# 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 ...

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 $$\operatorname{dn}(z,k)=\frac{\pi}{2K}\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\pi\frac{z}{K}}{\cosh n \pi ... 11 Hint: use the fact that \cot{x}-2\cot{2 x}=\tan{x} 11 Let$$ 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} $$from ... 10 Actually, this sum converges for every \alpha>0. Step I. The sequence s_n=\sum_{k=1}^n\sin kx is bounded. Indeed, if x=2m\pi, then s_n=0. If x\ne2m\pi, then \sin(x/2)\ne 0, and$$ 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*} ... 8 Heuristics: Say a_n is near zero but not zero, then \sin(x)=x-\frac16x^3+o(x^3) when x\to0 hencea_{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} $$so the nth partial sum is$$ \frac{1}{\sin \frac 12} \sin \frac ...

8

$$\tan rx=\frac{\binom r1\tan x-\binom r3\tan^3x+\cdots}{1-\binom r2\tan^2x+\cdots}\text{ (Proof below) }$$ 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) $$\tan2m\theta=\tan2mx=\frac{\binom {2m}1\tan x-\binom {2m}3\tan^3x+\cdots+\binom ... 8 Use:$$ \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.

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

Here is a rather elementary and simple way to sum this series: We know from De Moivre's Formula that $$(\cos\theta+\iota\sin\theta)^n=\cos n\theta+\iota\sin n\theta$$ Expanding the LHS using Binomial Theorem, $$\binom{n}{0}\cos^n\theta+\binom{n}{1}\cos^{n-1}\theta(\iota\sin\theta)+\cdots+\binom{n}{n}\iota^n\sin^n\theta=\cos n\theta+\iota\sin n\theta$$ ...

6

Hint Use the identity $$\cos\frac x{2^n}=\frac12\frac{\sin\frac x{2^{n-1}}}{\sin\frac{x}{2^n}}$$ and telescope.

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