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I am working through the following problem:

Let $P(x)$ be some arbitrary polynomial over the interval $[-1, +1]$. Then define $$A_n(P) = \int_{-1}^{+1} P(x)\cos{(n\pi x)}\,\mathrm{d}x$$

I am require to show that this coefficient (for $n > 0$) is in fact a polynomial in the variable $\frac{1}{n}$. My attempts so far have simply involved computing the integral using by parts, where the $u\cdot v$ term conveniently vanishes, and I am left with

$$A_n(P) = -\frac{1}{n\pi} \int_{-1}^{+1} P'(x)\sin{(n\pi x)} \, \mathrm{d}x$$

I am getting closer to the solution, or have I approached the problem in the wrong manner?

Chappers
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Victoria
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1 Answers1

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This is the correct technique. Persevere! One more integration by parts gives you $$ A_n(P) = \frac{1}{n^2\pi^2} \left[ P'(x)\cos{n\pi x} \right]_{-1}^1-\frac{1}{n^2\pi^2}\int_{-1}^1 P''(x) \cos{n\pi x} \, dx \\ = \frac{(-1)^n}{n^2\pi^2} (P'(1)-P'(-1))-\frac{1}{n^2\pi^2}\int_{-1}^1 P''(x) \cos{n\pi x} \, dx. $$ Now you can use induction, because $P''(x)$ has lower degree than $P(x)$. Because the degree jumps by two, you also have to check both degree one and degree zero to get both of the basis cases, however.

Chappers
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  • Thanks for the motivation - did you assume that n made the cosines equal to +1. – Victoria Oct 04 '15 at 23:28
  • Oh, whoops, brain fart. $\cos{n\pi}=(-1)^n$, of course... thanks. – Chappers Oct 04 '15 at 23:36
  • So, let me make sure understand the process - we can always find an expression for $A_n(P)$ that involves integrating a constant multiplied by a sine or a cosine, is that how you would do it? For example, if $P(x)$ is of degree 2, we can use the above expression in order to evaluate the coefficient? – Victoria Oct 04 '15 at 23:45
  • If $P$ has degree $k$, I have shown that it is equal to a (simple) polynomial in $1/n$, added to the Fourier transform of a polynomial of degree $k-2$. Therefore, provided that $k=0$ and $k=1$ give polynomials in $1/n$, I can use induction to show that every polynomial has Fourier transform a polynomial in $1/n$: suppose it is true for $k-2$, then the integration by parts shows it is true for $k$. – Chappers Oct 04 '15 at 23:56
  • Awesome answer! fr fr – tryst with freedom Sep 11 '23 at 19:10