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https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-8-techniques-of-integration-section-8-3-trigonometric-integrals-exercises-8-3-page-462/14 | 1,679,891,340,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00504.warc.gz | 852,970,992 | 13,824 | ## Thomas' Calculus 13th Edition
$\frac{\pi}{4}$
Recall the identity $\cos 2x=1-2\sin^{2} x$ which gives $\sin^{2}x=\frac{1-\cos 2x}{2}$. Therefore, $\int \sin^{2}x\,dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos 2x\,dx$ $= \frac{x}{2}-\frac{1}{4}\sin 2x+C$ Then $\int_{0}^{\pi/2} \sin^{2}x\,dx=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi/2}_{0}=(\frac{\pi}{4}-0)-0=\frac{\pi}{4}$ | 185 | 370 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2023-14 | latest | en | 0.32245 |
https://slideplayer.com/slide/5677869/ | 1,709,383,610,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00461.warc.gz | 539,631,567 | 22,106 | # Linear Inequalities in Two Variables
## Presentation on theme: "Linear Inequalities in Two Variables"— Presentation transcript:
Linear Inequalities in Two Variables
The graph of the solutions of a linear inequality in two variables is a half-plane. Procedure for Graphing Linear Inequalities 1. Rewrite the linear inequality as a linear equation. Change the inequality symbol ( < , > , , or ) to an equal sign. 2. The next step is to graph that line. If the inequality symbol has an equal sign ( or ), draw a solid line. If the inequality symbol does not have an equal sign (< , >), draw a dashed line. Pick any ordered pair that is not on the graphed line. This will be your test point. Substitute the coordinates of that ordered pair into the original inequality. If the inequality is true, shade the side of the line where the test point is. If the inequality is false, shade the other side of the line. Next Slide
x y 0 -4 Example 1. Graph: 3x +2y < 6 Solution:
x axis y axis Example 1. Graph: 3x +2y < 6 Solution: Write the inequality as a linear equation, then graph. 3x + 2y = 6 Pick a test point not on the line, say (0,0). Since this is true, we can shade the side where the test point is. x y 0 -4 3 0 y axis x axis
Write the inequality as a linear equation, and graph.
Example 2. Graph y > 2x Solution: x axis y axis Write the inequality as a linear equation, and graph. y = 2x Pick a test point, say (0,4). Since this is false, we must shade the side opposite the test point. Note: There is another method to determining which side of the line to shade. If the inequality is y> or y≥, shade above the line. If the inequality is y< or y≤, shade below the line. This example is y>2x, therefore we shade above the line. Using this method takes a little work to get y by itself but then you don’t have to worry about the test point. y axis x axis
Since this is false, shade on the opposite side of the test point.
x axis y axis Since this is false, shade on the opposite side of the test point. Note: If the inequality is x> or x≥, shade to the right of the line. If the inequality is x< or x≤, shade to the left of the line. This example is x≤−4, therefore we shade to the left of the line. x axis y axis
Because our inequality is y >, shade above the dashed line.
x axis y axis 1st, write the inequality as linear equation to graph the line. Either use the y intercept and the slope or choose values for x. If you choose values for x, always choose x=0 and also choose values that are multiples of the denominator in the fraction. Because our inequality is y >, shade above the dashed line. x axis y axis
The shaded area will be to the left.
x axis y axis Recall that the word “and” indicates the intersection of the solutions sets of each inequality. The shaded area will be to the left. Answer The shaded area will be below the line. The intersection is the area that satisfied both inequalities. (shaded twice) x axis y axis
In the previous sections, we solved systems of equations such as:
The solution set of the system is the intersection of the solution sets of the individual lines. (i.e., where the two lines meet.) The previous example contained the word “and”, which means the intersection. Another way of asking for the intersection is using a system such as: The solution set of the system is the intersection of the solution sets of the individual inequalities. (i.e., where the graph is shaded twice.) Next Slide
Example 6. Solve the following system by graphing.
Solution: x axis y axis Next Slide Since y >, shade above. The solution is the area which was shaded twice. We will darken that area with another color to show the intersection more clearly. Since y ≥, shade above.
Solve the following system by graphing.
Your Turn Problem #6. Solve the following system by graphing. x axis y axis Answer:
Example 7. Solve the following system by graphing.
Solution: x axis y axis Therefore, our graph will only be shaded in Quadrant I, the upper right-hand quadrant. The solution is the area in Quadrant I which satisfies all four inequalities, shown here in the light blue color. Easier to graph last two inequalities using the intercept method. Then use test points to determine shaded area. Next Slide
Your Turn Problem #7. Graph: Answer: The End. B.R. 1-25-07 y axis
x axis y axis (4,0) (0,4) (7,0) (0,6) Answer: The End. B.R.
Download ppt "Linear Inequalities in Two Variables"
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https://anhngq.wordpress.com/2009/02/11/solving-initial-value-problem-for-wave-equation-via-fourier-transform/ | 1,643,006,001,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00666.warc.gz | 154,939,951 | 34,080 | # Ngô Quốc Anh
## February 11, 2009
### Solving initial value problem for wave equation via Fourier transform
Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:43
In this topic, we will show you how can we use Fourier transform to solve initial value problem for wave equation in $\mathbb R$. Following is the problem
$\displaystyle \left\{ \begin{gathered} {u_{tt}} = {u_{xx}}, \qquad x \in \mathbb{R}, \hfill \\ u\left( {x,0} \right) = \varphi \left( x \right), \hfill \\ {u_t}\left( {x,0} \right) = \psi \left( x \right). \hfill \\ \end{gathered} \right.$
From the equation by taking Fourier transform to the both sides, we obtain
$\displaystyle \widehat{{u_{tt}}\left( {\eta ,t} \right)} - {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0$
This is an ODE, the solution is given by
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = A\sin \left( {\eta t} \right) + B\cos \left( {\eta t} \right)$
From the initial date we get
$\widehat{u\left( {\eta ,0} \right)} = \widehat{\varphi \left( \eta \right)}$
which implies that
$\widehat{\varphi \left( \eta \right)} = B$.
From the initial date we see that
$\widehat{{u_t}\left( {\eta ,0} \right)} = \widehat{\psi \left( \eta \right)} = A\eta$.
Thus, we obtain
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = \frac{{\widehat{\psi \left( \eta \right)}}} {\eta}\sin \left( {\eta t} \right) + \widehat{\varphi \left( \eta \right)}\cos \left( {\eta t} \right)$.
Note that
$\displaystyle \cos \left( {\eta t} \right) = \frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}, \quad \frac{{\sin \left( {\eta t} \right)}} {\eta } = \frac{{{e^{i\eta t}} - {e^{ - i\eta t}}}} {{2i\eta }} = \frac{1} {2}\int\limits_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta }$.
Moreover,
$\displaystyle \widehat{\delta ( {x - \alpha t} )} = \int\limits_{ - \infty }^\infty {{e^{ - i\eta x}}\delta ( {x - \alpha t} )dx} = {e^{ - i\alpha \eta t}}\int\limits_{ - \infty }^\infty {{e^{ - i\eta ( {x - \alpha t} )}}\delta ( {x - \alpha t} )dx} = {e^{ - i\alpha \eta t}}$.
Then
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = \left( {\frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}} \right)\widehat{\varphi \left( \eta \right)} + \left( {\frac{1} {2}\int\limits_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta } } \right)\widehat{\psi \left( \eta \right)}$.
Since
$\displaystyle \left( {\frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}} \right)\widehat{\varphi \left( \eta \right)} = \frac{{\widehat{\delta \left( {x + t} \right)*\varphi \left( x \right)} + \widehat{\delta \left( {x - t} \right)*\varphi \left( x \right)}}} {2}$
and
$\displaystyle \left( {\frac{1} {2}\int_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta } } \right)\widehat{\psi \left( \eta \right)} = \frac{1} {2}\int\limits_{ - t}^t {\widehat{\delta \left( {x + \theta } \right)*\psi \left( x \right)}d\theta }$
then
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = \frac{{\widehat{\delta \left( {x + t} \right)*\varphi \left( x \right)} + \widehat{\delta \left( {x - t} \right)*\varphi \left( x \right)}}} {2} + \frac{1} {2}\int\limits_{ - t}^t {\widehat{\delta \left( {x + \theta } \right)*\psi \left( x \right)}d\theta }$.
Thus,
$\displaystyle u\left( {x,t} \right) = \frac{{\delta \left( {x + t} \right)*\varphi \left( x \right) + \delta \left( {x - t} \right)*\varphi \left( x \right)}} {2} + \frac{1} {2}\int\limits_{ - t}^t {\delta \left( {x + \theta } \right)*\psi \left( x \right)d\theta }$
or equivalently,
$\displaystyle u\left( {x,t} \right) = \frac{{\varphi \left( {x + t} \right) + \varphi \left( {x - t} \right)}} {2} + \frac{1} {2}\int\limits_{x - t}^{x + t} {\psi \left( y \right)dy}$.
This is the so-call D’ Alembert formula.
1. Bài này nếu giải bằng hàm Green thì thế nào NQA nhỉ ?
Comment by viettran — March 24, 2009 @ 20:27
• Nếu thế cần phải tìm hàm Green đã, việc này chắc ko dễ 😦
Comment by Ngô Quốc Anh — December 14, 2009 @ 17:25
2. hi , my name nguyen , i come from USA , nice to meet you ,
Comment by vo khanh nguyen — December 31, 2009 @ 16:54
• Thanks for coming to my blog 🙂
Comment by Ngô Quốc Anh — January 2, 2010 @ 15:20
3. In the odd-dimensional space ($n \geq 3$)
http://arxiv.org/abs/0904.3252
and in the even case, implies by using Hadamard’s method of descent.
Comment by Tuan Minh — January 14, 2010 @ 23:40
• Aha, that’s a good and new work, thanks Minh 🙂
Comment by Ngô Quốc Anh — January 14, 2010 @ 23:44
4. In fact, this idea is mentioned in the well-known book of Stein in the case $n=3$ (by using FT/Bessel function), Stein also gave an exercise for the general case. The Torchinsky’s inverse formula for $\dfrac{\sin(R\xi)}{|\xi|}$ is nice!
Comment by Tuan Minh — January 15, 2010 @ 1:53
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,826 | 4,852 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 19, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2022-05 | latest | en | 0.541982 |
https://libraryguides.centennialcollege.ca/c.php?g=645085&p=5124707 | 1,726,203,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00606.warc.gz | 331,578,783 | 12,542 | # Math help from the Learning Centre
This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.
## Introduction
Question: Would you rather receive $10,000 a day for a month or one penny ($0.01) that doubles every day for a month?
You might choose the $10,000 at first, but you'll see that with the magic of compounding, the doubling penny option would earn you a whopping$10,737,418 while the $10,000 every day would only make you$310,000 richer. This means that the penny would earn almost 35X the amount that the $10,000 did! ## Terminology & Formulas Interest that is earned on both the initial principal amount and the interest accrued in previous periods during the investment/loan term is called compound interest. In other words, it can be seen as "interest on interest" - hence the name compound interest. Unlike simple interest, compound interest grows exponentially (while simple interest grows linearly), making it a common method of earning interest when investing or borrowing money. To calculate compound interest, we have to consider the original amount of money, called principal (present value), the time (period over which the money is being used), the length and number of compounding periods (details below), and the (compound) interest rate. Taking these factors into account, we can use the following formula to calculate the maturity/future value of an amount earning compound interest: $$S=P(1+i)^n$$, where: • $$S$$ is the maturity/future value, • $$P$$ is the principal/present value, • $$i$$ is the periodic rate of interest, and • $$n$$ is the total number of compounding periods. Tip: We can also rearrange the above formula to solve for $$P$$, the principal (present value): $$P=S(1+i)^{-n}$$ Compounding Periods The following table shows common compounding frequencies/periods, their durations, how many times each period occurs per year, and how to calculate the interest rates per compounding period given the annual interest rate: ## Application Examples The following examples will demonstrate how we can use the formulas introduced in the previous section. Example 1 Example: Marina is deciding between two investment opportunities for the next six years. Investment A pays 8.4% p.a. compounded bi-weekly and Investment B pays 7.5% p.a. compounded monthly. If she has$25,000 to invest, which investment option should she choose and how much more interest will she earn for that option?
Solution
First, we can identify the information that is given in the problem. For both options, we have:
$$P=25,000, t=6$$ years
Investment A Investment B Given information: $$i=\frac{0.084}{26}, n=26 \times 6 = 156$$ Calculation: $$S=25000(1+\frac{0.084}{26})^{156}=41349.63$$ Maturity Value: $$41,349.63$$ Interest Earned: $$41,349.63-25,000=$$ $$16,349.63$$ Given information: $$i=\frac{0.075}{12}=0.00625,n=12 \times 6=72$$ Calculation: $$S=25000(1+0.00625)^{72}=39152.94$$ Maturity Value: $$39,152.94$$ Interest Earned: $$39,152.94-25,000=$$ $$14,152.94$$
Comparing the two maturity values we get for both options, we find that Marina should choose Investment A, and she will earn $$16,349.63-14,152.94=$$ $$2196.69$$ more in interest with this option.
Example 2
Example: What amount must be invested at 4.75% compounded quarterly for 6 years to reach a maturity value of $15,000? Solution Since we're looking for the amount invested, we are determining the principal amount or $$P$$. First, we can identify the information that is given in the problem: $$S=15,000, i=\frac{0.0475}{4}=0.011875, n=6\times 4=24$$ Substituting these values into the compound interest formula gives: $$15000=P(1+0.011875)^{24}$$ Then, we can isolate for $$P$$: $$P=\frac{15000}{(1+0.011875)^{24}}$$ $$=11299.17$$ So, $$11, 299.17$$ must be invested to reach the required maturity value. Example 3 Example: Debts of$1250 due 6 months from today and \$4725 due 4 years from today are to be settled by a single payment 1.5 years from today. If money is worth 8.75% compounded bi-annually, what is the size of the single equivalent payment?
Solution
See the video below for the solution!
Tip: This solution will use the ideas of equivalent values and setting focal dates (To review these concepts, see Equivalent Values). | 1,118 | 4,377 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-38 | latest | en | 0.936535 |
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Activity Discussion Math Math Reply To: Math
• ### Aashutosh
Member
June 2, 2021 at 8:22 pm
0
Pythagoras theorem states that “The square of the hypotenuse side is equal to the sum of squares of the other two sides in a right-angle triangle“. The longest side of the right-angle triangle is called the hypotenuse. Because it is opposite to the angle of 90 degrees. The other two sides are perpendicular and base.
Let us consider the right-angle triangle ABC with 90 degrees at B and draw a perpendicular BD meeting AC at D as shown in the photo.
According to the question: △ADB ~ △ABC
Therefore, AD/AB=AB/AC (corresponding sides of similar triangles)
Or, AB^2 = AD × AC ……………………………..……..(1)
Also, △BDC ~△ABC
Therefore, CD/BC=BC/AC (corresponding sides of similar triangles)
Or, BC^2= CD × AC ……………………………………..(2)
Adding the equations (1) and (2) we get,
AB^2 + BC^2 = AD × AC + CD × AC
AB^2 + BC^2 = AC (AD + CD)
Since, AD + CD = AC
Therefore, AC^2 = AB^2 + BC^2
Hence, the Pythagoras theorem is proved.
+ | 306 | 1,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2023-40 | latest | en | 0.855592 |
https://themathmompuzzles.blogspot.com/2014/01/a-puzzle-that-99-will-fail-to-answer.html?showComment=1403624209783 | 1,618,493,083,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00121.warc.gz | 618,953,374 | 22,772 | ## Friday, January 10, 2014
### A puzzle that 99% will fail to answer.
This came from my son who got it from someone else at school.
I solved it quickly and texted him the answer just as I was getting into my car for work.
He wrote: "Wrong" and sent me a ridiculous solution.
I typed: "Ohh boy, how could you miss it! I will beat you up at home:)" and started driving.
Stuck in traffic I thought back to it and the 99% that are getting it wrong and suddenly realized that his answer is in fact correct!
I texted: "Apology! Now that my head has cooled off I see that you were right."
He wrote: "Hahahahahaha".
No one at my work got it right either. Could you?
There is at least two solutions. Let's call them: my and my son's. Both are kind of correct. But his is more original.
Your thoughts and suggestions are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.
Jody said...
26?
Mike said...
5=26! 2+5+3=10; 3+10+4=17; 4+17+5=26....
Leah said...
My first thought was to solve it as an algorithm and I got 5=26. x*x+1 (2*2+1=5, 3*3+1=10, 4*4+1=17)
But then, duh! 2=5, so 5=2! Admittedly, my son noticed that answer before me!
Anonymous said...
5=2; isn't an equal sign reflexive!?
Laura W said...
SteveGoodman18 said...
Most will answer 26, as the pattern appears to be 1 more than the square of the number. However, if the reflexive property holds, if 2 = 5, then 5 = 2.
tom said...
The two answers I like are both 26.
Solution #1 is "the square, plus one."
Solution #2 is just the series, +5, +7, and then +9.
renee said...
I'm going with 26 (x=y2 + 1), which I'm also guessing was your answer.
Well, the trap that you presumably fell into is noticing the pattern in each pair of numbers (call them A and B) - each 'B' number is 1 higher than the square of the 'A' number (so b = a^2 +1).
However, since the first thing we are told is '2 = 5' we therefore know that '5 = 2' ^_^
Jerome said...
2 or 26
25 because the pattern is n^2 + 1
2 because an equal sign means that if 2 is equivalent to 5 then 5 must be equivalent to 2.
I don't consider either one of them correct. Either the number system is getting tortured (2 and 5 are not equivalent) or the equal sign is. The whole premise of 2 being the same as 5 or 5 the same as 26 is a disobedience of the standard meaning of =.
But then I belong to the Bah Humbug school of thought. I might even be a charter member.
SN said...
Solution 1:
5 = 26 (square the number and add 1 to it)
Solution 2:
5 = 2 (from line 1)
Anonymous said...
