# Tag Info

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): ...

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 $$f(x) + \frac{f(x^2)}{2} + \frac{f(x^4)}{4} + \frac{f(x^8)}{8} + \ldots = \frac{x^2}{2} + \frac{x^4}{4} + \frac{x^6}{6} + \frac{x^8}{8} + ... 17 Of course there is. The fastest way to obtain it is to heuristically write \cos(\pi\ln n) as \frac12(n^{i\pi}+n^{-i\pi}). The answer is then given by$$\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 $$\operatorname{dn}(z,k)=\frac{\pi}{2K}\left[1+2\sum_{n=1}^{\infty}\frac{\cos n\pi\frac{z}{K}}{\cosh n \pi ... 11 The sequence (\sin n) doesn't converge to 0 so the given series is divergent. 11 Hint: use the fact that \cot{x}-2\cot{2 x}=\tan{x} 10 Let$$ 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*} ... 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

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

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 $$\begin{eqnarray} \sin \frac{\theta}{2} \cdot \sum_{k=1}^n \cos(k \theta) &=& \sum_{k=1}^n \left(f_{k+1} - ... 7 Consider$$\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)$$With that as your numerator, simply cancel the common factor (\cos x + \sin x) from numerator and denominator, leaving you with the desired 2\cos x. 6 HINT:$$\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$$so that inserting x=\tfrac\pi5=\pi-4\tfrac\pi5 yields the desired result. 6 A standard example is$$ 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 ...

5

Put $e^{2\pi i/11}=:\omega$ and $e^{i\theta}\omega^n=:z_n$. Then $$\sin\left(\theta+{2n\pi\over 11}\right)={\rm Im}(z_n)={1\over 2i}\bigl(e^{i\theta}\omega^n-e^{-i\theta}\omega^{-n}\bigr)$$ and $$\sin^{14}\left(\theta+{2n\pi\over 11}\right)={-1\over 2^{14}}\sum_{k=0}^{14}(-1)^k{14\choose k}e^{i(14-2k)\theta}\>\omega^{(14-2k)n}\ .$$ Now ...

5

There's one situation in which minutes and seconds are remarkably useful: doing celestial navigation without a calculator. It turns out that the circumference of the earth at the equator is just about 360 * 60 miles; it's close enough that folks defined the "Nautical mile" to be about 7/6 of a land-mile, so that there are exactly 360 * 60 nautical miles ...

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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