle v_{d}={\frac {V_{in}R_{in}}{R_{in}+R_{s}}}}
Since the dependent voltage source amplifies the voltage vd by Avo. We can work out the output voltage across RL once again using voltage divider rule.
{\displaystyle V_{out}={\frac {A_{vo}v_{d}R_{L}}{R_{L}+R_{out}}}}
Substituting equation 1 into 2.
{\displaystyle V_{out}={\frac {A_{vo}V_{in}R_{in}R_{L}}{(R_{L}+R_{out})(R_{in}+R_{s})}}}
Now if the properties of an Ideal op-amp are applied. The ideal properties of an op-amp are infinite input impedance and zero output impedance. Since Rin >>, much greater than, Rs, Rin/(Rin+Rs)
{\displaystyle \approx }
{\displaystyle V_{out}=A_{vo}V_{in}\,}
Which is basically the definition of an op-amp. But if the input impedance was not infinite and the output impedance nonzero then this would not be so.
Real Op-amps[edit | edit source]
Real op-amps are (normally) built as an integrated circuit, but occasionally with discrete transistors or vacuum tubes. A Real op-amp is an approximation of an Ideal op-amp. A real op-amp does not have infinite open loop gain, infinite input impedance nor zero output impedance. Real op-amps also create noise in the circuit, have an offset voltage, thermal drift and finite bandwidth.
An offset voltage
means that there exists a voltage vd when both inputs are grounded. This offset is called an input offset because the voltage vd is offset from its ideal value of zero volts. The input offset voltage is multiplied by the open loop gain to create an output offset voltage.
means that the characteristics of the op-amp change with temperature. That is the open loop gain, input and output impedances, offset voltages and bandwidth change as the temperature changes.
Op-amps are made up of transistors. Transistors can only respond at a certain rate because of certain capacitances that they have. This means that the op-amps cannot respond fast enough to frequencies above a certain level. This level is the bandwidth.
Modern integrated circuit MOSFET op-amps approximate closer and closer to these ideals in limited-bandwidth, large-signal applications at room temperature. When the approximation is reasonably close, we go ahead and call the practical device an 'op-amp', forget its limitations and use the thinking and formulae given in this article.
DC Behaviour[edit | edit source]
Open-loop gain is defined as the amplification from input to output without any feedback applied. For most practical calculations, the open-loop gain is assumed to be infinite; in reality, however, it is limited by the amount of voltage applied to power the operational amplifier, i.e. Vs+ and Vs- in the above diagram. Typical devices exhibit open loop DC gain ranging from 100,000 to over 1 million. This allows the gain in the application to be set simply and exactly by using negative feedback. Of course theory and practice differ, since op-amps have limits that the designer must keep in mind and sometimes work around.
AC Behaviour[edit | edit source]
The op-amp gain calculated at DC does not apply at higher frequencies. This effect is due to limitations within the op-amp itself, such as its finite bandwidth, and to the AC characteristics of the circuit in which it is placed. The best known stumbling-block in designing with op-amps is the tendency for the device to resonate at high frequencies, where negative feedback changes to positive feedback due to parasitic lowpasses.
Typical low cost, general purpose op-amps exhibit a gain bandwidth product of a few MHz. Specialty and high speed op-amps can achieve gain bandwidth products of 100s of MHz.
Op-amp Configurations[edit | edit source]
Linear Configurations
Non Linear Configurations
audio and video pre-amplifiers and buffers
Precision Peak Detector
Quick Design Process[edit | edit source]
Imagine we have a (weak) input signal, and we want to amplify it to generate a strong output signal; or we have several different voltages, and we want to add them together. For these purposes, we need op-amp amplifying and summing circuits. It seems that there is nothing new to say about these bare (usually single op-amp) circuits. Only, the innovator Dieter Knollman has suggested a new simpler design procedure in an EDN's article [1]. He has also placed his work on the web[2] (see also a circuit story[3] written after Dieter's work).
All inputs are ideal voltage sources.
Gain is defined as the gain from the ideal voltage source to the op-amp output.
The feedback resistor, Rf, connects from the op-amp output to the inverting input.
The input resistor, Ri, connects from the ideal voltage source input to the op-amp input. (This *includes* any source resistance).
positive-gain inputs connect (through that resistor) to V+
negative-gain inputs connect (through that resistor) to V−
First, specify the desired circuit gains for each input.
Then calculate the ground gain, as follows:
Daisy's theorem:
The sum of the gains = +1 in a properly-designed op-amp circuit.
Ground gain = 1 - ( the sum of the desired positive and negative gains).
Choose a feedback-resistor value. For example, let Rf=100 kiloOhms.
Next, calculate the resistor values for each input, including the ground-gain resistor, using
{\displaystyle Ri={\frac {Rf}{|gain|}}}
where |gain| is the absolute value of the desired gain.
Some op-amp circuits need a resistor to ground from the op-amp's inverting input. Others need a resistor to ground on the noninverting input. The sign of the ground gain determines where to place the ground resistor.
If the desired gains add up to one, a resistor to ground is unnecessary.
You will see below some examples that illustrate using the quick design procedure. First of all, answer the question, "What's the highest and lowest output voltage you want?" Make sure the power terminals of the op-amp are at least that high and low. (If you want it to swing from +5 V to -5 V, then an op-amp connected to 0 V and +10 V won't work).
The generic op-amp has two inputs and one output. (Some are made with floating, differential outputs.) The output voltage is a multiple of the difference between the two inputs:
floating outputs? Are you perhaps misunderstanding "fully differential amplifier" ?
G is the open-loop gain of the op-amp. The inputs are assumed to have very high impedance; negligible current will flow into or out of the inputs. Op-amp outputs have very low source impedance.
If an Operational Amplifier is used with positive feedback it can act as an oscillator.
Other Notation[edit | edit source]
The power supply pins (VS+ and VS−) can be labeled many different ways. For FET based op-amps, the positive, common drain supply is labeled VDD and the negative, common source supply is labeled VSS. For BJT based op-amps, the VS+ pin becomes VCC and VS− becomes VEE. They are also sometimes labeled VCC+ and VCC−, or even V+ and V−, in which case the inputs would be labeled differently. The function remains the same.
Fully Differential Amplifier[edit | edit source]
Basic Applications[edit | edit source]
... insert drawings of basic FDA amplifying circuits here ...
Fully Differential Amplifier versus Instrumentation Amplifier[edit | edit source]
An instrumentation amplifier has extremely high input impedance-- much higher than the input impedance of a fully differential amplifier (once all the feedback resistors are in place). So an instrumentation amplifier is better for measuring voltage inputs with unknown (and possibly time-varying) output resistance.
A fully differential amplifier is better than an instrumentation amplifier for precisely generating differential output voltages, with good rejection of differential noise on the input, output, and power lines.
Applying a Quick Design Procedure[edit | edit source]
Can we apply Dieter's procedure to a fully differential amplifier? We can, if we revise it a little:
Daisy's theorem revised:
The sum of the gains = 0 in a properly-designed fully differential amplifier circuit.
Here is the FDA quick design procedure:
Use equal feedback resistors Rf, one from the + output to the - input, and one from the - output to the + input. Connect the "common" input to the appropriate voltage source.
The sum of the gains = zero in a properly-designed fully differential amplifier circuit (note that this new formulation differs from Daisy's theorem, which claims that the sum is always 1).
Calculate resistor values for each input, using Ri = Rf / |desired gain|.
For inputs with positive gain, connect the calculated resistor value between that input and the + input on the amplifier.
For inputs with negative gain, connect the calculated resistor value between that input and the - input on the amplifier.
If all your inputs are differential pairs, then the sum of all the gains is now zero. Done. Otherwise add a ground resistor to bring the total gain to zero.
Example 7 (a classic design procedure):
Example 8 (a quick design procedure):
↑ Single-formula technique keeps it simple - EDN's article by Dieter Knollman, PhD, Lucent Technologies
↑ K9 analysis make analog circuit design and analysis doggone simple
↑ How to Simplify the Design of the Mixed Op-amp Voltage Summer (after EDN's Single-formula technique keeps it simple)
Wikipedia has related information at Fully differential amplifier and Instrumentation amplifier
Electronics/Amplifiers
Wikipedia:Operational amplifier
Voltage compensation reveals the philosophy behind op-amp inverting circuits with negative feedback
Sergio Franco, "Design with Operational Amplifiers and Analog Integrated Circuits," 3rd Ed.], McGraw-Hill, New York, 2002 ISBN 0-07-232084-2
Linear Databook published in 1982 by The National Semiconductor Corporation.
Semiconductors/Op-Amps
The Transfer Function of the Non-Inverting Summing Amplifier with “N” Input Signals
Retrieved from "https://en.wikibooks.org/w/index.php?title=Electronics/Op-Amps&oldid=3798246" |
Unitary_operator Knowpia
In functional analysis, a unitary operator is a surjective bounded operator on a Hilbert space that preserves the inner product. Unitary operators are usually taken as operating on a Hilbert space, but the same notion serves to define the concept of isomorphism between Hilbert spaces.
A unitary element is a generalization of a unitary operator. In a unital algebra, an element U of the algebra is called a unitary element if U*U = UU* = I, where I is the identity element.[1]
Definition 1. A unitary operator is a bounded linear operator U : H → H on a Hilbert space H that satisfies U*U = UU* = I, where U* is the adjoint of U, and I : H → H is the identity operator.
The weaker condition U*U = I defines an isometry. The other condition, UU* = I, defines a coisometry. Thus a unitary operator is a bounded linear operator which is both an isometry and a coisometry,[2] or, equivalently, a surjective isometry.[3]
An equivalent definition is the following:
Definition 2. A unitary operator is a bounded linear operator U : H → H on a Hilbert space H for which the following hold:
U is surjective, and
U preserves the inner product of the Hilbert space, H. In other words, for all vectors x and y in H we have:
{\displaystyle \langle Ux,Uy\rangle _{H}=\langle x,y\rangle _{H}.}
The notion of isomorphism in the category of Hilbert spaces is captured if domain and range are allowed to differ in this definition. Isometries preserve Cauchy sequences, hence the completeness property of Hilbert spaces is preserved[4]
The following, seemingly weaker, definition is also equivalent:
the range of U is dense in H, and
{\displaystyle \langle Ux,Uy\rangle _{H}=\langle x,y\rangle _{H}.}
To see that Definitions 1 & 3 are equivalent, notice that U preserving the inner product implies U is an isometry (thus, a bounded linear operator). The fact that U has dense range ensures it has a bounded inverse U−1. It is clear that U−1 = U*.
Thus, unitary operators are just automorphisms of Hilbert spaces, i.e., they preserve the structure (in this case, the linear space structure, the inner product, and hence the topology) of the space on which they act. The group of all unitary operators from a given Hilbert space H to itself is sometimes referred to as the Hilbert group of H, denoted Hilb(H) or U(H).
The identity function is trivially a unitary operator.
Rotations in R2 are the simplest nontrivial example of unitary operators. Rotations do not change the length of a vector or the angle between two vectors. This example can be expanded to R3.
On the vector space C of complex numbers, multiplication by a number of absolute value 1, that is, a number of the form eiθ for θ ∈ R, is a unitary operator. θ is referred to as a phase, and this multiplication is referred to as multiplication by a phase. Notice that the value of θ modulo 2π does not affect the result of the multiplication, and so the independent unitary operators on C are parametrized by a circle. The corresponding group, which, as a set, is the circle, is called U(1).
More generally, unitary matrices are precisely the unitary operators on finite-dimensional Hilbert spaces, so the notion of a unitary operator is a generalization of the notion of a unitary matrix. Orthogonal matrices are the special case of unitary matrices in which all entries are real. They are the unitary operators on Rn.
The bilateral shift on the sequence space ℓ2 indexed by the integers is unitary. In general, any operator in a Hilbert space which acts by permuting an orthonormal basis is unitary. In the finite dimensional case, such operators are the permutation matrices.
The unilateral shift (right shift) is an isometry; its conjugate (left shift) is a coisometry.
The Fourier operator is a unitary operator, i.e. the operator which performs the Fourier transform (with proper normalization). This follows from Parseval's theorem.
Unitary operators are used in unitary representations.
Quantum logic gates are unitary operators. Not all gates are Hermitian.
The linearity requirement in the definition of a unitary operator can be dropped without changing the meaning because it can be derived from linearity and positive-definiteness of the scalar product:
{\displaystyle {\begin{aligned}\|\lambda U(x)-U(\lambda x)\|^{2}&=\langle \lambda U(x)-U(\lambda x),\lambda U(x)-U(\lambda x)\rangle \\&=\|\lambda U(x)\|^{2}+\|U(\lambda x)\|^{2}-\langle U(\lambda x),\lambda U(x)\rangle -\langle \lambda U(x),U(\lambda x)\rangle \\&=|\lambda |^{2}\|U(x)\|^{2}+\|U(\lambda x)\|^{2}-{\overline {\lambda }}\langle U(\lambda x),U(x)\rangle -\lambda \langle U(x),U(\lambda x)\rangle \\&=|\lambda |^{2}\|x\|^{2}+\|\lambda x\|^{2}-{\overline {\lambda }}\langle \lambda x,x\rangle -\lambda \langle x,\lambda x\rangle \\&=0\end{aligned}}}
Analogously you obtain
{\displaystyle \|U(x+y)-(Ux+Uy)\|=0.}
The spectrum of a unitary operator U lies on the unit circle. That is, for any complex number λ in the spectrum, one has |λ| = 1. This can be seen as a consequence of the spectral theorem for normal operators. By the theorem, U is unitarily equivalent to multiplication by a Borel-measurable f on L2(μ), for some finite measure space (X, μ). Now UU* = I implies |f(x)|2 = 1, μ-a.e. This shows that the essential range of f, therefore the spectrum of U, lies on the unit circle.
A linear map is unitary if it is surjective and isometric. (Use Polarization identity to show the only if part.)
Quantum logic gate – Basic circuit in quantum computing
Unitary matrix – Complex matrix whose conjugate transpose equals its inverse
Unitary transformation – Endomorphism preserving the inner product
^ Doran & Belfi 1986, p. 55
^ Halmos 1982, Sect. 127, page 69
^ Conway 1990, Proposition I.5.2
^ Conway 1990, Definition I.5.1
Conway, J. B. (1990). A Course in Functional Analysis. Graduate Texts in Mathematics. Vol. 96. Springer Verlag. ISBN 0-387-97245-5.
Doran, Robert S.; Belfi (1986). Characterizations of C*-Algebras: The Gelfand-Naimark Theorems. New York: Marcel Dekker. ISBN 0-8247-7569-4.
Halmos, Paul (1982). A Hilbert space problem book. Graduate Texts in Mathematics. Vol. 19 (2nd ed.). Springer Verlag. ISBN 978-0387906850.
Lang, Serge (1972). Differential manifolds. Reading, Mass.–London–Don Mills, Ont.: Addison-Wesley Publishing Co., Inc. ISBN 978-0387961132. |
Michio Jimbo - Wikipedia
Michio Jimbo (神保 道夫, Jimbō Michio, born November 28, 1951) is a Japanese mathematician working in mathematical physics and is a professor of mathematics at Rikkyo University. He is a grandson of the linguist Kaku Jimbo [ja].[citation needed]
After graduating from the University of Tokyo in 1974, he studied under Mikio Sato at the Research Institute for Mathematical Sciences in Kyoto University. He has made important contributions to mathematical physics, including (independently of Vladimir Drinfeld) the initial development of the study of quantum groups, the development of the theory of
{\displaystyle \tau }
-functions for the KP (Kadomtsev–Petviashvili) integrable hierarchy, and other related integrable hierarchies ,[1][2] and development of the theory of isomonodromic deformation systems for rational covariant derivative operators.[3]
In 1993 he won the Japan Academy Prize for this work.[4] In 2010 he received the Wigner Medal.[citation needed]
with Tetsuji Miwa, Etsurō Date: Solitons – differential equations, symmetries and infinite dimensional algebras. Cambridge University Press 2000, ISBN 0-521-56161-2
with Tetsuji Miwa: Algebraic analysis of solvable lattice models. American Mathematical Society 1993, ISBN 0-8218-0320-4
Editor: Yang-Baxter Equation in integrable systems. World Scientific 1990, doi:10.1142/1021
^ E. Date, M. Jimbo, M. Kashiwara and T. Miwa, "Operator approach to the Kadomtsev-Petviashvili equation III". J. Phys. Soc. Jap. 50 (11): 3806–3812 (1981). doi:10.1143/JPSJ.50.3806.
^ M. Jimbo and T. Miwa, "Solitons and infinite-dimensional Lie algebras", Publ. Res. Inst. Math. Sci., 19(3):943–1001 (1983).
^ M. Jimbo, T. Miwa, and K. Ueno, "Monodromy Preserving Deformation of Linear Ordinary Differential Equations with Rational Coefficients I", Physica D, 2, 306–352 (1981)
^ List of Japan Academy Prize recipients.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Michio_Jimbo&oldid=994893071" |
The homogeneous plate ABCD is subjected to a biaxial loading asshown. It is know
The homogeneous plate ABCD is subjected to a biaxial loading asshown. It is known that \sigma_z=\sigma_o and that the change in length of the plate in
The homogeneous plate ABCD is subjected to a biaxial loading asshown. It is known that
{\sigma }_{z}={\sigma }_{o}
and that the change in length of the plate in the x direction must be zero, that is
{ϵ}_{x}=0
. Denoting E the modulus of elasticity and by v Poisson's ratio, determine (a) the required magnitude of
{\sigma }_{x}
, and (b) the ratio
\frac{{\sigma }_{o}}{{ϵ}_{z}}
a) Strain in x-direction,
{ϵ}_{x}=\frac{{\sigma }_{x}}{E}-V\frac{{\sigma }_{z}}{E}
But given,
{ϵ}_{x}=0,{\sigma }_{z}={\sigma }_{o}
\frac{{ϵ}_{x}}{E}-V\frac{{\sigma }_{o}}{E}=0
{\sigma }_{x}
b) Strain in Z-direction,
{ϵ}_{Z}=\frac{{\sigma }_{Z}}{E}-V\frac{{\sigma }_{x}}{E}
{ϵ}_{Z}=\frac{{\sigma }_{0}}{E}-V\frac{{\sigma }_{x}}{E}
\frac{{\sigma }_{0}}{{ϵ}_{z}}
{30}^{\circ }
ra\frac{d}{{s}^{2}}
-1.59\cdot {10}^{4}
ra\frac{d}{{s}^{2}}
The drag characteristics of a torpedo are to be studied in a water tunnel using a 1 : 5 scale model. The tunnel operates with fresh water at
{20}^{\circ }C
whereas the prototype torpedo is to be used in seawater at
{15.6}^{\circ }C
.To correctly simulate the behavior of the prototype moving with a velocity of 30 m/s, what velocity is required in the water tunnel?
A car is traveling at 50.0 km/h on a flat highway. If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?
To win a prize at the country fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should you choose to throw a rubber ball or a beanbag of equal size and weight? Explain. |
Is R an equivalence relation? If so, prove it discrete
Is R an equivalence relation? If so, prove it discrete math and if not, explain why it is not. Let R be a relation on Z defined by (x.y) \in R if and only if 5(x-y)=0. Formally state what it means for R to be a symmetric relation.
Let R be a relation on Z defined by (x.y)
\in
R if and only if 5(x-y)=0. Formally state what it means for R to be a symmetric relation. Is R an equivalence relation? If so, prove it discrete math and if not, explain why it is not.
Let R be a relation on Z defined by x, y
\in
Z. Then (x, y)
\in
R if 5(x-y)=0
Symmetric let
\left(x,y\right)\in Ri.e.5\left(x-y\right)=0
if R symmetric (x-y) Then
\left(y,x\right)\in R
means 5(y-x)=0
⇒y-x=0⇒y=x
5\left(x-y\right)=0⇒-5\left(y-x\right)=0
⇒5\left(y-x\right)=0⇒\left(y,x\right)\in R
so R is symmetric.
x\in Z,x-x=0⇒5\left(x-x\right)=0
\left(x,x\right)\in R
Transitive: Let
\left(x,y\right)\in R\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(y-z\right)\in R
5\left(x-y\right)=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}5\left(y-z\right)=0
add both
5\left(x-y\right)+5\left(y-z\right)=0
⇒5\left(x-y+y-z\right)=0
5\left(x-z\right)=0
i.e.\left(x,z\right)\in R
So, that R is an equivalence selation.
\cap \cup
\left(a,b\right)\in R
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
Prove by mathematical induction:
\mathrm{\forall }n\ge 1,\text{ }{1}^{3}+{2}^{3}+{3}^{3}+\cdots +{n}^{3}=\frac{{n}^{2}{\left(n+1\right)}^{2}}{4}
f\left(k+1\right)-f\left(k\right) |
Prove this statement : A non-zero integer a is said to divide an integer b if there exists an integer - Maths - Real Numbers - 12960109 | Meritnation.com
Prove this statement
Prove this statement : A non-zero integer a is said to divide an integer b if there exists an integer c such that b = ac.
\mathrm{Hi}, \phantom{\rule{0ex}{0ex}}\mathrm{It} \mathrm{is} \mathrm{self} \mathrm{proved} \mathrm{or} \mathrm{axioms} \mathrm{so} \mathrm{it} \mathrm{does} \mathrm{not} \mathrm{need} \mathrm{proof} \mathrm{but} \mathrm{here} \mathrm{is} \mathrm{the} \phantom{\rule{0ex}{0ex}}\mathrm{more} \mathrm{explained} \mathrm{form} \phantom{\rule{0ex}{0ex}}\mathrm{let} 15 \mathrm{is} \mathrm{divisible} \mathrm{by} 3 \mathrm{or} \mathrm{say} 3 \mathrm{divides} 15 \mathrm{then} \phantom{\rule{0ex}{0ex}}15 = 3×5 \mathrm{here} 5 \mathrm{is} \mathrm{c} \mathrm{and} \mathrm{a} = 3 \mathrm{and} \mathrm{b} \mathrm{is} 15 \phantom{\rule{0ex}{0ex}}\mathrm{and} \mathrm{also} \mathrm{it} \mathrm{is} \mathrm{not} \mathrm{divisible} \mathrm{by} 2 \mathrm{because} \phantom{\rule{0ex}{0ex}}\mathrm{there} \mathrm{is} \mathrm{no} \mathrm{integer} \mathrm{c} \mathrm{that} \mathrm{will} \mathrm{make} \phantom{\rule{0ex}{0ex}}15 = 2\mathrm{c} \mathrm{hold} \mathrm{true}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}} |
Magnetic Flux, Induction, and Ampere's Circuital Law: Level 4-5 Challenges Practice Problems Online | Brilliant
Consider a uniformly charged circular disk of radius
R
having charge
Q
rotating about a fixed point on its circumference with an angular velocity
\omega
. The magnetic field created at this point due to the motion is
k \times \dfrac{\mu_{0} Q \omega}{R}
k
In homogenous magnetic field of induction
B
thin charged ring is rotating around it's own axis. Mass of ring is
m
and charge is
q
. Find the velocity of precession ring's axis around magnetic field line passing through the centar of ring.