Maria.
I imagine that my answer is like yours. I got 19. Each number increases by 5 plus 2 above the preceding one.
Gurubandhu
Katrina said...
I solved it as 26. If you look at the sequence, the answer seems to be adding the previous equations numbers, plus the current equations number i.e. 2 = 5, 3 = 10 (2+3+5), 4=17 (3+4+10), 5 = 26 (4+5+17).
Showing this to my 11 yo, he said that 5 = 2 because it said it in the first equality.
Anonymous said...
Nicolas said...
I would be tempted to say that since the first line states 2=5, we should also have 5=2.
Anonymous said...
5 could equal 2 since you already said that once. But the next number could also be 26 since each number on the right is obtained by adding the two numbers on the left to it. 10 = 5 +(2+3), 17 = 10 + (3+4), so 26 = 17 + (4+5).
So 2 or 26, take your pick.
Anonymous said...
My answer is 26. And I got it by squaring the first number plus one.
Jessica K. via email
Anonymous said...
Kim via email:
Hi Maria,
Been busy, but saw today's email, and gave it a look. Plus, I see Ilya (who I know) is only one answer behind me!
I posted online, but I seem to recall that when I was doing these every week, my posts didn't always go through, so here's my answer:
26. In each case, the number to the right of the equals sign = the sum of the numbers in the equation above it and the number to the left of the equals sign (e.g., 2+5+3 = 10; 3+10+4=17, so 4+17+5=26)
BTW, my answer is also works for x^2+1.
Which is not a coincidence.
What we have is:
x --> x^2 + 1
so the next line is x+1 --> (x+1)^2 + 1 = x^2 + 2x + 2
and it would also equal (summing the three components), x + x^2 + 1 + (x+1) -> x^2 + 2x + 2!
So, I'm assuming that's what you got. What did your son get?
a bit later in another email:
The number on the left side is the number of prime numbers to add.
The first five prime numbers are 2, 3, 5, 7, 11
N sum of first n prime numbers
2 5
3 10
4 17
5 28
and a few hours later:
Still thinking about you. I have one other possible answer, and I bet THIS is your son's:
We know from the first equation that 2=5. Therefore, 5=2.
Susan said...
I keep coming up with 26.
A. If the column on the right is the addition of progression of odd numbers, starting with +5, then +7, the next would be 17+9, or 26.
B. If the pattern is Y=X^2 +1, then 5^2 + 1 is 26.
Jerome's wife said...
I tried both directions in attempting to come to the sums of each number; the horizontal and the vertical summations. The horizontal attempt fell through as I could not establish a pattern, but vertically, travelling downward from each number to the next I did see a pattern. The answer is 5 = 26. I noticed that from sum 5 to sum 10 were 5 steps or increments and from sum 10 to sum 17 were 7 steps or increments. Following through to arrive at the answer of 26 I added 9 steps or increments to 17. I knew this was correct because then testing the theory by noticing that there are 2 increments between 5 and 7 increments, it is logical to assume that there is also 2 more increments, to make the steps look like 5,7 and 9. Then 9 is correct by adding 17 to 9 equals 26.
Ilya said...
The notation with equality sign is somewhat strange, I interpreted it as "function of", i.e. F(2)=5, etc. One solution that fits is F(x)=x*x+1, leading to F(5)=26.
Anonymous said...
My first answer is 26. All the numbers are squared and increased by 1.
While trying to think of another answer, the only one that makes sense at the moment is: if 2 = 5 then 5 should be equal to 2.
Lulu
Annie said...
I would say there are two answers:
If it is looked at as a progression then the answer would be 26. (Add 5, then 7, then 9).
If each statement is looked at as a separate fact, then the answer would be 2, 2=5, so 5= 2!
Maria said...
Ok. As you see there are two very different answers. 26 and 2.
The easier one is 2 and it is harder to find it. But it is written up there on the first line that 2=5, so it should be that 5=2.
2 = 5
3 = 10
4 = 17
5 = ?
We see that mathematically it is wrong because 2 is not equal to 5 etc. So someone is lying: either numbers or the "=" sign. We are either in a kingdom where = is not equal but more like correspondence sign but numbers are real numbers. Or we are in a kingdom where numbers are not what we expect them to be but "=" is equal.
In the first kingdom, we all rushed to find the pattern that led us to the answer 26. And there are a few ways to find it!
In the second kingdom , "=" is "=" and if 2=5 then 5=2!
Pretty amazing that quite a few of you saw both answers. It is very difficult to change state of mind and move between those kingdoms. Those who got both answers will receive 2 puzzle points, those who got one will receive 1 point.
Gurubandhu - I am sorry but I do not understand the 19th. Your solution still comes to 26 if I understood it correctly.
Have a great week!
Jerome said...
I have to confess that I never would have found the reflexive property had you not said there were 2 answers. These problems you come up with are just wonderful even if I do crab about them sometimes.
Anonymous said...
32
mohamed el fouhil said...
If 2=5
then 5=2
mohamed el fouhil said...
If 2=5
Then 5=2
Anonymous said...
Interesting! I came up with 26, but, it's an odd use of equal signs, as has been pointed out. Missed the reflexivity solution completely.
As far as the "other solution" goes, I kept looking to the odd language on the original poster image. In fact if we're being quite logical about it, "99% will fail to answer" means that only 1% (presumably of people who see the puzzle) will actually hazard an answer, whether correct or not. It doesn't say that the 99% failed to answer _correctly_, after all.
I also notice that the puzzle starts with the word "if" but there's no "then," or perhaps the "then" is implied. Otherwise a wise guy could say "there's no answer because there's no question!"
All-in-all an interesting mishmash of conventions and assumption-baiting.
Margaret and Fiona
Unknown said...
In fact other than 2 and 26 even 24 is a possible answer!! Let me explain it..
2=5 -----> 2+3=5 and 3 is a prime number
3=10----->3+7=10 and 7 is a prime number and observe the pattern of prime numbers added here .. after 3, 5 is not added.
4=17----> 4+13=17 13 is a prime number and in between 7 and 13 , prime number 11 is left.
now, 5=?
5=5+19=24.. leaving prime number 17 in between 13 and 19..
But, answer 2 is more convincing. | 2,583 | 9,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2021-17 | latest | en | 0.977077 |
http://www.creatingrealmathematicians.com/year-planner-years-3-4 | 1,656,510,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00283.warc.gz | 76,605,995 | 13,942 | ### Year Planner. Years 3/4
Year Planner Years 3 & 4 Terms 1-4
Each term revisits concepts and teaching and learning takes individual students to higher level of understanding.
Individual student growth documented against VELS Progression Points for each topic each term.
Addition Strategies include Counting forwards on 100’s Chart, Open Number Lines, Partial Sums addition, Counting on Addition using Place Value (Jump strategy). Subtraction Strategies include Counting Backwards on 100’s Chart, Open Number Lines, Counting Up or Counting Back to find difference, Counting Up using Place Value. Multiplication Strategies include Making Arrays, Multiplication as Repeated Addition. Partial Products Multiplication and the Area Model for Multiplication (box method). Division Strategies include division as sharing, what to do with left overs, sharing one place value at a time (left to right). Inverse of multiplication (2x?=6). Fractions Making explicit links to division and multiplication. (½ of 6 is 3, 6÷2=3, 2x3=6, 3+3=6, 6-3-3=0). Denominator is number of groups shared between. Numerator number of groups selected. (Group is the whole & One is the whole). Shape Location Chance Data Measurement. Decimals & Percent Recognise and describe polygons students sort lines, shapes and solids according to key features. Recognise and describe the directions of lines as vertical, horizontal or diagonal. They use grid references (for example, B5 on a street directory) to specify location and compass bearings to describe directions. Locate and identify places on maps and diagrams. They give travel directions and describe positions using simple compass directions (for example, N for North) and grid references on a street directory. Investigate natural variability in chance events and order them from least likely to most likely. Compare the likelihood of everyday events (for example, the chances of rain and snow). They describe the fairness of events in qualitative terms (likely, unlikely) Conduct experiments and collect data to construct simple frequency They use simple two-way tables (karnaugh maps) to sort non-numerical data. Measure the attributes of everyday objects and events using formal (for example, metres and centimetres) and informal units(for example, pencil lengths). Recognise and name common three-dimensional shapes such as spheres, prisms and pyramid. Identify edges, vertices and faces. Use nets to create three-dimensional shapes and explore them by counting edges, faces and vertices They use local and larger-scale maps to locate places and describe suitable routes between them. Plan and conduct chance experiments (for example, using colours on a spinner) and display the results of these experiments. Interpret timetables and calendars in relation to familiar events. Tell the time using analogue and digital clocks and relate familiar activities to the calendar Visualise and draw simple solids as they appear from different positions. Use two-dimensional nets, cross-sections and simple projections to represent simple three-dimensional shapes. Recognise different types of data: non-numerical (categories), separate numbers (discrete) or points on an unbroken number line (continuous). Explore the concept of angle as turn (for example, using clock hands) and as parts of shapes and objects (for example, at the vertices of polygons). Recognise angles are the result of rotation of lines with a common end-point Investigate simple transformations (reflections, slides and turns) to create tessellations and designs. Follow instructions to produce simple tessellations (for example, with triangles, rectangles, hexagons) and puzzles such as tangrams. Use a column or bar graph to display the results of an experiment (for example, the frequencies of possible categories). · Contexts include Whole Number, Length, Time, Mass, Volume, Money. · Money used to develop concepts of Decimals within Operations. · All strategies emphasise use of Place Value components. · Teachers need to make explicit the links between the four operations and Fractions. · All activities emphasise the use of Number in student relevant authentic open contexts. · Differentiation built into every lesson extending understanding for all students.
Warm Up Activities emphasise the following throughout the year.
Patterns (Counting and Shape), Probability games, Automatic recall of simple number facts (Doubling, adding two one digit numbers, compliments of ten, odd and even, etc), Matching Analogue, Digital and Written time.
Problem Solving activities aligned to topics under investigation presented to students weekly to further develop Proficiency Strands of Australian Curriculum (Understanding, Fluency, Problem Solving & Reasoning). Resources include Exemplars, Pictures/Numbers/Words.com)
Literacy emphasised for each topic. Students to become fluent in vocabulary of each topic.
Written Share/Reflections of developed understandings are standard practice and completed at the end of every lesson.
All lessons to explicitly identify to students the Learning Intentions for each lesson.
## Mathematics - Level 3
### Learning focus
As students work towards the achievement of Level 3 standards in Mathematics, they recognise and explore patterns in numbers and shape. They increasingly use mathematical terms and symbols to describe computations, measurements and characteristics of objects.
In Number, students use structured materials to explore place value and order of numbers to tens of thousands. They skip count to create number patterns. They use materials to develop concepts of decimals to hundredths. They use suitable fraction material to develop concepts of equivalent fraction and to compare fraction sizes. They apply number skills to everyday contexts such as shopping. They extend addition and subtraction computations to three digit numbers. They learn to multiply and divide by single digit numbers.
In Space, students sort lines, shapes and solids according to key features. They use nets to create three-dimensional shapes and explore them by counting edges, faces and vertices. They visualise and draw simple solids as they appear from different positions. They investigate simple transformations (reflections, slides and turns) to create tessellations and designs. They explore the concept of angle as turn (for example, using clock hands) and as parts of shapes and objects (for example, at the vertices of polygons). They use grid references (for example, A5 on a street directory) to specify location and compass bearings to describe directions. They use local and larger-scale maps to locate places and describe suitable routes between them.
In Measurement, chance and data, students measure the attributes of everyday objects and events using formal (for example, metres and centimetres) and informal units(for example, pencil lengths). Students tell the time using analogue and digital clocks and relate familiar activities to the calendar. Students investigate natural variability in chance events and order them from least likely to most likely. Students conduct experiments and collect data to construct simple frequency graphs. They use simple two-way tables (karnaugh maps) to sort non-numerical data.
In Structure, students use structured material (in tens, hundreds and thousands) to develop ideas about multiplication by replication and division by sharing. They recognise the possibility of remainders when dividing. They learn to use number properties to support computations (for example, they use the commutative and associative properties for adding or multiplying three numbers in any order or combination). They investigate the distributive property to develop methods of multiplication and division by single digit whole numbers. They learn to use and describe simple algorithms for computations. They use simple rules to generate number patterns (for example, ‘the next term in the sequence is two more than the previous term’). They create and complete number sentences using whole numbers, decimals and fractions.
When Working mathematically, students use mathematical symbols (for example, brackets, division and inequality, the words and, or and not). Students develop and test ideas (conjectures) across the content of mathematical experience. For example:
• in Number, the size and type of numbers resulting from computations
• in Space, the effects of transformations of shapes
• in Measurement, chance and data, the outcomes of random experiments and inferences from collected samples.
Students learn to recognise practical applications of mathematics in daily life, including shopping, travel and time of day. They identify the mathematical nature of problems for investigation. They choose and use learned facts, procedures and strategies to find solutions. They use a range of tools for mathematical work, including calculators, computer drawing packages and measuring tools.
### National Statements of Learning
This learning focus statement, with the following elaboration, incorporates the Year 3 National Statement of Learning for Mathematics.
Elaboration:
They recognise angles … as parts of shapes and objects …
### Standards
#### Number
At Level 3, students use place value (as the idea that ‘ten of these is one of those’) to determine the size and order of whole numbers to tens of thousands, and decimals to hundredths. They round numbers up and down to the nearest unit, ten, hundred, or thousand. They develop fraction notation and compare simple common fractions such as 3/4 > 2/3 using physical models. They skip count forwards and backwards, from various starting points using multiples of 2, 3, 4, 5, 10 and 100.
They estimate the results of computations and recognise whether these are likely to be over-estimates or under-estimates. They compute with numbers up to 30 using all four operations. They provide automatic recall of multiplication facts up to 10 × 10.
They devise and use written methods for:
• whole number problems of addition and subtraction involving numbers up to 999
• multiplication by single digits (using recall of multiplication tables) and multiples and powers of ten (for example, 5 × 100, 5 × 70 )
• division by a single-digit divisor (based on inverse relations in multiplication tables).
They devise and use algorithms for the addition and subtraction of numbers to two decimal places, including situations involving money. They add and subtract simple common fractions with the assistance of physical models.
#### Space
At Level 3, students recognise and describe the directions of lines as vertical, horizontal or diagonal. They recognise angles are the result of rotation of lines with a common end-point. They recognise and describe polygons. They recognise and name common three-dimensional shapes such as spheres, prisms and pyramids. They identify edges, vertices and faces. They use two-dimensional nets, cross-sections and simple projections to represent simple three-dimensional shapes. They follow instructions to produce simple tessellations (for example, with triangles, rectangles, hexagons) and puzzles such as tangrams. They locate and identify places on maps and diagrams. They give travel directions and describe positions using simple compass directions (for example, N for North) and grid references on a street directory.
#### Measurement, chance and data
At Level 3, students estimate and measure length, area, volume, capacity, mass and time using appropriate instruments. They recognise and use different units of measurement including informal (for example, paces), formal (for example, centimetres) and standard metric measures (for example, metre) in appropriate contexts. They read linear scales (for example, tape measures) and circular scales (for example, bathroom scales) in measurement contexts. They read digital time displays and analogue clock times at five-minute intervals. They interpret timetables and calendars in relation to familiar events. They compare the likelihood of everyday events (for example, the chances of rain and snow). They describe the fairness of events in qualitative terms. They plan and conduct chance experiments (for example, using colours on a spinner) and display the results of these experiments. They recognise different types of data: non-numerical (categories), separate numbers (discrete), or points on an unbroken number line (continuous).They use a column or bar graph to display the results of an experiment (for example, the frequencies of possible categories).
#### Structure
At Level 3, students recognise that the sharing of a collection into equal-sized parts (division) frequently leaves a remainder. They investigate sequences of decimal numbers generated using multiplication or division by 10. They understand the meaning of the ‘=’ in mathematical statements and technology displays (for example, to indicate either the result of a computation or equivalence). They use number properties in combination to facilitate computations (for example, 7 + 10 + 13 = 10 + 7 + 13 = 10 + 20). They multiply using the distributive property of multiplication over addition (for example, 13 × 5 = (10 + 3) × 5 = 10 × 5 + 3 × 5). They list all possible outcomes of a simple chance event. They use lists, venn diagrams and grids to show the possible combinations of two attributes. They recognise samples as subsets of the population under consideration (for example, pets owned by class members as a subset of pets owned by all children). They construct number sentences with missing numbers and solve them.
#### Working mathematically
At Level 3, students apply number skills to everyday contexts such as shopping, with appropriate rounding to the nearest five cents. They recognise the mathematical structure of problems and use appropriate strategies (for example, recognition of sameness, difference and repetition) to find solutions.
Students test the truth of mathematical statements and generalisations. For example, in:
• number (which shapes can be easily used to show fractions)
• computations (whether products will be odd or even, the patterns of remainders from division)
• number patterns (the patterns of ones digits of multiples, terminating or repeating decimals resulting from division)
• shape properties (which shapes have symmetry, which solids can be stacked)
• transformations (the effects of slides, reflections and turns on a shape)
• measurement (the relationship between size and capacity of a container).
