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30

It is a simple trigonometric exercise to show that $$\frac{\sinh\pi n\sqrt2 -\sin\pi n\sqrt2}{\cosh\pi n\sqrt2 -\cos\pi n\sqrt2}= \Re \left(\coth \pi n z_0+z_0^2\coth\frac{\pi n}{z_0}\right),\tag{$\spadesuit$}$$ with $z_0=e^{i\pi/4}$. Recall the Ramanujan identity (several nice proofs of which may be found here): ...


30

For $x < 1$, we have the Taylor series expansion: $$f(x):= \frac{-1}{4} \log \left(- \frac{x - x^{-1}}{x + x^{-1}} \right) = \frac{x^2}{2} + \frac{x^6}{6} + \frac{x^{10}}{10} + \frac{x^{14}}{14} + \ldots $$ Then $$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + ...


17

Of course there is. The fastest way to obtain it is to heuristically write $\cos(\pi\ln n)$ as $\frac12(n^{i\pi}+n^{-i\pi})$. The answer is then given by $$\sum_{n=1}^{\infty}\frac{\cos\left(\pi\ln n\right)}{n^2}=\frac{\zeta(2+i\pi)+\zeta(2-i\pi)}{2}.$$


15

It doesn't. That series has a lot of terms near $\pm 1$; in particular, I can show it has infinitely many terms whose absolute value is larger than $\frac12$: The inequality $|\pi - p/q| < 1/q^2$ is satisfied by infinitely many pairs of positive integers $(p,q)$. (This is Dirichlet's approximation theorem.) Let $(p,q)$ be one such pair with $q > ...


13

Often when $\sin(x)$ and $x$ appear in a formula such as this, it is based on the fact that $$ \lim_{a\to0}\frac{\sin(ax)}{a}=x\tag{1} $$ Since the summands are $\dfrac{4^{-k}}{\cos^2(2^{-k}x)}$ , it appears that an identity involving a double angle formula is being exploited. Therefore, with an eye toward a telescoping sum, I started with $$ \begin{align} ...


12

Since: $$\arctan\frac{3n^2}{2n^4-1}=\arctan\frac{1}{n^2}+\arctan\frac{1}{2n^2}$$ then: $$\sum_{n=1}^{+\infty}\arctan\frac{3n^2}{2n^4-1} = \arg\prod_{n=1}^{+\infty}\left(1+\frac{i}{n^2}\right)+\arg\prod_{n=1}^{+\infty}\left(1+\frac{i}{2n^2}\right).\tag{1} $$ Since, by the Weierstrass product for the $\sinh$ function, $$\frac{\sinh(\pi z)}{\pi ...


12

$$\tan rx=\frac{\binom r1\tan x-\binom r3\tan^3x+\cdots}{1-\binom r2\tan^2x+\cdots}\text{ (Proof below) }$$ Now, if $\tan rx=\tan r\theta\implies rx=n\pi+r\theta$ where $n$ is any integer $\implies x=n\frac{\pi}r+\theta$ If $r$ is even, $=2m$(say) $$\tan2m\theta=\tan2mx=\frac{\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom ...


11

It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions. Indeed, Jacobi elliptic function $\operatorname{dn}(z,k)$ has Fourier series $$\operatorname{dn}(z,k)=\frac{\pi}{2K}\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\pi\frac{z}{K}}{\cosh n \pi ...


11

Vi├Ęte's formula gives: $$\prod_{k\geq 1} \cos\left(\frac{x}{2^k}\right)=\frac{\sin x}{x}\Rightarrow \prod_{k\geq 1} \sec\left(\frac{x}{2^k}\right)=\frac{x}{\sin x}\Rightarrow\sum_{k\geq 1} \ln \sec\left(\frac{x}{2^k}\right)=\ln x-\ln \sin x$$ Differentiate twice and you have your result.


