# Tag Info

29

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 $$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + ... 28 It is a simple trigonometric exercise to show that$$\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{$\spadesuit$}$$with z_0=e^{i\pi/4}. Recall the Ramanujan identity (several nice proofs of which may be found here): ... 17 Of course there is. The fastest way to obtain it is to heuristically write \cos(\pi\ln n) as \frac12(n^{i\pi}+n^{-i\pi}). The answer is then given by$$\sum_{n=1}^{\infty}\frac{\cos\left(\pi\ln n\right)}{n^2}=\frac{\zeta(2+i\pi)+\zeta(2-i\pi)}{2}.$$12 Since:$$\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

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 $$\operatorname{dn}(z,k)=\frac{\pi}{2K}\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\pi\frac{z}{K}}{\cosh n \pi ... 11 Hint: use the fact that \cot{x}-2\cot{2 x}=\tan{x} 11 Let$$ 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)= ...

10

The main building block of our solution will be the formula \begin{align*}\prod_{k=0}^{N-1}\left(x-e^{\frac{2k i\pi}{N}}\right)=x^N-1.\tag{0} \end{align*} It will be convenient to rewrite (0) for odd $N=2n+1$ in the form \begin{align*} \prod_{k=1}^{n}\left[x^2+1-2x\cos\frac{\pi k}{2n+1}\right]=\frac{x^{2n+1}-1}{x-1}. \tag{1} \end{align*} Replacing therein ...

9

$$\tan rx=\frac{\binom r1\tan x-\binom r3\tan^3x+\cdots}{1-\binom r2\tan^2x+\cdots}\text{ (Proof below) }$$ 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) $$\tan2m\theta=\tan2mx=\frac{\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom ... 9 Hint: consider$$x_{n}=a^n+b^n+c^nx_{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$$because you can see here lab bhattacharjee full ... 9 Recall that \cos(x)=(e^{ix}+e^{-ix})/2 and \sin(x)=(e^{ix}-e^{-ix})/2i, so your sum can be rewritten as$$\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*} ...

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

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 ...

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

$$\int_0^{\pi} \sin(nx)=\frac{1}{n}\left(-\cos(nx)\right|_0^{\pi} =\frac{1}{n}\left(1-(-1)^n\right)$$ 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

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 ...

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 $$\frac{1}{\sin \frac 12} \sin \frac ... 8 Use:$$ \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 ...

7

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, ...

7

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 ...

6

HINT: $$\frac{1}{\sin k^\circ\sin(k+1)^\circ}=\frac1{\sin1^\circ}\frac{\sin (k+1-k)^\circ}{\sin k^\circ\sin(k+1)^\circ}$$ $$=\frac1{\sin1^\circ}\cdot\frac{\cos k^\circ\sin(k+1)^\circ-\sin k^\circ\cos(k+1)^\circ}{\sin k^\circ\sin(k+1)^\circ}=\frac1{\sin1^\circ}\left(\cot k^\circ-\cot(k+1)^\circ\right)$$ Can you recognize Telescoping series / sum?

6

The trick is to turn it into a telescoping series. Since $$\tan (a - b) = \frac{\tan a - \tan b}{1+\tan a \tan b},$$ the thing to try is to write $\frac{4}{4n^2+3}$ in the form $\frac{c_n-c_{n+1}}{1+c_nc_{n+1}}$. A little experimentation leads to $$c_n = \frac{1}{n-\frac{1}{2}}.$$

6

You can also use that $$2\sin(x)\Bigl[\cos 2x+\cos 4x + \cos 6x+\cos 8x\Bigr] = \sin 9x-\sin x = 2\cos 5x\sin 4x$$ so that inserting $x=\tfrac\pi5=\pi-4\tfrac\pi5$ yields the desired result.

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