Students use calculators to explore number patterns and check the accuracy of estimations. They use a variety of computer software to create diagrams, shapes, tessellations and to organise and present data. | 2,940 | 14,866 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2022-27 | longest | en | 0.855621 |
https://schoollearningcommons.info/question/how-to-prove-a-2-b-2-a-2-b-2-ab-24945330-98/ | 1,632,715,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00295.warc.gz | 519,760,616 | 13,157 | ## How to prove (a /2+ b)^2(a /2- b)^2= ab
Question
How to prove (a /2+ b)^2(a /2- b)^2=
ab
in progress 0
1 month 2021-08-21T22:25:28+00:00 1 Answer 0 views 0
We can show that the right side of the equation is equal to the left side, by expanding it out and then factorising:
( a+b)2+(a−b)2
= (a+b)(a+b)+(a−b)(a−b)
= (a2+ab+ab+b2)+(a2−ab−ab+b2)
= a2+2ab+b2+a2−2ab+b2
= 2a2+2b2
= 2(a2+b2)
Therefore, it is shown that
2(a2+b2)=(a+b)2+(a−b)2 | 210 | 453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.694171 |
https://mathoverflow.net/questions/240166/inequalities-for-marginals-of-distribution-on-hyperplane | 1,718,476,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00696.warc.gz | 344,710,598 | 24,854 | # Inequalities for marginals of distribution on hyperplane
Let $H = \{ (a,b,c) \in \mathbb{Z}_{\geq 0}^3 : a+b+c=n \}$. If we have a probability distribution on $H$, we can take its marginals onto the $a$, $b$ and $c$ variables and obtain three probability distributions $\alpha$, $\beta$ and $\gamma$ in $\mathbb{R}^{n+1}$. The set of possible $\alpha$, $\beta$ and $\gamma$ is a polytope in $\mathbb{R}^{3n+3}$. Does anyone know a list of defining inequalities for it?
For those who don't like the probability language: Let $p_{ijk} \in \mathbb{R}_{\geq 0}^H$ with $\sum p_{ijk}=1$. Set $\alpha_i = \sum_{j} p_{ij(n-i-j)}$, $\beta_j = \sum_k p_{(n-j-k)jk}$ and $\gamma_k = \sum_i p_{i(n-i-k)k}$. What are the inequalities defining the possible vectors $\alpha$, $\beta$ and $\gamma$?
• The polytope in question (call it $P_n$) is the convex hull of $e_a + f_b + g_c$, where $e_0,\dots,e_n,f_0,\dots,f_n,g_0,\dots,g_n$ is a basis for ${\mathbb R}^{3n+3}$ and $(a,b,c)$ ranges over $H$. There appear to be some recursive relations that allow one to describe $P_{n+1}$ or $P_{2n}$ in terms of $P_n$ which may be helpful. Commented Jun 7, 2016 at 16:59
• The defining inequalities correspond to weights $r_0,\dots,r_n,s_0,\dots,s_n,t_0,\dots,t_n$ such that $r_a + s_b + t_c \geq 0$ for all $(a,b,c) \in H$, with equality holding in some "spanning" set of $H$. This looks somewhat combinatorially unappetising, though. Commented Jun 7, 2016 at 17:02
• For n=2. If the variables are $\alpha_0, \alpha_1, \alpha_2, \beta_0, \beta_1, \beta_2, \gamma_0, \gamma_1, \gamma_2$, then the polytope is defined by 6 inequalities and 4 equalities:$$0\le \alpha_2,\beta_2,\gamma_2,d,e,f;$$ $$\alpha_1=e+f;\ \ \beta_1=d+f;\ \ \gamma_1=d+e;$$ $$\alpha_0+\beta_0+\gamma_0=\alpha_2+\beta_2+\gamma_2+1$$ where $d,e,f$ abbreviate $\alpha_0-\beta_2-\gamma_2$, $\beta_0-\alpha_2-\gamma_2$, $\gamma_0-\alpha_2-\beta_2$.
– user44143
Commented Jan 15, 2018 at 5:18
While the literal question I asked is still open, Luke Pebody has proven an important relevant result: If $$\alpha_0 \geq \alpha_1 \geq \cdots \geq \alpha_n$$, and the same for $$\beta$$ and $$\gamma$$, and $$\sum \alpha_j + \sum \beta_j + \sum \gamma_k = 3$$, then there is a probability distribution $$\pi$$ with marginals $$\alpha$$, $$\beta$$, $$\gamma$$. This constructs a large polytope inside the one which this question asks for a characterization of. | 845 | 2,401 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.777733 |
https://en.m.wikipedia.org/wiki/Implicit_surface | 1,657,213,760,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00410.warc.gz | 288,786,052 | 23,959 | # Implicit surface
In mathematics, an implicit surface is a surface in Euclidean space defined by an equation
Implicit surface torus (R=40, a=15).
Implicit surface of genus 2.
Implicit non-algebraic surface (wineglass).
${\displaystyle F(x,y,z)=0.}$
An implicit surface is the set of zeros of a function of three variables. Implicit means that the equation is not solved for x or y or z.
The graph of a function is usually described by an equation ${\displaystyle z=f(x,y)}$ and is called an explicit representation. The third essential description of a surface is the parametric one: ${\displaystyle (x(s,t),y(s,t),z(s,t))}$, where the x-, y- and z-coordinates of surface points are represented by three functions ${\displaystyle x(s,t)\,,y(s,t)\,,z(s,t)}$ depending on common parameters ${\displaystyle s,t}$. Generally the change of representations is simple only when the explicit representation ${\displaystyle z=f(x,y)}$ is given: ${\displaystyle z-f(x,y)=0}$ (implicit), ${\displaystyle (s,t,f(s,t))}$ (parametric).
Examples:
1. plane ${\displaystyle x+2y-3z+1=0.}$
2. sphere ${\displaystyle x^{2}+y^{2}+z^{2}-4=0.}$
3. torus ${\displaystyle (x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(x^{2}+y^{2})=0.}$
4. Surface of genus 2: ${\displaystyle 2y(y^{2}-3x^{2})(1-z^{2})+(x^{2}+y^{2})^{2}-(9z^{2}-1)(1-z^{2})=0}$ (see diagram).
5. Surface of revolution ${\displaystyle x^{2}+y^{2}-(\ln(z+3.2))^{2}-0.02=0}$ (see diagram wineglass).
For a plane, a sphere, and a torus there exist simple parametric representations. This is not true for the fourth example.
The implicit function theorem describes conditions under which an equation ${\displaystyle F(x,y,z)=0}$ can be solved (at least implicitly) for x, y or z. But in general the solution may not be made explicit. This theorem is the key to the computation of essential geometric features of a surface: tangent planes, surface normals, curvatures (see below). But they have an essential drawback: their visualization is difficult.
If ${\displaystyle F(x,y,z)}$ is polynomial in x, y and z, the surface is called algebraic. Example 5 is non-algebraic.
Despite difficulty of visualization, implicit surfaces provide relatively simple techniques to generate theoretically (e.g. Steiner surface) and practically (see below) interesting surfaces.
## Formulas
Throughout the following considerations the implicit surface is represented by an equation ${\displaystyle F(x,y,z)=0}$ where function ${\displaystyle F}$ meets the necessary conditions of differentiability. The partial derivatives of ${\displaystyle F}$ are ${\displaystyle F_{x},F_{y},F_{z},F_{xx},\ldots }$ .
### Tangent plane and normal vector
A surface point ${\displaystyle (x_{0},y_{0},z_{0})}$ is called regular if and only if the gradient of ${\displaystyle F}$ at ${\displaystyle (x_{0},y_{0},z_{0})}$ is not the zero vector ${\displaystyle (0,0,0)}$ , meaning
${\displaystyle (F_{x}(x_{0},y_{0},z_{0}),F_{y}(x_{0},y_{0},z_{0}),F_{z}(x_{0},y_{0},z_{0}))\neq (0,0,0)}$ .
If the surface point ${\displaystyle (x_{0},y_{0},z_{0})}$ is not regular, it is called singular.
The equation of the tangent plane at a regular point ${\displaystyle (x_{0},y_{0},z_{0})}$ is
${\displaystyle F_{x}(x_{0},y_{0},z_{0})(x-x_{0})+F_{y}(x_{0},y_{0},z_{0})(y-y_{0})+F_{z}(x_{0},y_{0},z_{0})(z-z_{0})=0,}$
and a normal vector is
${\displaystyle \mathbf {n} (x_{0},y_{0},z_{0})=(F_{x}(x_{0},y_{0},z_{0}),F_{y}(x_{0},y_{0},z_{0}),F_{z}(x_{0},y_{0},z_{0}))^{T}.}$
### Normal curvature
In order to keep the formula simple the arguments ${\displaystyle (x_{0},y_{0},z_{0})}$ are omitted:
${\displaystyle \kappa _{n}={\frac {\mathbf {v} ^{\top }H_{F}\mathbf {v} }{\|\operatorname {grad} F\|}}}$
is the normal curvature of the surface at a regular point for the unit tangent direction ${\displaystyle \mathbf {v} }$ . ${\displaystyle H_{F}}$ is the Hessian matrix of ${\displaystyle F}$ (matrix of the second derivatives).
The proof of this formula relies (as in the case of an implicit curve) on the implicit function theorem and the formula for the normal curvature of a parametric surface.
## Applications of implicit surfaces
As in the case of implicit curves it is an easy task to generate implicit surfaces with desired shapes by applying algebraic operations (addition, multiplication) on simple primitives.
Equipotential surface of 4 point charges
### Equipotential surface of point charges
The electrical potential of a point charge ${\displaystyle q_{i}}$ at point ${\displaystyle \mathbf {p} _{i}=(x_{i},y_{i},z_{i})}$ generates at point ${\displaystyle \mathbf {p} =(x,y,z)}$ the potential (omitting physical constants)
${\displaystyle F_{i}(x,y,z)={\frac {q_{i}}{\|\mathbf {p} -\mathbf {p} _{i}\|}}.}$
The equipotential surface for the potential value ${\displaystyle c}$ is the implicit surface ${\displaystyle F_{i}(x,y,z)-c=0}$ which is a sphere with center at point ${\displaystyle \mathbf {p} _{i}}$ .
The potential of ${\displaystyle 4}$ point charges is represented by
${\displaystyle F(x,y,z)={\frac {q_{1}}{\|\mathbf {p} -\mathbf {p} _{1}\|}}+{\frac {q_{2}}{\|\mathbf {p} -\mathbf {p} _{2}\|}}+{\frac {q_{3}}{\|\mathbf {p} -\mathbf {p} _{3}\|}}+{\frac {q_{4}}{\|\mathbf {p} -\mathbf {p} _{4}\|}}.}$
For the picture the four charges equal 1 and are located at the points ${\displaystyle (\pm 1,\pm 1,0)}$ . The displayed surface is the equipotential surface (implicit surface) ${\displaystyle F(x,y,z)-2.8=0}$ .
### Constant distance product surface
A Cassini oval can be defined as the point set for which the product of the distances to two given points is constant (in contrast, for an ellipse the sum is constant). In a similar way implicit surfaces can be defined by a constant distance product to several fixed points.
In the diagram metamorphoses the upper left surface is generated by this rule: With
{\displaystyle {\begin{aligned}F(x,y,z)={}&{\Big (}{\sqrt {(x-1)^{2}+y^{2}+z^{2}}}\cdot {\sqrt {(x+1)^{2}+y^{2}+z^{2}}}\\&\qquad \cdot {\sqrt {x^{2}+(y-1)^{2}+z^{2}}}\cdot {\sqrt {x^{2}+(y+1)^{2}+z^{2}}}{\Big )}\end{aligned}}}
the constant distance product surface ${\displaystyle F(x,y,z)-1.1=0}$ is displayed.
Metamorphoses between two implicit surfaces: a torus and a constant distance product surface.
### Metamorphoses of implicit surfaces
A further simple method to generate new implicit surfaces is called metamorphosis of implicit surfaces:
For two implicit surfaces ${\displaystyle F_{1}(x,y,z)=0,F_{2}(x,y,z)=0}$ (in the diagram: a constant distance product surface and a torus) one defines new surfaces using the design parameter ${\displaystyle \mu \in [0,1]}$ :
${\displaystyle F(x,y,z)=\mu F_{1}(x,y,z)+(1-\mu )F_{2}(x,y,z)=0}$
In the diagram the design parameter is successively ${\displaystyle \mu =0,\,0.33,\,0.66,\,1}$ .
Approximation of three tori (parallel projection)
POV-Ray image (central projection) of an approximation of three tori.
### Smooth approximations of several implicit surfaces
${\displaystyle \Pi }$ -surfaces [1] can be used to approximate any given smooth and bounded object in ${\displaystyle R^{3}}$ whose surface is defined by a single polynomial as a product of subsidiary polynomials. In other words, we can design any smooth object with a single algebraic surface. Let us denote the defining polynomials as ${\displaystyle f_{i}\in \mathbb {R} [x_{1},\ldots ,x_{n}](i=1,\ldots ,k)}$ . Then, the approximating object is defined by the polynomial
${\displaystyle F(x,y,z)=\prod _{i}f_{i}(x,y,z)-r}$ [1]
where ${\displaystyle r\in \mathbb {R} }$ stands for the blending parameter that controls the approximating error.
Analogously to the smooth approximation with implicit curves, the equation
${\displaystyle F(x,y,z)=F_{1}(x,y,z)\cdot F_{2}(x,y,z)\cdot F_{3}(x,y,z)-r=0}$
represents for suitable parameters ${\displaystyle c}$ smooth approximations of three intersecting tori with equations
{\displaystyle {\begin{aligned}F_{1}=(x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(x^{2}+y^{2})=0,\\[3pt]F_{2}=(x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(x^{2}+z^{2})=0,\\[3pt]F_{3}=(x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(y^{2}+z^{2})=0.\end{aligned}}}
(In the diagram the parameters are ${\displaystyle R=1,\,a=0.2,\,r=0.01.}$ )
POV-Ray image: metamorphoses between a sphere and a constant distance product surface (6 points).
## Visualization of implicit surfaces
There are various algorithms for rendering implicit surfaces,[2] including the marching cubes algorithm.[3] Essentially there are two ideas for visualizing an implicit surface: One generates a net of polygons which is visualized (see surface triangulation) and the second relies on ray tracing which determines intersection points of rays with the surface.[4] The intersection points can be approximated by sphere tracing, using a signed distance function to find the distance to the surface.[5] | 2,740 | 8,899 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 59, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2022-27 | latest | en | 0.789309 |
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# How many joules of energy does a 100 watt light bulb have?
012
###### 2013-04-22 02:39:45
welll it really depends on the volasity of light energy but from my calculatios it is 900
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## Related Questions
there are 100 joules in an energy efficient light bulb 75 joules go towards the light and 25 joules go towards the heat
1) The power use of bulbs is specified in units of power, not energy. In other words, watts - which means joules per second. 2) Light-bulbs come in different capacities. 3) It is incorrect to say that the energy (or power) "is in" the light-bulb. The light-bulb simply transforms one type of energy to another, using electrical energy which is NOT stored in the light-bulb.
60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).
Watt means joules/second. It refers to the amount of energy a device uses, in this case. Multiply the power (in watts) by the time (in seconds) to get the energy (in joules).
It's 100 watts times 3600 seconds, that's 360,000 joules of energy. A joule is 1 watt for 1 second.
100 watts running for 60 seconds uses 100 x 60 Joules of energy. A Joule is the same as a watt-second.
Every second a 150 Watt bulb converts 150 Joules from electricity into heat and light. The number of Watts tells you how many Joules pass per second.
100 watts means 100 Joules/Second. So in 24 hours, the bulb would use 24*60*60*100 Joules. so that's 8,640,000 joules
Almost 90 % of electrical energy provided to an incandescent light bulb goes as heat and rest as light. A 100 Watt bulb puts out 100 Joules of heat per second. So - for one minute it would put out 6000 Joules (100 Watts X 60 seconds). 1 BTU (British Thermal Unit) of heat = 1055.056 Joules. So a 100 watt bulb, burning for one minute would put out 5.68 BTUs of heat. ( 6000 Joules / 1055.056 Joules) = 5.68 BTUs. Same bulb burning for one hour would generate 341 BTUs of heat.
A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.
You can't calculate how many volts with that information; you could calculate the energy - 60 watts for 15 minutes is equivalent to 54,000 joules.
1 watt= 1 joules/1 sec60 w = 60joules/minuteor=3600joules/hour
A light bulb produces light on demand when it is connected to a sufficient source of electrical energy. Many types of light bulbs also produce heat as a byproduct.
This is a very good question. A Watt is a unit of power, or energy with respect to (divided by) time, and is defined as 1 Joule per second. A 100 Watt light bulb consumes energy at a rate of 100 Joules (J) per second (s). There are 60 seconds in every minute, and 60 minutes in every hour. So also, there are 3600 seconds in every hour (60 X 60). 100 J/s X 3600s = 360,000 J (the s cancels out in the division) Your bulb consumes 360,000 Joules in one hour.
Your question is rather like asking "How many miles per hour do you do in a week?" You don't consume watts over time, it's a measure of how many joules of energy you consume over time.
Yes, there are many examples. Electrical energy to light energy (light bulb). Mechanical energy to heat energy (rub your hands together). Chemical energy to light energy (a glowstick), and on and on.
Joule is the unit of energy. Watt is the unit of power. One Joule provides one Watt for one second. A 100 Watt lamp uses 100 Joules per second.AnswerPower is the rate of doing work. Work is measured in joules, so power is measured in joules per second. However, in SI, a joule per second is given a special name: the watt. So a 100-W lamp consumes 100 J of energy per second.
Those numbers describe the power used by the two bulbs, in other words how many joules of electrical energy they use per second. The 100 watt bulb uses 40 watts more.
With current (as of 2013) technology, from best to worst efficiency, the light bulbs are basically:LED light bulbs (most efficient)Fluorescent lightsThe old-fashioned incandescent lights (worst)
In a traditional light bulb, the electrical energy is converted to heat. The filament gets hot and emits the thermal energy as light. The electrical energy itself is not directly converted to light but goes through the thermal energy stage.There are many kinds of lights and more complicated processes which are not described in this brief answer.
Energy (Joules) is equal to the mass multiplied by the speed of light squared (E=mc^2).
###### Home ElectricitySciencePhysicsUnits of MeasureMath and ArithmeticEnergyCalorie Count
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# Series Sum ISC 2014 Theory
A class SeriesSum is designed to calculate the sum of the following series:
Sum = x2 / 1! + x4 / 3! + x6 / 5! + … + xn / (n – 1)!
Some of the members of the class are given below:
Class name: SeriesSum
Data members/instance variables:
x: to store an integer number.
n: to store number of terms.
sum: double variable to store the sum of the series.
