Finding a bound for a Fourier coefficient

  • #1

Homework Statement



show that Ak will satisfy:

[tex]\left| A_{k} \right| \leq Mk^{-4} [/tex]

Homework Equations



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

given

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

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

The Attempt at a Solution



I differentiated 4 times using substitution, and came up with

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

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

Any comments or suggestions?
 

Answers and Replies

  • #2
Office_Shredder
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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
[tex] \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 [/tex]

I'm still not getting the inequality. Can someone please help me further?
 

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