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How find $\max _{z: \ |z|=1} \ f \left( z \right)$ for $f \left( z \right) = |z^3 - z +2|$ Let $f : C \mapsto R $, $f \left( z \right) = |z^3 - z +2|$. How find $\max _{z: \ |z|=1} \ f \left( z \right)$ ?

i am going to parametrize the unit circle by $z = \cos t + i \sin t.$ we have $$\begin{align}|z^3 - z + 2|^2 &= (\cos 3t - \cos t + 2)^2 +(\sin 3t - \sin t)^2 \\&=\cos^2 3t + \cos^2t+4-2\cos 3t \cos t+4\cos 3t-4\cos t \\ &+\sin^2 3t + \sin^2 t-2\sin 3t \sin t\\ &=6-2\cos 2t+4\cos 3t-4 \cos t\end{align}$$ the critical points of $ |z^3 - z + 2|$ are given by $$\begin{align}0 &=\sin 2t-3\sin 3t+\sin t \\ &= 2\sin t \cos t-3(3\sin t - 4 \sin^3 t) + \sin t\\ &=\sin t(2\cos t-9+12\sin^2 t+1)\\ &=-2\sin t(6\cos^2 t-\cos t-2)\\ &=-2\sin t(2\cos t+1)(3\cos t - 2) \end{align}$$ they are $$t = 0, 0.841, \pi, 2\pi/3, 4\pi/3, 5.442$$ by looking at these values, you can find the global max. it is at $t = 2\pi/3$ and the value is $$|z^3 - z + 2|^2 =6-2\cos 2t+4\cos 3t-4 \cos t\Big|_{t = 2\pi/3}=6+1+4+2=13.$$

Area of an equilateral triangle Prove that if triangle $\triangle RST$ is equilateral, then the area of $\triangle RST$ is $\sqrt{\frac34}$ times the square of the length of a side. My thoughts: Let $s$ be the length of $RT$. Then $\frac s2$ is half the length of $\overline{RT}$. Construct the altitude from the $S$ to side $\overline{RT}$. Call the intersection point $P$. Now, you have a right triangle whose sides are $|\overline{RP}| = \frac s2$ and $|\overline{RS}| = s$. By the Pythagorean Theorem, $|\overline{SP}| = \sqrt{s^2 - \frac14 s^2} = \sqrt{ \frac34 s^2} = \frac{\sqrt{3}}{2} s$. The area of the triangle is $$\frac12 \left(|\overline{SP}|\right)\left(|\overline{RT}|\right) = \left(\frac12 s\right) \left(\frac{\sqrt{3}}{2} s\right) = \frac{\sqrt{3}}{4}s^2,$$ as suggested.

From AoPS wiki, Method 1: Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is $\frac{s}{2}$. Using the Pythagorean theorem, we get $s^{2}=h^{2}+\frac{s^{2}}{4}$, where $h$ is the height of the triangle. Solving, $h=\frac{s \sqrt{3}}{2}$. (note we could use $30-60-90$ right triangles.) We use the formula for the area of a triangle, $\frac{1}{2} b h$ (note that $s$ is the length of a base), so the area is $$ \frac{1}{2}(s)\left(\frac{s \sqrt{3}}{2}\right)=\frac{s^{2} \sqrt{3}}{4} $$ Method 2: (warning: uses trig.) The area of a triangle is $\frac{a b \sin C}{2}$. Plugging in $a=b=s$ and $C=\frac{\pi}{3}$ (the angle at each vertex, in radians), we get the area to be $\frac{s^{2} \sin c}{2}=\frac{s^{2} \frac{\sqrt{3}}{2}}{2}=\frac{s^{2} \sqrt{3}}{4}$

Help in evaluating limit of the given function. Evaluate the limit $$\lim_{x\to \infty} \left(\cfrac{x^2+5x+3}{x^2+x+2}\right)^x $$ I'm not sure how to evaluate this limit. This is what I've done yet: $$\begin{align} \lim_{x\to \infty} \left\{1 + \left(\cfrac{x^2+5x+3}{x^2+x+2} - 1\right) \right\} ^x\\ =\ \lim_{x\to \infty} \left\{1 + \left(\cfrac{4x+1}{x^2+x+2}\right) \right\}^x\end{align}$$ Not sure where to proceed from here. Out of ideas! Any help will be greatly appreciated.

$$\begin{align}\lim_{x\to\infty}\left(\frac{x^2+5x+3}{x^2+x+2}\right)^x&=\lim_{x\to\infty}\left(1+\frac{1}{\frac{x^2+x+2}{4x+1}}\right)^{\frac{x^2+x+2}{4x+1}\cdot\frac{x(4x+1)}{x^2+x+2}}\\&=\lim_{x\to\infty}\left(\left(1+\frac{1}{\frac{x^2+x+2}{4x+1}}\right)^{\frac{x^2+x+2}{4x+1}}\right)^{\frac{4+\frac 1x}{1+\frac 1x+\frac{2}{x^2}}}\\&=e^4\end{align}$$ Here, note that $\lim_{x\to\infty}\frac{x^2+x+2}{4x+1}=+\infty$ and that $\lim_{y\to\infty}\left(1+\frac 1y\right)^y=e$.

range of $m$ such that the equation $|x^2-3x+2|=mx$ has 4 real answers. Find range of $m$ such that the equation $|x^2-3x+2|=mx$ has 4 distinct real solutions $\alpha,\beta,\gamma,\delta$ To show how I got the wrong answers. From $|x^2-3x+2|=mx$ I got the two case $x^2-3x+2=mx$ when $x>2 $ or $ x<1$ and $x^2-3x+2=-mx$ when $1<x<2$ also $m\neq0$ (because if $m=0$ , This will given only 2 answers not 4) try to find the first two answers $x^2-3x+2=mx$ when $x>2 $ or $ x<1$ $x^2-(3+m)x+2=0$ when $x>2 $ or $ x<1$ Use quadratic formula will given $x= \frac{3+m\pm \sqrt{(3+m)^2-4\times2}}{2}$ $x$ will be real number and have two answers if $\sqrt{(3+m)^2-4\times2} > 0$ $m^2+6m+11>0$, got that $-3-2\sqrt{2}<m<-3+2\sqrt{2}$ and on the another case where $x^2-3x+2=-mx$ when $1<x<2$ $x^2-(3-m)x+2=0$ when $1<x<2$ Use quadratic formula will given $x= \frac{3-m\pm \sqrt{(3-m)^2-4\times2}}{2}$ $x$ will be real number and have the others two answers if $\sqrt{(3-m)^2-4\times2} > 0$ $\sqrt{(3-m)^2-4\times2} > 0$, got that $3-2\sqrt{2}<m<3+2\sqrt{2}$ So, I believe that the answers should be $(-3-2\sqrt{2}<m<-3+2\sqrt{2}) \cup (3-2\sqrt{2}<m<3+2\sqrt{2})$ However, the book's right answer is $0 < m < 3-2\sqrt{2}$ Please show me the method to obtain the right answers.

There is some positive value $m$ such that $y=mx$ is tangent to $y=-(x^2-3x+2)$. This value must make $0$ the discriminant of the equation $$x^2-3x+2=-mx$$ That is, $$m^2-6m+1=0$$ The least root of this equation is $$3-2\sqrt2$$ So $0<m< 3-2\sqrt 2$.

How many ordered pairs are there in order for $\frac{n^2+1}{mn-1}$ to be an integer? For how many ordered pairs of positive integers like $(m,n)$ the fraction $\frac{n^2+1}{mn-1}$ is a positive integer?

We have: $$n^2+1=kmn-k$$ so we have $n$ divides $k+1$ we can write $k+1=nt$ so that $$n^2+1=(nt-1)(mn-1)$$ but if $m,t,n>1$ we have $(nt-1)(mt-1)\geq (2n-1)^2>n^2+1$impossible if either $t=1$ or $m=1$ in the two cases $n-1$ divides $n^2+1$ but we know that $n-1=\gcd(n^2+1,n-1)=1$ or $2$, so that $n=1$, $n=2$ or $n=3$ and here you have finite cases to find $m$

How to factor the polynomial $x^4-x^2 + 1$ How do I factor this polynomial: $$x^4-x^2+1$$ The solution is: $$(x^2-x\sqrt{3}+1)(x^2+x\sqrt{3}+1)$$ Can you please explain what method is used there and what methods can I use generally for 4th or 5th degree polynomials?

Actually you have: $$x^4-x^2+1=x^4+2x^2+1-3x^2=(x^2+1)^2-(\sqrt3 x)^2 $$ and use the identity $a^2-b^2=(a-b)(a+b)$

Derivatives - optimization (minimum of a function) For which points of $x^2 + y^2 = 25$ the sum of the distances to $(2, 0)$ and $(-2, 0)$ is minimum? Initially, I did $d = \sqrt{(x-2)^2 + y^2} + \sqrt{(x+2)^2 + y^2}$, and, by replacing $y^2 = 25 - x^2$, I got $d = \sqrt{-4x + 29} + \sqrt{4x + 29}$, which derivative does not return a valid answer. Where did I commit a mistake? Thank you!!

For better readability, $$S=\sqrt{29+4x}+\sqrt{29-4x}$$ $$\dfrac{dS}{dx}=\dfrac2{\sqrt{29+4x}}\cdot4-\dfrac2{\sqrt{29-4x}}\cdot4$$ For the extreme values of $S,$ we need $\dfrac{dS}{dx}=0\implies29+4x=29-4x\iff x=0$

Area enclosed by an equipotential curve for an electric dipole on the plane I am currently teaching Physics in an Italian junior high school. Today, while talking about the electric dipole generated by two equal charges in the plane, I was wondering about the following problem: Assume that two equal charges are placed in $(-1,0)$ and $(1,0)$. There is an equipotential curve through the origin, whose equation is given by: $$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 $$ and whose shape is very lemniscate-like: Is there a fast&tricky way to compute the area enclosed by such a curve? Numerically, it is $\approx 3.09404630427286$.

Here is another method based on the curve-linear coordinates introduced by Achille Hui. He introduced the following change of variables $$\begin{align} \sqrt{(x+1)^2+y^2} &= u+v\\ \sqrt{(x-1)^2+y^2} &= u-v \end{align} \tag{1}$$ Then solving for $x$ and $y$ we shall get $$\begin{align} x &= u v\\ y &= \pm \sqrt{-(u^2-1)(v^2-1)} \end{align} \tag{2}$$ required that $$-(u^2-1)(v^2-1) \ge 0 \tag{3}$$ It does not look familiar but in fact it is! Taking into account the equations $(2)$ and $(3)$, we can consider the following as a parameterization for the first quadrant of the $xy$ plane $$\boxed{ \begin{array}{} x=uv & & 1 \le u \lt \infty \\ y=\sqrt{-(u^2-1)(v^2-1)} & & 0 \le v \le 1 \end{array}} \tag{4}$$ I tried to draw the coordinate curves of this curve-linear coordinates and I just noticed that it is exactly the same as the Elliptic Coordinates and nothing else! You can show this analytically by the change of variables $$\begin{align} u &= \cosh p \\ v &= \cos q \end{align} \tag{5}$$ I leave the further details in this avenue to the reader. Let us go back to the problem of calculating the area. The equation of the $\infty$ curve was $$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 \tag{6}$$ so combining $(1)$ and $(6)$ leads to $$v=\pm \sqrt{u^2-u} \tag{7}$$ and hence the parametric equation of the $\infty$ curve in the first quadrant by considering $(4)$ and $(7)$ will be $$\boxed{ \begin{array}{} x=u\sqrt{u^2-u} & & 1 \le u \lt \phi \\ y=\sqrt{-(u^2-1)(u^2-u-1)} \end{array}} \tag{8}$$ and finally the integral for the area is $$\begin{align} \text{Area} &=4 \int_{0}^{\phi} y dx \\ &=4 \int_{1}^{\phi}y \frac{dx}{du}du \\ &=2 \int_{1}^{\phi} (4u-3)\sqrt{-u(u+1)(u^2-u-1)}du \\ &\approx 3.09405 \end{align} \tag{9}$$

How to find a basis of a linear space, defined by a set of equations Problem Find a basis of the intersection $P\cap Q$ of subspaces $P$ and $Q$ given by: $$ P: \begin{cases} x_1 - 2 x_2 + 2 x_4=0,\\ x_2 - x_3 + 2 x_4 = 0 \end{cases} \qquad Q: \begin{cases} -2 x_1 + 3 x_2 + x_3 -6 x_4=0,\\ x_1 - x_2 - x_3 + 4 x_4 = 0 \end{cases} $$ Attempted solution The intersection of these 2 sets can be written by joining the sets of equations into 1: $$ \begin{cases} x_1 - 2 x_2 + 2 x_4=0,\\ x_2 - x_3 + 2 x_4 = 0\\ -2 x_1 + 3 x_2 + x_3 -6 x_4=0,\\ x_1 - x_2 - x_3 + 4 x_4 = 0\\ \end{cases} $$ After solving it I got the following matrix: $$ \begin{pmatrix} 1 & 0 & -2 & 6\\ 0 & 1 & -1 & 2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{pmatrix} $$ So, the intersection is the set of vectors, satisfying these 2 equations: $$ \begin{cases} x_1 = 2 x_3 - 6 x_4,\\ x_2 = x_3 - 2 x_4\\ \end{cases} $$ We have 2 independent and 2 dependent variables. What to do next to find the basis?

Having: $$ \begin{cases} x_1 = 2 x_3 - 6 x_4,\\ x_2 = x_3 - 2 x_4\\ \end{cases} $$ We could set 1) For the first element of basis: \begin{split} x_3=1,\quad x_4=0:\\ x_1 = 2\cdot 1 - 6\cdot 0 = 2,\\ x2 = 1 - 2 \cdot 0 = 1 \end{split} So, we get: (2, 1, 1, 0). 2) Second element of basis \begin{split} x_3=0,\quad x_4=1:\\ x_1 = 2 \cdot 0 - 6\cdot 1 = -6,\\ x2 = 0 - 2 \cdot 1 = -2 \end{split} So, we get: (-6, -2, 0, 1). Solution So the basis is $(2, 1, 1, 0), \; (-6, -2, 0, 1)$

Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$ Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $ This is what I've tried yet: $$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\ =&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\tan x\right\}}{4\sin^4 x}\\ =& \cfrac{2x\tan x \left(\frac{\tan^2 x}{1 - \tan^2 x}\right) }{4\sin^4 x} \\ =& \cfrac{x\tan^3 x}{2\sin^4 x (1-\tan^2 x)} \\ =& \cfrac{\tan^3 x}{2x^3\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} \\ =& \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)}\end{align}$$ Taking limit of the above expression, we've : $$\lim_{x\to 0} \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} = \lim_{x\to 0} \cfrac{\cos^2x}{2\cos 2x} = \cfrac{1}{2} $$ Firstly, is my answer right or am I doing somewhere wrong? Secondly, this seems a comparatively longer method than expected for objective type questions. I'm seeking for a shortcut method for such type of questions. Is there any method I should've preferred?

Well, let's try something different from using power series expansions. Here, we simplify using trigonometric identities to reveal that $$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{2x}{\sin 4x }=\frac{1}{2\text{sinc}(4x)}$$ where the sinc function is defined as $\text{sinc}(x)=\frac{\sin x}{x}$. The limit as $x \to 0$ is trivial since $\text{sinc}(4x) \to 1$ . The limit is $1/2$ as expected. NOTE $1$: Establishing the identity Using standard trigonometric identities, we can write $$\begin{align} x\tan 2x-2x \tan x&=\frac{2x\sin x\cos x}{\cos 2 x}-2x\frac{\sin x}{\cos x}\\\\ &=2x \sin x \frac{\sin^2x}{\cos x\cos 2x}\\\\ &=2 \frac{\sin^4 x}{\text{sinc}( 4 x)} \end{align}$$ and $$(1-\cos 2x)^2=4\sin^4 x$$ Putting it together reveals that $$\frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}=\frac{1}{2\text{sinc}(4x)}$$ NOTE $2$: Series expansion is facilitated by simplifying using trigonometric identities We can use the Laurent series for the cosecant function $$\csc x=\frac1 x+\frac16 x+\frac{7}{360}x^3+\frac{31}{15120}x^5+O(x^7)$$ to establish that $$\begin{align} \frac{x\tan 2x-2x \tan x}{(1-\cos 2x)^2}&=\frac{2x}{\sin 4x }=2x\text{csc}(4x)\\\\ &=\frac12 +\frac43 x^2 +\frac{112}{45}x^4+\frac{3968}{945}x^6+O(x^7) \end{align}$$

Differentiate $\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$ with respect to $x$ Differentiate $$\sin^{-1}\left(\frac {\sin x + \cos x}{\sqrt{2}}\right)$$ with respect to $x$. I started like this: Consider $$\frac {\sin x + \cos x}{\sqrt{2}}$$, substitute $\cos x$ as $\sin \left(\frac {\pi}{2} - x\right)$, and proceed with the simplification. Finally I am getting it as $\cos \left(x - \frac {\pi}{4}\right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!

An alternative approach is to use Implicit Differentiation: \begin{equation} y = \arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \rightarrow \sin(y) = \frac{\sin(x) + \cos(x)}{\sqrt{2}} \end{equation} Now differentiate with respect to '$x$': \begin{align} \frac{d}{dx}\left[\sin(y) \right] &= \frac{d}{dx}\left[\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right] \\ \cos(y)\frac{dy}{dx} &= \frac{\cos(x) - \sin(x)}{\sqrt{2}} \\ \frac{dy}{dx} &= \frac{\cos(x) - \sin(x)}{\sqrt{2}\cos(y)} \end{align} Thus: \begin{equation} \frac{dy}{dx} = \frac{d}{dx}\left[\arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}} \right) \right] = \frac{\cos(x) - \sin(x)}{\sqrt{2}\cos\left(\arcsin\left(\frac{\sin(x) + \cos(x)}{\sqrt{2}}\right) \right)} \end{equation} Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.

show $\sum_{k=0}^n {k \choose i} = {n+1 \choose i+1}$ show for n $\geq i \geq 1 : \sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$ i show this with induction: for n=i=1: ${1+1} \choose {1+1}$ = $2 \choose 2$ = 1 = $0 \choose 1$ + $1 \choose 1$ = $\sum_{k=0}^1 {k \choose 1}$ now let $\sum_{k=0}^n {k \choose i} = $ ${n+1} \choose {i+1}$ for n+1: ${n+1+1} \choose {i+1}$ = $n+1 \choose i$ + $n+1 \choose i$ = $ n+1 \choose i$ + $\sum_{k=0}^n {k \choose i} = \sum_{k=0}^{n+1} {k \choose i} $ Is this the right way ?

