Question

(Fourier Series) I'm having trouble understanding how to set up a given Fourier Series, example put below.

Answer 1

This is it written up in LaTeX: http://latex.codecogs.com/png.latex?%5Ctext%7BWrite%20the%20Fourier%20cosine%20series%20for%20the%20function%7D%20%5C%5C%20%5C%5C%20f%28x%29%20%3D%20%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Blr%7D%20%5Ccos%28x%29%20%26%20%5Ctext%7Bwhen%7D%20%5C%200%3Cx%5Cleq%20%5Cpi/2%20%5C%5C%200%20%26%20%5Ctext%7Bwhen%7D%5C%20%5Cpi/2%20%5Cleq%20x%20%3C%20%5Cpi%20%5Cend%7Barray%7D%20%5Cright.

Answer 2

@ikram002p @perl

Answer 3

My confusion is whether L in this case (Where f(x) is a periodic function with period 2L, or -L<x<L)-whether it's pi/2 or not. half-range series tend to behave "oddly" and have a slightly different setup than normal full-range series-I understand this, but I don't totally grasp how to set them up well.

Answer 4

@SithsAndGiggles

Answer 5

The period here is \(\pi\).

Answer 6

So if I were to set up a fourier cosine series where the integral's bounds run from zero to L, would the bounds be from zero to pi/2, then?

Answer 7

Right, since \(2L=\pi\). Coefficients would be computed using the form
\[\frac{1}{L}\int_0^{2L}\cdots\]

Answer 8

Oh, so it'd be from zero to pi, lol. Thank you. I've been having this issue with some other problems and want to make absolutely sure I get this, should I post them here, or do you want me to open a new question?

Answer 9

Feel free. I have to go now, but I'll probably be back later.

Answer 10

Okay, sure thing; I'll just put it here for now. Thanks again for your help.

Answer 11

OS has been randomly refreshing on me, and I just lost everything I wrote...

https://s3.amazonaws.com/iedu-attachments-question/875d339debeca1a951697d60d25009b2_f80f1b83e31f1e8b2386caf91f5172df.png

^For this one, I think it's pi-periodic, thus L=pi/2, and the bounds of integration for a half-series would be from 0 to 2L or 0 to pi; the coefficient out front on a_0 or a_n would be 2/L=2/(pi/2)=4/pi, is that correct?

https://s3.amazonaws.com/iedu-attachments-question/46e9e4928e8049e2cbb40c6c23f1bb7e_42140d52559de40ede1cb4a1929123b8.png

^For this one, I'd think the exact same thing.

And something that confuses me...is that all of my book's half-range formula-rather than putting the upper bound to 2L, it just has it to L. Granted, it multiplies the whole expression by 2.

Answer 12

Yes and yes. It seems your text is assuming \(f(x)\) is symmetric across the period, which would only be okay if \(f\) is even (on the interval). For example, \(2\int_0^L\) would not work for \(f(x)=x^2\), but it would for \(f(x)=\left(x-\dfrac{\pi}{2}\right)^2\).

Answer 13

I get a little mixed up dealing with the notation for the coefficients because some people use a_0/2, and then use a constant *inside* that expression additionally, etc etc, if I could describe the fourier coefficients as multiplied by what constants (not writing it out, just like, "this thing multiplied by one half, regardless of formulae used"-how would I say that?

Answer 14

The way I remember the formulas are by fixing the coefficient of the integrals as \(\dfrac{1}{L}\), and that the constant term of the series is \(\dfrac{a_0}{2}\).

Answer 15

That's a better way to remember it IMO, thanks.

Answer 16

I'm still a little confused about what you said here, and I have to feel it's a typo or someting: http://i.imgur.com/QcH2kH9.png

If f(x)=x^2, f(x) is an even function (restating for my own sake), f(x)=f(-x), and is-at least on an interval centered on the origin, symmetric across the period, right? Wouldn't it be the exact opposite that would work, e.g. that integral would hold true for f(x)=x^2 and not f(x)=(x-pi/2)^2, with an interval centered on the origin?

Answer 17

My book, instead of integrating from 0 to 2L, for an even function over a symmetric interval (-L, L), multiplies the whole thing by 2, which I'm guessing is equivalent.

Answer 18

What I meant was that over the interval \([0,\pi]\), the function \(f(x)=x^2\) is not symmetric about \(x=\dfrac{\pi}{2}\) (whereas \(x^2\) is symmetric across \(x=0\), but only if \(0\) is the midpoint of the interval you're considering). This means you can't use the shortcut your text mentions, that
\[a_0=\frac{1}{L}\int_0^{2L}f(x)\,dx=\frac{2}{\pi}\int_0^{\pi}x^2\,dx\neq\frac{4}{\pi}\int_0^{\pi/2}x^2\,dx\]
but if \(f(x)=\left(x-\dfrac{\pi}{2}\right)^2\), you can:
\[a_0=\frac{2}{\pi}\int_0^{\pi}\left(x-\frac{\pi}{2}\right)^2\,dx=\frac{4}{\pi}\int_0^{\pi/2}\left(x-\frac{\pi}{2}\right)^2\,dx\]

Answer 19

So while \(x^2\) IS still an even function, it's not even relative to the axis \(x=\dfrac{\pi}{2}\)

Answer 20

The last thing you said is right, if \(f\) is even relative to the axis \(x=0\) (i.e. the typical symmetry seen in an even function). If \(f(x)\) is an even function defined over a symmetric interval, say \([-\pi,\pi]\), then indeed,
\[a_0=\frac{1}{L}\int_{-L}^Lf(x)\,dx=\frac{2}{L}\int_0^Lf(x)\,dx\]