B=0.1T
q=3C
m=0.01kg
by Вук Радовић
One day, I was out bicycling when a thunderstorm cropped up. While frantically pedaling home at a speed of
10~\mbox{km/hr}
, a wind blew a power line down and it brushed against the inside of my front wheel, putting a positive
1~\mbox{C}
charge on the wheel. I kept pedaling. If my bicycle wheels have radius
0.3~\mbox{m}
, what is the magnitude of the generated magnetic field in Teslas that I on my bike would measure at a distance of
0.1~\mbox{m}
perpendicular from the center of my front wheel?
Treat the wheels as thin conducting hoops.
You may assume the charge has spread out over the wheel but not left it yet.
My bike wheel was not slipping on the ground as I traveled.
The vacuum permeability is
\mu_0=4\pi \times 10^{-7}~\mbox{H/m}
You probably know Gauss Law very well. It states that if you take any closed surface, the electric flux through this surface is proportional to the total charge enclosed. Mathematically:
\Phi_{E}:=\oint \vec{E} \cdot d\vec{A}=\frac{Q_{enc}}{\epsilon_{0}}.
What about the magnetic flux? It turns out that
\Phi_{M}:=\oint \vec{B} \cdot d\vec{A}=0 \quad (\textrm{always!}).
This is a Law of Nature, equivalent to one of Maxwell's equations and it reflects the experimental fact that there are no magnetic charges . In particular,
\Phi_{M}=0
implies that not every magnetic field configuration can be realized in nature. For example, one can show that it is impossible to have a magnetic field that increases along the z-axis having only a z-component.
Consider an axially symmetric field with z-component (the field is symmetric about the z-axis) given by
B_{z}=B_{0}+ b z
B_{0}= 2~ \mu \mbox{T}
b=1 ~\mu \mbox{T/m}
Show that in addition to the z-component, this field must have a radial component
B_{r}
|B_{r}|
in Teslas at a point located
50~\textrm{cm}
away from the z-axis.
A wire is bent into the shape of an planar Archimedean spiral which in polar coordinates is described by the equation
r= b \theta.
The spiral has
N=100
turns and outer radius
R=10~\mbox{cm}
R
is the distance from point O to point T). Note that in the figure below we show a spiral having only 3 turns. The circuit is placed in a homogeneous magnetic field perpendicular to the plane of the spiral. The time dependence of the magnetic field induction is given by
B=B_{0}\cos(\omega t)
B_{0}=1 ~\mu \mbox{T}
\omega=2\times 10^{6}~\mbox{s}^{-1}
. Determine, the amplitude of the emf in Volts induced in the circuit. |
Sorting algorithms - Wikibooks, open books for an open world
< A-level Computing | AQA | Paper 1 | Fundamentals of algorithms(Redirected from A-level Computing/AQA/Problem Solving, Programming, Operating Systems, Databases and Networking/Programming Concepts/Insertion sort)
← Searching algorithms Sorting algorithms Optimisation algorithms →
1.2 Pseudocode implementation
Bubble Sort[edit | edit source]
Bubble sort is a simple sorting algorithm that works by repeatedly stepping through the list to be sorted, comparing each pair and swapping them if they are in the wrong order. The pass through the list is repeated until no swaps are needed, which indicates that the list is sorted. The algorithm gets its name from the way larger elements "bubble" to the top of the list. It is a very slow way of sorting data and rarely used in industry. There are much faster sorting algorithms out there such as insertion sort and quick sort which you will meet in A2.
Step-by-step example[edit | edit source]
Let us take the array of numbers "5 1 4 2 8", and sort the array from lowest number to greatest number using bubble sort algorithm. In each step, elements written in bold are being compared.
{\displaystyle \to }
( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps them since 5 > 1
{\displaystyle \to }
( 1 4 5 2 8 ), It then compares the second and third items and swaps them since 5 > 4
{\displaystyle \to }
( 1 4 2 5 8 ), Swap since 5 > 2
{\displaystyle \to }
( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.
The algorithm has reached the end of the list of numbers and the largest number, 8, has bubbled to the top. It now starts again.
{\displaystyle \to }
( 1 4 2 5 8 ), no swap needed
{\displaystyle \to }
{\displaystyle \to }
{\displaystyle \to }
{\displaystyle \to }
{\displaystyle \to }
{\displaystyle \to }
{\displaystyle \to }
Finally, the array is sorted, and the algorithm can terminate.
Pseudocode implementation[edit | edit source]
The algorithm can be expressed as:
We will now look at an example in Visual Basic using an array of people's heights. The following data set is being passed:
Sub bubbleSort(ByRef height() As integer)
For Count = 1 To MaxSize - 1
If height(Count + 1) < height(Count) Then
temp = height(Count)
height(Count) = height(Count + 1)
height(Count + 1) = temp
Loop Until swapped = False
'Print out the elements
For Count = 1 To MaxSize
Console.WriteLine(Count & ": " & height(Count))
Construct a trace table for the above code:
False 4 null 98 12 99 54
True 1 98 12 98
True 2 98 54 98 99
Show the following lists after one pass of bubble sort:
Sort into alphabetical order:
Henry, Cat, George, Mouse
Cat, George, Henry, Mouse
G, C, N, A, P, C
C, G, A, N, C, P
Sort into numerical order:
Show the following after 2 passes
Emu, Shrike, Gull, Badger
Emu, Gull, Badger, Shrike (Pass 1)
Emu, Badger, Gull, Shrike (Pass 2)
45, 32, 56, 12, 99 (Pass 1)
Let's look at a more complicated example, an array of structures, TopScores. The following data set is being passed:
Sub bubbleSort(ByRef TopScores() As TTopScore)
Dim temp As TTopScore
If TopScores(Count + 1).Score > TopScores(Count).Score Then
temp.Name = TopScores(Count).Name
temp.Score = TopScores(Count).Score
TopScores(Count).Score = TopScores(Count + 1).Score
TopScores(Count).Name = TopScores(Count + 1).Name
TopScores(Count + 1).Name = temp.Name
TopScores(Count + 1).Score = temp.Score
Console.WriteLine(Count & ": " & TopScores(Count).Name & " " & TopScores(Count).Score)
Exercise: Bubble Sort (Harder)
Draw a trace table to see if it works:
False 1 4 null null Michael 45 Dave 78 Gerald 23 Colin 75
True 1 4 Michael 45 Dave 78 Michael 45
True 3 4 Gerald 23 Colin 75 Gerald 23
True 2 4 Michael 45 Colin 75 Michael 45
1: Dave 78
2: Colin 75
3: Michael 45
4: Gerald 23
Insertion Sort[edit | edit source]
An example on insertion sort. Check each element and put them in the right order in the sorted list.
Unfortunately bubble sort is a very slow way of sorting data and very rarely used in industry. We'll now look at a much faster algorithm, insertion sort.
Insertion sort is a simple sorting algorithm: a comparison sort in which the sorted array (or list) is built one entry at a time. It is much less efficient on large lists than more advanced algorithms such as quicksort, heapsort, or merge sort and you may cover these at university. However, insertion sort provides several advantages:
efficient on small data sets
uses a fixed amount of memory when running
Insertion sort requires the use of two arrays, one ordered, and one unordered. Each repetition of the algorithm moves an item from the unordered list, into a sorted position in the ordered list, until there are no elements left in the unordered list.
Sorting is typically done in-place without needing extra memory. The resulting array after k iterations has the property where the first k + 1 entries are sorted. In each iteration the first remaining entry of the input is removed, inserted into the result at the correct position, thus extending the result:
Animation of the insertion sort sorting a 30 element array. Notice how the ordered array (left hand side) slowly consumes the unordered array (right hand side), until the entire data set is ordered
The following table shows the steps for sorting the sequence {5, 7, 0, 3, 4, 2, 6, 1}. For each iteration, the number of positions the inserted element has moved is shown in parentheses. Altogether this amounts to 17 steps.
5 7 0 3 4 2 6 1 (0)
// A[ i ] is added in the sorted sequence A[0, .. i-1]
// save A[i] to make a hole at index iHole
item ← A[i]
iHole ← i
// keep moving the hole to next smaller index until A[iHole - 1] is <= item
while iHole > 0 and A[iHole - 1] > item
// move hole to next smaller index
A[iHole] ← A[iHole - 1]
iHole ← iHole - 1
// put item in the hole
A[iHole] ← item
' a procedure to sort an array of integers
Describe the process of insertion sort
Insertion sort is a simple sorting algorithm: a comparison sort in which the sorted array (or list) is built one entry at a time.
Show the how insert sort would work on the following unordered array:
sort left hand side is underlined
Show how the insert sort would work on the following unordered array
G K L A J
A G K L J
A G J K L
Complete the trace table for the following code:
Retrieved from "https://en.wikibooks.org/w/index.php?title=A-level_Computing/AQA/Paper_1/Fundamentals_of_algorithms/Sorting_algorithms&oldid=4042502" |
JMSE | Free Full-Text | Modelling the Past and Future Evolution of Tidal Sand Waves
Effects of Various Fuels on Combustion and Emission Characteristics of a Four-Stroke Dual-Fuel Marine Engine
A Practical Trajectory Tracking Scheme for a Twin-Propeller Twin-Hull Unmanned Surface Vehicle
Krabbendam, J.
Nnafie, A.
Huib de Swart
Institute for Marine and Atmospheric Research, Utrecht University, Princetonplein 5, 3584 CC Utrecht, The Netherlands
Faculty of Engineering Technology, Twente University, De Horst 2, 7522 LW Enschede, The Netherlands
WaterProof B.V., IJsselmeerdijk 2, 8221 RC Lelystad, The Netherlands
Academic Editor: Michael G. Hughes
This study focuses on the hindcasting and forecasting of observed offshore tidal sand waves by using a state-of-the-art numerical morphodynamic model. The sand waves, having heights of several meters, evolve on timescales of years. Following earlier work, the model has a 2DV configuration (one horizontal and one vertical direction). First, the skill of the model is assessed by performing hindcasts at four transects in the North Sea where sand wave data are available of multiple surveys that are at least 10 years apart. The first transect is used for calibration and this calibrated model is applied to the other three transects. It is found that the calibrated model performs well: the Brier Skill Score is ‘excellent’ at the first two transects and ‘good’ at the last two. The root mean square error of calculated bed levels is smaller than the uncertainty in the measurements, except at the last transect, where the M
{}_{2}
is more elliptical than at the other three transects. The calibrated model is subsequently used to make forecasts of the sand waves along the two transects with the best skill scores. View Full-Text
Keywords: tidal sand waves; North Sea; Delft3D; numerical morphological modelling tidal sand waves; North Sea; Delft3D; numerical morphological modelling
Krabbendam, J.; Nnafie, A.; de Swart, H.; Borsje, B.; Perk, L. Modelling the Past and Future Evolution of Tidal Sand Waves. J. Mar. Sci. Eng. 2021, 9, 1071. https://doi.org/10.3390/jmse9101071
Krabbendam J, Nnafie A, de Swart H, Borsje B, Perk L. Modelling the Past and Future Evolution of Tidal Sand Waves. Journal of Marine Science and Engineering. 2021; 9(10):1071. https://doi.org/10.3390/jmse9101071
Krabbendam, Janneke, Abdel Nnafie, Huib de Swart, Bas Borsje, and Luitze Perk. 2021. "Modelling the Past and Future Evolution of Tidal Sand Waves" Journal of Marine Science and Engineering 9, no. 10: 1071. https://doi.org/10.3390/jmse9101071 |
Estimate Discrete-Time Grey-Box Model with Parameterized Disturbance - MATLAB & Simulink - MathWorks United Kingdom
Description of the SISO System
Estimating the Parameters of an idgrey Model
This example is based on a discrete, single-input and single-output (SISO) system represented by the following state-space equations:
\begin{array}{l}x\left(kT+T\right)=\left[\begin{array}{cc}par1& par2\\ 1& 0\end{array}\right]x\left(kT\right)+\left[\begin{array}{c}1\\ 0\end{array}\right]u\left(kT\right)+w\left(kT\right)\\ y\left(kT\right)=\left[\begin{array}{cc}par3& par4\end{array}\right]x\left(kT\right)+e\left(kT\right)\\ x\left(0\right)=x0\end{array}
where w and e are independent white-noise terms with covariance matrices R1 and R2, respectively. R1=E{ww'} is a 2–by-2 matrix and R2=E{ee'} is a scalar. par1, par2, par3, and par4 represent the unknown parameter values to be estimated.
Assume that you know the variance of the measurement noise R2 to be 1. R1(1,1) is unknown and is treated as an additional parameter par5. The remaining elements of R1 are known to be zero.
You can represent the system described in Description of the SISO System as an idgrey (grey-box) model using a function. Then, you can use this file and the greyest command to estimate the model parameters based on initial parameter guesses.
To run this example, you must load an input-output data set and represent it as an iddata or idfrd object called data. For more information about this operation, see Representing Time- and Frequency-Domain Data Using iddata Objects or Representing Frequency-Response Data Using idfrd Objects.
To estimate the parameters of a grey-box model:
Create the file mynoise that computes the state-space matrices as a function of the five unknown parameters and the auxiliary variable that represents the known variance R2. The initial conditions are not parameterized; they are assumed to be zero during this estimation.
R2 is treated as an auxiliary variable rather than assigned a value in the file to let you change this value directly at the command line and avoid editing the file.
function [A,B,C,D,K] = mynoise(par,T,aux)
R2 = aux(1); % Known measurement noise variance
A = [par(1) par(2);1 0];
C = [par(3) par(4)];
R1 = [par(5) 0;0 0];
[~,K] = kalman(ss(A,eye(2),C,0,T),R1,R2);
Specify initial guesses for the unknown parameter values and the auxiliary parameter value R2:
par1 = 0.1; % Initial guess for A(1,1)
par2 = -2; % Initial guess for A(1,2)
par3 = 1; % Initial guess for C(1,1)
par5 = 0.2; % Initial guess for R1(1,1)
Pvec = [par1; par2; par3; par4; par5]
auxVal = 1; % R2=1
Construct an idgrey model using the mynoise file:
Minit = idgrey('mynoise',Pvec,'d',auxVal);
The third input argument 'd' specifies a discrete-time system.
Estimate the model parameter values from data:
opt.Display = 'full';
Model = greyest(data,Minit,opt)
kalman (Control System Toolbox) | idgrey | greyest |
Factoring Polynomial Equations - Course Hero
College Algebra/Polynomial Operations and Theorems/Factoring Polynomial Equations
If a polynomial expression can be factored, then the zero product property (if
pq=0
p=0
q=0
) can be used to solve the related equation that is formed by setting the polynomial expression equal to zero. Each solution will need to be evaluated in the polynomial to determine if it is a true solution.
Solve a Polynomial Equation by Factoring
Factor to solve the equation:
x^4+x^3-6x^2=0
Start by factoring out the GCF (greatest common factor) of
x^2
\begin{aligned}x^4+x^3-6x^2&=0\\x^2(x^2+x-6)&=0\end{aligned}
Then factor the resulting trinomial.
x^2(x+3)(x-2)=0
Use the zero product property to identify solutions.
\boxed{\begin{aligned} x^2&=0\\x&=0\end{aligned}}\;\;\;\;\;\;\;\;\;\;\boxed{\begin{aligned}x+3&=0\\x&=0-3\\x&=-3\end{aligned}}\;\;\;\;\;\;\;\;\;\;\boxed{\begin{aligned}x-2&=0\\x&=0+2\\x&=2\end{aligned}}
Check the solutions. Substitute each solution into the original equation.
\boxed{\begin{aligned}x^4+x^3-6x^2&=0\\0^4+0^3-6(0)^2&\stackrel{?}{=}0\\0+0-0&\stackrel{?}{=}0\\0&=0\;\;\checkmark\end{aligned}}\;\;\;\;\;\;\;\;\;\;\boxed{\begin{aligned}x^4+x^3-6x^2&=0\\(-3)^4+(-3)^3-6(-3)^2&\stackrel{?}{=}0\\81-27-54&\stackrel{?}{=}0\\0&=0\;\;\checkmark\end{aligned}}
\boxed{\begin{aligned}x^4+x^3-6x^2&=0\\2^4+2^3-6(2)^2&\stackrel{?}{=}0\\16+8-24&\stackrel{?}{=}0\\0&=0\;\;\checkmark\end{aligned}}
x=-3
x=0
x=2
The related function of the polynomial is:
f(x)=x^4+x^3-6x^2
The graph of the related function shows that the solutions are the
x
-intercepts, which have the ordered pairs
(-3, 0)
(0, 0)
(2, 0)
<Polynomial Operations>Polynomial Theorems |
Inverse Trigonometric Identities | Brilliant Math & Science Wiki
Omkar Kulkarni, Pranjal Jain, Jimin Khim, and
Before reading this, make sure you are familiar with inverse trigonometric functions.
The following inverse trigonometric identities give an angle in different ratios. Before the more complicated identities come some seemingly obvious ones. Be observant of the conditions the identities call for.
\begin{array}{ l l l } \sin^{-1}(-x) &=-\sin^{-1}x,~&\lvert x\rvert\leq1 \\ \cos^{-1}(-x) &=\pi-\cos^{-1}x,~&\lvert x\rvert\leq1 \\ \tan^{-1}(-x) &=-\tan^{-1}x,~&x\in\mathbb{R} \\ \cot^{-1}(-x) &=\pi-\cot^{-1}x,~&x\in\mathbb{R} \\ \csc^{-1}x &=\sin^{-1}\left(\frac{1}{x}\right),~&\lvert x\rvert\geq1 \\ \sec^{-1}x &=\cos^{-1}\left(\frac{1}{x}\right),~&\lvert x\rvert\geq1 \\ \cot^{-1}x &=\tan^{-1}\left(\frac{1}{x}\right),~&x>0 \\ \cot^{-1}x &=\pi+\tan^{-1}\left(\frac{1}{x}\right),~&x<0 \\ \sin^{-1}x+\cos^{-1}x &=\frac{\pi}{2},~&\lvert x\rvert\leq1 \\ \csc^{-1}x+\sec^{-1}x &=\frac{\pi}{2},~&\lvert x\rvert\geq1 \end{array}
Now for the more complicated identities. These come handy very often, and can easily be derived using the basic trigonometric identities.
\begin{array}{ l l l } \sin^{-1}x&=\cos^{-1}\left(\sqrt{1-x^{2}}\right),~&x\geq0 \\ \cos^{-1}x&=\sin^{-1}\left(\sqrt{1-x^{2}}\right),~&x\geq0 \\ \cos^{-1}x&=\pi-\sin^{-1}\left(\sqrt{1-x^{2}}\right),~&x<0 \\ \tan^{-1}x+\tan^{-1}y&=\tan^{-1}\left(\frac{x+y}{1-xy}\right),~&xy<1 \\ \tan^{-1}x+\tan^{-1}y&=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right),~&xy>1 \\ \tan^{-1}x-\tan^{-1}y&=\tan^{-1}\left(\frac{x-y}{1+xy}\right) \\ \sin^{-1}x&=\tan^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right),~&x\in(0,1) \\ \cos^{-1}x&=\tan^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right),~&x\in(0,1) \\ \tan^{-1}x&=\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right),~&x>0 \\ \tan^{-1}x&=\cos^{-1}\left(\frac{1}{\sqrt{x^{2}+1}}\right),~&x>0 \end{array}
x
\sin\big(\cot^{-1}(1+x)\big)=\cos\big(\tan^{-1}(x)\big).
\begin{aligned} \cot^{-1}(1+x)&=\sin^{-1}\left(\frac{1}{\sqrt{1+(1+x)^{2}}}\right) \\ \tan^{-1}x&=\cos^{-1}\left(\frac{1}{\sqrt{x^{2}-1}}\right). \end{aligned}
\begin{aligned} \sin\big(\cot^{-1}(1+x)\big)&=\cos\big(\tan^{-1}(x)\big) \\ \sin\left(\sin^{-1}\bigg(\frac{1}{\sqrt{1+(1+x)^{2}}}\bigg)\right)&=\cos\left(\cos^{-1}\bigg(\frac{1}{\sqrt{x^{2}-1}}\bigg)\right) \\ \frac{1}{\sqrt{1+(1+x)^{2}}} &= \frac{1}{\sqrt{x^{2}+1}} \\ x^{2}+1&=(x+1)^{2}+1 \\ x^{2}+2x+1&=x^{2} \\ x&=-\frac{1}{2}.\ _\square \end{aligned}
Cite as: Inverse Trigonometric Identities. Brilliant.org. Retrieved from https://brilliant.org/wiki/inverse-trigonometric-identities/ |
Not to be confused with Image histogram.
The color histogram can be built for any kind of color space, although the term is more often used for three-dimensional spaces like RGB or HSV. For monochromatic images, the term intensity histogram may be used instead. For multi-spectral images, where each pixel is represented by an arbitrary number of measurements (for example, beyond the three measurements in RGB), the color histogram is N-dimensional, with N being the number of measurements taken. Each measurement has its own wavelength range of the light spectrum, some of which may be outside the visible spectrum.
If the set of possible color values is sufficiently small, each of those colors may be placed on a range by itself; then the histogram is merely the count of pixels that have each possible color. Most often, the space is divided into an appropriate number of ranges, often arranged as a regular grid, each containing many similar color values. The color histogram may also be represented and displayed as a smooth function defined over the color space that approximates the pixel counts.
Like other kinds of histograms, the color histogram is a statistic that can be viewed as an approximation of an underlying continuous distribution of colors values.
