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Use stirlings approximation to prove inequality. I have come across this statement in a text on finite elements. I can give you the reference if that will be useful. The text mentions that the inequality follows from Stirling's formula. I can't prove it to myself but I think it is true from checking values with Mathematica. Let $p,k\in \mathbb{N}$ with $k\le p$. Then we have $$\frac{(p-k)!}{(p+k)!} \le \left(\frac{\theta}{p}\right)^{2k},\, \theta = \left(\frac{e}{2}\right)^{k/p}.$$ Does anyone have an idea of how to get this inequality from Stirling's approximation? This is my attempt at a solution. To build intuition, let's look at the case $k=p$ we have $$ \begin{align*} \frac{(p-k)!}{(p+k)!} &= \frac{1}{(2p)!} \\ &\le \frac{1}{\sqrt{2\pi}} \frac{\exp(2p)}{(2p)^{2p+1/2} } \\ &= \frac{1}{\sqrt{2\pi}} \left(\frac{e}{2} \right)^{2p} \frac{1}{\sqrt{2}} \frac{1}{p^{2p}} \frac{1}{p^{1/2}}\\ & \le \frac{1}{\sqrt{2\pi}} \left( \frac{e}{2} \right)^{2p} \frac{1}{\sqrt{2}} \frac{1}{p^{2p}} \\ &= \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \left( \frac{\theta}{p} \right)^{2k}\\ &\le \left( \frac{\theta}{p} \right)^{2k} \end{align*} $$ And this is what we wanted to show. Now let's look at the case $k=p-1$, then we get $$ \begin{align*} \frac{(p-k)!}{(p+k)!} &= \frac{1}{(2p-1)!} \\ &\le \frac{1}{\sqrt{2\pi}} \frac{\exp( 2p-1)}{ (2p-1)^{2p-1/2}} \\ &= \frac{1}{\sqrt{2\pi}} \frac{\exp(2p-1)}{2^{2p-1/2}} \frac{1} {(p-1/2)^{2p-1/2} }\\ & = \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \left(\frac{e}{2} \right)^{2p-1} \frac{1}{(p-1/2)^{2p-1/2}}\\ &\le \frac{1}{2\sqrt{\pi}} \left( \frac{e}{2} \right)^{2p-1} \frac{1}{(p-1/2)^{2p-1}} \\ &\le ?? \end{align*}$$ This is where I am stuck. For the $k=p-1$ case, I can multiply and divide by $(\frac{e}{2})^{3-2/p}$ to get something of the correct form, but then it seems I would have to to prove that for all $p$ $$\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}} \le 1.$$ This seems like a dead end to me. And this is still just a special case of the general result. Any ideas?

For $\left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}} \le 1 $, $\begin{array}\\ (p-1/2)^{2p-1} &=p^{2p-1}(1-1/(2p))^{2p-1}\\ &\approx p^{2p-1}(1/e)(1-1/(2p))^{-1} \quad\text{since }(1-1/(2p))^{2p} \approx 1/e\\ \end{array} $ so $\begin{align*}\\ \left(\frac{e}{2}\right)^{3-2/p} \frac{1}{(p-1/2)^{2p-1}} &\approx \left(\frac{e}{2}\right)^{3-2/p}\frac1{p^{2p-1}}e(1-1/(2p))\\ &< \left(\frac{e}{2}\right)^{3-2/p}\frac1{p^{2p-1}}e\\ &< \left(\frac{e}{2}\right)^{3}\frac1{p^{2p-1}}e\\ &= \frac{e^4}{8p^{2p-1}}\\ &\ll 1 \quad\text{for } p > 5 \end{align*} $ So this inequality is very strongly true.

Math Subject GRE 1268 Question 55 If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$.

The integral being considered is, and is evaluated as, the following. \begin{align} I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\ &= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\ &= \left( \frac{1}{b} - \frac{1}{a} \right) \, \int_{1}^{\infty} \frac{dt}{t(1+t)} \mbox{ where $t = e^{bx}$ in the first and $t = e^{ax}$ in the second integral } \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \, \int_{1}^{p} \left( \frac{1}{t} - \frac{1}{1+t} \right) \, dt \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \, \left[ \ln(t) - \ln(1+t) \right]_{1}^{p} \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \left[ \ln\left( \frac{p}{1 + p}\right) + \ln 2 \right] \\ &= \left( \frac{1}{b} - \frac{1}{a}\right) \, \lim_{p \to \infty} \left[ \ln\left( \frac{1}{1 + \frac{1}{p}}\right) + \ln 2 \right] \\ &= \left( \frac{1}{b} - \frac{1}{a} \right) \, \ln 2 \end{align} Note: Originally the statement "This is valid if $a \neq b$" was given at the end of the solution. Upon reflection it is believed that the statement should have been "This is valid for $a,b \neq 0$".

$z^3=w^3 \implies z=w$? I've reached this in another problem I have to solve: $z,w \in \Bbb {C}$. $z^3=w^3 \implies z=w$? I've scratched my head quite a bit, but I completely forgot how to do this, I don't know if this is correct: $$ z^3=|z^3|e^{3ix}=|w^3|e^{3iy} $$ I know the absolute values are equal, so I get $3ix=3iy \implies x=y \implies z=w$. I'm not sure I solved this correctly, I know this is pretty basic, but I haven't done this stuff in a year...

We assume $z \ne 0 \ne w$; it is obvious that in the other circumstance $z = w = 0$ is the only possible solution. Now, since $z^3 = w^3 \Leftrightarrow z^3 - w^3 = 0, \tag{1}$ then $(z - w)(z^2 + zw + w^2) = z^3 - w^3 = 0; \tag{2}$ we see that if $z^2 + zw + w^2 \ne 0, \tag{3}$ then $z = w; \tag{4}$ furthermore, if (4) binds then $z^2 + zw + w^2 = 3z^2 \ne 0; \tag{5}$ thus $z = w \Leftrightarrow z^2 + zw + w^2 \ne 0; \tag{6}$ certainly $z = w$ is a possible solution to (1); (6) indicates that the any other prospects are to be found via the logically equivalent $z \ne w \Leftrightarrow z^2 + zw + w^2 = 0; \tag{7}$ we therefore scrutinize $z^2 + zw + w^2 = 0 \tag{8}$ for solutions of (1) other than $z = w$. Perhaps the most straightforward and clearest way to proceed is to exploit the assumption $w \ne 0$ and divide (8) through by $w^2$: $(\dfrac{z}{w})^2 + \dfrac{z}{w} + 1 = 0; \tag{9}$ if we now set $\omega = \dfrac{z}{w}, \tag{10}$ we find $\omega^2 + \omega + 1 = 0. \tag{11}$ (11) is readily recognized as the equation for the non-real cube roots of unity; indeed we have, taking for the moment $z = \omega$ and $w = 1$ in (2), $(\omega - 1)(\omega^2 + \omega + 1) = \omega^3 - 1 = 0, \tag{12}$ which shows that if $\omega^3 = 1, \;\; \omega \ne 1, \tag{13}$ then $\omega$ satisfies (11); it is easily seen from the quadratic formula that the two such $\omega$ are $\omega = -\dfrac{1}{2} \pm \dfrac{i}{2}\sqrt{3}; \tag{14}$ we recognize that $\cos \dfrac{2\pi}{3} = \cos -\dfrac{2\pi}{3} = -\dfrac{1}{2},$ $\sin \dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}; \tag{15}$ therefore, if $\omega \ne 1$ we have $\omega = \cos \dfrac{2\pi}{3} \pm i \sin \dfrac{2\pi}{3}$ $= e^{\pm 2\pi i /3} = e^{2\pi i / 3}, e^{4\pi i / 3}; \tag{16}$ now since $\dfrac{z}{w} = \omega = e^{\pm 2\pi i / 3}, \tag{17}$ it follows that, other than $z = w$, the solutions to (1) are $z = e^{\pm 2 \pi i /3} w. \tag{18}$ (18) is easily verified: $z^3 = (e^{\pm 2 \pi i / 3}w)^3 = w^3(e^{\pm 2\pi i / 3})^3$ $= w^3 e^{\pm 2 \pi i} = w^3, \tag{19}$ since $e^{\pm 2 \pi i} = 1$. We thus see that (1) entails precisely three possibilities: $z = w, z = e^{2\pi i /3} w, z = e^{- 2 \pi i /3} = e^{4 \pi i /3} w; \tag{20}$ it's really all about equations (11)-(12); and of course, there are the nearly "self-evident truths" of the generalization of these results to the case $z^n = w^n$ for $n \in \Bbb Z$.

Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit. Derivative of numerator in function is $$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$ and derivative of denominator is $$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$ Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit? Limit should be $L=4$

using Bernoulli $$x \to 0 \\ {\color{Red}{(1+ax)^n \approx 1+anx} } \\\sqrt{1-x^2} = (1-x^2)^{\frac{1}{2}} \approx 1-\frac{1}{2}x^2 \\ \sqrt{4-x^2}=\sqrt{4(1-\frac{x^2}{4}})=2(1-\frac{x^2}{4})^{\frac{1}{2}} \approx 2(1-\frac{1}{2} \frac{x^2}{4})=2-\frac{x^2}{4} $$ so by putting them in limit : $$\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} (1-(1-\frac{1}{2}x^2))}{\sqrt{1-x}(2-(2-\frac{x^2}{4}))} =\\ \lim_{x \to 0} \frac{\sqrt{2(2-x)} (\frac{1}{2}x^2)}{\sqrt{1-x}(\frac{x^2}{4})}=\\\lim_{x \to 0} \frac{\sqrt{2(2-x)} 2}{\sqrt{1-x}}=4 $$

The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ The range of the function $f:\mathbb{R}\to \mathbb{R}$ given by $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}$ contains $N$ integers. Find the value of $10N$. I tried to find the minimum and maximum value of the function.First i simplified the function. $f(x)=\frac{3+2 \sin x}{\sqrt{1+ \cos x}+\sqrt{1- \cos x}}=\frac{1+4\sin^2\left(\frac{x}{2}+\frac{\pi}{4}\right)}{2\sin \left(\frac{x}{2}+\frac{\pi}{4}\right)}$ Then i differentiated the function and equate it to zero to get the critical points. Critical point equations are $\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)=0$ $\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{1}{2},\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)=\frac{-1}{2}$ When i checked plotted the function on desmos.com graphing calculator,i found minimum value to be $0.5$ and maximum value to be $2.5$. Which i cannot get by my critical points.Where have i gone wrong?Please help me.

Put $\sqrt{1+\cos x}$ +$\sqrt{1-\cos x} = A$ $A^2 = 2\pm 2 \sin x ,\quad A^2 - 2 =\pm 2 \sin x$ $ -2\leq A^2 - 2\leq 2,\quad -2\leq A\leq2$ So $f(x) = \frac{5 - A^2}{A}$ or $\frac{A^2 + 1}{A}$ Find the minimum and maximum of $f(x)$ in the two conditions with $-2\leq A\leq 2$

How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$? How do you calculate $\lim_{z\to0} \frac{\bar{z}^2}{z}$? I tried $$\lim_{z\to0} \frac{\bar{z}^2}{z}=\lim_{\overset{x\to0}{y\to0}}\frac{(x-iy)^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}=\lim_{\overset{x\to0}{y\to0}}\frac{x^2-2xyi-y^2}{x+iy}\cdot\frac{x-iy}{x-iy} \\ \\ =\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$ And that I could not get out, can anyone help me?

From your calculation : $$=\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$ $$=\lim_{(x,y)\to (0,0)}\frac{x^3-3xy^2}{x^2+y^2}-i\lim_{(x,y)\to (0,0)}\frac{3x^2y-y^3}{x^2+y^2}$$ From here, show that both the limits are zero by changing polar form , $x=r\cos \theta$ , $y=r\sin \theta$. For the first limit, $$\left|\frac{r^3\cos^3\theta-3r^3\cos \theta\sin^2\theta}{r^2}\right|\le 4r<\epsilon$$whenever, $r^2<\epsilon^2/16$ i,e, whenever $|x|<\epsilon/\sqrt 8=\delta(say)$ , and $|y|<\epsilon/\sqrt 8=\delta(say)$. Similarly the second limit will be zero and hence the given limit will be $0$.

Prove $\begin{pmatrix} a&b\\ 2a&2b\\ \end{pmatrix} \begin{pmatrix} x\\y\\ \end{pmatrix}=\begin{pmatrix} c\\2c \\ \end{pmatrix}$ Prove that $$ \begin{pmatrix} a & b \\ 2a & 2b \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} c \\ 2c \\ \end{pmatrix} $$ has solution $$ \begin{pmatrix} x \\ y \\ \end{pmatrix} = \begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix} +t \begin{pmatrix} -b \\ a \\ \end{pmatrix} $$ and $t$ is real number. If I plug some numbers, all $y$ variables are vanish and the matrix has infinitely many solutions, but I have no idea how the solution could be into like that. Also I haven't reached Vector chapter. Any help is appreciated!

The fast approach, as I see it is (in case $a \neq 0$): you have $ax+by = c$ , just plug in $y=0+ta$, and you get $$ax + bta = c$$ $$x= \frac ca - bt$$. alternatively, in case you don't have the solution in advance, use same approach and you will get: $$ x = \frac ca - \frac ba y$$, so the general solution will be $$(x,y) = (\frac ca - \frac bat , t)$$ $$ (x,y) = (\frac ca, 0) + (-\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result.

Prove identity: $\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}$ Prove identity: $$\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}=\tan\frac{\alpha}{2}.$$ My work this far: we take the left side $$\dfrac{1+\sqrt{\frac{1-\cos2\alpha}{2}}-\sqrt{\frac{1+\cos2\alpha}{2}}}{1+\sqrt{\frac{1-\cos2\alpha}{2}}+\sqrt{\frac{1+\cos2\alpha}{2}}}=$$ then after some calulations I come to this $$=\frac{\sqrt{2} +\sqrt{1-\cos2\alpha}-\sqrt{1+\cos2\alpha}}{\sqrt{2} +\sqrt{1-\cos2\alpha}+\sqrt{1+\cos2\alpha}}$$ but now I'm stuck...

Notice, $$LHS=\frac{1+\sin\alpha-\cos\alpha}{1+\sin\alpha+\cos\alpha}$$ $$=\frac{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}-\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}{1+\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}+\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}}$$ $$=\frac{1+\tan^2\frac{\alpha}{2}+2\tan\frac{\alpha}{2}-1+\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}+2\tan\frac{\alpha}{2}+1-\tan^2\frac{\alpha}{2}}$$ $$=\frac{2\tan\frac{\alpha}{2}+2\tan^2\frac{\alpha}{2}}{2+2\tan\frac{\alpha}{2}}$$ $$=\tan\frac{\alpha}{2}\left(\frac{1+\tan\frac{\alpha}{2}}{1+\tan\frac{\alpha}{2}}\right)$$$$=\tan\frac{\alpha}{2}=RHS$$

Binomial expansion. Find the coefficient of $x$ in the expansion of $\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$. I've used the way that my teacher teach me. I've stuck in somewhere else. $\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6=\left(2-\frac{4}{x^3}\right)\left(x^6\left(1+\frac{2}{x^8}\right)^6\right)$ Can anyone teach me? Thanks in advance.

Notice, we have $$\left(x+\frac{2}{x^2}\right)^6=^6C_0x^{6}\left(\frac{2}{x^2}\right)^{0}+^6C_1x^{5}\left(\frac{2}{x^2}\right)^{1}+^6C_2x^{4}\left(\frac{2}{x^2}\right)^{2}+^6C_3x^{3}\left(\frac{2}{x^2}\right)^{3}+^6C_4x^{2}\left(\frac{2}{x^2}\right)^{4}+^6C_5x^{1}\left(\frac{2}{x^2}\right)^{5}+^6C_6x^{0}\left(\frac{2}{x^2}\right)^{6}$$ Now, we have $$\left(2-\frac{4}{x^3}\right)\left(x+\frac{2}{x^2}\right)^6$$ $$=2\left(x+\frac{2}{x^2}\right)^6-\frac{4}{x^3}\left(x+\frac{2}{x^2}\right)^6$$ Hence, the coefficient of $x$ in the above expansion $$=\text{coefficient of x in the expansion of}\ 2\left(x+\frac{2}{x^2}\right)^6-\text{coefficient of x in the expansion of}\ \frac{4}{x^3}\left(x+\frac{2}{x^2}\right)^6$$$$=2(0)-4(2\cdot ^6C_1)$$$$=-8(^6C_1)=-8(6)=-48$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{coefficient of}\ x=-48}}$$

Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$ Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$ My attempt So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$ $$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$ $$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$ Then I have $3$ limits to evaluate $$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$ $$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$ Now I'm having trouble with the last one which is $$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$ Thanks for any help.

Using L'Hospital's rule (since direct evaluation gives $\bigl(\frac{0}{0}\bigr)$ ), we have the following: $$\lim_{x \to 0} \frac{\cos x-\cos x +x\sin x}{3x^2}= \lim_{x \to 0} \frac{\sin x}{3x}.$$ We take the derivative of the numerator and denominator again: $$\lim_{x \to 0} \frac{\cos x}{3} = \frac{1}{3}.$$

Quadratic formula - check my simplificaiton I am trying to solve this equation using the quadratic formula: $$x^2 + 4x -1 = 0$$ I start by substituting the values into the quadratic formula: $$x = {-(4) \pm \sqrt {(4)^2 - 4(1)(-1)} \over 2}$$ which becomes $$x = {-4 \pm \sqrt{20} \over 2}$$ This is the answer the textbook that I am using gives but I would have thought I could have simplified this further to: $$x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$$ which becomes $$x = {-4 \pm 2 \sqrt 5 \over 2}$$ which becomes $$x = -2 \pm \sqrt 5$$ Am I right and if so, why would the textbook not have simplified it further?

Note that $$\frac{B+C}{A}=\frac{B}{A}+\frac{C}{A}$$ $$x=\frac{-4\pm 2\sqrt{5}}{2}=\frac{-4}{2}\pm\frac{2\sqrt{5}}{2}=-2\pm\sqrt{5}.$$

Algebraic Aproach For this word problem How can the followin question be solved algebraically? A certain dealership has a total of 100 vehicles consisting of cars and trucks. 1/2 of the cars are used and 1/3 of the trucks are used. If there are 42 used vehicles used altogether, how many trucks are there?

$ \newcommand{\xu}{x^{\text{used}}} \newcommand{\yu}{y^{\text{used}}} \newcommand{\xn}{x^{\text{new}}} \newcommand{\yn}{y^{\text{new}}} $ The key in converting text problems into algebraic expressions is to write an expression for every sentence or phrase which contains quantifiable information. For example, consider your problem A certain dealership has a total of $100$ vehicles consisting of cars and trucks. $1/2$ of the cars are used and $1/3$ of the trucks are used. If there are $42$ used vehicles used altogether, how many trucks are there? Let us disassemble it into set of statements working with on at a time: * * A certain dealership has a total of $100$ vehicles consisting of cars and trucks. Let $x, y$ be the total number of cars and trucks respectively. Then we write the first equation $$ x + y = 100 $$ * $1/2$ of the cars are used and $1/3$ of the trucks are used. Note that in this sentence we have two statemnts, so let us deal with them separately. Denote $\xn$ and $\yn$ the number of new cars and trucks, $\xu$, $\yu$ – number of used cars and trucks, then we can write the second and the third equations $1/2$ of the cars are used $(\dots)$ $$ \xu = \frac{1}{2} x $$ $(\dots)$ and $1/3$ of the trucks are used $$ \yu = \frac{1}{3} y $$ * If there are $42$ used vehicles used altogether, how many trucks are there? The last sentence contains one quantitate statement and states the question for problem If there are $42$ used vehicles used altogether, $( \dots ) $ $$\xu + \yu = 42$$ and states the question for problem how many trucks are there? which can be written in our notation as "find $y$"" $$ y \quad - \quad ? $$ Finally, combining items $1$ to $3$, we write the system of equations, which is the algebraic formulation of the original text problem: $ \begin{aligned} \text{Find } y \text{ given } \qquad \qquad \qquad \qquad \begin{cases} x+ y = 100, \\ \xu = \dfrac{1}{2} x, \\ \yu = \dfrac{1}{3} y, \\ \xu + \yu = 42, \end{cases} \end{aligned} $ Furthermore, we can simplify the system and write $$ \begin{cases} x + y = 100, \\ \dfrac{1}{2} x + \dfrac{1}{3} y = 42, \end{cases} $$ which is easy to solve: $$ \begin{cases} x + y = 100, \\ \dfrac{1}{2} x + \dfrac{1}{3} y = 42, \end{cases} \implies \begin{cases} \dfrac{1}{2} x + \dfrac{1}{2}y = 50, \\ \dfrac{1}{2} x + \dfrac{1}{3} y = 42, \end{cases} \implies \begin{cases} \left(\dfrac{1}{2} x - \dfrac{1}{2} x \right) + \left(\dfrac{1}{2} y - \dfrac{1}{3} y \right) = 50 - 42, \\ \dfrac{1}{2} x + \dfrac{1}{3} y = 42, \end{cases} \implies \dfrac{1}{6}y = 8 \implies y = 48 $$ Thus, the final answer is There are $48$ trucks at the dealership.

Perpendicular Bisector of Made from Two Points For a National Board Exam Review: Find the equation of the perpendicular bisector of the line joining (4,0) and (-6, -3) Answer is 20x + 6y + 29 = 0 I dont know where I went wrong. This is supposed to be very easy: Find slope between two points: $${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$ Obtain Negative Reciprocal: $${ m'=\frac{-10}{3}}$$ Get Midpoint fox X $${ \frac{-6-4}{2} = -5 }$$ Get Midpoint for Y $${ \frac{-0--3}{2} = \frac{3}{2} }$$ Make Point Slope Form: $${ y = m'x +b = \frac{-10}{3}x + b}$$ Plugin Midpoints in Point Slope Form $${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$ Evaluate b $${ b = \frac{109}{6}}$$ Get Equation and Simplify $${ y = \frac{-10}{3}x + \frac{109}{6}}$$ $${ 6y + 20x - 109 = 0 }$$ Is the problem set wrong? What am I doing wrong?

Notice, the mid=point of the line joining $(4, 0)$ & $(-6, -3)$ is given as $$\left(\frac{4+(-6)}{2}, \frac{0+(-3)}{2}\right)\equiv \left(-1, -\frac{3}{2}\right)$$ The slope of the perpendicular bisector $$=\frac{-1}{\text{slope of line joining}\ (4, 0)\ \text{&}\ (-6, -3)}$$ $$=\frac{-1}{\frac{-3-0}{-6-4}}=-\frac{10}{3}$$ Hence, the equation of the perpendicular bisector: $$y-\left(-\frac{3}{2}\right)=-\frac{10}{3}(x-(-1))$$ $$6y+9=-20x-20$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the perpendicular bisector:}\ 20x+6y+29=0}}$$

evaluate the integral Evaluate the integral from: $$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx$$ The way I approach this problem is $$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz$$ and $$ \text{Res}_{i\sqrt3}(f(z)) = \frac{e ^{-2\sqrt3}}{2\sqrt3}$$ Then, the integral will be: $$\frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz = \frac{1}{2} \cdot 2\pi i \cdot \frac{e^{-2\sqrt3}}{2\sqrt3} = \frac{\pi i e^{-2\sqrt3}}{2\sqrt3}$$ Is my approach correct? if not, can someone show me? Sorry because I just learn about residue theorem and don't know if my work is correct or not.

