Roots of $f(x)=a_0+a_1\cos x+a_2\cos 2x+\dots+a_n\cos nx$

Question

If $a_i$'s are nonzero real numbers such that $a_n > {\sum^{n-1}_{i=0}}|a_i|$ prove that the number of roots of $f(x)=a_0+a_1\cos x + a_2\cos 2x+\dots+a_n\cos nx$ is at least 2n.

Answer 1

The function $f$ is periodic with period $2\pi$. I claim that there are at least $2n$ roots in any period interval. To prove this it is enough to produce $2n$ points in such a period interval where $f$ has alternating signs. Then there will be an additional sign change between the last of the $2n$ points and the first one in the next copy of this interval.

Note that $$0\leq \left|\sum_{k=0}^{n-1} a_k\cos(kx)\right|\leq \sum_{k=0}^{n-1} |a_k|<a_n\qquad(x\in{\mathbb R}) .$$ For the points $$x_k:={k\pi\over n}\qquad(1\leq k\leq 2n)$$ one therefore has $$f(x_k)=a_n \bigl(\cos(k\pi)+\Theta_k\bigr)\ ,\quad |\Theta_k|<1\qquad(1\leq k\leq 2n)\ .$$ It follows that $${\rm sgn}\bigl(f(x_k)\bigr)=(-1)^k\qquad(1\leq k\leq 2n)\ .$$