2 Characteristics of a color histogram
3 Principles of the formation of a color histogram
5 Drawbacks and other approaches
6 Intensity histogram of continuous data
Color histograms are flexible constructs that can be built from images in various color spaces, whether RGB, rg chromaticity or any other color space of any dimension. A histogram of an image is produced first by discretization of the colors in the image into a number of bins, and counting the number of image pixels in each bin. For example, a Red–Blue chromaticity histogram can be formed by first normalizing color pixel values by dividing RGB values by R+G+B, then quantizing the normalized R and B coordinates into N bins each. A two-dimensional histogram of Red-Blue chromaticity divided into four bins (N=4) might yield a histogram that looks like this table:
blue 0-63 43 78 18 0
128-191 127 58 25 8
A histogram can be N-dimensional. Although harder to display, a three-dimensional color histogram for the above example could be thought of as four separate Red-Blue histograms, where each of the four histograms contains the Red-Blue values for a bin of green (0-63, 64-127, 128-191, and 192-255).
The histogram provides a compact summarization of the distribution of data in an image. The color histogram of an image is relatively invariant with translation and rotation about the viewing axis, and varies only slowly with the angle of view.[1] By comparing histograms signatures of two images and matching the color content of one image with the other, the color histogram is particularly well suited for the problem of recognizing an object of unknown position and rotation within a scene. Importantly, translation of an RGB image into the illumination invariant rg-chromaticity space allows the histogram to operate well in varying light levels.
A histogram is a graphical representation of the number of pixels in an image. In a more simple way to explain, a histogram is a bar graph, whose X-axis represents the tonal scale(black at the left and white at the right), and Y-axis represents the number of pixels in an image in a certain area of the tonal scale. For example, the graph of a luminance histogram shows the number of pixels for each brightness level(from black to white), and when there are more pixels, the peak at the certain luminance level is higher.
2. What is a color histogram?
A color histogram of an image represents the distribution of the composition of colors in the image. It shows different types of colors appeared and the number of pixels in each type of the colors appeared. The relation between a color histogram and a luminance histogram is that a color histogram can be also expressed as “Three Luminance Histograms”, each of which shows the brightness distribution of each individual Red/Green/Blue color channel.
Characteristics of a color histogram[edit]
A color histogram focuses only on the proportion of the number of different types of colors, regardless of the spatial location of the colors. The values of a color histogram are from statistics. They show the statistical distribution of colors and the essential tone of an image.
In general, as the color distributions of the foreground and background in an image are different, there might be a bimodal distribution in the histogram.
For the luminance histogram alone, there is no perfect histogram and in general, the histogram can tell whether it is over exposure or not, but there are times when you might think the image is over exposed by viewing the histogram; however, in reality it is not.
Principles of the formation of a color histogram[edit]
The formation of a color histogram is rather simple. From the definition above, we can simply count the number of pixels for each 256 scales in each of the 3 RGB channel, and plot them on 3 individual bar graphs.
In general, a color histogram is based on a certain color space, such as RGB or HSV. When we compute the pixels of different colors in an image, if the color space is large, then we can first divide the color space into certain numbers of small intervals. Each of the intervals is called a bin. This process is called color quantization. Then, by counting the number of pixels in each of the bins, we get the color histogram of the image.
The concrete steps of the principles can be viewed in Example 1.
Given the following image of a cat (an original version and a version that has been reduced to 256 colors for easy histogram purposes), the following data represents a color histogram in the RGB color space, using four bins.
Bin 0 corresponds to intensities 0-63
Bin 1 is 64-127
Bin 2 is 128-191 and Bin 3 is 192-255.
Color histogram of the above cat picture with x-axis being RGB and y-axis being the frequency.
A picture of a cat reduced to 256 colors in the RGB color space
Bin 0 Bin 0 Bin 0 7414
Bin 0 Bin 0 Bin 1 230
Bin 0 Bin 0 Bin 2 0
Bin 0 Bin 1 Bin 2 88
Bin 2 Bin 2 Bin 2 53110
Application in camera:
Nowadays, some cameras have the ability of showing the 3 color histograms when we take photos.
We can examine clips (spikes on either the black or white side of the scale) in each of the 3 RGB color histograms. If we find one or more clipping on a channel of the 3 RGB channels, then this would result in a loss of detail for that color.
To illustrate this, consider this example:
1. We know that each of the three R, G, B channels has a range of values from 0-255(8 bit). So consider a photo that has a luminance range of 0-255.
2. Assume the photo we take is made of 4 blocks that are adjacent to each other and we set the luminance scale for each of the 4 blocks of original photo to be 10, 100, 205, 245. Thus, the image looks like the first figure on the right.
3. Then, we over expose the photo a little, say, the luminance scale of each block is increased by 10. Thus, the luminance scale for each of the 4 blocks of new photo is 20, 110, 215, 255. Then, the image looks like the second figure on the right.
There is not much difference between figure 8 and figure 9, all we can see is that the whole image becomes brighter(the contrast for each of the blocks remain the same).
4. Now, we over expose the original photo again, this time the luminance scale of each block is increased by 50. Thus, the luminance scale for each of the 4 blocks of new photo is 60, 150, 255, 255. The new image now looks like the third figure on the right.
Note that the scale for last block is 255 instead of 295, for 255 is the top scale and thus the last block has clipped! When this happens, we lose the contrast of the last 2 blocks, and thus, we cannot recover the image no matter how we adjust it.
To conclude, when taking photos with a camera that displays histograms, always keep the brightest tone in the image below the largest scale 255 on the histogram in order to avoid losing details.
Drawbacks and other approaches[edit]
The main drawback of histograms for classification is that the representation is dependent of the color of the object being studied, ignoring its shape and texture. Color histograms can potentially be identical for two images with different object content which happens to share color information. Conversely, without spatial or shape information, similar objects of different color may be indistinguishable based solely on color histogram comparisons. There is no way to distinguish a red and white cup from a red and white plate. Put another way, histogram-based algorithms have no concept of a generic 'cup', and a model of a red and white cup is no use when given an otherwise identical blue and white cup. Another problem is that color histograms have high sensitivity to noisy interference such as lighting intensity changes and quantization errors. High dimensionality (bins) color histograms are also another issue. Some color histogram feature spaces often occupy more than one hundred dimensions.[2]
Some of the proposed solutions have been color histogram intersection, color constant indexing, cumulative color histogram, quadratic distance, and color correlograms. Although there are drawbacks of using histograms for indexing and classification, using color in a real-time system has several advantages. One is that color information is faster to compute compared to other invariants. It has been shown in some cases that color can be an efficient method for identifying objects of known location and appearance.
Further research into the relationship between color histogram data to the physical properties of the objects in an image has shown they can represent not only object color and illumination but relate to surface roughness and image geometry and provide an improved estimate of illumination and object color.[3]
Usually, Euclidean distance, histogram intersection, or cosine or quadratic distances are used for the calculation of image similarity ratings.[4] Any of these values do not reflect the similarity rate of two images in itself; it is useful only when used in comparison to other similar values. This is the reason that all the practical implementations of content-based image retrieval must complete computation of all images from the database, and is the main disadvantage of these implementations.
Another approach to representative color image content is two-dimensional color histogram. A two-dimensional color histogram considers the relation between the pixel pair colors (not only the lighting component).[5] A two-dimensional color histogram is a two-dimensional array. The size of each dimension is the number of colors that were used in the phase of color quantization. These arrays are treated as matrices, each element of which stores a normalized count of pixel pairs, with each color corresponding to the index of an element in each pixel neighborhood. For comparison of two-dimensional color histograms it is suggested calculating their correlation, because constructed as described above, is a random vector (in other words, a multi-dimensional random value). While creating a set of final images, the images should be arranged in decreasing order of the correlation coefficient.
The correlation coefficient may also be used for color histogram comparison. Retrieval results with correlation coefficient are better than with other metrics.[6]
Intensity histogram of continuous data[edit]
The idea of an intensity histogram can be generalized to continuous data, say audio signals represented by real functions or images represented by functions with two-dimensional domain.
{\displaystyle f\in L^{1}(\mathbb {R} ^{n})}
(see Lebesgue space), then the cumulative histogram operator
{\displaystyle H}
can be defined by:
{\displaystyle H(f)(y)=\mu \{x:f(x)\leq y\}}
{\displaystyle \mu }
is the Lebesgue measure of sets.
{\displaystyle H(f)}
in turn is a real function. The (non-cumulative) histogram is defined as its derivative.
{\displaystyle h(f)=H(f)'}
^ Shapiro, Linda G. and Stockman, George C. "Computer Vision" Prentice Hall, 2003 ISBN 0-13-030796-3
^ Xiang-Yang Wang, Jun-Feng Wu, and Hong-Ying Yang "Robust image retrieval based on color histogram of local feature regions" Springer Netherlands, 2009 ISSN 1573-7721
^ Anatomy of a color histogram; Novak, C.L.; Shafer, S.A.; Computer Vision and Pattern Recognition, 1992. Proceedings CVPR '92., 1992 IEEE Computer Society Conference on 15–18 June 1992 Page(s):599 - 605 doi:10.1109/CVPR.1992.223129
^ Integrated Spatial and Feature Image Systems: Retrieval, Analysis and Compression; Smith, J.R.; Graduate School of Arts and Sciences, Columbia University, 1997
^ Effectiveness estimation of image retrieval by 2D color histogram; Bashkov, E.A.; Kostyukova, N.S.; Journal of Automation and Information Sciences, 2006 (6) Page(s): 84-89
^ Content-Based Image Retrieval Using Color Histogram Correlation; Bashkov, E.A.; Shozda, N.S.; Graphicon proceedings, 2002 Page(s): [1] Archived 2012-07-07 at the Wayback Machine
3D Color Inspector/Color Histogram, by Kai Uwe Barthel. (Free Java applet.)
Stanford Student Project on Image Based Retrieval - more in depth look at equations/application
MATLAB/Octave code for plotting Color Histograms and Color Clouds - The source code can be ported to other languages
Retrieved from "https://en.wikipedia.org/w/index.php?title=Color_histogram&oldid=1054877317" |
zgribestika 2021-12-15 Answered
1.4{\left\{k\frac{W}{m}\right\}}^{2}
1.5×{10}^{11}m
(a) The given value represents the intensity
I=1.4k\frac{W}{{m}^{2}}
of the light. The intenity of a sinusoidal electromagnetic wave in vacuum is related to the electric-field amplitude
{E}_{max}
and the amplitude of magnetic field
{B}_{max}
and is given by the equation:
I=\frac{1}{2}{ϵ}_{0}c{E}_{max}^{2}
{ϵ}_{0}
is a electric constant,
c
is speed of light.
Now, solve the equation for
{E}_{max}
{E}_{max}=\sqrt{\frac{2I}{{ϵ}_{0}c}}
Plug the values for
I,ϵ,c
{E}_{max}=\sqrt{\frac{2I}{{ϵ}_{0}c}}
=\sqrt{\frac{2\left(1400\frac{W}{{m}^{2}}\right)}{\left(8.85×{10}^{-12}\frac{{C}^{2}}{N}×{m}^{2}\right)\left(3×{10}^{8}\frac{m}{s}\right)}}
The maximum electric field is related to the maximum magnetic field in form:
{B}_{max}=\frac{{E}_{max}}{c}
Now, plug the values into the equation
{B}_{max}=\frac{{E}_{max}}{c}=\frac{1026\frac{V}{m}}{3×{10}^{8}\frac{m}{s}}=3.42×{10}^{-6}T
(b) The intensity is propotional to
{E}_{max}^{2}
and represents the incident power
P
A
P=IA
The distance between the Earth and the Sun represents the radius of the path of the Earth around the Sun and it is considered to be sphere. So, the area is calculated:
A=4\pi {r}^{2}=4\pi {\left(1.5×{10}^{11}m\right)}^{2}=2.82×{10}^{23}{m}^{2}
Now, plug in the values:
P=IA=\left(1400\frac{W}{{m}^{2}}\right)\left(2.82×{10}^{23}{m}^{2}\right)=3.95×{10}^{26}W
The answer is: (a)
3.42×{10}^{-6}T
3.95×{10}^{26}W
\stackrel{\to }{B}
A 2.0 kg mass is projected from the edge of the top of a 20mtall building with a velocity of 24m/s at some unknown angle abovethe horizontal. Disregard air resistance and assume the ground islevel. What is the kinetic energy of the mass just before itstrikes the ground?
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal stretched string with a speed of 36.0 m/s. Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement. What is y (t) for a particle 1.35 m to the right of the origin? Round all numeric coefficients to exactly three significant figures. |
Architects, building designers, and building owners presently lack sufficient resources for thoroughly evaluating the economic impact of building integrated photovoltaics (BIPV). The National Institute of Standards and Technology (NIST) is addressing this deficiency by evaluating computer models used to predict the electrical performance of BIPV components. To facilitate this evaluation, NIST is collecting long-term BIPV performance data that can be compared against predicted values. The long-term data, in addition, provides insight into the relative merits of different building integrated applications, helps to identify performance differences between cell technologies, and reveals seasonal variations. This paper adds to the slowly growing database of long-term performance data on BIPV components. Results from monitoring eight different building-integrated panels over a 12-month period are summarized. The panels are installed vertically, face true south, and are an integral part of the building’s shell. The eight panels comprise the second set of panels evaluated at the NIST test facility. Cell technologies evaluated as part of this second round of testing include single-crystalline silicon, polycrystalline silicon, and two thin film materials: tandem-junction amorphous silicon
(2-a-Si)
and copper-indium-diselenide (CIS). Two
2-a-Si
panels and two CIS panels were monitored. For each pair of BIPV panels, one was insulated on its back side while the back side of the second panel was open to the indoor conditioned space. The panel with the back side thermal insulation experienced higher midday operating temperatures. The higher operating temperatures caused a greater dip in maximum power voltage. The maximum power current increased slightly for the
2-a-Si
panel but remained virtually unchanged for the CIS panel. Three of the remaining four test specimens were custom-made panels having the same polycrystalline solar cells but different glazings. Two different polymer materials were tested along with 6 mm-thick, low-iron float glass. The two panels having the much thinner polymer front covers consistently outperformed the panel having the glass front. When compared on an annual basis, the energy production of each polymer-front panel was 8.5% higher than the glass-front panel. Comparison of panels of the same cell technology and comparisons between panels of different cell technologies are made on daily, monthly, and annual bases. Efficiency based on coverage area, which excludes the panel’s inactive border, is used for most “between” panel comparisons. Annual coverage-area conversion efficiencies for the vertically-installed BIPV panels range from a low of 4.6% for the
2-a-Si
panels to a high of 12.2% for the two polycrystalline panels having the polymer front covers. The insulated single crystalline panel only slightly outperformed the insulated CIS panel, 10.1% versus 9.7%. |
Interior Angle Formula (Definition, Examples, & Video) // Tutors.com
Interior Angle Formula (Definition, Examples, Sum of Interior Angles)
Video Definition Sum of Interior Angles Finding Unknown Angles Regular Polygons
If you take a look at other geometry lessons on this helpful site, you will see that we have been careful to mention interior angles, not just angles, when discussing polygons. Every polygon has interior angles and exterior angles, but the interior angles are where all the interesting action is.
Identify interior angles of polygons
Recall and apply the formula to find the sum of the interior angles of a polygon
Recall a method for finding an unknown interior angle of a polygon
Calculate interior angles of polygons
Discover the number of sides of a polygon
From the simplest polygon, a triangle, to the infinitely complex polygon with
n
sides, sides of polygons close in a space. Every intersection of sides creates a vertex, and that vertex has an interior and exterior angle. Interior angles of polygons are within the polygon.
Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. Instead, you can use a formula that mathematically describes an interesting pattern about polygons and their interior angles.
Sum of Interior Angles Formula
This formula allows you to mathematically divide any polygon into its minimum number of triangles. Since every triangle has interior angles measuring
180°
, multiplying the number of dividing triangles times
180°
gives you the sum of the interior angles.
S = \left(n - 2\right) × 180°
S = sum of interior angles
n = number of sides of the polygon
Try the formula on a triangle:
S = \left(n - 2\right) × 180°
S = \left(3 - 2\right) × 180°
S = 1 × 180°
S = 180°
Well, that worked, but what about a more complicated shape, like a dodecagon?
[insert dodecagon drawing]
It has 12 sides, so:
S = \left(n - 2\right) × 180°
S = \left(12 - 2\right) × 180°
S = 10 × 180°
S = 1,800°
How do you know that is correct? Take any dodecagon and pick one vertex. Connect every other vertex to that one with a straightedge, dividing the space into 10 triangles. Ten triangles, each
180°
, makes a total of
1,800°
Finding an Unknown Interior Angle
The same formula,
S = \left(n - 2\right) × 180°
, can help you find a missing interior angle of a polygon. Here is a wacky pentagon, with no two sides equal:
[insert drawing of pentagon with four interior angles labeled and measuring 105°, 115°, 109°, 111°; length of sides immaterial]
The formula tells us that a pentagon, no matter its shape, must have interior angles adding to
540°
S = \left(n - 2\right) × 180°
S = \left(5 - 2\right) × 180°
S = 3 × 180°
S = 540°
So subtracting the four known angles from
540°
will leave you with the missing angle:
540° - 105° - 115° - 109° - 111° = 100°
The unknown angle is
100°
Finding Interior Angles of Regular Polygons
Once you know how to find the sum of interior angles of a polygon, finding one interior angle for any regular polygon is just a matter of dividing.
S
= the sum of the interior angles and
n
= the number of congruent sides of a regular polygon, the formula is:
\frac{S}{n}
Here is an octagon (eight sides, eight interior angles). First, use the formula for finding the sum of interior angles:
S = \left(n - 2\right) × 180°
S = \left(8 - 2\right) × 180°
S = 6 × 180°
S = 1,080°
Next, divide that sum by the number of sides:
measure of each interior angle
= \frac{S}{n}
= \frac{1,080°}{8}
measure of each interior angle =
= 135°
Each interior angle of a regular octagon is
= 135°
Finding the Number of Sides of a Polygon
You can use the same formula,
S = \left(n - 2\right) × 180°
, to find out how many sides
n
a polygon has, if you know the value of
S
, the sum of interior angles.
You know the sum of interior angles is
900°
, but you have no idea what the shape is. Use what you know in the formula to find what you do not know:
State the formula:
S = \left(n - 2\right) × 180°
Use what you know,
S = 900°
900° = \left(n - 2\right) × 180°
180°
\frac{900°}{180°} = \frac{\left(\left(n - 2\right) × 180°\right)}{180°}
No need for parentheses now
5 = n - 2
5 + 2 = n - 2 + 2
7 = n
The unknown shape was a heptagon!
Now you are able to identify interior angles of polygons, and you can recall and apply the formula,
S = \left(n - 2\right) × 180°
, to find the sum of the interior angles of a polygon. You also are able to recall a method for finding an unknown interior angle of a polygon, by subtracting the known interior angles from the calculated sum.
Not only all that, but you can also calculate interior angles of polygons using
\frac{S}{n}
, and you can discover the number of sides of a polygon if you know the sum of their interior angles. That is a whole lot of knowledge built up from one formula,
S = \left(n - 2\right) × 180° |
Model contact between two geometries - MATLAB
{f}_{n}
, which is aligned with the z-axis of the contact frame. This force pushes the geometries apart in order to reduce penetration.
The frictional force,
{f}_{f}
, which lies in the contact plane. This force opposes the relative tangential velocities between the geometries.
{f}_{n}=s\left(d\right)\cdot \left(k\cdot d+b\cdot {d}^{\text{'}}\right)
{f}_{n}
is the normal force applied in equal-and-opposite fashion to each contacting geometry.
d
is the penetration depth between two contacting geometries.
{d}^{\text{'}}
is the first time derivative of the penetration depth.
k
is the normal-force stiffness specified in the block.
b
is the normal-force damping specified in the block.
s\left(d\right)
is the smoothing function.
|{f}_{f}|=\mu \cdot |{f}_{n}|
{f}_{f}
is the frictional force.
{f}_{n}
\mu
is the effective coefficient of friction.
The effective coefficient of friction is a function of the values of the Coefficient of Static Friction, Coefficient of Dynamic Friction, and Critical Velocity parameters, and the magnitude of the relative tangential velocity. At high relative velocities, the value of the effective coefficient of friction is close to that of the coefficient of dynamic friction. At the critical velocity, the effective coefficient of friction achieves a maximum value that is equal to the coefficient of static friction. The graph shows the basic relationship in the typical case where
{\mu }_{static}
{\mu }_{dynamic}
. In this case, the model is able to approximate stiction with a higher effective coefficient of friction near small tangential velocities. |
Dear experts I want to make potash alum from scrap aluminium foil Please look at the following procedure that - Chemistry - Solutions - 11526885 | Meritnation.com
Dear experts I want to make potash alum from scrap aluminium foil. Please look at the following procedure that i have found on the internet and answer the questions that follows . ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?Procedure
? Clean a small piece of scrap aluminium with steel?
wool and cut it into very small pieces. Aluminium?
foil may be taken instead of scrap aluminium.
? Put the small pieces of scrap aluminium or?
aluminium foil(about 1.00g) into a conical flask and?
add about 50 ml of 4 M KOH solution to dissolve?
the aluminium.?
? The flask may be heated gently in order to facilitate?
dissolution. Since during this step hydrogen gas is?
evolved this step must be done in a well ventilated?
area.?
? Continue heating until all of the aluminium reacts.
? Filter the solution to remove any insoluble?
impurities and reduce the volume to about 25 ml by?
heating. ? ? ? ? ? Allow the filtrate to cool. Now add slowly 6 M?
H2SO4 until insoluble Al(OH)3 just forms in the?
? Gently heat the mixture until the Al(OH)3?
precipitate dissolves.?
? Cool the resulting solution in an ice-bath for about?
30 minutes whereby alum crystals should separate?
out. For better results the solution may be left?
overnight for crystallization to continue.?
? In case crystals do not form the solution may be?
further concentrated and cooled again.
? Filter the crystals from the solution using vacuum
pump, wash the crystals with 50/50 ethanol-water?
mixture. ? Continue applying the vacuum until the crystals?
appear dry.
