# How to calculate this fourier series?

So I've been asked to find the fourier series of the following function:

The method I use is explained in this old post: Express in terms of Legendre polynomials (Method 2 in the top answer)

How does the method used here differ when f(x) is set out like it is in my question and how does the "f(x+4)) = f(x)" affect my working out/answer? Or is there a completely new method I should use to tackle a function set out like this?

• Here, the main difficulty is that the period is 4 (not 1 or $2\pi$), so the Fourier basis is $e^{ni\pi {x\over 2}}$, or $\cos ( n\pi {x\over 2}), \sin ( n\pi {x\over 2})$, then you can use safely standard formulas. Apr 14, 2021 at 13:48

Your are over-complicating things. The fact that $$f(x+4)=f(x)$$ only implies that your fuction is $$4$$-periodic, and thus through a transformation of the variable

$$x\leftrightarrow \frac{2 x}{\pi}$$

you could consider the $$2\pi$$-periodic, piece-wise continuous function

$$g(x) = f\left(\frac{2 x}{\pi}\right),\quad x\in [-\pi,\pi]$$

on which you can apply the usual methods to calculate the Fourier coefficients. Note, of course, that the integrals should be split in two, given that $$f$$ is defined piece-wise.

Further explanation

The logic behind this variable transformation is that, for a $$T$$-periodic function defined on $$[-\frac{T}{2}, \frac{T}{2}]$$, you are looking for a $$k$$ satisfying $$k \pi = \frac{T}{2}$$ (so at $$x = \pi$$ you get $$k x = \frac{T}{2}$$). In your case $$\frac{T}{2} = 2$$ so $$k = \frac{2}{\pi}$$.

If your $$n$$th Fourier coefficient for a $$2\pi$$-periodic function $$f$$ is given by

$$c_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-i n x} dx$$

then you can calculate it too for you $$T$$-periodic function $$f$$ thanks to the previous variable transformation and applying the corresponding change of variables to the integral:

$$c_n = \frac{1}{2\pi}\int_{-\pi}^\pi f\left(\frac{T}{2\pi}x\right)e^{-i n x} dx = \frac{1}{T}\int_{-T/2}^{T/2} f\left(s\right)e^{-i n s (2 \pi / T)} ds$$

where the change of variables was $$\frac{T}{2 \pi} x \equiv s$$. If it is not clear for you why we had to perform the transformation $$x\leftrightarrow \frac{2 x}{\pi}$$, think about the fact that the Fourier coefficient is calculated as an integral from $$-\pi$$ to $$\pi$$ for a function whose period, in your case, is defined on $$[-2,2]$$, so we need to "tweak" our function for it to be periodic over $$[-\pi, \pi]$$.

• Ok, it makes sense but I have a couple questions, how did you get that transformation that turns x into 2x/pi? And am i now essentially working out the fourier series for 2x/pi? What happened to f(x) being 0 between -2 and 0 and f(x) being x if between 0 and 2? Apr 14, 2021 at 15:50
• @Maximus, look to the edited answer. Apr 14, 2021 at 19:21
• Thanks a lot, I think I get it now Apr 15, 2021 at 10:47