# Thread: Fourier Series - A few questions

1. ## Fourier Series - A few questions

Hi, I have a few questions about Fourier series.

I understand the derivation of the formulas for working out the coefficients in your series, but I have a few queries about your bounds of integration, and whether or not you need to change these bounds depending on what domain you are working in

For example, when I calculate the fourier series for $\displaystyle f(x) = 2x^2$ over the domain of $\displaystyle -1\leq x < 0$ and $\displaystyle f(x+1) = f(x), \forall x \in \mathbb{R}$.... I get the same as if my function was defined as $\displaystyle f(x) = 2x^2$ when $\displaystyle x \in [0,1)$. And then $\displaystyle f(x) = f(x+1),\forall x \in \mathbb{R}$ I believe these functions do NOT look the same.

Avoid my abuse of notation, but I only want to talk about bounds of integration....
When integrating we get $\displaystyle \int_{-L}^{L}=\int_{-\frac{1}{2}}^{\frac{1}{2}}$.

Now my professor has said that since the function is periodic we can 'shift' the bounds to make it easier, for example change it to $\displaystyle \int_0^{1}$.

The thing I'm having difficulty with is that these integrals do not evaluate to the same thing. Take the example of $\displaystyle f(x)=2x^2$. Now drawing the function over the x-axis, I can see that the area under the curve(s) from -1/2 to 1/2 is the same as the area under the curve from 0 to 1. So I can see the motivation there. Am I stuffing up here because in the case when you consider in the domain of $\displaystyle x\in [-1,0)$ that when you integrate from 0 to 1, if you look at the graph, it's actually the function $\displaystyle 2(1-x)^2$? Since you have shifted over, but you are still just integrating $\displaystyle 2x^2$ which may cause some trouble...

If this isn't it... i'm really stuck... because I end up getting the same Fourier series when the function is initially defined in 2 different domains. And these functions do NOT look the same!

Any help would be appreciated, thanks!

2. ## Re: Fourier Series - A few questions

Originally Posted by mathswannabe
Hi, I have a few questions about Fourier series.

I understand the derivation of the formulas for working out the coefficients in your series, but I have a few queries about your bounds of integration, and whether or not you need to change these bounds depending on what domain you are working in

For example, when I calculate the fourier series for $\displaystyle f(x) = 2x^2$ over the domain of $\displaystyle -1\leq x < 0$ and $\displaystyle f(x+1) = f(x), \forall x \in \mathbb{R}$.... I get the same as if my function was defined as $\displaystyle f(x) = 2x^2$ when $\displaystyle x \in [0,1)$. And then $\displaystyle f(x) = f(x+1),\forall x \in \mathbb{R}$ I believe these functions do NOT look the same.

Avoid my abuse of notation, but I only want to talk about bounds of integration....
When integrating we get $\displaystyle \int_{-L}^{L}=\int_{-\frac{1}{2}}^{\frac{1}{2}}$.

Now my professor has said that since the function is periodic we can 'shift' the bounds to make it easier, for example change it to $\displaystyle \int_0^{1}$.

The thing I'm having difficulty with is that these integrals do not evaluate to the same thing. Take the example of $\displaystyle f(x)=2x^2$. Now drawing the function over the x-axis, I can see that the area under the curve(s) from -1/2 to 1/2 is the same as the area under the curve from 0 to 1. So I can see the motivation there. Am I stuffing up here because in the case when you consider in the domain of $\displaystyle x\in [-1,0)$ that when you integrate from 0 to 1, if you look at the graph, it's actually the function $\displaystyle 2(1-x)^2$? Since you have shifted over, but you are still just integrating $\displaystyle 2x^2$ which may cause some trouble...

If this isn't it... i'm really stuck... because I end up getting the same Fourier series when the function is initially defined in 2 different domains. And these functions do NOT look the same!

Any help would be appreciated, thanks!
Your function is periodic so you can use any bounds that correspond to a periiod. But you have to use the right function values.

So:

$\displaystyle f(x)=2x^2,\ \ x\in [-1,0)$ and periodic with period $\displaystyle 1$

is:

$\displaystyle f(x)=2(x-1)^2,\ \ x \in [0,1)$ and periodic with period $\displaystyle 1$

CB