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Math Help - Parseval's identity: Real form

  1. #1
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    Parseval's identity: Real form

    \|f\|^2 = \sum_{n=-\infty}^{\infty}|c_n|^2 where c_n are the fourier coefficients.

    What does Parseval's identity look like if you use the coefficients from the real form of the fourier series, i.e. a_n = c_n + c_{-n}, b_n = i(c_n - c_{-n}) and f \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}(a_n\cos(n\Omega t) + b_n\sin(n\Omega t) )

    I've seen it used like this, but I can't figure out how you get here: \|f\|^2 = \left|\frac{a_0}{2}\right|^2 +\frac{1}{2}\sum_{n=1}^{\infty}(|a_n|^2+|b_n|^2)
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  2. #2
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    Quote Originally Posted by Mondreus View Post
    \|f\|^2 = \sum_{n=-\infty}^{\infty}|c_n|^2 where c_n are the fourier coefficients.

    What does Parseval's identity look like if you use the coefficients from the real form of the fourier series, i.e. a_n = c_n + c_{-n}, b_n = i(c_n - c_{-n}) and f \sim \frac{a_0}{2}+ \sum_{n=1}^{\infty}(a_n\cos(n\Omega t) + b_n\sin(n\Omega t) )

    I've seen it used like this, but I can't figure out how you get here: \|f\|^2 = \left|\frac{a_0}{2}\right|^2 +\frac{1}{2}\sum_{n=1}^{\infty}(|a_n|^2+|b_n|^2)


    As c_n=\frac{1}{2}(a_n\pm ib_n)\Longrightarrow |c_n|^2=\frac{1}{4}(a_n^2+b_n^2) ...

    Tonio
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  3. #3
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    Both a_n and b_n are complex numbers though, so I don't think what you posted holds in general. a_n^2 \neq |a_n|^2 in general as well.
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  4. #4
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    Quote Originally Posted by Mondreus View Post
    Both a_n and b_n are complex numbers though, so I don't think what you posted holds in general. a_n^2 \neq |a_n|^2 in general as well.
    If you write out the square integral of what you call the real form in extensive form you have terms of the form:

    \int_{t=-\pi/\Omega}^{\pi/\Omega} A_{n,m} \cos(n\Omega t) \sin( m\Omega t)\;dt=0

    \int_{t=-\pi/\Omega}^{\pi/\Omega} B_{n,m} \sin(n\Omega t) \sin( m\Omega t)\;dt=\frac{1}{2}B_{n,m}\delta_{n,m}

    \int_{t=-\pi/\Omega}^{\pi/\Omega} C_{n,m} \cos(n\Omega t) \cos( m\Omega t)\;dt=\frac{1}{2}C_{n,m}\delta_{n,m}

    CB
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    Quote Originally Posted by Mondreus View Post
    Both a_n and b_n are complex numbers though, so I don't think what you posted holds in general. a_n^2 \neq |a_n|^2 in general as well.

    This seems to be a problem of notation: I meant the real coefficients (real and imaginary parts) of the complex ones, and because of the same reason the coefficient 1/2 may change, according as what interval we're choosing to define our periodic functions on.

    Tonio
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    If you write out the square integral of what you call the real form in extensive form you have terms of the form:

    \int_{t=-\pi/\Omega}^{\pi/\Omega} A_{n,m} \cos(n\Omega t) \sin( m\Omega t)\;dt=0

    \int_{t=-\pi/\Omega}^{\pi/\Omega} B_{n,m} \sin(n\Omega t) \sin( m\Omega t)\;dt=\frac{1}{2}B_{n,m}\delta_{n,m}

    \int_{t=-\pi/\Omega}^{\pi/\Omega} C_{n,m} \cos(n\Omega t) \cos( m\Omega t)\;dt=\frac{1}{2}C_{n,m}\delta_{n,m}

    CB
    Sorry, I don't understand that notation.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by Mondreus View Post
    Sorry, I don't understand that notation.
    It is just a statement that the trig functions form an orthonormal basis for the space of square integrable functions on the interval (-\pi,pi].

    In particular I use Kronecker's delta:

    \delta_{a,b}=\left\{ \begin{array}{ll}<br />
1&, {\text{ if }} a=b\\<br />
0&, {\text{ if }} a \ne b \end{array} \right.

    CB
    Last edited by CaptainBlack; September 17th 2010 at 09:35 PM.
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