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28

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 $$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + ... 26 It is a simple trigonometric exercise to show that$$\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}.$$11 It will be helpful to start from an explanation of the origin and the proof of the Ramanujan identity. These are hidden (not very deeply) in the theory of elliptic functions. Indeed, Jacobi elliptic function \operatorname{dn}(z,k) has Fourier series$$\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

The sequence $(\sin n)$ doesn't converge to $0$ so the given series is divergent.

10

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. The sequence $s_n=\sum_{k=1}^n\sin kx$ is bounded. Indeed, if $x=2m\pi$, then $s_n=0$. If $x\ne2m\pi$, 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

Hint: use the fact that $\cot{x}-2\cot{2 x}=\tan{x}$

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 $$\frac{1}{\sin \frac 12} \sin \frac ... 8 I don't know the answer (I should have! -- see below), but in my opinion a lot of people are misinterpreting the question, so perhaps it is worth an answer to try to set this straight. Here is an analogy: trigonometric series is to Fourier series as power series is to Taylor series. In other words, a trigonometric series is just any series of the form ... 8 Heuristics: Say a_n is near zero but not zero, then \sin(x)=x-\frac16x^3+o(x^3) when x\to0 hence$$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$$or z-1=0. Since \begin{equation*} z=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}=e^{i2\pi /5}\ne 1 \end{equation*} is a root of z^5-1=0  and \begin{equation*} z^{k}=\cos \frac{2k\pi }{5}+i\sin \frac{2k\pi ... 7 Consider$$\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)$$Now, repeatedly multiplying and dividing by 2, we can reduce the above to, ... 7 You've done fine so far. Just factor out 2\cos x from the numerator:$$2 \cos^2 x + 2 \cos x \sin x = 2\cos x(\cos x + \sin x)$$With that as your numerator, simply cancel the common factor (\cos x + \sin x) from numerator and denominator, leaving you with the desired 2\cos x. 7 Use:$$ \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} - ...

6

A standard example is $$f(t)= \sum_{n>1} \frac{\sin(nt)}{\log(n)}$$ The conjugate of $f$ is a Fourier series but $f\not\in L^1(\mathbb{T})$ and hence is no Fourier series. For further explanation see Katznelson's book page 85. (Edit: If $f$ is not in $L^1(\mathbb{T})$ it is hard to define Fourier coefficients. Added 29/8 - 2010 Here is a screen dump from ...

6

$$\tan rx=\frac{\binom r1\tan x-\binom r3\tan^3x+\cdots}{1-\binom r2\tan^2x+\cdots}\text{ (Proof below) }$$ 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) $$\tan2m\theta=\tan2mx=\frac{\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom ... 6 HINT:$$\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)$$Can you recognize Telescoping series / sum? 6 Hint$$a_r=r(r+1)\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}=\frac{2r}{1-r^2+r^4}$$6 You can also use that$$2\sin(x)\Bigl[\cos 2x+\cos 4x + \cos 6x+\cos 8x\Bigr] = \sin 9x-\sin x = 2\cos 5x\sin 4x$$so that inserting x=\tfrac\pi5=\pi-4\tfrac\pi5 yields the desired result. 5 Here's a couple other trigonometric approaches:$$ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{4\pi}{5}\right) \ + \ \cos\left(\frac{6\pi}{5}\right) \ + \ \cos\left(\frac{8\pi}{5}\right)  = \ \left[ \cos\left(\frac{2\pi}{5}\right) \ + \ \cos\left(\frac{8\pi}{5}\right) \right] \ + \ \left[\cos\left(\frac{4\pi}{5}\right) \ + \ ...

5


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