Show that $x+e^{-Bx^2}\mbox{cos}(x)$ has only one root over all reals ($B>0$).

Question

Let $B>0$ and define for all $x\in \mathbb{R}$ the function $f(x)=x+e^{-Bx^2}\mbox{cos}(x)$. Prove that $f$ has exactly one root over $\mathbb{R}$.

My original attempt was to show that the function $g(x)=-e^{-Bx^2}\mbox{cos}(x)$ is contractive (in this sense) and the result would follow from the Bannach Fixed Point theorem. However, I posted this as a question (look here) and I've concluded that this idea is not the best one. Can anyone give me a tip about how to attack this problem? Thanks in advance!

Answer 1

$f(x)=0$ is equivalent to: $$ g(x) = x e^{Bx^2}+\cos x = 0.$$ However, $g(x)$ is an increasing function, since: $$ g'(x) = \left(e^{Bx^2}-\sin x\right) +2Bx^2 e^{Bx^2}>0 $$ for any $x$, since $e^{Bx^2}-\sin x\geq e^{Bx^2}-1\geq 0$ and $2Bx^2 e^{Bx^2}\geq 0$ with equality only in $x=0$.

So we have that $g(x)$ has at most one real zero, and it has for sure a real zero since: $$\lim_{x\to\pm\infty}g(x)=\pm\infty.$$

Answer 2

My idea started out as trying to apply the Banach fixed point theorem and then boiled down to a simpler idea.

Let $g(x)=-e^{-Bx^2} \cos(x)$. Note as you already have that $x$ is a root of $f$ if and only if it is a fixed point of $g$.

Now notice that $g(\mathbb{R}) \subseteq [-1,1]$ and $g([-1,1]) \subseteq [-1,0]$. We can't easily do much better than $g([-1,0]) \subset [-1,0]$, so for now we'll just say that any fixed point of $g$ must be in $[-1,0]$.

But on $(-1,0)$, $e^{-Bx^2}$ and $\cos(x)$ are both strictly increasing, so $f$ is strictly increasing there as well. So there is at most one root. Moreover $f$ is negative at $-1$ and positive at $0$, so by the intermediate value theorem there is at least one root. So there is exactly one root.