MHB Yes, $\cos \alpha x$ can be considered a $2\pi-$periodic function.

[Markov2]

1) Find the Fourier series of the $2\pi-$periodic function defined by $f(x)=2x,\,-\pi\le x<\pi.$

2) Use the Fourier series of $f(x)=\cos \alpha x,$ with $0\ne\alpha\in\mathbb R$ to show that $\displaystyle\cot \alpha \pi = \frac{1}{\pi }\left( {\frac{1}{\alpha } - \sum\limits_{n = 1}^\infty {\frac{{2\alpha }}{{{n^2} - {\alpha ^2}}}} } \right).$

Attempts:

1) I have $a_n=\displaystyle\frac1\pi\int_{-\pi}^\pi 2x \cos (nx)\,dx$ and $b_n=\displaystyle\frac1\pi\int_{-\pi}^\pi 2x \sin (nx)\,dx,$ and $a_0=\displaystyle\frac1\pi\int_{-\pi}^\pi 2x\,dx,$ so the Fourier series is $\displaystyle\frac{{{a_0}}}{2} + \sum\limits_{n = 1}^\infty {\left( {{a_n}\cos (nx) + {b_n}\sin (nx)} \right)} .$ Is this correct?

2) Do I use the standard period for $\cos\alpha x$ ? I mean $-\pi\le x<\pi$ then calculate the series as did in (1)? However I don't see how to prove the identity.

Thanks.

 

[Mathwebster]

For (2), yes do a Fourier series expansion for $\cos \alpha x$ then set $x = \pi$. See how that goes.

 

[Markov2]

Okay I can do that, but I don't see how to prove the identity though.
How about (1)? Is it correct?

 

[Mathwebster]

Yes, (1) looks good. Post your result for the Fourier series for $\cos \alpha x$ so we can get to the result.

 

[Markov2]

But do I consider $\cos\alpha x$ a $2\pi-$periodic function right?