Here is another approach: $$f(a)=\int_0^{\infty} \frac{x \cdot \sin(ax)}{x^2+3}dx$$ take a Laplace transform with respect to $a$ to obtain \begin{align} \mathcal{L}(f(a))&=\int_0^{\infty} \frac{x^2}{(x^2+3)(x^2+s^2)}dx\\ &=\frac{\pi}{2\sqrt3+2s} \end{align} now take an inverse Laplace to obtain $$f(a)=\frac{\pi}{2} e^{-\sqrt{3} a}$$ Therefore $f(2)=\frac{\pi}{2} e^{-2\sqrt{3}}$.

Question related to elliptical angles Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $ be an ellipse and $AB$ be a chord. Elliptical angle of A is $\alpha$ and elliptical angle of B is $\beta$. AB chord cuts the major axis at a point C. Distance of C from center of ellipse is $d$. Then the value of $\tan \frac{\alpha}{2}\tan \frac{\beta}{2}$ is $(A)\frac{d-a}{d+a}\hspace{1cm}(B)\frac{d+a}{d-a}\hspace{1cm}(C)\frac{d-2a}{d+2a}\hspace{1cm}(D)\frac{d+2a}{d-2a}\hspace{1cm}$ I dont know much about eccentric angles, so could not attempt this question. My guess is $\alpha+\beta=\frac{\pi}{2}$. I dont know this is correct or not. Please help me in solving this question.

Notice, we have the coordinates of the points $A$ & $B$ as follows $$A\equiv(a\cos\alpha, b\sin \alpha)$$ $$B\equiv(a\cos\beta, b\sin \beta)$$ Hence, the equation of the chord AB $$y-b\sin\beta=\frac{b\sin\alpha-b\sin \beta}{a\cos\alpha-a\cos \beta}(x-a\cos \beta)$$ $$y-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(x-a\cos \beta)$$ Since, chord AB intersects the x-axis at the point $C(d, 0)$, where $x=d$ & $y=0$ hence setting these values in the equation of the chord AB, we get $$0-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(d-a\cos \beta)$$ $$\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}=\frac{a\sin\beta}{a\cos \beta-d}$$ Now, set the values of $\sin \alpha=\frac{2\tan \frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$, $\sin \beta=\frac{2\tan \frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}$, $\cos \alpha=\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$, $\cos \beta=\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}$ I hope you can take it from here.

maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .

The Cauchy-Schwarz-inequality yields $$|a + b + c|^2 \le (a^2 + b^2 + c^2)\cdot 3 = 144$$ and therefore $a + b + c \le 12$. Plugging in $a = b = c = 4$ shows that this value is actually the maximum. Alternatively, you could use the convexity of the function $x \mapsto \sqrt{x}$. By Jensen's inequality we have $$a + b + c \le |a| + |b| + |c| = 3\left(\frac{1}{3}\sqrt{a^2} + \frac{1}{3}\sqrt{b^2} + \frac{1}{3}\sqrt{c^2}\right) \le 3 \sqrt{\frac{1}{3}a^2 + \frac{1}{3}b^2 + \frac{1}{3}c^2} = 12$$ Again, plugging in $a = b = c = 4$ yields the result.

$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful: $$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$ Thanks in advance!

Or without using L'Hospital, you can do this: $$ \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1+\sqrt [ 3 ]{ x } -1 }{ x-1 } } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1 }{ x-1 } } +\frac { \sqrt [ 3 ]{ x } -1 }{ x-1 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1 }{ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) } } +\frac { \sqrt [ 3 ]{ x } -1 }{ \left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 }+1 \right) } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { 1 }{ \sqrt { x } +1 } } +\frac { 1 }{ \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 }+1 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 1 }{ 2 } +\frac { 1 }{ 3 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 5 }{ 6 } $$ Identities used: * *$a^2-b^2 =(a-b)(a+b) $ *$a^3-b^3 = (a-b)(a^2+ab+b^2)$

Finding roots of the polynomial $x^4+x^3+x^2+x+1$ In general, how could one find the roots of a polynomial like $x^4+x^3+x^2+x^1+1$? I need to find the complex roots of this polynomial and show that $\mathbb{Q (\omega)}$ is its splitting field, but I have no idea of how to proceed in this question. Thanks in advance.

Since $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$, the roots are the fifth roots of $1$, excluding $1$. Note that the set of fifth roots of $1$ is a group of prime order, so it is cyclic and any element is a generator. Thus, if $\omega$ is any of the roots of the polynomials, the full set of roots is given by $\omega,\omega^2,\omega^3,\omega^4$. In particular all roots belong to $\mathbb{Q}[\omega]$. The roots are $$ \cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\\ \cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}\\ \cos\frac{6\pi}{5}+i\sin\frac{6\pi}{5}\\ \cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5} $$ However, you can also determine them “more explicitly”. Rewrite the equation as $$ x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0 $$ and recall that $$ x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 $$ so, setting $t=x+\dfrac{1}{x}$, the equation is $$ t^2+t-1=0 $$ Find the two roots $t_1$ and $t_2$; next solve the equations $$ x+\dfrac{1}{x}=t_1,\qquad x+\dfrac{1}{x}=t_2 $$

$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ I tried to solve it. $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac{3}{(2+\cos x)^2}dx$ But i could not solve further.Please help me in completing.

You may observe that: $$\frac{2 \cos (x)+1}{(\cos (x)+2)^2}=\frac{\cos (x)}{\cos (x)+2}+\frac{\sin ^2(x)}{(\cos (x)+2)^2}=\frac{\frac{d}{dx}\sin(x)}{\cos (x)+2}-\frac{\sin(x)\frac{d}{dx}(\cos x +2)}{(\cos (x)+2)^2}$$

Area of shaded region circle help Find the area of the shaded region Area of the sector is $240^\circ$ or $\frac{4\pi}{3}$ Next find $\frac{b\cdot h}{2}$ which is $\frac{2\cdot2}{2}$ which is $2$. Then subtract the former from the latter: $\frac{4\pi}{3} - 2$ Therefore the answer is $~2.189$? Is this correct?

Here, aperture angle $\theta=120^\circ=\frac{2\pi}{3}$ Area of shaded portion $$=\text{(area of the sector)}-\text{(area of isosceles triangle)}$$ $$=\frac{1}{2}(\theta)(r^2)-\frac{1}{2}(r^2)\sin\theta$$ $$=\frac{1}{2}\frac{2\pi}{3}(2)^2-\frac{1}{2}(2)^2\sin\frac{2\pi}{3}$$ $$=2.456739397$$ Edit: Note: Area of an isosceles triangle having each of the equal sides $a$ & an angle $\theta$ included between them then the area of the triangle $$=\frac{1}{2}(a)(a)\sin\theta=\frac{1}{2}a^2\sin\theta$$

Angle between segments resting against a circle Motivation: A couple of days ago, when I was solving this question, I had to consider a configuration like this Now, I didn't intentionally make those two yellow bars stand at what appears to be a $90^{\circ}$-angle but it struck me as an interesting situation, so much so that I thought the following question might be an interesting one to solve. The Question: Given a circle of radius $r$, a horizontal line a distance $c>r$ from the circle's centre, and two points $A$ and $B$ on that line located as indicated in the picture below, find the angle $\Theta$ as a function of the parameters given ($a,b,c,r$). The blue and red lines passing through $M$ are tangents to the circle, at $P$ and $Q$ respectively.

Set Cartesian coordinates on the plane so that the center of the circle is $(0,0)$ and the black line $\overline{AB}$ is parallel to the $x$-axis. The coordinates of $A$ are $(-(a - b), c)$ and the distance $OA$ is $\sqrt{(a - b)^2 + c^2}$. Therefore \begin{align} \angle OAB & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c^2}}{c} \right) = \arctan \left( \frac{c}{a-b} \right) & \text{and} \\ \angle OAQ & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c^2}}{r} \right). \end{align} The coordinates of $B$ are $(b, c)$ and the distance $OB$ is $\sqrt{b^2 + c^2}$, so \begin{align} \angle OBA & = \arcsin\left( \frac{\sqrt{b^2 + c^2}}{c} \right) = \arctan \left( \frac cb \right) & \text{and} \\ \angle OBP & = \arcsin\left( \frac{\sqrt{b^2 + c^2}}{r} \right). \end{align} Since $\angle BAM = \angle OAB + \angle OAQ$ and $\angle ABM = \angle OBA - \angle OBP$, and $\theta = \pi - \angle BAM - \angle ABM$, \begin{align} \theta & = \pi - (\angle OAB + \angle OAQ) - (\angle OBA - \angle OBP) \\ & = \pi - \arctan \frac{c}{a-b} - \arcsin \frac{\sqrt{(a - b)^2 + c^2}}{r} - \arctan \frac cb + \arcsin \frac{\sqrt{b^2 + c^2}}{r} . \end{align} Another approach: consider the figure below, which shows line $\overline{AB}$ and segments $\overline{OP}$ and $\overline{BP}$. It also shows the perpendicular from $O$ to $\overline{AB}$, which intersects $\overline{AB}$ at $C$ and $\overline{BP}$ at $R$. From the original problem statement, we have $OP = r$, $BC = |b|$, and $OC = c$. Let $OR = |u|$, with $u$ positive if $R$ is between $O$ and $C$ as shown. Then $CR = |c - u|$ and $PR = \sqrt{u^2 - r^2}$, and by similar triangles, $$ \frac{\sqrt{u^2 - r^2}}{r} = \pm\frac{c - u}{b}.$$ There are actually three cases represented here: * *$b > 0$, shown in the figure; *$-r < b < 0$, in which case $R$ is on the extension of $\overline{OC}$ beyond $C$, $u > c$, and $\triangle BCR$ has (positive) leg lengths $-b$ and $u - c$; and *$b < -r$, in which case $R$ is on the extension of $\overline{OC}$ beyond $O$, $u < -r$, and $\triangle BCR$ has (positive) leg lengths $-b$ and $c - u = c + |u|$. This is the case that requires the "$-$" option of the $\pm$ sign. In all three cases I assume $c > r$. Squaring both sides of this equation and rearranging terms appropriately, we get: $$ (b^2 - r^2)u^2 + 2cr^2 u - (b^2 + c^2)r^2 = 0. \tag 1$$ If $b^2 \neq r^2$ this is a quadratic equation in $u$, and it has roots $$ u = \frac{-cr^2 \pm br \sqrt{b^2 + c^2 - r^2}}{b^2 - r^2}.$$ We want the positive root if $b > -r$ and the negative root if $b < -r$, so $$ \angle CBP = \arccos \frac ru = \begin{cases} \arccos \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}} & \text{if $b > r$} \\ \arccos \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}} & \text{if $b < r$ and $b \neq -r$.} \end{cases}$$ But if $b = r$, then $ u = \dfrac{c^2+r^2}{2 c} $ and $$ \angle CBP = \arccos \frac ru = \arccos \frac{2cr}{c^2+r^2} ,$$ whereas if $b = -r$ then $\angle CBP = \frac\pi2$. And oh, look, all of these formulas apply to $\angle BAQ$ in the original figure if we substitute $b - a$ for $b$ (and $b - a < -r$ provided that $\angle BAQ$ is acute, as shown), so if we assume $b > -r$, \begin{align} \theta & = \pi - \angle BAQ - \angle CBP \\ & \begin{aligned} = \pi & - \arccos \dfrac{(a - b)^2 - r^2} {-cr + (a - b) \sqrt{(a - b)^2 + c^2 - r^2}} \\ & - \begin{cases} \arccos \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}} & \text{if $b > r$} \\ \arccos \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}} & \text{if $-r < b < r$} \\ \arccos \dfrac{2cr}{c^2+r^2} & \text{if $b = r$} \end{cases} \end{aligned} \\ & = \arcsin \dfrac{(a - b)^2 - r^2} {-cr + (a - b) \sqrt{(a - b)^2 + c^2 - r^2}} \\ & \qquad\qquad + \begin{cases} \arcsin \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}} & \text{if $b > r$} \\ \arcsin \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}} & \text{if $-r < b < r$} \\ \arcsin \dfrac{2cr}{c^2+r^2} & \text{if $b = r$} \end{cases} \end{align} So that's just two trig functions, though there are three cases depending on the value of $b$. As suggested in a comment, we could get this down to one trig function if we could find the three sides of $\triangle ABM$ without using trigonometry; but I think this would involve saying something about the triangles $\triangle OPM$ and $\triangle OQM$, and I do not yet see how to do it.

two variable nonhomogeneous inequality Let $$x\ge 0,y\ge 0,x\neq 1,y \neq 1$$Prove the inequality $$\dfrac {x}{(y-1)^2} +\dfrac {y}{(x-1)^2} \ge \dfrac {x+y-1}{(x-1)(y-1)} $$

hint: The endpoints case you can handle with ease, for more general case that: $x, y > 1\to x(x-1)^2 +y(y-1)^2 \geq (x+y-1)(xy-(x+y-1))\iff x(x^2-2x+1)+y(y^2-2y+1)\geq xy(x+y-1)-(x+y-1)^2\iff (x^3+y^3)-2(x^2+y^2)+(x+y)\geq xy(x+y-1)-(x^2+y^2+1+2xy-2x-2y)\iff f(x,y)=x^3+y^3-(x^2+y^2)+1+3xy-(x+y) -xy(x+y)\geq 0$. Taking partial derivatives: $f_x=3x^2-2x+3y-1-2xy-y^2 = 0=f_y=3y^2-2y+3x-1-x^2-2xy\to f_x-f_y = 0\to (x-y)(4(x+y)-5)=0\to x=y , x+y = \dfrac{5}{4}$. Consider each case separately will yield desire result. Note: If you want to avoid using calculus, there is another way ( I let you test it to see if it works) that you can prove this at this point: You prove : $f(x,y) \geq f(y,y)$, and $f(y,y) \geq 0$. The latter inequality is $(y-1)^2 \geq 0$. Another approach is let $a = x-1,b = y-1$, and assume $a,b > 0\to \dfrac{a+1}{b^2}+\dfrac{b+1}{a^2}\geq \dfrac{a+b+1}{ab}$.But this is quite simple...because $LHS =\dfrac{a}{b^2}+\dfrac{b}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{a^3+b^3}{a^2b^2}+\dfrac{2}{ab}\geq \dfrac{ab(a+b)}{a^2b^2}+\dfrac{2}{ab}=\dfrac{a+b+2}{ab} > RHS$

Division in Summations Suppose $a_n=\dfrac{2^n}{n(n+2)}$ and $b_n=\dfrac{3^n}{5n+18}$. I need to find the value of: $$\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{b_n}$$ I think this problem is meant for me to compute each sum differently and then divide. Is this some property of summations that we need to utilize here?

$$\begin{equation} \begin{split} \sum_{n=1}^\infty \frac{a_n}{b_n} & = \sum_{n=1}^\infty \frac{2^n(5n+18)}{3^nn(n+2)} \\ & = \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{5}{n+2} + 9\left(\frac{1}{n} - \frac{1}{n+2} \right) \right) \right] \\ & = \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{9}{n} - \frac{4}{n+2}\right) \right] \\ & = \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=1}^\infty \frac{2^{n+2}}{3^n(n+2)} \\ & = \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=3}^\infty \frac{2^{n}}{3^{n-2}n} \\ & = \sum_{n=1}^2 \frac{2^n}{3^{n-2}n} = \frac{2}{3^{-1}} + \frac{2^2}{3^{0}2} \\ & = 6 + 2 \\ & = 8 \end{split} \end{equation} $$

$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\ =\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$. Any hint will solve my problem.

$\bf{Another\; Solution::}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx.$$ Put $\displaystyle x = \frac{2}{t}\;,$ Then $\displaystyle dx = -\frac{2}{t^2}dt$ and Changing Limits, We get $$\displaystyle I = \int_{\infty}^{0}\frac{\ln\left(\frac{2}{t}\right)}{\frac{4}{t^2}+\frac{4}{t}+2}\cdot -\frac{2}{t^2}dt = \int_{0}^{\infty}\frac{\left(\ln 2-\ln t\right)}{t^2+2t+2}dt$$ Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$ and $$\displaystyle \bullet \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$ So $$\displaystyle I = \ln 2\int_{0}^{\infty}\frac{1}{x^2+2x+2}dx - \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx = \ln 2\int_{0}^{\infty}\frac{1}{(x+1)^2+1^2}dx-I$$ So we get $$\displaystyle I = \ln 2\times \left[\tan^{-1}\left(x+1\right)\right]_{0}^{\infty}-I$$ So we get $$\displaystyle I = \frac{\ln 2}{2} \times \left[\frac{\pi}{2}-\frac{\pi}{4}\right] = \frac{\pi}{8}\cdot \ln 2$$

Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct? \begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*} First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$. Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$ $$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$ I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$

$$\int \left(e^x\sin^2(x)\right)\text{d}x =$$ $$\int \left(e^x\left(\frac{1}{2}(1-\cos(2x))\right)\right)\text{d}x =$$ $$\frac{1}{2}\int \left(e^x-e^x\cos(2x)\right)\text{d}x =$$ $$\frac{1}{2} \left(\int \left(e^x\right) \text{d}x-\int \left(e^x\cos(2x)\right) \text{d}x\right) =$$ $$\frac{1}{2} \left(\int e^x \text{d}x-\int e^x\cos(2x) \text{d}x\right) =$$ $$\frac{1}{2} \left(e^x-\int e^x\cos(2x) \text{d}x\right) =$$ For the integrand $e^x\cos(2x)$, use the formula: $$\int\exp(\alpha x)\cos(\beta x)\text{d}x=\frac{\exp(\alpha x)(\alpha \cos(\beta x))+\beta\sin(\beta x)}{\alpha^2+\beta^2}$$ $$\frac{1}{2} \left(e^x-\frac{e^x(2\sin(2x)+\cos(2x))}{5}\right) + C =$$ $$-\frac{e^x(2\sin(2x)+\cos(2x)-5)}{10} + C $$

Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review: How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$? Answer is $2.73$ Simplify into Standard Form: $$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$ $$ a^2 = 1 $$ $$ b^2 = 2 $$ $$ c^2 = 5 $$ Hyperbola is Vertical: $$ C(-2,1) ; y = 1 $$ $$ y = 1 + \sqrt5 = 3.24 $$ $$ y = 1 - \sqrt5 = 1.24 $$ Both answers don't match; What am I doing wrong?

$\frac{(x + 2)^2}{2} - \frac{(y - 1)^2}{1}; C = (-2, 1)$ $C = \sqrt{2 + 1} = \sqrt{3}$ Answer: $1 + \sqrt(3) = 2.73$

Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$. Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first partials to zero we have \begin{align*} \frac{\partial L}{\partial x} &= 2x + \lambda y^2 z^4 = 0 \qquad (1) \\ \frac{\partial L}{\partial y} &= 2y + 2 \lambda x y z^4 = 0 \qquad (2) \\ \frac{\partial L}{\partial z} &= 2z + 4 \lambda x y^2 z^3 = 0 \qquad (3) \\ \frac{\partial L}{\partial \lambda} &= xy^2 z^4 - 32 = 0 \qquad (4) \end{align*} Now I'm having a hard time solving this system for $x,y$ and $z$. Here is what I did so far. From $(1)$ and $(2)$ we get \begin{align*} \frac{2x}{y^2 z^4} = - \lambda \qquad \text{and} \qquad \frac{1}{xz^4} = - \lambda \end{align*} Thus $\frac{2x}{y^2 z^4} = \frac{1}{xz^4} $ or $y^2 = 2x^2$ after simplification. Also, from $(2)$ and $(3)$ we can deduce that \begin{align*} \frac{1}{xz^4} = - \lambda = \frac{2z}{4xy^2 z^3} \end{align*} so that $2y^2 = z^2$ after simplification. Now I used all this and substituted it into $(4)$. This gave me \begin{align*} x(2x^2) (4y^4) - 32 = 0 \end{align*} or (since $y^4 = 4x^4)$ \begin{align*} 8x^3 (4x^4) - 32 = 0 \end{align*} This means that $32x^7 - 32 = 0$, so that $x = 1$. Then $y^2 = 2$, so that $y = \pm \sqrt{2}$. Then $z^2 = 4$, so that $z = \pm 2$. So I found the points $(x,y,z) = (1, \sqrt{2}, 2)$ and $(1, - \sqrt{2}, -2)$. They both give me the distance $\sqrt{x^2 + y^2 + z^2} = \sqrt{7}$, so I'm guessing they are equal? Is my reasoning correct?

A hint: Multiply $(1)$ by $x$, $(2)$ by $y$, and $(3)$ by $z$ and look at the three equations you got.

Prove that $(a+b)(b+c)(c+a) \ge8$ Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$, Prove that $(a+b)(b+c)(c+a)\ge8$. My attempt: By AM-GM inequality, we have $$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$ and similarly $$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$ $$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$ Multiplying $(1)$, $(2)$ and $(3)$ together we reach $$(a+b)(b+c)(c+a) \ge8abc.$$ Now, I need to show that $abc = 1$. Again by AM-GM inequality we have $$\frac{a+b+c}{3}\ge\sqrt[3]{abc} \implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc} \implies abc\le1$$ Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that? Other solutions to the above question are also welcomed.

From $abc(a+b+c)=3$ we have $$a^2+a(b+c)=\frac {3}{bc}.$$ Therefore we have $$(a+b)(b+c)(c+a)\geq 8\iff$$ $$(a+b)(a+c)\geq \frac {8}{b+c}\iff$$ $$a^2+a(b+c)\geq -bc+ \frac {8}{b+c)}\iff$$ $$(\bullet ) \quad \frac {3}{bc}\geq -bc+\frac {8}{b+c}.$$ Now $b+c\geq 2\sqrt {bc},$ so $$\frac {8}{b+c}\leq \frac {4}{\sqrt {bc}}.$$ So with $x=\sqrt {bc} $ we see that ($\bullet $ ) is satisfied if $\frac {3}{x^2}\geq -x^2+\frac {4}{x} $, or, equivalently, $\frac {3} {x^2}+x^2-\frac {4}{x}\geq 0 ,$ for all $x>0.$ We have $\frac {3}{x^2}+x^2-\frac {4}{x}=x^{-2}(x-1)(x^3+x^2+x-3). $ The terms $(x-1)$ and $(x^3+x^2+x-3)$ always have the same sign when $x\geq 0.$ (Case 1:$\;x\geq 1.$ Case 2: $\;0<x<1.$) Inelegant compared to the answer by Booldy, but it works.

Is the following solvable for x? I have the following equation and I was wondering if I can solve for x given that it appears both as an exponent and a base: $[\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}-0.5\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}.\frac{4.(\frac{x-N}{\sqrt {2}.D})}{\sqrt {\pi}.e^{-\frac{(x-N)^2}{2D^2}}+\sqrt {\pi.e^{-2\frac{(x-N)^2}{2D^2}}+16.\frac{(x-N)^2}{2D^2}}} ]+[ \frac{1}{\sqrt {2\pi}.D}.e^{-\frac{(x-N)^2}{2D^2}}-0.5\frac{1}{\sqrt {2\pi}.D}.e^{-\frac{(x-N)^2}{2D^2}}.\frac{4.(\frac{x-M}{\sqrt {2}.S})}{\sqrt {\pi}.e^{-\frac{(x-M)^2}{2S^2}}+\sqrt {\pi.e^{-2\frac{(x-M)^2}{2S^2}}+16.\frac{(x-M)^2}{2S^2}}}] = \frac{1}{v}$ In other words, can I find an expression for x as a function of all the other variables?