11

Let $$ f(x)=\prod_{n=0}^\infty\left(1-x^{2^n}\right)^{1/2^n}\tag{1} $$ and $$ g(x)=\prod_{n=0}^\infty\left(1+x^{2^n}\right)^{1/2^n}\tag{2} $$ Then $$ \begin{align} f(x)\,g(x) &=\prod_{n=0}^\infty\left(1-x^{2^{n+1}}\right)^{1/2^n}\\ &=\prod_{n=1}^\infty\left(1-x^{2^n}\right)^{2/2^n}\\ &=\left(\frac{f(x)}{1-x}\right)^2\tag{3} \end{align} $$ from ...


10

Actually, this sum converges for every $\alpha>0$. Step I. For every $x\in\mathbb R$, the sequence $s_n=\sum_{k=1}^n\sin kx$ is bounded. Indeed, if $x=m\pi$, then $s_n=0$. If $x\ne m\pi$, then $\sin(x/2)\ne 0$, and $$ s_n=\sum_{k=1}^n\sin kx=\mathrm{Im}\left(\mathrm{e}^{xi}+\mathrm{e}^{2xi}+\cdots\mathrm{e}^{nxi}\right)= ...


9

Recall that $\cos(x)=(e^{ix}+e^{-ix})/2$ and $\sin(x)=(e^{ix}-e^{-ix})/2i$, so your sum can be rewritten as $$\sum_{j=1}^k\biggl[\frac{e^{j\pi i/k}+e^{-j\pi i/k}}2\biggr]^n\,\biggl[\frac{e^{nj\pi i/k}-e^{-nj\pi i/k}}{2i}\biggr]\,.$$ Applying binomial theorem on the first factor (of each summand) your sum becomes $$\begin{align*} ...


9

We can use Riemann sums to evaluate the limit. We have $$\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=\sum_{k=1}^n\,n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\frac1n\to \int_0^1\frac{1}{1+x^2}\,dx=\pi/4$$ since we have $$\left(\frac{1}{1+(k/n)^2}\right)-\frac{1}{6n^2}\left(\frac{1}{1+(k/n)^2}\right)^3\le n\sin\left(\frac{1/n}{1+(k/n)^2}\right)\le ...


9

First notice that $\dfrac{7\pi }{8}=\pi-\dfrac{\pi}{8}$ and also $\dfrac{5\pi }{8}=\pi-\dfrac{3\pi}{8}$. Substituting these values in the original expression, you'll get: $$\left(1+\cos {\pi \over 8}\right)\left(1+\cos {3\pi \over 8}\right)\left(1-\cos {3\pi \over 8}\right)\left(1-\cos {\pi \over 8}\right)$$ Using the obvious algebraic identity, you get ...


9

Use: $$ \sin \frac{\theta}{2} \cdot \cos(k \theta) = \underbrace{\frac{1}{2} \sin\left( \left(k+\frac{1}{2}\right)\theta\right)}_{f_{k+1}} - \underbrace{\frac{1}{2} \sin\left( \left(k-\frac{1}{2}\right)\theta\right)}_{f_{k}} $$ Thus $$ \begin{eqnarray} \sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=& \sum_{k=1}^n \left(f_{k+1} - ...


8

I don't know the answer (I should have! -- see below), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight. Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series. In other words, a trigonometric series is just any series of the form ...


8

Hint: The partial sums have an explicit form, because there are the imaginary part of some geometric series. $$ \sum_{k=1}^n e^{ik} = \frac {1-e^{in}}{1-e^i} = \frac {e^{-in/2}-e^{in/2}}{e^{-i/2}-e^{i/2}} \frac{e^{in/2}}{e^{i/2}} = \frac{\sin \frac n2}{\sin \frac 12} e^{i(n-1)/2} $$ so the $n$th partial sum is $$ \frac{1}{\sin \frac 12} \sin \frac ...


8

We use this, which is somewhat complicated. Let $S$ be the product $\prod_{k=1}^{90} \sin k^\circ $. Then $S^2 = \prod_{k=1}^{179} \sin k^\circ = \frac{ 180} { 2^{179}}$ Hence $S = \sqrt{ 10} \frac{3}{2^{89} }$. I believe your method of using $\sin 2\theta$ repeatedly is better, in part because the proof of the quoted theorem is complicated.


