Suppose you have a function $F:[-\pi,\pi]\to [-\pi,\pi]$ that is strictly increasing, has a continuous derivative, and we know its Fourier coefficients (so we know also the Fourier coefficients of the derivative).
Is there a way to find the Fourier series of the inverse function $F^{-1}$?
(Note: the inverse, not the reciprocal)
I tried to compute them, but I get nested trigonometrical function like $\cos(n\cos(kx))$ and there seem not to be a wayout…
My approach:
suppose that
$$F(x) = a_0 + \sum a_n\cos(nx) + b_n\sin(nx) $$
and try to compute the coefficients of the inverse. If $F[-\pi,\pi] = [a,b]\subseteq [-\pi,\pi]$, then you get (up to constants)
$$\int_a^b F^{-1}(x)\cos(nx) dx = \int_a^b F^{-1}(F(y))\cos(nF(y)) dF(y)$$
$$
= \int_{-\pi}^\pi y\cos(nF(y)) F'(y)dy.
$$
Using the linearity of the integral, one can split the $F'$ into cosines and sines, so we have to compute, for example (and still up to constants)
$$
\int_{-\pi}^\pi y\cos(nF(y)) \cos(my)dy.
$$
and here is where I get stuck
Best Answer
For sake of simplicity I'll use the complex coefficient Fourier sum: $$F(x)=\sum_{n=-\infty}^\infty A_n e^{inx},\quad A_n=\frac1{2\pi}\int_{-\pi}^{\pi} F(x) e^{-inx}\; dx$$ Using your way we get to:$$F^{-1}(x)=\sum_{n=-\infty}^\infty B_n e^{inx},\quad B_n=\frac1{2\pi}\int_{-\pi}^{\pi} yF'(y) e^{-inF(y)}\; dy$$ With simple change(set $g(y)=\frac niF(y)$) we get $$B_n=\frac i{2\pi n}\int_{-\pi}^{\pi} yg'(y) e^{g(y)}\; dy$$
This is no nice form(that I know about) further than this, but if we assume a few more things, namely: $ yg'(y) e^{g(y)}\in C^{\omega}([-\pi,\pi])$ and the radius of convergence is for the Taylor series of it is at least $\pi$. Using the Taylor theorem we get:$$yg'(y) e^{g(y)}=e^{g(0)}g'(0)x+e^{g(0)}(g''(0)+g'(0))x^2+\frac12e^{g(0)}(g'''(0)+f'(0)^3+3g'(0)g''(0))x^3+O(x^4)$$