We can also use the recurrence from Pascal's Triangle and telescoping series: $$ \begin{align} \sum_{k=0}^n\binom{k}{i} &=\sum_{k=0}^n\left[\binom{k+1}{i+1}-\binom{k}{i+1}\right]\\ &=\sum_{k=1}^{n+1}\binom{k}{i+1}-\sum_{k=0}^n\binom{k}{i+1}\\ &=\binom{n+1}{i+1}-\binom{0}{i+1}\\ &=\binom{n+1}{i+1} \end{align} $$

For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$ $$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$ Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$ Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?

suppose the matrix is singular. then so is its transpose, which must annihilate some non-zero vector $(h,g,f)$ this gives three equations: $$ fa^2 + ga+ h =0 \\ fb^2 + gb+ h =0 \\ fc^2 + gc+ h =0 \\ $$ we know from the algebra of fields that a quadratic equation can have at most two roots, so the three values $a,b,c$ cannot all be different

Solve by using substitution method $T(n) = T(n-1) + 2T(n-2) + 3$ given $T(0)=3$ and $T(1)=5$ I'm stuck solving by substitution method: $$T(n) = T(n-1) + 2T(n-2) + 3$$ given $T(0)=3$ and $T(1)=5$ I've tried to turn it into homogeneous by subtracting $T(n+1)$: $$A: T(n) = T(n-1) + 2T(n-2) + 3$$ $$B: T(n+1) = T(n) + 2T(n-1) +3$$ $$A - B = T(n) - T(n+1) = T(n-1) + 2T(n-2) + 3 - (T(n) + 2T(n-1) + 3)$$ $$2T(n) - T(n+1) - T(n-1) + 2T(n-2) = 0$$ Assume $T(n) = x^{n}$ $$2x^{n} - x^{n+1} - x^{n-1} + 2x^{n-2} = 0$$ Dividing each side by $x^{n-2}$ leaves me with the impossible (beyond the scope of this class) equation: $$2x^2 - x^3 - x + 2 = 0$$ How can I solve this using substitution method?

suppose we look for a constant solution $t_n = a.$ then $a$ must satisfy $a = a+2a + 3.$ we pick $a = -3/2.$ make a change a variable $$a_n = t_n + 3/2, t_n = a_n - 3/2.$$ then $a_n$ satisfies the recurrence equation $$a_n= a_{n-1}+ 2a_{n-2}, a_0 = 9/2, a_1=13/2.$$ now look for solutions $$a_n = \lambda^n \text{ where }\lambda^2-\lambda - 2 = 0\to \lambda = 2, -1$$ the solution is $$a_n = c2^n + d(-1)^n, c+d = 9/2, 2c-d = 13/2$$ which gives you $$c = 11/3, d = 5/6.$$ finally $$t_n = \frac{11}3\, 2^n +\frac 56 \, (-1)^n -\frac32. $$

How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$ The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be. The full breakdown comes from this solution $$ \small\begin{align} \frac1{x^2-5x+6} &=\frac1{(x-2)(x-3)} =\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right) =\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\ &=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}} =\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n =\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n} \end{align} $$ Original image

$$ -\frac{1}{x-2} = \frac{1}{-(x-2)} = \frac{1}{-x+2} = \frac{1}{2-x}$$ and $$ \frac{1}{x - 3} = \frac{1}{-3 + x} = \frac{1}{-(3 - x)} = -\frac{1}{3-x}$$ Thus $$ -\frac{1}{x-2} + \frac{1}{x - 3} = \frac{1}{2-x} - \frac{1}{3-x} $$

Finding a nullspace of a matrix. I am given the following matrix $A$ and I need to find a nullspace of this matrix. $$A = \begin{pmatrix} 2&1&4&-1 \\ 1&1&1&1 \\ 1&0&3&-2 \\ -3&-2&-5&0 \end{pmatrix}$$ I have found a row reduced form of this matrix, which is: $$A' = \begin{pmatrix} 1&0&3&-2 \\ 0&1&-2&3 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}$$ And then I used the formula $A'x=0$, which gave me: $$A'x = \begin{pmatrix} 1&0&3&-2 \\ 0&1&-2&3 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$ Hence I obtained the following system of linear equations: $$\begin{cases} x_1+3x_3-2x_4=0 \\ x_2-2x_3+3x_4=0 \end{cases}$$ So I just said that $x_3=\alpha$, $x_4=\beta$ and the nullspace is: $$nullspace(A)=\{2\beta-3\alpha,2\alpha-3\beta,\alpha,\beta) \ | \ \alpha,\beta \in \mathbb{R}\}$$ Is my thinking correct? Thank you guys!

Since $x_1=2x_4-3x_3$ and $x_2=2x_3-3x_4\Rightarrow$ if $(x_1,x_2,x_3,x_4)\in$ nullspace($A$): $$(x_1,x_2,x_3,x_4)=(2x_4-3x_3,2x_3-3x_4,x_3,x_4)=x_3(-3,2,1,0)+x_4(2,-3,0,1)$$ So Nullspace$(A)=\langle (-3,2,1,0),(2,-3,0,1) \rangle$.

Inequality used in the proof of Kolmogorov Strong Law of Large Numbers I'm trying show that convergence follows. $$\sum_{k \geq 1} \frac{\sigma^2_k}{k^2} < \infty \Rightarrow \lim_{M \rightarrow \infty}\frac{1}{M^2}\sum_{k \leq M} \sigma^2_k=0.$$ Let's consider $D_k = \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2}$ para $k \geq 1$ and it is noted that for $k=1$ we have to: $$D_1 = \sum_{n \geq 1} \frac{\sigma_n^2}{n^2} < \infty \text{(by hypothesis)}.$$ or this I have proved that $\lim_{k \rightarrow \infty} D_k = \lim_{k \rightarrow \infty} \sum_{n \geq k} \frac{\sigma_n^2}{n^2} = 0$, Now note the following: \begin{eqnarray*} D_2 &=& \sum_{n \geq 2} \frac{\sigma_n^2}{n^2} = \frac{\sigma_2^2}{2^2} + \frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} \ldots\nonumber \\ D_3 &=& \sum_{n \geq 3} \frac{\sigma_n^2}{n^2} = \frac{\sigma_3^2}{3^2} + \frac{\sigma_4^2}{4^2} \ldots \nonumber\\ D_4 &=& \sum_{n \geq 4} \frac{\sigma_n^2}{n^2} = \frac{\sigma_4^2}{4^2} \ldots \nonumber\\ &\vdots& \\ \lim_{k \rightarrow \infty} D_k &=& \sum_{n \geq k} \frac{\sigma_n^2}{n^2} = 0. \nonumber \end{eqnarray*} then I considered a $M$ such that $M \geq 1$ and I then come to the following equation $$\frac{1}{M^2}\sum_{k=1}^M \sigma^2_k = \frac{1}{M^2} \sum_{k=1}^M k^2\left(D_k - D_{k+1}\right).$$ I noted in a book which obtained the following inequality $$\sum_{n \geq k} \displaystyle\frac{\sigma^2_k}{k^2} = \frac{1}{M^2} \sum_{k=1}^M k^2 \left(D_k - D_{k+1}\right)\text{(How could justify this inequality?)} \leq \frac{1}{M^2} \sum_{k=1}^M (2k - 1)D_k$$ It is easy to see that \begin{eqnarray} \sum_{n \geq k} \displaystyle\frac{\sigma^2_k}{k^2} &=& D_k - D_{k+1}\nonumber \\ &=& \sum_{n \geq k} \displaystyle\frac{\sigma_n^2}{n^2} - \sum_{n \geq k+1} \displaystyle\frac{\sigma_n^2}{n^2} \nonumber \\ &=& \left\{\displaystyle\frac{\sigma_k^2}{k^2} + \displaystyle\frac{\sigma_{(k+1)}^2}{(k+1)^2} + \ldots \right\} - \left\{\displaystyle\frac{\sigma_{(k+1)}^2}{(k+1)^2} + \displaystyle\frac{\sigma_{(k+2)}^2}{(k+1)^2} + \ldots \right\}. \end{eqnarray} where $$\lim_{k \rightarrow \infty} \frac{1}{M^2} \sum_{k=1}^M (2k - 1)D_k = 0$$ but I don´t know to justify that step. Could you please give me a suggestion on how to justify the last step. Thank you very much, for you help.

\begin{align*} \sum_{k=1}^M k^2(D_k-D_{k+1}) &= \sum_{k=1}^M k^2 D_k - \sum_{k=1}^M k^2 D_{k+1} \\ &= \sum_{k=1}^M k^2 D_k - \sum_{k=2}^{M+1} (k-1)^2 D_k \\ &= D_1 + \sum_{k=2}^M (k^2-(k-1)^2) D_k - M^2 D_{M+1} \\ &= D_1 + \sum_{k=2}^M (2k-1) D_k - M^2 D_{M+1} \\ &= \sum_{k=1}^M (2k-1) D_k - M^2 D_{M+1} \\ &\le \sum_{k=1}^M (2k-1) D_k \end{align*}

Find $x$ and $y$ If $\frac{\tan 8°}{1-3\tan^{2}8°}+\frac{3\tan 24°}{1-3\tan^{2}24°}+\frac{9\tan 72°}{1-3\tan^{2}72°}+\frac{27\tan 216°}{1-3\tan^{2}216°}=x\tan 108°+y\tan 8°$, find x and y. I am unable to simplify the first and third terms. I am getting power 4 expressions. Thanks.

HINT: $$\tan(3\cdot8^\circ)=\dfrac{3\tan 8^\circ-\tan^38^\circ}{1-3\tan^28^\circ}$$ Now, $$\frac{\tan 8^\circ}{1-3\tan^28^\circ}-y\tan8^\circ=\dfrac{(1-y)\tan 8^\circ-(-3y)\tan^38^\circ}{1-3\tan^28^\circ}$$ which will be a multiple of $\tan(3\cdot8^\circ)$ if $$\dfrac{1-y}{-3y}=\dfrac31\iff y=-\dfrac18$$ $$\implies\frac{\tan A}{1-3\tan^2A}-\left(-\dfrac18\right)\tan A=\dfrac38\tan3A$$ and I should leave it here.

First order differential equation integrating factor is $e^{\int\frac{2}{x^2-1}}$ So i got the first order ode $$(x^2-1)\frac{dy}{dx}+2xy=x$$ I divided both sides by $x^2-1$ $$\frac{dy}{dx}+\frac{2}{x^2-1}xy=\frac{x}{x^2-1}$$ in the form $y' + p(x)y = q(x)$ So that means the integrand is... $$e^{\int\frac{2}{x^2-1}}$$ But i'm not sure what to do i think the $\int\frac{2}{x^2-1}$ = $-\log{(x-1)}+4\log{(x+1)}$ So it's $$e^{-\log{(x-1)}+4\log{(x+1)}}$$ $$e^{\log{(x-1)^{-1}}+\log{(x+1)^4}}$$ $$\frac{1}{x-1}+(x+1)^4$$ Is this right? and then just multiply both sides by this?

your equation $$(x^2 - 1) \frac{dy}{dx} + 2x y = x $$ is an exact differential equation. the reason is it can be written as $$\frac d{dx}\left((x^2 - 1) y\right) = x $$ on integration gives you $$(x^2 - 1)y = \frac12 x^2 + c \to y = \frac{x^2}{2(x^2 - 1)} + \frac c{x^2 - 1}$$

Multivariable limit which should be simple ! How to calculate the following limit WITHOUT using spherical coordinates? $$ \lim _{(x,y,z)\to (0,0,0) } \frac{x^3+y^3+z^3}{x^2+y^2+z^2} $$ ? Thanks in advance

Let $\epsilon \gt 0$. If $(x,y,z)$ is close enough to $(0,0,0)$ but not equal to it, then $|x^3|\le \epsilon x^2$, with similar inequalities for $|y^3|$ and $|z^3|$. It follows that $$\frac{|x^3+y^3+z^3|}{x^2+y^2+z^2}\le \frac{|x^3|+|y^3|+|z^3|}{x^2+y^2+z^2}\le \frac{\epsilon (x^2+y^2+z^2)}{x^2+y^2+z^2}.$$

Proving vectors as a basis in $E^{m}$ Show that if the vectors $a_{1}$, $a_2$, $\cdots$, $a_m$, are a basis in $E^{m}$, the vectors $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$, also are a basis if and only if $y_{p,q} \neq 0$, where $y_{p,q}$ is defined by the following tableau: \begin{matrix} 1& 0& \cdots & 0& y_{1,m+1} & y_{1,m+2}& \cdots & y_{1n} & y_{10}\\ 0& 1& \cdot & 0&y_{2,m+1}& y_{2,m+2}& \cdots & y_{2n} & y_{20}\\ 0& 0& \cdot & 0& \cdot & \cdot & \cdot & \cdot & \cdot\\ \vdots& \vdots& \vdots & \vdots& \vdots& \vdots& \vdots & \vdots& \vdots\\ 0& 0& \cdot & 1& y_{m,m+1} & y_{m,m+2} & \cdots & y_{mn} & y_{m0} \end{matrix} Can the necessary and sufficient conditions be defined as follows. If $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$ are a basis in $E^{m}$ then $y_{p,q} \neq 0$ which implies to prove that they're LI (necessary condition) and if $y_{p,q} \neq 0$ then $a_{1}$, $a_2$, $\cdots$, $a_{p-1}$, $a_{q}, a_{p+1}, \cdots,a_{m}$ are a basis in $E^{m}$ (sufficient condition)? Does anyone have any idea to prove this? Any hint is welcome. Thanks.

Remark: Your problem has nothing to be with the rightmost column $(y_{1,0},y_{2,0},\dots,y_{m,0})^T$. I'll omit that column from the matrix. Settings Let $A = \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m&|\mathbf{a}_{m+1} & \cdots&\mathbf{a}_n\end{bmatrix}$ be the original coefficient matrix in $E^{m \times n}$, where $\mathbf{a}_j$ denotes the $j$-th column of $A$. By changing $A$ to the given matrix $\begin{bmatrix}I_m&|\mathbf{y}_{m+1} & \cdots&\mathbf{y}_n\end{bmatrix}$, where $\mathbf{y}_j$ denotes the $j$-th column of the given matrix (i.e. $\mathbf{y}_j = \mathbf{e}_j \;\forall j \in \{1,\dots,n\}$ and $\mathbf{y}_j^T = (y_{1,j},y_{2,j},\dots,y_{m,j}) \;\forall j \in \{n+1,\dots,m\}$), using row operations, the given matrix equals $B^{-1}A$, where $B := \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m\end{bmatrix}$ is formed by the given basis. To see this, use the fact that \begin{align} B^{-1} B &= I \\ B^{-1} \begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_m\end{bmatrix} &= \begin{bmatrix} \mathbf{e}_1 & \mathbf{e}_2 & \cdots & \mathbf{e}_m \end{bmatrix} \end{align} Therefore, we have $B^{-1}\mathbf{a}_j = \mathbf{y}_j \;\forall j \in \{1,2,\dots,n\}$. This is our starting point. Sufficient condition Assume that $y_{pq} \ne 0$. \begin{align} B^{-1}\mathbf{a}_q &= \mathbf{y}_q \\ \mathbf{a}_q &= B \mathbf{y}_q \\ &= \sum_{i = 1}^m y_{i,q} \mathbf{a}_i \\ &= y_{p,q} \mathbf{a}_p + \sum_{\substack{i = 1 \\ i \ne p}}^m y_{i,q} \mathbf{a}_i \\ \frac{1}{y_{p,q}} \mathbf{a}_q &= \mathbf{a}_p + \sum_{\substack{i = 1 \\ i \ne p}}^m \frac{y_{i,q}}{y_{p,q}} \mathbf{a}_i \\ \mathbf{a}_p &= \frac{1}{y_{p,q}} \mathbf{a}_q - \sum_{\substack{i = 1 \\ i \ne p}}^m \frac{y_{i,q}}{y_{p,q}} \mathbf{a}_i \end{align} Since it's given that $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_m \right\}$ is a basis for $E^m$, then $$\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \frac{1}{y_{p,q}} \mathbf{a}_q - \sum_{\substack{i = 1 \\ i \ne p}}^m \frac{y_{i,q}}{y_{p,q}} \mathbf{a}_i, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\},$$ is a basis for $E^m$. Hence $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \mathbf{a}_q, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\}$ is a basis for $E^m$. Necessary condition Suppose that $y_{pq} = 0$. Then from the section above, we have $$\mathbf{a}_q = y_{p,q} \mathbf{a}_p + \sum_{\substack{i = 1 \\ i \ne p}}^m y_{i,q} \mathbf{a}_i = \sum_{\substack{i = 1 \\ i \ne p}}^m y_{i,q} \mathbf{a}_i,$$ so $\mathbf{a}_q$ can be represented by $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\}$. In other words, the vector $\mathbf{a}_q$ has two different representations by the basis $\left\{ \mathbf{a}_1, \mathbf{a}_2, \dots, \mathbf{a}_{p - 1}, \mathbf{a}_q, \mathbf{a}_{p + 1}, \dots, \mathbf{a}_m \right\}$. This is a contradiction.

Maclurin Series. (Approximation) Given that $y=\ln \cos x$, show that the first non-zero terms of Maclurin's series for $y=-\frac{x^2}{2}-\frac{x^4}{12}$. Use this series to find the approximation in terms of $\pi$ for $\ln 2$. My question is how to determine value of $x$ which is suitable?

We have \begin{align} \ln(\cos(x)) & = \dfrac{\ln(\cos^2(x))}2 = \dfrac12 \cdot \ln(1-\sin^2(x)) = - \dfrac12 \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(x)}k\\ & = -\dfrac12 \sin^2(x) - \dfrac14 \sin^4(x) - \cdots\\ & = -\dfrac12 \left(x-\dfrac{x^3}{3!} + \mathcal{O}(x^5)\right)^2-\dfrac14 \left(x + \mathcal{O}(x^3)\right)^4 + \mathcal{O}(x^6)\\ & = -\dfrac12\left(x^2 - \dfrac{x^4}6\right) - \dfrac{x^4}4 + \mathcal{O}(x^6)\\ & = -\dfrac{x^2}2 - \dfrac{x^4}{12} + \mathcal{O}(x^6) \end{align} We need $\ln(2)$ to be expressed in terms of $\pi$, this means we need $\ln(\cos(x))$ to be some value related to $\ln(2)$. One such value is when $\cos(x) = 1/2$, which implies $$\ln(1/2) = \ln(\cos(2n\pi \pm \pi/3)) = -\dfrac{(2n\pi \pm \pi/3)^2}2-\dfrac{(2n\pi \pm \pi/3)^4}{12} + \mathcal{O}\left((2n\pi\pm \pi/3)^6\right)$$ Since typically we are after good approximation, we would like to have the least possible error term, which in turn forces $n=0$, which gives us that $$\ln(2) \approx \dfrac{\pi^2}{18} + \dfrac{\pi^4}{972}$$ Another way is to take $x=\pi/4$, which gives us that $$-\dfrac12\ln(2) = \ln(1/\sqrt{2}) = \ln(\cos(\pi/4))) \approx - \dfrac{\pi^2}{32} - \dfrac{\pi^4}{3072}$$ This gives us that $$\ln(2) \approx \dfrac{\pi^2}{16} + \dfrac{\pi^4}{1536} + \cdots$$

How to find cotangent? Need to find a $3\cot(x+y)$ if $\tan(x)$ and $\tan(y)$ are the solutions of $x^2-3\sqrt{5}\,x +2 = 0$. I tried to solve this and got $3\sqrt{5}\cdot1/2$, but the answer is $-\sqrt{5}/5$

Since by Vieta's formulas one has $$\tan x+\tan y=-\frac{-3\sqrt 5}{1}=3\sqrt 5,\ \ \ \tan x\tan y=\frac{2}{1}=2,$$ one has $$3\cot(x+y)=3\cdot\frac{1}{\tan(x+y)}=3\cdot\frac{1-\tan x\tan y}{\tan x+\tan y}=\frac{3(1-2)}{3\sqrt 5}=-\frac{\sqrt 5}{5}.$$

Proving simple trigonometric identity I need help with this one $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha $$ I tried moving sin a on the other side of the eqation $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0 $$ This are the operations I was able to do $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha*\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha - \sin \alpha = 0 $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha - \sin \alpha = 0 $$ I don't see what else I can do with this, so I tried to solve the left part of the equation. $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \tan^2\alpha} - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - cos\alpha} + \frac{\sin\alpha\cdot\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha = \sin \alpha $$ $$ \frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha = \sin \alpha $$ And I get to nowhere again. I have no other ideas, I didn't see some formula or something. Any help is appreciated.

$\left(\dfrac{\sin^2 \alpha}{\sin \alpha+\cos\alpha}-(\sin\alpha+\cos\alpha)\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\left(\dfrac{\sin^2 \alpha-\sin^2 \alpha+\cos^2\alpha}{\sin \alpha-\cos\alpha}\right)+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha}{\sin \alpha-\cos\alpha}+\dfrac{\sin \alpha+\cos\alpha}{1-\tan^2\alpha}\\=\dfrac{\cos^2\alpha-\sin^2\alpha+\sin^2\alpha-\cos^2\alpha}{(\sin \alpha-\cos\alpha)(1-\tan^2\alpha)}\\=0$

Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$ I broke the top into prime mods: $$x \equiv 8^{38} \pmod 3$$ $$x \equiv 8^{38} \pmod {70}$$ But $x \equiv 8^{38} \pmod {70}$ can be broken up more: $$x \equiv 8^{38} \pmod 7$$ $$x \equiv 8^{38} \pmod {10}$$ But $x \equiv 8^{38} \pmod {10}$ can be broken up more: $$x \equiv 8^{38} \pmod 5$$ $$x \equiv 8^{38} \pmod 2$$ In the end,I am left with: $$x \equiv 8^{38} \pmod 5$$ $$x \equiv 8^{38} \pmod 2$$ $$x \equiv 8^{38} \pmod 7$$ $$x \equiv 8^{38} \pmod 3$$ Solving each using fermat's theorem: * *$x \equiv 8^{38}\equiv8^{4(9)}8^2\equiv64 \equiv 4 \pmod 5$ *$x \equiv 8^{38} \equiv 8^{1(38)}\equiv 1 \pmod 2$ *$x \equiv 8^{38} \equiv 8^{6(6)}8^2\equiv 64 \equiv 1 \pmod 7$ *$x \equiv 8^{38} \equiv 8^{2(19)}\equiv 1 \pmod 3$ So now, I have four congruences. How can i solve them?

