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

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?

Answer 1

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}$.