Member functions:
SeriesSum(int xx, int nn): constructor to assign x = xx and n = nn.
double findFact(int m): to return the factorial of m using recursive technique.
double findPower(int x, int y): to return x raised to the power of y using recursive technique.
void calculate(): to calculate the sum of the series by invoking the recursive functions respectively.
void display(): to display the sum of the series.
(a) Specify the class SeriesSum, giving details of the constructor, double findFact(int), double findPower(int, int), void calculate() and void display(). Define the main() function to create an object and call the functions accordingly to enable the task.
(b) State the two differences between iteration and recursion.
``````import java.io.*;
class SeriesSum{
private int x;
private int n;
private double sum;
public SeriesSum(int xx, int nn){
x = xx;
n = nn;
sum = 0.0;
}
public double findFact(int m){
if(m <= 1)
return 1.0;
else
return m * findFact(m - 1);
}
public double findPower(int x, int y){
if(y == 0)
return 1.0;
else
return x * findPower(x, y - 1);
}
public void calculate(){
for(int i = 1; i <= n; i++)
sum += findPower(x, i * 2) / findFact(i * 2 - 1);
}
public void display(){
System.out.println("Sum = " + sum);
}
public static void main(String[] args)throws IOException{
System.out.print("x = "); | 452 | 1,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2021-25 | latest | en | 0.58354 |
https://plainmath.net/91156/distance-between-1-6-and-4-7 | 1,670,026,151,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00411.warc.gz | 499,317,596 | 11,477 | # Distance between (1,-6) and (4,7)?
Distance between (1,-6) and (4,7)?
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The formula for calculating the distance between two points is:
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Substituting the values from the points in the problem gives:
$d=\sqrt{{\left(4-1\right)}^{2}+{\left(7--6\right)}^{2}}$
$d=\sqrt{{3}^{2}+{13}^{2}}$
$d=\sqrt{178}$
$d\cong 13.342$ | 207 | 616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2022-49 | latest | en | 0.843472 |
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Pairs of primes (Posted on 2019-07-27)
Imagine a bag containing cards representing all n-digit odd numbers. A random card is drawn and two new numbers are created by preceding the drawn number by each of its even neighbors.
What is the probability that each of those 2 numbers is prime?
Examples:
For n=1 there are 5 cards i.e. 1,3,5,7 and 9. Clearly only numbers 3 and 9 qualifiy since fboth 23 and 43 are primes and so are 89 and 109 & there are no other answers. So for n=1 p=0.4 is the probability we were looking for.
For n=2 I will not provide the answer but will show you one of the qualifying numbers e.g. 69, since both 6869 and 7069 are prime.
Now evaluate the correct probabilities for n=2,3, ...8,9 (or as far as your resources allow) - and you will get a sequence for which you may be credited @ OEIS.
So this time you get a task both challenging and rewarding!
GOOD LUCK...
No Solution Yet Submitted by Ady TZIDON No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
re: via computer - the start of the soln | Comment 5 of 10 |
(In reply to via computer - the start of the soln by Steven Lord)
Your first three values agree with mine. Thereafter your probabilities go up while mine go down. Going up seems counterintuitive, as primes get rarer as the order of magnitude increases.
I found only 56 occurrences among the 4500 4-digit numbers. Your probability of .0376 implies .0376*4500 ~= 169. Perhaps you could modify your program to list the 169 found, to see if I missed some or some of yours involved non-primes.
I'm thinking that if the problem is on your side (too many found), the problem might lie in the primality test:
m=(1.*i)/(1.*j)
While m is defined as an integer, the decimal points make the division take place in floating point, or real. As the conversion is implicit, there's no "kind" specification. Perhaps in the division of 8-digit or higher numbers, the integer part of the quotient is not actually the true quotient, if only single precision is used.
Edited on July 28, 2019, 11:10 am
Posted by Charlie on 2019-07-28 07:58:12
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## Measurement and Central Tendency Assignment Help
The measurement of central tendency is recognized as the summary of statistic which represents the unique value or the center point of a dataset. These measures aim to indicate locations where most of the values in a distribution drop and they are also known as the central position of a distribution. The Measurement and Central Tendency is an essential aspect of statistics and students who are engaged in studying statistics often come across the task of completing assignments on this topic. When students find themselves helpless, they contact the writers of BookMyEssay for getting expert Measurement and Central Tendency assignment help. The assignment writing from our side tends to be absolutely flawless, and they help the students in securing impressive grades in their assignments. Moreover, the Measurement and Central Tendency assignment help from our side remains unique, and this makes our students confident and increases their belief in our services.
## Understanding the Concept of Measures of Central Tendency
In statistics, the three important common measures of Central Tendency are the arithmetic mean, the median, and the mode. Though all of these three measures do calculate the position of the central point, yet they make use of unique methods. However, selecting the most exceptional measure of central tendency is dependent on the kind of data you have. You can calculate the fundamental trend for either theoretical distribution, like the normal distribution or for a fixed set of values. This type of tendency is commonly contrasted with its variability or dispersion, and central tendency and dispersion are habitually branded characteristics of distributions. An analyst can judge whether or not a data has got a strong or a meek central tendency grounded in its dispersion.
Mean – The mean is considered the arithmetic average and it is also viewed as the measure of central tendency with which most of you are familiar. You can quickly calculate the mean, and for calculating your mean, you have to include all the values plus divide it by the observations present in your dataset. Calculating the mean consists of all the values present in the data and if you change any value, then the mean changes too.
Median – The median is considered the middle value and it is that value which splits the dataset into two halves. For finding out the median, you must order the data from the smallest to the largest and then discover the data point which has got a similar amount of values. The process of locating the mean does differ depending on whether or not your dataset has gained an even or odd number of values.
Mode – The mode is acknowledged as the value which happens in your data set most frequently. On a bar diagram, the mode is considered the highest bar, and if the data does have many values which are tied to occur most frequently, then you have got multimodal distribution. Again, if there is no repetition of value, then it can be said that the data doesn’t have any mode.
### Discovering the Best
When you have got asymmetrical distribution for incessant data then the mean, the median, plus the mode are similar. In these cases, the analysts use the mean as it involves all the data in the calculations. Again, when you have a skewed distribution, then the median is considered the best. And mode is used when categorical data is involved.
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https://arcade-book.readthedocs.io/en/latest/chapters/25_recursion/recursion.html | 1,539,869,992,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00098.warc.gz | 618,347,338 | 12,716 | # 28. Recursion¶
A child couldn't sleep, so her mother told her a story about a little frog,
who couldn't sleep, so the frog's mother told her a story about a little bear,
who couldn't sleep, so the bear's mother told her a story about a little weasel...
who fell asleep.
...and the little bear fell asleep;
...and the little frog fell asleep;
...and the child fell asleep.
Recursion is an object or process that is defined in terms of itself. Mathematical patterns such as factorials and the Fibonacci series are recursive. Documents that can contain other documents, which themselves can contain other documents, are recursive. Fractal images, and even certain biological processes are recursive in how they work.
## 28.1. Where is Recursion Used?¶
Documents, such as web pages, are naturally recursive. For example, Figure 20.1 shows a simple web document.
That web document can be contained in a “box,” which can help layout the page as shown in Figure 20.2.
This works recursively. Each box can contain a web page, that can have a box, which could contain another web page as shown in Figure 20.3.
Recursive functions are often used with advanced searching and sorting algorithms. We’ll show some of that here and if you take a “data structures” class you will see a lot more of it.
Even if a person does not become a programmer, understanding the concept of recursive systems is important. If there is a business need for recursive table structures, documents, or something else, it is important to know how to specify this to the programmer up front.
For example, a person might specify that a web program for recipes needs the ability to support ingredients and directions. A person familiar with recursion might state that each ingredient could itself be a recipes with other ingredients (that could be recipes.) The second system is considerably more powerful.
## 28.2. How is Recursion Coded?¶
In prior chapters, we have used functions that call other functions. For example:
Functions calling other functions
1 2 3 4 5 6 7 8 def f(): g() print("f") def g(): print("g") f()
It is also possible for a function to call itself. A function that calls itself is using a concept called recursion. For example:
Recursion
1 2 3 4 5 def f(): print("Hello") f() f()
The example above will print Hello and then call the f() function again. Which will cause another Hello to be printed out and another call to the f() function. This will continue until the computer runs out of something called stack space. When this happens, Python will output a long error that ends with:
RuntimeError: maximum recursion depth exceeded
The computer is telling you, the programmer, that you have gone too far down the rabbit hole.
## 28.3. Controlling Recursion Depth¶
To successfully use recursion, there needs to be a way to prevent the function from endlessly calling itself over and over again. The example below counts how many times it has been called, and uses an if statement to exit once the function has called itself ten times.
Controlling recursion levels
1 2 3 4 5 6 7 8 9 10 11 def f(level): # Print the level we are at print("Recursion call, level",level) # If we haven't reached level ten... if level < 10: # Call this function again # and add one to the level f(level+1) # Start the recursive calls at level 1 f(1)
Output
1 2 3 4 5 6 7 8 9 10 Recursion call, level 1 Recursion call, level 2 Recursion call, level 3 Recursion call, level 4 Recursion call, level 5 Recursion call, level 6 Recursion call, level 7 Recursion call, level 8 Recursion call, level 9 Recursion call, level 10
## 28.4. Recursion In Mathematics¶
### 28.4.1. Recursion Factorial Calculation¶
Any code that can be done recursively can be done without using recursion. Some programmers feel that the recursive code is easier to understand.
Calculating the factorial of a number is a classic example of using recursion. Factorials are useful in probability and statistics. For example:
Recursively, this can be described as:
Below are two example functions that calculate . The first one is non-recursive, the second is recursive.
Non-recursive factorial
1 2 3 4 5 6 7 # This program calculates a factorial # WITHOUT using recursion def factorial_nonrecursive(n): answer = 1 for i in range(2, n + 1): answer = answer * i return answer
Recursive factorial
1 2 3 4 5 6 7 # This program calculates a factorial # WITH recursion def factorial_recursive(n): if n == 1: return 1 elif n > 1: return n * factorial_recursive(n - 1)
The functions do nothing by themselves. Below is an example where we put it all together. This example also adds some print statements inside the function so we can see what is happening.
Trying out recursive functions
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 # This program calculates a factorial # WITHOUT using recursion def factorial_nonrecursive(n): answer = 1 for i in range(2, n + 1): print(i, "*", answer, "=", i * answer) answer = answer * i return answer print("I can calculate a factorial!") user_input = input("Enter a number:") n = int(user_input) answer = factorial_nonrecursive(n) print(answer) # This program calculates a factorial # WITH recursion def factorial_recursive(n): if n == 1: return 1 else: x = factorial_recursive(n - 1) print( n, "*", x, "=", n * x ) return n * x print("I can calculate a factorial!") user_input = input("Enter a number:") n = int(user_input) answer = factorial_recursive(n) print(answer)
Output
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 I can calculate a factorial! Enter a number:7 2 * 1 = 2 3 * 2 = 6 4 * 6 = 24 5 * 24 = 120 6 * 120 = 720 7 * 720 = 5040 5040 I can calculate a factorial! Enter a number:7 2 * 1 = 2 3 * 2 = 6 4 * 6 = 24 5 * 24 = 120 6 * 120 = 720 7 * 720 = 5040 5040
### 28.4.2. Recursive Expressions¶
Say you have a mathematical expression like this:
$$f_{n} = \begin{cases} 6 & \text{if } n = 1, \\ \frac{1}{2}f_{n-1}+4 & \text{if } n > 1. \end{cases}$$
Looks complicated, but it just means that if $$n=1$$ we are working with $$f_{1}$$. That function returns a 6.
For $$f_{2}$$ we return $$\frac{1}{2}f_{1}+4$$.
1 def f(n):
Then we need to add that first case:
1 2 3 def f(n): if n == 1: return 6
See how closely if follows the mathematical notation? Now for the rest:
1 2 3 4 5 def f(n): if n == 1: return 6 elif n > 1: return (1 / 2) * f(n - 1) + 4
Converting these types of mathematical expressions to code is straight forward. But we’d better try it out in a full example:
1 2 3 4 5 6 7 8 9 10 11 12 13 def f(n): if n == 1: return 6 elif n > 1: return (1 / 2) * f(n - 1) + 4 def main(): result = f(10) print(result) main()
## 28.5. Recursive Graphics¶
### 28.5.1. Recursive Rectangles¶
Recursion is great to work with structured documents that are themselves recursive. For example, a web document can have a table divided into rows and columns to help with layout. One row might be the header, another row the main body, and finally the footer. Inside a table cell, might be another table. And inside of that can exist yet another table.
Another example is e-mail. It is possible to attach another person’s e-mail to a your own e-mail. But that e-mail could have another e-mail attached to it, and so on.
Can we visually see recursion in action in one of our Pygame programs? Yes! Figure 19.4 shows an example program that draws a rectangle, and recursively keeps drawing rectangles inside of it. Each rectangle is 20% smaller than the parent rectangle. Look at the code. Pay close attention to the recursive call in the recursive_draw function.
recursive_rectangles.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 """ Recursive Rectangles """ import arcade SCREEN_WIDTH = 800 SCREEN_HEIGHT = 500 def draw_rectangle(x, y, width, height): """ Recursively draw a rectangle, each one a percentage smaller """ # Draw it arcade.draw_rectangle_outline(x, y, width, height, arcade.color.BLACK) # As long as we have a width bigger than 1, recursively call this function with a smaller rectangle if width > 1: # Draw the rectangle 90% of our current size draw_rectangle(x, y, width * .9, height * .9) class MyWindow(arcade.Window): """ Main application class. """ def __init__(self, width, height): super().__init__(width, height) arcade.set_background_color(arcade.color.WHITE) def on_draw(self): """ Render the screen. """ arcade.start_render() # Find the center of our screen center_x = SCREEN_WIDTH / 2 center_y = SCREEN_HEIGHT / 2 # Start our recursive calls draw_rectangle(center_x, center_y, SCREEN_WIDTH, SCREEN_HEIGHT) def main(): MyWindow(SCREEN_WIDTH, SCREEN_HEIGHT) arcade.run() if __name__ == "__main__": main()
### 28.5.2. Fractals¶
Fractals are defined recursively. Here is a very simple fractal, showing how it changes depending on how “deep” the recursion goes.
Here is the source code for the “H” fractal:
recursive_h.py
You can explore fractals on-line:
If you want to program your own fractals, you can get ideas of easy fractals by looking at Chapter 8 of The Nature of Code by Daniel Shiffman.
## 28.6. Recursive Mazes¶
There are maze generation algorithms. Wikipedia has a nice Maze generation algorithm article that details some. One way is the recursive division method.
The algorithm is described below. Images are from Wikipedia.
This method results in mazes with long straight walls crossing their space, making it easier to see which areas to avoid.
Here is sample Python code that creates a maze using this method:
Recursive Maze Example
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 import random # These constants are used to determine what should be stored in the grid if we have an empty # space or a filled space. EMPTY = " " WALL = "XXX" # Maze must have an ODD number of rows and columns. # Walls go on EVEN rows/columns. # Openings go on ODD rows/columns MAZE_HEIGHT = 51 MAZE_WIDTH = 51 def create_grid(width, height): """ Create an empty grid. """ grid = [] for row in range(height): grid.append([]) for column in range(width): grid[row].append(EMPTY) return grid def print_maze(maze): """ Print the maze. """ # Loop each row, but do it in reverse so 0 is at the bottom like we expect for row in range(len(maze) - 1, -1, -1): # Print the row/y number print(f"{row:3} - ", end="") # Loop the row and print the content for column in range(len(maze[row])): print(f"{maze[row][column]}", end="") # Go down a line print() # Print the column/x at the bottom print(" ", end="") for column in range(len(maze[0])): print(f"{column:3}", end="") print() def create_outside_walls(maze): """ Create outside border walls.""" # Create left and right walls for row in range(len(maze)): maze[row][0] = WALL maze[row][len(maze[row])-1] = WALL # Create top and bottom walls for column in range(1, len(maze[0]) - 1): maze[0][column] = WALL maze[len(maze[0]) - 1][column] = WALL def create_maze(maze, top, bottom, left, right): """ Recursive function to divide up the maze in four sections and create three gaps. Walls can only go on even numbered rows/columns. Gaps can only go on odd numbered rows/columns. Maze must have an ODD number of rows and columns. """ # Figure out where to divide horizontally start_range = bottom + 2 end_range = top - 1 y = random.randrange(start_range, end_range, 2) # Do the division for column in range(left + 1, right): maze[y][column] = WALL # Figure out where to divide vertically start_range = left + 2 end_range = right - 1 x = random.randrange(start_range, end_range, 2) # Do the division for row in range(bottom + 1, top): maze[row][x] = WALL # Now we'll make a gap on 3 of the 4 walls. # Figure out which wall does NOT get a gap. wall = random.randrange(4) if wall != 0: gap = random.randrange(left + 1, x, 2) maze[y][gap] = EMPTY if wall != 1: gap = random.randrange(x + 1, right, 2) maze[y][gap] = EMPTY if wall != 2: gap = random.randrange(bottom + 1, y, 2) maze[gap][x] = EMPTY if wall != 3: gap = random.randrange(y + 1, top, 2) maze[gap][x] = EMPTY # Print what's going on print(f"Top/Bottom: {top}, {bottom} Left/Right: {left}, {right} Divide: {x}, {y}") print_maze(maze) print() # If there's enough space, to a recursive call. if top > y + 3 and x > left + 3: create_maze(maze, top, y, left, x) if top > y + 3 and x + 3 < right: create_maze(maze, top, y, x, right) if bottom + 3 < y and x + 3 < right: create_maze(maze, y, bottom, x, right) if bottom + 3 < y and x > left + 3: create_maze(maze, y, bottom, left, x) def main(): # Create the blank grid maze = create_grid(MAZE_WIDTH, MAZE_HEIGHT) # Fill in the outside walls create_outside_walls(maze) # Start the recursive process create_maze(maze, MAZE_HEIGHT - 1, 0, 0, MAZE_WIDTH - 1) if __name__ == "__main__": main() | 3,794 | 13,212 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2018-43 | latest | en | 0.93039 |
https://socratic.org/questions/what-is-the-distance-between-4-3-and-2-4 | 1,579,754,743,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00204.warc.gz | 654,390,713 | 6,211 | # What is the distance between (-4, 3) and (-2, 4) ?
