For given 4 conditions, I have to find out what is $x(t)$ with period of 3, and I don't know if $x(t)$ is real or not.
For fourier coefficients $x_k$,
$$1.\ x_k=x_{k+2}$$
$$2.\ x_k=x_{-k}$$
$$3.\ \int_{-0.5}^{0.5}x(t)\, \mathrm dx\ = 1$$
$$4.\ \int_{1}^{2}x(t)\, \mathrm dx\ = 2$$
from condition 1 and 2, I've got this equations
fourier series expansion of $x(t)$: $$x(t)=\sum_{k=-\infty}^{\infty}\ x_k \exp(j{2\pi \over 3}kt)$$
by condition 2,
$$x(t)=x_0+\sum_{k=1}^{\infty}x_k(\exp(-j{2\pi \over 3}kt)+\exp(j{2\pi \over 3}kt))$$
and as $\exp(-j2\pi ft)+\exp(j2\pi ft)=2\cos(2\pi ft)$,
$$x(t)=x_0+2\sum_{k=1}^{\infty}x_k\cos({2\pi \over 3}kt)$$
by condition 1,
$$x(t)=x_0+2\sum_{m=1}^{\infty}\ (x_1\cos(2\pi{(2m-1) \over 3}t)+x_0\cos(2\pi{2m \over 3}t))$$
For condition 3,4 when I integrate $x(t)$, it gives me too complex sums.
$$\int_{-1/2}^{1/2}\cos(2\pi{2m-1 \over 3}t)\, \mathrm dt\ = {3 \over 2\pi (2m-1)}2\sin(\pi{2m-1 \over 3})$$ $$2\sin(\pi{2m-1 \over 3})={\sqrt3, 0, -\sqrt3, \sqrt3, 0, -\sqrt3, ...}$$ so, $\int_{-1/2}^{1/2}\sum_{m=1}^{\infty}\cos(2\pi{2m-1 \over 3}t) \mathrm dt\ ={3 \over 2\pi}\sum_{n=1}^{\infty}({\sqrt3 \over6n-5}+{0 \over6n-3}+{-\sqrt3 \over6n-1})=0.75$
This kind of summation comes out over and over. Of course, calculator can solve it because it converges. But this was problem from my mid-term exam and calculator was not allowed. This thought process(?) complicates the problem too much, so I wondered and asked here if there could be other way to solve this problem much simpler than this.
