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I am stuck on an excerise which says that prove the fourier transform $f(k)$ of a real function satisfied the condition $f(-k)=f*(-k)$. Where the astericks denotes the complex congugate.

I am beginning to think there is a typo as I am getting it to be $f(-k)=f*(k)$.

By def

$$f(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$$ LHS $$\int_{-\infty}^{\infty}f(x)(\cos kx+i\sin kx)dx$$

RHS $$\int_{-\infty}^{\infty}f(x)e*^{ikx}dx$$ $$..$$ $$=\int_{-\infty}^{\infty}f(x)(\cos kx-i\sin kx)dx$$ Sorry I dont have latex installed, Any help would be much appreciated

Matthew Cassell
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user147825
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  • Most likely a typo, why? would you write $f(-k) = f^(-k)$ instead of $f(k) = f^(k)$. – Jeb Dec 05 '14 at 18:37
  • Maybe the exercise has a typo? IMO f(-k) = f*(k) is right. – desos Dec 05 '14 at 18:40
  • I am not sure what is meant by your question other than that is the way the question was phrased – user147825 Dec 05 '14 at 18:40
  • You don't need to have latex installed. Please have a look at this site and try to edit your question so that other people can read it ;) (http://math.stackexchange.com/editing-help) – TheWaveLad Dec 05 '14 at 18:59

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Definition of Fourier transform: \begin{align*} \tilde{f}(k) &= \int_{-\infty}^\infty e^{-ikx} f(x) dx \\ &= \int_{-\infty}^\infty f(x) \cos kx dx - i \int_{-\infty}^\infty f(x) \sin kx dx \\ \Rightarrow \tilde{f}^*(k) &= \int_{-\infty}^\infty f(x) \cos kx dx + i \int_{-\infty}^\infty f(x) \sin kx dx \\ &= \int_{-\infty}^\infty e^{ikx} f(x) dx \\ &= \tilde{f}(-k) \end{align*}

So it seems that $\tilde{f}(-k) = \tilde{f}^*(k)$.

$\tilde{f}(-k) = \tilde{f}^*(-k)$ can't be true because it would imply that $\tilde{f}$ is always real.

desos
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  • Thank you for your reply, so would that imply the only way this can be satisfied is if say f(-k) =a where a is real and then the proof would hold? – user147825 Dec 05 '14 at 18:46