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For $f(x)=-x, 0\leq x \leq 1$, extend $f(x)$ into an even function into $-1 \leq x \leq 0$ and then regard $f(x)$ as a periodic function on $ -\infty < x < \infty $. Find the Fourier Series for this function.

So I think we have: $$f(x)=a_0 + \sum_{n=0}^{\infty} a_n \cos(\frac{n \pi x}{2})+b_n \sin(\frac{n \pi x}{2})$$ Since $f(x)$ is now even, the $b_n$ coefficients will be $0$.

So, calculating the coefficients: $$a_0=\frac{1}{2} \int_0^2(-x)dx=-4$$

$$a_n= \int_0^2(-x)\cos(\frac{n \pi x}{2})dx=[\frac{-2x}{n \pi}\sin(\frac{n \pi x}{2})]_0^2+\frac{2}{n \pi}\int_0^2 \sin(\frac{n \pi x}{2}) dx$$

$$=\frac{-4}{n^2 \pi^2}[\cos(\frac{n \pi x}{2})]_0^2=\frac{-4}{n^2 \pi^2}[(-1)^n-1]$$

For even n, this is $0$. So let $n=2m-1, m=1,2,...$

Hence the final answer is: $$f(x)=-4+\sum_{m=1}^{\infty}\frac{8}{(2m-1)^2\pi^2}\cos(\frac{(2m-1)\pi x}{2})$$

Is this right? I'm not sure if I extended the function correctly at the start. I drew a sketch of it and it kinda resembles lots of W's under the x-axis.

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When I want to check my computation of a Fourier series, I just ask Wolfram Alpha to plot it:

plot

The plot shows that you got the curve mostly right, but it's too low under the axis. Look at the constant coefficient: you forgot the factor of $\frac12$ for $a_0$. Otherwise correct.

The repeated-W shape (triangular wave) is to be expected because the even reflection of $-x$ is $\wedge$-shaped ($y=-|x|$ in a formula). Then this shape repeats periodically: $\wedge\!\!\wedge\!\!\wedge\!\!\wedge$

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  • $\begingroup$ thank you, and thank you for the advice about Wolfram Alpha! $\endgroup$ – Maths student May 25 '15 at 17:36

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