You made a small slip up when working $\bmod 2$, this is because $0^n$ is always congruent to $0$ no matter what congruence you are working with . I repeat this step once again. $8^{38}\equiv 3^{38}\equiv3^{9\cdot4}3^{2}\equiv(3^{4})^{9}3^2\equiv1^9\cdot3^2\equiv9\equiv 4\bmod 5$ $8^{38}\equiv0 \bmod 2$ since it is clearly even. $8^{38}\equiv(1)^{38}\equiv 1\bmod 7$ $8^{38}\equiv(-1)^{38}\equiv 1 \bmod 3$ So you have the following system of equations: $x\equiv4 \bmod 5$ $x\equiv 0 \bmod 2$ $x\equiv 1 \bmod 7$ $x\equiv 1 \bmod 3$ There are general ways to solve this, but it is possible to solve it step by step using basic substitutions. We start by writing $x=2k$ since $x\equiv 0 \bmod 2$ We now have $2k\equiv 1 \bmod 3$. Multiplying by two we get $4k\equiv 2\bmod 3$ since $4\equiv 1$ we have $k\equiv 2 \bmod 3$ so $k=3j+2$ and so $x=2(3j+2)=6j+4$ We have $6j+4\equiv 1 \bmod 7$ so $6j\equiv-3\equiv 4 \bmod 7$. Multiplying by $6$ we get $36j\equiv 24\equiv 3 \bmod 7$ Since $36\equiv 1 \bmod 7$ we have $j\equiv 3 \bmod 7$. So $x=6(7l+3)+4=42l+22$ We have $42l+22\equiv 4 \bmod 5$ from here $42l\equiv -18\equiv2 \bmod 5$ Since $42\equiv 2 \bmod 5$ we have $2l\equiv 2 \bmod 5$ so $l\equiv 1 \bmod 5$. So $l=5s+1$ and $x=42(5s+1)+22=210s+64$. So $8^{38}\equiv 64\bmod 210$ Second solution: Instead of using Fermat's theorem use Carmichael's theorem which says that if $a$ and $n$ are relatively prime then $a^{\lambda(n)}\equiv 1 \bmod n$. Using this and the fact that $\lambda(105)=12$ we get $8^{38}\equiv 8^{36}a^2\equiv (8^{12})^8a^2\equiv 1^38^2\equiv 8^2\equiv64\bmod 105$ Using this and the fact that $8^{38}$ is even we get $8^{38}\equiv 64\bmod 210$ Note: The theorem that tells us that we can separate congruences into congruences mod the power of primes dividing the number is called the Chinese Remainder theorem. This method assures us that the solution exists and that it is unique.

Solving an integral with trig substitution I'm looking to solve the following integral using substitution: $$\int \frac{dx}{2-\cos x}$$ Let $z=\tan\frac{x}{2}$ Then $dz=\frac 1 2 \sec^2 \frac x 2\,dx$ $$\sin x=\frac{2z}{z^2+1}$$ $$\cos x =\frac{1-z^2}{z^2+1}$$ $$dx=\frac{2\,dz}{z^2+1}$$ $$\int \frac{dx}{2-\cos x} = \int \frac{\frac{2\,dz}{z^2+1}}{2-\frac{1-z^2}{z^2+1}} =\int \frac{2\,dz}{3z^2+1}$$ But this is where things start to look at bit sticky. If I integrate this last fraction, then I get a very complex expression that seems to defeat the point of z-substitution. Any suggestions for where I may be going wrong? Thanks! Edit: Thank you for your feedback. I've completed my work as per your suggestions: $$\int \frac{2\,dz}{3z^2+1} = 2\cdot\left(\frac{\tan^{-1} \frac{z}{\sqrt{3}}}{\sqrt{3}} \right) = \frac{2\tan^{-1} \left(\sqrt{3}\tan{\frac{x}{2}}\right)}{\sqrt{3}}+c$$

HINT: $$\int \frac{1}{a^2+x^2}dx=\frac1a \arctan(x/a)+C$$

How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$ I have difficulty to evaluating above integrals. First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse. Using Mathematica I found the result $\dfrac{\pi}{48\sqrt{3}}$ I want to know the procedure of evaluating this integral.

Here is an approach. You may write $$\begin{align} \int_0^{\infty}\frac{x^4}{\left(x^4+x^2+1\right)^3}dx &=\int_0^{\infty}\frac{x^4}{\left(x^2+\dfrac1{x^2}+1\right)^3\,x^6}dx\\\\ &=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\frac{dx}{x^2} \\\\ &=\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}dx\\\\ &=\frac12\int_0^{\infty}\frac{1}{\left(x^2+\dfrac1{x^2}+1\right)^3}\left(1+\dfrac1{x^2}\right)dx\\\\ &=\frac12\int_0^{\infty}\frac{1}{\left(\left(x-\dfrac1{x}\right)^2+3\right)^3}d\left(x-\dfrac1{x}\right)\\\\ &=\frac12\int_{-\infty}^{+\infty}\frac{1}{\left(u^2+3\right)^3}du\\\\ &=\frac14\:\partial_a^2\left(\left.\int_{-\infty}^{+\infty}\frac{1}{\left(u^2+a\right)}du\right)\right|_{a=3}\\\\ &=\frac14\:\partial_a^2\left.\left(\frac{\pi}{\sqrt{a}}\right)\right|_{a=3}\\\\ &=\color{blue}{\frac{\pi }{48 \sqrt{3}}} \end{align}$$ as desired.

Find the range of a $4$th-degree function For the function $y=(x-1)(x-2)(x-3)(x-4)$, I see graphically that the range is $\ge-1$. But I cannot find a way to determine the range algebraically?

Non-calculus Approach/ Completing Square Approach Let $u=(x-2)(x-3)=x^2-5x+6$ $(x-1)(x-4)=x^2-5x+4=u-2$ So \begin{align} y&=(x-1)(x-2)(x-3)(x-4)\\&=u(u-2)\\&=u^2-2u\\&=(u-1)^2-1 \end{align} As $(u-1)^2\ge0$ for all $u\in{\Bbb{R}}$ $y\ge0-1=-1$

Show that $\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$ The question asks to prove that - $$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x) $$ I tried combining the first two or the last two fractions on the L.H.S to allow me to use the double angle formula and get $\sin 6x$ or $\sin 18x$ but that did not help at all. I'm pretty sure that if I express everything in terms of $x$, the answer will ultimately appear but I'm also certain that there must be another simpler way.

@MayankJain @user, I don't know that the case for $n=1$ is trivial, especially for someone in a trig class currently. I will offer a proof without induction using substitution instead. Mayank, to see that this is the case for $n=1$, convert the RHS of the equation as follows: $\dfrac{1}{2} \cdot \big{[}\dfrac{\sin 3x}{\cos 3x} - \dfrac{\sin x}{\cos x}]$ = (1) $\dfrac{1}{2} \cdot \big(\dfrac{\sin 3x \cos x - \cos 3x \sin x}{\cos 3x \cos x}\big{)} $ = (2) $ \dfrac{1}{2} \cdot \dfrac{\sin 2x}{\cos 3x \cos x}$ = (3) $\dfrac{1}{2} \cdot \dfrac{2\sin x \cos x}{\cos 3x \cos x}$ = $\dfrac{\sin x}{\cos 3x}$ This is pretty heavy on trig identities. We get equivalence (1) by multiplying out the fraction, equivalence (2) because $\sin(u - v) = \sin u \cos v - \cos u \sin v$, equivalence (3) because $\sin 2x = 2\sin x \cos x$, and, finally, (4) by cancellation. Now, if you are unfamiliar with induction, substitution will help here as an alternative method. Since you can now derive the equivalence for the 'trivial' case, set $3x = u$ for the second case, and $9x = v$ for the third. Then you already know $\dfrac{\sin u}{\cos 3u} = \dfrac{1}{2} \cdot (\tan 3u - \tan u)$, and similarly, $\dfrac{\sin v}{\cos 3v} = \dfrac{1}{2} \cdot (\tan 3v - \tan v)$. So now we can re-write the original expression $\dfrac{\sin x}{\cos 3x} + \dfrac{\sin u}{\cos 3u} + \dfrac{\sin v}{\cos 3v}$ as $\dfrac{1}{2} \cdot (\tan(3x) - \tan x + \tan(9x) - \tan(3x) + \tan(27x) - \tan(9x))$ and everything cancels out except the desired expression: $\dfrac{1}{2} \cdot (\tan(27x) - \tan(x))$.

How to solve the difference equation $u_n = u_{n-1} + u_{n-2}+1$ Given that: $$ \begin{equation} u_n=\begin{cases} 1, & \text{if $0\leq n\leq1$}\\ u_{n-1} + u_{n-2}+1, & \text{if $n>1$} \end{cases} \end{equation} $$ How do you solve this difference equation? Thanks EDIT: From @marwalix's answer: $$ u_n=v_n-1 $$ $$ \begin{equation} v_n=\begin{cases} 2, & \text{if $0\leq n\leq1$}\\ v_{n-1} + v_{n-2}, & \text{if $n>1$} \end{cases} \end{equation} $$ Characteristic equation of $v_n$ is $$ r^2=r+1 $$ Therefore, $$ r=\frac{1\pm\sqrt{5}}{2} $$ Therefore, the general solution for $v_n$ is $$ v_n=A\left(\frac{1+\sqrt{5}}{2}\right)^n+B\left(\frac{1-\sqrt{5}}{2}\right)^n $$ When $n=0$, $$ 2=A+B $$ When $n=1$, $$ 2=A\left(\frac{1+\sqrt{5}}{2}\right)+B\left(\frac{1-\sqrt{5}}{2}\right) $$ Therefore, $$ A=\frac{5+\sqrt{5}}{5} $$ $$ B=\frac{5-\sqrt{5}}{5} $$ Therefore, $$ u_n=\frac{5+\sqrt{5}}{5}\left(\frac{1+\sqrt{5}}{2}\right)^n+\frac{5-\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^n-1 $$

Write $u_n=v_n+a$ where $a$ is a constant. In that case the recurrence reads as follows $$v_n+a=v_{n-1}+v_{n-2}+2a+1$$ So if we chose $a=-1$ we are left with $$v_n=v_{n-1}+v_{n-2}$$ And we're back to a Fibonnacci type and in this case we have $v_0=v_1=2$

$\sin x + c_1 = \cos x + c_2$ While working a physics problem I ran into a seemingly simple trig equation I couldn't solve. I'm curious if anyone knows a way to solve the equation: $\sin(x)+c_1 = \cos(x)+c_2$ (where $c_1$ and $c_2$ are constants) for $x$ without using Newton's method or some other form of approximation.

$$ A\sin x + B\cos x = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\sin x+ \frac B {\sqrt{A^2+B^2}}\cos x \right) $$ Notice that the sum of the squares of the coefficients above is $1$; hence they are the coordinates of some point on the unit circle; hence there is some number $\varphi$ such that $$ \cos\varphi = \frac A {\sqrt{A^2+B^2}}\quad\text{and}\quad\sin\varphi=\frac B {\sqrt{A^2+B^2}}. $$ And notice that $\tan\varphi=\dfrac B A$, so finding $\varphi$ is computing an arctangent. Now we have $$ A\sin x + B\cos x = \sqrt{A^2+B^2}(\cos\varphi \sin x+ \sin\varphi \cos x) = \sqrt{A^2+B^2} \sin(x+\varphi). $$ Apply this to $\sin x - \cos x$, in which $A=1$ and $B=-1$, and you get $$ \sqrt{2} \sin(x+\varphi) = c_2-c_1. $$ So you just need to find an arcsine and an arctangent. And the arctangent is easy in this case.

Complex analysis, residues Find the residue at $z=0$ of $f(z)=\dfrac{\sinh z}{z^4(1-z^2)}$. I did \begin{align} \frac{\sinh z}{z^4(1-z^2)} & =\frac{1}{z^4}\left[\left(\sum_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!)}\right)\left(\sum_{n=0}^\infty z^{2n}\right)\right] \\[8pt] & =\frac{1}{z^4}\left[\left(z+\frac{z^3}{6}+\cdots\right)(1+z^2+z^4+\cdots)\right] \end{align} Then $\operatorname{Res}_{z=0}=1+1+\frac{1}{6}=\frac{13}{6}$ But the solutions say that $\operatorname{Res}_{z=0}=\frac{7}{6}$ Anyone can help me?

The residue at $z=0$ is the coefficient of the $\frac1z$ term in the expansion $$ \frac{\sinh(z)}{z^4(1-z^2)}=\frac1{z^4}\left(z+\frac{z^3}6+\frac{z^5}{120}+\dots\right)\left(1+z^2+z^4+\dots\right) $$ That is the coefficient of $z^3$ term in the expansion $$ \left(\color{#C00000}{z}+\color{#00A000}{\frac{z^3}6}+\frac{z^5}{120}+\dots\right)\left(\color{#00A000}{1}+\color{#C00000}{z^2}+z^4+\dots\right) $$ which is $$ \color{#C00000}{z\cdot z^2}+\color{#00A000}{\frac{z^3}6\cdot1}=\frac76z^3 $$

Why does $\frac{49}{64}\cos^2 \theta + \cos^2 \theta$ equal $\frac{113}{64}\cos^2 \theta $? I have an example: $$ \frac{49}{64}\cos^2 \theta + \cos^2 \theta = 1 $$ Then what happens next: $$ \frac{113}{64}\cos^2 \theta = 1 $$ Where has the other cosine disappeared to? What operation happened here? Any hints please.

$\dfrac{49}{64}\cos^2 \theta + \cos^2 \theta = \left(\dfrac{49}{64} +1\right) \cos^2 \theta = \dfrac{113}{64}\cos^2 \theta$

limits and infinity I'm having trouble wrapping my head around some of the 'rules' of limits. For example, $$ \lim_{x\to \infty} \sqrt{x^2 -2} - \sqrt{x^2 + 1} $$ becomes $$ \sqrt{\lim_{x\to \infty} (x^2) -2} - \sqrt{\lim_{x\to \infty}(x^2) + 1} $$ which, after graphing, seems to approach zero. My question is how do you know for sure the answer is zero without graphing? Thanks!

Hint: $$\left[\sqrt {x^2 - 2} - \sqrt{x^2 + 1}\right]\dot\, \frac{\sqrt {x^2 - 2} + \sqrt{x^2 + 1}}{\sqrt {x^2 - 2} + \sqrt{x^2 + 1}} = \frac{x^2 - 2 - (x^2 + 1)}{\sqrt {x^2 - 2} + \sqrt{x^2 + 1}}$$

$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$ Prove that $$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$$ I proved for two elements, ex, $\mathbb{Q}\left ( \sqrt{2},\sqrt{3}\right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3} \right )$, but I can't do it by the similar method.

It is worth noting from alex.jordan and achille hui's responses that if $\theta = \sqrt{2} + \sqrt{3} + \sqrt{5}$, we explicitly have $$\begin{align*} \sqrt{2} &= \tfrac{5}{3} \theta - \tfrac{7}{72} \theta^3 - \tfrac{7}{144} \theta^5 + \tfrac{1}{576} \theta^7, \\ \sqrt{3} &= \tfrac{15}{4} \theta - \tfrac{61}{24} \theta^3 + \tfrac{37}{96} \theta^5 - \tfrac{1}{96} \theta^7, \\ \sqrt{5} &= -\tfrac{53}{12} \theta + \tfrac{95}{36} \theta^3 - \tfrac{97}{288} \theta^5 + \tfrac{5}{576} \theta^7. \end{align*}$$

find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question: Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$? Key: I use $y = 12 - x$ and substitute into the equation, and derivative it. which I got this $$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$ However, after that. I don't know how to do next in order to find the minimum value. Please help!

the constraint is $x + y = 12.$ at a local extremum of $\sqrt{x^2 + 4} + \sqrt{y^2 + 9},$ the critical numbers satisfy $$dx + dy =0,\quad \frac{x\, dx}{\sqrt{x^2 + 4}} + \frac{y\, dy}{\sqrt{y^2 +9}} = 0 \to x\sqrt{y^2 + 9}=y\sqrt{x^2 + 4} $$ squaring the last equation we have $$9x^2 =4y^2 \to y = \pm\frac32x, x + y = 12 \implies x=24/5, y = 36/5,\\ \sqrt{x^2 + 4} + \sqrt{y^2+9} = 13$$ therefore the global minimum of $$\sqrt{x^2 + 4} + \sqrt{y^2+9} \text{ is } 13.$$

How to solve inequality for : $|7x - 9| \ge x +3$ How to solve inequality for : $|7x - 9| \ge x + 3$ There is a $x$ on both side that's make me confused...

We can solve the absolute value inequality by squaring both sides, then solving the resulting quadratic inequality. \begin{align*} |7x - 9| & \geq x + 3\\ |7x - 9|^2 & \geq (x + 3)^2\\ (7x - 9)^2 & \geq (x + 3)^2\\ 49x^2 - 126x + 81 & \geq x^2 + 6x + 9\\ 48x^2 - 132x + 72 & \geq 0\\ 4x^2 - 11x + 6 & \geq 0\\ 4x^2 - 8x - 3x + 6 & \geq 0\\ 4x(x - 2) - 3(x - 2) & \geq 0\\ (4x - 3)(x - 2) & \geq 0 \end{align*} Equality holds when $x = 3/4$ or $x = 2$. Since $(4x - 3)(x - 2)$ is continuous, the sign of the product can only change at one of the roots. We perform a line analysis. The sign of the product is the product of the signs of the factors. Hence $(4x - 3)(x - 2) \geq 0$ when $x \leq 3/4$ or when $x \geq 2$. Since $|7x - 9| = \sqrt{(7x - 9)^2}$, the steps are reversible. Hence, the solution set of the inequality $(4x - 3)(x - 2) \geq 0$ is the solution set of the absolute value inequality $|7x - 9| \geq x + 3$. Therefore, the solution set of the absolute value inequality is $$S = \left(-\infty, \frac{3}{4}\right] \cup [2, \infty)$$

How to compute determinant of $n$ dimensional matrix? I have this example: $$\left|\begin{matrix} -1 & 2 & 2 & \cdots & 2\\ 2 & -1 & 2 & \cdots & 2\\ \vdots & \vdots & \ddots & \ddots & \vdots\\ 2 & 2 & 2 & \cdots & -1\end{matrix}\right|$$ When first row is multiplied by $2$ and added to second, to $nth$ row, determinant is: $$\left|\begin{matrix} -1 & 2 & 2 & \cdots & 2\\ 0 & 3 & 6 & \cdots & 6\\ \vdots & \vdots & \ddots & \ddots & \vdots\\ 0 & 6 & 6 & \cdots & 3\end{matrix}\right|$$ Now using laplace expansion on first column: $$-\left|\begin{matrix} 3 & 6 & \cdots & 6\\ \vdots & \vdots & \ddots & \vdots \\ 6 & 6 & \cdots & 3\end{matrix}\right|$$ Is it possible to get recursive relation? What to do next? Thanks for replies.

Your recursive approach is fine; just follow it through. Let $D_n(a,b)$ be the determinant of the matrix with diagonal elements $a$ and all other elements $b$; clearly $D_1(a,b)=a$. For $n>1$, multiplying the first row by $-b/a$ and adding it to every other row gives $0$'s in the first column (except for an $a$ in the upper left), $a - b^2/a$ along the rest of the diagonal, and $b - b^2/a$ everywhere else. Using the Laplace expansion on the first column gives $$ D_n(a,b)=a D_{n-1}\left(a-\frac{b^2}{a}, b-\frac{b^2}{a}\right)=aD_{n-1}\left(\frac{(a-b)(a+b)}{a},\frac{b(a-b)}{a}\right). $$ Now, rescaling all $n-1$ remaining rows by $a/(a-b)$ gives $$ D_n(a,b)=\frac{(a-b)^{n-1}}{a^{n-2}}D_{n-1}(a+b,b). $$ Expanding this out, then, $$ D_n(a,b)=\frac{(a-b)^{n-1}}{a^{n-2}}\cdot\frac{a^{n-2}}{(a+b)^{n-3}}\cdot\frac{(a+b)^{n-3}}{(a+2b)^{n-4}}\cdots \frac{(a + (n-3)b)}{1}D_1(a+(n-1)b, n) \\ = (a-b)^{n-1}\left(a + (n-1)b\right), $$ since the product telescopes and all terms cancel except the first numerator. The solution to the original problem is $$ D_n(-1,2)=(-1-2)^{n-1}\left(-1 + (n-1)2\right)=(-3)^{n-1}(2n-3). $$

Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$. My Attempt: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty }\frac{2s}{s}=2$. Is that ok? Thanks.

