1. ## Fourier Series

I want to find the nth order Fourier approximation for $f(x)=x$. Since this function is odd, the projections on all cosines will be zero, hence it will be expressed through the sines only. So I just need to find the sine coefficients.

The problem is that I checked the answer to this problem in two different textbooks, the solutions in the books are totally different from each other.

• The first book calculates the coefficients like this:
$b_k= \frac{1}{\pi} \int^{2 \pi}_{0} f(x)sin(kx)= \frac{1}{\pi} \int^{2 \pi}_0 xsin(kx) dx =- \frac{2}{k}$

• The second book calculates the coefficients this way:
$b_k = (f(x),sin (k \pi x))= \int^1_{-1} x sin (k \pi x) dx= -x\frac{cos (k \pi x)}{k \pi} ]^1_{-1} + \int^1_{-1} \frac{cos (k \pi x)}{k \pi}dx$ $= -2 \frac{cos k \pi}{k \pi}=-(-1)^k \frac{2}{k \pi}$

So which one is correct? I mean, they both are probably correct but I don't understand why they use different methods and end up with different answers. Any explanation is very appreciated.

2. They are using two different "normalizations" because they are calculating the Fourier series over two different intervals.

Remember that you cannot actually find a Fourier series for "f(x)= x". You can only find Fourier series for periodic functions.

In the first case they are finding the Fourier series for the function that is equal to x for $0\le x\le 2\pi$ and then repeats that value with period $2\pi$.

In the second case they are finding the Fourier series for the function that is equal to x for $-1\le x\le 1$ and then repeats that value with period 2.