Largest root of a linear combination of Chebyshev polynomials

Question

I wonder if we can say something about roots of a linear combination of Chebyshev polynomials of the first kind. I have an example in my hand: $$(m+1)T_n(x)+(m-3)T_{n-2}(x)=0$$ for some $m>0$. I understand the root of $T_n(x)$ is given by $\cos\left(\frac{2k-1}{2n}\pi\right)$ for $k=1,\dots, n$. It does not look tractable to find an explicit form of the largest root. Is there a way to understand at least any bound on the largest root in terms of $n$ and $m$?

Answer 1

I'll assume $m > 3$ and $n > 3$.

The two greatest roots of $T_n(x)$ are $\cos(\pi/(2n))$ and $\cos(3\pi/(2n))$. If $n > 3$, $\cos(3\pi/(2n)) < \cos(\pi/(2(n-2))) < \cos(\pi/(2n))$. For $x > \cos(\pi/(2n))$ we have both $T_n(x) > 0$ and $T_{n-2}(x) > 0$, and for $\cos(3\pi/(2n)) < x < \cos(\pi/(2(n-2)))$ we have $T_n(x) < 0$ and $T_{n-2}(x) < 0$. So the greatest root of your expression is somewhere between $\cos(\pi/(2(n-2)))$ and $\cos(\pi/(2n))$.