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Math Help - Fourier series of f(x)=x between 0 and 2pi

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    Fourier series of f(x)=x between 0 and 2pi

    I am trying to expand f(x) = x into a real fourier series on the interval 0<=x<=2pi. The only examples I can find are on the interval -pi to pi, and when i try using the same method on my interval i come up with the answer zero, which is obviously wrong.

    any ideas...
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    Quote Originally Posted by johnbarkwith View Post
    I am trying to expand f(x) = x into a real fourier series on the interval 0<=x<=2pi. The only examples I can find are on the interval -pi to pi, and when i try using the same method on my interval i come up with the answer zero, which is obviously wrong.

    any ideas...
    Look at the Mathworld page what you need is about half way down.

    CB
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    i can see that you have to make a change of variables, but i am having trouble actually completing it.. could you give more detail...
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    Quote Originally Posted by johnbarkwith View Post
    I am trying to expand f(x) = x into a real fourier series on the interval 0<=x<=2pi. The only examples I can find are on the interval -pi to pi, and when i try using the same method on my interval i come up with the answer zero, which is obviously wrong.

    any ideas...
    Let \mathcal{F}(x)=x~\forall{x}\in(0.2\pi). Also let \mathcal{F}(x) posses the charcteristic that \mathcal{F}(x)\in{P}_{2\pi}\implies\mathcal{F}(x+2  \pi)=\mathcal{F}(x)~\forall{x}\in\mathbb{R}. Now note that \mathcal{F}(x)\in\mathcal{C}\left(x\ne{2\pi\mathbb  {N}}\right), and that \mathcal{F}(x)\in{D^1} or in other words it posses a left and right hand limit at all points. So now that we have established these conditions we can see that \mathcal{F}(x)\to{S_n(x)}~\forall{x}\ne{2\pi\mathb  b{N}}, where S_n(x) is the Fourier Series given by

    S_n(x)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\bigg[A_n\cos(nx)+B_n\sin(nx)\bigg]

    Where the coefficents are given by:

    \begin{aligned}A_0&=\frac{1}{\pi}\int_0^{2\pi}\mat  hcal{F}(x)dx\\<br />
&=\frac{1}{\pi}\int_0^{2\pi}xdx\\<br />
&=2\pi<br />
\end{aligned}


    \begin{aligned}A_n&=\frac{1}{\pi}\int_0^{2\pi}\mat  hcal{F}(x)\cos(nx)dx\\<br />
&=\frac{1}{\pi}\int_0^{2\pi}x\cos(nx)dx\\<br />
&=0\quad\forall{n}\in\mathbb{N}<br />
\end{aligned}


    \begin{aligned}A_n&=\frac{1}{\pi}\int_0^{2\pi}\mat  hcal{F}(x)\sin(nx)dx\\<br />
&=\frac{1}{\pi}\int_0^{2\pi}x\sin(nx)dx\\<br />
&=\frac{-2}{n}\quad\forall{n}\in\mathbb{N}<br />
\end{aligned}

    So we can see that

    S_n(x)=\pi-2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}

    And that

    \forall{x}\in(0,2\pi)~~x=\pi-2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}
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    Quote Originally Posted by johnbarkwith View Post
    I am trying to expand f(x) = x into a real fourier series on the interval 0<=x<=2pi. The only examples I can find are on the interval -pi to pi, and when i try using the same method on my interval i come up with the answer zero, which is obviously wrong.

    any ideas...
    One way of doing this without changing the interval is to realise that you are dealing with a periodic extension of your function. Then the function periodic with period 2 \pi defined as:

    f(x)=x, \ \ x \in [0,2 \pi)

    is the same thing as:

     <br />
f(x)=\begin{cases}x+2 \pi, & x \in [\pi,0) \\ <br />
x & x \in [0, \pi)\end{cases}<br />

    and now you can use your definition of a Fourier serier for a function of period 2 \pi on the interval [-\pi,pi), and because of the periodicity of a Fourier representation of a function this will give a series that is valid on any interval of length 2 \pi .

    CB
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    Let \mathcal{F}(x)=x~\forall{x}\in(0.2\pi). Also let \mathcal{F}(x) posses the charcteristic that \mathcal{F}(x)\in{P}_{2\pi}\implies\mathcal{F}(x+2  \pi)=\mathcal{F}(x)~\forall{x}\in\mathbb{R}. Now note that \mathcal{F}(x)\in\mathcal{C}\left(x\ne{2\pi\mathbb  {N}}\right), and that \mathcal{F}(x)\in{D^1} or in other words it posses a left and right hand limit at all points. So now that we have established these conditions we can see that \mathcal{F}(x)\to{S_n(x)}~\forall{x}\ne{2\pi\mathb  b{N}}, where S_n(x) is the Fourier Series given by

    S_n(x)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\bigg[A_n\cos(nx)+B_n\sin(nx)\bigg]

    Where the coefficents are given by:

    \begin{aligned}A_0&=\frac{1}{\pi}\int_0^{2\pi}\mat  hcal{F}(x)dx\\<br />
&=\frac{1}{\pi}\int_0^{2\pi}xdx\\<br />
&=2\pi<br />
\end{aligned}


    \begin{aligned}A_n&=\frac{1}{\pi}\int_0^{2\pi}\mat  hcal{F}(x)\cos(nx)dx\\<br />
&=\frac{1}{\pi}\int_0^{2\pi}x\cos(nx)dx\\<br />
&=0\quad\forall{n}\in\mathbb{N}<br />
\end{aligned}


    \begin{aligned}A_n&=\frac{1}{\pi}\int_0^{2\pi}\mat  hcal{F}(x)\sin(nx)dx\\<br />
&=\frac{1}{\pi}\int_0^{2\pi}x\sin(nx)dx\\<br />
&=\frac{-2}{n}\quad\forall{n}\in\mathbb{N}<br />
\end{aligned}

    So we can see that

    S_n(x)=\pi-2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}

    And that

    \forall{x}\in(0,2\pi)~~x=\pi-2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}
    Don't use a fancy F for a function name in harmonic analysis you will just get confused at some point with the fancy F you are going to use to denote the FT.

    Also you are using too much cryptic notation that can be explained better in English, and the relegation of discontinuities to end points of the interval is unnecessary (in fact undesirable). The only caveat is that the series does not converge to whatever functional value you have assigned at jump discontinuities but to the mean of the limit from the right and left.

    (If you are using Rudin as a reference, get another book you will never understand harmonic or any other branch of analysis if you start with Rudin, Rudin is for when you think you know a subject already)

    CB
    Last edited by CaptainBlack; November 20th 2008 at 09:39 PM.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Don't use a fancy F for a function name in harmonic analysis you will just get confused at some point with the fancy F you are going to use to denote the FT.

    Also you are using too much cryptic notation that can be explained better in English, and the relegation of discontinuities to end points of the interval is unnecessary (in fact undesirable). The only caveat is that the series does not converge to whatever functional value you have assigned at jump discontinuities but to the mean of the limit from the right and left.

    (If you are using Rudin as a reference, get another book you will never understand harmonic or any other branch of analysis if you start with Rudin, Rudin is for when you think you know a subject already)

    CB
    Yeah you are probably right about the F, for that is indeed the Fourier Transform F. As for not including the jump discontinuities I never said you cannot find the value for them, I just said that the trigonometric fourier series does not describe the function at those points, you would have to take the mean of the left and right hand limits to find that. And I appreciate the concern and I might consider it, but as of now I am finding Rudin enjoyable as well as thourough.

    Thanks CaptainBlack!
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