Let's simplify the equation a bit so we can better see its form: $$C_{1}e^{X_{1}} + C_{2}e^{X_{1}}\cdot \frac{X_{2}}{C_{3}e^{X_{3}} \sqrt{C_{4}e^{X_{3}} + C_{5}X_{3}}} + C_{6}e^{X_{4}} + C_{7}e^{X_{4}}\cdot \frac{X_{5}}{C_{8}e^{X_{6}} \sqrt{C_{9}e^{X_{6}} + C_{10}X_{6}}} = C_{11}$$ With constants as $C_{n}$ above and expressions involving x as $X_{n}$ above. Even if all of the constants were 1, we would get: $$e^{X_1} + e^{X_4} + X_{2}\frac{e^{X_1}e^{-X_{3}}}{\sqrt{e^{X_{3}} + X_{3}}} + X_{5}\frac{e^{X_4}e^{-X_{6}}}{\sqrt{e^{X_{6}} + X_{6}}} = 1$$ Which could, with some further magical simplifications (letting all of the expressions in x be the same), be written as $$e^{2X} + \frac{2X}{\sqrt{e^{X} + X}} = 1$$ Heck, even if the denominator was 1, we would get (letting $Y = 2X$): $$e^{Y} + Y = 1$$ Which in and of itself is not solvable for Y. All of this was just to show that even with certain simplifications your equation reduces to a form which is very much unsolvable, so it is extremely unlikely that your original equation is solvable for x.

$\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|dx$ MyAttempt $\int_{0}^{\pi}|\sqrt2\sin x+2\cos x|=\int_{0}^{\pi/2}\sqrt2\sin x+2\cos x dx+\int_{\pi/2}^{\pi}|\sqrt2\sin x+2\cos x| dx$ I could solve first integral but in second one,i could not judge the mod will take plus sign or minus sign or how to break it in further intervals?

$\displaystyle A \sin \ x + B \cos \ x = \sqrt{A^2 + B^2}\sin( x + \phi )$, where $\displaystyle \cos \phi = \frac{A}{\sqrt{A^2 + B^2}}$. In our case $\displaystyle \cos\phi=\frac{1}{\sqrt{3}} \Rightarrow \phi\in\left(0;\frac{\pi}{2}\right)$. Thus: $\displaystyle \begin{aligned}\int\limits_0^\pi\left|\sqrt{2}\sin x+2\cos x\right|dx&=\sqrt{6}\int\limits_0^\pi\left|\sin(x+\phi)\right|dx=\sqrt{6}\int\limits_0^{\pi-\phi}\sin(x+\phi)dx+\sqrt{6}\int\limits_{\pi-\phi}^\pi(-\sin(x+\phi))dx=\\&=\left.-\sqrt{6}\cos(x+\phi)\right|_0^{\pi-\phi}+\left.\sqrt{6}\cos(x+\phi)\right|_{\pi-\phi}^\pi=-\sqrt{6}\left(\cos\pi-\cos\phi\right)+\sqrt{6}\left(\cos(\pi+\phi)-\cos\pi\right )=\\&=\sqrt{6}+\sqrt{2}-\sqrt{2}+\sqrt{6}=2\sqrt{6}\end{aligned}$

$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$ $\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?

Let, $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt x)}{x^2-x+1}dx\tag 1$$ Now, using property of definite integral $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$ we get $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{(1-x)^2-(1-x)+1}dx$$ $$I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx\tag 2$$ Now, adding (1) & (2), we get $$I+I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})}{x^2-x+1}dx+\int_{0}^{1}\frac{\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$ $$2I=\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x})+\sin^{-1}(\sqrt {1-x})}{x^2-x+1}dx$$ $$I=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(\sqrt {x}\sqrt{1-1+x}+\sqrt{1-x}\sqrt{1-x})}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(x+(1-x)}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\sin^{-1}(1)}{x^2-x+1}dx$$ $$=\frac{1}{2}\int_{0}^{1}\frac{\frac{\pi}{2}}{x^2-x+1}dx$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{x^2-x+1}$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}}$$ $$=\frac{\pi}{4}\int_{0}^{1}\frac{dx}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt 3}{2}\right)^2}$$ $$=\frac{\pi}{4}\left[\frac{2}{\sqrt 3}\tan^{-1}\left(\frac{\left(x-\frac{1}{2}\right)}{\frac{\sqrt 3}{2}}\right)\right]_{0}^{1}$$ $$=\frac{\pi}{2\sqrt 3}\left[\frac{\pi}{6}+\frac{\pi}{6}\right]$$ $$=\frac{\pi^2}{6\sqrt 3}$$

$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$ My Attempt: $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$ Replacing $x$ by $1-x$,we get $$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$ Then I got stuck. Please help.

The integral can also be found using a self-similar substitution of $u = \dfrac{1 - x}{1 + x}$. Here we see that $x = \dfrac{1 - u}{1 + u}$ such that $dx = -\dfrac{2}{(1 + u)^2} \, du$. Writing the integral as $$I = -\int \frac{1 - x}{(1 + x) x \sqrt{x + 1 + \frac{1}{x}}} \, dx,$$ if we observe that $$x + 1 + \frac{1}{x} = \frac{u^2 + 3}{1 - u^2},$$ then under the self-similar substitution the integral becomes $$I = \int \frac{2u}{\sqrt{(1 - u^2)(3 + u^2)}} \, du,$$ or $$I = \int \frac{dt}{\sqrt{(1 - t)(3 + t)}},$$ after setting $t = u^2$. In the denominator, if we complete the square then integrate one has $$I = \int \frac{dt}{\sqrt{2^2 - (t + 1)^2}} = \sin^{-1} \left (\frac{t + 1}{2} \right ) + C,$$ or $$\int \frac{x - 1}{(x + 1) \sqrt{x^3 + x^2 + x}} \, dx = \sin^{-1} \left [\frac{1}{2} \left (\frac{1 - x}{1 + x} \right )^2 + \frac{1}{2} \right ] + C.$$

$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$ I tried to solve it. $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$ But this does not seem to be solving.Please help.

Please note that you have: $\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$, which on dividing numerator and denominator by $x^2$ becomes: $\int\frac{1+1/x^2}{(x^2+1/x^2)+3+3(x-1/x)}dx$=$\int\frac{1+1/x^2}{(x-1/x)^2+5+3(x-1/x)}dx$ Now put x-1/x =t so that (1+1/$x^2$)dx=dt and thus you get: $\int\frac{1}{t^2+5+3t}dt$;which can be rearranged to get:$\int\frac{1}{t^2+5+3t}dt$=$\int\frac{1}{(t+3/2)^2+11/4}dt$=(2/$\sqrt11$)arctan[(t+3/2)2/$\sqrt11$]+c=(2/$\sqrt11$)arctan[(x-1/x+3/2)2/$\sqrt11$]+c;where c is integration constant. PS:x>$0$ or x<$0$,only for these x , above solution is valid.At,x=$0$,arctan[(x-1/x+3/2)2/$\sqrt11$]becomes discontinuous so Fundamental theorem doesn't hold.

Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Two circles ,of radii $a$ and $b$,cut each other at an angle $\theta.$Prove that the length of the common chord is $\frac{2ab\sin \theta}{\sqrt{a^2+b^2+2ab\cos \theta}}$ Let the center of two circles be $O$ and $O'$ and the points where they intersect be $P$ and $Q$.Then angle $OPO'=\theta$ $\cos \theta=\frac{a^2+b^2-OO'^2}{2ab}$ $OO'^2=a^2+b^2-2ab\cos\theta$ In triangle $PO'Q$,angle $PO'Q=\pi-\theta$ $\cos(\pi-\theta)=\frac{b^2+b^2-l^2}{2b^2}$ Then i am stuck.Please help me to reach upto proof.

Let $A$ & $B$ be the centers of the circles with radii $a$ & $b$ respectively such that they have a common chord $MN=2x$ & intersecting each other at an angle $\theta$. Let $O$ be the mid-point of common chord $MN$. Point $O$ lies on the line AB joining the centers of circle then we have $$MO=ON=x$$ In right $\triangle AOM$ $$AO=\sqrt{AM^2-MO^2}=\sqrt{a^2-x^2}$$ Similarly, in right $\triangle BOM$ $$OB=\sqrt{BM^2-MO^2}=\sqrt{b^2-x^2}$$ $$AB=AO+OB=\sqrt{a^2-x^2}+\sqrt{b^2-x^2}$$ Now, applying cosine rule in $\triangle AMB$ as follows $$AB^2=AM^2+BM^2-2(AM)(BM)\cos (\angle AMB)$$ Setting the corresponding values, we get $$(\sqrt{a^2-x^2}+\sqrt{b^2-x^2})^2=a^2+b^2-2(a)(b)\cos (180^\circ-\theta)$$ $$a^2-x^2+b^2-x^2+2\sqrt{(a^2-x^2)(b^2-x^2)}=a^2+b^2+2ab\cos\theta$$ $$\sqrt{(a^2-x^2)(b^2-x^2)}=x^2+ab\cos\theta$$ $$(a^2-x^2)(b^2-x^2)=(x^2+ab\cos\theta)^2$$ $$a^2b^2-b^2x^2-a^2x^2+x^4=x^4+a^2b^2\cos^2\theta+2abx^2\cos\theta$$ $$(a^2+b^2+2ab\cos\theta)x^2=a^2b^2-a^2b^2\cos^2\theta$$ $$x^2=\frac{a^2b^2\sin^2\theta}{a^2+b^2+2ab\cos\theta}$$ $$x=\frac{ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$ Hence, the length of the common chord MN, $$MN=2x=\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Length of common chord}=\color{blue}{\frac{2ab\sin\theta}{\sqrt{a^2+b^2+2ab\cos\theta}}}}}$$

How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$ I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following: $$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$ Then by expanding each $e^{nx}$ term individually: $$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$ $$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$ $$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$ $$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$ So substituting $(2),(3),(4),(5)$ into $(1)$ i get: $$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$ Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.

$\begin{array}\\ f(x) &=\frac{1}{1+e^x}\\ &=\frac{1}{2+(e^x-1)}\\ &=\frac12\frac{1}{1+(e^x-1)/2}\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(e^x-1)^n}{2^n}\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n\frac{(x+x^2/2+x^3/6+...)^n}{2^n}\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+x/2+x^2/6+...)^n\\ &=\frac12\sum_{n=0}^{\infty}(-1)^n(x/2)^n(1+nx/2+...)\\ &=\frac12\left(1-\frac{x}{2}(1+\frac{x}{2}+...)+\frac{x^2}{4}(1+...)+...\right)\\ &=\frac12\left(1-\frac{x}{2}-\frac{x^2}{4} +\frac{x^2}{4}(1+...)+...\right)\\ &=\frac12\left(1-\frac{x}{2}+O(x^3)\right)\\ &=\frac12-\frac{x}{4}+O(x^3)\\ \end{array} $

Integrating $\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$ Integrating $$\int \sqrt{x+\sqrt{x^2+1}}\,\mathrm{d}x$$ Using substitution of $x=\tan \theta$, I got the required answer. But is there a more elegant solution to the problem?

HINT $$ \sqrt{ x + \sqrt{ x^2 + 1 } } = \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } } $$ That would be enough simple to solve the integral... We get $$ \begin{eqnarray} \int \sqrt{ x + \sqrt{ x^2 + 1 } } dx &=& \int \left\{ \sqrt{ \frac{ x + i }{ 2 } } + \sqrt{ \frac{ x - i }{ 2 } } \right\} d x\\\\ &=& \frac{4}{3} \left\{ \sqrt{ \frac{ x + i }{ 2 } }^3 + \sqrt{ \frac{ x - i }{ 2 } }^3 \right\} \quad \textrm{(*)}\\\\ &=& \bbox[16px,border:2px solid #800000] {\frac{4}{3} \sqrt{ x + \sqrt{ x^2 + 1 } } \left\{ x - \frac{1}{2} \sqrt{x^2+1}\right\}} \end{eqnarray} $$ (*) Where we have used $$ a^3 + b^3 = \Big( a + b \Big) \Big( a^2 + b^2 - a b \Big) $$

Prove that $\left|(|x|-|y|)\right|\leq|x-y|$ Prove that $\left|(|x|-|y|)\right|\leq|x-y|$ Proof: $$\begin{align} \left|(|x|-|y|)\right| &\leq|x-y| \\ {\left|\sqrt{x^2}-\sqrt{y^2}\right|}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{a^2}=|a|)$}\\ \sqrt{\left(\sqrt{x^2}-\sqrt{y^2}\right)^2}&\leq \sqrt{(x-y)^2} &\text{($\sqrt{a^2}=|a|)$} \\ \left(\sqrt{x^2}-\sqrt{y^2}\right)^2 &\leq (x-y)^2&\text{$(0\leq a\leq b\implies a^2\leq b^2)$}\\ \left(\sqrt{x^2}\right)^2-2\sqrt{x^2}\sqrt{y^2}+\left(\sqrt{y^2}\right)^2&\leq x^2-2xy+y^2 &\text{(Distribution)} \\ |x|^2-2|x||y|+|y|^2&\leq x^2-2xy+y^2 &\text{($\sqrt{a^2}=|a|)$}\end{align} \\ $$ From here, we can see that $|x|^2+|y|^2=x^2+y^2$, and it suffices to show that the following holds: $$-2|x||y|\leq -2xy.$$ When $x>0,y>0$ or $x<0,y<0$, the product $xy$ is positive, and so $-2|x||y|=-2xy$ holds. When $x<0,y>0$ or $x>0,y<0$, the product $xy$ is negative, and so $-2xy>0$, and it follows that: $$-2|x||y|<0<-2xy. $$ $\square$ Is this proof valid? I am unsure if I overlooked anything, also I have some specific questions about the proof (if it is valid up to this point). From the 3rd to the 4th inequality, it feels like I am making a leap in knowing that $a\leq b$ prior to proving what's to be proved by using the result $a^2\leq b^2$. Is this a problem? What are some alternative ways of showing this inequality?

Because of the triangle inequality, there is: $\left| x+y \right| \le \left| x \right| +\left| y \right| $. Using this fact: \begin{align} \left| x \right| &= \left| (x-y)+y \right| \\ &\le \left| x-y \right| +\left| y \right| \\ \left| x \right| -\left| y \right| &\le \left| x-y \right| \tag{1} \end{align} Proceeding similarly: \begin{align} \left| y \right| &= \left| (y-x)+x \right| \\ \left| y \right| &\le \left| y-x \right| +\left| x \right| \\ \left| y \right| -\left| x \right| &\le \left| y-x \right| \tag{2} \\ \end{align} And finally combining $(1)$ and $(2)$ :$$\left|(|x|-|y|)\right|\leq|x-y|$$

The number of ordered pairs of positive integers $(a,b)$ such that LCM of a and b is $2^{3}5^{7}11^{13}$ I started by taking two numbers such as $2^{2}5^{7}11^{13}$ and $2^{3}5^{7}11^{13}$. The LCM of those two numbers is $2^{3}5^{7}11^{13}$. Similarly, If I take two numbers like $2^{3-x}5^{7-y}11^{13-z}$ and $2^{3}5^{7}11^{13}$, the LCM is $2^{3}5^{7}11^{13}$. Here $0\le x\le3$, $0\le y\le7$, $0\le z\le13$ The number of ways of choosing 3 numbers $x,y,z$ is $^4C_1\cdot^8C_1\cdot^{14}C_1$ The above value is only for $b=2^{3}5^{7}11^{13}$.But now I have to consider another value of $b$ and the the solution becomes lengthy. Is this the correct approach?

Let $a = 2^{x_{1}}\cdot 5^{y_{1}}\cdot 11^{z_{1}}$ and $b = 2^{x_{2}}\cdot 5^{y_{2}}\cdot 11^{z_{2}}\;,$ Then Given $\bf{LCM(a,b)} = 2^{3}\cdot 5^{7}\cdot 11^{13}$ So Here $0\leq x_{1},x_{2}\leq 3$ and $0\leq y_{1},y_{2}\leq 7$ and $0\leq z_{1},z_{2}\leq 13.$ Here ordered pairs of $(x_{1},x_{2}) = \left\{(0,3),(1,3),(2,3),(3,3),((3,0),(3,1),(3,2)\right\}$ So Total $7$ ordered pairs Similarly ordered pairs for $(y_{1},y_{2}),$ We get $15$ ordered pairs.. Similarly ordered pairs for $(z_{1},z_{2}),$ We get $27$ ordered pairs.. So Total ordered pairs of $(a,b) = 7\times 15 \times 27$

Does $\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$ diverge? Does the limit of this summation diverge? $$\lim\limits_{n \to +∞} \sum_{k=1}^n \frac{n\cdot \ln (k)}{n^2+k^2}$$ Thanks!

$$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln k}}{{1 + {{(\frac{k}{n})}^2}}}} = \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln (\frac{k}{n})}}{{1 + {{(\frac{k}{n})}^2}}}} + \frac{1}{n}\sum\limits_{k = 1}^n {\frac{{\ln n}}{{1 + {{(\frac{k}{n})}^2}}}} $$ When $n \to \infty $, the first term is the definite integral $$\int_0^1 {\frac{{\ln x}}{{1 + {x^2}}}dx} $$ which obviously converge. To investigate the second term, we note that $$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{(\frac{k}{n})}^2}}}} } \right] = \int_0^1 {\frac{1}{{1 + {x^2}}}dx} $$ Hence your sequence diverges. PS: This method can be used to obtain estimate on your sequence, using $$\frac{\pi }{4} - \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{1 + {{(\frac{k}{n})}^2}}}} = \frac{1}{{4n}} + o(\frac{1}{n})$$ we have $$\sum\limits_{k = 1}^n {\frac{{n\ln k}}{{{n^2} + {k^2}}}} = - G + \frac{\pi }{4}\ln n - \frac{{\ln n}}{{4n}} + o(\frac{{\ln n}}{n})$$ with $G=-\int_0^1 {\frac{{\ln x}}{{1 + {x^2}}}dx}$ is the Catalan constant.

Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$ Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$ Numerically, it's about $$\approx 111.024457130115028409990464833072173251135063166330638343951498907293$$ or in a predicted closed form $$\frac{4 }{3}\pi ^3+32 \pi \log (2).$$ Ideas, suggestions, opinions are welcome, and the solutions are optionals. Supplementary question for the integrals lovers: calculate in closed form $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^3 \, dx.$$ As a note, it would be remarkable to be able to find a solution for the generalization below $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^n \, dx.$$

$$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$ By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we can write $$\operatorname{Li}_2\left(-\frac1{x^2}\right)=\int_0^1\frac{\ln u}{u+x^2}\ du$$ Then \begin{align} I&=-4\int_0^1\ln u\left(\int_0^\infty\frac{\ln\left(1+ \frac1{x^2}\right)}{u+x^2}\ dx\right)\ du\\ &=-4\int_0^1\ln u\left(\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)\ du, \quad \color{red}{\sqrt{u}=x}\\ &=-16\pi\int_0^1\ln x\left(\ln(1+x)-\ln x\right)\ dx\\ &=-16\pi\left(2-\frac{\pi^2}{12}-2\ln2-2\right)\\ &=\boxed{\frac43\pi^3+32\pi\ln2} \end{align} Proof of $\ \displaystyle\int_0^\infty\frac{\ln\left(1+ \frac1{x^2}\right)}{u+x^2}\ dx=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$ Let $$J=\int_0^\infty\frac{\ln\left(1+\frac1{x^2}\right)}{u+x^2}\ dx\overset{x\mapsto\ 1/x}{=}\int_0^\infty\frac{\ln(1+x^2)}{1+ux^2}\ dx$$ and $$J(a)=\int_0^\infty\frac{\ln(1+ax^2)}{1+ux^2}\ dx, \quad J(0)=0,\quad J(1)=J$$ \begin{align} J'(a)&=\int_0^\infty\frac{x^2}{(1+ux^2)(1+ax^2)}\ dx\\ &=\frac1{a-u}\int_0^\infty\left(\frac1{1+ux^2}-\frac1{1+ax^2}\right)\ dx\\ &=\frac1{a-u}*\frac{\pi}{2}\left(\frac1{\sqrt{u}}-\frac1{\sqrt{a}}\right)\\ &=\frac{\pi}{2}\frac{1}{\sqrt{u}a+u\sqrt{a}} \end{align} Then $$J=\int_0^1 J'(a)\ da=\frac{\pi}{2}\int_0^1\frac{da}{\sqrt{u}a+u\sqrt{a}}\\=\frac{\pi}{2}\left(\frac{2}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$$

bend measurement and calculating $\int_4^8 \sqrt{1+{\left(\frac{{x^2-4}}{4x}\right)^2}} $ How can i get the measure of this bend : $y=\left(\frac{x^2}{8}\right)-\ln(x)$ between $4\le x \le 8$. i solved that a bit according to the formula $\int_a^b \sqrt{1+{{f'}^2}} $:$$\int_4^8 \sqrt{1+{\left(\frac{x^2-4}{4x}\right)^2}} $$ $$= \cdots$$I don't know how I calculate this integration.the answer is: 6+ln2

You just need to expand the square: $$ \int_{4}^{8}{\sqrt{1+\left(\frac{x^2-4}{4x}\right)^2}dx}=\int_{4}^{8}{\sqrt{1+\frac{x^4-8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\frac{x^4+8x^2+16}{16x^2}}dx}=\int_{4}^{8}{\sqrt{\left(\frac{x^2+4}{4x}\right)^2}dx}=\int_{4}^{8}{\frac{x^2+4}{4x}dx}=\int_{4}^{8}{\frac{x}{4}+\frac{1}{x}dx}=\frac{8^2}{8}-\frac{4^2}{8}+\ln(8)-\ln(4)=6+\ln(2) $$

calculating the characteristic polynomial I have the following matrix: $$A=\begin{pmatrix} -9 & 7 & 4 \\ -9 & 7 & 5\\ -8 & 6 & 2 \end{pmatrix}$$ And I need to find the characteristic polynomial so I use det(xI-A) which is $$\begin{vmatrix} x+9 & -7 & -4 \\ 9 & x-7 & -5\\ 8 & -6 & x-2 \end{vmatrix}$$ Is there a way to calculate the determinate faster or is way is: $$(x+9)\cdot\begin{vmatrix} x-7 & -5 \\ -6 & x-2 \\ \end{vmatrix}+7\cdot\begin{vmatrix} 9 & -5 \\ 8 & x-2 \\ \end{vmatrix} -4\begin{vmatrix} 9 & x-7 \\ 8 & -6 \\ \end{vmatrix}=$$ $$=(x+9)[(x-7)(x-2)-30]+7[9x-18+40]-4[54-8x+56]=(x+9)[x^2-9x-16]+7[9x+22]-4[-8x+2]=x^3-2x+2$$

You could use $\displaystyle\begin{vmatrix}x+9&-7&-4\\9&x-7&-5\\8&-6&x-2\end{vmatrix}=\begin{vmatrix}x+2&-7&-4\\x+2&x-7&-5\\2&-6&x-2\end{vmatrix}$ $\;\;\;$(adding C2 to C1) $\displaystyle\hspace{2.6 in}=\begin{vmatrix}0&-x&1\\x+2&x-7&-5\\2&-6&x-2\end{vmatrix}$$\;\;\;$(subtracting R2 from R1) $\displaystyle\hspace{2.6 in}=\begin{vmatrix}0&0&1\\x+2&-4x-7&-5\\2&x^2-2x-6&x-2\end{vmatrix}$$\;\;$(adding x(C3) to C2)) $\displaystyle\hspace{2.6 in}=\begin{vmatrix}x+2&-4x-7\\2&x^2-2x-6\end{vmatrix}=x^3-2x+2$

How can the trigonometric equation be proven? This question : https://math.stackexchange.com/questions/1411700/whats-the-size-of-the-x-angle has the answer $10°$. This follows from the equation $$2\sin(80°)=\frac{\sin(60°)}{\sin(100°)}\times \frac{\sin(50°)}{\sin(20°)}$$ which is indeed true , which I checked with Wolfram. * *How can this equation be proven ?

$$\begin{align}\frac{\sin 80^\circ \sin 20^\circ\sin 100^\circ }{\sin 50^\circ}&=\frac{\sin 80^\circ \sin 20^\circ\cdot 2\sin 50^\circ \cos 50^\circ }{\sin 50^\circ}\\&=2\sin 80^\circ\sin 20^\circ \cos 50^\circ\\&=2\left(-\frac 12(\cos 100^\circ-\cos 60^\circ)\right)\cos 50^\circ\\&=(\cos 60^\circ-\cos 100^\circ)\cos 50^\circ\\&=\frac 12\cos 50^\circ-\cos 100^\circ\cos 50^\circ\\&=\frac 12\cos 50^\circ-\frac 12(\cos 150^\circ+\cos 50^\circ)\\&=-\frac 12\times\left(-\frac{\sqrt 3}{2}\right)\\&=\frac{\sqrt 3}{4}\\&=\frac 12\sin 60^\circ\end{align}$$

Inequality problem: Application of Cauchy-Schwarz inequality Let $a,b,c \in (1, \infty)$ such that $ \frac{1}{a} + \frac{1}{b} + \frac{1}{c}=2$. Prove that: $$ \sqrt {a-1} + \sqrt {b-1} + \sqrt {c-1} \leq \sqrt {a+b+c}. $$ This is supposed to be solved using the Cauchy inequality; that is, the scalar product inequality.

