Solving Fourier Question: -\pi < x< 0 & 0 < x < \pi

[manenbu]

Homework Statement

function is:
$$1, -\pi < x< 0$$
$$2, 0 < x < \pi$$

The Attempt at a Solution

What I get is this:

$$\frac{3}{2}+\frac{1}{\pi}\sum_{n=1}^\infty\frac{(-1)^n}{n}\sin{nx}$$

According to the answers in the exercise sheet, it should be:
$$\frac{3}{2}+\frac{2}{\pi}\sum_{n=1}^\infty\frac{1}{2n-1}\sin{(2n-1)x}$$

I don't understand why this is true. The 2n-1 hints that I should've got a $\pi/2$ somewhere inside the trig functions in the intergal, but I can't see where this is coming from.
Can anyone help me here?

 

[LCKurtz]

Homework Statement

function is:
$$1, -\pi < x< 0$$
$$2, 0 < x < \pi$$

The Attempt at a Solution

What I get is this:

$$\frac{3}{2}+\frac{1}{\pi}\sum_{n=1}^\infty\frac{(-1)^n}{n}\sin{nx}$$

According to the answers in the exercise sheet, it should be:
$$\frac{3}{2}+\frac{2}{\pi}\sum_{n=1}^\infty\frac{1}{2n-1}\sin{(2n-1)x}$$

I don't understand why this is true. The 2n-1 hints that I should've got a $\pi/2$ somewhere inside the trig functions in the intergal, but I can't see where this is coming from.
Can anyone help me here?

That is giving you every other term. You should probably have something like (-1)ⁿ-1 in the numerator which would drop out every other term. Maybe you overlooked a lower limit in your integral.

 

[manenbu]

Yes, apparently 2-1 became 0 in my first calculation. I got it correct now. Thanks!