TITLE: For what value of n this relation holds $\frac{x}{\pi}+\sum_{n \geq 1}\frac{2}{n\pi}\cos(n\pi)\sin(\pi x)=0$
QUESTION [1 upvotes]: I want the value of $n$ for which this relation must satisfy:
$$
\frac{x}{\pi}+\sum_{n \geq 1}\frac{2}{n\pi}\cos(n\pi)\sin(\pi x)=0
$$ How to solve this?

REPLY [0 votes]: You have
$$
\sum_{n \geq 1}\frac{2}{n\pi}\cos(n\pi)=\frac{2}{\pi}\sum_{n \geq 1}\frac{(-1)^n}{n}=-\frac{2}{\pi}\ln 2
$$ Hence, your equation is
$$
\frac{x}{\pi}-\frac{2}{\pi}\sin(\pi x)\ln 2=0
$$
which is equivalent to
$$
x-2\ln 2\:\sin(\pi x)=0
$$
Setting $u=\pi x$, you get
$$
\sin u=\alpha \:u
$$ which admits a transcendental solution $u:=u ( \alpha )$ with $
\alpha=\dfrac{1}{2\pi\ln 2}$.
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