We apply the Cauchy Schwarz Inequality to the vectors $x = \left({\sqrt {\dfrac{a-1}{a}} , \sqrt {\dfrac{b-1}{b}} , \sqrt {\dfrac{c-1}{c}}}\right) $ and $y = \left({\dfrac{1}{\sqrt{bc}},\dfrac{1}{\sqrt{ac}}, \dfrac{1}{\sqrt{ab}} }\right)$ in $\Bbb R^3$. Then, $x \cdot y \le \lVert x\rVert \lVert y\rVert$ yields, $$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}}{\sqrt{abc}} \le \sqrt{\dfrac{a-1}{a} + \dfrac{b-1}{b} + \dfrac{c-1}{c}} \times \sqrt{\dfrac{1}{bc} + \dfrac{1}{ac} + \dfrac{1}{ab}}$$ Now manipulating the $\sqrt{abc}$ across the sign we get $$ \sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \le \sqrt{\dfrac{a-1}{a} + \dfrac{b-1}{b} + \dfrac{c-1}{c}} \times \sqrt{a+ b+c} $$ So, $$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}}{\sqrt{a+ b+c}} \le \sqrt{\left({1 - \dfrac 1 a}\right) + \left({1 - \dfrac 1 b}\right) + \left({1 - \dfrac 1 c}\right) }$$ $$ \dfrac{\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1}}{\sqrt{a+ b+c}} \le \sqrt{3 - \left({\dfrac 1 a + \dfrac 1 b + \dfrac 1 c}\right)} = \sqrt{3-2} = 1$$ $\mathscr{Q.E.D}$

How to calculate the closed form of the Euler Sums We know that the closed form of the series $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}\left( {\begin{array}{*{20}{c}} {2n} \\ n \\ \end{array}} \right)}}} = \frac{1}{3}\zeta \left( 2 \right).$$ but how to evaluate the following series $$\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n\left( {\begin{array}{*{20}{c}} {2n} \\ n \\ \end{array}} \right)}}} ,\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^2}\left( {\begin{array}{*{20}{c}} {2n} \\ n \\ \end{array}} \right)}}} .$$

Using the main result proved in this question, $$ \sum_{n\geq 1}\frac{x^n}{n^2 \binom{2n}{n}}=2 \arcsin^2\left(\frac{\sqrt{x}}{2}\right)\tag{1}$$ it follows that: $$ \sum_{n\geq 1}\frac{H_n}{n^2 \binom{2n}{n}}=\int_{0}^{1}\frac{2 \arcsin^2\left(\frac{\sqrt{x}}{2}\right)-\frac{\pi^2}{18}}{x-1}\,dx \tag{2}$$ and: $$ \sum_{n\geq 1}\frac{H_n}{n\binom{2n}{n}}=\int_{0}^{1}\frac{\frac{2\sqrt{x}\arcsin\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{4-x}}-\frac{\pi}{3\sqrt{3}}}{x-1}.\tag{3}$$ The RHS of $(2)$ equals: $$ \int_{0}^{\pi/6}\frac{8\sin(t)\cos(t)}{4\sin^2(t)-1}\left(2t^2-\frac{\pi^2}{18}\right)\,dt=-\int_{0}^{\pi/6}4t\cdot\log(1-4\sin^2(t))\,\tag{4}$$ that is computable in terms of dilogarithms and $\zeta(3)$. In fact, we have: $$ \sum_{n\geq 1}\frac{H_n}{n^2\binom{2n}{n}}=\frac{\pi}{18\sqrt{3}}\left(\psi'\left(\frac{1}{3}\right)-\psi'\left(\frac{2}{3}\right)\right)-\frac{\zeta(3)}{9},\tag{5}$$ where $\psi'\left(\frac{1}{3}\right)-\psi'\left(\frac{2}{3}\right)=9\cdot L(\chi_3,2)$ with $\chi_3$ being the non-principal character $\!\!\pmod{3}$. Now, I am working on $(3)$ that should be similar. It boils down to: $$-16\int_{0}^{\pi/6}\log(1-4\sin^2(t))\left(\frac{t}{\cos^2(t)}+\tan(t)\right)\,dt=\\=\frac{2\pi^2}{3}+2\log^2(3)-2\log^2(4)-4\,\text{Li}_2\left(\frac{3}{4}\right)-16\int_{0}^{\pi/6}\log(1-4\sin^2(t))\left(\frac{t}{\cos^2(t)}\right)\,dt$$ and the last integral depends on $$ \int_{0}^{\frac{1}{\sqrt{3}}}\arctan(z)\log(1-3z^2)\,dz,\qquad \int_{0}^{\frac{1}{\sqrt{3}}}\arctan(z)\log(1+z^2)\,dz,$$ so $(2)$ is a complicated expression involving $\log(2),\log(3),\pi$ and their squares, the previous $L(\chi_3,2)$ and $\text{Li}_2\left(\frac{3}{4}\right)$.

Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem. The question simply reads : prove the following using induction: $$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$

If you wish a more direct application of induction, we have that if $$ 1^2+2^2+\cdots+(n-1)^2 < \frac{n^3}{3} $$ then $$ 1^2+2^2+\cdots+n^2 < \frac{n^3+3n^2}{3} < \frac{n^3+3n^2+3n+1}{3} = \frac{(n+1)^3}{3} $$ Similarly, if $$ 1^2+2^2+\cdots+n^2 > \frac{n^3}{3} $$ then $$ 1^2+2^2+\cdots+(n+1)^2 > \frac{n^3+3n^2+6n+3}{3} > \frac{n^3+3n^2+3n+1}{3} = \frac{(n+1)^3}{3} $$

Largest Number that cannot be expressed as 6nm +- n +- m I'm looking to find out if there is a largest integer that cannot be written as $6nm \pm n \pm m$ for $n,m$ elements of the natural numbers. For example, there are no values of $n,$m for which $6nm \pm n \pm m = 17.$ Other numbers which have no solution are $25$, $30$, and $593$. Can all numbers above a certain point be written as $6nm \pm n \pm m$ or will there always be gaps? Thanks.

This question is equivalent to the Twin Prime Conjecture. Let's look at the three cases: $$6nm+n+m =\frac{(6n+1)(6m+1)-1}{6}\\ 6nm+n-m = \frac{(6n-1)(6m+1)+1}{6}\\ 6nm-n-m = \frac{(6n-1)(6m-1)-1}{6} $$ So if $N$ is of one of these forms, then: $$6N+1=(6n+1)(6m+1)\\ 6N-1=(6n-1)(6m+1)\\ 6N+1=(6n-1)(6m-1)$$ If $6N-1$ and $6N+1$ are both primes, then there is no non-zero solutions. If $6N-1$ is composite, its factorization is of the form $(6n-1)(6m+1)$. If $6N+1$ is composite, its factorization is either of the form $(6n-1)(6m-1)$ or $(6n+1)(6m+1)$. So the question is equivalent to the Twin Prime Conjecture. I don't think this will be solved on this site, unless it is just a reference to a resolution of the conjecture elsewhere. Currently, the state of knowledge about twin primes is thin. For example, when $N=31$, $6N-1=5\cdot 37 = (6\cdot 1-1)(6\cdot 6+1)$. So with $m=1,n=6$, we get $31 = 6\cdot 1\cdot 6 +1 -6$. Alternative, $6N+1=11\cdot 17=(6\cdot 2-1)(6\cdot 3-1)$. So $(m,n)=(2,3)$ and $31=6\cdot2\cdot 3 -2-3$.

Calculate simple expression: $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ Tell me please, how calculate this expression: $$ \sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} $$ The result should be a number. I try this: $$ \frac{\left(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}\right)\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)}{\left(\sqrt[3]{\left(2 + \sqrt{5}\right)^2} - \sqrt[3]{\left(2 + \sqrt{5}\right)\left(2 - \sqrt{5}\right)} + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}\right)} = $$ $$ = \frac{2 + \sqrt{5} + 2 - \sqrt{5}}{\sqrt[3]{\left(2 + \sqrt{5}\right)^2} + 1 + \sqrt[3]{\left(2 - \sqrt{5}\right)^2}} $$ what next?

$(\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} )^3 \\ =(\sqrt[3]{2 + \sqrt{5}})^3+(\sqrt[3]{2 - \sqrt{5}} )^3+3(\sqrt[3]{2 + \sqrt{5}} ) (\sqrt[3]{2 - \sqrt{5}} )(\sqrt[3]{2 + \sqrt{5}} +\sqrt[3]{2 - \sqrt{5}} ) $ S0 $s^3=4-3s$ From this we get S = 1

Find $\sum\limits_{n=1}^{\infty}\frac{n^4}{4^n}$ So, yes, I could not do anything except observing that in the denominators, there is a geometric progression and in the numerator, $1^4+2^4+3^4+\cdots$. Edit: I don't want the proof of it for divergence or convergence only the sum.

\begin{align} n^4 = {} & \hphantom{{}+{}} An(n-1)(n-2)(n-3) \\ & {} + B n(n-1)(n-2) \\ & {} + C n(n-1) \\ & {} + D n \end{align} Find $A,B,C,D$. Then \begin{align} n^4 x^n & = An(n-1)(n-2)(n-3) x^n + \cdots \\[10pt] & = A\frac{d^4}{dx^4} x^n + B \frac{d^3}{dx^3} x^n + \cdots \end{align} Then add over all values of $n$, and then find the derivatives. Finally, apply this to the case where $x=\dfrac 1 4$.

Solve logarithmic equation $\log_{\frac{x}{5}}(x^2-8x+16)\geq 0$ Find $x$ from logarithmic equation: $$\log_{\frac{x}{5}}(x^2-8x+16)\geq 0 $$ This is how I tried: $$x^2-8x+16>0$$ $$ (x-4)^2>0 \implies x \not = 4$$ then $$\log_{\frac{x}{5}}(x^2-8x+16)\geq \log_{\frac{x}{5}}(\frac{x}{5})^0 $$ because of base $\frac{x}{5}$, we assume $x \not\in (-5,5)$, then $$x^2-8x+16 \geq 1$$ $$ (x-3)(x-5) \geq 0 \implies$$ $$ \implies x \in {(- \infty,-5) \cup (5, \infty)} \cap x\not = 4 $$ But this is wrong, because the right solution is $$x \in {(3,4) \cup (4,6)} $$ I'm sorry if I used the wrong terms, English is not my native language.

Given $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\;,$$ Here function is defined when $\displaystyle \frac{x}{5}>0$ and $\displaystyle \frac{x}{5}\neq 1$ and $(x-4)^2>0$. So we get $x>0$ and $x\neq 5$ and $x\neq 4$ If $$\displaystyle \; \bullet\; \frac{x}{5}>1\Rightarrow x>5\;,$$ Then $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\Rightarrow (x^2-8x+16)\geq 1$$ So we get $$\displaystyle x^2-8x+15\geq 0\Rightarrow (x-3)(x-5)\geq 0$$ So we get $x>5$ If $$\displaystyle \; \bullet 0<\frac{x}{5}<1\Rightarrow 0<x<5\;,$$ Then $$\displaystyle \log_{\frac{x}{5}}(x^2-8x+16)\geq 0\Rightarrow (x^2-8x+16)\leq 1$$ So we get $$\displaystyle (x-3)(x-5)\leq 0$$ So $$3\leq x<5-\left\{4\right\}$$ So our final Solution is $$\displaystyle x\in \left[3,4\right)\cup \left(4,5\right)\cup \left(5,\infty\right)$$

Another messy integral: $I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$ I found the following question in a practice book of integration:- $Q.$ Evaluate $$I=\int \frac{\sqrt{2-x-x^2}}{x^2}\ dx$$ For this I substituted $t^2=\frac {2-x-x^2}{x^2}\implies x^2=\frac{2-x}{1+t^2}\implies 2t\ dt=\left(-\frac4{x^3}+\frac 1{x^2}\right)\ dx$. Therefore $$\begin{align}I&=\int\frac {\sqrt{2-x-x^2}}{x^2}\ dx\\&=\int \left(\frac tx\right)\left(\frac{2t\ dt}{-\frac4{x^3}+\frac 1{x^2}}\right)\\&=\int \frac{2t^2\ dt}{\frac{x-4}{x^2}}\\&=\int \frac{2t^2(1+t^2)\ dt}{{x-4}\over{2-x}}\\&=\int \frac{2t^2(1+t^2)(5+4t^2-\sqrt{8t^2+9})\ dt}{\sqrt{8t^2+9}-(8t^2+9)}\end{align}$$ Now I substituted $8t^2+9=z^2 \implies t^2=\frac {z^2-9}8 \implies 2t\ dt=z/4\ dz$. So, after some simplification, you get $$\begin{align}I&=-\frac1{512}\int (z^2-9)(z+1)(z-1)^2\ dz\end{align}$$ I didn't have the patience to solve this integration after all these substitutions knowing that it can be done (I think I have made a mistake somewhere but I can't find it. There has to be an $ln(...)$ term, I believe). Is there an easier way to do this integral, something that would also strike the mind quickly? I have already tried the Euler substitutions but that is also messy.

Let $$\displaystyle I = \int \frac{\sqrt{2-x-x^2}}{x^2}dx = \int \sqrt{2-x-x^2}\cdot \frac{1}{x^2}dx\;, $$ Now Using Integration by parts $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\int\frac{1+2x}{2\sqrt{2-x-x^2}}\cdot \frac{1}{x}dx $$ So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\underbrace{\int\frac{1}{\sqrt{2-x-x^2}}dx}_{J}-\underbrace{\int\frac{1}{x\sqrt{2-x-x^2}}dx}_{K}$$ So for Calculation of $$\displaystyle J = \int\frac{1}{\sqrt{2-x-x^2}}dx = \int\frac{1}{\sqrt{\left(\frac{3}{2}\right)^2-\left(\frac{2x+1}{2}\right)^2}}dx$$ Now Let $\displaystyle \left(\frac{2x+1}{2}\right)=\frac{3}{2}\sin \phi\;,$ Then $\displaystyle dx = \frac{3}{2}\cos \phi d\phi$ So we get $$\displaystyle J = \int 1d\phi = \phi+\mathcal{C_{1}} = \sin^{-1}\left(\frac{2x+1}{3}\right)+\mathcal{C}$$ Similarly for calculation of $$\displaystyle K = \int \frac{1}{x\sqrt{2-x-x^2}}dx$$ Put $\displaystyle x=\frac{1}{u}$ and $\displaystyle dx = -\frac{1}{u^2}dt$ So we get $$\displaystyle K = -\int\frac{1}{\sqrt{2u^2-u-1}}du = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\left(u-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}}dx$$ So we get $$\displaystyle J = -\frac{\sqrt{2}}{3}\ln\left|\left(u-\frac{1}{4}\right)+\sqrt{\left(u-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C_{2}}$$ So we get $$\displaystyle J = -\frac{\sqrt{2}}{3}\ln\left|\left(\frac{1}{x}-\frac{1}{4}\right)+\sqrt{\left(\frac{1}{x}-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C_{2}}$$ So $$\displaystyle I = -\frac{\sqrt{2-x-x^2}}{x}-\sin^{-1}\left(\frac{2x+1}{3}\right)+\frac{\sqrt{2}}{3}\ln\left|\left(\frac{1}{x}-\frac{1}{4}\right)+\sqrt{\left(\frac{1}{x}-\frac{1}{4}\right)^2-\left(\frac{3}{4}\right)^2}\right|+\mathcal{C}$$

Determining values of a coefficient for which a system is and isn't consistent. Given the system : \begin{array}{ccccrcc} x & + & 2y & + & z & = & 3 \\ x & + & 3y & - & z & = & 1 \\ x & + & 2y & + & (a^2-8)z & = & a \end{array} Find values of $a$ such that the system has a unique solution, infinitely many solutions, or no solutions. I begin by placing the system into an augmented matrix. $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 1 & 2 & a^2-8 & a \\ \end{array} \right] $ I then perform operations on the matrix. $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 1 & 2 & a^2-8 & a \\ \end{array} \right] $ $: (R_3-R_1)\rightarrow$ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 0 & 0 & a^2-9 & a-3 \\ \end{array} \right] $ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 1 & 3 & -1 & 1 \\ 0 & 0 & a^2-9 & a-3 \\ \end{array} \right] $ $: (R_2-R_1; R_3/(a-3))\rightarrow$ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & a+3 & 1 \\ \end{array} \right] $ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 2 & 1 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & a+3 & 1 \\ \end{array} \right] $ $: (R_1-2R_2)\rightarrow$ $\displaystyle \left[ \begin{array}{rrr|r} 1 & 0 & 5 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & a+3 & 1 \\ \end{array} \right] $ Truth be told, at this point, I'm not too clear on how to progress (or if my approach is even ideal for this issue). I desire to find intervals across all of $a\in\mathbb{R}$ that explain where $a$ causes the system to have a unique, infinite, or inconsistent solution set.

What strikes me immediately is that if $a^2-8 = 1$, the first and last equations have the same LHS. Since choosing $a=3$ makes these equations identical, this becomes only two equations in three unknowns, so there are an infinite number of solutions. On the other hand, if you choose $a=-3$, then the first and third equations are contradictory, so there are no solutions in this case. I will leave the discussion of other values of $a$ to others.

Finding the $n$th derivative of trigonometric function.. My maths teacher has asked me to find the $n$th derivative of $\cos^9(x)$. He gave us a hint which are as follows: if $t=\cos x + i\sin x$, $1/t=\cos x - i\sin x$, then $2\cos x=(t+1/t)$. How am I supposed to solve this? Please help me with explanations because I am not good at this. And yes he's taught us Leibniz Theorem. Thanks.

De Moivre taught us that if $t=\cos x + i\sin x$ then $t^n = \cos(nx) + i\sin(nx)$ and $t^{-n} = \cos(nx) - i\sin(nx)$ so $$ t^n + \frac 1 {t^n} = 2\cos(nx). $$ Then, letting $s=1/t$, we have \begin{align} & (2\cos x)^9 =(t+s)^9 \\[10pt] = {} & t^9 + 9t^8 s + 36t^7 s^2 + 84 t^6 s^3 + 126 t^5 s^4 + 126 t^4 s^5 + 84 t^3 s^6 + 36 t^2 s^7 + 9 t s^8 + s^9 \\ & {}\qquad \text{(binomial theorem)} \\[10pt] = {} & t^9 + 9t^7 + 36 t^5 + 84 t^3 + 126 t + 126 \frac 1 t + 84 \frac 1 {t^3} + 36 \frac 1 {t^5} + 9 \frac 1 {t^7} + \frac 1 {t^9} \\[10pt] = {} & \left( t^9 + \frac 1 {t^9} \right) + 9\left( t^7 + \frac 1 {t^7} \right) + 36\left( t^5 + \frac 1 {t^5} \right) + 84 \left( t^3 + \frac 1 {t^3} \right) + 126\left( t + \frac 1 t \right) \\[10pt] = {} & 2\cos(9x) + 18 \cos(7x) + 72\cos(5x) + 168\cos(3x) + 252\cos x. \end{align} Now find the first, second, third, etc. derivatives and see if there's a pattern that continues every time you differentiate one more time.

Find the sum of the following series to n terms $\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$ Find the sum of the following series to n terms $$\frac{1}{1\cdot3}+\frac{2^2}{3\cdot5}+\frac{3^2}{5\cdot7}+\dots$$ My attempt: $$T_{n}=\frac{n^2}{(2n-1)(2n+1)}$$ I am unable to represent to proceed further. Though I am sure that there will be some method of difference available to express the equation. Please explain the steps and comment on the technique to be used with such questions. Thanks in advance !

The $n$th term is $$n^2/(4n^2-1) =$$ $$ \frac{1}{4}.\frac {(4n^2-1)+1} {4n^2-1}=$$ $$\frac{1}{4} + \frac{1}{4}.\frac {1}{4n^2-1}=$$ $$\frac{1}{4}+ \frac {1}{4}. \left(\frac {1/2}{2n-1}- \frac {1/2}{2n+1}\right).$$ Is this enough?

Inequation: quadratic difference equations Given: $$\frac{(x - 3)}{(x-4)} > \frac{(x + 4)}{(x + 3)}$$ Step 1: $$(x + 3)(x - 3) > (x + 4)(x - 4)$$ Step2 : Solving step 1: $$x^2 - 3^2 > x^2 - 4^2$$ *Step 3: $ 0 > -16 + 9$ ??? As you see, I can delete the $x^2$, but there is no point in doing that. What should be the next step?

What you did in step 1 amounts to multiply both sides by $(x-4)(x+3)$. Unfortunately, you have to reverse the inequation if this expression is negative, and leave it as is if it is positive. And as you don't know the sign of this product… You can simplify solving this inequation writing both sides in canonical form: $$\frac{x-3}{x-4}=1+\frac1{x-4}>\frac{x+4}{x+3}=1+\frac1{x+3}\iff\frac1{x-4}>\frac1{x+3}$$ Multiplying both members by $(x-4)^2(x-3)^2$ (which is positive on the domain of the inequation), we obtain: $$(x-4)(x+3)^2> (x-4)^2(x+3)\iff7(x-4)(x+3)>0\iff\begin{cases}x<-3\\\text{or}\\x>4\end{cases}$$

Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28 This is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents. Prove that if an integer $a$ is both a square and a cube then $a \equiv 0,1,9, \textrm{ or } 28 (\textrm{ mod}\ 36)$ An outline of the proof I have is Any such integer $a$ has $a = x^2$ and $a = y^3$ for some integers $x,y$ Then by the Division Algorithm, $x = 36s + b$ for some integers $s,b$ with $0 \le b \lt 36$ and $y = 36t + c$ for some integers $t,c$ with $0 \le c \lt 36$ Using binomial theorem, it is easy to show that $x^2 \equiv b^2$ and $y^3 \equiv c^3$ Then $a \equiv b^2$ and $a \equiv c^3$ By computer computation (simple script), the intersection of the possible residuals for any value of $b$ and $c$ in the specified interval is 0,1,9,28 These residuals are possible but not actual without inspection which shows $0^2 = 0^3 \equiv 0$ , $1^2 = 1^3 \equiv 1$ , $27^2 = 9^3 \equiv 9$, and $8^2 = 4^3 \equiv 28$ $\Box$ There is surely a more elegant method, can anyone hint me in the right direction.

