Complex fourier Series

Question

I would like to ask some help on complex fourier series. For this fourier series, its quite hard for me because it look confusing when solving it. This is my question

1) Find the complex form of the Fourier Series expansion of

$$f(x)=\cos ax, \quad -\pi< x <\pi. $$

Someone please show me some work done and steps to solve it so that I can understand better. I'm a beginner in maths.

Answer 1

Expand $\cos (ax) = \frac{1}{2} (e^{iax}+e^{-iax})$.

Then $\hat{f_n} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} \cos (ax) e^{-inx} dx = \frac{1}{2 \pi} \frac{1}{2}\int_{-\pi}^{\pi} (e^{iax}+e^{-iax}) e^{-inx} dx = \frac{1}{4 \pi} \int_{-\pi}^{\pi} (e^{i(a-n)x}+e^{-i(a+n)x}) dx$.

This is a straightforward integration. Take care when $a$ is an integer (in which case the Fourier series is trivially obtained from the above expansion).

If $a \in \mathbb{Z}$, then the formula for $\cos$ gives $f_n = \frac{1}{2} \delta_{|n||a|}$ (ie, $f_n = \frac{1}{2}$ iff $n = \pm a$).

If $a \notin \mathbb{Z}$, then $\hat{f_n} = \frac{1}{4 \pi} \int_{-\pi}^{\pi} (e^{i(a-n)x}+e^{-i(a+n)x}) dx = \frac{1}{4 \pi}( \left.\frac{e^{i(a-n) x}}{i (a-n)}\right|_{-\pi}^{\pi} + \left.\frac{e^{-i(a+n) x}}{-i (a+n)}\right|_{-\pi}^{\pi})$, noting that $e^{i n \pi} = (-1)^n$, we have

$$ \hat{f_n} = \frac{1}{4 \pi i} (-1)^n( \frac{e^{i a\pi} - e^{-i a\pi}}{a-n} - \frac{e^{-i a\pi} - e^{i a\pi}}{a+n}) \\ = \frac{1}{4 \pi i} (-1)^n (e^{i a\pi} - e^{-i a\pi})(\frac{1}{a-n} + \frac{1}{a+n})\\ = \frac{1}{2 \pi}(-1)^n \sin ( a \pi) \frac{2 a}{a^2-n^2} \\ = \frac{1}{ \pi}(-1)^{n+1} \sin ( a \pi) \frac{a}{n^2-a^2} $$

Answer 2

$$c_n=\frac{1}{2\pi}\int_{-\pi}^\pi\cos ax\,\,e^{-inx}dx$$

We make integration by parts twice to

$$I:=\int_{-\pi}^\pi\cos ax\,\,e^{-inx}dx=\left.\frac{i}{n}\cos ax\,\,e^{-inx}\right|_{-\pi}^\pi+\frac{ai}{n}\int_{-\pi}^\pi\sin ax\,\,e^{-inx}dx=$$

$$=\left.-\frac{a}{n^2}\sin ax\,\,e^{-inx}\right|_{-\pi}^\pi+\frac{a^2}{n^2}I$$

I think you can handle it from here.

Answer 3

Complex form of Fourier series is given by $$f(x) \sim \sum\limits_{n=-\infty}^{\infty}c_n(f)e^{inx}.$$ You need to find Fourier coefficients $$c_n(f)=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(x)e^{-inx}\;dx\,.$$ Thus, simply integrate by parts (twice)$$c_n(f)=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}f(x)e^{-inx}\;dx=\frac{1}{2\pi}\int\limits_{-\pi}^{\pi}\mathrm{cos}(ax)e^{-inx}\;dx.$$