How to express this Fourier coefficient without trig function?

[zenterix]

Homework Statement

Consider the expression

$$b_n=\frac{4}{n\pi}\left (1-\cos{\left (n\frac{\pi}{2}\right )}\right )$$

This is the $n$-th Fourier coefficient for sine factors for the function

$$f(t)=\begin{cases} 2\ \ \ \ \ 0<t<\pi/2 \\ 0\ \ \ \ \ \frac{\pi}{2}<t<\pi\end{cases}$$

Relevant Equations

Can we express this without the cosine?

Here is a little table I made with the values of $b_n$ for $n=1,2,3,4,5,6$.

Is there a way to write a formula for $b_n$ not involving a trigonometric function?

 

[Hill]

How about complex exponentials?

 

[zenterix]

The coefficients for the cosine factors are

$$a_n=\frac{4}{n\pi}\sin{\left (n\frac{\pi}{2}\right )}$$

and noting that as $n$ varies the $\sin$ is $1,0,-1,0,1,-1,0,\ldots$ we can write

$a_{2k+1}=\frac{4(-1)^{k+1}}{\pi(2k+1)}$ for $k=0,1,2,\ldots$.

I'm looking for something similar.

 

[zenterix]

I think what I am looking for is something like

$$b_{2k}=\frac{4}{2k\pi}(1-(-1)^k)\tag{1}$$

for $k=1,2,3,\ldots$

and

$$b_{2k+1}=\frac{4}{(2k+1)\pi}\tag{2}$$

for $k=0,1,2,3,\ldots$.

We could further break down (1) as

$$b_{2+4k}=b_{2(2k+1)}=\frac{4\cdot 2}{2(1+2k)\pi}=\frac{4}{(2k+1)\pi}\tag{3}$$

$$b_{4k}=0$$

Oh well, given all these cases, not sure if this is useful at all at this point.

 

[zenterix]

This is all part of the first problem of this problem set.

I just looked at the solution and noticed I confused something though it is my impression that the problem set itself is the one that wasn't specific enough.

A function $f$ was defined as

$$f(t)=\begin{cases} 2\ \ \ \ \ 0<t<\pi/2 \\ 0\ \ \ \ \ \frac{\pi}{2}<t<\pi\end{cases}$$

and now I see that they explicitly said it has period $2\pi$. I had understood that they were giving me one period's worth of function.

So the function looks like this according to the solutions

That being said, with only the definition and the information that it has (minimal) period $2\pi$, the function could also be

right?

 

[Hill]

right?

No. It also says that the function is even.

 

[zenterix]

oh, right!