What you did is correct, but yes, a lot of the work (especially the computer check) could have been avoided. Firstly, if $a$ is both a square and a cube, then it is a sixth power. This is because, for any prime $p$, $p$ divides $a$ an even number of times (since it is a square), and a multiple of 3 number of times (since it is a cube), so $p$ divides $a$ a multiple of 6 number of times altogether, and since this is true for any prime $p$, $a$ is a perfect sixth power. So write $a = z^6$. Next, rather than working mod $36$, it will be nice to work mod $9$ and mod $4$ instead; this is equivalent by the chinese remainder theorem. So: * *Modulo $9$, $z^6 \equiv 0 \text{ or } 1$. You can see this just by checking every integer or by applying the fact that $\varphi(9) = 6$. *Modulo $4$, $z^6 \equiv 0 \text{ or } 1$. This is easy to see; $0^6 = 0$, $1^6 = 1$, $(-1)^6 = 1$, and $2^6 \equiv 0$. So $a = z^6$ is equivalent to $0$ or $1$ mod $4$ and mod $9$. By the chinese remainder theorem, this gives four possibilities: * *$a \equiv 0 \pmod{4}, a \equiv 0 \pmod{9} \implies a \equiv 0 \pmod{36}$ *$a \equiv 0 \pmod{4}, a \equiv 1 \pmod{9} \implies a \equiv 28 \pmod{36}$ *$a \equiv 1 \pmod{4}, a \equiv 0 \pmod{9} \implies a \equiv 9 \pmod{36}$ *$a \equiv 1 \pmod{4}, a \equiv 1 \pmod{9} \implies a \equiv 1 \pmod{36}$.

Find the general integral of $ px(z-2y^2)=(z-qy)(z-y^2-2x^3).$ $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $ Find the general integral of the linear PDE $ px(z-2y^2)=(z-qy)(z-y^2-2x^3). $ My attempt to solve this is as follows: $ p=\frac{\partial z}{\partial x} $ and $ q=\frac{\partial z}{\partial y} $ $$px(z-2y^2)+qy(z-y^2-2x^3)=z(z-y^2-2x^3)$$ \begin{align*} \text{The Lagrange's auxiliary equation is:} \frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}=\frac{dz}{z(z-y^2-2x^3)} \end{align*} Now consider the 2nd and 3rd ratios, \begin{align*} \frac{dy}{y(z-y^2-2x^3)} & =\frac{dz}{z(z-y^2-2x^3)}\\ \implies \frac{dy}{y} & =\frac{dz}{z}\\ \implies \ln(y) & =\ln(z)+\ln(c_1)\\ \implies \frac{y}{z} & =c_1. \end{align*} But I am unable to get the 2nd integral surface. Kindly, help me. Thanks in advance.

Your calculus is correct. A first family of characteristic curves comes from $\frac{dx}{x(z-2y^2)}=\frac{dy}{y(z-y^2-2x^3)}$ which solution is $z=\frac{1}{c_1}y=c'_1y$ $$\frac{z}{y}=c'_1$$ A second family of characteristic curves comes from $$\frac{dx}{x(c'_1y-2y^2)}=\frac{dy}{y(c'_1y-y^2-2x^3)}$$ The solution of this ODE is : $y=\frac{c'_1}{2}\pm \sqrt{x^3+\frac{(c'_1)^2}{4}+c_2}$ $\left(y-\frac{c'_1}{2}\right)^2-x^3-\frac{(c'_1)^2}{4}=c_2$ $y^2-c'_1y -x^3=c_2$ $$y^2-z-x^3=c_2$$ The general solution of the PDE expressed on the form of implicit equation is : $$\Phi\left(\:\left(\frac{z}{y}\right)\:,\:\left(y^2-z-x^3\right)\: \right)=0$$ where $\Phi$ is an arbitrary function of two variables. Or equivalently : $$z=y\:F(y^2-z-x^3)$$ where $F$ is an arbitrary function. $\Phi$ or $F$ has to be determined to fit some boundary condition (not specified in the wording of the question).

Simple limit of a sequence Need to solve this very simple limit $$ \lim _{x\to \infty \:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$ I know how to solve these limits: by using $a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!

Can you help with O-symbols? It's all right here? $$f(x) = \sqrt[3]{3x^2}\left(1 + \frac{4}{9x} + O\left(\frac{1}{x^2}\right) - 1 - \frac{1}{x} -O \left(\frac{1}{x^2}\right)\right)= \sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right). $$ Hence $$\lim _{x\to \infty }\sqrt[3]{3x^2} \left(\frac{-5}{9x} + \frac{1}{18x^2} \right)= \lim_{x \to \infty} \sqrt[3]{3}{x^{-1/3}}^{\to 0} \left( - \frac{5}{9}+\frac{1}{18x}^{\to 0} \right) = 0.$$

Prove that $g$ is continuous at $x=0$ Given, $g(x) = \frac{1}{1-x} + 1$. I want to prove that $g$ is continuous at $x=0$. I specifically want to do an $\epsilon-\delta$ proof. Related to this Is $g(x)\equiv f(x,1) = \frac{1}{1-x}+1$ increasing or decreasing? differentiable $x=1$?. My work: Let $\epsilon >0$ be given. The challenging part for me is to pick the $\delta$. Let $\delta = 2 \epsilon$ and suppose that $|x-0| = |x| < \delta$ and $|\frac{1}{1-x}| < \frac{1}{2}$. So $$|g(x) - g(0)|$$ $$=|\frac{2-x}{1-x} - 2|$$ $$=|\frac{x}{1-x}|$$ $$=|\frac{1}{1-x}||x|$$ $$<\frac{1}{2} 2 \epsilon = \epsilon$$. So $g$ is continuous at $x=0$. Is my proof correct? EDIT: Rough work on how I picked $\delta$. Suppose $|x-0|<\delta$ and since $\delta \leq 1$, we have $|x|<1$, so $-1<x<1$ Then this implies $0<1-x<2$. The next step I'm not too sure of, I have $0<\frac{1}{1-x}<\frac{1}{2} \implies |\frac{1}{1-x}|<\frac{1}{2}$.

Hint. See my comment to your post. After doing your algebra, as you have shown, $$\left|\dfrac{1}{1-x}+1-2\right| = \left|x\right|\left|\dfrac{1}{1-x}\right|\text{.}$$ We have $|x| < \delta$. Let $\delta = 1/2$, then $$|x| < \delta \implies -\delta < x < \delta \implies 1-\delta < 1-x < 1 + \delta \implies \dfrac{1}{1+\delta} < \dfrac{1}{1-x} < \dfrac{1}{1-\delta}= 2\text{,}$$ thus implying $\left|\dfrac{1}{1-x}\right| < 2$ (notice $\dfrac{1}{1+\delta} = \dfrac{1}{1.5} \in (-2, 2)$, so we can make this claim). Choose $\delta := \min\left(\dfrac{\epsilon}{2}, \dfrac{1}{2}\right)$.

Evaluate $\int \tan^6x\sec^3x \ \mathrm{d}x$ Integrate $$\int \tan^6x\sec^3x \ \mathrm{d}x$$ I tried to split integral to $$\tan^6x\sec^2x\sec x$$ but no luck for me. Help thanks

Here is what I got from Wolfram Alpha, condensed a little bit $$\int tan^6(x)sec^3(x)dx$$ $$= \int (sec^2-1)^3sec^3(x)dx$$ $$= \int \Big(sec^9(x) -3sec^7(x)+3sec^5(x)-sec^3(x)\Big)$$ $$= \int sec^9(x)dx -3\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$ Since $\int sec^m(x) = \frac{sin(x)sec^{m-1}(x)}{m-1} + \frac{m-2}{m-1}\int sec^{m-2}(x)dx$ $$\Rightarrow \frac{1}{8}tan(x)sec^7(x) -\frac{17}{18}\int sec^7(x)dx+3\int sec^5(x)dx-\int sec^3(x)dx$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{48}\int sec^5(x)dx-\int sec^3(x)dx$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{64}\int sec^3(x)dx$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{128}[tan(x)sec(x)-\int sec(x)dx]$$ $$ = \frac{1}{8}tan(x)sec^7(x) -\frac{17}{48}tan(x)sec^5(x)+\frac{59}{192}tan(x)sec^3(x)-\frac{5}{128}tan(x)sec(x)-\frac{5}{128}log[tan(x)+sec(x)]$$ $$ = \frac{1}{384}\bigg(48tan(x)sec^7(x) -136tan(x)sec^5(x)+118tan(x)sec^3(x)-15tan(x)sec(x)-15log[tan(x)+sec(x)]\bigg)$$

Assume this equation has distinct roots. Prove $k = -1/2$ without using Vieta's formulas. Given $(1-2k)x^2 - (3k+4)x + 2 = 0$ for some $k \in \mathbb{R}\setminus\{1/2\}$, suppose $x_1$ and $x_2$ are distinct roots of the equation such that $x_1 x_2 = 1$. Without using Vieta's formulas, how can we show $k = -1/2$ ? Here is what I have done so far: $(1-2k)x_1^2 - (3k+4)x_1 + 2 = 0$ $(1-2k)x_2^2 - (3k+4)x_2 + 2 = 0$ $\to (1-2k)x_1^2 - (3k+4)x_1 + 2 = (1-2k)x_2^2 - (3k+4)x_2 + 2$ $\to (1-2k)x_1^2 - (3k+4)x_1 = (1-2k)x_2^2 - (3k+4)x_2$ $\to (1-2k)x_1^2 - (3k+4)x_1 = (1-2k)(1/x_1)^2 - (3k+4)(1/x_1)$ $\to (1-2k)[x_1^2 - (1/x_1)^2] - (3k+4)[x_1 - (1/x_1)] = 0$ $\to (1-2k)[x_1 - (1/x_1)][x_1 + (1/x_1)] - (3k+4)[x_1 - (1/x_1)] = 0$ $\to (1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$ or $[x_1 - (1/x_1)] = 0$ $\to (1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$, $x_1 = 1$ or $x_1 = -1$ Since the later two cases violate the distinct roots assumption, we have: $(1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$ This gives: The answer is $x_1 = \frac{5-i\sqrt{39}}{8}$ or $x_1 = \frac{5+i\sqrt{39}}{8}$. How do I get that (without Vieta's)?

The problem is find $k$ not $x_1$ isn't it? If so take the equation you arrived at before you started doing computer algebra: $$(1-2k)[x_1 + (1/x_1)] - (3k+4) = 0$$ and multiply by $x_1$ to get $$(1-2k)x_1^2 - (3k+4) x_1 + (1 -2k) = 0$$ but you know that $$(1-2k)x_1^2 - (3k+4) x_1 + 2 = 0$$ so you must have $1 - 2k = 2$, i.e., $k = -\frac{1}{2}$. (Using the Vieta formulas which just amount to $(x - x_1)(x - x_2) = x^2 - (x_1 + x_2)x + x_1x_2$ for a quadratic seems a much simpler approach to me.)

Application of Differentiation Prove that if the curve $y = x^3 + px + q$ is tangent to the x-axis, then $$4p^3 + 27q^2 = 0$$ I differentiated $y$ and obtained the value $3x^2 + p$. If the curve is tangent to the x-axis, it implies that $x=0$ (or is it $y = 0$?). How do I continue to prove the above statement? Thanks. If I substitute in $x=0$, I will obtain $y= q$? Are my above steps correct? Please guide me. Thank you so much!

Notice, we have $$y=x^3+px+q$$ $$\frac{dy}{dx}=3x^2+p$$ Since, the x-axis is tangent to the curve at some point where $y=0$ & slope $\frac{dy}{dx}=0$ hence, we have $$\left(\frac{dy}{dx}\right)_{y=0}=0$$ $$3x^2+p=0\iff x^2=\frac{-p}{3}\tag 1$$ Now, at the point of tangency with the x-axis we have $$y=0\iff x^3+px+q=0$$ $$x^3+px+q=0$$ $$(x^3+px)=-q$$$$ (x^3+px)^2=(-q)^2$$ $$x^6+2px^4+p^2x^2=q^2$$ $$(x^2)^3+2p(x^2)^4+p^2x^2=q^2$$ Setting the value of $x^2$ from (1), we get $$\left(\frac{-p}{3}\right)^3+2p\left(\frac{-p}{3}\right)^2+p^2\left(\frac{-p}{3}\right)=q^2$$ $$-\frac{p^3}{27}+\frac{2p^3}{9}-\frac{p^3}{3}=q^2$$ $$-\frac{4p^3}{27}=q^2\iff -4p^3=27q^2$$ $$\color{red}{4p^3+27q^2=0}$$

Troubles with solving $\sqrt{2x+3}-\sqrt{x-10}=4$ I have been trying to solve the problem $\sqrt{2x+3}-\sqrt{x-10}=4$ and I have had tons problems of with it and have been unable to solve it. Here is what I have tried-$$\sqrt{2x+3}-\sqrt{x-10}=4$$ is the same as $$\sqrt{2x+3}=4+\sqrt{x-10}$$ from here I would square both sides $$(\sqrt{2x+3})^2=(4+\sqrt{x-10})^2$$ which simplifies to $$2x+3=16+x-10+8\sqrt{x-10}$$ I would then isolate the radical $$x-3=8\sqrt{x-10}$$ then square both sides once again $$(x-3)^2=(8\sqrt{x-10})^2$$ which simplifies to $$x^2-6x+9=8(x-10)$$ simplified again $$x^2-6x+9=8x-80$$ simplified once again $$x^2-14x+89=0$$ this is where I know I have done something wrong because the solution would be $$14 \pm\sqrt{-163 \over2}$$ I am really confused and any help would be appreciated

Note that when you square something like $a\sqrt{b}$ you get $a^2b$. Thus, you should get: $\begin{align} x^2-6x+9 &= 64(x-10)\\ x^2-6x+9 &= 64x-640\\ x^2-70x+649 &= 0\\ (x-11)(x-59) &= 0\\ \therefore \boxed{x=11,59}. \end{align}$

Evaluation of $\int_0^\infty \frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx$ Evaluation of $\displaystyle \int_0^\infty \frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx$ $\bf{My\; Try::}$ Let $$I= \int_{0}^{\infty}\frac{1}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx \tag 1 $$ Let $\displaystyle x=\frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}$ and changing limit, we get So $$I = \int_0^\infty \frac{t^{102}}{\left[t^4+(1+\sqrt{2})t^2+1\right]\cdot \left[t^{100}-t^{98}+\cdots+1\right]}dt$$ So $$\displaystyle I = \int_0^\infty \frac{x^{102}}{\left[x^4+(1+\sqrt{2})t^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]}dx \tag 2$$ So we get $$2I = \int_0^\infty \frac{1+x^{102}}{\left[x^4+(1+\sqrt{2})x^2+1\right]\cdot \left[x^{100}-x^{98}+\cdots+1\right]} \,dx$$ Now Using Geometric Progression series, We can write $$ 1-x^2+x^4-\cdots-x^{98}+x^{100} = \left(\frac{x^{102+1}}{1+x^2}\right)$$ so we get $$2I = \int_0^\infty \frac{1+x^2}{x^4+ax^2+1}dx\;,$$ Where $a=(\sqrt{2}+1)$ So we get $$2I = \int_0^\infty \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+\left(\sqrt{a+2}\right)^2} dx = \frac{1}{\sqrt{a+2}}\left[\tan^{-1}\left(\frac{x^2-1}{x\cdot \sqrt{a+2}}\right)\right]_0^\infty$$ So we get $$2I = \frac{\pi}{\sqrt{a+2}}\Rightarrow I = \frac{\pi}{2\sqrt{3+\sqrt{2}}}$$ My Question is can we solve the Integral $\bf{\displaystyle \int_0^\infty \frac{1+x^2}{x^4+ax^2+1}dx}$ Using any other Method Means Using Complex analysis or any other. Thanks.

I think your way of evaluating the integral is very short and elegant. Nevertheless, you can do partial fraction decomposition. You'll end up with (as long as I did not do any mistakes) $$ \frac{1}{\sqrt{2+a}}\biggl[\arctan\Bigl(\frac{\sqrt{2-a}+2x}{\sqrt{2+a}}\Bigr)+\arctan\Bigl(\frac{\sqrt{2-a}-2x}{\sqrt{2+a}}\Bigr)\biggr] $$ as a primitive. Edit That way of writing it is not so good, since $a>2$. It is better to write $$ x^4+ax^2+1=\Bigl(x^2+\frac{a}{2}-\frac{\sqrt{a^2-4}}{2}\Bigr)\Bigl(x^2+\frac{a}{2}+\frac{\sqrt{a^2-4}}{2}\Bigr) $$ and then do partial fraction decomposition. The resulting primitive then reads (watch out for typos!) $$ \begin{aligned} \frac{1}{\sqrt{2}\sqrt{a^2-4}}\biggl[&\frac{2-a+\sqrt{a^2-4}}{\sqrt{a-\sqrt{a^2-4}}}\arctan\Bigl(\frac{\sqrt{2}x}{\sqrt{a-\sqrt{a^2-4}}}\Bigr)\\ &\qquad+\frac{a-2+\sqrt{a^2-4}}{\sqrt{a+\sqrt{a^2-4}}}\arctan\Bigl(\frac{\sqrt{2}x}{\sqrt{a+\sqrt{a^2-4}}}\Bigr)\biggr]. \end{aligned} $$

Evaluate limit without L'hopital $$\lim_{x\to2}\dfrac{\sqrt{x^2+1}-\sqrt{2x+1}}{\sqrt{x^3-x^2}-\sqrt{x+2}}$$ Please help me evaluate this limit. I have tried rationalising it but it just can't work, I keep ending up with $0$ at the denominator... Thanks!

$$\displaystyle \lim_{x\rightarrow 2}\frac{\sqrt{x^2+1}-\sqrt{2x+1}}{\sqrt{x^3-x^2}-\sqrt{x+2}}\times \frac{\sqrt{x^2+1}+\sqrt{2x+1}}{\sqrt{x^2+1}+\sqrt{2x+1}}\times \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^3-x^2}+\sqrt{x+2}}$$ So we get $$\displaystyle \lim_{x\rightarrow 2}\left[\frac{x^2-2x}{x^3-x^2-x-2}\times \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^2+1}+\sqrt{2x+1}}\right]$$ So we get $$\displaystyle \lim_{x\rightarrow 2}\frac{x(x-2)}{(x-2)\cdot (x^2+x+1)}\times \lim_{x\rightarrow 2} \frac{\sqrt{x^3-x^2}+\sqrt{x+2}}{\sqrt{x^2+1}+\sqrt{2x+1}}$$ So we get $$\displaystyle = \frac{2}{7}\times \frac{4}{2\sqrt{5}} = \frac{4}{7\sqrt{5}}$$

Evaluating $\int \frac{\sin x}{\sin 4x}\mathrm dx$ $$\int \frac{\sin x}{\sin 4x}\mathrm dx$$ I have tried to do this by expanding $\sin 4x$ but it was worthless.

Using $\displaystyle \bullet\; \sin 2x = 2\sin x\cdot \cos x$ and $\; \bullet\; \cos 2x = 1-2\sin^2 x$ Let $$\displaystyle I = \int\frac{\sin x}{2\sin 2x\cdot \cos 2x}dx = \int\frac{\sin x}{4\sin x\cos x\cdot \cos 2x}dx = \frac{1}{4}\int\frac{1}{\cos x\cdot \cos 2x}dx$$ Now $$\displaystyle I = \frac{1}{4}\int\frac{\cos x}{\cos^2 x\cdot (1-2\sin^2x)}dx = \frac{1}{4}\int\frac{\cos x}{(1-\sin^2x)\cdot (1-2\sin^2x)}dx$$ Now Put $\sin x= t\;,$ Then $\cos xdx = dt$ So we get $$\displaystyle I = \frac{1}{4}\int\frac{1}{(1-t^2)(1-2t^2)}dt = \frac{1}{4}\int\frac{1}{(2t^2-1)(t^2-1)}dt$$ So we get $$\displaystyle I = -\frac{1}{4}\int\left[\frac{2}{(2t^2-1)}-\frac{1}{t^2-1}\right] = -\frac{1}{4}\int\frac{1}{t^2-\left(\frac{1}{\sqrt{2}}\right)^2}dt+\frac{1}{4}\int\frac{1}{t^2-1}dt$$ So we get $$\displaystyle I = -\frac{1}{4}\cdot \frac{1}{\sqrt{2}}\ln\left|\frac{\sqrt{2}t-1}{\sqrt{2}t+1}\right|+\frac{1}{4}\cdot \frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+\mathcal{C}$$ So we get $$\displaystyle I = -\frac{1}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1}\right|+\frac{1}{8}\ln\left|\frac{\sin x-1}{\sin x+1}\right|+\mathcal{C}$$

Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$ $$\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$$ So $$\frac{1-\sqrt[3]{4-3x}}{x-1} \cdot \frac{1+\sqrt[3]{4-3x}}{1+\sqrt[3]{4-3x}}$$ Then $$\frac{1-(4-3x)}{(x-1)(1+\sqrt[3]{4-3x})}$$ That's $$\frac{3\cdot \color{red}{(x-1)}}{\color{red}{(x-1)}(1+\sqrt[3]{4-3x})}$$ Finally $$\frac{3}{(1+\sqrt[3]{4-3x})}$$ But this evaluates to $$\frac{3}{2}$$ When the answer should be $$1$$ Where did I fail?

