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

Question

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

Answer 1

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 $1/x$.

Answer 2

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