Dec 5, 2015
$\sqrt{5}$
#### Explanation:
The distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$
is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
In this case
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$
The relationship between the point can be seen in the image below: | 206 | 535 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-05 | longest | en | 0.762427 |
http://www.physicsforums.com/showthread.php?t=165482 | 1,369,332,317,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703662159/warc/CC-MAIN-20130516112742-00098-ip-10-60-113-184.ec2.internal.warc.gz | 649,327,166 | 11,733 | ## Change of integration order?
1. The problem statement, all variables and given/known data
On this site,
http://www.math.ohio-state.edu/~gerl..._with_function
they changed integration order with e^(-ikt) integrated wrt t to t is now a constant and it's now integrated wrt k. How can one do this?
As stated from this sentence "By interchanging integration order and letting one has"
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
$$f(x) \propto \int e^{ikx} \int e^{-ikt} f(t) dt dk = \int \int e^{ikx} e^{-ikt} f(t) dt dk = \int \int e^{ikx} e^{-ikt} f(t) dk dt$$ There's nothing wrong with changing the order of integration here, think back of integrals as Riemann sums, changing the order in which you sum terms doesn't matter. (You were going to integrate both exp(ikx) and exp(-ikt) over k anyway, so why not do that first and then integrate over t on which only the latter term depends).
Right, but why is the f (hat) function in terms of t? f is in terms of x and k only.
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## Change of integration order?
k and t are "dummy" variables. They can call them whatever they wish!
So only x counts. This brings about the question, what is the point of the fourier transform.
The point? Um, well, from my relatively limited experience in partial differential equations, in Fourier space differentiation and antidifferentiation break down into multiplication and division. Which is always good I suppose.
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Quote by pivoxa15 So only x counts. This brings about the question, what is the point of the fourier transform.
I'm not sure what you mean by that. Certainly in the integral you show, f is a function of x only. However, you have not said what it is you want to do with the Fourier Transform (the function inside the integral). Surely "the point of the Fourier transform" depends on that!
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Take equation (234) from here and there, replace $$\hat{f}(k)$$ by equation (233).
Then you almost have equation (235), except that you have to change
the order of integration.
Quote by pivoxa15 Right, but why is the f(hat) function in terms of t? f is in terms of x and k only.
The function $$\hat{f}(k)$$ is indeed a function of k, but it is not the function f, whose variable is x. Maybe you are a little confused about the meaning of $$\hat{f}(k)$$.
Think of the Fourier expansion as sort of linear combination of basis-vectors.
Let me try to explain:
Take the vectors in $$R^3$$.
Every vector can be represented as a linear combination of the basis-vectors $$(1,0,0)$$, $$(0,1,0)$$ and $$(0,0,1)$$.
For example:
Take the vector $$v=(9,8,4)$$. This vector can be written as a linear combination of the basisvectors vectors
$$u_1=(1,0,0)$$
$$u_2=(0,1,0)$$
$$u_3=(0,0,1)$$
in the following way:
$$(9,8,4) = 9 \cdot (1,0,0) + 8 \cdot (0,1,0) + 4 \cdot (0,0,1)$$
$$(9,8,4) = 9 \cdot u_1 + 8 \cdot u_2 + 4 \cdot u_3$$
Let's call the numbers 9,8,4 in front of the basis-vectors coefficients.
$$(9,8,4) = c_1 \cdot u_1 + c_2 \cdot u_2 + c_3 \cdot u_3$$
where $$c_1 = 9$$, $$c_2=8$$ and $$c_3=4$$
What if I choose different basis vectors, for example:
$$b_1=(3,0,0)$$
$$b_2=(0,2,0)$$
$$b_3=(0,0,2)$$
Then our vector $$v=(9,8,4)$$ can be written as:
$$(9,8,4) = 3 \cdot b_1 + 4 \cdot b_2 + 2 \cdot b_3$$
Thus, the coefficients are 3,4,2 for our new basis-vectors
$$b_1$$,$$b_2$$ and $$b_3$$.
In general, you can write a vector $$v$$ as a linear combination of
basis-vectors $$b_k$$, where in front of the basis-vectors you have the coefficients $$c_k$$.
$$v = \sum_{k=1}^{3} c_k b_k$$
In our last example we had
$$b_1=(3,0,0)$$
$$b_2=(0,2,0)$$
$$b_3=(0,0,2)$$
together with the coefficients:
$$c_1=3$$
$$c_2=4$$
$$c_3=2$$
Just check the formula
$$v = \sum_{k=1}^{3} c_k b_k$$
by plugging in the above values:
$$v = c_1 \cdot b_1 + c_2 \cdot b_2 + c_3 \cdot b3$$
$$v = 3 \cdot b_1 + 4 \cdot b_2 + 2 \cdot b_3$$
$$=3 \cdot (3,0,0) + 4 \cdot (0,2,0) + 2 \cdot (0,0,2)$$
----------
So, every vector can be represented by some basis vectors $$b_k$$
as
$$v = \sum_{k=1}^{3} c_k b_k$$
More general, if we have a vector with n entries instead of 3, we write
$$v = \sum_{k=1}^{n} c_k b_k = c_1 b_1 + c_2 b_2 + ... + c_n b_n$$
Now, let's make a step from the discrete to the continuous case.
Say we have a function $$f(x)$$ and we also want to ask, whether it's possible to represent $$f(x)$$ as a linear combination of ''basis-vectors''.
Question: Is it possible to write
$$f(x) = \sum_{k=1}^n c_k b_k$$
Let us be more specific and ask: Can I write $$f(x)$$ as a sum
of $$b_k = e^{ikx}$$? So my new basis-vectors are $$b_k=e^{ikx}$$.
The question then becomes:
Question: Is it possible to write
$$f(x) = \sum_{k=1}^n c_k e^{ikx}$$
Indeed, it is possible, with a correction for the values of k.
Instead of going from k=1 to n, we use infinitely many basis-vectors, and
write $$k=- \infty$$ to $$k=+ \infty$$.
Corrected version:
$$f(x) = \sum_{k=-\infty}^{+\infty} c_k e^{ikx}$$
The only question is, how do the coefficients
$$c_k$$ look like?
Have a look at Wikipedia
or here on page 2 of the pdf. It shows how the coefficients can be calculated.
To finally get to your Fourier integral, we replace the sum by an integral
$$f(x) = \int_{-\infty}^{+\infty} c(k) e^{ikx} dk$$
Now, on your ohio-website, $$c(k)$$ is $$\hat{f}(k)$$, see equation (234) on the ohio-website.
Thus, $$\hat{f}(k)$$ plays the role of the coefficients.
----------
Note 1 (on how to get from the Fourier series to the Fourier integral):
A Fourier series can sometimes be used to represent a function over an interval. If a function is defined over the entire real line, it may still have a Fourier series representation if it is periodic. If it is not periodic, then it cannot be represented by a Fourier series for all x. In such case we may still be able to represent the function in terms of sines and cosines, except that now the Fourier series becomes a Fourier integral. The motivation comes from formally considering Fourier series for functions of period 2T and letting T tend to infinity.
The quote is taken from here.
Also see here
for the transition from the Fourier series to the Fourier integral.
Note 2: Summary:
In summary, consider the Fourier integral as a linear combination of basis vectors $$e^{ikx}$$ with the
coefficients $$c(k)$$ (or $$\hat{f}(k)$$).
Note 3 (on applications of the Fourier transform:
Have a look at What is a Fourier Transform and what is it used for?.
Applications of Fourier Transform
in Communications, Astronomy, Geology and Optics
Note 4 (functions as basis-vectors?)
You might ask "Why can I consider the functions $$e^{ikx}$$ as
basis-vectors?"
The functions $$e^{ikx}$$ fulfill some properties similar to those of
basis vectors (1,0,0),(0,1,0),(0,0,1)from $$R^3$$.
The functions are orthonormal, that is they are
orthogonal to each other and they are normalized to 1 (delta-function?).
See
Fourier Analysis on page 3.
here on page 8.
Orthonormal functions: Definition on Wolfram mathworld
http://mathworld.wolfram.com/OrthonormalBasis.html
Note 5: Some examples of coefficients
Examples of Fourier transforms can be found here
Note 6: Java Applets
Approximation of a function by a Fourier transform
This applet shows you how you can approximate a function by a Fourier transform. The more
coefficients you use, the better the approximation becomes.
Applet: Rectangular pulse approximation by Fourier Transform
This applet shows how you approximate a rectangular shaped pulse. If you don't use enough
"basis-vectors" (bandwidth is limited), then the pulse will not look rectangular anymore. This has applications in
electronics where you want to transmit a signal but the bandwidth is limited.
I see. From all the Fourier transforms I have seen they use x as the integration variable in f (hat) which is confusing. f(hat) is not related to x in f(x) the original function but merely computes the coefficients of the basis vectors e^(ikx) for each k from -infinity to +infinity.
One thing I find fascinating is that there is a dirac delta function lurking inside every fourier transform. | 2,454 | 8,324 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2013-20 | latest | en | 0.91773 |
https://math.stackexchange.com/questions/2373613/calculating-the-number-of-irreducible-polynomials-over-a-finite-field | 1,561,366,504,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999298.86/warc/CC-MAIN-20190624084256-20190624110256-00201.warc.gz | 516,993,475 | 35,932 | # Calculating the number of irreducible polynomials over a finite field
I am trying to find the number of irreducible polynomials of degree $n$ over $\mathbb{F}_p$. Here is what I have done:
(1). Let $K=\mathbb{F}_{p^n}$. Let $M(n,p)$ the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_p$.
(2). Any root of a monic irreducible polynomial of degree $n$ is a primitive element of $K|\mathbb{F}_p$. Conversely, any primitive element of $K|\mathbb{F}_p$ is a root of a monic irreducible polynomial of degree $n$. Therefore, $$n\cdot M(n,p)=\text{ Number of primitive elements of }K|\mathbb{F}_p$$
(3). But we know that $K^{\times}$ is a cyclic group of order $p^n-1$. The primitive elements of $K|\mathbb{F}_p$ are precisely the generators of the cyclic group $K^{\times}$. The number of such generators are $\varphi(p^n-1)$.
(4). Combining (2) and (3) we conclude that $$M(n,p)=\frac{\varphi(p^n-1)}{n}$$
(5). Any irreducible polynomial of degree $n$ is a nonzero constant multiple of a monic irreducible polynomial of degree $n$. The number of such constants is $p-1$. Therefore the number of such polynomials is $$(p-1)\cdot M(n,p)=\frac{(p-1)\varphi(p^n-1)}{n}$$
So the final answer is $$\frac{(p-1)\varphi(p^n-1)}{n}$$
But my answer does not match with the usual way of solving this question which involves the Mobius inversion formula. What am I doing wrong?
First of all, very well-asked question! which makes it much easier to find the most helpful response.
It seems that you are conflating two different properties into one:
• a "primitive element" of a field extension $L/K$, which is an element $\alpha\in L$ such that $L=K(\alpha)$;
• a "primitive element" of a cyclic group, which is a generator of that group.
These two concepts do not correspond exactly with each other.
For example, consider $\Bbb F_9$, which is a degree-2 extension of $\Bbb F_3$. There are 6 elements of $\Bbb F_9$ that are not in $\Bbb F_3$, and any of them serves as a primitive element for the extension $\Bbb F_9/\Bbb F_3$ (in the first sense above). On the other hand, the multiplicative group $\Bbb F_9^\times$ is cyclic of order 8, and thus has $\phi(8)=4$ generators; all of these are primitive elements in the field-extension sense, but there are two other primitive elements that are not generators of the multiplicative group.
We can be super concrete here. Write $\Bbb F_9$ as $\Bbb F_3[x]/\langle x^2+1\rangle$, valid since $x^2+1$ is a degree-2 irreducible polynomial over $\Bbb F_3$. Then the elements of $\Bbb F_9$ are $a+bx$ where $a,b\in\{0,1,2\}$, with addition and multiplication defined as usual for polynomials except that $x^2=-1=2$. In this field, $x$ (or, more pedantically, the image of $x$ under the quotient map from $\Bbb F_3[x]$ to $\Bbb F_3[x]/\langle x^2+1\rangle$) is certainly a primitive element for the field extension. However, $x^4 = (x^2)^2 = 2^2 = 1$, and so $x$ does not generate the multiplicative group $\Bbb F_9^\times$. | 889 | 2,985 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-26 | longest | en | 0.844805 |
http://www.multiplication.com/learn/more/7/x/9 | 1,527,073,417,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00329.warc.gz | 443,452,347 | 8,807 | # Learn a Fact: 7 x 9
Follow steps 1-6 to master this fact.
## More Tips to Remember 7 x 9 = 63
Step 1Put your hands on the table in front of you. Step 2Your fingers represent the numbers 1 through 10. Step 3Each finger to the left of the curled finger represents 10. Since one finger is to the left of the curled finger, there are 6 tens or 60. (The first number of the answer is 6)Each finger to the right of the curled finger represents one. Count 1, 2, 3. (Or 61, 62, 63) 7 x 9 = 63
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8
X 0
0 | 204 | 595 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.807995 |
https://everything.explained.today/Superquadrics/ | 1,627,082,261,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00631.warc.gz | 264,441,268 | 6,311 | In mathematics, the superquadrics or super-quadrics (also superquadratics) are a family of geometric shapes defined by formulas that resemble those of ellipsoids and other quadrics, except that the squaring operations are replaced by arbitrary powers. They can be seen as the three-dimensional relatives of the superellipses. The term may refer to the solid object or to its surface, depending on the context. The equations below specify the surface; the solid is specified by replacing the equality signs by less-than-or-equal signs.
The superquadrics include many shapes that resemble cubes, octahedra, cylinders, lozenges and spindles, with rounded or sharp corners. Because of their flexibility and relative simplicity, they are popular geometric modeling tools, especially in computer graphics.
Some authors, such as Alan Barr, define "superquadrics" as including both the superellipsoids and the supertoroids.[1] [2] However, the (proper) supertoroids are not superquadrics as defined above; and, while some superquadrics are superellipsoids, neither family is contained in the other.Comprehensive coverage of geometrical properties of superquadrics and a method of their recovery from range images is covered in a monograph.[3]
## Formulas
### Implicit equation
The surface of the basic superquadric is given by
\left|x\right|r+\left|y\right|s+\left|z\right|t=1
where r, s, and t are positive real numbers that determine the main features of the superquadric. Namely:
• less than 1: a pointy octahedron modified to have concave faces and sharp edges.
• exactly 1: a regular octahedron.
• between 1 and 2: an octahedron modified to have convex faces, blunt edges and blunt corners.
• exactly 2: a sphere
• greater than 2: a cube modified to have rounded edges and corners.
• infinite (in the limit): a cube
Each exponent can be varied independently to obtain combined shapes. For example, if r=s=2, and t=4, one obtains a solid of revolution which resembles an ellipsoid with round cross-section but flattened ends. This formula is a special case of the superellipsoid's formula if (and only if) r = s.
If any exponent is allowed to be negative, the shape extends to infinity. Such shapes are sometimes called super-hyperboloids.
The basic shape above spans from -1 to +1 along each coordinate axis. The general superquadric is the result of scaling this basic shape by different amounts A, B, C along each axis. Its general equation is
\left| x A
\right|r+\left|
y B
\right|s+\left|
z C
\right|t=1.
### Parametric description
Parametric equations in terms of surface parameters u and v (equivalent to longitude and latitude if m equals 2) are
\begin{align} x(u,v)&{}=Ag\left(v,
2 r
\right)g\left(u,
2 r
\right)\\ y(u,v)&{}=Bg\left(v,
2 s
\right)f\left(u,
2 s
\right)\\ z(u,v)&{}=Cf\left(v,
2 t
\right)\\ &-
\pi 2
\lev\le
\pi 2
,-\pi\leu<\pi, \end{align}
where the auxiliary functions are
\begin{align} f(\omega,m)&{}=sgn(\sin\omega)\left|\sin\omega\right|m\\ g(\omega,m)&{}=sgn(\cos\omega)\left|\cos\omega\right|m \end{align}
and the sign function sgn(x) is
sgn(x)=\begin{cases} -1,&x<0\\ 0,&x=0\\ +1,&x>0. \end{cases}
## Plotting code
The following GNU Octave code generates a mesh approximation of a superquadric:function superquadric(epsilon,a) n = 50; etamax = pi/2; etamin = -pi/2; wmax = pi; wmin = -pi; deta = (etamax-etamin)/n; dw = (wmax-wmin)/n; [i,j] = meshgrid(1:n+1,1:n+1) eta = etamin + (i-1) * deta; w = wmin + (j-1) * dw; x = a(1) .* sign(cos(eta)) .* abs(cos(eta)).^epsilon(1) .* sign(cos(w)) .* abs(cos(w)).^epsilon(1); y = a(2) .* sign(cos(eta)) .* abs(cos(eta)).^epsilon(2) .* sign(sin(w)) .* abs(sin(w)).^epsilon(2); z = a(3) .* sign(sin(eta)) .* abs(sin(eta)).^epsilon(3);
mesh(x,y,z);end | 1,087 | 3,767 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2021-31 | latest | en | 0.921658 |
https://www.physicsforums.com/threads/method-od-undetermined-coeff-help-needed.168892/ | 1,511,344,729,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00073.warc.gz | 808,944,168 | 15,156 | # Method od undetermined coeff help needed.
