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this may sound like a dumb question but are fourier coefficients always symmetric?

ie $\hat{f}(n) = \hat{f}(-n)$?

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You should consider the Fourier coefficients of $e^{ix}$, for example. – Akhil Mathew Apr 1 '11 at 3:07
up vote 4 down vote accepted

Consider $f(x)=e^{ix}$, or polynomials in this $f$.

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ok i guess i have another dumb question: if i take the derivative of $e^{inx}$, do i get $ine^{inx}$ or $ne^{inx}$ – jack Apr 1 '11 at 3:21
@jack: The chain rule still applies to complex valued functions, and it's easy to show that the derivative of $x\mapsto inx$ is $in$. – Jonas Meyer Apr 1 '11 at 3:24
so i guess it should be $ine^{inx}$, is that right? – jack Apr 1 '11 at 3:28
Yes.${}{}{}{}{}$ – Jonas Meyer Apr 1 '11 at 3:33

The answer is no, as Jonas Meyer pointed out. However, if $f$ is a real-valued function, then the real parts of the Fourier coefficients will be symmetric, and the imaginary parts antisymmetric:



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Hence for real, even functions equality holds. The equality also holds for some complex functions. – Douglas Zare Apr 1 '11 at 4:53

You might like to consider the difference: $$\hat{f}(n)-\hat{f}(-n)=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{-int}dt-\frac{1}{2\pi}\int_{-\pi}^\pi f(t)e^{int}dt=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)(e^{-int}-e^{int})dt$$ Next, note from the Euler equations that $e^{-int}-e^{int}=-2i\cdot \sin(nt)$.

Hence $\hat{f}(n)=\hat{f}(-n)$ (for some $n$) means that $f$ is orthogonal to (or rather, that $f$ is annihilated by) the function $t\mapsto\sin(nt)$ (for some $n$).

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