This problem came to mind in conjunction with two earlier ones [1] [2].
Let $f(x)$ be positive square-integrable function on $[0,2\pi]$ with Fourier series $\sum\limits_{n=-\infty}^{\infty} c_k e^{i k x}$. Then the principal square root $\sqrt{f(x)}$ is well-defined on $[0,2\pi]$. How do the Fourier coefficients of $\sqrt{f(x)}$ on this interval depend on $c_k$?
As indicated in the comments, there is no formula giving the Fourier coefficients of $\sqrt f$ in terms of the coefficients of $f$. Let $f(x)=1+t\,\cos x$ for $|t|\le1$. It is clear that $f(x)\ge0$ if $x\in[-\pi,\pi]$. For simplicity I will consider real instead of Fourier coefficients. Since $f$ is even, all sine Fourier coefficients vanish. The Fourier series of $f$ has only two coefficients different from cero, but $$ \int_{-\pi}^\pi \sqrt{f(x)}\,\cos(n\,x)\,dx=2 \pi \sqrt{1-t} \, _3\tilde{F}_2\left(-\frac{1}{2},\frac{1}{2},1;1-n,n+1;\frac{2 t}{t-1}\right). $$ Computation done in Mathematica.