# Fourier coefficients of even and odd functions

I am unable to confirm the following. If a signal is odd, then does the following hold? $$a_k = -a_{-k}$$ If a signal is even, then: $$a_k = a_{-k}$$ If the aforementioned qualities hold, then is the following true? If a signal is odd and the period is $2k$, then $a_k = -a_{-k} = 0$.

Let $f$ be $p$-periodic and $f \in L^1[-p/2,p/2]$. Then the Fourier coefficients of $f$ are $$\hat{f}(k) = \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi k t/p} \,dt$$ If $f$ is odd then the change of variable $u=-t$ gives (we can keep the interval of integration $[-p/2,p/2]$ by $p$-periodicity) \begin{align} \hat{f}(k) &= -\frac{1}{p} \int_{-p/2}^{p/2} (-f(-t)) e^{i2\pi k t/p} \,dt \\ &= -\frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi (-k) t/p} \,dt \\ &= -\hat{f}(-k) \end{align}

If $f$ is even then the same change of variable gives \begin{align} \hat{f}(k) &= \frac{1}{p} \int_{-p/2}^{p/2} f(-t) e^{i2\pi k t/p} \,dt \\ &= \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi (-k) t/p} \,dt \\ &= \hat{f}(-k) \end{align}

I don't think your third affirmation holds.

Take $f(t):=\sin(\pi t)$. Then $f$ is odd and of period $2\cdot1$. Yet, we can calculate $\hat{f}(1)=-i/2 \neq 0$.

• But logically speaking, if the period is 2k, then, the value at -1 and 1 should be the same. But them being odd requires that the signs of the values are opposite. Hence, the only value that satisfies both is 0, right? Or am I erring in my logic? – Jonathan Feb 16 '17 at 8:40
• @Christian Hmmm yes I think you are correct, my example is not an odd signal. Let me rethink this. – NeedForHelp Feb 16 '17 at 8:43
• @Christian Do you agree with my new counter-example? – NeedForHelp Feb 16 '17 at 8:49
• Oh, I see. Just to confirm, the notation with caret sign over the function signifies the Fourier coefficient at k=1 right? Not the actual function itself? – Jonathan Feb 16 '17 at 9:04
• @Christian The Fourier coefficients will satisfy the odd property $\hat{f}(k)=-\hat{f}(-k)$, as I showed. However they won't be periodic in general. That's why it doesn't work. – NeedForHelp Feb 16 '17 at 9:19