I think you overlooked this multiplication $(1+(4-3x)^{1/3})(1-(4-3x)^{1/3})$ which equals $1-(4-3x)^{2/3}$ not $1-(4-3x)$

Integral $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$ Here is an integral I derived while evaluating another. It appears to be rather tough, but some here may not be so challenged :) Show that: $$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\frac{11}{9}\zeta(3)-\frac{\pi}{72\sqrt{3}}\left(5\psi_{1}\left(\frac13\right)+4\psi_{1}\left(\frac23\right)-3\psi_{1}\left(\frac56\right)\right)$$ $$=\frac{11}{9}\zeta(3)+\frac{4\pi^{3}}{27\sqrt{3}}-\frac{2\pi}{9\sqrt{3}}\psi_{1}\left(\frac13\right)=\frac{11}{9}\zeta(3)-\frac{4\pi}{9}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)$$ $$=\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)-2\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)-\frac{4\pi}{9}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)$$ I attempted all kinds of 'starts' to no satisfactory end, but things look promising. There are some mighty sharp folks here that may be better at deriving the solution. I thought perhaps the identity: $$\frac{\log^{2}(1-(x-x^{2}))}{x}=2\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}x^{n}(1-x)^{n+1}$$ or the Beta function could be used if given enough ingenuity. This led me to the no-less-imposing Euler/reciprocal of central binomial coefficients sum below. It would be great to just show the middle sum is equivalent to the right sum: $$1/4\sum_{n=1}^{\infty}\frac{H_{n}n\Gamma^{2}(n)}{(n+1)(2n+1)\Gamma(2n)}=1/2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(2n+1)\binom{2n}{n}}=1/3\zeta(3)-2/3\sum_{n=1}^{\infty}\frac{1}{n^{3}\binom{2n}{n}}$$ Is there a general form for $$\sum_{n=1}^{\infty}\frac{H_{n}}{\binom{2n}{n}}x^{n}?$$ I tried starting with the identity: $$\sum_{n=1}^{\infty}\frac{\Gamma^{2}(n)}{\Gamma(2n)}x^{n-1}=\frac{4\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}}$$ and using various manipulations to hammer into the needed form. This, too, turned monstrous. There appears to be a relation to Clausen functions (as with other log integrals such as $\int_{0}^{1}\frac{\log(x)}{x^{2}-x+1}dx$), to wit: I use Cl for sin and CL for cos Clausen functions $$\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})}{k^{2}}=\frac{\sqrt{3}}{72}\left(\psi_{1}(1/6)+\psi_{1}(1/3)-\psi_{1}(2/3)-\psi_{1}(5/6)\right)$$ $$=\frac{\sqrt{3}}{6}\psi_{1}(1/3)-\frac{\pi^{2}\sqrt{3}}{9}$$ and $$\operatorname{Cl}_{3}\left(\frac{\pi}{3}\right)-\operatorname{Cl}_{3}\left(\frac{2\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})}{k^{3}}-2\sum_{k=1}^{\infty}\frac{\cos(\frac{2\pi k}{3})}{k^{3}}=\frac{11}{9}\zeta(3)$$ Another approach. I also broke the integral up as such: $$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\int_{0}^{1}\frac{\log^{2}(1-xe^{\frac{\pi i}{3}})}{x}dx+2\int_{0}^{1}\frac{\log(1-xe^{\pi i/3})\log(1-xe^{-\pi i/3})}{x}dx+\int_{0}^{1}\frac{\log^{2}(1-xe^{-\pi i/3})}{x}dx$$ The middle integral right of the equal sign is the one that has given me the fit. I think this is a fun and head-scratchin' integral that has led me to other discoveries. Maybe a generalization could be obtained with other powers of log such as n = 3, 4, etc. I wonder if they can also be evaluated in terms of Clausens and then into closed forms involving $\zeta(n+1)$ and derivatives of digamma, $\psi_{n-1}(z)?$. Another easier one is $$\int_{0}^{1}\frac{\log(x^{2}-x+1)}{x}dx=\frac{-\pi^{2}}{18}=\frac{-1}{3}\zeta(2)?$$

Asset at our disposal: $$\sum\limits_{n=0}^{\infty} \frac{x^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}} = 4(\arcsin (x/2))^2$$ Differentiation followed by the substitution $x \to \sqrt{x}$ gives: $\displaystyle \sum\limits_{n=0}^{\infty} \frac{x^{n}}{(2n+1)\binom{2n}{n}} = \frac{2\arcsin (\sqrt{x}/2)}{\sqrt{x}\sqrt{1-(\sqrt{x}/2)^2}}$ Thus, we split the series as: $$ \sum\limits_{n=0}^{\infty} \frac{H_n}{(n+1)(2n+1)\binom{2n}{n}} \\= \sum\limits_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}} - \sum\limits_{n=0}^{\infty} \frac{1}{(n+1)^2(2n+1)\binom{2n}{n}}$$ The first series can be dealt with using, $\displaystyle\frac{H_{n+1}}{n+1} = -\int_0^1 x^n\log(1-x)\,dx$ \begin{align*}\sum\limits_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}&= -\sum\limits_{n=0}^{\infty} \int_0^1 \frac{x^n\log(1-x)}{(2n+1)\binom{2n}{n}}\,dx\\ &= -2\int_0^1 \frac{\arcsin (\sqrt{x}/2)\log (1-x)}{\sqrt{x}\sqrt{1-(\sqrt{x}/2)^2}}\,dx\\ &= -8\int_0^{1/2} \frac{\arcsin x \cdot \log (1-4x^2)}{\sqrt{1-x^2}}\,dx\\ &= -8\int_0^{\pi/6} \theta \log (1-4\sin^2 \theta)\,d\theta\\ &= -8\int_0^{\pi/6} \theta \log \left(4\sin\left(\theta + \frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}-\theta\right)\right) \end{align*} Using the Fourier Series, $\displaystyle \log (2\sin \theta) = -\sum\limits_{n=1}^{\infty} \frac{\cos 2n\theta}{n}$ we get: \begin{align*}&\int_0^{\pi/6} \theta\log \left(2\sin\left(\frac{\pi}{6}+\theta\right)\right)\,d\theta \\&= -\sum\limits_{n=1}^{\infty} \int_0^{\pi/6} \frac{\theta\cos \left(\dfrac{n\pi}{3}+2n\theta\right)}{n}\,d\theta\\&= -\frac{\pi}{12}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2}-\frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{\cos (2n\pi/3)}{n^3} +\frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^3}\end{align*} and, \begin{align*}&\int_0^{\pi/6} \theta\log \left(2\sin\left(\frac{\pi}{6}-\theta\right)\right)\,d\theta \\&= -\sum\limits_{n=1}^{\infty} \int_0^{\pi/6} \frac{(\pi/6 - \theta)\cos \left(2n\theta\right)}{n}\,d\theta\\&= -\frac{1}{4}\zeta(3)+\frac{1}{4}\sum\limits_{n=1}^{\infty}\frac{\cos (n\pi/3)}{n^3}\end{align*} Hence, $$\sum\limits_{n=0}^{\infty}\frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}} = -\frac{2}{9}\zeta(3) + \frac{2\pi}{3}\sum\limits_{n=1}^{\infty}\frac{\sin (2n\pi/3)}{n^2}$$ Similarly we may deal with the second series: \begin{align*}\sum\limits_{n=0}^{\infty} \frac{1}{(n+1)^2(2n+1)\binom{2n}{n}} &= 8\int_0^{1/2} \frac{\arcsin^2 (x)}{x}\,dx \\&= -4\zeta(3)+4\sum\limits_{n=1}^{\infty}\frac{\cos (n\pi/3)}{n^3}+\frac{4\pi}{3}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2}\end{align*} Combining the results we get: \begin{align*}\sum\limits_{n=1}^{\infty} \frac{H_n}{(n+1)(2n+1)\binom{2n}{n}} &= \frac{22}{9}\zeta(3) - \frac{2\pi}{3}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2} \\&= \frac{22}{9}\zeta(3) - \frac{\pi}{9\sqrt{3}}\left(\psi'\left(\frac{1}{3}\right) - \psi'\left(\frac{2}{3}\right)\right)\end{align*}

Tile a 1 x n walkway with 4 different types of tiles... Suppose you are trying to tile a 1 x n walkway with 4 different types of tiles: a red 1 x 1 tile, a blue 1 x 1 tile, a white 1 x 1 tile, and a black 2 x 1 tile a. Set up and explain a recurrence relation for the number of different tilings for a sidewalk of length n. b. What is the solution of this recurrence relation? c. How long must the walkway be in order have more than 1000 different tiling possibilities? This is a problem on my test review and I have no idea how to approach it. We did a similar example in class but only using 1x1 tiles that were all the same (no separate tile colors or sizes). Any help/hints would be appreciated. Thanks in advance! My initial thought is something along the lines of finding all the ways to use the 1 x 1 tiles then multiplying that by 3 to consider each color variant (don't know how the 2x1 factors in to this though).

Call the number of tilings of length $n$ $t_n$, then to get a tiling of length $n$, you take one of length $n - 1$ and add a red, a white or a blue tile (3 ways); add a black tile to one of length $n - 2$. I.e.: $\begin{equation*} t_{n + 2} = 3 t_{n + 1} + t_n \end{equation*}$ Directly we find $t_0 = 1$, $t_1 = 3$. Define the generating function: $\begin{equation*} T(z) = \sum_{n \ge 0} t_n z^n \end{equation*}$ Take the recurrence, multiply by $z^n$ and sum over $n \ge 0$, recognize resulting sums: $\begin{align*} \sum_{n \ge 0} t_{n + 2} z^n &= 3 \sum_{n \ge 0} t_{n + 1} z^n + \sum_{n \ge 0} t_n z^n \\ \frac{T(z) - t_0 - t_1 z}{z^2} &= 3 \frac{A(z) - t_0}{z} + A(z) \end{align*}$ Solve for $A(z)$, split into partial fractions: $\begin{align*} T(z) &= \frac{1}{1 - 3 z - z^2} \\ &= \frac{2 (\sqrt{13} + 3)}{\sqrt{13}} \cdot \frac{1}{1 + \frac{2}{3 + \sqrt{13}} z} + \frac{2 (\sqrt{13} - 3)}{\sqrt{13}} \cdot \frac{1}{1 + \frac{2}{3 - \sqrt{13}} z} \end{align*}$ Need to extract the coefficients from these geometric series: $\begin{equation*} [z^n] T(z) = \frac{2 (\sqrt{13} + 3)}{\sqrt{13}} \cdot \left( \frac{2}{3 + \sqrt{13}} \right)^n + \frac{2 (\sqrt{13} - 3)}{\sqrt{13}} \cdot \left( \frac{2}{3 - \sqrt{13}} \right)^n \end{equation*}$ Note that: $\begin{align*} \frac{2 (\sqrt{13} + 3)}{\sqrt{13}} &= 2.8321 \\ \frac{2 (\sqrt{13} - 3)}{\sqrt{13}} &= 1.1679 \\ \frac{2}{3 + \sqrt{13}} &= 0.30278 \\ \frac{2}{3 - \sqrt{13}} &= 3.3028 \end{align*}$ Note that already by $n = 1$ the first term is less than 1, so a very good approximation is $t_n = 1.1679 \cdot 3.3028^n$. To get $t_n = 1000$, you need: $\begin{align*} 1000 &= 1.1679 \cdot 3.3028^n \\ n &= 5.65 \end{align*}$ Thus you need at least length 6.

If $x^2+y^2+xy=1\;,$ Then minimum and maximum value of $x^3y+xy^3+4\;,$ where $x,y\in \mathbb{R}$ If $x,y\in \mathbb{R}$ and $x^2+y^2+xy=1\;,$ Then Minimum and Maximum value of $x^3y+xy^3+4$ $\bf{My\; Try::} $Given $$x^2+y^2+xy=1\Rightarrow x^2+y^2=1-xy\geq 0$$ So we get $$xy\leq 1\;\;\forall x\in \mathbb{R}$$ and $$x^2+y^2+xy=1\Rightarrow (x+y)^2=1+xy\geq0$$ So we get $$xy\geq -1\;\;\forall x\in \mathbb{R}$$ So we get $$-1\leq xy\leq 1$$ $$\displaystyle f(x,y) = xy(x^2+y^2)+4 = xy(1-xy)+4 = (xy)-(xy)^2+4 = -\left[(xy)^2-xy-4\right]$$ So $$\displaystyle f(x,y) = -\left[\left(xy-\frac{1}{2}\right)^2-\frac{17}{4}\right] = \frac{17}{4}-\left(xy-\frac{1}{2}\right)^2$$ So $$\displaystyle f(x,y)_{\bf{Min.}} = \frac{17}{4}-\left(-1-\frac{1}{2}\right)^2 = 2\;,$$ which is occur when $xy=-1$ But I did not understand how can i calculate $f(x,y)_{\bf{Max.}}$ Plz Help me, Thanks

Using your second last line, $$f(x,y) = \frac{17}{4} - (xy-\frac 12)^2 $$ now let $\displaystyle xy=u$, $x^2 + y^2 + xy = 1$ becomes $(x+y)^2 = 1+u$ Therefore $x,y$ are roots of the quadratic $k^2 \pm \sqrt(1+u) k + u = 0.$ If $x, y$ are real, discriminant is non negative, solving this gets $\displaystyle u\leq \frac{1}{3}$ therefore $\displaystyle xy\leq \frac{1}{3}.$ Minimum value of $f(x,y)$ occurs when $\displaystyle \left(xy-\frac{1}{2}\right)^2$ is minimum. This occurs when $\displaystyle xy=\frac{1}{3}$ as shown above. Therefore, max value $\displaystyle = \frac{17}{4} - \left(\frac{1}{3}-\frac{1}{2}\right)^2 = \frac{38}{9},$ which is what Wolfram Alpha says

A non-linear homogeneous diophantine equation of order 3 I'm a math teacher and one of my student have come to me with several questions. One of them is the following; Prove that there is no positive integral solution of the equation $$x^2y^4+4x^2y^2z^2+x^2z^4=x^4y^2+y^2z^4.$$ I have tried several hours but failed. Help me if you have any opinion. Note. It can be factored so that written equivalently $$f(x,y,z)f(x,-y,z)=0$$ where $f(x,y,z)=xy(x+y)+z^2(x-y)$. So, it is equivalent to prove that $f(x,y,z)=0$ implies that $xyz=0$ over integers.

Let $f(x,y,z) = xy(x+y) +z^2(x-y)$. As you already have in the post, we need to prove that $f(x,y,z)=0$ implies $xyz=0$ over integers. First, rewrite $f(x,y,z) = y x^2 + (y^2+z^2) x - z^2 y$ and consider $f(x,y,z)=0$ as a quadratic equation in $x$. Then the discriminant of the quadratic equation is $$ D=(y^2 + z^2)^2 + 4y^2z^2.$$ For the equation to have integer solution $x$, we must have that $D$ is a perfect square. We prove that $D$ cannot be a perfect square if $yz\neq 0$. This is a Diophantine equation $$x^4 + 6x^2y^2 + y^4 = z^2.$$ Then we need to prove that $xy=0$. This equation is described in Mordell's book 'Diophantine Equations', page 17. The idea is first proving that $$ x^4 - y^4 = z^2, \ \ (x,y)=1$$ gives $xyz=0$. This was proven there by using Pythagorian triple and infinite descent. As a corollary, we have $$ x^4 + y^4 = 2z^2, \ \ (x,y)=1$$ has only integer solutions $x^2=y^2=1$. To prove this, note that $x, y$ are both odd and $$ z^4 - x^4y^4 = \left(\frac{x^4-y^4}{2}\right)^2.$$ Then substituting $x+y$ for $x$, and $x-y$ for $y$, we have $$ (x+y)^4 + (x-y)^4 = 2( x^4 + 6x^2y^2 + y^4) = 2z^2.$$ Thus, we see that the Diophantine equation $$ x^4 + 6x^2y^2 + y^4 = z^2$$ gives $(x+y)^2=(x-y)^2$. Therefore, $xy=0$. Going back to the original problem, we now have that $D$ is not a perfect square if $yz\neq 0$. Hence, any integer solution to $f(x,y,z)=0$ must satisfy $yz=0$.

Stuck solving a logarithmic equation $$\log _{ 2 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } -\log _{ 2 }{ 2x } $$ Steps I took: $$\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } =\log _{ 4 }{ 4x^{ 6 } } -\frac { \log _{ 4 }{ 2x } }{ \log _{ 4 }{ 2 } } $$ $$2\log _{ 4 }{ 2x } +2\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$ $$4\log _{ 4 }{ 2x } =\log _{ 4 }{ 4x^{ 6 } } $$ At this point I get stuck I don't think turning this into $\log _{ 4 }{ (2x)^{ 4 } } =\log _{ 4 }{ 4x^{ 6 } } $ is the right answer. It leads to: $16x^{ 4 }=4x^{ 6 }$ and this has what seem to be extraneous solutions.

$$ \log_{2} 2x = \log_{4} 4x^{6} - \log_{2}2x \quad \text{iff} \quad \frac{\log 2x}{\log 2} = \frac{\log 4x^{6}}{\log 4} - \frac{\log 2x}{\log 2};\\ \frac{\log 2x}{\log 2} = \frac{\log 4x^{6}}{\log 4} - \frac{\log 2x}{\log 2} \quad \text{iff} \quad 2\log 2x = \log 4x^{6} - 2\log 2x;\\ 2\log 2x = \log 4x^{6} - 2\log 2x \quad \text{iff} \quad \log (2x)^{4} = \log 4x^{6};\\ \log (2x)^{4} = \log 4x^{6} \quad \text{iff} \quad x = 2. $$

Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$ Steps I took: $$\frac { \log _{ 2 }{ x^{ 2 } } }{ \log _{ 2 }{ 8 } } =\log _{ 2 }{ x-1 } $$ $$\frac { \log _{ 2 }{ x^{ 2 } } }{ 3 } =\log _{ 2 }{ x-1 } $$ $$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$ $$2\log _{ 2 }{ x } =3\log _{ 2 }{ x-1 } $$ $$\log _{ 2 }{ x } =\frac { 3 }{ 2 } \log _{ 2 }{ x-1 } $$ $$\log _{ 2 }{ x } =\log _{ 2 }{ (x-1)^{ \frac { 3 }{ 2 } } } $$ This method seems to be very inefficient and I don't know how I would go from here. Can someone please point me in the right direction. Hints only please. No actual solution.

Notice, $\ \ \large \log_{a^n}(b)=\frac{1}{n}\log_a(b)$ Now, we have $$2\log_8x=\log_2x-1$$ $$2\log_{2^3}x=\log_2x-1$$ $$\frac{2}{3}\log_{2}x=\log_2x-1$$ $$\frac{1}{3}\log_{2}x=1$$ $$\log_2x=3\implies x=2^3=\color{red}{8}$$

The possible number of blue marbles is A boy has a collection of blue and green marbles. The number of blue marbles belong to the sets $\{2,3,4,\ldots,13\}$. If two marbles are chosen simultaneously and at random from this collection, then the probability that they have different colour is $\frac{1}{2}$. The possible number of blue marbles is $$(A)\ 2\hspace{1cm}(B)\ 3\hspace{1cm}(C)\ 6\hspace{1cm}(D)\ 10$$ Let there are $x$ blue and $y$ green marbles in the collection. Since two marbles are chosen chosen simultaneously and at random from this collection, the possible color combinations are blue-green, green-green, blue-blue. But now i am stuck. Please help me.

Suppose, there are $m$ blue and $n$ green marbles. There are $\binom{m+n}{2}=\frac{(m+n)(m+n-1)}{2}$ ways to choose $2$ marbles. There are $mn$ ways to choose $2$ marbles with different colors. The probability of getting two marbles with different colours is therefore $$\frac{2mn}{(m+n)(m+n-1)}$$ So, the probability is $\frac{1}{2}$, if and only if $$(m+n)(m+n-1)=4mn$$ holds. $1)\ m=2\ :\ n^2+3n+2=8n$ has no solution in $\mathbb N$ $2)\ m=3\ :\ n^2+5n+6=12n$ has the solutions $1$ and $6$. $3)\ m=6\ :\ n^2+11n+30=24n$ has the solutions $3$ and $10$. $4)\ m=10\ :\ n^2+19n+90=40n$ has the solutions $6$ and $15$.

An example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge Give an example of a series $\sum a_n$ that converges conditionally but $\sum a_n^3$ does not converge conditionally. I've come up with an example. $\frac{1}{\sqrt[3]2}-\frac{1}{2\sqrt[3]2}-\frac{1}{2\sqrt[3]2}+\frac{1}{\sqrt[3]3}-\frac{1}{3\sqrt[3]3}-\frac{1}{3\sqrt[3]3}-\frac{1}{3\sqrt[3]3}+\cdots$. While the sum of the cubes is $\frac{1}{2}-\frac{1}{8\cdot 2}-\frac{1}{8\cdot 2}+\frac{1}{3}-\frac{1}{27\cdot 3}-\frac{1}{27\cdot 3}-\frac{1}{27\cdot 3}+\cdots$ Now the series seems to converge to 0, however, I cannot show using an epsilon argument that it does. Also, the sum of the cubes looks like $\frac{1}{4}\cdot \frac{1}{2}+\frac{8}{9}\cdot \frac{1}{3}+ \frac{15}{16}\cdot \frac{1}{4}+\cdots$, so I can see that it diverges, but likewise, cannot supply this with a rigorous argument. I would greatly appreciate it if anyone can help me with this part.

Consider , $a_n=\frac{(-1)^n}{n}$. Then $a_n$ is conditionally convergent. But , $a_n^3=\frac{(-1)^n}{n^3}$ is NOT conditionally convergent ; as it is absolutely convergent.

What is the name for the mathematical property involving addition or subtraction of fractions over a common denominator? For any number $x$ where $x\in\Bbb R$ and where $x\ne0$, what is the mathematical property which states that: $${1-x^2\over x} = {1\over x} - {x^2\over x}$$

Just the computation rules for fractions: $$ \frac{a - b}{c} = \frac{a + (-b)}{c} \\ \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c} \\ \frac{(-b)}{c} = -\frac{b}{c} $$ for $a = 1$, $b = -x^2$ and $c = x \ne 0$.

Is $\frac{4x^2 - 1}{3x^2}$ an abundancy outlaw? Let $\sigma(x)$ denote the sum of the divisors of $x$. For example, $\sigma(6) = 1 + 2 + 3 + 6 = 12$. We call the ratio $I(x) = \sigma(x)/x$ the abundancy index of $x$. A number $y$ which fails to be in the image of the map $I$ is said to be an abundancy outlaw. My question is: Is $$\frac{4x^2 - 1}{3x^2}$$ an abundancy outlaw? In this paper, it is mentioned that if $r/s$ is an abundancy index with $\gcd(r,s) = 1$, then $r \geq \sigma(s)$. Note that $$\gcd(4x^2 - 1, 3x^2) = \gcd(x^2 - 1, 3x^2) = 1$$ if $x \not\equiv 1 \pmod 3$. Hence, if $(4x^2 - 1)/3x^2$ is an abundancy index with $x \not\equiv 1 \pmod 3$, then $4x^2 - 1 \geq \sigma(3x^2)$. Here is where I get stuck. (I would like to affirmatively answer my question via the contrapositive.) Lastly, if $(4x^2 - 1)/3x^2$ were an abundancy index $I(y)$, then $${3x^2}\sigma(y) = y(4x^2 - 1),$$ so that $${3x^2}\left(\sigma(y) - y\right) = y(x^2 - 1).$$ Since $\gcd(x^2, x^2 - 1) = 1$, I am sure that $x^2 \mid y$ and $(x^2 - 1) \mid \left(3(\sigma(y) - y))\right)$. So I can write $$y = kx^2$$ and $$3(\sigma(y) - y) = m(x^2 - 1).$$ I am not too sure though, how this can help in answering my question. Update [October 10, 2015 - 1:05 PM]: Additionally, $I(y) = (4x^2 - 1)/3x^2 < 4/3$, so that $3(\sigma(y) - y) < y$.

A partial answer. Let $I(y)=\frac{4x^2 - 1}{3x^2}$. If $\gcd(4x^2 - 1, 3x^2) = 1$ then $3$ is a factor of $y$ contradicting the fact that $I(y)<\frac{4}{3}$. Therefore $\gcd(4x^2 - 1, 3x^2) = 3$.

Incorrect General Statement for Modulus Inequalities $$|x + 1| = x$$ This, quite evidently, has no solution. Through solving many inequalities, I came to the conclusion that, If, $$|f(x)| = g(x)$$ Then, $$f(x) = ±g(x)$$ And this was quite successful in solving many inequalities. However, applying the above to this particular inequality: $$|x + 1| = x$$ $$x + 1 = ±x$$ When ± is +, there is no solution. However, when ± is -: $$x + 1 = -x$$ $$2x = -1$$ $$x = -\frac{1}{2}$$ Which does not make sense. What is wrong with the supposedly general statement that seemed to work flawlessly?

The definition of the absolute value is $$ \lvert x \rvert = \begin{cases} x & \text{if $x\geqslant 0$} \\ -x & \text{if $x < 0$}, \end{cases} $$ so $$ \lvert f(x) \rvert = g(x) \iff \begin{cases} f(x) = g(x) \\ f(x) \geqslant 0 \end{cases} \qquad\text{or}\qquad \begin{cases} -f(x) = g(x) \\ f(x) < 0. \end{cases} $$ In particular, $$ \begin{align*} \lvert x+1 \rvert = x &\iff \begin{cases} x + 1 = x \\ x + 1 \geqslant 0 \end{cases} \qquad\text{or}\qquad \begin{cases} - (x + 1) = x \\ x + 1 < 0 \end{cases} \\ &\iff \begin{cases} 1 = 0 \\ x \geqslant -1 \end{cases} \qquad\text{or}\qquad \begin{cases} x = -1/2 \\ x < - 1. \end{cases} \end{align*} $$ As $1 = 0$ and $-1/2 < -1$ are not true, we conclude that the equation $\lvert x+1 \rvert = x$ has no real solutions.

