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Fourier coefficients

  1. Dec 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ##f## be a ##2\pi## periodic function. Let ##\hat{f}(n)## be the fourier coefficient of ##f## defined by
    $$
    \hat{f}(n)=\frac{1}{2\pi}\int_{a}^{b}f(x)e^{-inx}dx.
    $$
    for ##n\in\mathbb{N}##. If ##\overline{\hat{f}(n)}=\hat{f}(-n)## show that ##f## is real valued.

    3. The attempt at a solution
    Suppose for contradiction that ##f## is not real valued. Then
    $$
    \overline{\hat{f}(n)}=\frac{1}{2\pi}\overline{\int_{a}^{b}f(x)\cos nxdx}=\frac{1}{2\pi}\left[\overline{\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx}\right]\\
    =\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx-i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]
    $$
    and
    $$
    \hat{f}(-n)=\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]
    $$
    so ##\hat{f}(n)\not=\hat{f}(-n)## a contradiction. Can anyone verify my proof for me? Thanks!
     
    Last edited: Dec 18, 2013
  2. jcsd
  3. Dec 18, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Note that [itex]f[/itex] is real valued by assumption!
     
  4. Dec 18, 2013 #3
    Woops! This was an if and only if problem and I was having trouble with the converse part. Sorry for the confusion.
     
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