Fourier Series of a function not centered at zero

[MedPhysKid235]

Homework Statement

f(x)=x on [0,2)

Homework Equations

Fourier Series is given as:
f(x)=a₀/2 + _n=1^∞∑(a_n*cos(nπx/L) + b_n*sin(nπx/L)
a₀=1/L*_-L^L∫f(x)dx

The Attempt at a Solution

Basically what I am being taught is that we take the Period, T, to be equal to 2L so, T=2L
In this case T=2 and L=1. My issue arises when looking at my limits of integration. If the function was centered at zero and the range was [-1, 1) that would be fine, but in this case it isn't. So doing the integral from -L to L doesn't make sense to me for this question since the function from 0 to 2 which I want to represent by the Fourier series is different than f(x) on -1 to 1.

I guess you could say this is more of a situational based question rather than how to actually solve it, I am just trying to understand which Fourier series equations to use and how to plug the given functions into them, as this was something that was very poorly taught to me and I can't find much information on this situation.

Thanks in advance!

 

[LCKurtz]

The way your problem is given, there could be many answers. If you just care that you have a FS that represents $x$ on $[0,2)$, you could, for example, use $f(x) = |x|$ on $[-2,2]$ and expand it in a half-range cosine expansion. Or you could use $f(x) = x$ on $[-2,2]$ and use a half-range sine expansion. Or if you actually want what you have given to be a full period, you could use $T = 2$ and simply take $a_n = \frac 2 T \int_0^T x \cos(\frac{2n\pi x}{T})$ and similarly for $a_0$ and $b_n$.

In terms of $L = \frac T 2$ your $a_n = \frac 1 L \int_0^{2L} x \cos(\frac{n\pi x}{L})$. Note that this last integral could be over any period, but you want to use a period where you have the correct formula for $f(x)$.

 

[vela]

Remember that the function is assumed to be periodic. I've attached a plot of two periods of the function based on your interpretation of the problem. You should be able to convince yourself that integrating from -1 to 0 gives you the same result as integrating from 1 to 2 since the function of interest, sine, and cosine are all periodic. You can integrate over any interval of length 2L, so choose the one where it's easiest to write down the integrals.

 

[MedPhysKid235]

ahhhh, so I can integrate over any part of the periodic function. Thanks so much for the help, I'll let you know how I make out.

 

[LCKurtz]

ahhhh, so I can integrate over any part of the periodic function. Thanks so much for the help, I'll let you know how I make out.

That is generally true for any periodic function. Say $f(x)$ has period $P$, so $f(x+P)=f(x)$ for all $x$. Then $\int_t^{t+P}~f(x)~dx$ is independent of $t$. So it doesn't matter where the integral starts. While that may seem "obvious", if you want to prove that is true, call that expression $H(t)$ and calculate $H'(t)$.