complex fourier series problem Cn term

Question

can anyone help with this complex fourier series problem (in details please specially the Cn term) (as in the attached picture)here is the problem

Answer 1

Ok, now I see your problem. There are three mistakes in your computation.

1.

Like ProfessorVector mentioned: You have a sign mistake in your cosine identity. It is $\cos(x)=\frac12(e^{xi}+e^{-xi})$.

2.

On sheet 2 you change $\frac{2(j+2jn)}{-(1-4n^2)}$ to $\frac{2(j+2n\pi)}{-(1-4n^2)}$. How did $j$ become $\pi$? That's wrong. Further, you are not consequent in your computation. So started with $\int\ldots$ and in the middle part, you included the factor $\frac1{4\pi}$ out of the blue. The right solution of your sheet $2$ should be $$ \int_{-\pi}^\pi e^{(1/2-n)tj}~dt=\frac{2(1+2n)j}{-(1-4n^2)}(-1)^n\left(e^{\frac{\pi}2j}-e^{-\frac{\pi}2j}\right). $$ If you use the identities $e^{\frac{\pi}2j}=j$ and $e^{-\frac{\pi}2j}=-j$ you get $$ \frac{2(1+2n)j}{-(1-4n^2)}(-1)^n\left(e^{\frac{\pi}2j}-e^{-\frac{\pi}2j}\right)=\frac{2(1+2n)j}{-(1-4n^2)}(-1)^n2j=(-1)^n\frac{4(1+2n)}{1-4n^2}. $$ It is more useful to get $e^{\frac{\pi}2j}-e^{-\frac{\pi}2j}=2j$ insteat of $2\sinh\left(\frac{\pi}2j\right)$.

3.

On sheet 3, your computation seems right, but when you included the result on sheet 4, you changed again $j$ to $\pi$. And using the identities as above you should get $$ \int_{-\pi}^\pi e^{-(1/2+n)tj}~dt=(-1)^n\frac{4(1-2n)}{1-4n^2}. $$ Finally you can combine the results and you get \begin{align} C_n&=\frac1{4\pi}\left(\int_{-\pi}^\pi e^{(1/2-n)tj}+\int_{-\pi}^\pi e^{-(1/2+n)tj}\right)\\ &=\frac1{4\pi}\left((-1)^n\frac{4(1+2n)}{1-4n^2}+(-1)^n\frac{4(1-2n)}{1-4n^2}\right)\\ &=\frac{(-1)^n}{\pi}\left(\frac{1+2n}{1-4n^2}+\frac{1-2n}{1-4n^2}\right)\\ &=\frac{(-1)^n\cdot 2}{\pi(1-4n^2)} \end{align}