Question

(Fourier Series) I'm trying to calculate an unusual piecewise-defined function. I've realized that it's an even function, symmetric across its period, and that I can integrate it as a half-range series, but because of the way it's piecewise-defined I'm not sure how to integrate it. More info below.

Answer 1

\[ f(x) = \left\{
\begin{array}{lr}
0, & -2<x<-1 \\
2+x, & -1<x<0 \\
2-x, & 0<x<1 \\
0, & 1<x<2
\end{array}
\right. \]

Answer 2

What I'd typically do in this situation with an even function like this, even if it has the zero (negligible) portions, is to take the function split across zero and integrate it from zero to half the period, times two. The problem is that while f(x) *is* an even function and can in theory be integrated like \[a_0 = \frac{1}{L} \int\limits_{-L}^{L}f(x)dx=\frac{2}{L}\int\limits_{0}^{L}f(x)dx\]
I'm not sure which f(x) I should use, or if I have to do something different. I'm leaning towards integrating w.r.t. the function from 0 to 1, just due to the bounds matching that portion-would that work?

Answer 3

@SithsAndGiggles

Answer 4

are you sure the bounds are 0 to 1 because it ought to be -1 to 1,-2pi to 2pi

Answer 5

That's not how these work. I'm guessing you're thinking of a specific Fourier Series, but no, a generalized fourier series can be whatever bounds you want as long as it is one period interval of a periodic function.

Answer 6

Which function you use depends on how you split the interval. You can integrate either of two ways:
\[a_0=\frac{1}{2}\int_{-2}^2f(x)\,dx=\frac{1}{2}\left.\begin{cases}\displaystyle2\int_0^2f(x)\,dx=2\int_0^1(2+x)\,dx\\\\
\displaystyle2\int_{-2}^0f(x)\,dx=2\int_{-1}^0(2-x)\,dx
\end{cases}\right\}=\frac{3}{2}\]

Answer 7

So, since I'm treating it like a half-range series, I'd just do the ones in the bounds of integration, e.g. where f(x) is defined from 0 to 1, right?

Answer 8

Exactly right

Answer 9

Although, to be more precise, the half-range is \([0,2]\), but the function is zero over \([1,2]\).

Answer 10

Yep, just meant that when integrated over that portion, it'll be zero, so the bounds can be from 0 to 1 instead. The change in bounds doesn't change the value of the integral being multiplied by two though, right? e.g. L is still 2, just like you said, even though the bounds of integration simplify from 0 to L instead, to 0 to (L-1)=1.

Answer 11

Right, we have \(\int_{-2}^2=2\int_0^2=2\int_0^1\).