Fourier coefficients of even and odd functions

Question

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 $a_k = -a_{-k} = 0$.

Answer 1

Your first two affirmations hold.

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 $\hat{f}(1)=-i/2 \neq 0$.