# $\sin(x) = \sum{a_n \sin(n \log(x))+b_n \cos(n \log(x))}$

Can $f(x)=\sin(x), x>0$ be represented by the series $\sum{a_n \sin(n \log(x))+b_n \cos(n \log(x))}$ ?

note: this is in continuation of another question that I asked earlier here

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This depends on what you understand by represent. Consider the function $\sin(e^t)$ and its Fourier expansion on the interval $[0,2\pi]$: $$\sin(e^t)\sim\frac{a_0}{2}+\sum_{n=1}^\infty(a_n\cos(nt)+b_n\sin(nt),$$ where $$a_n=\frac{1}{\pi}\int_0^{2\pi}\sin(e^t)\cos(nt)dt,\quad b_n=\frac{1}{\pi}\int_0^{2\pi}\sin(e^t)\sin(nt)dt.$$ Standard convergence theorems show that the series converges pointwise to $\sin(e^t)$ in $(0,2\pi)$ (with uniform convergence on compact subintervals.) Fot $t=0$ and $t=2\pi$ it converges to $(\sin(1)+\sin(e^{2\pi})/2=0.9151\dots$ Morever, $$\lim_{N\to\infty}\int_0^{2\pi}\Bigl|\sin(e^t)-\Bigl(\frac{a_0}{2}+\sum_{n=1}^N(a_n\cos(nt)+b_n\sin(nt)\Bigr)\Bigr|^2dt=0.$$

Now do the change of variable $x=e^t$. You get that $$\frac{a_0}{2}+\sum_{n=1}^N(a_n\cos(n\log x)+b_n\sin(n\log x)$$ converges to $\sin x$ pointwise on $(1,e^{2\pi})$ and $$\lim_{N\to\infty}\int_1^{e^{2\pi}}\Bigl|\sin x-\Bigl(\frac{a_0}{2}+\sum_{n=1}^N(a_n\cos(n\log x)+b_n\sin(n\log x)\Bigr)\Bigr|^2\frac{dx}{x}=0.$$

This all amounts to the fact that the functions $\{\sin(n\log x),\cos(n\log x)\}$ are a complete orthogonal system on the interval $(1,e^{2\pi})$ with respect to the weight $1/x$.

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The series would satisfy (if the sum is s(x) ): $$s(e^{2 \pi} ) = s(1)$$ This is not the case for the sine function. Therefore, it can't be done.

EDIT: fixed 0 to 1

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 Well, @yaakov: what do you mean by $s(0)$? The logarithmic function doesn't take values at x=0. – awllower Apr 8 '11 at 14:36 @awllower: my bad, of course I meant 1 – yaakov Apr 9 '11 at 20:50 It's ok, and then it helps clarify the mysterious term--$representation$. – awllower Apr 11 '11 at 12:01