$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$ How to find the solution of this trigonometric equation $$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$$ I have used the formulae $$\cos(x+60^\circ)\cos(x-60^\circ)=\cos^2 60^\circ - \sin^2x$$ $$\sin(x+60^\circ)\sin(x-60^\circ)=\sin^2x-\sin^260^\circ$$ How to move further? What is the least positive value of $x$?

$$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$$ $$\frac{\cot(x+110^\circ)}{\cot x} = \cot(x+60^\circ)\cdot \cot(x-60^\circ)$$ $$\frac{\cos(x+110^\circ)\cdot \sin x}{\sin (x+110^\circ)\cdot \cos x} = \frac{\cos(x+60^\circ)\cdot \cos(x-60^\circ)}{\sin(x+60^\circ)\cdot \sin(x-60^\circ)}$$ Now Using Componendo and Dividendo, We get $$\frac{\cos(x+110^\circ)\cdot \sin x+\sin(x+110^\circ)\cdot \cos x}{\cos(x+110^\circ)\cdot \sin x-\sin(x+110^\circ)\cdot \cos x} = \frac{\cos(x+60^\circ)\cdot \cos(x-60^\circ)+\sin(x+60^\circ)\cdot \sin(x-60^\circ)}{\cos(x+60^\circ)\cdot \cos(x-60^\circ)-\sin(x+60^\circ)\cdot \sin(x-60^\circ)}$$ $$-\frac{\sin \left[(x+110^\circ)+x\right]}{\sin\left[(x+110^\circ)-x\right]} = \frac{\cos[(x+60^\circ)-(x-60^\circ)]}{\cos[(x+60^\circ)+(x-60^\circ)]}$$ So we get $$-\frac{\sin(2x+110^\circ)}{\sin (110^\circ)} = \frac{\cos (120^\circ)}{\cos (2x)}$$ $$\sin (2x+110^\circ)\cdot \cos (2x) = \frac{1}{2}\sin (110^0)$$ So we get $$2\sin (2x+110^\circ)\cdot \cos (2x) = \sin(110^0)$$ $$\sin(4x+110^\circ)+\sin(110^\circ) = \sin (110^\circ)$$ So we get $$\sin(4x+110^\circ) =0\Rightarrow 4x+110^0 = n\times 180^\circ\;,$$ Where $n\in \mathbb{Z}$

How many roots are rational? If $P(x) = x^3 + x^2 + x + \frac{1}{3}$, how many roots are rational? EDIT: $3x^3 + 3x^2 + 3x + 1 = 0$, if any rat roots then, $x = \pm \frac{1}{1, 3} = \frac{-1}{3}, \frac{1}{3}$, and none of these work. Complete?

The polynomial $3P(x) = 3x^3 + 3x^2 + 3x + 1$ has integer coefficients. So, if $p/q$ with $\operatorname{gcd}(p,q) = 1$ is a rational root, necessary $p$ divides the constant term $1$, and $q$ divides the dominant coefficient $3$. So we have four candidates $-1$, $-1/3$, $1/3$ and $1/3$. Clearly, the roots of $P(x)$ are negative, so we just have to test $-1$ and $-1/3$. The polynomial $P(x)$ has no rational roots.

Using arithmetic mean>geometric mean Prove that if a,b.c are distinct positive integers that $$a^4+b^4+c^4>abc(a+b+c)$$ My attempt: I used the inequality A.M>G.M to get two inequalities First inequality $$\frac{a^4+b^4+c^4}{3} > \sqrt[3]{a^4b^4c^4}$$ or $$\frac{a^4+b^4+c^4}{3} > abc \sqrt[3]{abc}$$ or new -- first inequality $$\frac{a^4+b^4+c^4}{3abc} > \sqrt[3]{abc}$$ second inequality: $$\frac{a+b+c}{3} > \sqrt[3]{abc}$$ I am seeing the numbers in the required equation variables here but am not able to manipulate these to get the inequality I want?? Please direct me on which step should I take after this??

I just want to add another way to solve the problem. $$\begin{align} & a^4 + b^4 + c^4 = (a^2)^2+(b^2)^2+(c^2)^2 \\& \ge a^2 b^2 + b^2c^2 + c^2a^2 \text{ (From Cauchy-Schwarz)}\\&= (ab)^2 + (bc)^2 + (ca)^2 \\ & \ge abbc + bcca + caab\ (\because a^2 + b^2 + c^2 \ge ab + bc + ac) \\ &= abc(a+b+c), \end{align}$$

Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular. Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular. Let the equation to the tangent be $$y=mx+\sqrt{a^2m^2+b^2} $$ This has to roots for $m$ that i.e $m_1$ and $m_2$. perpendicular to this line passing through $(0,0)$ is $$my+x=0 $$ slope is $\frac{-1}{m}$ so for perpendiculars $$\frac{(-1)(-1)}{m_1m_2} =-1 $$ so $m_1m_2=-1 $ from equation of tangent $y=mx+\sqrt{a^2m^2+b^2} $ $$m^2(x^2-a^2)-2mxy+y^2-b^2=0 $$and hence locus is $$\frac{y^2-b^2}{x^2-a^2}=-1 $$ But this is not the correct answer. The answer is $b^4x^2+a^4y^2=a^2b^2(a^2+b^2) $. What's the error ?

Let $(x_1,y_1)$ be the generic point of the locus. It is well known that $$\frac {xx_1}{a^2}+\frac {yy_1}{b^2}=1$$ represents the line passing through the points of contact of the tangents from $(x_1,y_1)$. Now consider the equation $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\left(\frac {xx_1}{a^2}+\frac {yy_1}{b^2}\right)^2=0$$ It is satisfied by the coordinates of the center and the points of contact. Since it can be written $$x^2 \left(\frac {x_1^2}{a^4}-\frac 1{a^2}\right) + y^2 \left(\frac {y_1^2}{b^4}-\frac 1{b^2}\right) + \frac {2\,x\,y\,x_1y_1}{a^2b^2}=0$$ it is quadratic homogeneous so represents a pair of lines (degenerate conic) , clearly the lines joining the points of contact to the centre. It is not difficult to prove that the lines are mutually perpendicular iff the sum of the coefficients of $x^2$ and $y^2$ is zero, that is $$\frac {x_1^2}{a^4}+\frac {y_1^2}{b^4}=\frac 1{a^2}+\frac 1{b^2}$$

representation of a complex number in polar from Write the following in polar form: $\frac{1+\sqrt{3}i}{1-\sqrt{3}i}$ $$\frac{1+\sqrt{3}i}{1-\sqrt{3}i}=\frac{1+\sqrt{3}i}{1-\sqrt{3}i}\cdot\frac{1+\sqrt{3}i}{1+\sqrt{3}i}=\frac{(1+\sqrt{3}i)^2}{1^2+(\sqrt{3})^2}=\frac{1+2\sqrt{3}i-3}{4}=\frac{-2+2\sqrt{3}i}{4}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$$ $|z|=r=\sqrt{(-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$ arg$z$=$tan^{-1}(\frac{\sqrt{3}}{2}\cdot -2)=-\frac{\pi}{3}$ acorrding to Wolfram $\theta=0$

HINT: $$\frac{1+i\sqrt{3}}{1-i\sqrt{3}}=$$ $$\left|\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right|e^{\arg\left(\frac{1+i\sqrt{3}}{1-i\sqrt{3}}\right)i}=$$ $$\frac{|1+i\sqrt{3}|}{|1-i\sqrt{3}|}e^{\left(\arg\left(1+i\sqrt{3}\right)-\arg\left(1-i\sqrt{3}\right)\right)i}=$$ $$\frac{\sqrt{1^2+\left(\sqrt{3}\right)^2}}{\sqrt{1^2+\left(\sqrt{3}\right)^2}}e^{\left(\tan^{-1}\left(\frac{\sqrt{3}}{1}\right)--\tan^{-1}\left(\frac{\sqrt{3}}{1}\right)\right)i}=$$ $$\frac{\sqrt{4}}{\sqrt{4}}e^{\left(\tan^{-1}\left(\frac{\sqrt{3}}{1}\right)+\tan^{-1}\left(\frac{\sqrt{3}}{1}\right)\right)i}=$$ $$\frac{2}{2}e^{\frac{2\pi}{3}i}=$$ $$1\cdot e^{\frac{2\pi}{3}i}=$$ $$e^{\frac{2\pi}{3}i}$$

Taylor Series Expansion of $ f(x) = \sqrt{x} $ around $ a = 4 $ guys. Here's the exercise: find a series representation for the function $ f(x) = \sqrt{x} $ around $ a = 4 $ and find it's radius of convergence. My doubt is on the first part: I can't seem to find a pattern. $ \circ f(x) = \sqrt x \rightarrow f(4) = 2 \\ \circ f'(x) = \frac{1}{2} \cdot \frac{1}{\sqrt x} \rightarrow f'(4) = \frac{1}{4} \\ \circ f''(x) = \frac{1}{2} \cdot \left( -\frac{1}{2} \right) \cdot x^{-\frac{3}{2}} \rightarrow f''(4) = \frac{-1}{4 \cdot 8} = \frac{-1}{4 \cdot (2 \cdot 4^1)} = \frac{-1}{2 \cdot 4^2} \\\\ \circ f^{(3)}(x) = -\frac{1}{4} \cdot \left(- \frac{3}{2} \right) \cdot x^{-\frac{5}{2}} \rightarrow f^{(3)}(4) = \frac{3}{4 \cdot 32} = \frac{3}{4 \cdot (4^2 \cdot 2)} = \frac{3}{4^3 \cdot 2} \\\\ \circ f^{(4)}(x) = \frac{3}{8} \cdot \left(- \frac{5}{2} \right) \cdot x^{-\frac{7}{2}} \rightarrow f^{(4)}(4) = \frac{-3 \cdot 5}{4 \cdot 128} = -\frac{3 \cdot 5}{4 \cdot (4^3 \cdot 2)} = -\frac{3 \cdot 5}{4^4 \cdot 2} $ hmm.. Can I say then that: $ f(x) = 2 + \frac{(x-4)}{4} + \frac{1}{2} \cdot \sum_{n=2}^{\infty} \frac{(-1)^{n-1} \cdot (1 \cdot 3 \cdot 5 \dots (2n-3)) \cdot (x-4)^n}{4^n} $ and using the ratio test: $ \lim_{n \to \infty} \left| \frac{(x-4)^{n+1} \cdot (2n-1)}{4^{n+1}} \cdot \frac{4^n}{(x-4)^n \cdot (2n-3)} \right| = \frac{|x-4|}{4} \rightarrow -4 < x-4 < 4 \therefore 0 < x < 8 $ and thus R = 4. But here's the thing: the answer on the book is: $ 2 + \frac{(x-4)}{4} + 2 \cdot \sum_{n=2}^{\infty} (-1)^{n-1} \cdot \frac{1 \cdot 3 \cdot 5 \dots (2n-3) \cdot (x-4)^n}{2 \cdot 4 \cdot 6 \dots (2n) \cdot 4^n} $ Am I missing something? Thanks in advance, guys!

You were on the right track. We have for $n\ge 2$ $$\begin{align} f^{(n)}(x)&=(-1)^{n-1}\frac12 \frac12 \frac32 \frac52 \cdots \frac{2n-3}{2}x^{-(2n-1)/2}\\\\ &=(-1)^{n-1}\frac{(2n-3)!!}{2^n}x^{-(2n-1)/2}\tag 1 \end{align}$$ where $(2n-3)!! = 1\cdot 3\cdot 5\cdot (2n-3)$ is the double factorial of $2n-3$. Evaluating $(1)$ at $x=4$ yields $$f^{(n)}(4)=(-1)^{n-1}\frac{2(2n-3)!!}{2^n4^n}$$ Therefore, we have $$\begin{align} f(x)&=f(4)=f'(4)(x-4)+2\sum_{n=2}^{\infty}(-1)^{n-1}\frac{(2n-3)!!}{(2^n)(4^n)n!}(x-4)^n\\\\ &=2+\frac{(x-4)}{4}+2\sum_{n=2}^{\infty}(-1)^{n-1}\frac{(2n-3)!!}{(2n)!!(4^n)}(x-4)^n\\\\ \end{align}$$ where $(2^n)n!=(2n)!!=2\cdot 4\cdot 6\cdots 2n$ and again $(2n-3)!!=1\cdot3\cdot 5\cdots(2n-3)$. And we are done!

Find $2^{3^{100}}$ (mod 5) and its last digit The question is to find $2^{3^{100}} \pmod 5$ and its last digit. I think we have to find $2 \pmod 5$ and $3^{100}\pmod 5$ separately, right? $$2 = 2 \pmod 5$$ $$3^4 = 1 \pmod 5$$ $$3^{100} = 1 \pmod 5$$ $$2^{3^{100}} = 2^1 = 2 \pmod 5$$ Is this solution correct? And to find the last digit do we just solve modulo $10$?

Yes, you solve modulo $10$ for the last digit. Note that the number is even, so its last digit is taken from $2, 5, 6, 8$--it's not divisible by $5$ so the last digit cannot be $0$. Your original solution is a bit off, though $3^4\equiv 1\mod 5$, but in the exponent it's every $4$ which gives $1$, so we want $3^2\equiv 1\mod 4$. Then $3^{100}\equiv 1\mod 4$. So $$2^{3^{100}}=2^{4k+1}\equiv (16)^k\cdot 2\mod 5$$ which gives the same result, but by sounder methods. Since it is $2$ mod $5$ as you have said, the only choices are $2$ and $7$. $7$ is odd, so it must be $2$.

(Combinatorial) proof of an identity of McKay Lemma 2.1 of this paper claims that for integer $s>0$ and $v \in \mathbb{N}$, we have $$ \sum_{k=1}^s \binom{2s-k}{s} \frac{k}{2s-k} v^k (v-1)^{s-k} = v \sum_{k=0}^{s-1} \binom{2s}{k} \frac{s-k}{s} (v-1)^k $$ The author gives a combinatorial interpretation of the left hand side in terms of closed walks on a class of graphs that are locally acyclic in a particular sense, but omits the proof of equality. I've hit a wall trying to prove this, so I was wondering what ideas others have. Combinatorial proofs welcomed!

Suppose we seek to verify that $$\sum_{k=1}^n {2n-k\choose n} \frac{k}{2n-k} v^k (v-1)^{n-k} = v \sum_{k=0}^{n-1} {2n\choose k} \frac{n-k}{n} (v-1)^k.$$ Now the LHS is $$\frac{1}{n}\sum_{k=1}^n {2n-k-1\choose n-1} k v^k (v-1)^{n-k}.$$ Re-write this as $$\frac{1}{n}\sum_{k=1}^n {2n-k-1\choose n-k} k v^k (v-1)^{n-k}.$$ Introduce $${2n-k-1\choose n-k} = \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n-k+1}} (1+w)^{2n-k-1} \; dw.$$ Observe that this zero when $k\gt n$ so we may extend $k$ to infinity to obtain for the sum $$\frac{1}{n} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (1+w)^{2n-1} \sum_{k\ge 1} k v^k (v-1)^{n-k} \frac{w^k}{(1+w)^k}\; dw \\ = \frac{(v-1)^n}{n} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (1+w)^{2n-1} \frac{vw/(v-1)/(1+w)}{(1-vw/(v-1)/(1+w))^2} \; dw \\ = \frac{(v-1)^n}{n} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n+1}} (1+w)^{2n} \frac{vw(v-1)}{((v-1)(1+w)-vw)^2} \; dw \\ = v\frac{(v-1)^{n+1}}{n} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n}} (1+w)^{2n} \frac{1}{(-1-w+v)^2} \; dw \\ = v\frac{(v-1)^{n-1}}{n} \frac{1}{2\pi i} \int_{|w|=\epsilon} \frac{1}{w^{n}} (1+w)^{2n} \frac{1}{(1-w/(v-1))^2} \; dw.$$ Extracting the coefficient we obtain $$v\frac{(v-1)^{n-1}}{n} \sum_{q=0}^{n-1} {2n\choose q} \frac{(n-1-q+1)}{(v-1)^{n-1-q}} \\ = v \sum_{q=0}^{n-1} {2n\choose q} \frac{(n-1-q+1)}{n} (v-1)^q \\ = v \sum_{q=0}^{n-1} {2n\choose q} \frac{(n-q)}{n} (v-1)^q.$$ This concludes the argument.

Square root of complex number. The complex number $z$ is defined by $z=\frac{9\sqrt3+9i}{\sqrt{3}-i}$. Find the two square roots of $z$, giving your answers in the form $re^{i\theta}$, where $r>0$ and $-\pi <\theta\leq\pi$ I got the $z=9e^{\frac{\pi}{3}i}$. So I square root it, it becomes $3e^{\frac{\pi}{6}i}$. But the given answer is $3e^{-\frac{5}{6}\pi i}$. Why?

Notice, $$z=\frac{9\sqrt 3+9i}{\sqrt 3-i}$$ $$=\frac{9(\sqrt 3+i)(\sqrt 3+i)}{(\sqrt 3-i)(\sqrt 3+i)}$$ $$=\frac{9(\sqrt 3+i)^2}{3-i^2}=\frac{9(2+2i\sqrt 3)}{3+1}$$$$=9\left(\frac{1}{2}+i\frac{\sqrt 3}{2}\right)=9\left(\cos\frac{\pi}{3}+i\sin \frac{\pi}{3}\right)=9e^{i\pi/3}$$ hence, the square roots of $z$ are found as follows $$z^{1/2}=\sqrt{9\left(\cos\frac{\pi}{3}+i\sin \frac{\pi}{3}\right)}$$ $$=3\left(\cos\left(2k\pi+\frac{\pi}{3}\right)+i\sin \left(2k\pi+\frac{\pi}{3}\right)\right)^{1/2}$$$$=3\left(\cos\left(\frac{6k\pi+\pi}{6}\right)+i\sin \left(\frac{6k\pi+\pi}{6}\right)\right)$$ where, $k=0, 1$ Setting $k=0$, we get first square root $$z^{1/2}=3\left(\cos\left(\frac{6(0)\pi+\pi}{6}\right)+i\sin \left(\frac{6(0)\pi+\pi}{6}\right)\right)=3\left(\cos\left(\frac{\pi}{6}\right)+i\sin \left(\frac{\pi}{6}\right)\right)=\color{red}{e^{\frac{\pi}{6}i}}$$ Now, setting $k=1$, we get second square root $$z^{1/2}=3\left(\cos\left(\frac{6(1)\pi+\pi}{6}\right)+i\sin \left(\frac{6(1)\pi+\pi}{6}\right)\right)$$ $$=3\left(-\cos\left(\frac{\pi}{6}\right)-i\sin \left(\frac{\pi}{6}\right)\right)$$ $$=3\left(\cos\left(-\frac{5\pi}{6}\right)+i\sin \left(-\frac{5\pi}{6}\right)\right)=\color{red}{e^{\frac{-5\pi}{6}i}}$$

Finding the reflection that reflects in an arbitrary line y=mx+b How can I find the reflection that reflects in an arbitrary line, $y=mx+b$ I've examples where it's $y=mx$ without taking in the factor of $b$ But I want to know how you can take in the factor of $b$ And after searching through for some results, I came to this matrix which i think can solve my problems. But it doesn't seem to work. $$ \begin{bmatrix} x' \\ y' \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1-m^2}{1 + m^2} & \frac{-2m}{1 + m^2} & \frac{-2mb}{1 + m^2} \\ \frac{-2m}{1 + m^2} & \frac{m^2-1}{1 + m^2} & \frac{2b}{1 + m^2} \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ 1 \end{bmatrix}. $$ The example I tried to use using this matrix is the point $(0,8)$ reflected on $y=-\frac{1}{2}x+2$. The result I get from that matrix is $[6.4,-0.6,0]$. The actual answer should be $[-4.8, -1.6]$ , according to Geogebra

One way to do this is as a composition of three transformations: * *Translate by $(0,-b)$ so that the line $y=mx+b$ maps to $y=mx$. *Reflect through the line $y=mx$ using the known formula. *Translate by $(0,b)$ to undo the earlier translation. The translation matrices are, respectively, $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -b \\ 0 & 0 & 1 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} $$ and the matrix of the reflection about $y=mx$ is $$ \frac{1}{1 + m^2} \begin{pmatrix} 1-m^2 & 2m & 0 \\ 2m & m^2-1 & 0 \\ 0 & 0 & 1 + m^2 \end{pmatrix}. $$ Applying these in the correct sequence, the transformation is $$ \frac{1}{1 + m^2} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1-m^2 & 2m & 0 \\ 2m & m^2-1 & 0 \\ 0 & 0 & 1 + m^2 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & -b \\ 0 & 0 & 1 \end{pmatrix} $$ $$ = \frac{1}{1 + m^2} \begin{pmatrix} 1-m^2 & 2m & -2mb \\ 2m & m^2-1 & 2b \\ 0 & 0 & 1 + m^2 \end{pmatrix}. $$ This is much like the matrix you found, but the entries that you set to $\frac{-2m}{1+m^2}$ are instead $\frac{2m}{1+m^2}$. Setting $m=-\frac12$, $b=2$, the matrix is $$ \frac45 \begin{pmatrix} \frac34 & -1 & 2 \\ -1 & -\frac34 & 4 \\ 0 & 0 & \frac54 \end{pmatrix} = \begin{pmatrix} 0.6 & -0.8 & 1.6 \\ -0.8 & -0.6 & 3.2 \\ 0 & 0 & 1 \end{pmatrix} $$ and applying this to the point $(0,8)$ we have $$ \begin{pmatrix} 0.6 & -0.8 & 1.6 \\ -0.8 & -0.6 & 3.2 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 \\ 8 \\ 1 \end{pmatrix} = \begin{pmatrix} -4.8 \\ -1.6 \\ 1 \end{pmatrix}, $$ that is, the reflection of $(0,8)$ is $(-4.8, -1.6)$, so the matrix multiplication has the desired effect.

Finding the coordinate C. A triangle $A$, $B$, $C$ has the coordinates: $A = (-1, 3)$ $B = (3, 1)$ $C = (x, y)$ $BC$ is perpendicular to $AB$. Find the coordinates of $C$ My attempt: Grad of $AB$ = $$\frac{3-1}{-1-3} = -0.5$$ Grad of $BC = 2$ ($-0.5 \times 2 = -1$ because AB and BC are perpendicular). Equation of $BC$ $(y-1) = 2(x-3)$ $y = 2x - 5$ Equation of $AC$ $(y-3) = m(x--1)$ $y = mx+m+3$ I do not know how to proceed further. Please help me out.

The equation of $AC$ is $x-3y=-10$ as slope of any line is $-\frac{a}{b}$ where $a$ is $x$-coordinate and $b$ is $y$-coordinate so slope is $-\left(\frac{-1}{3}\right)$. $C$ is the point where $AC$ and $BC$ so meet we have two simultaneous equations $2x-y=5$ and $x-3y=-10$ solving them you get $x=5$ and $y=5$.

Laplace's equation after change of variables Show that if $u(r, \theta)$ is dependent on $r$ alone, Laplace's equation becomes $$u_{rr} + \frac{1}{r}u_r=0.$$ My first reaction is to replace $r=x$ and $\theta=y$, but obviously it does not work. Then I recall $x=r\cos \theta$ and $y=r\sin \theta$. Then I obtain the following: $$v(r, \theta) = u(r\cos \theta, r\sin \theta).$$ Then I start to differentiate it, but what is $u_r$ and $u_{rr}$? Can anyone give me some hints to move on?

