Complete graphically and analytically the function $f(t)$ so that the coefficients of the exponential Fourier series are pure imaginary: $$f(t)=\begin{cases}2t+1&\text{if $0\leq t\leq2$},\\\underline{\phantom{2t+1}}&\text{if $\underline{\phantom{0}} <t<\underline{\phantom{2}}$},\end{cases}\quad\text{with $f(t)=f(t+\underline{\phantom{4}})$}.$$
Find analytically and graphically the value of the coefficient $c_0$ and indicate if the Fourier series has only odd frequencies except for the independent term of the series.
Let's sketch the graph of $f$:
If the coefficients of the exponential Fourier series are pure imaginary then $\Re(c_n)=0$, so $a_n=0$ and the function is odd.
With this information we can extend the graph of $f$:
We observe that $$\begin{cases}f(2)=5\\f(4)=9\end{cases}\equiv\begin{cases}2a+b=5\\4a+b=9\end{cases}\implies a=2,b=1,$$ so $f(t)=2t+1$ for $t\in[2,4]$ and so $$f(t)=\begin{cases}2t+1&\text{if $0\leq t\leq2$},\\2t+1&\text{if $2\leq t\leq4$}\end{cases}=2t+1\quad\text{with $f(t)=f(t+4)$},$$ the period is $T=4$ and $\omega=\frac{2\pi}{T}=\frac{\pi}{2}$.
$$c_0=\frac{1}{T}\int_0^Tf(t)\,\mathrm dt=\frac14\int_0^4(2t+1)\,\mathrm dt=5.$$ Graphically:
My question is: how to know if the Fourier series has or not only odd frequencies except for the independent term of the series?
We know that the Fourier series has only odd frequencies because $f$ is an odd function, but I am puzzled by the phrase "except for the independent term of the series" (which I know is $c_0=5$, but $5$ is an odd number!, so the answer would be... no?).
Thanks!
Every constant function is even by definition, since $f(x)=f(-x)=c,$ for any constant $c.$