Hint The numerator is $H_n$ and the denominator is $H_{2n+1}-\frac12H_n$. Also, $$\frac{H_n}{H_{2n+1}-\frac12H_n}=\frac1{-\frac12+\frac{H_{2n+1}}{H_n}}$$ and $$H_n\sim\ln n$$

The perpendicular distance from the origin to point in the plane The plane $3x-2y-z=-4$ is passing through $A(1,2,3)$ and parallel to $u=2i+3j$ and $v=i+2j-k$. The perpendicular distance from the origin to the plane is $r.n = d$ but how to determine the point (Call it N) on the plane and what's the coordinate of the point N?

There is a general procedure & formula derived in Reflection formula by HCR to calculate the point of reflection $\color{blue}{P'(x', y', z')}$ of the any point $\color{blue}{P(x_{o}, y_{o}, z_{o})}$ about the plane: $\color{blue}{ax+by+cz+d=0}$ & hence the foot of perpendicular say point $N$ is determined as follows $$\color{blue}{N\equiv\left(\frac{x_{o}+x'}{2}, \frac{y_{o}+y'}{2}, \frac{z_{o}+z'}{2}\right)}$$ Where $$\color{red}{x'=x_{o}-\frac{2a(ax_{o}+by_{o}+cz_{o}+d)}{a^2+b^2+c^2}}$$ $$\color{red}{y'=y_{o}-\frac{2b(ax_{o}+by_{o}+cz_{o}+d)}{a^2+b^2+c^2}}$$ $$\color{red}{z'=z_{o}-\frac{2c(ax_{o}+by_{o}+cz_{o}+d)}{a^2+b^2+c^2}}$$ As per your question, the foot of perpendicular $N$ drawn from the origin $\color{blue}{(0, 0, 0)\equiv(x_{o}, y_{o}, z_{o})}$ to the given plane: $\color{blue}{3x-2y-z+4=0}$ is determined by setting the corresponding values in the above expression as follows $$\color{}{x'=0-\frac{2(3)(3(0)-2(0)-(0)+4)}{(3)^2+(-2)^2+(-1)^2}}=-\frac{24}{14}=-\frac{12}{7}$$ $$\color{}{y'=0-\frac{2(-2)(3(0)-2(0)-(0)+4)}{(3)^2+(-2)^2+(-1)^2}}=\frac{16}{14}=\frac{8}{7}$$ $$\color{black}{z'=0-\frac{2(-1)(3(0)-2(0)-(0)+4)}{(3)^2+(-2)^2+(-1)^2}}=\frac{8}{14}=\frac{4}{7}$$ Now, setting these values, we get foot of perpendicular $N$ $$N\equiv\left(\frac{0+\left(-\frac{12}{7}\right)}{2}, \frac{0+\frac{8}{7}}{2}, \frac{0+\frac{4}{7}}{2}\right)$$ $$\color{blue}{N\equiv\left(-\frac{6}{7}, \frac{4}{7}, \frac{2}{7}\right)}$$

Solving easy first-order linear differential question. Question Solve $y'=2x(1+x^2-y)$. My attempt Rearranging gives $y'+2xy=2x(1+x^2)$. Thus, the integrating factor is $e^{\int2x\,dx}=e^{x^2}$ and multiplying the equation throughout by this gives $e^{x^2}y'+2xe^{x^2}y=2xe^{x^2}(1+x^2)\Rightarrow\dfrac{d}{dx}{e^{x^2}y}=2xe^{x^2}+2x^3e^{x^2}$ which is separable. Then, $e^{x^2}y=\int2xe^{x^2}\,dx+2\int x^3e^{x^2}\,dx=e^{x^2}+2\int x^3e^{x^2}\,dx$ Is this correct and is there some easy way of working out $\int x^3e^{x^2}\,dx$? Wolfram has it as $e^{x^2}(x^2-1)$. Maybe I write $2x^3e^{x^2}$ as $x(2x^2e^{x^2})$?

make a change of variable $$1+x^2 - y = u, \quad y =1+x^2 - u, y' = 2x-u' $$ then the de $y' = 2x(1+x^2 - y)$ is turned into $$2x-u' = 2xu $$ multiplying by $e^{x^2},$ we get $$e^{x^2}(u'+2xu) = \left(e^{x^2}u\right)' = 2xe^{x^2}$$ on integration yields $$e^{x^2}u = e^{x^2} + c\to u = 1+ce^{-x^2},\quad y = x^2 -ce^{-x^2} $$

Evaluate $\lim\limits_{x\to\infty}(\sin\sqrt{x+1}-\sin\sqrt{x})$ When using Maclaurin series, the limit is $$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x+1}+\sqrt{x}}=0$$ If we expand the expression with two limits $$\lim\limits_{x\to\infty}\sin\sqrt{x+1}-\lim\limits_{x\to\infty}\sin\sqrt{x}$$ it diverges. Which solution is right?

Hint: Mean value theorem implies there exist $c\in]x,x+1[$ such that $$\sin(\sqrt{x+1})-\sin(\sqrt{x})=\frac{\sqrt{x+1}-\sqrt{x}}{2\sqrt{c}}\cos\sqrt{c}$$ Then $$\left|\sin(\sqrt{x+1})-\sin(\sqrt{x})\right|\le\left(\frac{\sqrt{x+1}}{2\sqrt{x}}-\frac{\sqrt{x}}{2\sqrt{x+1}}\right)|\cos\sqrt{c}|\le\frac{1}{2}\left(\sqrt{1+\frac{1}{x}}-\sqrt{\frac{1}{1+1/x}}\right)$$

What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational? Here is my favorite: Theorem: $\sqrt{2}$ is irrational. Proof: $3^2-2\cdot 2^2 = 1$. (That's it) That is a corollary of this result: Theorem: If $n$ is a positive integer and there are positive integers $x$ and $y$ such that $x^2-ny^2 = 1$, then $\sqrt{n}$ is irrational. The proof is in two parts, each of which has a one line proof. Part 1: Lemma: If $x^2-ny^2 = 1$, then there are arbitrarily large integers $u$ and $v$ such that $u^2-nv^2 = 1$. Proof of part 1: Apply the identity $(x^2+ny^2)^2-n(2xy)^2 =(x^2-ny^2)^2 $ as many times as needed. Part 2: Lemma: If $x^2-ny^2 = 1$ and $\sqrt{n} = \frac{a}{b}$ then $x < b$. Proof of part 2: $1 = x^2-ny^2 = x^2-\frac{a^2}{b^2}y^2 = \frac{x^2b^2-y^2a^2}{b^2} $ or $b^2 = x^2b^2-y^2a^2 = (xb-ya)(xb+ya) \ge xb+ya > xb $ so $x < b$. These two parts are contradictory, so $\sqrt{n}$ must be irrational. Two things to note about this proof. First, this does not need Lagrange's theorem that for every non-square positive integer $n$ there are positive integers $x$ and $y$ such that $x^2-ny^2 = 1$. Second, the key property of positive integers needed is that if $n > 0$ then $n \ge 1$.

$$\boxed{\text{If the boxed statement is true, then the square root of two is irrational.}}$$ Lemma. The boxed statement is true. Proof. Assume for a contradiction that the boxed statement is false. Then it has the form "if $S$ then $T$" where $S$ is false, but a conditional with a false antecedent is true. Theorem. The square root of two is irrational. Proof. * *The boxed statement is true. (By the Lemma.) *If the boxed statement is true, then the square root of two is irrational. (This is the boxed statement itself.) *The square root of two is irrational. (Modus ponens.)

Proving a formula using another formula These questions are from the book "What is Mathematics": Prove formula 1: $$1 + 3^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$$ formula 2: $$1^3 + 3^3 + \cdots + (2n+1)^3 = (n+1)^2(2n^2+4n+1)$$ Using formulas 4 and 5; formula 4: $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$ formula 5: $$1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$ My approach for proving the first one was to do something like a subtraction between formula 4 and 1, substituting the value of $n $ in formula 4 for $2n + 1$. I was left with a formula which I proved by mathematical induction gives me $2^2 + 4^2 + \cdots + (2n)^2$. I am wondering if this is the correct approach before doing the second proof? Is there a better and simpler way of doing this?

I would go like this: $(2k+1)^2 = 4k^2+4k+1 \Rightarrow 1+3^2+5^2+\cdots + (2n+1)^2=\displaystyle \sum_{k=0}^n (2k+1)^2=4\displaystyle \sum_{k=0}^n k^2 + 4\displaystyle \sum_{k=0}^n k + (n+1)= 4\cdot\dfrac{n(n+1)(2n+1)}{6}+4\cdot\dfrac{n(n+1)}{2}+(n+1)=...$

Simplify $\prod_{k=1}^5\tan\frac{k\pi}{11}$ and $\sum_{k=1}^5\tan^2\frac{k\pi}{11}$ My question is: If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\frac{3\pi}{11}+\tan^2\frac{4\pi}{11}+\tan^2\frac{5\pi}{11}=Y$ then find $5X^2-Y$. I couldn't find any way to simplify it. Please help. Thanks.

The main building block of our solution will be the formula \begin{align*}\prod_{k=0}^{N-1}\left(x-e^{\frac{2k i\pi}{N}}\right)=x^N-1.\tag{0} \end{align*} It will be convenient to rewrite (0) for odd $N=2n+1$ in the form \begin{align*} \prod_{k=1}^{n}\left[x^2+1-2x\cos\frac{\pi k}{2n+1}\right]=\frac{x^{2n+1}-1}{x-1}. \tag{1} \end{align*} Replacing therein $x\leftrightarrow -x$ and multiplying the result by (1), we may also write \begin{align*} \prod_{k=1}^{n}\left[\left(x^2-1\right)^2+4x^2\sin^2\frac{\pi k}{2n+1}\right]=\frac{1-x^{4n+2}}{1-x^2}. \tag{2} \end{align*} * *Setting in (1) $x=-i$, we get $$\left(2i\right)^n\prod_{k=1}^n\cos\frac{\pi k}{2n+1}=\frac{i^{2n+1}-1}{i-1} \qquad \Longrightarrow\qquad \prod_{k=1}^n2\cos\frac{\pi k}{2n+1}=1.$$ *Setting in (2) $x=1$ and computing the corresponding limit on the right, we get $$\prod_{k=1}^n2\sin\frac{\pi k}{2n+1}=\left[\lim_{x\to 1}\frac{1-x^{4n+2}}{1-x^2}\right]^{\frac12}=\sqrt{2n+1}.$$ *Combining the two results yields $$\boxed{\quad\prod_{k=1}^n\tan\frac{\pi k}{2n+1}=\sqrt{2n+1}\quad}$$ and to find $X$, it suffices to set $n=5$. *To find $Y$, let us rewrite (1) in the form (set $x=-e^{i\gamma}$) $$\prod_{k=1}^n \left[2\cos\gamma+2\cos\frac{\pi k}{2n+1}\right]=\frac{\cos\frac{\left(2n+1\right)\gamma}{2}}{\cos\frac{\gamma}{2}}$$ Taking the logarithm and differentiating twice with respect to $\gamma$, we find $$\sum_{k=1}^{n}\frac{1}{\left(\cos\gamma+\cos\frac{\pi k}{2n+1}\right)^2} =-\frac{1}{\sin\gamma}\frac{\partial}{\partial \gamma}\left(\frac{1}{\sin\gamma}\frac{\partial}{\partial \gamma}\ln \frac{\cos\frac{\left(2n+1\right)\gamma}{2}}{\cos\frac{\gamma}{2}}\right).\tag{3}$$ *Computing the right side of (3) and setting therein $\gamma=\frac{\pi}{2}$, we finally arrive at $$\sum_{k=1}^{n}\frac{1}{\cos^2\frac{\pi k}{2n+1}}=2n(n+1)\qquad \Longrightarrow\quad \boxed{\quad\sum_{k=1}^{n}\tan^2\frac{\pi k}{2n+1}=n(2n+1)\qquad}$$ This yields $Y=55$.

Sum with Generating Functions Find the sum $$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$ How can I use generating functions to solve this?

If $f(z) = \sum\limits_{n=0}^\infty a_n z^n$ converges for $|z| < \rho$, then for any $m \ge 0$, $$\frac{z^m}{m!} \frac{d^m}{dz^m} f(z) = \sum_{n=0}^\infty a_m\binom{n}{m} z^n \quad\text{ for } |z| < \rho.$$ Apply this to $$f(z) = \frac{1}{1-z} = \sum_{n=0}^\infty z^n,$$ we get $$\sum_{n=2}^\infty \binom{n}{2} z^n = \frac{z^2}{2!}\frac{d^2}{dz^2}\frac{1}{1-z} = \frac{z^2}{(1-z)^3}$$ Substitute $z = \frac14$ will give you $$\sum_{n=2}^\infty \binom{n}{2} z^n = \frac{\frac14^2}{(1-\frac14)^3} = \frac{4}{27}$$

An inequality with $\sum_{k=2}^{n}\left(\frac{2}{k}+\frac{H_{k}-\frac{2}{k}}{2^{k-1}}\right)$ show that $$\sum_{k=2}^{n}\left(\dfrac{2}{k}+\dfrac{H_{k}-\frac{2}{k}}{2^{k-1}}\right)\le 1+2\ln{n}$$where $ n\ge 2,H_{k}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{k}$ Maybe this $\ln{k}<H_{k}<1+\ln{k}$?

We have, by partial summation: $$\begin{eqnarray*} \sum_{k=2}^{n}\frac{H_k}{2^{k-1}}&=&H_n\left(1-\frac{1}{2^{n-1}}\right)-\sum_{k=2}^{n-1}\left(1-\frac{1}{2^{k-1}}\right)\frac{1}{k+1}\\&=&1-\frac{H_n}{2^{n-1}}+\sum_{k=2}^{n}\frac{2}{k\, 2^{k-1}}\tag{1}\end{eqnarray*}$$ hence: $$ \sum_{k=2}^{n}\frac{H_k-\frac{2}{k}}{2^{k-1}}=1-\frac{H_n}{2^{n-1}}\tag{2}$$ and: $$\sum_{k=2}^{n}\left(\frac{2}{k}+\frac{H_k-\frac{2}{k}}{2^{k-1}}\right)\leq 2 H_n-1\leq 2\log n+1.\tag{3}$$

How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$ How to find $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$ I tried using L Hospital's rule but its not working!Help please!

Before using L'Hospital, turn the products to sums $$\frac{\cos(ax)-\cos(bx)\cos(cx)}{\sin(bx)\sin(cx)}=\frac{2\cos(ax)-\cos((b-c)x)-\cos((b+c)x)}{\cos((b-c)x)-\cos((b+c)x)}.$$ Then by repeated application $$\frac{2a\sin(ax)-(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)}{(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)},$$ and $$\frac{2a^2\cos(ax)-(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}{(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}.$$ The limit is $$\frac{2a^2-(b-c)^2-(b+c)^2}{(b-c)^2-(b+c)^2}=-\frac{a^2-b^2-c^2}{2bc}.$$

How prove this $\cot(\pi/15)-4\sin(\pi/15)=\sqrt{15}$ I need some help with this demonstration, please I have tried with some identities but nothing. I wanted to use this $$\sin(\pi/15)\cdot \sin(2\pi/15)\cdots\sin(7\pi/15)=\sqrt{15}$$

We may prove: $$ \cos\frac{\pi}{15}-4\sin^2\frac{\pi}{15}=\sqrt{15}\sin\frac{\pi}{15} $$ by squaring both sides. By setting $\theta=\frac{\pi}{15}$, that leads to: $$ \frac{13}{2}-2\cos(\theta)-\frac{15}{2}\cos(2\theta)+2\cos(3\theta)+2\cos(4\theta) = \frac{15}{2}-\frac{15}{2}\cos(2\theta)$$ or to: $$ -\cos(\theta)+\cos(3\theta)+\cos(4\theta) = \frac{1}{2} $$ so we just have to prove that $\cos(\theta)$ is a root of: $$ p(x) = 16x^4+8x^3-16x^2-8x+1.$$ That easily follows from: $$ \Phi_{30}(x) = x^8+x^7-x^5-x^4-x^3+x+1.$$

Evaluating a sum (begginer) I am stuck on this evalutaion number. So the sum is $$\sum_{k=1}^n{1\over (k+1)^2}-{1\over k^2}$$ I can't find a way :(. Do we use Comparasion or partial sums? Thanks for the help!

$$\sum_{k=1}^n\ \left(\frac{1}{(k+1)^2} - \frac{1}{k^2}\right)$$ $S_1 = \frac{1}{2^2} - \frac{1}{1^2}$ $S_2 =\left(\frac{1}{2^2} - \frac{1}{1^2}\right)+\left(\frac{1}{3^2}-\frac{1}{2^2}\right)$ $S_3 = \left(\frac{1}{2^2} - \frac{1}{1^2}\right)+\left(\frac{1}{3^2}-\frac{1}{2^2}\right)+\left(\frac{1}{4^2}-\frac{1}{3^2}\right)$ . . . $S_n = -1+\frac{1}{(n+1)^2} = \sum_{k=1}^n\ \left(\frac{1}{(k+1)^2} - \frac{1}{k^2}\right)$ $\\$ From here you should be able to see what the sum goes to as $n \to \infty$.

How do I solve $y''+y'+7y=t$? How do I solve $y''+y'+7y=t$ where $y(0)=0$ and $y'(0)=0$ $(t\geq 0)$? I tried to solve this by Laplace transformation, but I couldn't find the inverse of $1/(s^2(s^2+s+7))$. How would I solve this?

You have the Laplace transform, so lets do a partial fraction decomposition. We want to find $a,b,c,d$ such that $$\frac{1}{s^2(s^2 + s +7)} = \frac{a}{s} + \frac{b}{s^2} + \frac{cs + d}{s^2 + s + 7}.$$ So we have $$as(s^2 + s +7) + b(s^2 + s +7) + cs^3 + ds^2 = 1.$$ Collecting coefficients we have $$ (a+c)s^3 + (a+b+d)s^2 + (7a+b)s + 7b =1$$ So after working your way through this, you should find, $$a = -\frac{1}{49},b = \frac{1}{7}, d= -\frac{6}{49}, c = \frac{1}{49}.$$ The final trick is to complete the square so that you have $$ s^2+s +7 = \left(s+\frac{1}{2}\right)^2 +\left(\frac{3\sqrt{3}}{2}\right)^2.$$ So $$\frac{1}{s^2(s^2 + s +7)} = \frac{1}{49}\left(\frac{-1}{s} + \frac{7}{s^2} + \frac{s+\frac{1}{2} }{\left(s+\frac{1}{2}\right)^2 +\frac{27}{4}}+\frac{13}{3\sqrt{3}}\frac{\frac{3\sqrt{3}}{2}}{\left(s+\frac{1}{2}\right)^2 +\frac{27}{4}}\right).$$ From this point the solution can be read off tables. Edit: I'd like to add some insight into wythagoras' answer. Let's suppose you have some function $y$. Suppose then you differentiate it a couple of times and you end up with a polynomial. Then it follows that $y$ had to be a polynomial to start with. Remember that differentiating polynomials gives you another polynomial of one less order. Hence, if i add a polynomial to it's derivatives, then what comes out the other end must be the same order as what i started with. So when i have $y'' + y'+7y =t$ then i know that i must have a polynomial, and that the highest order of that polynomial must be of order $t$. So we are well justified assuming that the particular solution has the form $y_p= bt +a$.

Finding a polynomial by divisibility Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$? My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?

One way you could express this idea is through functions $g,h$ where $$f(x) = g(x)(x^2+1) \\ f(x)+1 = h(x)(x^3+x^2+1)$$ This means that $$g(x)(x^2+1)+1 =h(x)(x^3+x^2+1)$$ or that $$h(x) = \frac{g(x)(x^2+1)+1}{x^3+x^2+1}$$ So you could begin by choosing any polynomial $g$ you want so long as it forces $h$ to be a polynomial. Then $f$ will be a polynomial.