? Determine the mass of alum crystals. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?#Respected experts please tell me 1)how shall i make 50 ml of 4M KOH solution ? If i add 11.22 gram of KOH pellets in 50ml of water will it do ? If not please elaborate each step clearly on how i can make it . ? ? ? 2) after filtering the solution of aluminium and koh how am i supposed to reduce it to?25.ml?... am i supposed to reduce it by heating ? ?3) how shall i make 6 M of h2so4 solution .... the procedure doesn't provide me with the amount of solution to be prepared in milli literes . Please tell me how i can make it . ? ? ? Last time i followed this procedure my crystals were not formed . I have to make this investigatory project as soon as possible so please help . My teacher said the koh pellets are very expensive , i have wasted them once and i need them to be perfect this time . Also how long am i supposed to keep the solution fir it to make the crystals ?
\mathrm{Use} \mathrm{the} \mathrm{formula} \mathrm{of} \mathrm{Molarity} \mathrm{to} \mathrm{prepare} 50 \mathrm{ml} \mathrm{of} 4\mathrm{M} \mathrm{KOH}\phantom{\rule{0ex}{0ex}}\mathrm{We} \mathrm{know} \mathrm{that},\phantom{\rule{0ex}{0ex}}\mathrm{Molarity}=\frac{\mathrm{mass}}{\mathrm{molar} \mathrm{mass}}×\frac{1000}{\mathrm{Volume} \mathrm{of} \mathrm{solution}}\phantom{\rule{0ex}{0ex}}\mathrm{or}, 4=\frac{\mathrm{mass}}{56}×\frac{1000}{50}\phantom{\rule{0ex}{0ex}}\mathrm{or}, \mathrm{mass} \mathrm{of} \mathrm{KOH}=\frac{4×56×50}{1000}=11.2 \mathrm{g}\phantom{\rule{0ex}{0ex}}\mathrm{So}, 11.2 \mathrm{g} \mathrm{of} \mathrm{KOH} \mathrm{must} \mathrm{be} \mathrm{taken} \mathrm{to} \mathrm{prepare} 50 \mathrm{ml} \mathrm{of} 4\mathrm{M} \mathrm{KOH}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now}, \mathrm{further} \mathrm{you} \mathrm{can} \mathrm{reduce} \mathrm{the} \mathrm{filtered} \mathrm{solution} \mathrm{to} 25 \mathrm{ml} \mathrm{by} \mathrm{heating} \mathrm{the} \mathrm{solution}. \mathrm{Heat} \mathrm{the} \mathrm{solution} \mathrm{till} 25 \mathrm{ml}. \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}6 \mathrm{M} {\mathrm{H}}_{2}{\mathrm{SO}}_{4} \mathrm{can} \mathrm{also} \mathrm{be} \mathrm{prepared} \mathrm{by} \mathrm{the} \mathrm{above} \mathrm{method}. \mathrm{You} \mathrm{just} \mathrm{need} \mathrm{to} \mathrm{know} \mathrm{the} \mathrm{amount} \mathrm{of} \mathrm{solution} \mathrm{you} \mathrm{need} \mathrm{to} \mathrm{make}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{You} \mathrm{can} \mathrm{keep} \mathrm{the} \mathrm{solution} \mathrm{overnight} \mathrm{if} \mathrm{crystals} \mathrm{are} \mathrm{not} \mathrm{formed}. \mathrm{You} \mathrm{can} \mathrm{also} \mathrm{concentrate} \mathrm{the} \mathrm{solution} \mathrm{more} \mathrm{by} \mathrm{heating} \mathrm{and} \mathrm{cool} \mathrm{it} \mathrm{if} \mathrm{crystal} \mathrm{doesn}\text{'}\mathrm{t} \mathrm{appear}. |
A box contains 11 marbles, 7 red, and 4 green.
fumefluosault7pa 2022-02-12 Answered
Finn Lucero
Red marbles
=7
Green marbles
=4
Total no of marbles
=11
Number of Marbles removed
=5
Marbles in the box
=11-5=6
Probability of drawing one green marble is
=0.5
Among the remaining 6 marbles, x number of marbles are green.
Then Probability of drawing one green marble is = number of green marbles / Total Number of marbles.
{P}_{1\text{ }\text{green}}=\frac{x}{6}
{P}_{1\text{ }\text{Green}}=0.5
\frac{x}{6}=0.5
x=0.5×6=3
After 5 marbles are removed from the box, the number of green marbles remaining is 3.
It means out of 4 green marbles, one green marble has been removed.
Out of 5 marbles one marble is green. Then the other four must be red.
Then number of red marbles removed were
=4
An urn contains 3 one-dollar bills, 1 five-dollar bill and 1 ten-dollar bill. A player draws bills one at a time without replacement from the urn until a ten-dollar bill is drawn. Then the game stops. All bills are kept by the player. Determine the probability of the game stopping at the second draw.
If volume is high this week, then next week it will be high with a probability of 0.9 and low with a probability of 0.1.
If volume is low this week then it will be high next week with a probability of 0.5.
Assume that state 1 is high volume and that state 2 is low volume.
If the volume this week is high, what is the probability that the volume will be high two weeks from now?
Toss 5 coins 20 times simultaneously and observe the number of heads that will occur. The possible values of X or the number of heads in 5 coins tossed are 0, 1, 2, 3, 4, and 5. |
Area of a Trapezium | Brilliant Math & Science Wiki
Andy Hayes and Ashish Menon contributed
A trapezium, also known as a trapezoid, is a quadrilateral in which a pair of sides are parallel, but the other pair of opposite sides are non-parallel. The area of a trapezium is computed with the following formula:
\text{Area}=\frac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.
The parallel sides are called the bases of the trapezium. Let
b_1
b_2
be the lengths of these bases. The distance between the bases is called the height of the trapezium. Let
h
be this height. Then this formula becomes:
\text{Area}=\frac{1}{2}(b_1+b_2)h
Given a trapezium, let
b_1
b_2
be the lengths of the bases, and let
h
be the height. Draw a segment parallel to the bases that is halfway between the bases. This divides the trapezium into two trapeziums, each with the same height of
\frac{1}{2}h.
Labeling the angles of these trapeziums:
Note the following congruences and identities due to the fact that the bases are parallel:
\begin{aligned} m \angle 4 + m \angle 5 &= 180^\circ \\ m \angle 1 + m \angle 7 &= 180^\circ \\ \angle 2 &\cong \angle 6 \\ \angle 3 &\cong \angle 8 \end{aligned}
Now rotate the top trapezoid and place it adjacent to the bottom trapezoid in the following way:
Due to the aforementioned congruences and identities, this shape is a parallelogram. The length of its base is
(b_1+b_2),
and its height is
\frac{1}{2}h.
This parallelogram has the same area as the trapezoid, so the area of the trapezoid is
\text{Area}=\frac{1}{2}(b_1+b_2)h.\ _\square
Consider a trapezium
ABCD
AB \parallel CD
AB=10\text{ cm}
CD=\text{ cm}
and they are separated by a distance of
4\text{ cm}
ABCD
We know that the area of a trapezium =
\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}
\dfrac {1}{2}×(10+5)×4
\dfrac {1}{2}×15×4
30\text{ cm}^{2}
_\square
Find the area of a trapezium with parallel lines of
9\text{ cm}
7\text{ cm}
, and a height of
3\text{ cm}
\dfrac {1}{2}×(9 \text{ cm} + 7 \text{ cm}) × 3 \text{ cm}
\dfrac {1}{2} ×(16 \text{ cm})×(3 \text{ cm})
\dfrac {1}{2}×48 \text{ cm}^2
24 \text{ cm}^2
_\square
Cite as: Area of a Trapezium. Brilliant.org. Retrieved from https://brilliant.org/wiki/area-of-a-trapezium/ |
Waves/Total Internal Reflection - Wikibooks, open books for an open world
Waves/Total Internal Reflection
Waves : Geometrical Optics
Total Internal Reflection[edit | edit source]
When light passes from a medium of lesser index of refraction to one with greater index of refraction, Snell's law indicates that the ray bends toward the normal to the interface. The reverse occurs when the passage is in the other direction. In this latter circumstance a special situation arises when Snell's law predicts a value for the sine of the refracted angle greater than one. This is physically untenable. What actually happens is that the incident wave is reflected from the interface. This phenomenon is called total internal reflection. The minimum incident angle for which total internal reflection occurs is obtained by substituting
{\displaystyle \theta _{R}=\pi /2}
into equation (3.2), resulting in
{\displaystyle \sin(\theta _{I})=n_{R}/n_{I}\quad {\mbox{(total internal reflection)}}.}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Waves/Total_Internal_Reflection&oldid=3253249" |
Remote Sensing-Based Accounting of Reservoir’s Water Storage for Water Scarcity Mitigation: A Case Study for Small and Medium Irrigation Dams in Vietnam
Hung X. Dinh*, Thanh T. Hoang, Lan T. Ha, Tuan V. Nguyen, Thanh C. Pham, Minh C. Nguyen, Hiep T. Luong
Institute of Water Resources Planning, Hanoi, Vietnam.
Abstract: Integrated water resources management requires consistent and accurate data on available water storage in reservoirs as well as water stress level. Vietnam is enduring a significant deficit in collecting necessary information to manage its water resources in that manner. While reservoirs are abundant, the majority of them were constructed a long ago and often lack of regular and adequate measurement on storage volume. Furthermore, the condition of water stress is often missing or remains bias leading to certain risks in reservoir operation, e.g. during water scarcity period. This paper presents how remote sensing data can be used to acquire needed information that is fundamental to understand water resources conditions. The results indicated that Sentinel-1 and Moderate Resolution Imaging Spectroradiometer (MODIS) can be applied to determine water surface area and water stress, through vegetation health index (VHI). This information is deemed necessary to improve water resources monitoring and management and hence, ensure long-term drought resilience and water and food security.
Keywords: Remote Sensing, Water Scarcity Management, Reservoir Volume
Reservoirs play an important role in provision of water during dry season and regulation of excessive flows during wet seasons (Donchyts et al., 2016). Monitoring of reservoir storage provides water managers with the ability to enhance water supply for irrigation, fisheries, hydropower, eco-tourism besides reduced risks from water-related disasters. On the other hand, the deficit of knowledge on water storage will lead to reduction in water supply capacity and potentially contribute to water scarcity and pollution. Furthermore, increased inflow into lakes might lead to overflow and flooding in the downstream area, hence impacting livelihoods (Vanthof & Kelly, 2019; Guo et al., 2017). Additional surface water storage capacity enhances the flexibility of a basin to retain and release water.
While Vietnam has more than 7800 reservoirs for water supply and electricity generation, it is struggling to ensure adequate reservoir monitoring to satisfy the increasing of water demands. Hence, a continuous monitoring and data collection of water storage in lakes are fundamental for water use planning. Nevertheless, measuring water storage in lakes, reservoirs and wetlands is still a challenge with basic information regarding geometry. Volume-level curve is missing in most of small and medium size reservoirs. While installing sensors is still costly for dam managers, dam monitoring using remote sensing (Duan & Bastiaanssen, 2017) or hydrological model (Ha et al., 2018) proves to be an effective tool.
To account for water availability for understanding the impacts of reservoir on water scarcity mitigation, reservoir’s storage must be monitored systematically. While water level can be observed using water level scale, measurement of storage is more challenging as this information is often missing, especially for small and old dams. Li and Sheng (2012) used Landsat data and SRTM and ASTER DEM from SRTM to monitor storage variations in reservoirs. Nevertheless, SRTM and ASTER are only available for bathymetry measured before 2001-2008 (ASTER) and 200 (SRTM), hence are valid only for reservoirs constructed after that period. Cai et al. (2016) used Moderate Resolution Imaging Spectroradiometer (MODIS) to monitor water storage of large lakes and reservoirs in Yangtze River Basin for a consecutive period of 15 years from 2000-2014. Under this study, 230 lakes and reservoirs were monitored. Due to the coarse resolution of MODIS (250 m), the method proved to be successful to assess trend in huge bundle of reservoirs larger than 8 km2 in surface area.
Many satellites remote sensing surface water mapping studies and applications focus on the use of optical sensors, such as Landsat (Crétaux et al., 2015), Sentinel (Markert et al., 2020), the Moderate Resolution Imaging Spectroradiometer (MODIS) (Vanthof & Kelly, 2019), Wardlow and Egbert, 2010, Pervez et al., 2014 and the Visible Infrared Imaging Radiometer Suite (VIIRS) (Usman et al., 2015; Biggs et al., 2006). These optical water mapping methods include spectral information and thresholds (Sruthi & Aslam, 2015). With areas heavily affected by cloud, Synthetic aperture radar (SAR) data have been considered a promising tool to monitor water dynamics, because they are cloud-proof. With the recent launch of the Sentinel-1 satellite and the free access to its products, SAR data have both high temporal and spatial resolution, which can be used to extract surface water extent and assess its dynamics more efficiently. In addition, radar altimeter satellites are cloud-proof and can also be used to monitor surface water area.
In addition to meteor-hydrological data, mapping of water scarcity areas also required information. Traditional assessment involves locally collected data at field level and from census sources. Determination of vegetation growth and level of water scarcity derived from satellite for basin scale cultivation is deemed a modern and appropriate method of mapping drought and assessing reservoir irrigation efficiency. The Moderate Resolution Imaging Spectroradiometer (MODIS) presents a considerably accurate and operational dataset to classify irrigated area at reasonable resolution. Usman et al. (2015) applied MODIS NDVI to detect temporal changes for irrigated cropland in Pakistan. Biggs et al. utilized MODIS to map land cover and irrigated pixels for small scale farming in India using MODIS and Landsat time series. Sruthi and Aslam, 2015 published drought maps for Raichur District, India using similar approaches. Luong et al. (2020) mapped irrigated paddy rice and drought using MODIS time series. This paper presents how remote sensing data can be used to 1) derive information of reservoir storage for irrigation dams and 2) combine remote sensing based-drought information for operation of dams and reservoirs.
The Central Vietnam is a large area with complex terrain and extensive river network. Numerous reservoirs were constructed to improve water supply for agriculture production and other users such as industry, domestic etc. Ninh Thuan province (Figure 1) was selected as a case study due to having a large amount of small and medium size dams that need water storage accounting. Ninh Thuan is also prone to regular drought.
Figure 1. Four reservoirs and their locations under this study.
Sentinel-1 data was used in this study to extract information on water storage. The Sentinel-1 sensor is a C-band SAR that operates in multiple acquisition modes at different ground sampling distances (GSD). For this study we used the Sentinel-1 Interferometric Wide (IW) swath mode at the 10 m GSD, which offers single and dual polarization options of vertical transmitting with vertical receiving (VV) and vertical transmitting with horizontal receiving (VH) dual-polarization. VV polarization data was used in this study. Two pre-processed versions of the ESA Copernicus Open Access Hub Sentinel 1 Level-1 IW Ground Range Detected (GRD) dataset were used for this study. This first version used, provided through GEE, was the Sentinel-1 Level 1 GRD ARD derived from the ESA data on Copernicus.
Information on drought was derived using Vegetation Health Index (VHI) from MODIS, which can be used to monitor the growth of vegetation cover and water stress. VHI value reflects the concentration of green biome in vegetation leaf and hence, indicates vegetation health and dynamics. Regional and global spatially determination of irrigated lands conducted earlier using MODIS dataset Luong et al. (2020). In this study, VHI dataset was calculated from MODIS Global MOD13C2. MOD13C2 is cloud-free spatial composites of the gridded monthly 1-km geographic (lat/lon) Climate Modeling Grid (CMG). Cloud-free global coverage is achieved by replacing clouds with the historical MODIS time series climatology record.
2.3. Calculation of Reservoir Volume
Detailed calculation step to derived reservoir storage is illustrated in Figure 2. Remote sensing data from Sentinel-1 was processed in Google Earth Engine to extract water surface area (F). Water area was subsequently combines with water level (Z) monitored at the reservoirs to estimate water volume. This information together provides knowledge on reservoir remaining storage capacity.
Reservoir volume calculation is estimated using following equation:
Figure 2. Step-wise calculation of reservoir storage volume.
{W}_{t+1}=\frac{\left({F}_{i}+{F}_{i+1}+\sqrt{{F}_{i}\cdot {F}_{i+1}}\right)\cdot \left({Z}_{i+1}-{Z}_{i}\right)}{3}
{W}_{i+1},{W}_{i}
: Reservoir storage at the time step i and i + 1
{F}_{i+1},{F}_{i}
: Reservoir surface area the time step i and i + 1 calculated from Sentinel-1
{Z}_{i+1},{Z}_{i}
: Reservoir water level at the time step i and i + 1 measured at the reservoir
The bathymetry for seven reservoirs was provided in Figure 3 as immersion frequency (%). DEM data from ALOS and SRTM was shown as comparison of the result. It is shown that reservoir volume can be estimated using Sentinel-1 with good agreement.
Figure 4 indicates VHI calculated for the year 2020, which is a drought year. The value was normalized from 0 - 100 to reflect better the water stress level with 100 is the highest level while 0 describes no water stress at all. The VHI time series for 2020 show two distinguished periods following a wet-dry climatic condition. The period from March until May is the most critical period of water scarcity while the remaining months enjoy a more relaxed water resources condition.
Figure 3. Reservoirs storage and bathymetry derrived from remote sensing and DEM for Cho Mo, Song Sat, Song Trau and Lanh Gia reservoir.
Figure 4. VHI calculated for Ninh Thuan province during the drought year of 2020.
This study presents a comprehensive and step-wise calculation of reservoir storage and implication to water scarcity management in Central Vietnam. It was demonstrated that remote sensing-derived reservoir information and water stress level has a good potent and agreement with theory and field observation. The data on reservoir storage offers exciting insights for reservoir operator and water users to have continuous information on available water for their use and therefore, promote more effective water resource management strategies in the study area.
While this study shows specific results from satellite data, the role of in-situ measurement should not be underestimated. The new generation of elevation data available through altimetry sensors will provide a medium to substitute field measurement. Furthermore, is should be noted that all analyses strongly rely on the quality of the satellite data and algorithm to extract necessary information map.
This research was funded by Ministry of Agriculture and Rural Development (MARD) through the research “Study on the application of remote sensing technology to account and monitor of water resources in irrigation and hydropower dams for agriculture production in drought-affected provinces in South Centre and Central Highlands” conducted during 2019-2021 period.
Cite this paper: Dinh, H. , Hoang, T. , Ha, L. , Nguyen, T. , Pham, T. , Nguyen, M. and Luong, H. (2021) Remote Sensing-Based Accounting of Reservoir’s Water Storage for Water Scarcity Mitigation: A Case Study for Small and Medium Irrigation Dams in Vietnam. Journal of Geoscience and Environment Protection, 9, 89-97. doi: 10.4236/gep.2021.911007.
[1] Biggs, T. W., Thenkabail, P. S., Gumma, M. K., Scott, C. A., Parthasaradhi, G. R., & Turral, H. N. (2006). Irrigated Area Mapping in Heterogeneous Landscapes with MODIS Time Series, Ground Truth and Census Data, Krishna Basin, India. International Journal of Remote Sensing, 27, 4245-4266.
[2] Cai, X., Feng, L., Hou, X. et al. (2016). Remote Sensing of the Water Storage Dynamics of Large Lakes and Reservoirs in the Yangtze River Basin from 2000 to 2014. Scientific Reports, 6, Article No. 36405.
[3] Crétaux, J.-F., Biancamaria, S., Arsen, A., Bergé-Nguyen, M., & Becker, M. (2015). Global Surveys of Reservoirs and Lakes from Satellites and Regional Application to the Syrdarya River Basin. Environmental Research Letters, 10, Article ID: 015002.
[4] Donchyts, G., Baart, F., Winsemius, H. et al. (2016). Earth’s Surface Water Change over the Past 30 Years. Nature Climate Change, 6, 810-813.
[5] Duan, Z., & Bastiaanssen, W. (2017). Evaluation of Three Energy Balance-Based Evaporation Models for Estimating Monthly Evaporation for Five Lakes Using Derived Heat Storage Changes from a Hysteresis Model. Environmental Research Letters, 12, Article ID: 024005.
[6] Guo, M., Li, J., Sheng, C., Xu, J., & Wu, L. (2017). A Review of Wetland Remote Sensing. Sensors (Basel, Switzerland), 17, 777.
[7] Ha, L. T., Bastiaanssen, W. G. M., Van Griensven, A., Van Dijk, A. I. J. M., & Senay, G. B. (2018). Calibration of Spatially Distributed Hydrological Processes and Model Parameters in SWAT Using Remote Sensing Data and an Auto-Calibration Procedure: A Case Study in a Vietnamese River Basin. Water, 10, 212.
[8] Li, J. L., & Sheng, Y. W. (2012). An Automated Scheme for Glacial Lake Dynamics Mapping Using Landsat Imagery and Digital Elevation Models: A Case Study in the Himalayas. International Journal of Remote Sensing, 33, 5194-5213.
[9] Luong, C., Ha, L., Pham, T., Dinh, H., Hoang, T., Nguyen, H., Nguyen, T., & Nguyen, P. (2020). Spatio-Temporal Determination of Irrigated Paddy Rice Pixels Using Evapotranspiration and Vegetation Indices: A Case Study for Ca River Basin in Vietnam. Journal of Geoscience and Environment Protection, 8, 94-107.
[10] Markert, K. N., Markert, A. M., Mayer, T., Nauman, C., Haag, A., Poortinga, A., Bhandari, B., Thwal, N. S., Kunlamai, T., Chishtie, F., Kwant, M., Phongsapan, K., Clinton, N., Towashiraporn, P., & Saah, D. (2020). Comparing Sentinel-1 Surface Water Mapping Algorithms and Radiometric Terrain Correction Processing in Southeast Asia Utilizing Google Earth Engine. Remote Sensing, 12, 2469.
[11] Pervez, M. S., Budde, M., & Rowland, J. (2014). Mapping Irrigated Areas in Afghanistan over the Past Decade Using MODIS NDVI. Remote Sensing of Environment, 149, 155-165.
[12] Sruthi, S., & Aslam, M. A. (2015). Agricultural Drought Analysis Using the NDVI and Land Surface Temperature Data; a Case Study of Raichur District. Aquatic Procedia, 4, 1258-1264.
[13] Usman, M., Liedl, R., Shahid, M. A. et al. (2015). Land Use/Land Cover Classification and Its Change Detection Using Multi-Temporal MODIS NDVI Data. Journal of Geographical Sciences, 25, 1479-1506.
[14] Vanthof, V., & Kelly, R. (2019). Water Storage Estimation in Ungauged Small Reservoirs with the TanDEM-X DEM and Multi-Source Satellite Observations. Remote Sensing of Environment, 235, Article ID: 111437.
