# Finding a bound for a Fourier coefficient

1. Nov 5, 2008

### Somefantastik

1. The problem statement, all variables and given/known data

show that Ak will satisfy:

$$\left| A_{k} \right| \leq Mk^{-4}$$

2. Relevant equations

$$A_{k} = \frac{2}{L}\int^{L}_{0} \phi(x) sin \left( \frac{k \pi x}{L} \right) dx$$

given

$$\phi(x) \in C^{4} ([0,L]) \and\ \phi^{(p)}(0) = \phi^{(p)}(L) = 0, p = 0, 1, 2, 3$$

in this notation, \phi is continuous and differentiable up to 4th derivative over [0,L], and (p) denotes the order of the derivative.

3. The attempt at a solution

I differentiated 4 times using substitution, and came up with

$$\frac{L^{4}}{k^{4}\pi^{4}} \int^{L}_{0} \phi^{(4)}(x)sin \left( \frac{k \pi x}{L}\right) dx$$.

I was told using the Weierstrass M-test (or Comparison test?) would help find solution, but I'm not sure how to proceed.

2. Nov 5, 2008

### Office_Shredder

Staff Emeritus
You would use this result to apply the Weierstrass M-test, but I don't see how the M-test would give the inequality to you.

On the other hand, you're nearly there. The fourth derivative of phi is continuous, and sin(k*pi*x/L) is bounded by one, so you can bound the integral above by a constant that doesn't depend on k.

3. Nov 5, 2008

### Somefantastik

$$\frac{L^{4}}{k^{4}\pi^{4}} \left| \int^{L}_{0} \phi^{(4)}(x)sin \left(\frac{k \pi x}{L} \right) dx \right| \leq M$$