Need a hint to evaluate $\lim_{x \to 0} {\sin(x)+\sin(3x)+\sin(5x) \over \tan(2x)+\tan(4x)+\tan(6x)}$ I know that $\sin A + \sin B + \sin C = 4\cos({A \over 2})\cos({B \over 2})\cos({C \over 2})$ when $A+B+C=\pi$. If ${x \to 0}$ then I have a half circle, right? If it is right then I have $\tan(2x) + \tan(4x) + \tan(6x)=\tan(2x)\tan(4x)\tan(6x)$. I got stuck at $${4\cos({x \over 2})\cos({3x \over 2})\cos({5x \over 2}) \over \tan(2x)\tan(4x)\tan(6x)}$$ Is this correct? Do you have a hint to help me?

In what follows I will provide full details of the idea of Abhishek Parab. First divide top and bottom by $x,$ then \begin{eqnarray*} \frac{\sin x+\sin 3x+\sin 5x}{\tan 2x+\tan 4x+\tan 6x} &=&\frac{\left( \dfrac{\sin x+\sin 3x+\sin 5x}{x}\right) }{\left( \dfrac{\tan 2x+\tan 4x+\tan 6x}{x}\right) } \\ && \\ &=&\frac{\left( \dfrac{\sin x}{x}+3\left( \dfrac{\sin 3x}{3x}\right) +5\left( \dfrac{\sin 5x}{5x}\right) \right) }{\left( 2\left( \dfrac{\tan 2x}{% 2x}\right) +4\left( \dfrac{\tan 4x}{4x}\right) +6\left( \dfrac{\tan 6x}{6x}% \right) \right) } \end{eqnarray*} Now using the standard limits \begin{equation*} \lim_{u\rightarrow 0}\frac{\sin u}{u}=\lim_{u\rightarrow 0}\frac{\tan u}{u}=1 \end{equation*} the limit follows \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{\sin x+\sin 3x+\sin 5x}{\tan 2x+\tan 4x+\tan 6x} &=&\frac{\left( \lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}+3\left( \lim\limits_{3x\rightarrow 0}\dfrac{\sin 3x}{3x}\right) +5\left( \lim\limits_{5x\rightarrow 0}\dfrac{\sin 5x}{5x}\right) \right) }{\left( 2\left( \lim\limits_{2x\rightarrow 0}\dfrac{\tan 2x}{2x}\right) +4\left( \lim\limits_{4x\rightarrow 0}\dfrac{\tan 4x}{4x}\right) +6\left( \lim\limits_{6x\rightarrow 0}\dfrac{\tan 6x}{6x}\right) \right) } \\ &=& \\ &=&\frac{\left( 1+3\left( 1\right) +5\left( 1\right) \right) }{\left( 2\left( 1\right) +4\left( 1\right) +6\left( 1\right) \right) }=\frac{3}{4}. \end{eqnarray*}

Partition Generating Function a) Let $$P(x)=\sum_{n=0}^{\infty} p_nx^n=1+x+2x^2+3x^3+5x^4+7x^5+11x^6+\cdots$$ be the partition generating function, and let $Q(x)=\sum_{n=0}^{\infty} q_nx^n$, where $q_n$ is the number of partitions of $n$ containing no $1$s. Then $\displaystyle\frac{Q(x)}{P(x)}$ is a polynomial. What polynomial is it? b) Let $P(x)$ be the partition generating function, and let $R(x)=\sum_{n=0}^{\infty} r_nx^n$, where $r_n$ is the number of partitions of $n$ containing no $1$s or $2$s. Then $\displaystyle \frac{R(x)}{P(x)}$ is a polynomial. What polynomial is it? (Put answer in expanded form) How can I start this problem?

Hint: Try to understand why $$P(x)=\frac{1}{\prod_{n=1}^{\infty}\left(1-x^n\right)},$$ and what are the corresponding expressions for $Q(x)$ and $R(x)$. Illustration: \begin{align*} \frac{1}{1-x}\cdot \frac{1}{1-x^2} \cdot \frac{1}{1-x^3}\cdot\ldots =\,\Bigl(&1+x^1+x^{1+1}+x^{1+1+1}+\ldots\Bigr)\\ \times\,&\Bigl(1+x^2+x^{2+2}+x^{2+2+2}+\ldots\Bigr)\\ \times\,&\Bigl(1+x^3+x^{3+3}+\ldots\Bigr)\times \ldots\end{align*} \begin{align*} =1+x^1+\left(x^{1+1}+x^2\right)+\left(x^{1+1+1}+x^{2+1}+x^3\right)+\\+\left(x^{1+1+1+1}+x^{2+1+1}+x^{2+2}+x^{3+1}+x^4\right)+\ldots\end{align*}

Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form. $$ \sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2 $$ Background To give you some background I am trying to derive $MaxInt(n) = 2^n-1$ which describes that the maximum integer which can be created using the two's complement where $n$ is the number of bits the integer is encoded by. How two's complement encodes numbers with 4 bits is explained by the images below: Therefore: $$ MaxInt(n) = \sum_{i=0}^{n-2} 2^i = (2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2} ) $$ Maybe there is a way of integrating this or simplifying it but I figured that this is a similar problem to sum to N where $$ \frac{T(n) + T(n)}{2} = T(n) = \sum_{i=1}^{n} n-i+1 = \sum_{i=1}^{n} i $$ So following this logic $MaxInt(n)$ is also equal to: $$ MaxInt(n) = \frac{MaxInt(n) + MaxInt(n)}{2} $$ Since $$ (2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2}) = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0) $$ Then $$ MaxInt(n) = \sum_{i=0}^{n-2} 2^{n-2-i} = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0) $$ Putting it all together: $$ MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{n-2-i} + \sum_{i=0}^{n-2} 2^i}{2} $$ $$ MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i}{2} $$ Which is when I got stuck, cheating with wolfrom alpha I found that $$ \sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2 $$ But I don't know why. If you see a better alternative way (i.e. not using sum to N method) of deriving $MaxInt(n) = 2^n-1$ please let me know. Thanks for reading.

The thing you've asked to show isn't too hard: \begin{align} \sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right) &= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \sum_{i=0}^{n-2}\left(2^i\right) \\ &= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(1 + 2 + \ldots + 2^{n-2}\right) \\ &= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(2^{n-1} - 1\right) \\ \end{align} where that last thing comes form the formula for the sum of a geomtric series, which I think you probably know. Now let's simplify the left-hand term... \begin{align} \sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right) &= \sum_{i=0}^{n-2} \left(2^{-i+n-2}\right) + \left(2^{n-1} - 1\right) \\ &= \sum_{i=0}^{n-2} \left(2^{(n -2)-i)}\right) + \left(2^{n-1} - 1\right) \\ &= \left(2^{n -2} + 2^{n-3} + \ldots + 2^0\right) + \left(2^{n-1} - 1\right) \\ \end{align} which we recognize as another geometric series, written in reverse order; the sum there gives us \begin{align} \sum_{i=0}^{n-2} \left(2^{-i+n-2} + 2^i\right) &= \left(2^{n -2} + 2^{n-3} + \ldots + 2^0\right) + \left(2^{n-1} - 1\right) \\ &= \left(2^{n -1} - 1\right) + \left(2^{n-1} - 1\right) \\ &= 2 \cdot 2^{n -1} - 2 \\ &= 2^{n} - 2. \end{align} Quick proof for the geometric series: If we expand $$ U = (1 - a) (1 + a + a^2 + \ldots a^k) $$ with the distributive law, and then gather like terms via the commutative law for addition, we get this: \begin{align} U &= (1 - a) (1 + a + a^2 + \ldots + a^{k-1} + a^k)\\ &= 1 \cdot (1 + a + a^2 + a^{k-1} + \ldots a^k) - a \cdot (1 + a + a^2 + \ldots + a^{k-1} + a^k)\\ &= (1 + a + a^2 + \ldots + a^{k-1} + a^k) - (a + a^2 + a^3 + \ldots + a^k + a^{k+1})\\ &= 1 + (a + a^2 + \ldots a^k) - (a + a^2 + a^3 + \ldots + a^k) - a^{k+1})\\ &= 1 - a^{k+1}) \end{align} so we have that $$ 1-a^{k+1} = (1-a) (1 + a + \ldots + a^k) $$ hence (for $a \ne 1$), $$ 1 + a + \ldots + a^k = \frac{1-a^{k+1}}{1-a}, $$ which is the formula for the sum of a finite geometric series whose ratio is not 1. For an infinite series whose ratio has absolute value less than 1, the infinite sum turns out to be $\frac{1}{1-a}$, by the way, but this requires a careful definition of a sum for an infinite series.

When does equality hold in this inequality? The following inequality can be proven as follows: Let $n\geq3$ and $0=a_0<a_1<\dots<a_{n+1}$ such that $a_1a_2+a_2a_3+\dots+a_{n-1}a_n=a_na_{n+1}$. Show that \begin{equation*} \frac{1}{{a_3}^2-{a_0}^2}+\frac{1}{{a_4}^2-{a_1}^2}+\dots+\frac{1}{{a_{n+1}}^2-{a_{n-2}}^2}\geq\frac{1}{{a_{n-1}}^2}. \end{equation*} Solution: The expression on the left-hand side can be rewritten as $$ \frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2} + \frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2} + \cdots + \frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2}. $$ Applying the Cauchy-Schwarz inequality then yields $$ \begin{align*} &\frac{a_1^2 a_2^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2} + \frac{a_2^2 a_3^2}{a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2} + \cdots + \frac{a_{n-1}^2 a_n^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2} \\ & \ge \frac{\left( a_1 a_2 + a_2 a_3 + \cdots + a_{n-1} a_n \right)^2}{a_1^2 a_2^2 a_3^2 - a_0^2 a_1^2 a_2^2 + a_2^2 a_3^2 a_4^2 - a_1^2 a_2^2 a_3^2 + \cdots + a_{n-1}^2 a_n^2 a_{n+1}^2 - a_{n-2}^2 a_{n-1}^2 a_n^2} \\ & = \frac{a_n^2 a_{n+1}^2}{a_{n-1}^2 a_n^2 a_{n+1}^2 - a_0^2 a_1^2 a_2^2} \ge \frac{1}{a_{n-1}^2}. \end{align*} $$ When does equality hold?

Deduced from the equality condition of Cauchy-Schwarz inequality, we have: $$ \forall 0 \leqslant k \leqslant n-1, \, \, a_{k+3}^2 = a_k^2 + c$$ where $c > 0$ is a constant. So the equation has 3 possiblities that depends on $n$. For example, if $n = 3K + 1$: $$\sum_{k=0}^{K-1} (a_0 + kc)(a_1 + kc) + (a_1 + kc)(a_2 + kc) + (a_2 + kc)(a_0 + (k+1)c) = (a_1 + Kc)(a_2 + Kc) - (a_0 + Kc)(a_1 + Kc).$$ We can solve the above equation with respect to $c$(quadratic) and see what the roots are. If one of the roots is positive then the equality holds.

How prove $\frac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$ Let $a,b,c\in R$,and such $ab+bc+ac=0,a+b+c\neq 0$ show that $$\dfrac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$$

Hint: Express the numerator in terms of elementary symmetric polynomials, as the denominator and constraint already are. You get $$(a-b)^4+(b-c)^4+(c-a)^4 \\= 2(a+b+c)^4 - 12(ab + bc + ca) (a+b+c)^2 + 18(ab+bc+ca)^2$$

High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine: Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).

First lets consider $$\frac{b-2a\cos\gamma}{a\sin\gamma}$$ The numerator looks similar to the RHS of the cosine law, but not quite. It would be nice to see $-2ab\cos\gamma$ instead of $-2a\cos\gamma$. So lets just multiply by $b$. Then the numerator would look like this $$b^2-2ab\cos\gamma$$ Now all that is missing is the $a^2$, so just add it $$a^2+b^2-2ab\cos\gamma$$ Now my expression is equal to $c^2$. Easy, right? It would be nice if math was really like this, but unfortunately, we can't just add and multiply arbitrary constants whenever it seems convenient to do so. We can, however, add zero (e.g. $+\phi-\phi$) and multiply by one (e.g. $\frac{\phi}{\phi}$). $$\frac{b-2a\cos\gamma}{a\sin\gamma} = \frac{b-2a\cos\gamma}{a\sin\gamma}\cdot\frac{b}{b}=\frac{b^2-2ab\cos\gamma}{ab\sin\gamma}=\frac{a^2+b^2-2ab\cos\gamma-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}$$ We have just added zero ($a^2-a^2$), so next, lets multiply by one ($\frac{c}{c}$), so we can isolate $\frac{c}{\sin\gamma}$. $$\frac{c^2-a^2}{ab\sin\gamma}=\frac{c^2-a^2}{ab\sin\gamma}\cdot\frac{c}{c}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}$$ Keeping in mind, of course that $$\Psi=\frac{c}{\sin\gamma}=\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}$$ Then our expression reduces to $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\cos\alpha}{b\sin\alpha}+\frac{a-2c\cos\beta}{c\sin\beta}=\frac{c^2-a^2}{abc}\cdot\frac{c}{\sin\gamma}+\frac{a^2-b^2}{abc}\cdot\frac{a}{\sin\alpha}+\frac{b^2-c^2}{abc}\cdot\frac{b}{\sin\beta}$$ $$=\frac{\Psi}{abc}\big((c^2-a^2)+(a^2-b^2)+(b^2-c^2)\big)=0$$

Sums $a_k=[\frac{2+(-1)^k}{3^k}, (\frac{1}{n}-\frac{1}{n+2}), \frac{1}{4k^2-1},\sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}]$ Determien the sums of the following series'. 1:$\sum_{k=0}^\infty \frac{2+(-1)^k}{3^k}$ 2:$\sum_{k=0}^\infty (\frac{1}{n}-\frac{1}{n+2})$ 3:$\sum_{k=0}^\infty \frac{1}{4k^2-1}$ 4:$\sum_{k=0}^\infty \sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}$ I have always been bad with infinite series', so I decided to practice them strongly for the next couple of days. I found an exercise in a textbook and kinda troubling with some of them. Here's what I was thinking of doing. 1: I thought about dividing this into two sums, $\sum_{k=0}^\infty \frac{2}{3^k}+\sum_{k=0}^\infty \frac{(-1)^k}{3^k}$, which should be okay since they converge, right? Anyway, then I can see the geometric series there, giving me $3+\frac{3}{4}=3,75$. Is that correct? 2: I think that's what's called a telescope series? So I tried writing down some of the first numbers to see where it starts canceling each other out: $\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1 }{ 4}+\frac{ 1}{ 3}-\frac{ 1}{ 5}+\frac{1 }{ 4}-\frac{1 }{6 } + ...$. Since $\frac{1}{n}$ converges to zero for $n\to \infty$, the sum of this infinite series is $\frac{3}{2}$? About 3 and 4. I'm pretty much lost there. Can't seem to find an approach to find the sum. Any tips?

$(1)$ Looks good to me. $(2)$ You are correct that this is a telescoping series and your answer is correct (but the index should start at $k=1$, not $k=0$) You could also rewrite the series as $$\begin{align}\sum_{k=1}^\infty \frac{1}{k}-\frac{1}{k+2} = \sum_{k=1}^\infty \frac{1}{k}-\sum_{k=1}^\infty \frac{1}{k+2} \\ = \left(1+\frac{1}{2}+ \sum_{k=3}^\infty \frac{1}{k}\right)-\sum_{k=3}^\infty \frac{1}{k} \\ = 1+\frac{1}{2}\end{align}$$ to more rigorously see that the series converges to $\frac{3}{2}$. $(3)$ Write the fraction $\frac{1}{4k^2-1}$ as $\frac{1}{(2k-1)(2k+1)}$, then use partial fraction decomposition to find constants $A,B$ where $\frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1}+ \frac{B}{2k+1}$ You may again need to invoke an argument about telescoping series. For $(4)$, is that really the binomial coefficient $\binom{k}{l} = \frac{k!}{l!(k-l)!}$? I did not encounter that in my calc class unit on infinite series. However, start writing out a few terms of the series to get a handle on what the general behavior is. That is, fix $k=0$ and find a value. Then set $k=1$ to get a few more values, etc.

evaluate $\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx $ This is supposed to be a very easy integral, however I cannot get around. Evaluate: $$\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx$$ What I did is: $$\int_{0}^{2\pi}\frac{dx}{\cos x + \sin x +2} = \int_{0}^{2\pi} \frac{dx}{\left ( \frac{e^{ix}-e^{-ix}}{2i}+ \frac{e^{ix}+e^{-ix}}{2} +2\right )}= \oint_{|z|=1} \frac{dz}{iz \left ( \frac{z-z^{-1}}{2i} + \frac{z+z^{-1}}{2}+2 \right )}$$ Although the transformation is correct I cannot go on from the last equation. After calculations in the denominator there still appears $i$ and other equations of that. How can I proceed?

Via corindo's rearrangement: we have $$ I = \int_0^{2 \pi} \frac{1}{\sqrt{2}\cos(x - \pi/4) + 2}\,dx =\\ \frac 1{\sqrt{2}} \int_0^{2 \pi} \frac{1}{\cos(x - \pi/4) + \sqrt{2}}\,dx = \\ \frac 1{\sqrt{2}} \int_0^{2 \pi} \frac{1}{\cos(x) + \sqrt{2}}\,dx =\\ \frac 1{\sqrt{2}} \oint_{|z| = 1} \frac{1}{iz((z + z^{-1})/2 + \sqrt{2})}\,dx = \\ \frac {\sqrt{2}}{i} \oint_{|z| = 1} \frac{1}{(z^2 + 2\sqrt{2}z + 1)}\,dz $$ Note the integrand has one pole inside the unit circle, namely $$ z_{0} = 1-\sqrt{2} $$

Rotation of a line by a matrix Give the equation of the line $\ell'$ that is obtained by rotating $\ell$: $x+2y=5$ by an angle of $\theta=\frac{1}{2}\pi$ with center point $O(0,0)$. The rotation matrix is $\left.\begin{pmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{pmatrix}\right|_{\alpha=\frac{1}{2}\pi}=\begin{pmatrix}0 & -1 \\ 1&0\end{pmatrix}$. Two points on $\ell$ are $(1,2)$ and $(3,1)$, whose images are respectively $\begin{pmatrix}0 & -1 \\ 1&0\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}-2\\1\end{pmatrix}$ and $\begin{pmatrix}0 & -1 \\ 1&0\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}-1\\3\end{pmatrix}$. Hence, the gradient of $\ell'$ is $\dfrac{\Delta y}{\Delta x}=2$. Therefore, the equation of $\ell'$ is $y=2(x+2)+1\iff \boxed{y=2x+5}$. I have two questions: my book says the answer should be $y=2x+6$. If I am wrong, what did I wrong? Second, is there anyone who suggests a more elegant of faster method for this problem?

I notice that the point $(1,2)$ is on the original line. After a rotation by $\pi/2$, this point becomes $(-2,1)$. I also notice that the original slope was $-1/2$. A rotation by $\pi/2$ has the effect of finding a line perpendicular to our starting line, so it will have slope $2$. The equation of a line with slope $2$ passing through $(-2,1)$ is $y - 1 = 2(x + 2)$, or rather $y = 2x + 5$.

Is the following a conic section All vectors are in $\mathbb{R}^3$ and only $\mathbf{r} = \left[ x; y; z \right]$ is unknown. My question is does the following system define a conic section in the $x-y$ plane and, if so, how can I find it: $$ \begin{align} \mathbf{r}^\mathrm{T} \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] & = 0 \\ \\ \mathbf{v}_1^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_1} {\Vert \mathbf{r} - \mathbf{r}_1 \Vert } + \mathbf{v}_2^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_2} {\Vert \mathbf{r} - \mathbf{r}_2 \Vert } & = c \end{align} $$ If either $\Vert \mathbf{v}_1 \Vert = 0$ or $\Vert \mathbf{v}_2 \Vert = 0$, then the above is the intersection of a cone and the $z=0$ plane. Likewise if $\mathbf{r} = \mathbf{r}_1$ or $\mathbf{r} = \mathbf{r}_2$. However, I have been unable to figure out what the above represents in the general case.

No, these are not conic sections in general. Your first equation forces $z = 0$. For \begin{align*} \mathbf{r}_1 = \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \\ \mathbf{r}_2 = \mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \text{,} \end{align*} with $c = 0$, your second equation reduces to $$ \frac{x}{\sqrt{|x|^2+|y-1|^2}} + \frac{y}{\sqrt{|x-1|^2+|y|^2}} = 0 \text{.} $$ This is not the equation of a conic section. It is the line $y = 1-x$ for $x \in (-\infty, 0] \cup [1,\infty)$ joined by the arc of the circle centered at $(1/2,1/2)$ with radius $1/2$ for the angles in the interval $[3\pi/4, 7\pi/4]$. Changing only $c=3/2$, it's not a conic section. Changing $\mathbf{v}_1 = \begin{pmatrix}0\\2\\0\end{pmatrix}$ and setting $c = 1/2$, the graph loses its symmetry. The $r_i$ seems to control where the pieces of the piecewise curve meet. Many choices of $c$ give no solutions (which is another thing that cannot happen with a planar section of nappes). Why would you believe there is a short description of what is represented by varying the parameters in your equation?