[15] Wardlow, B. D., & Egbert, S. L. (2010). A Comparison of MODIS 250-m EVI and NDVI Data for Crop Mapping: A Case Study for Southwest Kansas. International Journal of Remote Sensing, 31, 805-830. |
PLEASE SHOW ME THE FOLLOWING GRAPHS WITH THE SHADED PORTION : 1) x=0 [ x is greater than or equal - Maths - Linear Inequalities - 7762447 | Meritnation.com
Region for x\ge 0\phantom{\rule{0ex}{0ex}}
As we know that x =0 is nothing but y axis and if we are asked for x greater than or equal to 0 then,
it will be right to y- axis and this region can be shown as,
Here the shaded portion is the required portion.
For y
\ge 0
as we know that y =0 is x -axis so y greater than or equal to zero will be a region above x axis and can be given as,
for the region x
\ge 0 \mathrm{and} \mathrm{y}\ge 0
Now this is nothing but intersection of first part and second part and we can see that x and y are both positive in first quadrant and so is the intersection of first part and second part, |
The Laplace transform function for the output voltage of a
The Laplace transform function for the output voltage of a network is expressed
The Laplace transform function for the output voltage of a network is expressed in the following form:
{V}_{0}\left(s\right)=\frac{12\left(s+2\right)}{s\left(s+1\right)\left(s+3\right)\left(s+4\right)}
Determine the final value of this voltage; that is,
{\upsilon }_{0}\left(t\right)
t\to \mathrm{\infty }
We will use the expression for final value of f(t):
f\left(\mathrm{\infty }\right)=\underset{s\to o}{lim}sF\left(s\right)
{\upsilon }_{0}\left(\mathrm{\infty }\right)=\underset{s\to o}{lim}s\left(\frac{12\left(s+2\right)}{s\left(s+1\right)\left(s+3\right)\left(s+4\right)}\right)
=\underset{s\to o}{lim}\left(\frac{12\left(s+2\right)}{s\left(s+1\right)\left(s+3\right)\left(s+4\right)}\right)
=\frac{\left(12\right)\left(2\right)}{\left(1\right)\left(3\right)\left(4\right)}
=2
f\left(x\right)=\frac{1}{1}+x+{x}^{2}
. find the maclaurin polynomial
\frac{dS}{dr}=kS
Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming.(Write your answer as a function of s.)
L\left\{{e}^{-t}\cdot {e}^{t}\mathrm{cos}\left(t\right)\right\}
Find the inverse Laplace transform of the given function by using the convolution theorem.
F\left(s\right)=\frac{s}{\left(s+1\right)\left({s}^{2}+4\right)}
f\left(t\right)={e}^{3t}\mathrm{cos}5t-{e}^{-t}\mathrm{sin}2t
\frac{d}{dt}u=f\left(t,x\right)\text{ }\text{and}\text{ }\frac{d}{dt}x=g\left(t,u\right)
u\left(t+h\right)=u\left(t\right)+hf\left(t,x\left(t\right)\right)
x\left(t+h\right)=x\left(t\right)+hg\left(t,u\left(t\right)\right) |
Andy Hayes, Muhammad Rasel Parvej, Hua Zhi Vee, and
Discrete mathematics is the study of mathematical structures that are countable or otherwise distinct and separable. Examples of structures that are discrete are combinations, graphs, and logical statements. Discrete structures can be finite or infinite. Discrete mathematics is in contrast to continuous mathematics, which deals with structures which can range in value over the real numbers, or have some non-separable quality.
Since the time of Isaac Newton and until quite recently, almost the entire emphasis of applied mathematics has been on continuously varying processes, modeled by the mathematical continuum and using methods derived from the differential and integral calculus. In contrast, discrete mathematics concerns itself mainly with finite collections of discrete objects. With the growth of digital devices, especially computers, discrete mathematics has become more and more important.
Discrete structures can be counted, arranged, placed into sets, and put into ratios with one another. Although discrete mathematics is a wide and varied field, there are certain rules that carry over into many topics. The concept of independent events and the rules of product, sum, and PIE are shared among combinatorics, set theory, and probability. In addition, De Morgan's laws are applicable in many fields of discrete mathematics.
Often, what makes discrete mathematics problems interesting and challenging are the restrictions that are placed on them. Although the field of discrete mathematics has many elegant formulas to apply, it is rare that a practical problem will fit perfectly to a specific formula. Part of the joy of discovering discrete mathematics is to learn many different approaches to problem-solving, and then be able to creatively apply disparate strategies towards a solution.
Combinatorics is the mathematics of counting and arranging. Of course, most people know how to count, but combinatorics applies mathematical operations to count things that are much too large to be counted the conventional way.
At a local deli, the following options are given for a sandwich:
Bread types: White, Rye, Wheat
Cheese Types: Swiss, Cheddar, Havarti, Provolone
Meat Types: Roast Beef, Turkey, Ham, Corned Beef, Pulled Pork
If a customer chooses exactly one of each type of item, then how many possible sandwiches can be made?
A more specific type of arrangement is a permutation. A permutation is an arrangement of objects with regard to order.
At the start of a horse race, there are 12 distinct horses in the field. 3 horses can place at the end of the race, and it matters what order the horses placed in. For example, if horses
\text{A},
\text{B},
\text{C}
placed, then it would matter which horse came in
1^\text{st}, 2^\text{nd},
3^\text{rd}.
\text{ABC}
would be different than
\text{ACB}.
How many possible fields of placed horses are there?
A combination (not to be confused with combinatorics) is another type of arrangement that is related to permutations. A combination is an arrangement of objects without regard to order.
There are 12 distinct players about to play a pick-up game of football. The two team captains are Brandon and Meredith (included in the 12). They will each take turns selecting a player to be on their team until all players are selected.
How many ways can the players be divided into teams?
As a field of mathematics, combinatorics is nearly as broad as discrete mathematics. Other topics within combinatorics include
derangements: a permutation such that no object is in its original spot in the order;
rectangular grid walks: determining the number of ways a rectangular lattice can be traversed;
distribution of objects into bins: determining how objects can be grouped into bins.
Set theory is the branch of mathematics that is concerned about collections of objects. Sets can be discrete or continuous; discrete mathematics is primarily concerned with the former. At a basic level, set theory is concerned with how sets can be arranged, combined, and counted.
The cardinality of a finite set is the number of elements in that set. For a given set
A,
its cardinality is denoted by
|A|.
What is the cardinality of the set of prime numbers less than 25?
The set of prime numbers less than 25 is
\{2,3,5,7,11,13,17,19,23\}.
There are 9 elements in this set, so the cardinality is 9.
_\square
Cardinality can also be extended to infinite sets. Although this kind of cardinality cannot be counted, each cardinality can be compared with another cardinality.
A
B
be sets. Their cardinalities are compared as follows:
If there exists a bijection between
A
B,
|A|=|B|.
If there exists an injective function from
A
B,
but no bijective function, then
|A|<|B|.
Show that the set of integers and the set of even integers have the same cardinality.
It might seem strange that these sets have the same cardinality. After all, the even integers are more "rare." However, these sets are both infinite. Therefore, the "common sense" thinking about finite sets must be discarded. Instead, the goal is to obtain a bijective function from the set of integers to the set of even integers:
f(n)=2n, \ n \in \mathbb{Z}.
The function above gives a one-to-one correspondence between each integer
n
and each even integer
2n.
Since the bijection is established, the set of integers and the set of even integers have the same cardinality.
_\square
A complement of a set
A
A.
The study of set complements gives a number of efficient methods to calculate cardinalities of finite sets. For example, one can efficiently obtain the cardinality of a set that contains "at least one" element of another set.
David is the leader of the David Committee. He wants to appoint 3 people to be on the Head Council. He has to choose from 9 applicants, three of whom are Tommy, Jack, and Michael. In how many ways can he choose the people to be on the Council, so that at least one of Tommy, Jack, and Michael is chosen?
The union and intersection give ways to describe how sets can be combined.
A positive integer less than 1000 is not only a perfect square but also a perfect cube.
De Morgan's laws give identities for the complements of unions and intersections.
The principle of inclusion and exclusion, or PIE, gives a method to find the union or intersection of more than two sets.
How many integers from
1
10^6
(inclusive) are neither perfect squares nor perfect cubes nor perfect fourth powers?
Graphs are useful for representing all kinds of real-world problems.
P
\big\lfloor 10^{10} P \big\rfloor?
\frac{1}{2}
A probability is a number, between 0 and 1 inclusive, that represents the likelihood of an event. Discrete probability is a probability based on discrete sets of outcomes. The most basic type of probability is a uniform probability. If each outcome in a set is equally likely, then the probability of an event is equal to a ratio of cardinalities.
S
be a sample space of outcomes. If each outcome in this set is equally likely, then the probability of an event
A
S
P(A)=\frac{|A|}{|S|}.
Many of the rules of probability are analogous to the rules of combinatorics. The probabilistic rules of product, sum, and complement work similarly to those same rules from combinatorics. In addition, the structure of the probabilistic principle of inclusion and exclusion is the same as PIE for sets.
\dfrac{121}{243}
\dfrac{122}{243}
\dfrac{124}{243}
\dfrac{125}{243}
\frac{2}{3}
A discrete probability distribution is a function that takes a numerical outcome as an argument and gives a probability as a result. Discrete probability distributions can be created using the rules and guidelines described above. There are also some discrete probability distributions that show up in many problems:
Geometric Distribution: Given repeated trials in which the probability of success is the same each time, this gives the probability that the first success will occur on a certain trial. Example: You roll a dice until you roll a 6. What is the probability that the first 6 will occur on the third roll?
Binomial Distribution: Given a certain number of trials in which the probability of success is the same each time, this gives the probability of a certain number of successes. Example: You flip a coin 10 times. What is the probability that there will be exactly 5 heads?
Poisson Distribution: Given a time period in which an event occurs a certain average number of times, this gives the probability that the event will occur a specific number of times. Example: A fast food drive-through gets 3 customers per minute. What's the probability they will get 4 customers in the next minute?
Although basic probabilities are based on discrete sets, the concept of probability can be extended to continuous sets by using concepts from calculus.
A point is marked at random on a unit line segment.
Find the expected value of the sum of the squares of the lengths of the two parts.
A statistic is a number used to describe a set of data or a probability distribution. Statistics is widely used in many fields outside of mathematics, from biology to politics to sports. The power of statistics lies in taking a massive, varied set of data and making sense out of it. Furthermore, statistics has the power to quantify confidence in those findings. Of course, the usefulness of statistics is not without controversy, but an understanding of its theoretical underpinnings can help one avoid its misuse.
One major kind of statistic is a measure of central tendency. A measure of central tendency is a number which describes what a value of a probability distribution or data set will tend to. An expected value is the theoretical long-run average outcome of a probability experiment when it is performed many times.
A game costs $150 to play. In this game, you roll a fair six-sided die repeatedly until each of all the six numbers has been rolled at least once. You are then paid 10 times the number of rolls you made.
For example, if the rolls were 3, 5, 4, 3, 2, 5, 1, 4, 1, 3, 6, then you would get
(10)(11) = 110
Including the price to play, what is your expected value in this game?
Somewhat related to the expected value is the mean. The mean is the average value of a set of numerical data.
What is the mean of the first 100 positive integers?
Another major kind of statistic is a measure of variation. A measure of variation is a number which describes the distribution of a probability distribution or data set. The standard deviation of a probability distribution is a number that represents how much the outcomes differ from the expected value. Likewise, the standard deviation of a data set is a number that represents how much the elements of the set differ from the mean.
\sigma=\frac{d}{6}\sqrt{3(n^2-1)}
\sigma=\frac{d(n-1)}{6}\sqrt 3
\sigma=\frac{d}{3}\sqrt{6(n^2+1)}
The elements of set
S
of real numbers with cardinality
n
d
. Express the population standard deviation of set
S
n
and
An example of set
S
\text{{2, 5, 8, 11, 14, 17}}
. It has cardinality 6, and its elements form an AP with common difference 3.
Although discrete statistics are based off of discrete events and probability distributions, these same concepts can be extended to continuous events and probability distributions using concepts from calculus.
Main Article: Bijection, Injection, and Surjection
A bijection is a relationship between two sets such that each element in a set is paired with exactly one element in the other set, and vice versa. Bijections can be applied to problem solving by establishing a bijection between a set that is difficult to enumerate and a discrete stucture that is well understood. By establishing a bijection, one can take advantage of the known formulas and theorems that the discrete structure affords.
The three Molloy siblings, April, Bradley, and Clark, have integer ages that sum to 15. How many possible distribution of ages are there?
Note: It is possible that an age can be 0, which means that the child was just born.
One can establish a bijection between the set of distributions of ages and a set of combinations. Consider the arrangement of stars and bars below:
\star \star \mid \star \star \star \star \mid \star \star \star \star \star \star \star \star \star
This arrangement corresponds to the following distribution of ages: April - 2, Bradley - 4, Clark - 9. Note that there are 15 stars and 2 bars in the arrangement above. This gives a total of 17 objects, 2 of which are bars. Placing the bars in different spots among the 17 placements will give a new distribution of ages. Thus, a bijection can be established between the set of distributions of ages and the set of combinations of 2 objects out of 17.
The number of distributions of ages is
\binom{17}{2}=136.\ _\square
A parking lot has 10 empty spaces in a row.
6 cars arrive, each of which fills exactly 1 parking spot, chosen at random from among the available spaces. Robbie then arrives in his pick-up truck, which requires 2 empty adjacent spaces to park.
If the probability that Robbie will be able to park is
\frac{a}{b},
and
b
a+b?
A proposition is a statement that can either be true or false. Propositional logic aims to outline the rules of how these statements can be altered and combined.
Which of the following statements are true and which are false, knowing that the entire set is uncontradictory?
S1. Statements 2 and 3 are either both true either both false.
S2. Exactly one of the statements 4 and 5 is true.
S5. Statements 1 and 3 are of the same type (both true or both false).
S6. Exactly one statement from statements 2 and 5 is true.
Write the answer as the concatenation of the digits 1 and 0 for the truth values of the statements (true and false) starting from S1 to S6, where for the value true corresponds 1 and for the value false corresponds 0. For example, if the first 2 statements would be true and the rest false, the answer would be 110000.
If the correct answer begins with some number of leading 0s, remove it from writing the answer. For example, if the answer is 001100, write 1100 anyway.
Similarly, Boolean algebra outlines the operations defined on variables that can take the values of true (1) or false (0). Boolean algebra is used to design computer circuits through logic gates, which take signal(s) as inputs and return a signal as an output.
Cite as: Discrete Mathematics. Brilliant.org. Retrieved from https://brilliant.org/wiki/discrete-mathematics/ |
Let p(x) be a nonzero polynomial of degree less than
Let p(x) be a nonzero polynomial of degree less than 1992 having no nonconstant factor in common with x^3 for polynomials f (x) and g(x).
Let p(x) be a nonzero polynomial of degree less than 1992 having no nonconstant factor in common with
{x}^{3}
for polynomials f (x) and g(x). Find the smallest possible degree of f (x)?
\frac{{d}^{1992}}{{dx}^{1992}}\left(\frac{p\left(x\right)}{{x}^{3}-x}\right)=\frac{f\left(x\right)}{g\left(x\right)}
Given p(x) be a nonzero polynomial of degree less than 1992 having no non constant factor in common with
{x}^{3}-x
\frac{{d}^{1992}}{{dx}^{1992}}\left(\frac{p\left(x\right)}{{x}^{3}-x}\right)=\frac{f\left(x\right)}{g\left(x\right)}
f\left(x\right)
g\left(x\right)
To find the smallest possible degree of
f\left(x\right)
BY division algorithm we know that,
f\left(x\right)
g\left(x\right)
are any two polynomials with
g\left(x\right)\ne 0
f\left(x\right)=g\left(x\right).q\left(x\right)+r\left(x\right)
r\left(x\right)=0
r\left(x\right)
q\left(x\right)
r\left(x\right)
Hence p(x) can be written as
p\left(x\right)=\left({x}^{3}-x\right)q\left(x\right)+r\left(x\right)
Here, we can see that the degree of
r\left(x\right)
is 2 and the degree of
q\left(x\right)
is less than 1989
\frac{{d}^{1992}}{{dx}^{1992}}\left(\frac{p\left(x\right)}{{x}^{3}-x}\right)=\frac{{d}^{1992}}{{dx}^{1992}}\left(\frac{r\left(x\right)}{{x}^{3}-x}\right)
\frac{r\left(x\right)}{{x}^{3}-x}
Using partial fractions, we have
\frac{r\left(x\right)}{{x}^{3}-x}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x+1}
\frac{{d}^{1992}}{{dx}^{1992}}\left(\frac{r\left(x\right)}{{x}^{3}-x}\right)=1992!\left(\frac{A}{{\left(x-1\right)}^{1993}}+\frac{B}{{x}^{1993}}+\frac{C}{{\left(x+1\right)}^{1993}}
=1992!\left(\frac{A{x}^{1993}{\left(x+1\right)}^{1993}+B{\left(x-1\right)}^{1993}{\left(x+1\right)}^{1993}+C{\left(x-1\right)}^{1993}{x}^{1993}}{{\left(x-1\right)}^{1993}{x}^{1993}{\left(x+1\right)}^{1993}}\right)
4.50\cdot {10}^{4}
\theta
{13.0}^{\circ }
\theta
Use the given information to evaluate each expression. Answer as simplified fraction if possible.
1\right)\mathrm{tan}\theta =2.6
\mathrm{tan}\left(-\theta \right)=
2\right)\mathrm{cos}\theta =-0.65
\mathrm{cos}\left(-\theta \right)=
3\right)\mathrm{csc}\left(-\theta \right)=-\frac{3}{2}
\mathrm{csc}\theta =
4\right)\mathrm{sec}\left(-\theta \right)=\frac{6}{5}
\mathrm{cos}\theta =
5\right)\mathrm{cos}\left(-\theta \right)=-\frac{3}{5}\text{ }\mathrm{&}\text{ }\mathrm{sin}\left(-\theta \right)=\frac{4}{5}
\mathrm{tan}\theta =
6\right)\mathrm{cos}\theta =-\frac{12}{13}\text{ }\mathrm{&}\text{ }\mathrm{sin}v=\frac{5}{13}
\mathrm{cot}\left(-\theta \right)=
A skateboarder, starting from rest, rolls down a 12.0m ramp. Whenshe arrives at the bottom of the ramp her speed is 7.70 m/s. (a)Determine the magnitude of her acceleration, assumed to be constant. (b) If the ramp is inclined at 25.0 degrees with respectto the ground, what is the component of her acceleration that isparallel to the ground?
G=\left\{x\in R\text{ }\text{such that}x\ne 0\right\}
e=-\frac{1}{2}
\mathrm{\infty } |
Determination of Palladium II in 5% Pd/BaSO 4by ICP-MS with Microwave Digestion, and UV-VIS Spectrophotometer
Determination of Palladium II in 5% Pd/BaSO4 by ICP-MS with Microwave Digestion, and UV-VIS Spectrophotometer
Y. Yildiz , M. Kotb, A. Hussein, M. Sayedahmed*, M. Rachid, M. Cheema
Department of Science, Al-Ghazaly High School, Wayne, NJ, USA
Determination palladiums have been reported 5% (w/w) Pd/BaSO4 known as Rosenmund Catalyst. The determination of palladium II known as Rosenmund Catalyst is always an expensive procedure usually involving procedures such as flame atomic absorption spectrophotometry, emission spectrometry, and many spectrophotometric methods. In this study, palladium II in 5% Pd/BaSO4, was synthesized and employed to develop an extractive UV-Visible Spectrophotometric, and an inductively coupled plasma mass spectrometry ICP/MS methods for the determination of palladium II. Specification for Pd is 4.85% to 5.10%; the result was 4.97% for the UV-Visible spectrophotometer and 4.90% for the ICP/MS. Both results meet the requirements.
Palladium Determination, 5% Palladium Barium Sulfate, Inductively Coupled Plasma Mass Spectrometry, UV-Visible Spectrophotometer, Microwave Digestion
Palladium, platinum, rhodium, ruthenium, iridium and osmium form a group of elements referred to as the “platinum group elements” (PGMs). These have similar chemical properties, but palladium has the lowest melting point and is the least dense of them. Palladium founds in a variety of ores. It is found to some degree in all platinum ores and is present in a number gold, nickel and copper ore bodies [1] - [10] . Palladium is a broadly useful silver white, nobleductile metal [11] discovered in 1803 by William Hyde Wollaston. Palladium is a chemical element having an atomic weight of 106.4, atomic number of 46, and the symbol Pd. Palladium compounds are more stable unlike other platinum metals. Palladium is soluble in concentrated HNO3. Palladium gives stable amine, nitrite, cyanide, chloride, bromide, and iodide complexes. Palladium group metals especially palladium is very important to industry [12] . Palladium is generally resistant to corrosion by most single acids, alkalies and aqueous solutions of simple salts. It is not attacked at room temperature by non-oxidizing acids such as sulfuric, hydrochloric, hydrofluoric, acetic and oxalic acids. Strongly oxidizing acids such as nitric acid and hot sulfuric acid attack palladium, as do ferric chloride and hypochlorite solutions.
Palladium is not tarnished by dry or moist air at ordinary temperatures [13] [14] [15] . Palladium is used in many applications because of its noble metal characteristics and often these can be provided most economically in the form of a coating. The catalyst stored indefinitely in well-sealed containers. The activation energy value for all Pd/BaSO4 catalyst was around 49.1 kJ/mol.
Palladium on Barium Sulfate 5% Pd/BaSO4 an appropriate form of carbon when used to catalyze the hydrogenation of acyl chlorides to aldehydes, the Rosenmund reduction [16] , useful catalyst for many other hydrogenations. The palladium catalyzed hydrogenation of an acid chloride to an aldehyde is known as the Rosenmund reduction (Figure 1 and Figure 2).
Figure 1. The catalytic hydrogenation of acid chlorides allows the formation of aldehydes.
Figure 2. Mechanism of the rosenmund reduction.
The Pd catalyst must be poisoned, for example with BaSO4, because the untreated catalyst is too reactive and will give some over reduction. Some of the side products can be avoided if the reaction is conducted in strictly anhydrous solvents.
