Proving that $\displaystyle\|\frac{\partial ^n}{\partial x^n}\frac{1}{x^2+1}\|_\infty =o\left({2^n (n+1)!} \right)$

Question

Prove that $\displaystyle\left\|\frac{\partial ^n}{\partial x^n}\left(\frac{1}{x^2+1}\right)\right\|_\infty =o\left({2^n (n+1)!} \right)$

Notations

Here, $\| \cdot\|_\infty$ denotes the sup-norm over $[-1,1]$

$$\frac{\partial ^n}{\partial x^n}\left(\frac{1}{x^2+1}\right)\text{ denotes the $n$-th derivative of the function }\frac{1}{x^2+1}$$

Motivation

I'm self-studying about Runge Phenomenon and what happens at Tchebychev nodes.

Thoughts

Note that getting a closed form of the $n$-th derivative here is non-trivial.

Here's the plot of the sequence $\displaystyle \frac{\|\frac{\partial ^n}{\partial x^n}\left(\frac{1}{x^2+1}\right)\|_\infty}{2^n (n+1)!} $

It converges fast to $0$.

Question How to prove this by computations ?

According to Mathematica, $\displaystyle \frac{\partial ^n}{\partial x^n}\left(\frac{1}{x^2+1}\right) =(-2)^n x^n \left(x^2+1\right)^{-n-1}\, _2F_1\left(\frac{1-n}{2},-\frac{n}{2};-n;\frac{1}{x^2}+1\right)$

Therefore, it suffices to prove that $\displaystyle \|_2F_1\left(\frac{1-n}{2},-\frac{n}{2};-n;\frac{1}{x^2}+1\right)\|_\infty = o((n+1)!)$

I have no experience with hypergeometric functions, and I can't prove this last statement.

Thanks for your suggestions.

Answer 1

Note that for a real $x$ we have $$ f(x)=\frac{1}{1+x^2}={\rm Re}\left(\frac{1}{1+ix}\right) $$ So $$ f^{(n)}(x)={\rm Re}\left(\frac{(-i)^n n!}{(1+ix)^{n+1}}\right) $$ Thus, for $x\in\Bbb{R}$ we have $$ |f^{(n)}(x)|\leq \frac{n!}{|1+ix|^{n+1}}=\frac{n!}{(1+x^2)^{(n+1)/2}}\leq n! $$ (with equality at $x=0$ if $n$ is even.) Thus, in fact we have a much stronger result than the proposed one, namely $$ \frac{1}{n!}\sup_{x\in\Bbb{R}}|f^{(n)}(x)|\leq 1. $$

Answer 2

Use Cauchy's integral formula. The function $f(x) = 1/(1+x^2)$ is analytic in $C\setminus \{\pm i\}$. In particular, it is analytic in every unit ball with a purely real center.

For a given $x \in R$, using $C = \partial B_{1/2}(x)$, by Cauchy's integral formula we have $$f^{(n)}(x) = \frac{n!}{2 \pi i} \oint_C \frac{1}{1+z^2} \frac{1}{(x-z)^{n+1}} dz = \frac{2^{n+1} n!}{2 \pi i} \int_0^{2 \pi} \frac{i e^{ti}/2}{1+ (x+e^{ti}/2)^2} dt$$ since $|x-z|=1/2$.

It's easy to show that the integrand above is bounded in magnitude by $1$ so the integral is bounded in norm by $2 \pi$. It follows $|f^{(n)}(x)| \leq 2^n n!$ for all real $x$. Then $|f^{(n)}(x)| / (2^n (n+1)!) \leq 1/(n+1)$.

We can do away with the $2^n$ factor for all but $x=0$ since a ball of radius $1$ will never hit any poles otherwise. Continuity should allow us to fill in this hole and get the same bound for all $x$.