Minimum value of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{24}{5\sqrt{5a+5b}}$ Let $a\ge b\ge c\ge 0$ such that $a+b+c=1$ Find the minimum value of $P=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\dfrac{24}{5\sqrt{5a+5b}}$ I found that the minimum value of $P$ is $\dfrac{78}{5\sqrt{15}}$ when $a=b=\dfrac{3}{8};c=\dfrac{1}{4}$ And this is my try Applying AM-GM inequality, we get: $\dfrac{b+c}{a}+\dfrac{5}{3}\ge2\sqrt{\dfrac{5}{3}}.\sqrt{\dfrac{b+c}{a}}$ This implies $\sqrt{\dfrac{a}{b+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3a}{3+2a}$ Similarly, $\sqrt{\dfrac{b}{a+c}}\ge2\sqrt{\dfrac{5}{3}}.\dfrac{3b}{3+2b}$ We need to prove that: $2\sqrt{\dfrac{5}{3}}\left(\dfrac{3a}{3+2a}+\dfrac{3b}{3+2b}\right)+\dfrac{24}{5\sqrt{5a+5b}}\ge\dfrac{78}{5\sqrt{15}}$ But I have no idea how to continue. Who can help me or have any other idea?

You can use calculus to find this. $f(a,b) = \sqrt{\frac{a}{1-a}} + \sqrt{\frac{b}{1-b}} + \frac{24}{5\sqrt{5(a + b)}}$ Now, let us first find the critical point, $(x,y)$, of this function, where $f_a(x,y) = 0$ and $f_b(x,y) = 0$. Taking partial derivatives and setting them to 0, you will see that the critical point is at $a = b$. So we now have a 1D function, $g(a) = f(a,a) = 2\sqrt{\frac{a}{1 - a}}+ \frac{24}{5\sqrt{10 a}}$. Then differentiate $g$ with respect to $a$, set it equal to 0, and you get that $a = 3/8$, so then $b = 3/8$ and $c = 1/4$. You can make this more concrete by constructing the Hessian matrix of $f(a,b)$ and verifying that this is indeed the minimum (check that the determinant is >0 and that $f_{aa} > 0$ at this critical point).

Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system $$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$ I have solved it by performing the substitutions $x-y=u$ and $xy=v$. Then I multiplied the first equation by $6^u$ and used $a^2-b^2=(a+b)(a-b)$ to find $$(3^u+2^u)(3^u-2^u) = 65 \cdot 6^{u-2}.$$ By inspection I found $u=2$ and $v=120$. I solved the original system in $x,y$ and got the answers. Is there another quicker way to solve this without resorting to this sort of ninja inspection? I have found a second solution by solving $a^u +1/a^u = 65/36$, which assures $u=2$ but takes much more time. Could there be a third way faster than these?

Consider the equation $$\left( \dfrac{3}{2} \right)^{x-y} - \left( \dfrac{2}{3} \right)^{x-y} = \dfrac{65}{36}.$$ Let $u=2^{x-y}$ and $v=3^{x-y}$ then we have $36u^2+65uv-36v^2=0.$ Hence $$u=\dfrac49v,\,\,\,\text{or}\,\,\,\,u=-\dfrac94v$$ For real solutions, take the first one and then $\dfrac{u}{v}=\left(\dfrac23\right)^2\implies x-y=2.$ Now $$xy-x+y=118\implies (x+1)(y-1)=117\implies(y-1)(y+3)=9\times13=(-9)\times(-13).$$ Hence $x=12,\,\,\ y=10$ and $x=-10,\,\,\ y=-12$ are the only real solutions for given non-linear system.

Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$ My attempt is $$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$ $$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$ then what I will do??

$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}\\ \implies x^3=2-\color{red}{\sqrt3}+2+\color{red}{\sqrt3}+3(2-\sqrt3)(2+\sqrt3)(\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}})\\ \implies x^3=4+3.1.\color{red}{x}\\ \implies x^3-4-3x=0$

Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve: If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$ I tried to substitute the value of x in the expression, but that comes out to be very big.

Since everyone else has answered with various shortcut methods, let me just show you how you could simply substitute in directly and get an answer cleanly without things ever getting 'too big'. We know that $x=2+\sqrt{3}$, so we can square this using the usual $a^2+2ab+b^2$ binomial formula: $x^2=2^2+2(2)(\sqrt3)+(\sqrt3)^2$ $= 4+4\sqrt3+3$ $=7+4\sqrt{3}$. Now, $\dfrac1{x^2}$ is $\dfrac1{7+4\sqrt3}$; we can use the usual method for rationalizing the denominator to handle this, by multiplying by $\dfrac{7-4\sqrt3}{7-4\sqrt3}$. This gives $\dfrac1{x^2}$ $=\dfrac{7-4\sqrt3}{(7+4\sqrt3)(7-4\sqrt3)}$ $=\dfrac{7-4\sqrt3}{7^2-(4\sqrt3)^2}$ $=\dfrac{7-4\sqrt3}{7^2-4^2\cdot 3}$ $=\dfrac{7-4\sqrt3}{49-48}$ $=7-4\sqrt3$. The most important lesson here is that whenever you have an expression of the form $x=a+b\sqrt{c}$, then the powers of $x$ — both positive and negative — will never get too messy; they'll all always be of the form $p+q\sqrt{c}$ for some rational $p$ and $q$. In this case, $p$ and $q$ were integers even for the negative powers of $x$; that won't always be the case, but they'll always be rational.

Finding the sum of the series $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$ Deteremine the sum of the series $$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \ldots$$ So I first write down the $n^{th}$ term $a_n=\frac{\frac{n(n+1)}{2}}{n!}=\frac{n+1}{2(n-1)!}$. So from there I can write the series as $$1+\frac{3}{2}+\frac{4}{2\times 2!}+\ldots +\frac{n+1}{2(n-1)!}+\ldots $$ I am quite sure I can do some sort of term by term integration or differentiation of some standard power series and crack this. Any leads?

HINT: Split $$\frac{n+1}{2(n-1)!}= \frac{1}{2(n-2)!}+\frac{1}{(n-1)!}$$ and use the fact that $\sum_{n=0}^{\infty} \frac{1}{n!}=e$

Product of Matrices I Given the matrix \begin{align} A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right) \end{align} consider the first few powers of $A^{n}$ for which \begin{align} A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right) \hspace{15mm} A^{2} = \left( \begin{matrix} 7 & 6 \\ 9 & 10 \end{matrix} \right) \hspace{15mm} A^{3} = \left( \begin{matrix} 25 & 26 \\ 39 & 38 \end{matrix} \right). \end{align} Notice that the first rows have the values $(1,2)$, $(7,6)$, $(25,26)$ of which the first and second elements are rise and fall in a cyclical pattern. The same applies to the bottom rows. * *Is there an explanation as to why the numbers rise and fall in order of compared columns? *What is the general form of the $A^{n}$ ?

In answer to part 2, the general form is $$M^n=\frac{1}{5}\begin{pmatrix} 2\times 4^n+3 \times(-1)^n & 2\times 4^n-2(-1)^n\\3\times 4^n+3(-1)^{n+1} & 3\times 4^n+2(-1)^{n}\\\end{pmatrix}$$ But you know this already because it has already been posted! (it just took me longer to write out in MathJax)

How can I prove irreducibility of polynomial over a finite field? I want to prove what $x^{10} +x^3+1$ is irreducible over a field $\mathbb F_{2}$ and $x^5$ + $x^4 +x^3 + x^2 +x -1$ is reducible over $\mathbb F_{3}$. As far as I know Eisenstein criteria won't help here. I have heard about a criteria of irreducibility which sounds like "If $f(x)$ divides $x^{p^n-1}-1$ (where $n=\deg(f))$, $p$ is from $\mathbb F_{p}$ then $f(x)$ is irreducible." But I don't know how to show if $f(x)$ can be divided by this or not.

I think that the criterion that you allude to is the following: Assume that $X^p - a$ has no roots in $\Bbb F _{p^n}$. Then $X^{p^m} - a$ is irreducible in $\Bbb F _{p^n}[X]$ $\forall m \ge 1$. In any case, you can't use it here. 1) Let us see how to prove that the second polynomial (call it $P$) is reducible. First, none of $0, 1, 2$ is a root, so $P$ has no factor of degree $1$ (and, therefore, no factor of degree $4$). Let us see whether $P$ can be written as a product of two polynomials, of degree $2$ and, respectively, $3$. The long method is to multiply the polynomials $aX^2 + bX + c$ and $eX^3 + fX^2 + gX + h$, equate the coefficients of equal powers etc... A shorter approach is the following: let us list all the polynomials of degree $2$ and check whether they divide $P$. Since $P$ has no linear factor, it is enough to list only those polynomials of degree $2$ that are irreducible. Note that no such polynomial may end in $0$, so the constant term is $1$ or $2$. Concerning the leading coefficient, it is enough to consider only monic polynomials. So far we get the polynomials $X^2 + aX + 1$, $X^2 + aX +2$. A polynomial of degree $2$ is irreducible if it has no roots, i.e. if its discriminant is not a perfect square. Since the only perfect squares in $\Bbb F _3$ are $0$ and $1$, you want $a$ such that the discriminant should be $2$. In the first case, the discriminant is $a^2 - 4$, so you want $a$ such that $a^2 - 4 = 2$, so $a=0$. In the second case, the discriminant is $a^2 - 8$, so you want $a$ such that $a^2 - 8 = 2$, i.e. $a^2 = 1$, i.e. $a=1$ or $a=2$. So, the only monic irreducible polynomials of degree $2$ are $X^2 + 1$, $X^2 + X + 2$, $X^2 + 2X +2$. Let us see which one divides our polynomial. Note that $P = X^3 (X^2+1) + X^2 (X^2+1) + X-1$, so when you divide $P$ by $X^2 +1$ you get the remainder $X-1$, so $X^2-1 \not| P$. Finally, try to divide $P$ by the last two polynomials. $X^2 + 2X +2$ will turn out to be a factor. 2) Concerning the first polynomial (call it $Q$), the approach will be similar. First, note that it has no roots, so it has no linear factor. Therefore, we are going to look only for irreducible factors of degree $2, \dots, 5$. In order to be irreducible, these potential factors must have the constant term $1$. Looking for irreducible polynomials of degree $2$, these must look like $X^2 +aX +1$. Clearly, $a=1$ gives the only irreducible one. For degree $3$, you want those polynomials $X^3 + aX^2 + bX +1$ that have no linear factor; since $0$ cannot be a root, you also do not want $1$ to be so, therefore you want $1+a+b+1 \ne 0$, which means $a+b =1$, so the only possibilities are $X^3 + X^2 +1$ and $X^3 +X+1$. In degree $4$, you want those polynomials $X^4 + aX^3 + bX^2 + cX +1$ that have no roots (so $1+a+b+c+1 \ne 0$, i.e. $a+b+c=1$) and that have no irreducible factor of degree $2$, i.e. that are not divided by $X^2+X+1$ (found above). A reducible factor of degree $4$ having no root would have to be $(X^2+X+1)^2 = X^4 + X^2 +1$. Therefore, the only irreducible polynomials of degree $4$ remain $X^4 + X^3 +1$, $X^4+ X+1$ and $X^4+ X^3 + X^2 + X + 1$. Finally, the reducible polynomials $x^5 + aX^4 + bX^3 +cX^2 + dX +1$ of degree $5$ are those that have roots (i.e. $a+b+c+d=0$) and those that can be divided by $X^2+1$. Performing long division by $X^2+1$, you get the remainder $(b+d+1)x + (a+c+1)$, so in order to get the reducible polynomials impose $a+b+c+d = 0, \; b+d+1 = 0, \; a+c+1 = 0$. Solve this system (it will have several solutions); the polynomials that are not among these solutions are the irreducible ones of degree $5$. Now that you've listed all the irreducible polynomials of degree $\le 5$, check (by performing long division or by computing the greatest common divisor) which ones divide $Q$. None will, so $Q$ is irreducible. Below is the proof of the irreducibility criterion mentioned at the beginning of my post. Notice that $X^{p^m} - a$ has at least one root $x$ in some algebraic closure $K$ of $\mathrm F_{p^m}$; if $y \in K$ is another root, it follows that $x^{p^m} = y^{p^m}$ and, since $r \mapsto r^{p^m}$ is an automorphism of $K$ (because the Frobenius map $r \mapsto r^p$ is), it follows that $x=y$. It follows that $X^{p^m} - a$ has exactly one root $x \in K$, of multiplicity $p^m$. If $g \in \mathrm F_{p^m} [X]$ is the minimal polynomial of $x$, then $X^{p^m} - a = g^s$; since $p^m = s \deg g$, it follows that $s = p^t$. Let $b = -g(0)$ and assume $t>0$. Evaluating $X^{p^m} - a = g^s$ in $0$, and assuming $t>0$, we get $a = (b^{p^{t-1}})^p$ (because $-1 = (-1)^s$ in characteristic $p>0$), which would imply that $X^p - a$ has the root $b^{p^{t-1}} \in \mathrm F _{p^m}$, which would contradict the hypothesis of the criterion. It follows that $t=0$, so that $s=1$, therefore $X^{p^m} - a$ is the minimal polynomial of $a$, therefore irreducible by the definition of the concept of "minimal polynomial".

Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$ I really tried, but I couldn't, help guys?

Hint: $$\sum_{n=1}^{\infty} \frac{n^2}{2^n} = \sum_{i=1}^{\infty} (2i - 1) \sum_{j=i}^{\infty} \frac{1}{2^j}.$$ Start by using the geometric series formula on $\displaystyle \sum_{j=1}^{\infty} \frac{1}{2^j}$ to simplify the double series into a singular series. Then you will have a series that looks like $\displaystyle \sum_{i=1}^{\infty} \frac{i}{2^i}$. Just as I broke your initial series with quadratic term $n^2$ into a double series with linear term $i$, you can break this series with linear term $i$ into a double series with constant term $c$. In a clearer form: \begin{align} &\frac{1}{2} + &\frac{4}{4} + &\frac{9}{8} + &\frac{16}{16} + \dots =\\ &\frac{1}{2} + &\frac{1}{4} + &\frac{1}{8} + &\frac{1}{16} + \dots +\\ & &\frac{3}{4} + &\frac{3}{8} + &\frac{3}{16} + \dots + \\ & & & \frac{5}{8} + &\frac{5}{16} + \dots + \\ & & & &\frac{7}{16} + \dots + \\ \end{align} Notice that each of the sums are geometric series, which can be evaluated easily. Sorry about the weird formatting, I don't know where these spaces are coming from.

Approximating $\tan61^\circ$ using a Taylor polynomial centered at $\frac \pi 3$ : how to proceed? Here's what I have so far... I wrote a general approximation of $f(x)=\tan(x)$ , which then simplified a bit to this: $$\tan \left(\frac{61π}{180}\right) + \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right) + \tan\left(\frac{61π}{180}\right) \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right)^2 $$ Thing is, I'm not seeing anything obvious to do next... any hints/suggestions on how to proceed in my approximation? Thanks in advance!

There are many ways to approximate (even very accurately) functions close to a point. The simplest is Taylor expansion; in the case of the tangent, assuming $b<<a$, the expansion is $$\tan(a+b)=\tan (a)+ \left(\tan ^2(a)+1\right)b+ \left(\tan ^3(a)+\tan (a)\right)b^2+$$ $$ \left(\tan ^4(a)+\frac{4 \tan ^2(a)}{3}+\frac{1}{3}\right)b^3+ \left(\tan ^5(a)+\frac{5 \tan ^3(a)}{3}+\frac{2 \tan (a)}{3}\right)b^4+$$ $$ \left(\tan ^6(a)+2 \tan ^4(a)+\frac{17 \tan ^2(a)}{15}+\frac{2}{15}\right)b^5+O\left(b^6\right)$$ Applied to $a=\frac \pi 3$, it gives $$\tan(\frac \pi 3+b)=\sqrt{3}+4 b+4 \sqrt{3} b^2+\frac{40 b^3}{3}+\frac{44 b^4}{\sqrt{3}}+\frac{728 b^5}{15}+O\left(b^6\right)$$ Using $b=\frac \pi {180}$ and the successive orders the approximate value is $$1.80186397765$$ $$1.80397442904$$ $$1.80404531673$$ $$1.80404767396$$ $$1.80404775256$$ while, for twelve significant digits, the exact value should be $$1.80404775527$$ Edit The following is just for your curiosity Another way is to use Pade approximant (these are ratios of polynomials); built at $a=\frac \pi 3$, the simplest would be $$P_{(1,1)}(x)=\frac{(x-\frac{\pi }{3})+\sqrt{3}}{1-\sqrt{3} \left(x-\frac{\pi }{3}\right)}$$ which gives for $x=\frac {61\pi} {180}$ $$1.80404021672$$ Similarly $$P_{(2,2)}(x)=\frac{-\frac{\left(x-\frac{\pi }{3}\right)^2}{\sqrt{3}}+(x-\frac{\pi }{3})+\sqrt{3}}{-\frac{1}{3} \left(x-\frac{\pi }{3}\right)^2-\sqrt{3} \left(x-\frac{\pi }{3}\right)+1}$$ which gives $$1.80404775512$$

Show that in any triangle, we have $\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$ Show that in any triangle, we have $$\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$$ where $R$ is the circumradius of the triangle. Here is my work: We know that $A+B+C=180^\circ$, so $C=180^\circ -(A+B)$. Plugging this in, we get that $\sin C=\sin (A+B)$ and $\cos C = -\cos (A+B)$. When we plug this into the equation we get, $$\frac{a\sin A+b\sin B+c\sin (A+B)}{a\cos A+b\cos B-c\cos (A+B)}.$$ If we expand out $c\sin (A+B)$ and $c\cos (A+B)$, we get $$\frac{\sin A+b \sin B+c \cos A\cos B - c\sin A\sin B}{a\cos A+b\cos B-c\cos A\cos B+c\sin A\sin B}.$$ Using the Extended Law of Sines, we can use $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$, and $\sin C=\frac{c}{2R}$. How can I continue on?

Since the sine theorem implies: $$\sum_\text{cyc}a\sin A = \frac{1}{2R}\sum_\text{cyc}a^2 \tag{1}$$ we just need to prove: $$ \sum_\text{cyc} a \cos A = \frac{abc}{2R^2}=\frac{2\Delta}{R}\tag 2$$ that is trivial since twice the (signed) area of the triangle made by $B,C$ and the circumcenter $O$ is exactly $aR\cos A$:

Square roots equations I had to solve this problem: $$\sqrt{x} + \sqrt{x-36} = 2$$ So I rearranged the equation this way: $$\sqrt{x-36} = 2 - \sqrt{x}$$ Then I squared both sides to get: $$x-36 = 4 - 4\sqrt{x} + x$$ Then I did my simple algebra: $$4\sqrt{x} = 40$$ $$\sqrt{x} = 10$$ $$x = 100$$ The problem is that when I go back and plug my $x$-value into the equation, it doesn't work. $$\sqrt{100} + \sqrt{100-36} = 2$$ $$10+8 = 2$$ Which is obviously wrong.

