1. ## Fourier series representation

okay, so i was given the piecewise function f(t)= k if -pi/2< x< pi/2
0 if pi/2 < x < 3pi/2

i computed the series to be

$Sf(t)= \frac{k}{2} + \sum_{n = 1}^\infty\frac{2k}{(2n-1)\pi } {(-1)}^{ n+1} \cos((2n-1)t)$

which i think is right,

but now i need to show that $\frac{\pi}{4 } = [1-1/3+1/5-1/7+........]$

i know it is probably pretty simple but iam a bit stuck so what steps should i take to show this.

2. Originally Posted by olski1
okay, so i was given the piecewise function f(t)= k if -pi/2< x< pi/2
0 if pi/2 < x < 3pi/2

i computed the series to be

$Sf(t)= \frac{k}{2} + \sum_{n = 1}^\infty\frac{2k}{(2n-1)\pi } {(-1)}^{ n+1} \cos((2n-1)t)$

which i think is right,

but now i need to show that $\frac{\pi}{4 } = [1-1/3+1/5-1/7+........]$

i know it is probably pretty simple but iam a bit stuck so what steps should i take to show this.
Try integrating from -pi to pi.

CB

3. Alternatively, evaluate it at t=0.

4. If you let t=0, you can arrange it so you end up with pi/4 in the LHS and corresponding infinite series in the RHS.

5. why t = 0?

6. okay so let me see if i have it right,

i set my fourier series of f(t) = k, set t =0 because it is within the domain of k
and do some re arranging and get the sum = pi/4.

and i got the correct answer.

is that correct? and is that why i choose t=0?

7. Yes, it's a common trick to evaluate infinite series using the fourier representation of a function and for a certain t. In this case we chose t=0 because it makes the cos term become 1 for all values of n. Another choice could have been 2pi, but then the LHS wouldn't be defined.