Did I set this Fourier series up correctly?

snesnerd
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If you take the Fourier series of a function $f(x)$ where $0 < x < \pi$, then would $a_{0}$, $a_{n}$, and $b_{n}$ be defined as,

$a_{0} = \displaystyle\frac{1}{\pi}\int_{0}^{\pi}f(x)dx$

$a_{n} = \displaystyle\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos(nx)dx$

$b_{n} = \displaystyle\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx$

assuming that I am using the definition of a Fourier series in the following way:

$f(x) = \displaystyle\frac{a_{0}}{2} + \sum_{k=1}^{\infty}a_{n}\cos\left(\frac{n\pi x}{L}\right) + b_{n}\sin\left(\frac{n\pi x}{L}\right)$

I only ask because I found this article:

http://people.uncw.edu/hermanr/mat463/ODEBook/Book/Fourier.pdf

On page 163 he defines the Fourier series from $[0,L]$ but in his definition, he has

$\cos\left(\frac{2n\pi x}{L}\right)$ and $\sin\left(\frac{2n\pi x}{L}\right)$ in his formulation. Is the way I set it up correct or his way? I have seen it both ways. I have also seen $a_{0}$ defined with the constant $\displaystyle\frac{2}{\pi}$ in front of the integral at times too so I am unsure if mine is correct or I need to add the two.
 
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Hmmmm it did not translate my latex writing. I apologize for that.
 
snesnerd said:
Hmmmm it did not translate my latex writing. I apologize for that.

Edit your post and put double pound signs ## around inline tex and double dollar signs $$ around displayed tex. Then use the preview button before posting it.
 
Given f: [0,L] \to \mathbb{R} there are three ways to express it as a Fourier series.

First, you can expand it as is, and you'll have in general both sines and cosines of frequencies 2\pi n/L for each positive integer n (because when x = L you want 2\pi n x/L = 2n\pi).

Secondly, you can extend it to [-L,L] as an even function by setting f(-x) = f(x) for each x \in [0,L] and using the series of that. Because it's an even function the series will only have cosine terms, and the frequencies are \pi n/L for each positive integer n (because when x = \pm L you want \pi nx/L = \pm n\pi).

Thirdly, you can extend it to [-L,L] as an odd function setting f(-x) = -f(x) for each x \in [0,L] and using the series of that. Because it's an odd function the series will only have sine terms, and the frequencies are \pi n/L for each positive integer n (because when x = \pm L you want \pi nx/L = \pm n\pi).
 
pasmith said:
Given f: [0,L] \to \mathbb{R} there are three ways to express it as a Fourier series.

Actually, there are infinitely many ways, of which, admittedly, most are pointless. But you could define ##f(x)## pretty much any way you want on ##(-L,0)## and extend periodically from ##(-L,L)##.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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