Prove that a function with $\cos(x)$ has zeros The Problem
I'm trying to find the zeros of this function in the $[0, +\infty]$ interval:
$$f(x) = e^{-x^2} - \cos(x)$$
What I've tried
I know that $\cos(x)$ is a periodic function limited by $[-1, 1]$ and that $e^{-x^2}$ asymptotically approaches $0$ as $x \rightarrow +\infty$, therefore, both $\cos(x)$ and $e^{-x^2}$ have equal values in many points in the $[0, +\infty]$ interval . But i don't know how to manipulate the function formula in order to find the infinite amount of zeros of this in function in the $[0, +\infty]$. Any ideas on how can i do that? I've thought about reviewing some materials related to numerical analysis ( like the method of successive approximations ) but found nothing that could help me.
 A: Consider that you look for the zeros of function
$$f(x)=e^{-x^2}-\cos(x)$$
The first iterate of Newton method gives
$$x_1^{(n)}=(2n+1)\frac \pi 2+\frac 1{(2n+1)\pi-(-1)^n \exp\left(\Big[(2n+1)\frac \pi 2\Big]^2\right)}$$
For the second root (after the trivial $x=0$), this gives
$$x_1=\color{red}{4.712388980611567581644190927}14$$ while the solution is
$$x=  \color{red}{4.71238898061156758164419092765}$$
For the third root, the difference is $5.25\times 10^{-79}$ and for the fourth it is $7.24\times 10^{-156}$.
Doing the same with Householder method, the second root will be in absolute error of $1.80\times 10^{-46}$.
A: For $x\ge 1$, you know that $0\le e^{-x^2}\le 1/e<1$.
Now, you also know that $\cos(2k\pi)=1$ and $\cos((2k+1)\pi)=-1$ for all $k\in\Bbb N$
So, you can apply Bolzano's Theorem to $f(x):=e^{-x^2}-\cos x$ in each interval $[2k\pi,(2k+1)\pi]$.
A: The Gaussian exponential is so quickly decaying that the solutions are essentially the roots of the cosine, except for $x=0$ and $x=1.4474142\cdots$.
For $x=\dfrac{3\pi}2$, we already have $e^{-x^2}<3\cdot10^{-10}$.

It is an easy matter to show that every neighborhood of $(n+\frac12)\pi$ of width, say $0.1$ for $n>1$ contains exactly one root.
