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Compute the coefficients of $\sum\limits_{n=1}^\infty A_n\sin\frac{n\pi x}{l}=\frac{x^2-x}{4}.$

If I already computed the coefficient for $\sum\limits_{n=1}^\infty A_n\sin\frac{n\pi x}{l}=x^2$ and $\sum\limits_{n=1}^\infty A_n\sin\frac{n\pi x}{l}=x,$ can I just divide both series by 4 and then subtract them and the sum would be the coefficient $A_n$ of $\sum\limits_{n=1}^\infty A_n\sin\frac{n\pi x}{l}=\frac{x^2-x}{4}$ ,

Am I right?

If I am not right, then is there an easier way to compute the coefficient other than compute the complicated integral ?

1 Answers1

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The coefficients of $f(x)=\sum_{n=1}^\infty A_n\sin \tfrac{n \pi x}{l}$ are $A_n=\frac{1}{l}\int_{-l}^l f(x)\sin \tfrac{n \pi x}{l}\; dx$.

The coefficients of $g(x)=\sum_{n=1}^\infty B_n\sin \tfrac{n \pi x}{l}$ are $B_n=\frac{1}{l}\int_{-l}^l g(x)\sin \tfrac{n \pi x}{l}\; dx$.

Thus, the coefficients of $\alpha f(x)+\beta g(x)=\sum_{n=1}^\infty C_n\sin \tfrac{n \pi x}{l}$ are \begin{align*}C_n&=\frac{1}{l}\int_{-l}^l (\alpha f(x)+\beta g(x))\sin \tfrac{n \pi x}{l}\; dx\\ &=\alpha\frac{1}{l}\int_{-l}^l f(x)\sin \tfrac{n \pi x}{l}\; dx+\beta\frac{1}{l}\int_{-l}^l g(x)\sin \tfrac{n \pi x}{l}\; dx\\ &=\alpha A_n+\beta B_n\end{align*}

Taking $f(x)=x^2$, $g(x)=x$, $\alpha=\tfrac{1}{4}$ and $\beta=-\tfrac{1}{4}$ we conclude that

  • the coefficients of $\sum_{n=1}^\infty C_n\sin\tfrac{n\pi x}{l}=\frac{x^2-x}{4}$ are $C_n=\frac{A_n-B_n}{4}$

where

  • $A_n$ are the coefficients of $\sum_{n=1}^\infty A_n\sin\frac{n\pi x}{l}=x^2$
  • $B_n$ are the coefficients of $\sum_{n=1}^\infty B_n\sin\frac{n\pi x}{l}=x$.

So, yes. You can.

Pedro
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  • ah ok, at first I thought it was a crazy idea, but it wasn't, thanks! –  Apr 29 '18 at 01:22
  • Will the integral be always linear (independently of the problem)? –  Apr 29 '18 at 01:29
  • @Isa You are welcome. Actually, this kind of idea (i.e. reduce a problem to a form which has already been solved) is very common in mathematics (by the way, it is a source of jokes like this or this). With respect to the question, yes (in the sense that the second formula here holds provided that $f,g$ are integrable and $\alpha,\beta$ are constants). See this. – Pedro Apr 29 '18 at 02:09