2.1. Determination of Palladium by ICP-MS
・ NexIon 300X ICP-MS, Inductively Coupled Plasma computer-controlled sequential emission spectrometer with interelement and background correction capabilities, and provisions for interfacing to a printer and an auto sampler [17] .
・ Ethos Plus Microwave
・ ETHOS One Closed Vessel Microwave Digestion System, with temperature control and rotating turntable, well ventilated with corrosion-resistant cavity.
・ Microwave digestion vessels for water samples, Teflon, capable of holding ~75 milliliters (mL), designed “for temperatures up to 260˚C with self-regulating pressure control
・ Digestion vessels for soil samples, capable of holding ~250 mL
・ Watch glasses or vapor recovery device
・ Glass dispensers, 2-liter (L), 1-L, or 1-gallon, checked quarterly for accuracy
・ Graduated Cylinder, Class A, 50 mL
・ Volumetric flasks, Class A, assorted volumes
・ Balance, top-loading, capable of reading to 0.01 grams (g), for weighing digestion vessels before and after digestion
・ Henke SASS plastic syringes
・ Corning SCFA 0.45 microns (µm) filters
・ Argon Plasma Support Gas in pressurized cylinders.
2.1.2. Reagents and Solutions
・ Concentrated nitric acid, Seastar Chemicals. 67% - 70% (w/w HNO3), purified by re-distilled, ≥99.999% trace metals basis.
・ Nitric acid, 2 percent(%) volume to volume (v/v), for the preparation of working standards, also to be used for the initial calibration blank.
・ Deionized (DI) water, Type I Deionized water, for the preparation of all reagents and calibration standards and as dilution water.
・ Hydrogen peroxide solution contains inhibitor, 30 wt% in H2O, Sigma-Aldrich, ACS grade.
・ Palladium, 5% Pd/BaSO4 (Palladium on Barium Sulfate) obtained from Sigma-Aldrich.
2.1.3. Microwave Digestion
200 mg of sample weighed and placed in each digestion vial. 8 mL of nitric acid (HNO3) and 2 mL hydrogen peroxide (H2O2) have been added to each vial including a blank. Tighten the vials in the vessels, twist the screw on top by hand then use the teardrop racket to tighten further. No sample was added in a vessel for the blank preparation. For the blank, line up the holes and ensure that the tube goes in. Place them in microwave. After digestion, take out the vials from the ETHOS One Closed Vessel Microwave Digestion System. The teardrop racket was used to loosen the vials from the vessels. All the liquid from each digestion vial were transferred into 50 mL centrifuge tubes. About 10 mL deionized (DI) water, Type I Water was added to wash and rinse the vials to make the total volume 20 mL using a 5 mL pipet (Figure 3).
Table 1. Intensity of Palladium Standards.
Figure 3. Calibration curve of Palladium by ICP-MS.
Table 2. Intensity of Palladium samples.
Table 3. Intensity of Palladium Samples (Cont.).
Table 4. Intensity of Palladium Samples (Cont).
2.2. Determination of Palladium by UV-Vis Spectrophotometer Method
UV/Visible spectrophotometers are widely used by many laboratories including those in academia and research as well as industrial quality assurance. The technique is mainly used quantitatively. The absorbance spectra for all measurements were carried out using a Shimadzu 1601 PC double beam UV-VIS Spectrophotometer, with 1 cm quartz cells and 2.0 nm fixed slit width. The spectrophotometer was connected to a computer, loaded with Shimadzu UVPC software, and equipped with an Epson LQ-850 printer. 1 [18] [19] .
・ 5% Pdbasis BaSO4 obtained from Sigma-Aldrich
・ Deionized water, on the day of use. High-purity deionized water was obtained by Aries High Purity Water System, Aries Filter Works.
・ Diluted hydrochloric acid TS (1:1, or 6 N)
・ Palladium, 5% PdBaSO4 (Palladium on Barium Sulfate) obtained from Sigma-Aldrich.
・ Palladium standard stock solution for AA, Lot # BCBM7956V, c(HCl) = 5% (W/w)
・ Palladium standard solutions: Serial palladium standards were prepared by 5.0 mL of 1000 mg/LPd AA Stock standard in a 100 mL VF, QF water (50 mg/L); 5.0 mL 1000 mg/L in 50 mL VF QF DI water (100 mg/L), 5 mL 1000 mg/L in 25 mL VF QF DI water (200 mg/L).
543.4 mg sample was leached and warmed with diluted HCl, dissolved PdSO4; filtered, washed and brought to 200 mL in volumetric flask. A solution was in dilution also prepared 50 mL VF, pipette to 100 mL QFH2O. DF = 2.
All absorption measurements were made in a Shimadzu 1601 PC souble beam UV-VIS Spectrophotometer equipped with 1.0 cm quartz cells. The instrumental parameters were optimized and the best results were obtained in the wavelength at 453 nm (Figure 4).
Table 5. Calibration Curve Data.
Table 6. Absorbance of Palladium 5% Samples.
Figure 4. Calibration Curve of Palladium by UV-Vis.
QA/QC Study
Calibration Blank: A calibration blank is a sample of analyte-free media that can be used along with prepared standards to calibrate the instrument. A calibration blank may also be used to verify absence of instrument contamination [20] .
Calibration Curve: A plot of instrument response to an analyte versus known concentrations or amounts of analyte standards. Calibration standards are prepared by successively diluting a standard solution to produce working standards which cover the working range of the instrument. Standards should be prepared at the frequency specified in the appropriate method. The calibration standards should be prepared using the same type of acid or solvent and at the same concentration as the samples following sample preparation [20] .
Laboratory Control Sample (LCS): The Laboratory Control Sample (LCS) is analyzed to assess general method performance based on the ability of the laboratory to successfully recover target analytes from a control matrix. Aqueous and solid LCSs is obtained from an independent source, and was prepared with each analytical batch of samples using the same preparation method as that employed for the samples. Percent recovery (%R) must be within 75% - 125% and calculated as:
\%R=\frac{\left(LCS-B\right)}{SA}\times 100
where: LCS = LCS result, B = Blank result, SA = spiked amount [20] .
Laboratory Duplicate: The analysis or measurements of the variable of interest performed identically on two sub-samples of the same sample, usually taken from the same container. The results from duplicate analyses are used to evaluate analytical or measurement precision and include variability associated with sub-sampling and the matrix, but not the precision of field sampling, preservation, or storage internal to the laboratory [20] .
Matrix Spike (MS)/Matrix Spike Duplicate (MSD): Matrix spikes are aliquots of samples to which known concentrations of certain target analytes have been added before sample preparation. Matrix spike-with compounds spiked into at a known level. Matrix spike duplicates are additional replicates of matrix spike samples that are subjected to the sample preparation and analytical scheme as the original sample. Analysis of spiked samples ensures a positive value, allowing for estimation of analytical precision [20] . Spiked sample percent recovery (%R) must be within 75% - 125% and calculated as [20] :
\%R=\frac{\left(SSR-SR\right)}{SA}\times 100
where; SSR = Spiked sample result, SR = Sample result, SA = Spiked amount
Relative Percent Difference (RPD): The Relative Percent Difference (RPD) and between matrix spike and matrix spike duplicates (MS/MSD) samples were ±20% and calculated as:
\%RPD=\frac{\left(S-D\right)}{\left(S+D\right)/2}\times 100
where: S = %R for matrix spike sample
After microwave digestion intensity of palladium (II) sample solutions and the intensity of palladium standards (0.0, 0.1, 1.0, 2.0, 5.0, 10.0, 20.0, 50.0, and 100.0 ppb) were measured by ICP/MS for the calibration curve (Table 1, Table 2, Table 3 and Table 4). Average results was 4.90% for the ICP/MS. The palladium (II) standard has been added into the sample for percent recovery. %RMS and %RMSD and %RSD have been shown in Table 7.
After extraction 5% Pd on BaSO4 catalyst, PdSO4 present in solution was measured at 453 nm. By UV-VIS Spectrophotometry. The absorbance readings plotted against the concentrations of palladium (II) to obtain the calibration curve. Table 5 and Table 6. Average results was 4.97% for the UV-Visible spectrophotometer. 5 ppm palladium (II) standards was added for the % recovery (MS/MSD), The results obtained are given in Table 7 and show that both instrument method can be successfully determined by UV-VIS and ICP/MS.
Table 7. Results of Palladium 5% by ICP/MS and UV-Vis.
Inductively Coupled PlasmaMass Spectrometry ICP/MS and Direct Spectrophotometric method was developed for estimation of Palladium (II) and successfully used for quantitative extraction of Palladium in 5% Pd/BaSO4 at acidic conditions. Since the equilibration time is very less; the method is quick and applicable for determination of Pd (II) from different synthetic mixtures and catalysts. The results obtained are given in Table 7 and show that 5% (w/w) Pd (II) can be successfully determined by both methods.
We would like to thank you to Dr. Gregory Edens, Bloomfield College of New Jersey for his valuable information to parts of this paper.
Yildiz, Y., Kotb, M., Hussein, A., Sayedahmed, M., Rachid, M. and Cheema, M. (2019) Determination of Palladium II in 5% Pd/BaSO4 by ICP-MS with Microwave Digestion, and UV-VIS Spectrophotometer. American Journal of Analytical Chemistry, 10, 127-136. https://doi.org/10.4236/ajac.2019.104011
1. More, P.S. and Sawant, A.D. (1994) Isonitroso-4-Methyl-2-Pentanone for Solvent Extraction and Spectrophotometric Determination of Palladium (II) at Trace Level. Analytical Letters, 27, 1737-1748. https://doi.org/10.1080/00032719408007432d
2. Abu-Baker, M.S. (1996) Indian Journal of Chemistry Section A, 35, 69.
3. Dakshinmoorthy, A., Singh, R.K. and Iyer, R.H. (1994) Journal of Radioanalytical and Nuclear Chemistry, 17, 327.
4. Chakkar, A.K., Kakkar, L.R. and Fresenius, J. (1994) Extractive Spectrophotometric Determination of Palladium Using 2-(2-hydroxyimino-1-oxoethyl)furan. Analytical Chemistry, 350, 127-131. https://doi.org/10.1007/BF00323173
5. Zhu, Y.R. and Yang, L. (1993) Effect of Surfactant on the Spectrophotometric Determination of Palladium Complexed with 4-Nitrobenzenediazobenzene and 1,10- Phenanthroline. Analytical Letters, 26, 309-323. https://doi.org/10.1080/00032719308017387
6. Jha, A. and Mishra, R.K. (1993) J. Chin, Soc., 40, 351.
7. Fuji, Z., Bincai, W., Hengchauan, L. and Cheng, W. (1993) Characterization of Three New Phosphonic (Arsonic) Acid Type Thiazolylazo Reagents and Application to Spectrophotometric Determination of Microamounts of Palladium. Microchemical Journal, 48, 104. https://doi.org/10.1006/mchj.1993.1077
8. Sakuraba, S., Oguna, K. and Fresenius, J. (1994) Spectrophotometric Determination of Palladium(II) with Phenylfluorone in the Presence of Hexadecylpyridinium Bromide. Analytical Chemistry, 349, 523-526. https://doi.org/10.1007/BF00323986
9. Mathew, V.J. and Khopkar, S.M. (1997) Hexaacetato Calix(6)arene as the Novel Extractant for Palladium. Talanta, 44, 1699-1703. https://doi.org/10.1016/S0039-9140(97)00013-1
10. The International Nickel Company (1966) Palladium the Metal Its Properties and Applications.
11. Wise, E.M. Palladium-Recovery, Properties and Applications. Academic Press, New York.
12. Sathe, G.B., Valda, V.V. and Ravindra, G. (2015) Extractive Spectrophotometric Determination of Palladium (II) Using Novel Salen Ligand. International Journal of Advanced Research, 3, 699-704.
13. Betteridge, W. and Rhys, D.W. (1962) High-Temperature Oxidation of Palladium Metals and Their Alloys. 1st International Congress on Metallic Corrosion, London, 10-15 April 1961, 186-192.
14. Chaston, J.C. (1965) Reaction of Oxygen with the Platinum Metals III—The Oxidation of Palladium. Platinum Metals Review, 9, 126-129.
15. Wise, E.M. and Vines, R.F. (1948) U.S. Pat. 2457021.
16. Rosenmund, K.W. CB 1018, 51, 585.
17. Standard Operating Procedures. Determination of Metals by Inductively Coupled Plasma (ICP) Methods. EPA/SW-846 Methods 30115/3050B/6010B.
18. Yildiz, Y., Jan, A. and Yildiz, B. (2017) Determination of Tin in Trityl Candesartan by UV-VIS Spectrophotometer Using Phenylfluorone. World Journal of Applied Chemistry, 2, 134-139.
19. Yildiz, Y., Jan, A. and Chien, H.-C. (2018) Determination of Tin in Trityl Candesartan by Flame Atomic Absorption Spectrophotometry (AAS). International Journal of Chemical Studies, 6, 2687-2691.
20. Project Quality Assurance and Quality Control EPA. SW-846 Update V, Revision 2, July 2014. |
Value and Risk Practice Problems Online | Brilliant
Most people think of value as being measured in terms of money (e.g., dollars). In other words, people and investors make decisions to maximize the expected value of their money. While this is generally true, it is potentially misleading because it does not account for risk.
In a one-time bet, a fair coin is flipped. If it is heads, the player doubles their life savings and gets an additional $1. If the coin is tails, they lose all of their assets (their entire life savings, home, etc.). Would the average adult human take this bet?
The last question illustrated that “value” is often more complicated than an expected value calculation. Which of the following curves is a good depiction of an average individual’s “happiness” as a function of their wealth?
A trading firm has the utility function
U(w) = \sqrt{w}
w
is the wealth of the firm, in dollars. Currently, the firm is worth $100,000,000, so their happiness is
U(100,000,000) = \sqrt{100,000,000} = 10,000,
and they always want to maximize their expected happiness.
They are offered a risky bet which will succeed with probability
p,
doubling the wealth of the firm. However, if it fails, the firm will go bankrupt.
Find the smallest probability under which they would take this bet, and then round that probability to the nearest 10%.
Hint: Find their expected utility based on the possibilities after this bet, and determine the conditions under which it exceeds their current utility of
10,000.
As the last few questions have illustrated, there is an aspect of diminishing returns for wealth; e.g., the happiness gained from each additional $1 tends to decrease as a person becomes wealthier.
Similarly, there is also an effect of time on the value of money. The last two questions begin to explore this effect.
Under typical circumstances, what is the most that an individual would pay right now to receive $1,000 one year from now?
A little bit less than $1,000 Exactly $1,000 A little bit more than $1,000
The previous question illustrated the idea that “money now is worth more than money later”. A fundamental reason behind this is that money can be invested - in financial assets or elsewhere - so that it grows over time. This is why lenders are paid interest.
If money is lent with an annual interest rate of 1% compounded continuously, about how long would it take for the money to double? |
Revision as of 17:05, 6 January 2022 by Eragon4 (talk | contribs) (→Pokémon GO: Home)
{\displaystyle HP={\Biggl \lfloor }{{\Biggl (}(Base+DV)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {STATEXP}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times Level \over 100}{\Biggr \rfloor }+Level+10}
{\displaystyle OtherStat={\Biggl \lfloor }{{\Biggl (}(Base+DV)\times 2+{\biggl \lfloor }{\tfrac {{\bigl \lceil }{\sqrt {STATEXP}}{\bigr \rceil }}{4}}{\biggr \rfloor }{\Biggr )}\times Level \over 100}{\Biggr \rfloor }+5}
{\displaystyle HP={\Bigl \lfloor }{(2\times Base+IV+\lfloor {\tfrac {EV}{4}}\rfloor )\times Level \over 100}{\Bigr \rfloor }+Level+10}
{\displaystyle OtherStat={\Biggl \lfloor }{\biggl (}{\Bigl \lfloor }{(2\times Base+IV+\lfloor {\tfrac {EV}{4}}\rfloor )\times Level \over 100}{\Bigr \rfloor }+5{\biggr )}\times Nature{\Biggr \rfloor }}
{\displaystyle Stat=(base+IV)\times cpMult}
If a Pokémon is transferred from Pokémon GO to Pokémon: Let's Go, Pikachu! and Let's Go, Eevee!, the IVs will be recalculated directly based off the IVs it had in Pokémon GO.
{\displaystyle 2\times IV_{HP}+1}
{\displaystyle 2\times IV_{Attack}+1}
{\displaystyle 2\times IV_{Defense}+1} |
An oil drilling company ventures into various locations, and its success or failure is independent from one location to
An oil drilling company ventures into various locations, and
An oil drilling company ventures into various locations, and its success or failure is independent from one location to another. Suppose the probability of a success at any specific location is 0.30. What is the probability that a driller drills 10 locations and finds 1 success? The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller's prospects for bankruptcy? i. ii.
Probability of success at any specific location i.e.,
p=0.30
i) To find the probability that a driller drills 10 locations and finds 1 success:
Let X be a random variable which denotes the number of successes among 10 drills. Because trials are independent, X follows a Binomial distribution with parameters
n=10\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p=0.30
Probability mass function of Binomial variable X is given as:
P\left(X=x\right){=}^{n}{C}_{x}{p}^{x}{\left(1-p\right)}^{n-x}
P\left(X=1\right){=}^{10}{C}_{1}{\left(0.30\right)}^{1}{\left(1-0.30\right)}^{10-1}
=\frac{10!}{1!\left(10-1\right)!}×0.30×{\left(0.70\right)}^{9}
=\frac{10×9!}{9!}×0.30×0.040353607
=10×0.0121060821
=0.121060821
\approx 0.1211
Thus, the probability that a driller drills 10 locations and finds 1 success is 0.1211
ii) Let X represent the number of drills required to get first success.
Find the probability that the driller prospects for bankruptcy.
In other words, that there are no successes
\left(X=0\right)
in 10 trials:
P\left(X=0\right){=}^{10}{C}_{0}{\left(0.30\right)}^{0}{\left(1-0.30\right)}^{10-0}
=\frac{10!}{0!\left(10-0\right)!}×1×{\left(0.70\right)}^{10}
=\frac{10!}{10!}×1×0.0282475249
=0.0282475249
\approx 0.0282
Suppose the probability of a success at any specific location is 0.25.
i) What is the probability that a driller drills 10 locations and finds 1 success?
P\left(\text{one success in 10}\right)=\left(10C1\right){\left(0.25\right)}^{1}{\left(0.75\right)}^{9}=0.1877
ii) The driller feels that he will go bankrupt if he drills 10 times before the first success occurs. What are the driller is prospects for bankruptcy?
P\left(\text{no success in 10 trials}\right)={0.75}^{10}=0.0563
Chances are 563 in 10000 that he will go bankrupt.
- Consider that drilling in a location is a trial.
- Probability of success in each trial is p=0.25,
- Probability of failure is q=1-p=1-0.25=0.75
a) Let random variable X represent the number of success among 10 drills.
Because trials are independent, X has binomial distribution with parameters n=10 and p=0.25.
Probability mass function of X is
P\left(X=x\right)=Bin\left(x;10,0.25\right)
=\left(\begin{array}{c}10\\ x\end{array}\right)\left(0.25{\right)}^{x}\left(0.75{\right)}^{10-x},\text{ }where\text{ }x=0,1,2,...,10
Lets calculate the probability the driller drills at at 10 locations and has 1 success.
b) Let random variable X represent the number of drills required to get first success.
Now, lets find the probability that the driller prospects for bankruptcy, in other words, that there are no successes in 10 trials.
P\left(X=0\right)=\left(\begin{array}{c}10\\ 0\end{array}\right)\left(0.25{\right)}^{0}\left(0.75{\right)}^{10}
If X has a binomial distribution,
n=5\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p=0.4
. Determine the probabilities of
X=0
a). -1.2321
d) 0.1534621
In Exercises, X denotes a binomial random variable with parameters n and p. For each exercise, indicate whicharea under the appropriate normal curve would be determined to approximate the specified binomial probability.
P\left(7\le X\le 10\right)
P\left(X\le 4\right)
Convert the binomial probability to a normal distribution probability using continuity correction. P (x = 45).
Tell, what is the factorial of 10? |
What is the advantage of saying that to multiply a
What is the advantage of saying that to multiply a number by 100, we s
We have to show the advantage of multiply a number by 100, we shift the digits two places to the left, instead of saying move the decimal places to the right
We will understand this statement by an example, normally we multiply any number with for simple calculation
x=r\mathrm{cos}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y=r\mathrm{sin}
\underset{\left(x,y\right)\to \left(0,0\right)}{lim}\frac{{x}^{2}-{y}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}
P\left(x\right)=-12{x}^{2}+2136x-41000
{x}^{2}ydy-x{y}^{2}dx+3ydy-2xdx=0
{X}_{1},{X}_{2},\stackrel{˙}{s},{X}_{n}
\left(X>2000\right)\ge 0.95
g\left(r,t\right)=t\mathrm{ln}r+11r{t}^{7}-5\left({8}^{r}\right)-tr
{g}_{r}
{g}_{rr}
{g}_{rt}
{g}_{t}
{g}_{t}
p=126-0.5x
C=50x+39.75 |
About Sovryn Governance | S O V R Y N
Sovryn is a smart contract system built on RSK (https://www.rsk.co). The Sovryn smart contracts have many parameters that can be modified, and the smart contracts themselves can also be modified. This raises the issue of governance: what the parameters and smart contracts of the Sovryn smart contract system are that can be governed, what the processes are for modifying these parameters and smart contracts, and how to modify the governance process itself.
Note: This page intends to describe how Sovryn governance currently works at a medium level of detail. While this page may not have all of the information about Sovryn governance, we have strived to ensure that the information that is here is 100% accurate. We welcome feedback with any additions or corrections to this content.
For the canonical description of Sovryn governance, you will have to find and read the source code of the relevant smart contracts in their current state on the RSK blockchain.
¶ System contracts, parameters, and their owners
All ownable contracts are currently owned by the Exchequer multisig, except Staking and FeeSharingProxy which are owned by the Owner Governor (see Owner Governor subsection below).
Sovryn smart contracts directory:
https://docs.google.com/spreadsheets/d/1SSqQJ4HNrIo8jRghKBa4ao_aAt4TJxP7jafRV-l0YcM/edit#gid=294704344
List of ownable contracts:
https://docs.google.com/document/d/1gGY4Rua_FVBZCJCftzf14cD4c6kqg6VTr9g-9-uDCA0/edit
Sovryn contracts and their modifiable parameters:
https://docs.google.com/document/d/1DoTxAlOFE27HZ7BPcMv_wHX8scqmVwjpKTxlcdeuvHI/edit
¶ Governance contracts and their settings
¶ Owner Governor: upgrade contract logic
GovernorOwner is an instance of GovernorAlpha.
