# Help with computing/understanding Fourier Sine Expansion.

1. Mar 3, 2013

### LoganS

1. Find the Fourier sine expansion of $\phi(x)=1$. This was posted in Calculus and Beyond thread, but I realized that this thread may be more appropriate.

2. The attempt at a solution.
I start with $\phi(x)=A_1sin(\pi x)+A_2sin(2\pi x)+\cdots+A_nsin(n\pi x),$ and then add multiply by $A_msin(m\pi x)$ term on each side and integrate from 0 to 1.
So I have $\int\phi(x)A_msin(m\pi x)=\int\left(A_1sin(\pi x)A_msin(m\pi x)+A_2sin(2\pi x)A_msin(m\pi x)+\cdots+A_nsin(n\pi x)A_msin(m\pi x)\right).$
I know that due to orthogonality you can discard the terms where m is not equal to n (but I don't really understand why so if you can explain this I would appreciate it).
Discarding those terms and using a trig relation I get, $\int\phi(x)A_msin(m\pi x)=1/2\int\left(cos((m-n)\pi x)-cos((m+n)\pi x)\right).$
I then solve the integral and try to get $A_m=\frac{4}{\pi}\left(\frac{1}{m}\right),$ but I don't know why the answer has only the odd terms.
The book answer is $1=\frac{4}{\pi}\left(sin(\pi x)+\frac{1}{3}sin(3\pi x)+\frac{1}{5}sin(5\pi x) +\cdots\right).$

Any and all help is greatly appreciated.

Last edited: Mar 3, 2013
2. Mar 6, 2013

### pasmith

The starting point is
$$\phi(x) = \displaystyle \sum_{n=1}^{\infty} a_n \sin(n\pi x)$$

To work out $a_n$, multiply by $\sin (m\pi x)$ and integrate between 0 and 1:
$$\int_0^1 \phi(x) \sin(m\pi x)\,\mathrm{d}x = \sum_{n=1}^\infty a_n \int_0^1 \sin(n \pi x) \sin(m\pi x)\,\mathrm{d}x$$

We need to calculate $I = \int_0^1 \sin(n\pi x)\sin(m \pi x)\,\mathrm{d}x$. This we do by parts:
$$I = \int_0^1 \sin(n\pi x)\sin(m \pi x)\,\mathrm{d}x \\ = \left[ -\frac{1}{m\pi} \sin(n\pi x)\cos(m\pi x)\right]_0^1 + \frac{n}{m}\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\mathrm{d}x \\ = \frac{n}{m}\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\mathrm{d}x$$
where the last follows because $\sin(0) = \sin(m\pi) = 0$. Now we can integrate by parts again:
$$I = \frac{n}{m}\int_0^1 \cos(n\pi x)\cos(m\pi x)\,\mathrm{d}x \\ = \frac{n}{m}\left[ \frac{1}{m\pi} \cos(n\pi x)\sin(m \pi x)\right]_0^1 + \frac{n^2}{m^2} \int_0^1 \sin(n\pi x)\sin(m\pi x)\,\mathrm{d}x \\ = \frac{n^2}{m^2} I$$

So $(1 - n^2/m^2)I = 0$. It follows then that if $n \neq m$ then $I = 0$. If $n = m$ then the integrand is $\sin^2(m\pi x)$ which is non-negative and not identically zero, so its integral must be strictly positive. So we have, using the identity $\sin^2 \theta = \frac12 (1 - \cos 2\theta)$:
$$\int_0^1 \phi(x) \sin(m\pi x)\,\mathrm{d}x = a_m \int_0^1 \sin(m \pi x) \sin(m\pi x)\,\mathrm{d}x \\ = a_m \int_0^1 \sin^2(m \pi x)\,\mathrm{d}x \\ = \frac12 a_m \int_0^1 1 - \cos(2m \pi x)\,\mathrm{d}x\\ = \frac12 a_m$$
so that
$$a_m = 2 \int_0^1 \phi(x) \sin(m\pi x)\,\mathrm{d}x.$$

$$a_m = 2\int_0^1 \sin(m\pi x)\,\mathrm{d}x = -\frac{2}{m\pi} \left[\cos(m\pi x)\right]_0^1 \\ = - \frac{2}{m\pi}( \cos(m\pi) - 1) \\ = - \frac{2}{m\pi}( (-1)^m - 1) \\ = \frac{2}{m\pi}(1 - (-1)^m)$$
If $m$ is even, $(-1)^m = 1$ so $a_m = 0$. If $m$ is odd, $(-1)^m = -1$ and $a_m = 4/(m\pi)$, which is the given answer.