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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} + ...


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 $$\sum_{n=1}^{\infty}\frac{\cos\left(\pi\ln n\right)}{n^2}=\frac{\zeta(2+i\pi)+\zeta(2-i\pi)}{2}.$$


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 $x=2m\pi i$, then $s_n=0$. If $x\ne2m\pi i$, 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

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 ...


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

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

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

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


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

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.


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

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

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

$$\frac z{e^z-1}+\frac z 2=1+\sum_{n=2}^\infty\frac{B_n}{n!}z^n$$ 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 $$\tan(z)=\cot(z)-2\cot(2z).$$


5

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?


5

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

Let $\zeta_n = e^{2\pi i/n} = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$ be a primitive $n^{\rm th}$ root of unity, so in particular, the roots of $z^5 - 1$ are $\zeta_5^k$ for $k = 0, 1, 2, 3, 4$. Since the sum of the roots of a polynomial of degree $n$ is equal to the negative of the coefficient of the degree $n-1$ term, it follows that $$\sum_{k=0}^4 ...


5

Let $\omega = \exp\left(\frac{2\pi i}{5}\right) = \cos\left(\frac{2\pi}{5}\right) + \sin\left(\frac{2\pi}{5}\right)$. Then $\omega$ is a fifth root of unity ($\omega^5 = 1$). Then $$1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0.$$ By taking the real part of both sides (after applying De Moivre's Theorem), we obtain $$1 + \cos\left(\frac{2\pi}{5}\right) + ...


5

Solve the equation $z^5=1$, for $z\in\mathbb{C}$. The solutions are $1$, $e^{i2\pi/5}$, $e^{i4\pi/5}$, $e^{i6\pi/2}$, $e^{i8\pi/2}$. Since those are the fifth roots of unity their sum is $0$. Use Euler's formula, $$e^{ix}=\cos x+ i\sin x,$$ and consider that since the sum of the roots is zero, then so must be its real part. Taking the real part of the roots ...


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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5

Hint $$a_r=r(r+1)$$ $$\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}=\frac{2r}{1-r^2+r^4}$$


5

[The proof has been completed following the strategy suggested in the earlier revisions of the answer.] For $x = \frac{2u}{n}$ the limit of $a_n$ is $\frac{\sin^2 u}u$ and one can take $u$ to maximize this. Let $M$ be the maximum value of $(\sin^2{u})/u$. We have proved that $\limsup a_n \geq M$ and want to show the opposite inequality to prove $\lim a_n = ...


5

$$\tan rx=\frac{\binom r1\tan x-\binom r3\tan^3x+\cdots}{1-\binom r2\tan^2x+\cdots}\text{ (Proof below) }$$ 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

Hint You could use $$\int _0^{\pi }\:\sum _{n=0}^{\infty \:}\frac{n\cdot \sin \left(nx\right)}{e^n}dx=\sum _{n=0}^{\infty \:}\int _0^{\pi }\:\frac{n\cdot \sin \left(nx\right)}{e^n}dx=\sum _{n=0}^{\infty \:}e^{-n} (1-\cos (\pi n))$$ At this point, I stop since appeared Pranav Arora's very nice answer.


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

Using the fact that $$\frac{\sin 2nx}{\sin x} = 2 \cos(2n-1)x + 2 \cos(2n-3)x + \ldots + 2 \cos x$$ we have $$ \begin{align} \int_{0}^{\pi /2} \frac{\sin 2nx}{\sin x} \ dx &= 2 \int_{0}^{\pi/2} \left(\cos x + \cos 3x + \ldots + \cos(2n-1) \ x \right) \ dx \\ &= 2 \left(\sin x + \frac{1}{3} \sin (3x) + \ldots + \frac{1}{2n-1} \sin(2n-1) x ...


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 ...



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