Showing $f(x)=\text{csch}(x^{-2})$ is infinitely differentiable

Question

For a problem I am trying to solve I need to show that the function $f(x)=\text{csch}(x^{-2})$ with $f(0)=0$ is infinitely differentiable in $\mathbb R$. (this is just a conjecture at the moment). It is clearly infinitely differentiable for $x \ne 0$. The only difficulty is at the origin. I can only think of coming up with a general expression for the $n$th derivative and taking the limit, but this would be very painful and there is probably a better way that I cannot see.

Answer 1

Try a proof by induction, starting with the base case being $n=1$ (the first derivative) and then showing that $f^{(k)}(x) = \frac{d^k}{dx^k} \text{csch}(x^{-2})$ implies $f^{(k+1)}(x) = \frac{d^{k+1}}{dx^{k+1}} \text{csch}^{-2})$: $$f^{(k+1)}(x) = \frac{d}{dx}f^{(k)}(x).$$ If you can come up with a general formulat for the $k$th derivative of $\text{csch}(x^{-2})$, then you can easily prove that for all integer $n$, $f(x)$ is infinitely differentiable.

Answer 2

The function $g(u) = 2u/(1-u^2)$ is $C^\infty$ on $(-1,1).$ Your function equals $g(e^{-1/x^2}),$ which is the composition of $C^\infty$ functions, hence is $C^\infty.$

Added later: Define $h(x)=e^{-1/x^2}, x\ne 0,$ $h(0)=0.$ Sketch of proof that $h\in C^\infty(\mathbb R):$

1. Show by induction that there exist polynomials $p_n,n=0,1,2,\dots,$ such that

$$h^{(n)}(x) = p_n(1/x)e^{-1/x^2},\,\, x\ne 0.$$

2. Show that for any polynomial $p,$

$$\lim_{x\to 0} p(1/x)e^{-1/x^2} =0.$$

1., 2. lead to a quick proof that $h^{(n)}(0)=0$ for all $n,$ and hence that $h\in C^\infty(\mathbb R).$