8

We use the Poisson summation formula. Define $f(x) \equiv \sin(\pi x) / (\pi x)$. Then the sum we are trying to solve is $$g(y) = \sum_{n=-\infty}^\infty f(n-y) \, .$$ The Poisson summation formula converts the sum over values of $f$ to a sum over values of the Fourier transform of $f$. Poisson summation Note that $g(y)$ is periodic with period $1$. The ...


8

Hint: consider $$x_{n}=a^n+b^n+c^n$$ $$x_{n+3}=(a+b+c)x_{n+2}-(ab+bc+ac)x_{n+1}+abcx_{n}$$ let $a=\tan^2{20}=\tan^2{\dfrac{\pi}{9}},b=\tan^2{40}=\tan^2{\dfrac{2\pi}{9}},c=\tan^2{80}=\tan^2{\dfrac{4\pi}{9}}$ and you can add $d=\tan^2{\dfrac{3\pi}{9}}=3$ It is easy to find $$a+b+c=30,ab+bc+ac=27, abc=3$$ because you can see here lab bhattacharjee full ...


8

$$\int_0^{\pi} \sin(nx)=\frac{1}{n}\left(-\cos(nx)\right|_0^{\pi} =\frac{1}{n}\left(1-(-1)^n\right)$$ Hence, $$\int_0^{\pi} \sum_{n=0}^{\infty} \frac{n\sin(nx)}{e^n}=\sum_{n=0}^{\infty} \frac{1-(-1)^n}{e^n}=\left(\sum_{n=0}^{\infty}\frac{1}{e^n}\right)-\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{e^n}\right)$$ ...


8

$$\frac z{e^z-1}+\frac z 2=1+\sum_{n=2}^\infty\frac{B_n}{n!}z^n$$ Replace $z$ with $2iz$ to get $$\color{red}{z\cot(z)}=\frac{iz(e^{iz}+e^{-iz})}{e^{iz}-e^{-iz}}=1+\sum_{n=2}^\infty\frac{B_n}{n!}(2iz)^n=1+\sum_{n=1}^\infty\frac{B_{2n}}{(2n)!}(-1)^n(2z)^{2n}$$ Now use following trigonometric formula $$\tan(z)=\cot(z)-2\cot(2z).$$


8

Consider $$\prod_{k = 0}^{n - 2}\cos(2^k \theta)$$ Multiplying numerator and denominator by $2\sin(\theta)$ we get, $$\frac{2\sin(\theta)\cos(\theta)}{2\sin(\theta)}\prod_{k = 1}^{n - 2} \cos(2^k\theta) = \frac{\sin(2\theta)}{2\sin(\theta)}\prod_{k = 1}^{n - 2} \cos(2^k\theta)$$ Now, repeatedly multiplying and dividing by 2, we can reduce the above to, ...


8

If $z^{5}-1=0$, then $$(z^{4}+z^{3}+z^{2}+z+1)(z-1)=0,$$ because $z^{5}-1=(z^{4}+z^{3}+z^{2}+z+1)(z-1)$. This implies that $$ z^{4}+z^{3}+z^{2}+z+1=0$$ or $z-1=0$. Since \begin{equation*} z=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}=e^{i2\pi /5}\ne 1 \end{equation*} is a root of $z^5-1=0 $ and \begin{equation*} z^{k}=\cos \frac{2k\pi }{5}+i\sin \frac{2k\pi ...


7

Heuristics: Say $a_n$ is near zero but not zero, then $\sin(x)=x-\frac16x^3+o(x^3)$ when $x\to0$ hence $$a_{n+1}=a_n-\frac16a_n^3+o(a_n^3).$$ If $a_n\sim c/n^b$, this imposes that $c^2=3$ and $b=\frac12$, hence the idea that $a_n\sim\alpha\sqrt{\alpha/n}$ with $\alpha=\pm1$ depending on $a_1$. Idea of the full proof: Change variables and consider ...



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