You have $$ \frac{\partial f}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial f}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial f}{\partial y} = f_x \cos{\theta} + f_y \sin{\theta}. $$ Then differentiating again (and using that $\partial \theta/\partial r=0$), $$ f_{rr} = \cos{\theta} (\cos{\theta} \partial_x+\sin{\theta} \partial_y)f_x + \sin{\theta} (\cos{\theta} \partial_x+\sin{\theta} \partial_y)f_y \\ f_{xx} \cos^2{\theta} + f_{yy}\sin^2{\theta} + 2f_{xy} \sin{\theta}\cos{\theta} $$ Similarly, we can show that $$ \partial_{\theta} = -r\sin{\theta} \partial_x + r\cos{\theta}\partial_y, $$ so $$ f_{\theta\theta} = \partial_{\theta}(-r\sin{\theta} f_x + r\cos{\theta}f_y) = r^2[(-\cos{\theta} f_x -\sin{\theta} f_y ) +\sin^2{\theta} f_{xx}+\cos^2{\theta}f_yy -2\sin{\theta}\cos{\theta} f_{xy}] $$ Thus $$ f_{rr} + \frac{1}{r^2}f_{\theta\theta} + \cos{\theta} f_x + \sin{\theta} f_y = f_{xx}+f_{yy} $$ But the last terms on the left are just $r^{-1}\partial_r $, so $$ f_{xx} + f_{yy} = f_{rr}+\frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta}. $$ That's a bit of a mess, though. Instead, we can use $u(r,\theta)=f(r)$, and $r=\sqrt{x^2+y^2}$, and compute $$ \Delta f = \frac{\partial^2}{\partial x^2} f(\sqrt{x^2+y^2})+\frac{\partial^2}{\partial y^2} f(\sqrt{x^2+y^2}) \\ = \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2+y^2}} f'(\sqrt{x^2+y^2}) \right) + \frac{\partial}{\partial y} \left( \frac{y}{\sqrt{x^2+y^2}} f'(\sqrt{x^2+y^2}) \right) \\ = \frac{y^2}{(x^2+y^2)^{3/2}}f'(\sqrt{x^2+y^2}) + \left(\frac{x}{\sqrt{x^2+y^2}}\right)^2 f''(\sqrt{x^2+y^2}) + \frac{x^2}{(x^2+y^2)^{3/2}}f'(\sqrt{x^2+y^2}) + \left(\frac{y}{\sqrt{x^2+y^2}}\right)^2 f''(\sqrt{x^2+y^2}) \\ = f''(\sqrt{x^2+y^2})+\frac{1}{\sqrt{x^2+y^2}}f'(\sqrt{x^2+y^2}) = u_{rr}+\frac{1}{r}u_r. $$

Can this function be integrated? Can't seem to figure out this integral! I'm trying to integrate this but I think the function can't be integrated? Just wanted to check, and see if anyone is able to find the answer (I used integration by parts but it doesn't work). Thanks in advance; the function I need to integrate is $$\int\frac{x}{x^5+2}dx$$

Probably you'd want to use the calculus of residues to do this. But below I do it using first-year calculus methods. The cumbersome part may be the algebra, and that's what I concentrate on here. \begin{align} x^5 + 2 & = \left( x+\sqrt[5]{2} \right) \underbrace{\left( x - \sqrt[5]{2} e^{i\pi/5}\right)\left( x - \sqrt[5]{2} e^{-i\pi/5}\right)}_\text{conjugates}\ \underbrace{\left( x - \sqrt[5]{2} e^{i3\pi/5}\right)\left( x - \sqrt[5]{2} e^{-i3\pi/5}\right)}_\text{conjugates} \\[10pt] & = \left( x+\sqrt[5]{2} \right) \left( x^2 - 2 \sqrt[5]{2} \cos\frac{\pi} 5 + \sqrt[5]{2}^2 \right)\left( x^2 - 2\sqrt[5]{2}\cos\frac{3\pi}5 + \sqrt[5]{2}^2 \right) \end{align} Then use partial fractions. That the polynomials $x^2 - 2\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2$ and $x^2 - 2\sqrt[5]{2}\cos\frac{3\pi} 5 + \sqrt[5]{2}^2$ cannot be factored using real numbers can be seen from the way we factored them above. So you'll have $$ \cdots + \frac{Bx+C}{x^2 - 2\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2} + \cdots $$ etc. and you'll need to find $B$ and $C$. Let $u=x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2$ then $du = (2x + \sqrt[5]{2}\cos\frac{\pi}5)\,dx$ and so $$ \frac{Bx + C}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx = \underbrace{\frac{\frac B 2 \left( 2x+\sqrt[5]{2}\cos\frac{\pi}5 \right)}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx}_\text{Use the substitution for this part.} + \underbrace{\frac{\left( 1 - \frac B2 \right) \sqrt[5]{2}\cos\frac{\pi} 5}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + 1}\,dx}_\text{See below for this part.} $$ To integrate $\frac{\left( 1 - \frac B2 \right) \sqrt[5]{2}\cos\frac{\pi} 5}{x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2}\,dx$, complete the square: \begin{align} & x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2 = \left( x^2 - 2x\sqrt[5]{2}\cos\frac{\pi} 5 + \sqrt[5]{2}^2\cos^2\frac{\pi}5 \right) + \sqrt[5]{2}^2 \sin^2\frac{\pi} 5 \\[10pt] = {} & \left( x - \sqrt[5]{2}\cos\frac{\pi} 5 \right)^2 + \sqrt[5]{2}^2\sin^2 \frac{\pi} 5 \\[10pt] = {} & \left(\sqrt[5]{2}^2 \sin^2 \frac{\pi} 5 \right) \left( \left( \frac{x-\sqrt[5]{2}\cos\frac{\pi} 5}{\sqrt[5]{2}\sin\frac{\pi} 5} \right)^2 + 1 \right) = (\text{constant})\cdot(w^2 + 1) \\[10pt] & \text{and } dw = \frac{dx}{\sqrt[5]{2}\sin\frac{\pi}5}. \end{align} So you get an arctangent from this term.

Can one do anything useful with a functional equation like $g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$? I got $$g(x^2) = \frac{4x^2-1}{2x^2+1}g(x)$$ as a functional equation for a generating function. Is there a way to get a closed form or some asymptotic information about the Taylor coefficients from such an equation? Here g(0) = 1, g'(0) = 0, and g''(0) = 4. Edit: Thanks, everyone; as has been pointed out , I've just made a mistake in obtaining the recurrence relation. I'll leave this as-is instead of editing because I think some of the comments will be helpful to others and if I change it now they won't make sense.

I get that the only solution is $g(x) = 0$ if we can write $g(x) =\sum_{n=0}^{\infty} a_n x^n $. Here is my proof: We have $g(x^2) = \frac{4x^2-1}{2x^2+1}g(x) $ or $(2x^2+1)g(x^2) = (4x^2-1)g(x) $. From this, as copper.hat pointed out, $g(0) = 0$. If $g(x) =\sum_{n=1}^{\infty} a_n x^n $, (since $g(0) = 0$) the left side is $\begin{align*} \sum_{n=1}^{\infty} (2x^2+1)a_nx^{2n} &=\sum_{n=1}^{\infty} 2a_nx^{2n+2}+\sum_{n=1}^{\infty} a_nx^{2n}\\ &=\sum_{n=2}^{\infty} 2a_{n-1}x^{2n}+\sum_{n=1}^{\infty} a_nx^{2n}\\ &=a_1x^2+\sum_{n=2}^{\infty} (2a_{n-1}+a_n)x^{2n}\\ \end{align*} $ and the right side is $\begin{align*} \sum_{n=1}^{\infty} (4x^2-1)a_nx^{n} &=\sum_{n=1}^{\infty} 4a_nx^{n+2}-\sum_{n=1}^{\infty} a_nx^{n}\\ &=\sum_{n=3}^{\infty} 4a_{n-2}x^{n}-\sum_{n=1}^{\infty} a_nx^{n}\\ &=-a_1x-a_2x^2+\sum_{n=3}^{\infty} (4a_{n-2}-a_n)x^{n}\\ \end{align*} $ Equating coefficients, $-a_1x = 0$, $-a_2x^2 = a_1x^2$, and $2a_{n-2}+a_n =4a_{2n-2}-a_{2n} $ and $0 =4a_{2n-1}-a_{2n+1} $. Therefore $a_1 = a_2 = 0$ and $a_{2n} =4a_{2n-2}-2a_{n-2}-a_n $ and $a_{2n+1} =4a_{2n-1} $. From this second recurrence, since $a_1 = 0$, $a_{2n+1} = 0$ for all $n$. From $a_{2n} =4a_{2n-2}-2a_{n-2}-a_n $, for $ n=2, a_4 =4a_2-2a_0-a_2 =0 $, for $ n=3, a_6 =4a_4-2a_2-a_3 =0 $. By strong induction, all the $a_{2n} = 0$, so the only solution is $g(x) = 0$.

Proof that given equation(quartic) doesn't have real roots $$ (x^2-9)(x-2)(x+4)+(x^2-36)(x-4)(x+8)+153=0 $$ I need to prove that the above equation doesn't have a real solution. I tried breaking it up into an $(\alpha)(\beta)\cdots=0$ expression, but no luck. Wolfram alpha tells me that the equation doesn't have real roots, but I'm sure there's simpler way to solve this than working trough the quartic this gives.

Your polynomial is $$P(x) = ({x^2} - 9)(x - 2)(x + 4) + ({x^2} - 36)(x - 4)(x + 8) + 153\tag{1}$$ Now consider theses $$\eqalign{ & f(x) = ({x^2} - 9)(x - 2)(x + 4) \cr & f({x \over 2}) = \left( {{{\left( {{x \over 2}} \right)}^2} - 9)} \right)\left( {\left( {{x \over 2}} \right) - 2} \right)\left( {\left( {{x \over 2}} \right) + 4} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over 4}\left( {{x^2} - 36} \right){1 \over 2}\left( {x - 2} \right){1 \over 2}\left( {x - 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = {1 \over {16}}\left( {{x^2} - 36} \right)\left( {x - 2} \right)\left( {x - 8} \right) \cr}\tag{2}$$ combine $(1)$ and $(2)$ to get $$P(x) = f(x) + 16f({x \over 2}) + 153\tag{3}$$ Next, we try to find the range of $f(x)$. For this purpose, consider this $$\eqalign{ & f(x) = ({x^2} - 9)(x - 2)(x + 4) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {x^4} + 2{x^3} - 17{x^2} - 18x + 72 \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\left( {{x^2} + x - 9} \right)^2} - 9 \cr}\tag{4}$$ Now, by $(4)$ you can conclude that $$\left\{ \matrix{ f(x) \ge - 9 \hfill \cr f({x \over 2}) > - 9 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\left\{ \matrix{ f(x) > - 9 \hfill \cr f({x \over 2}) \ge - 9 \hfill \cr} \right.\tag{5}$$ Notice the equality signs! Can you figure out why this happens? Then using $(5)$ you can conclude that $$\left\{ \matrix{ f(x) \ge - 9 \hfill \cr 16f({x \over 2}) > - 144 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,or\,\,\,\,\,\,\,\left\{ \matrix{ f(x) > - 9 \hfill \cr 16f({x \over 2}) \ge - 144 \hfill \cr} \right.\tag{6}$$ and then summing up either of the relations $(6)$ will lead to $$f(x) + 16f({x \over 2}) > - 153\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,f(x) + 16f({x \over 2}) + 153 > 0\,\,\,\,\,\, \to \,\,\,\,\,\,\,P(x) > 0\tag{7}$$ I think we are done now! :)

show that for any prime p: if $p|x^4 - x^2 + 1$, with $x \in \mathbb{Z}$ satisfies $p \equiv 1 \pmod{12}$? show that for any prime p: if $p|x^4 - x^2 + 1$ satisfies $p \equiv 1 \pmod{12}$ I suppose that if $p$ divides this polynomial we can see that: $x^4 - x^2 + 1 = kp$ for some $k \in \mathbb{N}$. But then $x^4 - x^2 + 1 \equiv 0 \pmod{p}$. This means that $(2x^2 - 1)^2 \equiv 4(x^4 - x^2 +1) - 3 \equiv - 3 \pmod {p}$. So $-3$ must be a quadratic residue for this to have a solution, but $\left(\frac{-3}{p}\right) = 1$ if $p \equiv +- 1\pmod{3}$. Is there something i am doing wrong here?

If $p\mid x$, then $p\mid x^4-x^2+1\implies p\mid 1$, contradiction. Therefore $p\nmid x$. $(2x^2-1)^2\equiv -3\pmod{p}$ and $\left(\left(x^2-1\right)x^{-1}\right)^2\equiv -1\pmod{p}$. We can't have $p=3$, because $2x^2-1\equiv 0\pmod{3}$ has no solutions. We also can't have $p=2$, because $x^4-x^2+1$ is always odd, so $2$ can't divide it. By Quadratic Reciprocity we get $p\equiv 1\pmod{3}$ and $p\equiv 1\pmod{4}$, respectively, so $p\equiv 1\pmod{12}$. Alternatively, $\left(\left(x^2+1\right)x^{-1}\right)^2\equiv 3\pmod{p}$ and $\left(\left(x^2-1\right)x^{-1}\right)^2\equiv -1\pmod{p}$ and again we can't have $p=2$ or $p=3$, because $\left(x^2+1\right)x^{-1}\equiv 0\pmod{3}$ has no solutions and $x^4-x^2+1$ is always odd. By Quadratic Reciprocity we get $p\equiv \pm1\pmod{12}$ and $p\equiv 1\pmod{4}$, respectively, so $p\equiv 1\pmod{12}$.

How to find the determinant of this $5 \times 5$ matrix? How can I find the determinant of this matrix? I know in matrix $3 \times 3$ $$A= 1(5\cdot 9-8\cdot 6)-2 (4\cdot 9-7\cdot 6)+3(4\cdot 8-7\cdot 5) $$ but how to work with a $5\times 5$ matrix?

Multiplying the 1st row by $3$ and then adding it to the 4th row, and then multiplying the 3rd row of the resulting matrix by $-\frac 1 4$ and adding it to the 5th row, we obtain $$\det \begin{bmatrix} 1 & 2 & 3 & 4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ -3 & -6 & -9 & -12 & 4\\ 0 & 0 & 1 & 1 & 1\end{bmatrix} = \det \begin{bmatrix} 1 & 2 & 3 & 4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0 & 7\\ 0 & 0 & 1 & 1 & 1\end{bmatrix} = \det \begin{bmatrix} 1 & 2 & 3 & 4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0 & 7\\ 0 & 0 & 0 & 1 & 1\end{bmatrix}$$ We have obtained a block upper triangular matrix. Hence, $$\det \begin{bmatrix} 1 & 2 & 3 & 4 & 1\\ 0 & -1 & 2 & 4 & 2\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 0 & 7\\ 0 & 0 & 0 & 1 & 1\end{bmatrix} = \det \begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 2\\ 0 & 0 & 4\end{bmatrix} \cdot \det \begin{bmatrix} 0 & 7\\ 1 & 1\end{bmatrix} = (-4) \cdot (-7) = 28$$

Calculating two specific limits with Euler's number I got stuck, when I were proving that $$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \frac {5}{2}$$ $$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \frac {1}{3}$$ First one I tried to solve like $$\lim_{n \to \infty} \frac {\sqrt[2]{(n^2+5)}-n}{\sqrt[2]{(n^2+2)}-n} = \lim_{n \to \infty} \frac {n\sqrt[2]{(1+\frac {5}{n^2})}-n}{n\sqrt[2]{(1+\frac{2}{n^2})}-n}= \lim_{n \to \infty} \frac {\sqrt[2]{(1+\frac {5}{n^2})}-1}{\sqrt[2]{(1+\frac{2}{n^2})}-1}$$ and now I think, that this one sholud go like $$\lim_{n \to \infty} \frac{\frac{5}{n^2}}{\frac{2}{n^2}}=\frac{5}{2} $$ but I have no idea how to prove this. In the second one I made $$\lim_{n \to \infty}n(\sqrt[3]{(n^3+n)}-n) = \lim_{n \to \infty}n(n\sqrt[3]{(1+\frac{1}{n^2}}-n)= \lim_{n \to \infty}n^2(\sqrt[3]{(1+\frac{1}{n^2}}-1)= \lim_{n \to \infty}n^2(e^{\frac{1}{3n^2}}-1) $$ And now I do not know what to do next... I would be really grateful, for any help, or prompt, how to solve these ones (or information, where is the mistake).

$$(a)\;\;\lim_{n\rightarrow \infty}\frac{\sqrt{n^2+5}-n}{\sqrt{n^2+2}-n} =\lim_{n\rightarrow \infty}\frac{\sqrt{n^2+5}-n}{\sqrt{n^2+2}-n}\times \frac{\sqrt{n^2+5}+n}{\sqrt{n^2+5}+n}\times \frac{\sqrt{n^2+2}+n}{\sqrt{n^2+2}+n} $$ So we get $$=\lim_{n\rightarrow \infty}\frac{5}{2}\times \frac{\sqrt{n^2+2}+n}{\sqrt{n^2+5}+n} = \frac{5}{2}\lim_{n\rightarrow \infty}\frac{n(\sqrt{1+\frac{2}{n^2}}+1)}{n(\sqrt{1+\frac{5}{n^2}}+1)}=\frac{5}{2}$$ ..................................................................................................................................................................... $$(b) \lim_{n\rightarrow \infty}n\left[\sqrt[3]{n^3+n}-n\right]$$ Now Put $\displaystyle n=\frac{1}{y}\;,$ When $n\rightarrow \infty\;,$ Then $y\rightarrow 0$ So we get $$\lim_{y\rightarrow 0}\frac{(1+y^2)^{\frac{1}{3}}-1}{y^2}$$ Now Let $(1+y^2)^{\frac{1}{3}}=A$ and $1=B\;,$ Then $A^3-B^3 = 1+y^2-1=y^2$ So we get $$\lim_{y\rightarrow 0}\frac{A-B}{A^3-B^3} = \lim_{y\rightarrow 0}\frac{A-B}{(A-B)(A^2+B^2+AB)} = \lim_{y\rightarrow 0}\frac{1}{A^2+B^2+AB}$$ So we get $$\lim_{y\rightarrow 0}\frac{1}{(1+y^2)^{\frac{2}{3}}+1^2+(1+y^2)^{\frac{1}{3}}} = \frac{1}{3}$$

In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? In the real number system,the equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1$ has how many solutions? I tried shifting the second term to the rhs and squaring.Even after that i'm left with square roots.No idea how to proceed.Help!

\begin{align} &\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}} = 1\\ \implies &\sqrt{(x-1)-4\sqrt{x-1} + 4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1\\ \implies &\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=1\\ \implies &|\sqrt{x-1}-2| + |\sqrt{x-1}-3| = 1\tag{1} \end{align} This calls for casework: 1. $\quad\sqrt{x-1}\geq 3$ $(1)\implies \sqrt{x-1}-2 + \sqrt{x-1}-3 = 1\implies \sqrt{x-1} = 3\implies x=10$ 2. $\quad 2\leq\sqrt{x-1} < 3$ $(1)\implies \sqrt{x-1}-2 - \sqrt{x-1}+3 = 1\implies 1 = 1$ and thus all $x$ such that $2\leq\sqrt{x-1} < 3$, i.e. $x\in [5,10\rangle$ 3. $\sqrt{x-1} < 2$ $(1)\implies -\sqrt{x-1}+2 - \sqrt{x-1}+3 = 1\implies \sqrt{x-1} = 2 \implies x = 5$, but this $x$ doesn't satisfy our condition 3 (although it does satisfy condition 2, and is already included as a solution) Taking union, we conclude that any $x\in[5,10]$ is a solution to the equation.

In a triangle prove that $\sin^2({\frac{A}{2}})+\sin^2(\frac{B}{2})+\sin^2(\frac{C}{2})+2\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})= 1$ Let ABC be a triangle. Thus prove that $$\sin^2\left({\frac{A}{2}}\right)+\sin^2\left(\frac{B}{2}\right)+\sin^2\left(\frac{C}{2}\right)+2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)= 1$$ How do i go about solving this?

Let $x=\frac{A}{2}$, $y=\frac{B}{2}$, $z=\frac{C}{2}$, $x+y+z=\pi/2$ $$\sin^2(x)+\sin^2(y)+\sin^2(z)+2\sin(x)\sin(y)\sin(z)\\=\sin^2(x)+\sin^2(y)+\sin^2(\pi/2-x-y)+2\sin(x)\sin(y)\sin(\pi/2-x-y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+2\sin(x)\sin(y)\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+(\cos(x-y)-\cos(x+y))\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos(x-y)\cos(x+y)\\=\frac{1-\cos(2x)}{2}+\frac{1-\cos(2y)}{2}+\cos(x-y)\cos(x+y)\\=1-\frac{\cos(2x)+\cos(2y)}{2}+\cos(x-y)\cos(x+y)\\=1-\cos(x-y)\cos(x+y)+\cos(x-y)\cos(x+y)\\=1$$

Evaluate the Integral $\int \sqrt{1-4x^2}\ dx$ $\int \sqrt{1-4x^2}\ dx$ I am confused as I get to the end. Why would I use a half angle formula? And why is it necessary to use inverses?

When you're doing the trigonometric substitution, you write $x=a\sin\theta$, which is good; you should also remember how to get back from $\theta$ to $x$, that is, $$ \theta=\arcsin\frac{x}{a}=\arcsin\frac{x}{1/2}=\arcsin(2x) $$ which actually should be the starting point, because it guarantees the angle $\theta$ is between $-\pi/2$ and $\pi/2$. When you arrive to $$ \frac{1}{4}\theta+\frac{1}{8}\sin(2\theta)= \frac{1}{4}\theta+\frac{1}{4}\sin\theta\cos\theta $$ you indeed need to get back to $x$. Since $-\pi/2\le\theta\le\pi/2$, you know $\cos\theta\ge0$ and so $$ \cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-4x^2} $$ In conclusion your integral is $$ \frac{1}{4}\arcsin(2x)+\frac{1}{2}x\sqrt{1-4x^2} $$ However, there's no need for trigonometric substitutions. Consider $$ \int\sqrt{1-t^2}\,dt= \int\frac{1-t^2}{\sqrt{1-t^2}}\,dt= \int\frac{1}{\sqrt{1-t^2}}\,dt+\int\frac{-t^2}{\sqrt{1-t^2}}\,dt $$ The first one is immediate; the second one can be computed with integration by parts: $$ \int t\frac{-t}{\sqrt{1-t^2}}\,dt= t\sqrt{1-t^2}-\int\sqrt{1-t^2}\,dt $$ All in all, we have $$ \int\sqrt{1-t^2}\,dt= \arcsin t+t\sqrt{1-t^2}-\int\sqrt{1-t^2}\,dt $$ so we can transport the integral from the right-hand side to the left-hand side and get $$ \int\sqrt{1-t^2}\,dt= \frac{1}{2}\arcsin t+\frac{1}{2}t\sqrt{1-t^2} $$ For your integral use the substitution $2x=t$.

Integrating $\sqrt{x^2+a^2}$ I'm trying to integrate this function wrt $x$, substituting $x = a \tan \theta$ $$ \int \sqrt{x^2+a^2} dx = a^2 \int \frac {d\theta}{\cos^3\theta} = $$ $$= a^2 \cdot \frac 12 \left( \tan\theta \sec\theta + \ln\lvert \tan\theta + \sec\theta \rvert \right) = \frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| \frac x a + \frac{\sqrt{a^2+x^2}}{a} \right| \right)$$ But WolframAlpha says it should be $$= \frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)$$ What am I doing wrong?

$$\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| \frac x a + \frac{\sqrt{a^2+x^2}}{a} \right| \right)\\=\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)-\frac{a^2\ln|a|}{2} $$ and $$\frac 12 \left( x \sqrt{a^2 + x^2} + a^2 \ln \left| x + \sqrt{a^2+x^2} \right| \right)$$ are different by a constant, so nothing is wrong!