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Did I set this Fourier series up correctly?

  1. Nov 6, 2014 #1

    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.
     
  2. jcsd
  3. Nov 6, 2014 #2
    Hmmmm it did not translate my latex writing. I apologize for that.
     
  4. Nov 6, 2014 #3

    LCKurtz

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    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.
     
  5. Nov 6, 2014 #4

    pasmith

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    Given [itex]f: [0,L] \to \mathbb{R}[/itex] 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 [itex]2\pi n/L[/itex] for each positive integer [itex]n[/itex] (because when [itex]x = L[/itex] you want [itex]2\pi n x/L = 2n\pi[/itex]).

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

    Thirdly, you can extend it to [itex][-L,L][/itex] as an odd function setting [itex]f(-x) = -f(x)[/itex] for each [itex]x \in [0,L][/itex] and using the series of that. Because it's an odd function the series will only have sine terms, and the frequencies are [itex]\pi n/L[/itex] for each positive integer [itex]n[/itex] (because when [itex]x = \pm L[/itex] you want [itex]\pi nx/L = \pm n\pi[/itex]).
     
  6. Nov 6, 2014 #5

    LCKurtz

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    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)##.
     
    Last edited: Nov 6, 2014
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