1. May 5, 2007
### HappMatt
1. The problem statement, all variables and given/known data
Ive been working on this problem for way too long and i can get it almost right except im missing a factor of (1/12)e^4x and for the life of me i cant figure out why. Ive been mostly trying to use the methode of undeermined coefficients to no luck and so have also tried variation of parameters and still something is not going right. I have run the equation through my 89 and i have the answer i just cant seem to get it by hand.
2. Relevant equations
y''-2y'-8y=3*e^(4x)-5x^2
complimentary solution=yc=C1*e^(4x)+C28*e^(-2x)
as for finding the particular solution(yp) i think this is where my problem is.
the actual solution as given by my ti 89 is y=((x/2)
+C1-(1/12))*e^(4x)+C2*e^(-2x)+(5x^2)/8-(5x)/16+15/64
3. The attempt at a solution
using yp=Axe^(4x) +Bx^2+Cx+d i got it all right except that im missing the (1/12)*e^4x value and ive tried to many varriations to list them all here but i have litterally spent hours on this and i probally should have posted this back when i was still at only 3 hours worth of time into it but im well beyond that now and its time for a little rest so i can wake up and keep hammering at this one till i get it.
2. May 5, 2007
### Pseudo Statistic
Just a word of advice.
When dealing with undetermined co-efficients, you'd probably find it best to, rather than tackle both right hand terms (3e^(4x) - 5x^2) at the same time, deal with them as two seperate problems; that is, guess their particular solution for:
y'' - 2y' - 8y = 3e^(4x)
Then do the same for y'' - 2y' - 8y = -5x^2 and add up your solutions.
However, it seems like that's not the problem you're having; when you say you're missing this "1/12 e^4x", that doesn't quite add up, since 1/12 e^4x contributes to the homogeneous solution.
That is, unless you're given some initial conditions you're not telling us about? ;)
3. May 5, 2007
### HappMatt
thanks for the advice Psuedo as far as seperating the particular solution Ive tried that and recived the same results. As for the (1/12)*e^(4x) Im mostly sure that is part of the particular solution since the homogeneous solution is just C2e^4x +C1e^(-2x)
4. May 5, 2007
### HappMatt
tried this probleb again for the 10000th and i finally got the answer using variation of parameters.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 714 | 2,485 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-47 | longest | en | 0.930269 |
https://www.physicsforums.com/threads/why-did-they-use-only-meter-resistance-for-tau-of-inductor.848049/ | 1,529,703,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00008.warc.gz | 867,525,402 | 16,258 | # Homework Help: Why did they use only meter resistance for tau of inductor
1. Dec 13, 2015
### nickmanc86
1. The problem statement, all variables and given/known datahttp://imgur.com/uLZdBJC
27. The switch has been closed for
about 1 h. It is then opened at the time defined as t 0 s.
a. Determine the time required for the current iL to drop to
10 μA.
b. Find the voltage VL at t 10μs
c. Calculate VL at t 5tau.
2. Relevant equations
So I completely worked through this problem. Got Eth (20V) and Rth (1.66MΩ). Used 5H for the inductance and plugged into the formula for IL getting iL = 20V/1.66MΩ * e -t/tau where I assumed tau to be 5H/1.66MΩ= 3μs. However the answer comes out incorrect .55μs.
Looking at the solution for the book it calculates tau using only the resistance of the meter 10MΩ and gets a tau of 5μs. This of course gives the same answer as the back of the book. I do not understand why they use only the meters resistance for the tau calculation.
For part B i use the simplified release formula for VLof VL = -Vi * e-t/tau . I simply took my Vi to be Eth or 20V with my tau of 3μs and solved for VL.
However they proceed to use Vi = I*Rm where Vi = (20/1.66MΩ)*10MΩ =120V giving them VL = 120V * e-t/tau using their tau of 5μs.
I do not understand why they calculate tau using only Rmeter and then subsequently find Vi using Eth/Rth * Rmeter.
3. My Work
A)
Rth = 2MΩ||10MΩ = 1.66MΩ
Eth = 24V*10MΩ/12MΩ = 20V
Tau = L/R = 5H/1.66MΩ = 3μs
Plugged into
IL = E/R * e-t/tau -> IL = .55μs
B)
Vi = 20V
Tau = L/R = 5H/1.66MΩ = 3μs
Plugged into
VL = -Vi *e-t/tau = -0.713V
2. Dec 13, 2015
### Staff: Mentor
Are you sure the problem statement is that the switch has been closed and is then opened at t=0? I could be wrong, but that will result in a *very* large voltage being developed across the inductor, limited only by the parasitic capacitance of the inductor and the circuit (which are not specified).
The problem would make much more sense if the switch were closed at t=0...
3. Dec 13, 2015
### nickmanc86
I double checked the problem in the book and it definitely states the switch was closed for 1h and then opened at a time t = 0s. I suppose it is possible there is a typo in the book (it apparently has many). However, correct or not, they proceed to solve it as stated and end up with those answers. Their initial voltage for part B is 120V which is high relative to the voltage source. Unfortunately out of my realm of understanding (basic circuits FTW).
4. Dec 13, 2015
### Staff: Mentor
When the switch opens the only available path for the current is via the 10 MΩ meter...
5. Dec 13, 2015
### nickmanc86
Oh my, of course. Wow I feel dumb now. Thank you very much that definitely makes sense. | 828 | 2,745 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-26 | latest | en | 0.92068 |
https://convertoctopus.com/274-feet-to-yards | 1,619,167,956,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00456.warc.gz | 294,645,168 | 7,562 | ## Conversion formula
The conversion factor from feet to yards is 0.33333333333333, which means that 1 foot is equal to 0.33333333333333 yards:
1 ft = 0.33333333333333 yd
To convert 274 feet into yards we have to multiply 274 by the conversion factor in order to get the length amount from feet to yards. We can also form a simple proportion to calculate the result:
1 ft → 0.33333333333333 yd
274 ft → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 274 ft × 0.33333333333333 yd
L(yd) = 91.333333333333 yd
The final result is:
274 ft → 91.333333333333 yd
We conclude that 274 feet is equivalent to 91.333333333333 yards:
274 feet = 91.333333333333 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 0.010948905109489 × 274 feet.
Another way is saying that 274 feet is equal to 1 ÷ 0.010948905109489 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred seventy-four feet is approximately ninety-one point three three three yards:
274 ft ≅ 91.333 yd
An alternative is also that one yard is approximately zero point zero one one times two hundred seventy-four feet.
## Conversion table
### feet to yards chart
For quick reference purposes, below is the conversion table you can use to convert from feet to yards
feet (ft) yards (yd)
275 feet 91.667 yards
276 feet 92 yards
277 feet 92.333 yards
278 feet 92.667 yards
279 feet 93 yards
280 feet 93.333 yards
281 feet 93.667 yards
282 feet 94 yards
283 feet 94.333 yards
284 feet 94.667 yards | 461 | 1,669 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.767378 |
https://math.stackexchange.com/questions/639972/method-of-characteristics-and-initial-value-problem | 1,643,212,302,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00375.warc.gz | 446,520,607 | 33,802 | # Method of Characteristics and Initial Value Problem
$u_t + 3u_x = 2t$, $u(x,0)=\sin(x/2)$.
I used the method of characteristics to get the answer, $u(x,t)=t^2 + 2\sin^{-1}(x-3t)$.
Does this satisfy the initial condition? I checked for the first equation and it does; however I do not think it satisfies the intial value when $t=0.$ Am I correct in saying so?
• Yes you are corect to saying that it does not satisfy the IC. It is clear when you put $t=0$ in the solution you have found. Jan 16 '14 at 3:48
The solution is $u=\frac{2}{3}\cos(\frac{x}{2})+\sin(\frac{x-3t}{2})-\frac{2}{3}\cos(\frac{x-3t}{2})$.
I think your u(x,t) is not correct: apart from some factors of 2 different, I have $\sin$ and not $\sin^{-1}$.
Also, worth recognizing that you can solve this with just change of variables, $x \rightarrow x + 3 t$.
answer for above question is $$u(x,t)=t^2+\sin(x−3t/2)$$ | 286 | 886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-05 | latest | en | 0.899414 |
https://www.khanacademy.org/math/mr-class-10/x5cfe2ca097f0f62c:surface-area-and-volume/x5cfe2ca097f0f62c:areas-of-combination-of-plane-figures/v/area-of-shaded-region | 1,686,266,349,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00718.warc.gz | 885,971,197 | 60,522 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Class 10 (Marathi)
### Course: Class 10 (Marathi)>Unit 11
Lesson 6: Areas of combination of plane figures
# Area of a shaded region
Here's a fun one: find the area of a shaded region where you first determine the area of a square and then the area of a circle. Created by Sal Khan.
## Want to join the conversation?
• This may not sound very smart but why did you multiple 3*3
• Sal multiplied 3 and 3 because the formula for getting area is A = r^2 pi. If our radius is 3, and if part of the formula is r^2, to get the radius to the second power you multiply 3 and 3 .
• at what was that green thing
• It is due to an incomplete answer. Once you finish typing your answer, assuming it is an acceptable form for the particular question, the green guy goes away:)
• Why he didn't multiply it by 4 like:
100 - 4(3^2)pi? isn't this going to give us all four sides? o.O #confused
PS: Oh I get it, I get it now :D. (3^2)pi will give us the entire area of full circle :D
• What if your book doesn't give the area of the shape?
• what is this website he's using called?
• it still doesn't make sence. | 350 | 1,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-23 | latest | en | 0.935451 |
https://www.physicsforums.com/threads/physics-problem-with-spring-constant-and-kinetic-friction.790440/ | 1,508,825,320,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828178.96/warc/CC-MAIN-20171024052836-20171024072836-00327.warc.gz | 965,717,625 | 17,924 | # Physics Problem with Spring Constant and Kinetic Friction
1. Jan 3, 2015
### Q7heng
1. The problem statement, all variables and given/known data
A person attaches a spring to an cubic object that weighs 36kg and pulls this object along a table made of material X horizontally with a steady speed of 1.3m/s. The spring stretches a distance of 3.4cm. Find the ratio of the spring constant to the kinetic friction coefficient.
2. Relevant equations
I suppose FkkN
Fspring=-kx
F=ma
Not sure if any other ones are needed, but I couldn't find a way to solve this and get a reasonable answer.
2. Jan 3, 2015
### Bystander
What does this information tell you?
3. Jan 4, 2015
### Q7heng
I'm not sure, but I tried to apply it to the F=MA formula... Am I on track?
4. Jan 4, 2015
### Bystander
Okay, we'll work it through that way: "Steady speed" means what in terms of "A?"
5. Jan 4, 2015
### Q7heng
Acceleration=0 at steady speed, but the trouble I'm having is converting everything to Newtons, since spring constant is Newtons/meter, and kinetic friction coefficient is Force of Kinetic Friction/Normal Force. The problem didn't give any units related to those calculations, maybe there is a way to convert it but I have yet figured it out/learned it.
6. Jan 4, 2015
### Bystander
This looks perfectly useful. What's the normal force?
Nothing wrong with this.
You aren't required to generate a numerical answer for every problem on the planet. Sometimes it's just a matter of coming up with a symbolic expression.
7. Jan 4, 2015
### haruspex
All good so far. Now, what is the relationship between F, Fk and Fspring?
8. Jan 5, 2015
### Q7heng
F=Fk+Fspring right?
If that is so then:
F=MA, A=0, and F=0
So 0=Fk+Fspring
0=-kx+μkN
kx=μkN
k/μk=N/x, since we are trying to find the ratio between the spring constant and the kinetic friction coefficient
Since N, normal force, is 36*9.8N right now, and x, distance, is 3.4cm or 0.034m right now, then it is 36*9.8/0.034=10,376.47, which is the ratio of spring constant to kinetic friction, is that correct?
9. Jan 5, 2015
### haruspex
Yes, that all looks right. But you should include the units in the answer.
10. Jan 5, 2015
Thanks!!! | 647 | 2,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2017-43 | longest | en | 0.911221 |
https://www.queryhome.com/puzzle/12075/if-3-4-3-4-4-35-5-4-4-6-4-12-7-4-60-then-8-4 | 1,542,088,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00308.warc.gz | 970,446,273 | 31,146 | # If 3+4=3; 4+4=35; 5+4=4; 6+4=12; 7+4=60 then 8+4=?
2,945 views
If
3+4=3
4+4=35
5+4=4
6+4=12
7+4=60
Then
8+4=?
posted Jan 10, 2016
Ans will be 5
Look
3+4=3 in this
Multiply these 3 and 4 and when u get the two digit ans, square the both, higher digit square will be substracted by lower digit square
Now 3*4=12
2^ - 1^ = 4-1=3
Here ^ is used by me as square
4+4=35
4*4=16
6^ - 1^=36-1=35
Solve all like this and finally
8+4
8*4=32
3^ - 2^=9-4=5
And 5 is ans
Ans given by 8601084829
Krishna P. Sharma
answer Aug 13, 2016 by anonymous
Wow wonderful, touch one to crack
Similar Puzzles
If
2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72
then
7+8=??
–1 vote
If 2+3=3, 4+4=5, 4+5=7, 5+7=10, 5+8=12, 6+5=8
then (3+7)+2 = ? | 374 | 716 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-47 | latest | en | 0.568094 |
https://www.exactlywhatistime.com/time-from/107-weeks-from-today | 1,718,895,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00886.warc.gz | 665,817,556 | 7,701 | # What's the date 107 weeks from today?
## Thursday July 09, 2026
0
107 weeks from today will be 09 Jul 2026, a Thursday. For half-year projections, double-check whether 107 weeks forward remains within 2024. It seems that 107 weeks from now will bring us back to pre 2023. Please include this into our planning as it impacts both the calendar and fiscal year transitions. For extensive calculations like this, I begin by segmenting the year, then multiple 107 by days to get 749 total days ahead. Then either count 749 from June (but that will take forever). You can also estimate 107 weeks to 3.566666666666667 months, count from July and get closer to Thursday July 09, 2026.
## How we calculated 107 weeks from today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how adding 107 weeks to today's date gets calculated on each visit:
• Started with date inputs: starting point: 20 Jun, Units to add: 107 weeks, and year: 2024
• Noted your current time of year: 10 days in middle of June
• Added 107 weeks from current day: 20 Jun, factoring in there are 10 days left in before July
• Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the 2024 fiscal calendar.
### Tips to get your solution: July 09
Thursday Thursday July 09, 2026 is the 190 day of the year or 52.05% through 2026.
• Current date: 20 Jun
• Day of the week: Thursday
• New Date: Thursday July 09, 2026
• New Date Day of the week: Thursday
• June is the end of Q2 so if you're counting dates ahead, you might end up in H2 of the fiscal year.
• This calculation crosses at least one month. Remeber, this will change our day of the week.
• The solution crosses into a different year..
## Ways to calculate 107 weeks from today
1. Just calculate it: Start with a time from today calculator. 107 weeks is easiest solved on a calculator. For ours, we've already factored in the days in + all number of days in each month and the number of days in 2024. Simply add your weeks and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use June's calendar: Begin by identifying on a calendar, note that it’s Thursday, and the total days in July (trust me, you’ll need this for smaller calculations) and days until next year (double trust me, you'll need this for larger calculations). From there, count forward 107 times by weeks, adding weeks from 20 Jun.
3. Use Excel: Regardless of unit type, I use day calculations here. Type =TODAY()+107 into the cell. If you want to add weeks, multiply your day by 7 and months/years will take their own calculation due to the changing days of the week. To find 107 weeks workdays, convert to days but use =WORKDAY(TODAY(), [number of days], [holidays]) into the cell. [number of days] is how many working days you want to add, and [holidays] is an optional range of cells that contain dates of holidays to exclude.
## 107 working weeks from today
107 weeks is Thursday July 09, 2026 or could be Monday May 03, 2027 if you only want workdays. This calculation takes 107 weeks and only adds by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Thursday July 09, 2026 date.
Work weeks Solution
Monday
Tuesday
Wednesday
Thursday
July 09
Friday
Saturday
Sunday
## In 107 weeks, the average person Spent...
• 160885.2 hours Sleeping
• 21391.44 hours Eating and drinking
• 35053.2 hours Household activities
• 10426.08 hours Housework
• 11504.64 hours Food preparation and cleanup
• 3595.2 hours Lawn and garden care
• 62916.0 hours Working and work-related activities
• 57882.72 hours Working
• 94733.52 hours Leisure and sports
• 51411.36 hours Watching television
## What happened on July 09 (107 weeks from now) over the years?
### On July 09:
• 1952 American educator, Presbyterian minister and TV host Fred Rogers (24) weds Sara Joanne Byrd
• 1877 First ever Wimbledon tennis championship begins - first official lawn tennis tournament - men's singles only | 1,041 | 4,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-26 | latest | en | 0.937656 |
https://mocktestpro.in/mcq/fluid-mechanics-mcq-surface-tension-capillarity-vapour-pressure-and-cavitation/ | 1,685,258,701,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00447.warc.gz | 444,016,778 | 23,678 | Engineering Questions with Answers - Multiple Choice Questions
# Fluid Mechanics MCQ – Surface Tension, Capillarity, Vapour Pressure and Cavitation
1 - Question
Calculate the magnitude of capillary effect in millimeters in a glass tube of 7mm diameter, when immersed in mercury. The temperature of the liquid is 25℃ and the values of surface tension of mercury at 25℃ is 0.51 N/m. The angle of contact for mercury is 130°.
a) 140
b) 280
c) 170
d) 210
Explanation: Capillarity rise or fall
h=4*cosθ*σ/ρ*g*d
=4*cos130*0.51/13600*9.81*0.007
=140 mm.
2 - Question
Determine the minimum size of glass tube that can be used to measure water level if the capillary rise in the tube is restricted to 5mm. Consider surface tension of water in contact with air as 0.073 N/m
a) 5.95mm
b) 11.9mm
c) 2.97mm
d) 4.46mm
Explanation: d=4*cosθ*σ/ρ*g*h
=4*1*0.073/1000*9.81*0.005
=5.95mm.