You can find a flatenized version of GovernorAlpha here.
You can find the main ABI of GovernorAlpha contract here.
The mainnet address of GovernorOwner (instance of GovernorAlpha) is: 0x6496df39d000478a7a7352c01e0e713835051ccd.
Voting token: SOV (0xefc78fc7d48b64958315949279ba181c2114abbd)
Quorum: >20%
Support: >70%
Min. Proposal Stake: 1%
Execution delay: TimelockOwner (48 hours)
Vote duration: 2880 blocks (approx. 1 day)
Contract call limit: 10 calls
Guardian: Exchequer Multisig
¶ Admin Governor: modify settings
GovernorAdmin is an instance of GovernorAlpha.
The mainnet address of GovernorAdmin (instance of GovernorAlpha) is: 0xff25f66b7d7f385503d70574ae0170b6b1622dad.
Quorum: >5%
Execution delay: TimelockAdmin (24 hours)
¶ TimelockOwner
The mainnet address of TimelockOwner is: 0x967c84b731679E36A344002b8E3CE50620A7F69f.
¶ TimelockAdmin
The mainnet address of TimelockAdmin is: 0x6c94c8aa97C08fC31fb06fbFDa90e1E09529FB13.
Staking period length: Two weeks, repeating bi-weekly starting from SOV TGE
Voting weight equation:
f(x)=V*({m^{2}}-{x^{2}})/{m^{2}}+1
Voting power equation:
VP(x) = f(x) * s
¶ Staking overview
SOV holders must stake their SOV to gain voting power in Bitocracy. Voting power is determined using a quadratic formula based on the amount and duration of SOV staked.
Staking periods are divided into bi-weekly intervals, beginning at the time of the SOV token generation event and repeating every two weeks from then on. Users will acquire voting power from their stake and begin their staking duration when the staking period after they started staking begins. Once their staking duration begins, users can withdraw their staked SOV early if they are willing to take a slashing penalty. Users can also delegate their voting power to other addresses.
In addition to earning voting power, SOV holders who stake their SOV become eligible for protocol fee-sharing, receiving a proportional share (as measured by voting power) of all tokens that have been collected in the fee-sharing pool. Tokens collected in the fee-sharing pool come from protocol fees plus all slashed SOV from early unstaking penalties.
¶ Calculating voting weight and voting power
The stake weighting function is based on a quadratic formula, using the amount and duration of SOV staked.
Let m be the maximum staking period (1092 days) ,
x
be the number of days passed since the beginning of the period, and
V
be the maximum voting weight (currently 9). The voting weight at a given time
x
f(x)=V*({m^{2}}-{x^{2}})/{m^{2}}+1
+1
is added in the end to shift the weights to lie in
[1, V+1]
[0, V]
Users will usually not stake the maximum duration, so
x
has to be computed based on the remaining days until unstaking:
x = m-RemainingDays
The user’s voting power
VP
x
is the product of their stake
s
and their voting weight:
VP(x) = f(x) * s
The total voting power equals the sum of the voting power of all users. However, since we can’t iterate over an array with unlimited size, we need to compute it differently. Instead of summing up the user voting powers, we introduce a mapping, which stores the total amount of SOV to be unstaked on a given day, and call it stakedUntil.
Whenever a user stakes SOV, not only is their staking balance updated, but also stakedUntil[unstakingDay]. Whenever a user increases their staking balance, the mapping also needs to be increased. Whenever a user increases their staking time, their stake is subtracted from stakedUntil[previousUnstakingDay] and added to stakedUntil[newUnstakingDay].
Now, we can compute the voting power of all SOV staked until a given point in time in a single operation. The daily voting power DVP is given by:
DVP(x) = f(x) *\sum_{u} s_{u}(x)
\sum_{u}s_{u}(x)
is the content of stakedUntil[x].
The total voting power can then be computed by summing up the daily voting powers of the next
m
TVP=\sum_{x=0}^{m}*DVP(x)
This process requires
m
iterations. With a maximum staking duration of 3 years, this would cost approximately 1,000,000 gas. In order to save gas costs, we decided to stake in bi-weekly periods instead. As a consequence, voting weights are only adjusted every 2 weeks and we need a maximum of 78 iterations (instead of 1095).
The stakedUntil mapping needs to be checkpointed to allow the computation of the total voting power for a point of time in the past. The same is true for the user stakes.
¶ Delegating
Once their SOV is staked, users are able to delegate their voting power to another address. Users may want to do this if, for example, they are unable to be actively involved in governance and want a more active or qualified individual/group to vote on their behalf, or if they prefer to maintain transfer authority for their SOV on one address but voting authority on a different address.
Given the possibility of delegation, a user can have a potentially large number of addresses delegating voting power to their address, with each of these other addresses staking for different durations. The voting power of a delegatee is computed the same way the total voting power is computed: as a sum of the voting powers per unstaking day.
¶ Early unstaking penalty
Users who unstake before the end of their staking duration are subject to a token slashing penalty of up to 30%. This incentivizes users who stake to maintain their commitment to the protocol. Slashing penalties are deducted from the staking balance and sent to the fee-sharing pool and redistributed to all other staked SOV holders.
The penalty chart for early unstaking is as follows:
Early unstaking penalty
Thank you to the following contributors for their help with this documentation:
Ponjinge |
chemical bonding - molecular orbital theory | Brilliant Math & Science Wiki
chemical bonding - molecular orbital theory
Molecular orbital (MO) theory is a method for predicting molecular bonds and structure in which electrons are not assigned to individual bonds between atoms – as in valence shell electron pair repulsion (VESPR) theory – but as interacting with the nuclei in the molecule as a whole. Electrons are allowed to move around atomic nuclei in trajectories explained by mathematical functions called orbitals. As atoms bond to form molecules, their atomic orbitals combine to form molecular orbitals. Atomic orbital functions are described by Schrödinger’s wave equation. The Schrödinger wave equation is a linear equation and thus molecular orbitals can be described by simple addition of the atomic orbitals; the linear combinations of atomic orbitals (LCAO).
With advances that have been made in computational chemistry, the LCAO approximations can be further refined by applying the density functional theory or Hartree–Fock models to the Schrödinger equation.
Linear combinations of atomic orbitals
Bonding, anti-bonding, non-bonding molecular orbitals
A molecular orbital is really a mathematical function that describes the wave-like behavior of an electron in a molecule and used to determine the probability of finding an electron in any particular location. Molecular orbitals are delocalized over the entire molecule; they can surround many atoms in a molecule and thus can contain many valence electrons, therefore any electron in a molecule may be found anywhere in that molecule. At the same time, in accordance with Hund’s rule and the Pauli exclusion principle, each MO can contain only 2 electrons each with opposite spin.
When two atoms bond, the electrons occupy a molecular orbital whose wave function is analogous to that of an atomic orbital. For example, in the case of a diatomic molecule, LCAO declares that the molecular wave function can be built as a linear combination of the wavefunctions for each atom. Thus, two molecular orbital wavefunctions are formed
\ce{Ψ\,=\,c_{1}ψ_{1}\,+\,c_{2}ψ_{1}}
\ce{Ψ} *{=\,c_{1}ψ_{1}\, -\,c_{2}ψ_{1}}
\ce{ψ_{1}}
\ce{ψ_{2}}
are the atomic wavefunctions for atoms 1 and 2;
\ce{c_{1}}
\ce{c_{2}}
are variable coefficients that change depending on the energies of the atomic orbitals.
\ce{Ψ}
\ce{Ψ}
* represent the two independent quantum states for electrons in the diatomic molecule - the molecular wavefunctions for the bonding and antibonding molecular orbitals which will be described in the next section.
Molecular wavefunctions are thought of as combinations of atomic wavefunctions. When two atoms come into close proximity their orbitals overlap producing an area of high electronic probability density and a molecular orbital is formed where the atoms are held together by electrostatic attraction between the positively charged nuclei and negatively charged electrons.
These combinations of wavefunctions can be constructive (in-phase) or destructive (out of phase). When they are constructive, regions of higher probability of electron density are produced and when destructive, there is no chance of finding an electron in that region.
constructive, in-phase
destructive, out-of-phase
The term phase is a consequence of the mathematical description of the wavefunction. Phase is usually discussed in terms of + sign or – sign or in different colors. The sign of the phase doesn’t have meaning except to distinguish constructive or destructive interference when combining atomic orbitals to form molecular orbitals.
Bonding molecular orbitals are formed when atomic orbitals have constructive interaction and have lower energy than the individual atomic orbitals. Bonding MOs are stabilizing since the bound atoms have less energy than unbound atoms. Anti-bonding MOs are formed when atomic orbitals have destructive interaction – there is a nodal plane where the wavefunction of the anti-bonding orbital is zero between the two atoms and there is no probability of finding an electron.
EXAMPLE. Constructive and destructive interaction of
\ce{H_{2}}
When atomic orbitals come into proximity and overlap, the overlap in-phase (+ with +) or out-of-phase (- with +). When the s orbitals of
\ce{H_{2}}
overlap, their linear combination can be an in-phase
\ce{σ_{s}}
bonding molecular orbital or an out-of-phase
\ce{σ_{s}}
* anti-bonding molecular orbital.
Anti-bonding MOs have decreased electron density and higher energy than the individual atomic orbitals. Non-bonding MOs are formed when atomic orbitals are have no interaction beause their symmetries are not compatible. Therefore, the original energy of the individual atomic orbitals is retained. As seen in the example above, when two atomic s-orbitals are combined, two types of molecular orbitals are formed. The in-phase, constructive form produces a low energy, bonding, molecular orbital denoted
\ce{σ_{s}}
with the electrons attracted to both atomic nuclei and the electron probability density concentrated between the atomic nuclei. This is the single bond seen in a Lewis structure and VSEPR, it corresponds to the sharing of one pair of electrons. The electron probability density concentrated between the two nuclei is the essential component of a σ-bond. On the other hand, the out-of-phase, destructive form produces a high energy, anti-bonding molecular orbital denoted
\ce{σ_{s}}
* that forms a node of low-to-zero electronic probability density between the atomic nuclei. With the electrons in
\ce{σ_{s}}
* so far away from the center of the atomic nuclei, the attraction between the nuclei and the electrons pulls the nuclei apart. An antibonding molecular orbital weakens the atomic bond because it has higher energy than the two atoms separately. When a nodal plane is formed in an anti-bonding orbital it means that the orbital’s electron density is mostly outside the bonding region and the nuclei are thus pulled away from each other. If the bonding orbitals are filled, then electrons will start occupying anti-bonding orbitals. As anti-bonding orbitals have higher energy than bonding orbitals, bonding is not favored.
EXAMPLE. LCAO and MO for hydrogen and helium.
\ce{H_{2}}
H has one proton, no neutrons, and one electron. When hydrogen atoms are in close proximity atomic orbitals (1s) combine to become molecular orbitals resulting in
\ce{H_{2}}
with one bonding molecular orbital of lower energy (1σ) and one anti-bonding molecular orbital of higher energy (1σ*). Because the bonding molecular orbital is more stable and lower energy than the individual atomic orbitals,
\ce{H_{2}}
is more stable than two isolated H atoms. An antibonding orbital with two electrons in it would be less stable and does not form.
MO diagram for
\ce{H_{2}}
Each H atom has one electron (blue up and down arrows for spin up and down) in the 1s-orbital. Remember that the Pauli exclusion principle states that two electrons can only share an orbital if their spins are opposite. Since
\ce{H_{2}}
has only two electrons, both electrons will enter the lower energy σ-bonding orbital upon
\ce{H_{2}}
formation. There are no anti-bonding electrons so
\ce{H_{2}}
is a stable molecule.
\ce{He_{2}}
Helium’s atomic number is 2 so it has two electrons in each atom, both in the 1s-orbital. To form a molecule, the four electrons must be placed in bonding and anti-bonding orbitals as in this MO diagram:
\ce{He_{2}}
Since both bonding σ and anti-bonding σ* orbitals would be filled upon formation of
\ce{He_{2}}
and anti-bonding orbitals are higher in energy than bonding orbitals,
\ce{He_{2}}
will not form as two separate He atoms are more stable.
What’s interesting is that
\ce{He_{2}^{+}}
does form! This is because
\ce{He_{2}^{+}}
is short one electron meaning only one electron will be in the anti-bonding σ* orbital. There is less energy when two electrons are in the bonding orbital and one in the anti-bonding orbital than when there are two separate helium atoms and so formation of
\ce{He_{2}^{+}}
is favoured.
Shape of molecular orbitals.
The wave function for p-orbitals is not spherical as for s-orbitals, but with two lobes that are opposite in-phase. When lobes of the same phase overlap, constructive interference occurs and there is increased electron density. When regions of opposite phase overlap, destructive interference occurs decreasing electron density and nodes are formed. When p-orbitals overlap end to end, σ and σ* orbitals are formed. Two overlapping
\ce{p_{x}}
orbitals form the low-energy, bonding
\ce{σ_{px}}
. Electrons in the bonding orbital interact with both nuclei and help hold the two atoms together. The overlapping
\ce{p_{x}}
orbitals can also form high-energy, anti-bonding
\ce{σ_{px}}
* . The x subscript denotes the x-axis in a cartesian coordinate system and is used for visualization purposes. In the same way p-orbitals can form on y- and z-axis.
If the p-orbitals overlap side-by-side, π-orbital and π-orbital are formed. This is a double bond between two atoms sharing of two pairs of electrons, as seen in the Lewis approach. One pair of electrons in a σ-bond and the other pair in a π-bond. It follows that a triple bond made up of three shared electron pairs, forming one σ-bond and two π-bonds. The d-orbitals and *f-orbitals combine to form bonding and anti-bonding orbitals in the shape depicted in the figure above.
Combining p-orbitals.
In molecular orbital theory, an electron stabilizes bonding interactions if it is in a bonding orbital and destabilizes bonding effects if it is in an antibonding orbital. The bond strength of a molecule (due to electrons being found in bonding or anti-bonding orbitals) can be found by calculating the bond order that results from filling the molecular orbitals:
bond\, order\, = \frac{1}{2} (bonding\, electrons\, - anti\, bonding\, electrons)
EXAMPLES. Calculating bond orders.
What is the bond order for hydrogen gas
\ce{H_{2(g)}}
Answer: Hydrogen atoms each have one electron and this electron is in the s-orbital. Each orbital can hold up to two electrons. When two hydrogen atoms bond together to form
\ce{H_{2(g)}}
, each atom contributes to complete the others’ s-orbital and thus two bonding orbitals are formed. Since no electrons are left unpaired or forced to move to a higher energy level orbital, no antibonding orbitals are formed. The bonding order can then be found to be
bond\, order\, = \frac{1}{2} (2 - 0) = 1
What is the bond order for acetylene
\ce{C_{2}H_{2}}
Answer: This molecule is a bit more involved than
\ce{H_{2(g)}}
so lets draw the Lewis structure:
There are two types of bonds in this molecule, the H-C and the triple bond between the carbon atoms. The bond order for H-C = 1 for the same reason that the bond order between H-H was 1. To find the bond order for the triple bond we see that
bond\, order\, for\, the\, triple\, bond\, = \frac{1}{2} (6 - 0) = 3
You can always draw MO diagrams if that method of visualization helps determine how many bonding and anti-bonding electrons there are.
The Lewis approach to chemical bonding defines bond order as the number of chemical bonds in a molecule. So, in Lewis structures, a single bond has bond order = 1, a double bond has bond order = 2, and a triple bond has bond order = 3. Molecular orbital theory is more accurate in its description of electron distribution but the resulting bond order is usually the same as both methods describe the same phenomenon.
Cite as: chemical bonding - molecular orbital theory. Brilliant.org. Retrieved from https://brilliant.org/wiki/chemical-bonding-molecular-orbital-theory/ |
Random variables X and Y have joint PDF f_{X,Y}(x,y)=\begin{cases}12e^{-(
Random variables X and Y have joint PDFf_{X,Y}(x,y)=\begin{cases}12e^{-(
{f}_{X,Y}\left(x,y\right)=\left\{\begin{array}{l}12{e}^{-\left(3x+4y\right)},\text{ }x\ge 0,y\ge 0\\ 0,\text{ }otherwise\end{array}
P\left[max\left(X,Y\right)\le 0.5\right]
P\left[max\left(X,Y\right)\le 0.5\right]
P\left[max\left(X,Y\right)\le 0.5\right]=P\left(X\le 0.5,Y\le 0.5\right)
=P\left(X\le 0.5\right)×P\left(Y\le 0.5\right)
P\left(X<0.5\right)={\int }_{0}^{0.5}3{e}^{-3x}dx
=3{\left[-\frac{{e}^{-3x}}{3}\right]}_{0}^{0.5}
=1-{e}^{-1.5}
P\left(Y<0.5\right)={\int }_{0}^{0.5}4{e}^{-4y}dy
=4{\left[-\frac{{e}^{-4y}}{4}\right]}_{0}^{0.5}
=1-{e}^{-2}
P\left[max\left(X,Y\right)\le 0.5\right]=P\left(X\le 0.5\right)×P\left(Y\le 0.5\right)
=07769×0.8647
\stackrel{\to }{B}
4.50\cdot {10}^{4}
\theta
{13.0}^{\circ }
\theta
\frac{m}{{s}^{2}}
\frac{m}{{s}^{2}}
A wagon with two boxes of Gold, having total mass 300 kg, is cutloose from the hoses by an outlaw when the wagon is at rest 50m upa 6.0 degree slope. The outlaw plans to have the wagon roll downthe slope and across the level ground, and then fall into thecanyon where his confederates wait. But in a tree 40m from thecanyon edge wait the Lone Ranger (mass 75.0kg) and Tonto (mass60.0kg). They drop vertically into the wagon as it passes beneaththem. a) if they require 5.0 s to grab the gold and jump out, willthey make it before the wagon goes over the edge? b) When the twoheroes drop into the wagon, is the kinetic energy of the system ofthe heroes plus the wagon conserved? If not, does it increase ordecrease and by how much?
{f}_{XY}\left(x,y\right)=\frac{6}{7}\left({x}^{2}+\frac{xy}{2}\right)\text{ }0<x<1,0<y<2
P\left(X<1,Y>1\right)
X=0.5
P\left(Y<1/X=0.5\right)
f\left(x\right)={\left(x+1\right)}^{2}-2
. Give the (a) vertex, (b) axis, (c) domain, and (d) range.
Then determine (e) the largest open interval of the domain over which the function is increasing and (f) the largest open interval over which the function is decreasing.
(a) The vertex is (-1,-2).
Use the graphing tool to graph the function.
(Type an equation.) |
Out of the following four numbers, the largest is (1) 3210 (2) 7140 (3) 17105 (4) 3184 ( How - Maths - Relations and Functions - 10200815 | Meritnation.com
\mathrm{As}, {3}^{210}={\left({3}^{2}\right)}^{105}={9}^{105}\phantom{\rule{0ex}{0ex}}⇒{17}^{105}>{9}^{105}\phantom{\rule{0ex}{0ex}}⇒{17}^{105}>{3}^{210}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Since}, {7}^{140}={\left({7}^{4}\right)}^{35}={784}^{35}\phantom{\rule{0ex}{0ex}}\mathrm{and} {17}^{105}={\left({17}^{3}\right)}^{35}={4913}^{35}\phantom{\rule{0ex}{0ex}}\mathrm{As}, {4913}^{35}>{784}^{35}\phantom{\rule{0ex}{0ex}}⇒{17}^{105}>{7}^{140}\phantom{\rule{0ex}{0ex}}\mathrm{Also}, 4913>3184\phantom{\rule{0ex}{0ex}}⇒{17}^{105}>3184\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So}, {17}^{105} \mathrm{is} \mathrm{the} \mathrm{largest} \mathrm{of} \mathrm{all}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{option} \mathrm{is} \left(3\right).
Anirudh Sharma answered this
Sourav.......
the correct ans. is (3) 17105 |
Quantum Entanglement | Brilliant Math & Science Wiki
Matt DeCross, July Thomas, Tiffany Wang, and
Quantum entanglement occurs when a system of multiple particles in quantum mechanics interact in such a way so that the particles cannot be described as independent systems but only as one system as a whole. Measurement (e.g. of the spin of an entangled electron) may instantaneously affect another electron's spin at an arbitrarily distant location, apparently (but not actually) faster than the lightspeed limit of special relativity. The fact that electron spin measurements can be highly correlated, violating Bell's inequality, is one of the cornerstone experimental results in the modern theory and interpretation of quantum mechanics.
The properties of quantum entanglement may engender quantum teleportation, where the state of one entangled particle is sent from one location to another without moving the particle. This phenomenon may prove extremely useful in the nascent field of quantum computing, where manipulating quantum states without losing information by exposing them to the environment is highly valued.
In quantum entanglement, measurements of the spin state of one particle influence the spin state of another arbitrarily distant particle with which the first is entangled.
EPR Paradox and Stern-Gerlach Experiments
Measurement of Electron Spin
Entanglement and Tensor Products
The original Stern-Gerlach experiment examined the behavior of silver atoms in magnetic fields. Magnetic dipoles couple to magnetic fields; they feel a force
F = \nabla (\vec{\mu} \cdot \vec{B})
\mu
is the magnetic dipole moment and
B
is the magnetic field vector. Since silver atoms only have one valence electron with zero orbital angular momentum (a
5s
electron), any coupling of silver atoms to magnetic fields is due to an intrinsic magnetic dipole moment or spin of the electron. Since only the behavior of the valence electron matters, the rest of this article will refer to electrons where silver atoms were actually used originally.
In the Stern-Gerlach experiment, the magnetic field gradient
\nabla B
is aligned along the z-axis and set to some constant
C
so that electrons are deflected in the magnetic field according to the magnetic dipole moment:
F = C\mu
. If the electrons are spin-up in the z-direction, they are deflected upwards, whereas if the electrons are spin-down in the z-direction, they are deflected downwards. This experiment provided the first evidence for the quantization of spin: on a detecting screen behind the magnets, only two peaks appear rather than a continuous spectrum, corresponding to the two possible quantized values of spin rather than a continuous distribution of possible magnetic dipole moments.
