are fourier coefficients always symmetric?

Question

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

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

You should consider the Fourier coefficients of $e^{ix}$, for example. — Akhil Mathew, Apr 01 '11 at 03:07

score 5 · Answer 1 · answered Apr 01 '11 at 04:49

5

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:

$\operatorname{Re}(\hat{f}(n))=\operatorname{Re}(\hat{f}(-n))$

$\operatorname{Im}(\hat{f}(n))=-\operatorname{Im}(\hat{f}(-n))$

answered Apr 01 '11 at 04:49

Skatche

1,522

Hence for real, even functions equality holds. The equality also holds for some complex functions. – Douglas Zare Apr 01 '11 at 04:53

score 4 · Accepted Answer · answered Apr 01 '11 at 03:07

4

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

answered Apr 01 '11 at 03:07

Jonas Meyer

53,602

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 01 '11 at 03:21
1

@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 01 '11 at 03:24
so i guess it should be $ine^{inx}$, is that right? – jack Apr 01 '11 at 03:28
Yes.${}{}{}{}{}$ – Jonas Meyer Apr 01 '11 at 03:33

score 0 · Answer 3 · answered Apr 01 '11 at 05:31

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 $n$).

are fourier coefficients always symmetric?

3 Answers3