3 - Question
A An oil of vicosity 7 poise is used for lubrication between shaft and sleeve. The diameter of shaft is 0.6 m and it rotates is 360 rpm. Calculate the power lost in oil for a sleeve length of 160mm. The thickness of oil film is 1.0mm
a) 25.31 kW
b) 50.62 kW
c) 37.97 kW
d) 12.65 kW
Explanation: Power lost= torque * angular velocity
= shear stress * area* radius* angular velocity
Shear Stress = viscosity* velocity gradient
Power lost= 7916.8*3.142*0.3*0.3*0.3*2*3.142*60
= 25.31 kW.
4 - Question
Find the capillarity rise or fall if a capillary tube of diameter .03m is immersed in hypothetical fluid with specific gravity 6.5, surface tension 0.25 N/m and angle of contact 147°.
a) 0.44mm fall
b) 0.88mm fall
c) 0.44mm rise
d) 0.88mm rise
Explanation: h=4*cosθ*σ/ρ*g*d
=4*cos147*0.25/6.5*1000*9.81*0.03
=-0.44 mm i.e 0.44 mm fall.
5 - Question
Will capillary rise occur and if it occurs what will be capillary rise if glass capillarity tube is immersed in water and experiment is carried out by astronauts in space.
a) Capillarity rise will not occur
b) Capillarity rise will occur infinitely and will come out in form of fountain
c) Capillarity rise will occur finitely and will be the whole length of tube
d) None of the mentioned
Explanation: Capillary rise is given by
h=4*cosθ*σ/ρ*g*d
hence rise is inversely proportional to g
In space g is 0 m/s2
Hence, capillarity rise will occur finitely and will be the whole length of tube.
6 - Question
The surface tension of fluid in contact with air at 25℃ is 0.51N/m. The pressure inside a droplet is to be 0.05 N/cm2 greater than outside pressure. Determine the diameter of the droplet of water.
a) 4.08mm
b) 8.16mm
c) 2.04mm
d) None of the mentioned
Explanation: P=4*σ/d
d= 4*.51/500
=4.08 mm.
7 - Question
If a fluid of certain surface tension and diameter is used to create a soap bubble and a liquid jet. Which of the two, bubble or liquid jet, will have greater pressure difference on the inside and outside.
a) Liquid jet
b) Soap bubble
c) Both will have same pressure differrence
d) None of the mentioned
Explanation: For soap bubble,
P=8*σ/d
For liquid jet,
P=2*σ/d
Hence, soap bubble will be having more pressure difference.
8 - Question
Capillarity fall is reduced if we take the appartus (capillary tube immersed in fluid having acute angle of contact) considerable distance inside the earth( i.e below the earth crust).
a) True
b) False
Explanation: Capillary rise is given by
h=4*cosθ*σ/ρ*g*d
Inside the earth, g (acceleration due to gravity) decreases. Hence, capillary rise will increase compared to that on the earth’s surface.
9 - Question
For liquid fluids will capillarity rise (or fall) increase or decrease with rise in temperature.
a) Increase
b) Decrease
c) Remain constant
d) First decrease then increase
Explanation: Capillary rise is given by
h=4*cosθ*σ/ρ*g*d
As temperature increases, σ(surface tension) decreases. Therefore, correspondingly rise(or fall) will decrease as their is direct proportional relation between the two.
10 - Question
Cavitation is more pronounced in rough pipes than smooth surfaced pipes.
a) True
b) False | 1,199 | 4,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2023-23 | latest | en | 0.793148 |
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# If x^2 - 3x + k = -30, what is the value of k?
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If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 06:34
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If x^2 - 3x + k = -30, what is the value of k?
(1) 2 is an element of the solution set of the given equation.
(2) 5 is an element of the solution set of the given equation.
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 09:10
Bunuel wrote:
If x^2 - 3x + k = -30, what is the value of k?
(1) 2 is an element of the solution set of the given equation.
(2) 5 is an element of the solution set of the given equation.
x^2 - 3x + k = -30
we need value of x to find the value of k
from 1:
x=2
we can find the value of k by replacing the value of x in given eq.
sufficient.
from 2:
x=5
we can find the value of k by replacing the value of x in given eq.
sufficient
hence D
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 09:47
x^2 - 3x + k = -30
x^2 - 3x + (k + 30) = 0
sum of roots = 3
product of roots = (k+30)
1) 2 + a = 3 => a = 1.
So, k+30 = 2 => k = 28
2) 5 + b = 3 => b = -2.
So, k + 30 = -10 => k = -40
EIther sufficient.
(D)
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Joined: 04 Apr 2018
Posts: 21
Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 23:21
(1). We know that 2 is the solution set, hence simply put that in the given equation & "k" can easily be deduced. Hence, SUFFICIENT.
(2). Similarly, this is also SUFFICIENT.
Hence, both are individually self sufficent to asnwer the question. Hence D
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Joined: 24 Apr 2016
Posts: 30
Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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11 Apr 2018, 06:04
1
This one stumped me for awhile and I'm still not 100% sure. I thought for DS questions 1) and 2) are hints for the same answer. But if X=2 then k=-28. But if X=5 then k= -40. Am I calculating this incorrectly?
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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11 Apr 2018, 06:07
Don't know lol....
Math Expert
Joined: 02 Sep 2009
Posts: 56276
Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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11 Apr 2018, 06:08
dracobook wrote:
This one stumped me for awhile and I'm still not 100% sure. I thought for DS questions 1) and 2) are hints for the same answer. But if X=2 then k=-28. But if X=5 then k= -40. Am I calculating this incorrectly?
Sent from my Pixel 2 XL using GMAT Club Forum mobile app
You are right. The question is flawed. Archived.
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink] 11 Apr 2018, 06:08
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https://byjusexamprep.com/syllogism-for-ssc-exams-i-03d9a2c3-b92c-11e5-a60c-80817d4bb946 | 1,631,876,941,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00592.warc.gz | 193,134,483 | 76,029 | Tricks to solve Syllogism easily in Reasoning Section!
By Parnab Mallick|Updated : January 6th, 2021
The Reasoning section of SSC Exams includes questions from the topic “Syllogisms”. Syllogism is very important not only for SSC CGL but for all the SSC Exams. Students generally find difficulty in understanding Syllogism. In this article, we have explained the basics of syllogism with the help of Venn diagram. Sit with pen and paper while going through this article.
The Reasoning section of SSC Exams includes questions from the topic “Syllogisms”. Syllogism is very important not only for SSC CGL but for all the SSC Exams. Students generally find difficulty in understanding Syllogism. In this article, we have explained the basics of syllogism with the help of Venn diagram. Sit with pen and paper while going through this article.
Syllogism Concepts for SSC Exams
Syllogism comes under verbal reasoning section and is frequently asked in many competitive exams. These types of questions contain two or more statements and these statements are followed by the number of conclusions. You have to find which conclusions logically follows from the given statements.
The best method of solving the Syllogism’s problem is through Venn Diagrams. There are four ways in which the relationship could be made.
Category 1
All A are B – Means the whole circle representing A lies within the circle representing B.
Here we can also make a conclusion: Some B are A. Some A are B.
For example: All boys are men.
Here we can also make a conclusion: Some men are boys. Some boys are men.
All apples are fruits.
Here we can also make a conclusion: Some fruits are apples. Some apples are fruits.
Category 2
No A is B – means that circles representing A and B does not intersect at all.
For example : No ball is bat.
No door is wall.
Category 3
Some A are B
Means that some part of the circle represented by A is within the circle represented by B.
This type of (category 3) statement gives the following conclusions:
(i) Some A are B also indicates that - Some A are not B
(ii) Some A are B also indicates that – All A are B.
(iii) Some A are B also indicates that – All B are A.
(iv) Some A are B also indicates that – All A are B and All B are A.
For e.g.: Some mobiles are phones.
(i)
Category 4.
Some A are not B
Means that some portion of circle A has no intersection with circle B while the remaining portion of circle A is uncertain whether this portion touches B or not.
(i) Some A are not B also indicates that – Some A are B.
(ii) Some A are not B also indicates that – No A is B.
Types of Sentences Conclusions All A are B Some B's are A Some A's are B Some A are B All A are B All B are A Some B are A Some A are not B No A is B No B is A Some A are not B Some A are B All B are A No A is B
Important Points –
1. At least statement – At least statement is same as some statement.
For ex:
Statement: All kids are innocent.
Here we can make conclusion: At least some innocent are kids (Some innocent are kids).
2. Some not statement: Some not statement is opposite to “All type” statement. If All being true then Some not being false
For e.g.
1. Statement: Some pens are pencils. No pencils are jug. Some jug is pens.
Here we can make a conclusion: Some pens are not pencils, which is true. In above figure, green shaded part shows; some pens are not pencils, because in statement it is already given No pencils is jug.
Complementary Pairs: (Either & or) – Either and or cases only takes place in complementary pairs.
Conclusions: (i) Some A are B. (ii) No A are B.
From the given above conclusions, it is easy to understand that one of the given conclusions must be true, which is represented by option either (i) or (ii). These types of pairs are called complementary pairs.
Note: ‘All A are B’ & ‘Some A are not B’ are also complementary pairs.
Note: It is important to note that, in complementary pairs, one of the two conclusions is true and other will be false simultaneously.
For example –
Statement: All A are B. Some B are C.
Conclusion: I. All C are A. II. Some C are not A.
Here we can make a conclusion, either I or either II follows.
Possibility cases in Syllogism – In possibilities cases, we have to create all possibilities to find whether the given conclusion is possible or not. If it is possible and satisfies the given statement than given conclusion will follow otherwise conclusion will not follow.
1. E.g.
Statement: All A are B. Some B are C.
Conclusion: All A being C is a possibility.
Conclusion is true.
Possibility figure –
2. E.g.
Statements: No stone is a white. Some white are papers.
Conclusions: I. All stones being paper is a possibility.
Possibility figure:
Conclusion is true.
3. E.g.
Statements: Some mouse is cat.
All mouse are pets. No pet is animal.
Conclusions: I. All mouse being animal is a possibility.
The conclusion is false because possibility figure is not possible.
If we say all mouse being an animal is a possibility is true than given statements No pet is animal will be wrong. Here in the statement, it is given No pet is animal and All mouse is pet. So we can make also conclusion here that no mouse are animal is true.
Important Rule:
The statement itself is not a conclusion – Conclusion has to be different from the statement.
E.g. Statement - All A are B
Conclusion - All are B. (invalid) Conclusion does not follow.
Conclusion - Some A are B (follow) Conclusion follows.
Note: If statement and conclusion are same then the conclusion does not follow. This rules also follow in possibilities case
You can now take up the below-given Reasoning Quiz on Syllogisms to clear your doubts related to the topic:
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Parnab Mallick is an educator and mentor with an expertise in SSC and Railway exams. He tries to make students’ life easy by guiding them the right path and knowledge to cater to their dream govt. job. He lives with the notion ‘We all have equal opportunity to be unequal’
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SSE dayaram meenaJul 15, 2021 | 1,569 | 6,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-39 | longest | en | 0.958912 |
http://mathhelpforum.com/pre-calculus/119405-quadratic-factors-cubic-polynomial.html | 1,526,903,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122758-00285.warc.gz | 185,826,809 | 12,127 | 1. Quadratic factors of a cubic polynomial
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
2. Hello Detanon
Originally Posted by Detanon
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
Suppose the other factor is $\displaystyle (rx+s)$. Then:
$\displaystyle (x^2+bx-2)(rx+s) = x^3+(2b-1)x^2 -p$
Comparing coefficients:
$\displaystyle x^3: r = 1$
So:
$\displaystyle (x^2+bx-2)(x+s) = x^3+(2b-1)x^2 -p$
Comparing coefficients:
$\displaystyle x^2: (s+b) = (2b-1) \Rightarrow s = b-1$ (1)
$\displaystyle x: -2+sb = 0 \Rightarrow sb = 2$ (2)
So, from (1) and (2):
$\displaystyle b(b-1)=2$
$\displaystyle \Rightarrow b^2-b-2=0$
$\displaystyle \Rightarrow (b-2)(b+1)=0$
$\displaystyle \Rightarrow b = 2, -1$
$\displaystyle \Rightarrow s = 1, -2$
Constant term:
$\displaystyle p = 2s$
$\displaystyle \Rightarrow p = 2, -4$
3. Sorry, but I really dont get that.
I dont get why you suppose the other facotr is (rx+s)
and what does $\displaystyle x^3: r$
mean?
Is there a different method to do it?
4. Hello Detanon
Originally Posted by Detanon
Sorry, but I really dont get that.
I dont get why you suppose the other facotr is (rx+s)
Because when one factor of a cubic is a quadratic, the other is linear; i.e. it is of the form $\displaystyle (rx + s)$ for some values of $\displaystyle r$ and $\displaystyle s$.
and what does $\displaystyle x^3: r$mean?
It doesn't mean anything. You have taken it out of context. Read the whole thing. I am comparing coefficients. (Do you know what thay means?) First I am looking at the coefficient of $\displaystyle x^3$. Later on, I look at the coefficients of $\displaystyle x^2, x$ and the constant. I have used a colon (':') each time to show which coefficient I am looking at.
Is there a different method to do it?
No, not a method that is as sensible and straightforward as the one I have shown you.
Read it through again - it's really very simple.
5. I still dont get it.. Is there another method where dividing is involved? I thnik thats the one I get.
6. Hello Detanon
I really don't know what your problem is here. Multiply out the brackets from
$\displaystyle (x^2+bx-2)(rx+s)$
and then note that the result has to be identically equal to
$\displaystyle x^3 +(2b-1)x^2 -p$
(Can you do that?)
In this way you get the equations for $\displaystyle r, s, b$ and $\displaystyle p$ that I have shown you. I repeat: if you understand the basic rules of algebra, then trust me, this is really very simple. (And certainly much easier than trying to do an algebraic long division.)
7. Originally Posted by Detanon
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
If $\displaystyle x^2 + bx - 2$ is a factor then it should be obvious, after a little thought, that the other factor has to be $\displaystyle x + \frac{p}{2}$. This is the only way that you can get the $\displaystyle x^3$ term and the constant term of $\displaystyle -p$.
So $\displaystyle x^3 + (2b - 1) x^2 - p = (x^2 + bx - 2)\left(x + \frac{p}{2}\right)$. Expand this out. You require the coefficient of $\displaystyle x^2$ to be $\displaystyle 2b - 1$ and the coeffcient of $\displaystyle x$ to be zero:
$\displaystyle \frac{p}{2} + b = 2b - 1 \Rightarrow p = 2b - 2$ .... (1)
$\displaystyle \frac{bp}{2} - 2 = 0 \Rightarrow bp = 4$ .... (2)
Solve equations (1) and (2) simultaneously for b and p.
For checking purposes:
Spoiler:
b = 2, -1 and p = 2, -4
8. Where do you get
Constant term:
p = 2s
from??
9. Hello Detanon
Originally Posted by Detanon
Where do you get
Constant term:
p = 2s
from??
The constant term when we expand:
$\displaystyle (x^2+bx-2)(rx+s)$ is $\displaystyle -2s$, and this is equal to the constant term in $\displaystyle x^3 +(2b-1)x^2 -p$
Hence $\displaystyle p = 2s$ | 1,240 | 4,011 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-22 | latest | en | 0.836116 |
https://math.stackexchange.com/questions/3172943/number-of-matrices-with-each-row-and-column-having-exactly-one-1 | 1,561,542,965,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000266.39/warc/CC-MAIN-20190626094111-20190626120111-00226.warc.gz | 512,959,215 | 35,924 | Number of matrices with each row and column having exactly one 1.
Consider a square matrix of order $$n = 5$$ such that $$a_{ij} = 0 ~ \forall ~ i+j = n+1; a_{ij} \in \{0,1\}$$. In each row as well as in each column there is only one non zero element. Then number of such matrices is?
First we note that right diagonal has only $$0$$s.
Then I tried it this way: We choose a place for one among each row and mark the other places in that column and same row as forbidden (i.e. no more one's).
So for first column we have 4 choices then 3 choices then 2 then 1 and then 2s.
Thus, Number of ways = $$4\times 3 \times 2\times 1\times 2 = 48$$
But its erroneous because the number of choices change if we place 1 above 0 in each attempt.
What's the correct way to solve this question?
Flip the array uoside down. Then $$a_{ii}=0$$, so the positions of the nonzero $$a_{ij}$$ form a derangement of the numbers from $$1$$ to $$5$$.
• Suppose the nonzero cells are $a_{1a},a_{2b},a_{3c},a_{4d}$ and $a_{5e}$. Then $a,b,c,d,e$ are five different numbers, and $1\neq a,2\neq b,3\neq c,4\neq d,5\neq e$. – Empy2 Apr 3 at 8:47
• The only one in row 1 is $a_{1a}$. The only one in column $a$ is $a_{1a}$. – Empy2 Apr 3 at 9:49 | 406 | 1,221 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-26 | longest | en | 0.864078 |
http://mathoverflow.net/questions/61125/system-of-two-second-order-differential-equations | 1,469,681,712,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00085-ip-10-185-27-174.ec2.internal.warc.gz | 160,288,893 | 16,449 | # system of two second order differential equations
Hi everyone, I know that this system dont have analytical solutions. I want to get numerical solutions, but in function of some constants $A_i$. Mathematica can help me, but if somebody have idea? This equations describe a physical model
$$A_6 x + A_4 (y')^2 - 2 A_2 x'' - A_3xy'' + A_4yy'' = 0$$
$$A_5 - A_3 (x')^2 - A_3xx'' + A_4yx'' - 2A_1y'' = 0$$
$'=(d/dt)$, $''=(d^2/dt^2)$, $A_i$-known constants. The initial conditions are:
$$x(0)=a, y(0)=0, x'(0)=0, y'(0)=0$$
Thank you in advance!!!