Method $\#1:$ As for real $a,\sqrt a\ge0\ \ \ \ (1)$ $(\sqrt x+\sqrt{36-x})^2=36+2\sqrt{x(36-x)}\ge36$ $\implies\sqrt x+\sqrt{36-x}\ge6\ \ \ \ (2)$ or $\sqrt x+\sqrt{36-x}\le-6\ \ \ \ (3)$ Finally $(1)$ nullifies $(3)$ Method $\#2:$ WLOG let $\sqrt x=6\csc2y$ where $0<2y\le\dfrac\pi2\implies\sqrt{x-36}=+6\cot2y$ $\sqrt x+\sqrt{36-x}=6\cdot\dfrac{1+\cos2y}{\sin2y}=6\cot y$ Now $0<2y\le\dfrac\pi2\implies0<y\le\dfrac\pi4\implies\cot0>\cot y\ge\dfrac\pi4=1$ as $\cot y$ is decreasing in $\left[0,\dfrac\pi2\right]$

this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$ show that: $$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$ Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$ so it suffices to prove that $$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$ then the problem is solved. I stuck in here

with loss of generality, assume $x\ge y $ consider $f(x, y, z) = (x^2 + x + 1)(y^2 + y + 1) (z^2 + z + 1) - 9 (z(x+y) + xy)$ * *we first show that $$f(x, y, z) \ge f((x+y)/2, (x+y)/2, z)$$ since $$f(x,y,z) - f((x+y)/2, (x+y)/2, z) = (x-y)((z^2 + z + 1)(z -2 - 2xy) + 9)$$ we simply show that $$(z^2 + z + 1)(z -2 - 2xy) + 9 \ge 0$$ while $z^2 + z + 1 > 0$, $$(z^2 + z + 1)(z - 2 - 2xy) + 9 \ge (z^2 + z + 1)(z - 2 - \frac{(3 - z)^2}{2}) + 9\\ = \frac{1}{2}(18 - (z^2 + z + 1)(z^2 - 8z +13))$$ and it is quite easy to show that when $0\le z\le 3$, $$(z^2 + z + 1)(z^2 - 8z +13)\le 18$$ with maximum at $z = 1$. *If we achieve minimum of $f(x, y, z)$ at any point other than $(1,1,1)$, use the above argument, we can show a contradiction.

How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, : $$ x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\ x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z} $$ Wolfram Alpha, here the link, instead, gives the following results: $ x_{1} = 2\pi n - 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}\\ x_{2} = 2\pi n + 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z} $ Now, supposing that my solutions are correct, this means that there must be a relation between: $\pi - \arccos(\frac{\sqrt{6}}{3})$ and $- 2\arctan(\sqrt{5 + 2\sqrt{6}})$ or between: $\pi + \arccos(\frac{\sqrt{6}}{3})$ and $+ 2\arctan(\sqrt{5 + 2\sqrt{6}})$ or viceversa. But, given the solutions I have found, how can I prove that they are effectively the same as the solutions Wolfram found? Mathematically? P.S.: I have found out, by looking at the graphs of $y_{1} = \pi - \arccos(\frac{\sqrt{6}}{3})$ and $y_{2} = 2\arctan(\sqrt{5 + 2\sqrt{6}})$ e.g., that: $\pi - \arccos(\frac{\sqrt{6}}{3}) = 2\arctan(\sqrt{5 + 2\sqrt{6}})$ Then, of course: $\pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi \\ \pi - \arccos(\frac{\sqrt{6}}{3}) + 4\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi \\ ... every\,\,360°n, n \in \mathbb{Z}$ And that: $\pi + \arccos(\frac{\sqrt{6}}{3}) = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi\\ \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi\\ \pi + \arccos(\frac{\sqrt{6}}{3}) + 4\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 6\pi\\ ... \pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ n\pi\\ $ So we can say that $\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi$ differs from $-2\arctan(\sqrt{5 + 2\sqrt{6}}) + n\pi$ by just one lap ($2\pi$), otherwise they can be safely considered the same (for all integers). So how to rewrite $\arccos$ in terms of $\arctan$? Thanks for the attention!

We define $y = \arccos \left( \dfrac{\sqrt{6}}{3} \right)$. Since $0 < \dfrac{\sqrt{6}}{3} < 1$, we have $0 < y < \dfrac{\pi}{2}$, and $\dfrac{\pi}{2} < \pi - y < \pi$. Therefore, \begin{equation*} \tan \left( \frac{\pi - y}{2} \right) = \sqrt{\frac{1 - \cos (\pi - y)}{1 + \cos (\pi + y)}} = \sqrt{\frac{1 + \cos y}{1 - \cos y}} = \sqrt{\dfrac{1 + \frac{\sqrt{6}}{3}}{1 - \frac{\sqrt{6}}{3}}} = \sqrt{\frac{3 + \sqrt{6}}{3 - \sqrt{6}}} = \sqrt{5 + 2\sqrt{6}}. \end{equation*} Then we have \begin{equation*} \pi - y = 2 \arctan \left( \sqrt{5 + 2\sqrt{6}} \right), \end{equation*} that is \begin{equation*} \pi - \arccos \left( \frac{\sqrt{6}}{3} \right) = 2 \arctan \left( \sqrt{5 + 2\sqrt{6}} \right), \end{equation*} By the same token, we have $\pi < \pi + y < \dfrac{3}{2} \pi$. Therefore, \begin{equation*} \begin{split} \tan \left( \frac{\pi + y}{2} \right) &= -\sqrt{\frac{1 - \cos (\pi - y)}{1 + \cos (\pi + y)}} = -\sqrt{\frac{1 + \cos y}{1 - \cos y}} \\ &= -\sqrt{\dfrac{1 + \frac{\sqrt{6}}{3}}{1 - \frac{\sqrt{6}}{3}}} = -\sqrt{\frac{3 + \sqrt{6}}{3 - \sqrt{6}}} = -\sqrt{5 + 2\sqrt{6}}. \end{split} \end{equation*} Then we have \begin{equation*} \pi + y = 2 \arctan \left( -\sqrt{5 + 2\sqrt{6}} \right) = -2\arctan \left( \sqrt{5 + 2\sqrt{6}} \right), \end{equation*} that is \begin{equation*} \pi + \arccos \left( \frac{\sqrt{6}}{3} \right) = -2 \arctan \left( \sqrt{5 + 2\sqrt{6}} \right). \end{equation*} From this exercise, you can easily prove a general trigonometric relation between $\arccos$ and $\arctan$.

Sum of Two Continuous Random Variables Consider two independent random variables $X$ and $Y$. Let $$f_X(x) = \begin{cases} 1 − x/2, & \text{if $0\le x\le 2$} \\ 0, & \text{otherwise} \end{cases}$$.Let $$f_Y(y) = \begin{cases} 2-2y, & \text{if $0\le y\le 1$} \\ 0, & \text{otherwise} \end{cases}$$. Find the probability density function of $X + Y$. Can anyone show me a step by step solution to this problem? I've been applying this theorem to solve the problem with limited success

Let $Z=X+Y$. We know $0\leq Z\leq 3$ but for the density of $Z$ we have three cases to consider due to the ranges of $X$ and $Y$. If $0\leq z\leq 1$: \begin{eqnarray*} f_Z(z) &=& \int_{x=0}^{z} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=0}^{z} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=0}^{z} \\ &=& 2z - \dfrac{3}{2}z^2 + \dfrac{1}{6}z^3. \end{eqnarray*} If $1\lt z\leq 2$: \begin{eqnarray*} f_Z(z) &=& \int_{x=z-1}^{z} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=z-1}^{z} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=z-1}^{z} \\ &=& \dfrac{7}{6} - \dfrac{1}{2}z. \end{eqnarray*} If $2\lt z\leq 3$: \begin{eqnarray*} f_Z(z) &=& \int_{x=z-1}^{2} f_X(x)f_Y(z-x)\;dx \\ &=& \int_{x=z-1}^{2} (1-x/2)(2-2y)\;dx \\ &=& \left[ 2x-2zx+zx^2/2+x^2/2-x^3/3 \right]_{x=z-1}^{2} \\ &=& \dfrac{9}{2} - \dfrac{9}{2}z + \dfrac{3}{2}z^2 - \dfrac{1}{6}z^3. \end{eqnarray*}

Find the roots of the summed polynomial Find the roots of: $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$ I got that: $$\frac{1 - x^8}{1-x} - x^6 - x = 0$$ But that doesnt make it any easier.

$$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$ $$(x^7 + x^4) + (x^5 + x^3) + x^2 + 1 = 0$$ $$x^4(x^3 + 1) + x^3(x^2 + 1) + x^2 + 1 = 0$$ $$x^4(x^3 + 1) + (x^2 + 1)(x^3 + 1) = 0$$ $$(x^3 + 1)(x^4+x^2 + 1) = 0$$ from there we can continue with factoring $$x^3+1=x^3+1^3=(x+1)(x^2-x+1)$$ and $$x^4+x^2 + 1 =(x^2)^2+2x^2+1-x^2=(x^2+1)^2-x^2=$$ $$=(x^2+1-x)(x^2+1+x)$$

Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. The Statement of the Problem: Let $X$ have pdf $$f_X(x) = \begin{cases} \frac{1}{4} & 0<x<1 \\ \frac{3}{8} & 3<x<5 \\ 0 & \text{otherwise} \end{cases}$$ (a) Find the cumulative distribution function of $X.$ (b) Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. Hint: Consider three cases: $1/5 \le y \le 1/3, 1/3 \le y \le 1,$ and $ y \ge 1.$ Where I Am: I think I did part (a) correctly. I did the following: $$F_X(x) = \begin{cases} \frac{1}{4}x +c_1 & 0<x<1 \\ \frac{3}{8}x + c_2 & 3<x<5 \\ 0 & \text{otherwise} \end{cases}$$ $$F(0)=0=\frac{1}{4}(0)+c_1 \implies c_1 = 0 $$ $$F(5)=1=\frac{3}{8}(5)+c_2 \implies c_2 = -\frac{7}{8} $$ Therefore: $$F_X(x) = \begin{cases} \frac{1}{4}x & 0<x<1 \\ \frac{1}{4} & 1<x<3 \\ \frac{3}{8}x - \frac{7}{8} & 3<x<5 \\ 1 & x > 5 \end{cases}$$ If that's not right, however, please let me know. Now, for part (b), I got a little lost. Here's what I did: $$ \text{Let } g(x) = \frac{1}{x} \implies g'(x) = -\frac{1}{x^2} $$ Then: $$ f_Y(y)=\frac{f_X(x)}{\lvert g'(x) \rvert} = \frac{f_X(\frac{1}{y})}{\lvert g'(\frac{1}{y})\rvert} $$ Therefore: $$f_Y(y) = \begin{cases} \frac{1}{4} & 0<\frac{1}{y}<1 \\ \frac{3}{8} & 3<\frac{1}{y}<5 \\ 0 & \text{otherwise} \end{cases}$$ and taking reciprocals and flipping inequalities... $$f_Y(y) = \begin{cases} \frac{1}{4} & y \ge 1 \\ \frac{3}{8} & \frac{1}{5} \le y \le \frac{1}{3} \\ 0 & \frac{1}{3} \le y \le 1 \end{cases}$$ This, however... doesn't seem right. For example, what is $f_Y(y)$ when $y \in [0, \frac{1}{5}]$? Is it just $0$? I know I did something wrong here, but I can't quite figure out what exactly. If anybody could help me out, I'd appreciate it. EDIT: "Second" Attempt... $$F_X \left( \frac{1}{y} \right) = \begin{cases} \frac{1}{4}\left( \frac{1}{y} \right) & y \ge 1 \\ \frac{1}{4} & \frac{1}{3} \le y \le 1 \\ \frac{3}{8}\left( \frac{1}{y} \right) - \frac{7}{8} & \frac{1}{5} \le y \le \frac{1}{3} \\ 0 & \text{otherwise} \end{cases}$$ Therefore: $$ f_Y(y)=\frac{d}{dy}F_X \left( \frac{1}{y} \right)= \begin{cases} -\frac{1}{4}\left( \frac{1}{y^2} \right) & y \ge 1 \\ -\frac{3}{8}\left( \frac{1}{y^2} \right) & \frac{1}{5} \le y \le \frac{1}{3} \\ 0 & \text{otherwise} \end{cases} $$

Crude sketch of CDF of Y based on a simulation. Perhaps helpful as a check on your work.

Prove that $\sum_{i=1}^{i=n} \frac{1}{i(n+1-i)} \le1$ $$f(n)=\sum_{i=1}^{i=n} \dfrac{1}{i(n+1-i)} \le 1$$ For example, we have $f(3)=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot1}=\dfrac{11}{12}\lt 1$ If true, it can be used to prove: Proving $x\ln^2x−(x−1)^2<0$ for all $x∈(0,1)$ Also, can you prove $f(n)\ge f(n+1)$?

For second part $f(n)-f(n+1)= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)}] -\frac{1}{n+1}$ $= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -\frac{1}{n(n+1)}]$ $= \sum\limits_{i=1}^n [\frac{1}{i(n+1-i)}-\frac{1}{i(n+2-i)} -(\frac{1}{n}-\frac{1}{n+1})]$ $= \sum\limits_{i=1}^n [(\frac{1}{i(n+1-i)}-\frac{1}{n})-(\frac{1}{i(n+2-i)} -\frac{1}{(n+1)})]$ $=\sum\limits_{i=1}^n [\frac{n-in-i+i^2}{i.n(n+1-i)}-\frac{n+1-in-2i+i^2}{i(n+1)(n+2-i)}]$ $\leq\sum\limits_{i=1}^n [ \frac{n-in-i+i^2}{i(n+1)(n+2-i)}-\frac{n+1-in-2i+i^2}{i(n+1)(n+2-i)}]$ $=\sum\limits_{i=1}^n \frac{i-1}{i(n+1)(n+2-i)} \geq0$ , as $(i-1) \geq 0$ for $1 \leq i \leq n$

find the integral using Integration by partial Fractions Here is my work for this problem...just wanted a check over and see if i missed anything Original Problem: $\int$ $\frac6{x^3-3x^2}$ F 6/x^3-3x^2= F 6/x^2(x-3) 6/x^2(x-3)= Ax+B/x^2+C/(x-3) C/x^2(x-3)= (A+C)x^2+(B-3A)x-3B B=-2 A=-2/3 C=1/3 F C/x^2-3x^2 dx= F(-2/3)x-2/x^2+ F 2/3(x-3) F -2/3x-2/x^2+2/3(x-3) F 6/x^3-3x^2=-2/3 ln (x)+2/x+2/3 ln(x-3)

We have the expansion $$\frac{1}{x^3-3x^2}=\frac{A+Bx}{x^2}+\frac{C}{x-3}\tag 1$$ Multiplying both sides of $(1)$ by $x-3$ and letting $x\to 3$ reveals that $C=1/9$. Multiplying both sides of $(1)$ by $x^2$ and letting $x\to 0$ reveals that $A=-1/3$. Multiplying both sides of $(1)$ by $x^2$, taking a derivative with respect to $x$, and letting $x\to 0$ reveals that $B=-1/9$. Thus, we have $$\begin{align}\int\frac{6}{x^3-3x^2}\,dx&=\int\left(\frac{-2}{x^2}+\frac{-2/3}{x}+\frac{2/3}{x-3}\right)dx\\\\ &=\frac{2}{x}-\frac23\log |x|+\frac23 \log |x-3|+C \end{align}$$

Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$ L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$ Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.

$$\lim_{x\to 0} \frac{1}{x}-\frac{\cos(x)}{\sin(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{\tan(x)}=\lim_{x\to 0} \frac{1}{x}-\frac{1}{x+\frac{x^3}{3}+O(x^5)}=\lim_{x\to 0}\frac{1}{x}-\frac{1}{x}=0$$

Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that $$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$ (is it correct?) but I don't get how to find the upper bound.

Suppose $\dfrac{x-y}{x+y} = c$, where $c \neq -1$. Then $$\begin{eqnarray} x - y &=& c(x + y) \\ x - cx &=& y + cy \\ (1 - c)x &=& (1 + c)y \end{eqnarray}$$ Therefore $y = \dfrac{1-c}{1+c} x$. But if $\dfrac{x-y}{x+y} = -1$, then $x - y = -1(x + y)$, and from this we conclude that $x = 0$ and $y$ can be anything you want. So you can set $\dfrac{x-y}{x+y}$ to any value you like by choosing suitable values of $x$ and $y$. There are no limits on $\dfrac{x-y}{x+y}$.

Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$ Given the Fibonacci, tribonacci, and tetranacci numbers, $$F_n = 0,1,1,2,3,5,8\dots$$ $$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$ $$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$ and so on, how do we show that, $$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$ $$\sum_{n=0}^{\infty}\frac{T_n}{10^n} = \frac{100}{889}$$ $$\sum_{n=0}^{\infty}\frac{U_n}{10^n} = \frac{1000}{8889}$$ or, in general, $$\sum_{n=0}^{\infty}\frac{S_n}{p^n} = \frac{(1-p)p^{k-1}}{(2-p)p^k-1}$$ where the above were just the cases $k=2,3,4$, and $p=10$? P.S. Related post.

The difference equations given by the suggested series are: \begin{align} F_{n+2} &= F_{n+1} + F_{n} \\ T_{n+3} &= T_{n+2} + T_{n+1} + T_{n} \\ \tag{1} U_{n+4} &= U_{n+3} + U_{n+2} + U_{n+1} + U_{n} \end{align} and so on. In general they take on the form \begin{align}\tag{2} \phi_{n+m} = \sum_{k=0}^{m-1} \phi_{n+m-k-1}, \end{align} where $\phi_{0}, \phi_{1}, \phi_{2}, \cdots $ are the initial values. By considering the generating function defined by \begin{align} f_{m}(t) = \sum_{n=0}^{\infty} \phi_{n+m} \, t^{n} \end{align} then it is readily found that \begin{align} f_{m}(t) &= \frac{1}{ 2 - \sum_{k=0}^{m} t^{k}} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \\ &= \frac{1 - t}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \tag{3} \end{align} if $t \to 1/t$ then \begin{align} f_{m}(t) &= \frac{t^{m} (t - 1)}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, \frac{1}{t^{k}} \right] \end{align} When $t = 10$ this reduces to \begin{align}\tag{4} f_{m}\left(\frac{1}{10}\right) = \frac{9}{(10)^{m+1}- 2 \, (10)^{m} + 1} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, (10)^{m-k} \right] \end{align} As an example let $m=3$, which corresponds to the Tribonacci series, to obtain \begin{align} f_{3}\left(\frac{1}{10}\right) &= \sum_{n=0}^{\infty} \frac{T_{n}}{(10)^{n}} = \frac{9 \, (10)^{3}}{10^{4} - 2 \cdot 10^{3} + 1} \cdot \left(\frac{1}{10}\right) = \frac{100}{889}. \end{align}

The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$ My try:I have just normally squared the expression and got $u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$ $u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$ I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?

Write $$\cos^2x=\frac{1+\cos 2x}{2}$$ and $$\sin^2x=\frac{1-\cos 2x}{2}$$ Then, we have $$u=\sqrt{A+B\cos 2x}+\sqrt{A-B\cos 2x}\tag 1$$ where $$A=\frac{a^2+b^2}{2}$$ $$B=\frac{a^2-b^2}{2}$$ Taking the derivative of u in $(1)$ and setting the derivative equal to zero reveals $$\frac{-B\sin 2x}{\sqrt{A+B\cos 2x}}+\frac{B\sin 2x}{\sqrt{A-B\cos 2x}}=0$$ whereupon solving reveals that either $\sin 2x=0$ or $\cos 2x=0$. When $\cos 2x=0$, $$\bbox[5px,border:2px solid #C0A000]{u=\sqrt{2(a^2+b^2)} \,\,\text{is the maximum}}$$ and when $\sin 2x =0$, $$\bbox[5px,border:2px solid #C0A000]{u=|a|+|b|\,\,\,\text{is the minimum}}$$

Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$ Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$ I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this. Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!

Since $\cos(2x)=2\cos^2(x)-1$, we have $$\cos^2(x)=\frac {\cos(2x)+1}2$$ Therefore, $$\cos^2(\theta)=\frac {\cos(2\theta)+1}{2}$$ $$\cos^2(\theta+120)=\frac{\cos(2\theta+240^\circ)+1}{2}$$ $$\cos^2(\theta-120)=\frac{\cos(2\theta-240^\circ)+1}{2}$$ So the original equation become: $$\frac{\cos(2\theta)+1+cos(2\theta+240^\circ)+1+\cos(2\theta-240^\circ)+1}2.$$ Using the sum formula, we get $$\frac{3+\cos(2\theta)-\frac12\cos(2\theta)-\sin(2\theta)\sin(240^\circ)-\frac12\cos(2\theta)+\sin(2\theta)\sin(240^\circ)}2$$ $$\Longrightarrow \frac{3+\cos(2\theta)-\cos(2\theta)}2\Longrightarrow\frac32.$$ Hope this is helpful.

Nature of the roots of quadratic equation Here is the problem that I need to prove: If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$ Here is what I did: \begin{align*} p(2x-1)=3(x^2+1) \\ 3x^2 - 2px + (p+3)=0 \\ b^2 - 4ac = 4(p^2-3(p+3)) \end{align*} By inspection I can see that $p^2 > -3(p+3)$ for almost all values of $p \ $, therefore $p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$ If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore $\ p^2-3(p+3) \geq 0$ Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$. Thank you

You have $$3x^2 - 2px + (p+3) = 0.$$ Given that $x$ is real the quadratic needs to have a discriminant $\Delta \ge 0$. So $$\Delta = 4p^2 - 12(p+3) \geq 0.$$ Dividing by $4$ yields $$\bbox[10px, border: blue 1px solid]{p^2 - 3(p+3) \ge 0.}$$ as required.

