# Fourier series with trig parameters

Suppose I want to write the function $x \sin(t)$ as the series over the interval $x \in (0,\pi)$

$$x\sin(t) = \sum_{n=1}^{\infty}(a_n \cos(t) + b_n \sin(t) )\sin(nx)$$

Then would the coefficients $a_n$ and $b_n$ be just simply

$$a_n =\frac{1}{\pi \cos(t)} \int_{0}^{\pi}x\sin(nx) \;\mathrm{d}x = \frac{\tan(t)}{\pi} \int_{0}^{\pi}x\sin(nx) \;\mathrm{d}x$$

$$b_n =\frac{1}{\pi \sin(t)} \int_{0}^{\pi}x\sin(t)\sin(nx) \;\mathrm{d}x = \frac{1}{\pi} \int_{0}^{\pi}x\sin(nx) \;\mathrm{d}x$$

Could i just treat the trig functions as parameters?

Thanks

EDIT: The bounty is supposed to read "not enough attention". I may have misselcted the wrong item when I set the bounty

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Expand $x$ as usual, and then multiply by $\sin\,t$. – J. M. Aug 3 '12 at 21:54
Right, I could cancel out that sine for $b_n$. But is my theory right? – Hawk Aug 3 '12 at 21:55
I'm afraid a lot went wrong in your question as it stands on 08/06/12. – Christian Blatter Aug 6 '12 at 20:59
@ChristianBlatter, what do you mean by that? – Hawk Aug 6 '12 at 22:47
How does $1/\cos t$ becomes $\tan t$? – enzotib Aug 10 '12 at 21:28

## 1 Answer

(Cf. my comment above)

The only sense one can give the formula $$x\sin(t) = \sum_{n=1}^{\infty}(a_n \cos(t) + b_n \sin(t) )\sin(x)$$ is the following: Assume that the (constant) series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ both converge with sum $a$ and $b$ respectively. Then one would have the equation $$x\sin t=(a\cos t + b\sin t)\sin x$$ connecting the variables $t$ and $x$. Nothing about Fourier series here.

The formula $$a_n =\frac{1}{\pi \cos(t)} \int_{0}^{\pi}x\sin(x) \;\mathrm{d}x = \frac{\tan(t)}{\pi} \int_{0}^{\pi}x\sin(x) \;\mathrm{d}x$$ and the similar formula for $b_n$ doesn't make any sense whatsoever. Note that the letter $n$ isn't even appearing on the RHS.

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Does it improve anything if I add back "n"? I'll do it. I may have forgotten about that. – Hawk Aug 7 '12 at 18:56