Experimental setup of a Stern-Gerlach experiment. An electron source beams electrons into a magnetic gradient, where they are deflected according to their intrinsic magnetic moment. Only two peaks are detected beyond the magnets, corresponding to the two possible spin values.
The entirety of a Stern-Gerlach experiment can be thought of as a black box which is able to measure whether an entering electron is spin-up or spin down.
In the EPR thought experiment, Einstein, Podolsky, and Rosen considered a system of two electrons which are sent in opposite directions through two Stern-Gerlach apparatuses, measuring each of their spins. If the electrons are entangled, then when electron 1 is measured to be spin-up, electron 2 is measured to be spin-down a large percentage of the time, and vice versa. The precise meaning of "entangled" in quantum mechanics is discussed below mathematically. The result seemed to imply that at one Stern-Gerlach apparatus, the measurement result was being communicated instantly to the other apparatus. Although EPR did not perform the experiment themselves, actual experimental results are consistent with their mathematics. The nonlocality of the EPR paradox would be in violation of the principles of relativity, since it seems like information is being communicated to a distant location faster than the speed of light.
Setup of EPR's thought experiments: spin-entangled electrons are sent in opposite directions to two distant Stern-Gerlach apparatuses, where their spins are measured, revealing remarkable correlations.
EPR tried to avoid this conclusion by asserting that there must have been some local hidden variables: at the electron source, some variable that cannot be measured was determining the outcomes of spin measurements when the entangled electrons were produced. They took this to mean that the theory of quantum mechanics was not complete, since the spin and spatial wavefunctions of the electrons did not include the hidden variables and so could not account for all the physics that they would observe. However, in the 1960s, Bell's theorem showed that this was not the case: no local hidden variables theory could possibly account for the degree of correlation present in measurement of spin-entangled electrons.
Entanglement usually originates on the subatomic scale via processes that produce two electrons simultaneously in a correlated way, although complicated systems of fiber optics can be used to produce e.g. systems of two photons with entangled polarizations.
A single electron spin in quantum mechanics is described by a vector in a two-dimensional vector space. The basis vectors are taken to be the eigenvectors of the
\hat{S}_z
operator giving the spin of the electron along the
z
axis, and are labeled
|0\rangle
|1\rangle
|+\rangle
|-\rangle
|\uparrow\rangle
|\downarrow\rangle
depending on context. In this article, the latter notation will be used. In general, an arbitrary electron spin state can be a superposition of the "spin-up" and "spin-down" states:
|\Psi\rangle = c_1 |\uparrow\rangle + c_2 |\downarrow\rangle,
c_1
c_2
are some complex constants normalized so that
|c_1|^2+|c_2|^2 = 1
. This notation suggests that a single-spin state can be written as a vector:
|\Psi \rangle = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}.
The spin states are defined so that the spin operator in the z-direction,
\hat{S}_z = \frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},
\pm\frac{\hbar}{2}
corresponding to the eigenvectors
|\uparrow\rangle
|\downarrow\rangle
. This operator acts on the vector representation of spin states by simple matrix multiplication.
\frac12
\frac14
\frac34
\frac{\sqrt{3}}{2}
\large |\Psi\rangle = \frac{i}{2} |\uparrow\rangle + \dfrac{\sqrt{3}}{2} |\downarrow\rangle.
z
-\dfrac{\hbar}{2}
In a two-particle system, the spins of two electrons can be described jointly using tensor product spaces. The tensor product of two electron spin states lives in a four-dimensional vector space, with basis vectors labeled:
|\uparrow\rangle\otimes|\uparrow\rangle, \quad |\uparrow\rangle\otimes|\downarrow\rangle, \quad |\downarrow\rangle\otimes|\uparrow\rangle, \quad |\downarrow\rangle\otimes|\downarrow\rangle.
The tensor product of two states
|x\rangle
|y\rangle
obeys the following rules:
1) Scaling the tensor product is equivalent to scaling either state:
c(|x\rangle \otimes |y\rangle) = (c|x\rangle) \otimes |y\rangle = |x\rangle \otimes (c|y\rangle).
2) The tensor product is distributive in both the first and second slots:
\begin{aligned} &|x\rangle \otimes (|y_1\rangle + |y_2\rangle) = |x\rangle \otimes |y_1\rangle + |x\rangle\otimes |y_2\rangle \\ &(|x_1\rangle + |x_2\rangle) \otimes |y\rangle = |x_1\rangle \otimes |y\rangle + |x_2\rangle\otimes |y\rangle \end{aligned}
An arbitrary tensor product state for a two-spin system can be written out in the four-dimensional basis as:
|\Psi\rangle = c_{11} |\uparrow\rangle\otimes|\uparrow\rangle + c_{12} |\uparrow\rangle\otimes|\downarrow\rangle + c_{21} |\downarrow\rangle\otimes|\uparrow\rangle + c_{22} |\downarrow\rangle\otimes|\downarrow\rangle.
\frac{1}{\sqrt{2}} \vert\uparrow\uparrow\rangle
\frac{1}{\sqrt{2}} (\vert\uparrow\uparrow\rangle + \vert\downarrow\uparrow\rangle)
\frac{1}{\sqrt{2}} \vert\uparrow\downarrow\rangle
\frac{1}{\sqrt{2}} (\vert\uparrow\uparrow\rangle + \vert\uparrow\downarrow\rangle)
|\Psi_1\rangle = |\uparrow\rangle
|\Psi_2 \rangle = \frac{1}{\sqrt{2}} (|\uparrow\rangle + |\downarrow\rangle)
|\Psi_1 \rangle \otimes |\Psi_2\rangle
|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle
This notation suggests that it is more convenient to represent a two-spin state as a matrix, in analogue to the vector representation of a one-spin state:
|\Psi \rangle = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix}.
To perform a measurement on a two-spin system, one needs to specify the operators acting on each independent space. For instance, to measure the spin of just the first particle, use the operator written as
\hat{S}_z \otimes I
, which measures the spin of the first particle in the z-direction and acts on the second space as the identity. To measure the spins in the z-direction of both particles simultaneously, use
\hat{S}_z \otimes \hat{S}_z
Having defined this notation, it is now possible to state an explicit mathematical criterion of entanglement that is useful for proving things in quantum information theory.
A two-electron spin state is a product state if it can be written as the tensor product of exactly two independent single-spin states.
Some states are obviously product states:
|\uparrow\rangle \otimes |\downarrow\rangle
, for instance, is clearly the product of
|\uparrow\rangle
|\downarrow\rangle
. This state, therefore, can be described just by taking each independent single-spin state independently and ignoring the other. A state such as
|\uparrow\rangle \otimes |\downarrow\rangle + |\downarrow\rangle \otimes |\uparrow\rangle
, however, is clearly not a product state. Not all cases are so obvious, however, which is why formulating a precise mathematical criterion is important.
The intuition above suggests the definition of entanglement:
A two-electron spin state is entangled if it is not a product state, that is, if the matrix representation
M
of the state does not satisfy
\det M = 0
For states of more than two electrons, this condition is actually too strong -- if
\det M \neq 0
, then all of the electron spins are entangled. But it's enough for just two of the electrons to be entangled in a multiparticle state for the whole state to be considered entangled. The more accurate condition for a general multiparticle state is that
\text{rank } M \geq 2
, since this is equivalent to entanglement in a two-dimensional subspace. Note that this condition reduces to
\det M \neq 0
for a two-electron spin state.
Challenge problem: prove the determinant condition for entanglement described above
|\Psi\rangle
|\Psi\rangle = \frac{1}{\sqrt{2}} (|\downarrow\downarrow\rangle + |\downarrow\uparrow\rangle).
|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle
|\Psi\rangle = \frac{1}{\sqrt{5}}|\uparrow\uparrow\rangle +\frac{\sqrt{2}}{\sqrt{5}} |\downarrow\uparrow\rangle - \frac{\sqrt{2}}{\sqrt{5}} |\downarrow\downarrow\rangle
|\Psi\rangle
|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle
The reason for this definition of entanglement is as described above. If a two-spin state can be written as a product state, it is as if each spin is independent. Measurements on each spin do not affect the other. But if a two-spin state is entangled, the fact that the the state is not a product state means that collapsing to one eigenvector in one space causes collapse in the other space as well. Measurements on one spin thus affect any entangled spins instantaneously by influencing what measurements of the other spin are possible, regardless of the spatial location of the other electron! It is important to note, however, that this is not in violation of special relativity: information cannot be sent faster than lightspeed via entanglement; the effects of entanglement can only be seen after the fact when two observers meet up to compare spin measurements.
Consider the entangled state given by:
|\Psi_E \rangle =\frac{1}{\sqrt{2}} ( |\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle \otimes |\uparrow\rangle),
and the non-entangled state given by
|\Psi_N \rangle = \frac{1}{\sqrt{2}}( |\uparrow\rangle \otimes |\uparrow\rangle + |\uparrow\rangle \otimes |\downarrow\rangle).
Show that measuring the spin of the first particle for
|\Psi_E\rangle
affects the probabilities of finding the second particle in each possible state, while the same measurement for
|\Psi_N\rangle
does not affect the results.
For the entangled state, before measurement the second particle has an equal chance to be found in either the spin-up or spin-down states. After measurement of the first particle's spin with the operator
\hat{S}_z \otimes I
, if the first electron is found in the spin-up state,
|\Psi_E\rangle
|\uparrow \rangle \otimes |\downarrow\rangle
and the second electron must be found in the spin-down state. If the first electron is found in the spin-down state the reverse is true; thus, measurement of the state of the first electron has affected the overall state of the system.
For the non-entangled state, the probability of finding the first particle in the spin-up state is one. After measurement, the state does not change, and the second particle is still in an equal superposition of both spin states. Note that there also exist non-entangled states for which measuring the first particle would change the overall spin wavefunction without changing the state of the second particle.
Cite as: Quantum Entanglement. Brilliant.org. Retrieved from https://brilliant.org/wiki/quantum-entanglement/ |
Topology/Continuity and Homeomorphisms - Wikibooks, open books for an open world
Topology/Continuity and Homeomorphisms
< Topology
← Quotient Spaces Continuity and Homeomorphisms Separation Axioms →
Continuity is the central concept of topology. Essentially, topological spaces have the minimum necessary structure to allow a definition of continuity. Continuity in almost any other context can be reduced to this definition by an appropriate choice of topology.
{\displaystyle X,Y}
be topological spaces.
{\displaystyle f:X\to Y}
{\displaystyle x\in X}
if and only if for all open neighborhoods
{\displaystyle B}
{\displaystyle f(x)}
{\displaystyle A}
{\displaystyle x}
{\displaystyle A\subseteq f^{-1}(B)}
{\displaystyle f:X\to Y}
is continuous in a set
{\displaystyle S}
if and only if it is continuous at all points in
{\displaystyle S}
{\displaystyle f:X\to Y}
is said to be continuous over
{\displaystyle X}
if and only if it is continuous at all points in its domain.
{\displaystyle f:X\to Y}
is continuous if and only if for all open sets
{\displaystyle B}
{\displaystyle Y}
{\displaystyle f^{-1}(B)}
is also an open set.
{\displaystyle \Rightarrow }
{\displaystyle f:X\to Y}
{\displaystyle B}
{\displaystyle Y}
. Because it is continuous, for all
{\displaystyle x}
{\displaystyle f^{-1}(B)}
{\displaystyle x\in A\subseteq f^{-1}(B)}
, since B is an open neighborhood of f(x). That implies that
{\displaystyle f^{-1}(B)}
{\displaystyle \Leftarrow }
The inverse image of any open set under a function
{\displaystyle f}
{\displaystyle Y}
{\displaystyle X}
{\displaystyle x}
{\displaystyle X}
. Then the inverse image of any neighborhood
{\displaystyle B}
{\displaystyle f(x)}
{\displaystyle f^{-1}(B)}
, would also be open. Thus, there is an open neighborhood
{\displaystyle A}
{\displaystyle x}
{\displaystyle f^{-1}(B)}
. Thus, the function is continuous.
If two functions are continuous, then their composite function is continuous. This is because i{\displaystyle f}
{\displaystyle g}
have inverses which carry open sets to open sets, then the inverse
{\displaystyle g^{-1}(f^{-1}(x))}
would also carry open sets to open sets.
{\displaystyle X}
have the discrete topology. Then the map
{\displaystyle f:X\rightarrow Y}
is continuous for any topology on
{\displaystyle Y}
{\displaystyle X}
have the trivial topology. Then a constant map
{\displaystyle g:X\rightarrow Y}
{\displaystyle Y}
Homeomorphism[edit | edit source]
When a homeomorphism exists between two topological spaces, then they are "essentially the same", topologically speaking.
{\displaystyle X,Y}
be topological spaces
{\displaystyle f:X\to Y}
is said to be a homeomorphism if and only if
{\displaystyle f}
{\displaystyle f}
is continuous over
{\displaystyle X}
{\displaystyle f^{-1}}
{\displaystyle Y}
If a homeomorphism exists between two spaces, the spaces are said to be homeomorphic
If a property of a space
{\displaystyle X}
applies to all homeomorphic spaces to
{\displaystyle X}
, it is called a topological property.
A map may be bijective and continuous, but not a homeomorphism. Consider the bijective map
{\displaystyle f:[0,1)\rightarrow S^{1}}
{\displaystyle f(x)=e^{2\pi ix}}
mapping the points in the domain onto the unit circle in the plane. This is not a homeomorphism, because there exist open sets in the domain that are not open in
{\displaystyle S^{1}}
, like the set
{\displaystyle \left[0,{\frac {1}{2}}\right)}
Homeomorphism is an equivalence relation
Prove that the open interval
{\displaystyle (a,b)}
{\displaystyle \mathbb {R} }
Establish the fact that a Homeomorphism is an equivalence relation over topological spaces.
(i)Construct a bijection
{\displaystyle f:[0,1]\to [0,1]^{2}}
(ii)Determine whether this
{\displaystyle f}
Retrieved from "https://en.wikibooks.org/w/index.php?title=Topology/Continuity_and_Homeomorphisms&oldid=2983710"
Book:Topology |
A binomial probability experiment is conducted with the given parameters.
A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials
A binomial probability experiment is conducted with the given parameters. Compute the probability of x successes in the n independent trials of the experiment. n=20, p=0.7, x=19 P(19)=
Compute P(X) using the binomial probability formula. Then determine whether the normal distriution can be used to estimate this probability. If so, approximate P(X) using the normal distriution and compare the result with exact probability.
n=47,p=0.6
X=34
n=47,p=0.6
X=34
, use the binomial probability formula to find P(X).
(Round to four decimal plases as needed.)
Can be normal distriution be used to approximate this probability?
A. Yes, because
\sqrt{np\left(1-p\right)}\ge 10
B. No, because
\sqrt{np\left(1-p\right)}\le 10
C. No, because
np\left(1-p\right)\le 10
D. Yes, because
np\left(1-p\right)\ge 10
A firm reports that 25 percent of its accounts receivable from other business firms are overdue due to economic conditions. If an accountant takes a random sample of seven such accounts, determine the probability of each of the following events by use of the formula for binomial probabilities:
(i) none of the accounts is overdue, and, (ii) exactly two accounts are overdue,
A retailer is receiving a large shipment of media players. In order to determine whether she should accept or reject the shipment, she tests a sample of media players; if she finds at least one defective player, she will reject the entire shipment. If 0.5% of the media players are defective, what is the probability that she will reject the shipment if
a)she tests fifteen media players.
b)she tests thirty media players.
In an experiment, there are n independent trials. For each trial, there are three outcomes, A, B, and C. For each trial, the probability of outcome A is 0.80; the probability of outcome B is 0.10; and the probability of outcome C is 0.10. Suppose there are 10 trials.
(1) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type B, and one of type C? Explain.
a) Yes. Each outcome has a probability of success and failure.
b) No. A binomial probability model applies to only two outcomes per trial.
c) Yes. A binomial probability model applies to three outcomes per trial.
d) No. A binomial probability model applies to only one outcome per trial.
(2) Can we use the binomial experiment model to determine the probability of four outcomes of type A and six outcomes that are not of type A? Explain.
a) Yes. Assign outcome B to "success" and outcomes A and C to "failure."
c) Yes. Assign outcome C to "success" and outcomes A and B to "failure."
d) Yes. Assign outcome A to "success" and outcomes B and C to "failure."
What is the probability of success on each trial?
n=5\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }p=0.4
X=0 |
anon
Defenitely something we need to look at. We managed to work around this in SBT with some dirty tricks (and it still happens but very rarely)
not nice when some code does a casting of a Reporter to some concrete type it expects it to be
I’ve moved the compiler version and clippy version around (couldn’t try the 0.3.0 suggestion as the code is build using 2.12 and used dependencies only available in 2.12 so …) it always comes down to https://github.com/scala/scala/blob/v2.12.1/src/compiler/scala/tools/nsc/Global.scala#L1519 or similar in the compiler during the presentation compiler run.
should InjectReporter be a GlobalPhase or is it somehow ending up in the list improperly?
higher up in the calls stack interactive.Global there’s comments suggesting that sbt does inject phases into compilation and there’s some need for the compiler to be careful about this
In the end I can compile with clippy enabled but not use the presentation compiler … which breaks ensime
@crispywalrus not sure what's the intended meaning of GlobalPhase - but maybe that's something clippy could extend/implement?
Our phases are quite simple
I don’t know what clippy is doing or needs to do in this case. Thing is that it all works when using the compiler proper but not the presenation compiler. I’m convinced this is some subtle difference between the presentation compiler and the compiler proper.
Is there a plan to support Scala 2.13?
Vladyslav Pekker
@agilesteel
Is this room still alive? :) Scala 2.13 is a good question. Is there an answer?
ok I found this softwaremill/scala-clippy#64
@adamw @kciesielski is anyone alive here? :)
I was wondering how to do sth like clippyColorType := Some(ClippyColor.None) without the sbt plugin... I figured out this much: scalacOptions ++= Seq("-P:clippy:colors=true") and then tried P:clippy:color:type=none and a few other variations without luck... any help?
Found it... https://github.com/softwaremill/scala-clippy/blob/master/plugin-sbt/src/main/scala/com/softwaremill/clippy/ClippySbtPlugin.scala#L86
Hi there, sorry for long silence, I'm going to start implementing a version for 2.13 in a few days
@kciesielski Is there a plan to support 2.13? :)
Unfortunately it's still pending :( |
Precise Determination of Uranium Isotopes in Suez Canal Sediment
The streambed sediments of the Suez Canal have been analyzed for determining the natural radionuclides and long-lived radionuclides such as uranium by gamma and alpha spectrometric techniques. The specific activities of 238U series, 232Th series and 40K (Bq/kg) were measured by gamma spectrometry based on Hyper-Pure Germanium detector (HPGe). The average specific activities of 226Ra(238U) series, 232Th series and 40K were ranged from 3.04 ± 1.10 to 14.70 ± 1.24 Bq/kg, from 1.12 ± 0.66 to 16.10 ± 1.30 and from 77 ± 4.90 to 350.50 ± 8.90 Bq/kg respectively. The concentration of 238U and 234U in the streambed sediments are ranged from 3.24 ± 0.21 to 13.34 ± 0.61 ppm and from 3.18 ± 0.02 to 13.77 ± 0.03 ppm in dry weight respectively. 234U/238U ratios of the sediments are relatively lower than unity in many locations indicating the preferential uranium leaching process. The results with the high ratios for 234U/238U were observed in the sediment collected from Port Said. This may be attributed to the sorption of uranium by sediment which has a relatively high content of organic matter. The geochemical behavior of sediment, the chemistry of uranium and the flow rates of water are considered as the most important factors controlling uranium isotopic composition of the streambed sediment. The result of radioactivity in sediment samples can be used to distinguish any future changes due to non-nuclear industries on the Suez Canal area.
Sediment, Uranium, Isotopic Ratios, Natural Radioactivity, Suez Canal, Egypt
{A}_{\text{Nuclide}}=\frac{{N}_{\text{Nuclide}}-{N}_{\text{Background}}}{m\times t\times {Y}_{\text{Tracer}}\times eff}
\text{MDA}=\frac{2.71+3.29\sqrt{{N}_{p}}}{k\ast V\ast t}\left[\text{Bq}/\text{cpm}\right]
k=\frac{N}{t\ast A}
Abdellah, W.M. (2019) Precise Determination of Uranium Isotopes in Suez Canal Sediment. Journal of Analytical Sciences, Methods and Instrumentation, 9, 30-41. https://doi.org/10.4236/jasmi.2019.92004
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Compute aerodynamic forces and moments using aerodynamic coefficients, dynamic pressure, center of gravity, center of pressure, and velocity - Simulink - MathWorks Australia
{C}_{s←b}=\left[\begin{array}{ccc}\mathrm{cos}\left(\alpha \right)& 0& \mathrm{sin}\left(\alpha \right)\\ 0& 1& 0\\ -\mathrm{sin}\left(\alpha \right)& 0& \mathrm{cos}\left(\alpha \right)\end{array}\right]
{C}_{w←s}=\left[\begin{array}{ccc}\mathrm{cos}\left(\beta \right)& \mathrm{sin}\left(\beta \right)& 0\\ -\mathrm{sin}\left(\beta \right)& \mathrm{cos}\left(\beta \right)& 0\\ 0& 0& 1\end{array}\right]
{C}_{w←b}=\left[\begin{array}{ccc}\mathrm{cos}\left(\alpha \right)\mathrm{cos}\left(\beta \right)& \mathrm{sin}\left(\beta \right)& \mathrm{sin}\left(\alpha \right)\mathrm{cos}\left(\beta \right)\\ -\mathrm{cos}\left(\alpha \right)\mathrm{sin}\left(\beta \right)& \mathrm{cos}\left(\beta \right)& -\mathrm{sin}\left(\alpha \right)\mathrm{sin}\left(\beta \right)\\ -\mathrm{sin}\left(\alpha \right)& 0& \mathrm{cos}\left(\alpha \right)\end{array}\right]
{F}_{A}^{w}\equiv \left[\begin{array}{c}-D\\ -C\\ -L\end{array}\right]={C}_{w←b}\cdot \left[\begin{array}{c}{X}_{A}\\ {Y}_{A}\\ {Z}_{A}\end{array}\right]\equiv {C}_{w←b}\cdot {F}_{A}^{b} |
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