this may sound like a dumb question but are fourier coefficients always symmetric?
ie $\hat{f}(n) = \hat{f}(-n)$?
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Consider $f(x)=e^{ix}$, or polynomials in this $f$. |
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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: $\operatorname{Re}(\hat{f}(n))=\operatorname{Re}(\hat{f}(-n))$ $\operatorname{Im}(\hat{f}(n))=-\operatorname{Im}(\hat{f}(-n))$ |
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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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