How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question: $ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $ I have partially solved this:- $$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$ $$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$ $$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$ $$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$ $$ \sin(18^\circ) - \cos(36^\circ) $$ At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.

$\sin(5\cdot78^\circ)=\sin(360^\circ+30^\circ)=\sin30^\circ$ $\sin\{5(-66^\circ)\}=\sin(-360^\circ+30^\circ)=\sin30^\circ$ If $\sin5x=\sin30^\circ\implies5x=n180^\circ+(-1)^n30^\circ$ where $n$ is any integer $\implies x=n72^\circ+6^\circ$ where $n\equiv-2,-1,0,1,2\pmod5$ Again, $\sin5x=16\sin^5x-20\sin^3x+5\sin x$ So, the roots of $\displaystyle16\sin^5x-20\sin^3x+5\sin x=\dfrac12$ are $\sin\left(n72^\circ+6^\circ\right)$ where $n\equiv-2,-1,0,1,2\pmod5$ Using Vieta's formula, $\displaystyle\sum_{n=-2}^2\sin\left(n72^\circ+6^\circ\right)=0$ $n=-2\implies$ $-2\cdot72^\circ+6^\circ=-138^\circ\implies\sin(-138^\circ)=-\sin(138^\circ)=-\sin(180^\circ-42^\circ)=-\sin42^\circ$ $n=-1\implies$ $-1\cdot72^\circ+6^\circ=-66^\circ\implies\sin(-66^\circ)=-\sin66^\circ$ $n=0\implies ?$ $n=1\implies ?$ and $n=2\implies\sin\left(2\cdot72^\circ+6^\circ\right)=\sin150^\circ=\dfrac12$ Do you see the destination?

Minor mistake computing $\int \frac{1}{x^3+2x^2-3x} \; dx$? I'm trying to compute: $$\int \frac{1}{x^3+2x^2-3x} \; dx$$ Until now, I did the following: Factoring: $$x^3+2x^2-3x=x(x-1)(x+3)$$ To obtain the parcial fractions: $$\frac{a}{x}+\frac{b}{x+3}+\frac{c}{x-1}=\frac{1}{x(x-1)(x+3)}$$ $$a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$$ $$-3 a + 2 a x - b x + 3 c x + a x^2 + b x^2 + c x^2 = 1$$ This give me the system: $$\begin{eqnarray*} {a+b+c}&=&{0} \\ {2a-b-3c}&=&{0} \\ {-3a}&=&{1} \end{eqnarray*}$$ Solving it yields: $$a= -\cfrac{1}{3}\quad b=\cfrac{5}{6} \quad c=-\cfrac{1}{3}$$ And then: $$\int \frac{-1/3}{x}+\frac{5/6}{x+3}+\frac{-1/2}{x-1} \; dx$$ $$\int \frac{-1/3}{x} \; dx+ \int \frac{5/6}{x+3}dx+ \int \frac{-1/2}{x-1} \; dx + C$$ $$\frac{-1}{3}\int \frac{1}{x} \; dx+ \frac{5}{6} \int \frac{1}{x+3}dx+ \frac{-1}{2} \int \frac{1}{x-1} \; dx + C$$ Which should be: $$\frac{-1}{3} \log(x)+ \frac{5}{6} \log(x+3)+ \frac{-1}{2} \log(x-1)+ C$$ But Mathematics gives me: $$\frac{1}{4} \log (1-x)-\frac{\log (x)}{3}+\frac{1}{12} \log (x+3)$$ But I have no idea of what I did wrong.

It should be $2a-b+3c=0$, not $2a-b-3c=0$. As a side note, if you want to solve $a(x-1)(x+3)+bx(x-1)+cx(x+3)=1$, then it's a lot easier to plug in $x=0, 1, -3$ instead of expanding.

Find the $\int \frac{(1-y^2)}{(1+y^2)}dy$ $\int \frac{(1-y^2)}{(1+y^2)}dy$ first I tried to divide then I got 1-$\frac{2y^2}{1+y^2}$ and i still can't integrate it.

Hint: $$ \int \frac{1-y^2}{1+y^2}dy = \int \frac{1}{1+y^2}dy-\int \frac{y^2}{1+y^2}dy$$ Note (using long division or otherwise): $$\int \frac{y^2}{1+y^2}dy = \int dy -\int \frac{1}{1+y^2}dy$$ Therefore:$$ \int \frac{1-y^2}{1+y^2}dy = 2\int\frac{1}{1+y^2}dy -\int dy$$ The solution should be straight forward from here.

Range of an inverse trigonometric function Find the range of $f(x)=\arccos\sqrt {x^2+3x+1}+\arccos\sqrt {x^2+3x}$ My attempt is:I first found domain, $x^2+3x\geq0$ $x\leq-3$ or $x\geq0$...........(1) $x^2+3x+1\geq0$ $x\leq\frac{-3-\sqrt5}{2}$ or $x\geq \frac{-3+\sqrt5}{2}$...........(2) From (1) and (2), domain is $x\leq-3$ or $x\geq0$ but could not solve further..Any help will be greatly appreciated.

You do not have the correct domain. We must also have $-1\leq\sqrt{x^2+3x+1}\leq1$ and $-1\leq\sqrt{x^2+3x}\leq1$. In other words, $\sqrt{x^2+3x+1}\leq1$ and $\sqrt{x^2+3x}\leq1$, since they are positive. Thus $x^2+3x+1\leq1$ (squaring is allowed since both are positive), or $x^2+3x\leq0$, this gives $-3 \leq x \leq 0$. Together with $x \leq -3$ or $x \geq 0$, we get that the only numbers that give a well defined value are $x=0$ or $x=-3$. This gives $\arccos(\sqrt{1})+\arccos(\sqrt{0})$ in both cases, so the range is $\frac{1}{2}\pi$.

Finding $P(X < Y)$ where $X$ and $Y$ are independent uniform random variables Suppose $X$ and $Y$ are two independent uniform variables in the intervals $(0,2)$ and $(1,3)$ respectively. I need to find $P(X < Y)$. I've tried in this way: $$ \begin{eqnarray} P(X < Y) &=& \int_1^3 \left\{\int_0^y f_X(x) dx\right\}g_Y(y) dy\\ &=& \frac{1}{4} \int_1^3 \int_0^y dx dy\\ &=& \frac{1}{4} \int_1^3 y dy\\ &=& \frac{1}{8} [y^2]_1^3\\ &=& 1 \end{eqnarray} $$ But I'm suspicious about this result. It implies that $X<Y$ is a sure event, which is not at all true.

Answer: Divide the regions of X with respect to Y for the condition $X<Y$. For $0<X<1$, $P(X<Y) = \frac{1}{2}$ For $1<X<2$ and $1<Y<2$ $P(X<Y) = \int_{1}^{2}\int_{x}^{2} \frac{1}{2}\frac{1}{2}dydx = \frac{1}{8}$ For $1<X<2$, and $2<Y<3$ $P(X<Y) = \frac{1}{2}.\frac{1}{2}=\frac{1}{4}$ Thus $P(X<Y) = \frac{1}{2}+ \frac{1}{8}+\frac{1}{4} = \frac{7}{8}$

remainder of $a^2+3a+4$ divided by 7 If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7 (A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$ if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$ if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod 7$ thus, we can say that any number, $a$ that divided by 7 has remainder of 6, the remainder of $a^2 + 3a + 4$ is 2. is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)

If the remainder when $a$ divided by $7$ is $b$, then $a = 7n+b$ for some integer $n$. Hence, $a^2+3a+4 = (7n+b)^2+3(7n+b)+4$ $= 49n^2 + 14nb + b^2 + 21n + 3b + 4$ $= 7(7n^2+2nb+3n) + (b^2+3b+4)$. So, the remainder when $a^2+3a+4$ is divided by $7$ will be the same as the remainder when $b^2+3b+4$ is divided by $7$. For the specific case when $b = 6$, we get that $a^2+3a+4 = 7(7n^2+12n+3n)+58$ $= 7(7n^2+12n+3n+8)+2$. So the remainder when $a^2+3a+4$ is divided by $7$ is $2$.

Is the inequality $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ true? I'm having some trouble deciding whether this inequality is true or not... $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$

$|x - y|^2 = (x - y)^2 $ Let $\sqrt[3]{x} = a , \sqrt[3]{y} = b$ I will study this $$| a^2 - b^2 | \leq \sqrt[3]{|a^3-b^3|^2} $$ L.H.S $$|a^2 - b^2|^3 = |a-b|^3 \cdot |a+b|^3= \color{red}{|a-b|^2}\cdot |a^2-b^2|\cdot |a^2 +2ab+b^2|$$ R.H.S $$|a^3 - b^3|^2 = \color{red}{|a-b|^2} \cdot |a^2 +ab + b^2|^2 $$ So our problem reduced into studying if $$|a^2-b^2|\cdot |a^2 +2ab+b^2| \leq |a^2 +ab + b^2|^2$$ If $a,b >0$. Then L.H.S $$\color{red}{a^4 +2a^3b} - 2ab^3 -b^4$$ R.H.S $$\color{red}{a^4 +2a^3b} + 3a^2b^2 +2ab^3 + b^4$$ which is absolutely bigger than L.H.S Hope it will help you..

Evaluate $\iint_{R}(x^2+y^2)dxdy$ $$\iint_{R}(x^2+y^2)dxdy$$ $$0\leq r\leq 2 \,\, ,\frac{\pi}{4}\leq \theta\leq\frac{3\pi}{4}$$ My attempt : Jacobian=r $$=\iint_{R}(x^2+y^2)dxdy$$ $$x:=r\cos \theta \,\,\,,y:=r\cos \theta$$ $$\sqrt{x^2+y^2}=r$$ $$\int_{\theta=\pi/4}^{\theta=3\pi/4}\bigg[\int_{r=0}^{r=2}\bigg(r^2\bigg)dr\bigg]d\theta$$ $$.....=\boxed{\frac{4\pi}{3}}$$ Is it correct? >

Switching to polar coordinates, the Jacobian is given by $ |J|$ where $$ J = \dfrac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \dfrac{\partial x}{\partial r} & \dfrac{\partial y}{\partial r} \\ \dfrac{\partial x}{\partial \theta} & \dfrac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix} = r$$ Therefore, your double integral is given by $$ \begin{aligned} \iint_{R} \left( x^2 + y^2 \right) \text{ d}x \text{ d}y & = \int_{\pi/4}^{3\pi/4} \int_{0}^{2} \left( (r\cos\theta)^2 + (r\sin\theta)^2 \right) |J| \text{ d}r \text{ d}\theta \\ & = \int_{\pi/4}^{3\pi/4} \int_{0}^{2} r^2 |r| \text{ d}r \text{ d}\theta \end{aligned}$$ and since $r \in \left[0,2\right]$, $|r| = +r$ so the integrand is $r^3$. I leave the rest to you.

Solve the equation $4\sqrt{2-x^2}=-x^3-x^2+3x+3$ Solve the equation in $\Bbb R$: $$4\sqrt{2-x^2}=-x^3-x^2+3x+3$$ Is there a unique solution $x=1$? I have trouble when I try to prove it. I really appreciate if some one can help me. Thanks!

It can be seen that $4 \sqrt{2 - x^2} = (1+x)(3 - x^2)$ leads to the expanded form, after squaring both sides and combining terms, $$ x^6 + 2 x^5 - 5 x^4 - 12 x^3 + 19 x^2 + 18 x - 23 = 0 .$$ It is readily identified that $x=1$ is a solution and can then be factored out leading to $$ (x-1)( x^5 + 3 x^4 - 2 x^3 - 14 x^2 + 5x + 23) = 0.$$ The polynomial of order $5$ has one real solution and $4$ complex solutions.

Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).

Partial fractions: $$\frac{1}{n(n+1)^2(n+2)} = \frac{1}{2} \left( \frac{1}{n} - \frac{1}{n+2} \right) - \frac{1}{(n+1)^2},$$ the first part being telescoping, and the last part being related to $\zeta(2)$.

Limit behavior of a definite integral that depends on a parameter. Let $A>0$ and $0\le \mu \le 2$. Consider a following integral. \begin{equation} {\mathcal I}(A,\mu) := \int\limits_0^\infty e^{-(k A)^\mu} \cdot \frac{\cos(k)-1}{k} dk \end{equation} By substituting for $k A$ and then by expanding the cosine in a Taylor series about zero I have shown that: \begin{equation} {\mathcal I}(A,\mu) = \frac{1}{\mu} \sum\limits_{n=1}^\infty \frac{(1/A)^n}{n!} \cos\left(\frac{\pi}{2} n\right) \cdot \Gamma\left(\frac{n}{\mu}\right) \end{equation} Unfortunately the series on the right hand side above does not converge for small values of $A$. My question is therefore how do we find the small-$A$ behavior of ${\mathcal I}(A,\mu)$ ?

Here we provide an answer for $\mu=2q/p$ where $p$ is a positive integer and $p+1\le 2q \le 2 p$. By using the multiplication theorem for the Gamma function we have shown that: \begin{eqnarray} &&{\mathcal I}(A,\mu) =\\ &&\frac{1}{\mu} \sqrt{\frac{(2\pi)^{2q-p}}{p 2q}} \sum\limits_{r=1}^q \left(\frac{p^{p/q}}{(2q)^2} \frac{(-1)}{A^2}\right)^r \cdot \frac{\prod\limits_{k=0}^{p-1} \Gamma(\frac{r}{q}+\frac{k}{p})}{\prod\limits_{k=0}^{2q-1} \Gamma(\frac{2r+1}{2q} + \frac{k}{2q})} \cdot F_{p+1,2q} \left[\begin{array}{rr} 1 & \left\{\frac{r}{q}+\frac{k}{p} \right\}_{k=0}^{p-1} \\ & \left\{\frac{2r+k}{2q}\right\}_{k=1}^{2q} \end{array}; \left(\frac{p^{p/q}}{(2q)^2} \frac{(-1)}{A^2}\right)^q\right] \end{eqnarray} Now we use the asymptotic behaviour of hypergeometric functions, see for example Wolfram's site. To make things simpler we plug the expression above into Mathematica and expand it using the Series[] command. The result is the following: \begin{eqnarray} {\mathcal I}(A,\frac{2}{1}) &=& \log(A) - \frac{1}{2} \gamma + A^2 + O(A^4)\\ {\mathcal I}(A,\frac{3}{2}) &=& \log(A) - \frac{1}{3} \gamma + A^{3/2} \frac{1}{2} \sqrt{\frac{\pi}{2}} + O\left((A^{3/2})^2\right) \\ {\mathcal I}(A,\frac{4}{3}) &=& \log(A) - \frac{1}{4} \gamma + A^{4/3} \frac{9 \sqrt{3} \pi }{280 \Gamma \left(-\frac{10}{3}\right)} + O\left((A^{4/3})^2\right) \end{eqnarray} As it seems the function in question is a linear combination of $\log(A)$ and some function that is analytic in $A^\mu$. It would be nice to find the full series expansion of that later function. Since using Mathematica is cheating here we explicitely treat the case of $\mu=2$. In this case we have: \begin{eqnarray} {\mathcal I}(A,2) &=& -\frac{\, _2F_2\left(1,1;\frac{3}{2},2;-\frac{1}{4 A^2}\right)}{4 A^2} \\ &=& -\frac{1}{8 A^2} \int\limits_0^1 \int\limits_0^1 (1-t)^{-1/2} \exp\left(-\frac{t \cdot t_1}{4 A^2}\right) dt dt_1 \\ &=&-\frac{1}{2} \int\limits_0^{\frac{1}{4 A^2}} \left(1 - 4 A^2 t\right)^{-1/2} \left(\frac{1- e^{-t}}{t}\right) dt \end{eqnarray} Now we expand the first factor in the integrand in a Taylor series. Therefore we have: \begin{eqnarray} -2 {\mathcal I}(A,2) &=& \log \left(\frac{1}{4 A^2}\right)+\Gamma \left(0,\frac{1}{4 A^2}\right)+\gamma +\\ &&\sum\limits_{n=1}^\infty \binom{-1/2}{n} (-1)^n \left[\frac{1}{n} - (4 A^2)^n \gamma(n,\frac{1}{4 A^2})\right] \end{eqnarray} Now, since \begin{equation} \sum\limits_{n=1}^\infty \binom{-1/2}{n} \frac{(-1)^n}{n} = \log(4) \end{equation} we are getting the final answer: \begin{eqnarray} {\mathcal I}(A,2) = \log(A) - \frac{\gamma}{2} - \frac{1}{2} \Gamma(0,\frac{1}{4 A^2}) - \frac{1}{2} \sum\limits_{n=1}^\infty \binom{-1/2}{n} (-1)^n (4 A^2)^n \gamma(n,\frac{1}{4 A^2}) \end{eqnarray} Note that our conjecture was not quite true. Indeed our function (at least in the case $\mu=2$) is a linear combination of a natural logarithm, and a function that has a nontrivial Laurant expansion about the origin(ie has all possible both positive and negative powers in the expansion).

A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?

Notice, we have $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\tag 1$$ let $$\sqrt{x^2+5} + \sqrt{x^2-3} = y\tag 2$$ Now, multiplying both (1) & (2), we get $$(\sqrt{x^2+5} - \sqrt{x^2-3} )(\sqrt{x^2+5} + \sqrt{x^2-3})=2y$$ $$(\sqrt{x^2+5})^2-(\sqrt{x^2-3})^2=2y$$ $$x^2+5-(x^2-3)=2y$$ $$8=2y\implies y=4$$ Hence, $$\bbox[5px, border:2px solid #C0A000]{\sqrt{x^2+5} + \sqrt{x^2-3} = 4}$$

find the complex number $z^4$ Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$. Find $c+d$. I got that the distance is: $$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$ So I need to maximize the distance between those two points. $|z^2| = 25$ means since: $z^2 = a^2 - b^2 + 2abi$ that: $625 = (a^2 - b^2)^2 + 4a^2b^2 = a^4 + b^4 + 2a^2b^2$ But that doesnt help much.

Let $\theta$ denote the argument of $z$, then the arguments of $(1+2i)z^3$ and $z^5$ are respectively $3\theta +\arctan(2)$ and $5\theta$. Now imagine that these points lie on concentric circles, so in order to maximize the distance, the difference between the arguments must be $\pi$ (or $-\pi$ it won't matter): $$5\theta - 3\theta -\arctan(2)=\pi$$ $$\tan(5\theta - 3\theta -\arctan(2))=0$$ $$\tan(2\theta -\arctan(2))=0=\frac{\tan(2\theta)-2}{1+2\tan(2\theta)}$$ $$\tan(2\theta)=2$$ $$\theta=\frac{1}{2}\arctan(2)\tag{A}$$ $$\tan(\theta)=\frac{1}{2}(\sqrt{5}-1)=\frac{b}{a}$$ $$b^2=\frac{1}{2}(3-\sqrt{5})a^2$$ $$a^2+b^2=25$$ $$a^2=\frac{25+5\sqrt{5}}{2}$$ $$b^2=\frac{25-5\sqrt{5}}{2}$$ ALSO, to get c and d directly, multiply both sides of $(\text{A})$ by 4: $$4\theta=2\arctan(2)$$ $$tan(4\theta)=-\frac{4}{3}=\frac{d}{c}$$ and we have that $$\sqrt{c^2+d^2}=5^4$$ Finally, solve for $c$ and $d$.

Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is (A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$ (B)$\frac{1}{2}(x^2+y^2+z^2)$ (C)$\frac{1}{2}(xy+yz+zx)$ (D)$\frac{1}{2}(x+y+z)\sqrt{x^2+y^2+z^2}$ I tried applying Heron's formula but calculations are very messy and simplification is difficult.I could not think of any other method to find this area.Can someone assist me in solving this problem.

hint: Let $a = \sqrt{x^2+y^2}, b = \sqrt{y^2+z^2}, c = \sqrt{z^2+x^2} \to a^2 = x^2+y^2, b^2 = y^2+z^2, c^2= z^2+x^2 $. Use this and Cosine Law to find $\cos^2 A$, then $\sin^2 A$, and use $S^2 = \dfrac{b^2c^2\sin^2 A}{4}$, to find $S^2$ and then take square-root to get back $S$.