I've been working on an exercise and have gotten stuck:

Suppose that $T(x)=\sum_{n=0}^N a_n\cos(2\pi nx)+b_n\sin(2\pi nx)$ is non-negative on $[0,1]$. Show that there exist $c_0,...,c_N\in \mathbb{C}$ such that $$ T(x)=\left|\sum_{n=0}^{N}c_n \exp(2\pi inx) \right|^2. $$

My strategy was to expand the argument using $|z|^2=z\bar{z}$. Equating coefficients and working from the largest value $N$ first, I obtained $$ a_N \cos(2N\pi x)+b_N\sin(2N\pi x)=c_N\bar{c_0}\exp(2N\pi ix)+\overline{c_N\bar{c_0}}\exp(-2N\pi ix) $$ If we call $c_N\bar{c_0}=u-iv$, for instance, then we get equality if $u=a_N/2$ and $v=b_N/2$. Thus, upon choosing $c_0$, for instance, one can solve for $c_N$ to get these terms to match up. Now, looking at the $N-1$ term, we get $$ a_N \cos(2N\pi x)+b_N\sin(2N\pi x)=(c_N\bar{c_1}+c_{N-1}\bar{c_0})\exp(2N\pi ix)+\overline{(c_N\bar{c_1}+c_{N-1}\bar{c_0})}\exp(-2N\pi ix) $$ Again, calling the coefficient $c_N\bar{c_1}+c_{N-1}\bar{c_0}=u-iv$, we get equality if $u=a_{N-1}/2,v=b_{N-1}/2$. Since $c_N$ and $c_{N-1}$, are known, this comes down to solving $$ c_N\bar{c_1}+c_{N-1}\bar{c_0}=a_{N-1}/2-ib_{N-1}/2 $$ Equating real and imaginary parts, this gives us an inhomogeneous system of 2 linear equations to solve. This is where I'm stuck; I haven't used the (clearly necessary) hypothesis that $T\geq 0$, and I'm assuming it will somehow imply that the system has a solution (in which case we could iterate this method and get systems of 2 equations for smaller values of $n$ as well), but I can't see how. I also suspect that the freedom to choose $c_0$ might be important. I'd appreciate any advice on if this method seems viable, or if there is an easier way to do this exercise.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.