-
What is the "physical model" this system describes; i.e., where does it come from? Knowing this might help answer your question. – drbobmeister Apr 10 '11 at 18:03
## 2 Answers
You can reduce the number of parameters quite a bit, for starters. Set $$x = \alpha u , \; y = \beta v, \; t = \gamma \tau$$
and write $\dot w = \frac{d}{d \tau} w, \ddot w = \frac{d^2}{d\tau^2} w$. By choosing the constants $\alpha, \beta, \gamma$ properly, you should be able to nondimensionalize the system to something like $$c u + (\dot v)^2 - \ddot u - d u \ddot v + v \ddot v = 0$$ $$1 - (\dot u)^2 - u \ddot u + d^{-1} \ddot u v - \ddot v = 0$$ $$u(0) = \tilde a, \; v(0) = \dot u(0) = \dot v(0) = 0.$$ So there are only three independent constants in the system, not 7.
-
Thank you a lot professor Engler. In my new post - New system of two second order differential equations I got maybe simple system but I am not sure can I get a solutions like this way. If you can help me it will be great. In some way, if the number of constants can reduce, can I final get the solutions [x(...,t),y(,...t)]? Thank you again! – reptil Apr 13 '11 at 8:37
If you can take this problem into consideration to reduce number of constants to get semi-analytical solutions for my new post, I will be very grateful (New system of two second order differential equations). Of course, to lead system to may cause long with analytical aid. But I have 5 constants and Runge-Kutte can not help me. Just for special cause, where constants have numerical value. Thank you professor Engler. – reptil Apr 13 '11 at 14:17
It is unlikely that there is an analytic solution but you may be able to make some progress by rewriting as a first-order system. For example, with the equations as in Hans Engler's answer, you can define $w=\dot{u}$ and $z=\dot{v}$, and get a system of equations
$\dot{u} = w$
$\dot{v} = z$
$\dot{w} = \frac{cu + z^2 + (v-du)(1-w^2)}{1-(du-v)(u-v/d)}$
$\dot{z} = \frac{(v/d-u)(cu+z^2) + 1 - w^2}{1-(du-v)(u-v/d)}$
with
$u(0)=\tilde{a}$, $v(0)=w(0)=z(0)=0$.
This looks more complicated than the original set of equations, but you have the advantage that it's first-order and autonomous, and hence amenable to the techniques applicable to first-order autonomous nonlinear dynamical systems, such as linearization about fixed points, analysis of periodic orbits, energy theorems etc.
-
Thank you for good suggestion. If you can describe some of method on simple example or take me to some literature. In my new post - New system of two second order differential equations I got maybe simple system but I am not sure can I get a solutions like this way. If you can help me it will be great. In some way, if the number of constants can reduce, can I final get the solutions [x(...,t),y(,...t)]? Thank you a lot! – reptil Apr 13 '11 at 8:36
What do you think to find dw/dz and eliminate t? Can you help me to describe my new system of two second order differential equations to find w and z like here? Yes iti is trivial but I didn't understand how Hans Engler got just 1 in front of almost all (du/dt,d^2v/dt^2...) – reptil Apr 14 '11 at 11:19 | 1,122 | 3,683 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2016-30 | latest | en | 0.850732 |
https://math.stackexchange.com/questions/1416175/ternary-expansion-and-cantor-set | 1,563,608,528,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00339.warc.gz | 474,247,133 | 35,396 | # Ternary expansion and Cantor set
If $x$ has a ternary expansion $\sum \limits_{k=1}^{\infty}\dfrac{c_k}{3^k}$ where each $c_k\in \{0,2\}$ then $x$ belongs to Cantor set.
Proof: Suppose $x$ has a ternary expansion $\sum \limits_{k=1}^{\infty}\dfrac{c_k}{3^k}$ where each $c_k\in \{0,2\}$. We will show $x\in C$ by induction. Clearly $x\in C_0=[0,1]$ since $0\leqslant x \leqslant 1$. Next, if $c_1=0$, then $$x\leqslant (0.0222\dots)_3=(0.1)_3=\frac{1}{3}$$ On the other hand, if $c_1=2$, then $$x\geqslant (0.2000\dots)_3=(0.2)_3=\frac{2}{3},$$and hence, $x\in C_1$. Now we approach the inductive step.
Assume $x\in C_k$ for some $k\in \mathbb{N}$. Then $x$ is in any of the $2^k$ subintervals of $C_k$, each of the length $\frac{1}{3^k}$. WLOG, say $x\in [a,b]$. Since we remove the interior of the middle-thirds, we have two disjoint closed intervals, $$\left[a, a+\frac{1}{3^{k+1}}\right] \quad\text{and} \quad \left[b-\frac{1}{3^{k+1}}, b\right]$$If $c_{k+1}=0$, then $x=\sum \limits_{n=1}^{\infty}\dfrac{c_n}{3^n}\leqslant \sum \limits_{n=1}^{k}\dfrac{c_n}{3^n}+\dfrac{1}{3^{k+1}}$ and $x\in [a,b]$.
Why in this case the right side of inequality is $\leqslant a+\dfrac{1}{3^{k+1}}$?
• Because $\sum\limits_{n=k+2}^\infty\frac2{3^n}=\frac1{3^{k+1}}$. – Did Aug 31 '15 at 20:12
• @Did, I know this. I am asking why $\sum \limits_{n=1}^{k}\dfrac{c_n}{3^n}\leqslant a$ – ZFR Aug 31 '15 at 20:21
• Actually, by construction, $\sum\limits_{n=1}^k\frac{c_n}{3^n}=a$... – Did Aug 31 '15 at 20:25
• Is it also possible to argue that $x$ is removed in the $k$-th iteration only if $\alpha_k = 1$ for all ternary expansions of $x$? – David Nov 5 '15 at 19:37 | 711 | 1,659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2019-30 | latest | en | 0.526084 |
https://blog.c0nrad.io/posts/scattering-angles/ | 1,611,528,601,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00301.warc.gz | 236,031,974 | 5,498 | ## Learning and learning
Aug 30, 2020 - 3 minute read
# Scattering Angles
Calculating scattering angles from fully elastic spherical collision.
## Background
In one of my first blog posts, I talked about the difficulty I had with calculating the resulting angles from a collision of two spheres.
I could bounce objects off each other, but not at an angle.
Well, fast forward to today (5 months later), I’m now in school for physics (undergrad)! And a little bit better prepared to tackle the problem.
I reached out to my professor of classical mechanics about the above problem, and he suggested I read Ch. 14 of Taylor’s Mechanics on Collision Theory. The chapter was pretty interesting, and it was stuff I wanted to learn sooner or later. I’m hoping soon I can start modeling particle collisions (a long term goal of mine, I’m about halfway through “Elementary Particles” by Griffiths).
Anyways, in Taylor’s book, example 14.5 talks about hard sphere scattering, and it turns out that the angle of incidence and reflection are equal. Kind of reminds me of bouncing photons off a medium.
To be honest, I don’t quite understand why the angles should be equal. There’s probably some beautiful action/Lagrangian minimization that shows that it must be the path. I’ll solve that another day.
But, poking around some more, I found out that wikipedia lists the equations I needed:
$$\bm{v_1 \prime} = \bm{v_1} - \frac{2m_2}{m_1 + m_2} \frac{ < \bm{v_1} - \bm{v_2}, \bm{x_1} - \bm{x_2}>}{ || \bm{x_1} - \bm{x_2} ||^2} ( \bm{x_1} - \bm{x_2})$$
$$\bm{v_2 \prime} = \bm{v_1} - \frac{2m_1}{m_1 + m_2} \frac{ < \bm{v_2} - \bm{v_1}, \bm{x_2} - \bm{x_1}>}{ || \bm{x_2} - \bm{x_1} ||^2} ( \bm{x_2} - \bm{x_1})$$
With this, I could put a bunch of balls into a container and watch them bounce! I think I might use this in the future to do some Statistical Mechanics/Boltzman entropy modeling fun later:
## Scattering
Anyways, I figured it’d be fun to build something similar to Rutherford’s Scattering Experiment:
(Code)
It turned out pretty nice.
From this I was able to plot the scattering angle against the impact parameter (b, how much offset the two items are).
(Code)
Pretty neat stuff. It’s cool how often stuff bounces back. I imagine this is probably similar to what Rutherford saw all those years ago.
### What was difficult
• In terminal, character height is twice as much as character width, so the points appears to move much faster in one direction. So instead the renderer now treats one character height as 2 units of length
• I’m using a very simple time-stepping integrator, and some of the incident angles are pretty sensitive, so if the ball moves to fast I’ll get nonsense collision angles since the circles will be on top of each other
• The correct thing to do would be to reverse the simulation till the two spheres are at the point of collision, and then calculate the angles
• Being lazy, I just set the timestep to 1000 frames/second, in the future I’ll go back and do it correct
Otherwise not much. I wish I understood the math behind the collisions better.
### What went well
• I really prefer doing graphics in terminal compared to rendering in 3d with webgl, being able to quickly iterate is very nice.
### Future
Probably nothing for this project. Up next I want to calculate the Clebsh-Gordan coefficients. Eventually I’ll come back to this for elementary particle collisions. First will probably be relativistic collisions. | 877 | 3,469 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-04 | longest | en | 0.899301 |
https://ddxart.co.uk/the-goldbach-conjecture/ | 1,631,869,714,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00578.warc.gz | 235,816,442 | 8,881 | # The Goldbach Conjecture
Mathematicians like to prove things. Things that most of us would take for granted, statements like 1+1=2 are of no use to a mathematician unless it they can be ‘proved’ – demonstrated to be true in every circumstance. In Bertrand Russell and Alfred North Wallace’s great work, The Principles of Mathematics, several hundred pages are devoted to proving that 1+1=2. Why? Because in other places mathematicians will want to use the statement that 1+1=2 as part of the proof of something far more complex, and any proof is only as good as the parts that make it up. The history of mathematics is littered with proofs that relied on something that was ‘self-evident’, only to have it turn out years later that the proof is useless because the self-evident something turned out not always to be true.
It’s every mathematician’s dream to produce a proof of something that has never been proved before (or to demolish someone else’s proof). Many of the propositions that have been put forward but not yet proved are exceedingly complex but some are so simple that you’d think it would take no more than a few minutes to work out why they ‘must’ be true.
So it was that back in the 18th century, a German mathematician named Christian Goldbach suggested that every even number other than 2 could be produced by adding together two prime numbers – numbers that can only be divided by themselves and 1 without leaving a remainder. It doesn’t take long to demonstrate that it is true for even numbers of reasonable size and, in the centuries following, Goldbach’s ‘conjecture’ was found to be true of bigger and bigger even numbers. The advent of computers has led to the analysis of numbers of ridiculous size being considered and, yes, every one of them can be produced by adding together two prime numbers.
But no-one has ever proved that it will be true for every conceivable number, and there are plenty of cases of statements that turned out to be true over and over again, with bigger and bigger numbers, only for an exception to be found – often after many years.
So two centuries later Goldbach’s Conjecture remains just that, a conjecture. In practice, unless you’re going to be working with numbers that have 20 or more digits, you needn’t worry, because that’s how far the conjecture has been shown to be true. But don’t try telling a mathematician that the conjecture ‘must’ be true after all this time – or they may ask you to prove it!
Anyway this little illustration is hardly the most visually exciting of the collection but it does make the point – for the first few even numbers at least – and there are some interesting patterns if you look. Simply follow the lines from any pair of prime numbers, one on each side, and where they meet they are added together to produce an even number.
Before you ask, the number 1 isn’t included since mathematicians don’t consider it a prime number – you can work out for yourself why if you look back at the definition of a prime number given above.
And finally, a little problem to ponder. If the Goldbach Conjecture is ever proved to be true, it will also prove that any number is actually the sum of three prime numbers. Can you see how? | 695 | 3,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-39 | latest | en | 0.973207 |
https://physics.com.hk/2021/08/10/ex-1-21-the-dumbbell-3-2/ | 1,675,619,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00356.warc.gz | 477,335,245 | 16,816 | # Ex 1.21 The dumbbell, 3.2
Structure and Interpretation of Classical Mechanics
.
c. Make a change of coordinates to a coordinate system with center of mass coordinates $\displaystyle{x_{cm}}$, $\displaystyle{y_{cm}}$, angle $\displaystyle{\theta}$, distance between the particles $\displaystyle{c}$, and tension force $\displaystyle{F}$. Write the Lagrangian in these coordinates, and write the Lagrange equations.
~~~
[guess]
\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 y_1}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 x_1}{m_0 + m_1} \\ \end{aligned}}
\displaystyle{ \begin{aligned} \ddot y_{cm} &= \frac{F \sin \theta - F \sin \theta}{m_0 + m_1} = 0 \\ \ddot x_{cm} &= \frac{F \cos \theta - F \cos \theta}{m_0 + m_1} = 0 \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} y_0 &= y_{cm} - \frac{m_1}{M} c(t) \sin \theta \\ x_0 &= x_{cm} - \frac{m_1}{M} c(t) \cos \theta \\ y_1 &= y_{cm} + \frac{m_0}{M} c(t) \sin \theta \\ x_1 &= x_{cm} + \frac{m_0}{M} c(t) \cos \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} x_1 - x_0 &= \frac{m_0}{M} c(t) \cos \theta + \frac{m_1}{M} c(t) \cos \theta \\ &= c(t) \cos \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} \dot x_1 - \dot x_0 &= \dot c(t) \cos \theta - c(t) \dot \theta \sin \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= \ddot c(t) \cos \theta - \dot c(t) \dot \theta \sin \theta - \dot c(t) \dot \theta \sin \theta - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
\displaystyle{ \begin{aligned} y_1 - y_0 &= c(t) \sin \theta \\ \dot y_1 - \dot y_0 &= \dot c(t) \sin \theta + c(t) \dot \theta \cos \theta \\ \ddot y_1 - \ddot y_0 &= \ddot c(t) \sin \theta + \dot c(t) \dot \theta \cos \theta + \dot c(t) \dot \theta \cos \theta + c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}
When $\displaystyle{\dot c(t) = 0}$ and $\displaystyle{\ddot c(t) = 0}$,
\displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
\displaystyle{ \begin{aligned} \ddot y_1 - \ddot y_0 &= ... \\ - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} \tan \theta &= \frac{ c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta }{- c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta} \\ &= \frac{ \ddot \theta - \dot \theta^2 \tan \theta }{- \ddot \theta \tan \theta - \dot \theta^2} \\ \end{aligned}}
\displaystyle{ \begin{aligned} \tan \theta \left( - \ddot \theta \tan \theta - \dot \theta^2 \right) &= \ddot \theta - \dot \theta^2 \tan \theta \\ &... \\ 0 &= \ddot \theta (1 + \tan^2 \theta) \\ \ddot \theta &= 0 \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
Let $\displaystyle{\frac{1}{\mu} = \left( \frac{1}{m_1} + \frac{1}{m_1} \right)}$ and since $\displaystyle{\ddot \theta = 0}$,
\displaystyle{ \begin{aligned} - \frac{1}{\mu} F \sin \theta &= - c(t) \dot \theta^2 \sin \theta \\ - \frac{1}{\mu} F \cos \theta &= - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
Since $\displaystyle{\sin \theta}$ and $\displaystyle{\cos \theta}$ cannot be both zero at the same time,
\displaystyle{ \begin{aligned} - \frac{1}{\mu} F &= - c(t) \dot \theta^2 \\ \end{aligned}}
Put $\displaystyle{c(t) = l}$,
\displaystyle{ \begin{aligned} \frac{1}{\mu} F &= l \dot \theta^2 \\ \dot \theta^2 &= \frac{1}{l \mu} F \\ \end{aligned}}
[guess]
— Me@2021-08-08 05:41:21 PM
.
. | 1,750 | 4,361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 29, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2023-06 | latest | en | 0.356996 |
http://math.stackexchange.com/questions/265897/whether-twin-primes-satisfy-this-one | 1,466,820,119,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783391766.5/warc/CC-MAIN-20160624154951-00135-ip-10-164-35-72.ec2.internal.warc.gz | 188,713,330 | 18,105 | # Whether twin primes satisfy this one?
It seems that difference of squares of any twin primes $+1$ will always lead to
number which might be
a) A square of a twin prime
b) Itself a twin prime
$C$ = ($A^2$-$B^2$ )+$1$ ------> $(1)$
Where
$C$ --- > might be a twin prime or square of a twin prime,
$A$ and $B$ are twin primes where $A$ is > $B$
My questions is whether eqn ($1$) is true?
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When you say $A$ and $B$ are twin primes with $A > B$, do you mean $A = B + 2$, or that $A$ can be any twin prime greater than $B$? – Michael Albanese Dec 27 '12 at 12:57
If $A$, $B$ are twin primes, they differ by 2, so the conjecture seems to be that
if $C = 2(A + B) + 1$, with $A$, $B$ twin primes, then $C$ is either a twin prime or square of a twin prime.
That's quickly falsified by taking $A = 101, B = 103$. For then $C = 409$ which is neither a twin prime nor the square of one.
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Similarly 137, 139, 553 and many (most?) larger examples – Henry Dec 27 '12 at 13:02
$\,(103^2-101^2)+1=409\,$ , which is neither of (a)-(b), though it is a prime.
$\,(4801^2-4799^2)+1=19201=7\cdot 13\cdot 211\,$ , which is neither of (a)-(b) and not even a prime
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Again the number 19201 is a product of primes? – Shan Dec 27 '12 at 13:14
Well, any natural number is a prime or a product of primes except $\,1\,$....is this what you actually meant to ask? – DonAntonio Dec 27 '12 at 13:15
I agree..need to work more on this.. – Shan Dec 27 '12 at 13:19
Some counterexamples:
$$\begin{array}{c|c} p & (p+2)^2-p^2+1 \\\hline 137 & 7^1 \times 79^1 \\ 179 & 7^1 \times 103^1 \\ 197 & 13 \times 61^1 \\ 227 & 11^1\times 83^1 \\ 269 & 23^1\times 47^1 \\ 347 & 7^1\times 199^1 \\ 431 & 7^1\times 13^1\times 19^1 \end{array}$$
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