Bounding the $n$-th derivatives of $\frac{1-\cos(x)}{x^2}$

Question

Define the smooth map $f : \mathbb{R} \rightarrow \mathbb{R}$ by $f(x) := \frac{1-\cos(x)}{x^2} = -\sum\limits_{k=1}^\infty \frac{(-1)^k}{(2k)!} x^{2k-2}$.

I am looking for a nice bound on $|f^{(n)}(x)|$ for any $n \in \mathbb{N}_0$ and $x \in \mathbb{R}$. Some very preliminary simulations seem to suggest something like $|f^{(n)}(x)| \leq \min(\frac{1}{2},\frac{2}{x^2})$.

This seems easy at first glance, since we can just give an explicit formula for the $n$-th derivative with \begin{align*} & f^{(n)}(x) = \sum\limits_{k=0}^n {n \choose k} \partial_x^k[1-\cos(x)] \, \partial_x^{n-k}[x^{-2}] \ , \end{align*} which is a relatively simple formula, but using this formula to explain the behaviour for $x\rightarrow 0$ is already difficult.

Here is a plot containing $f$(blue), $f^{(1)}$(orange), $f^{(2)}$(green), $f^{(3)}$(purple), $f^{(4)}$(brown) and $\min(\frac{1}{2},\frac{2}{x^2})$(red):

Any help is much apprechiated!

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Edit: Thank you for all the helpful answers! I used the idea by Fedor Petrov to adjust your bounds to my needs. I arrive at \begin{align*} & |f^{(n)}(x)| \leq \min\Big( \frac{1}{(n+1)(n+2)}, \frac{6}{x^2} \Big) \ . \end{align*} Since the first bound $|f^{(n)}(x)| \leq \frac{1}{(n+1)(n+2)}$ was already shown by multiple answers and since $f$ is symmetric, it suffices to show $|f^{(n)}(x)| \leq \frac{6}{x^2}$ for $x \geq \sqrt{6(n+1)(n+2)}$. For $n \geq 2$ we have the following calculations:

Since \begin{align*} & f^{(n)}(x) = \Re\bigg( \int_0^1 (1-t) (it)^n e^{itx} \, dt \bigg) \ , \end{align*} integrating by parts twice yields \begin{align*} & f^{(n)}(x) = \Re\bigg( \Big[ (1-t) (it)^n \frac{-i e^{itx}}{x} \Big]_0^1 - \int_0^1 \frac{(it)^n (n(1-t)-t)}{t} \frac{-i e^{itx}}{x} \, dt \bigg)\\ & = \Re\bigg( 0 + i \int_0^1 \frac{(it)^n (n(1-t)-t)}{t} \frac{e^{itx}}{x} \, dt \bigg)\\ & = \Re\bigg( i \Big[ \frac{(it)^n (n(1-t)-t)}{t} \frac{-i e^{itx}}{x^2} \Big]_0^1 - i \int_0^1 \frac{n(it)^n (n(1-t)-t-1)}{t^2} \frac{-i e^{itx}}{x^2} \, dt \bigg)\\ & = \Re\bigg( i^2 i^n \frac{e^{ix}}{x^2} + \frac{i^2 n^2}{x^2} \int_0^1 \frac{1-t}{t^2} (it)^n e^{itx} \, dt - \frac{i^2 n}{x^2} \int_0^1 \frac{t+1}{t^2} (it)^n e^{itx} \, dt \bigg) \end{align*} and we for $n \geq 2$ get the bound \begin{align*} & |f^{(n)}(x)| \leq \frac{1}{x^2} + \frac{n^2}{x^2} \underbrace{\int_0^1 \frac{1-t}{t^2} t^n \, dt}_{=\frac{1}{n(n-1)}} + \frac{n}{x^2} \underbrace{\int_0^1 \frac{t+1}{t^2} t^n \, dt}_{= \frac{1}{n} + \frac{1}{n-1}} \leq \frac{6}{x^2} \ . \end{align*} It remains to check the cases $n=0$ and $n=1$. We calculate \begin{align*} & |f^{(0)}(x)| = \bigg|\Re\bigg( \int_0^1 (1-t) e^{itx} \, dt \bigg)\bigg| = \bigg|\Re\bigg( \frac{ix-e^{ix}+1}{x^2} \bigg)\bigg| \leq \frac{2}{x^2} \end{align*} and \begin{align*} & |f^{(n)}(x)| = \bigg|\Re\bigg( i \int_0^1 (1-t) t e^{itx} \, dt \bigg)\bigg| = \bigg|\Re\bigg( -i \frac{x+e^{ix}(x+2i)-2i}{x^3} \bigg)\bigg| \leq \frac{1}{x^2} + \frac{4}{x^3} \overset{x \geq 1}{\leq} \frac{5}{x^2} \ . \end{align*}

Answer 1

We have $$f(x)=\int_0^1(1-t)\cos(tx)dt$$ (It's the integral formula for the remainder of Taylor expansion of $\cos(x)$). Differentiating under the integral sign you get $$f^{(n)}(x)=\text{Re} \int_0^1(1-t) (it)^n e^{itx}dt$$ so that for all $n$ $f^{(n)}(x) =o(1)$ as $|x|\to\infty$ by Riemann-Lebesgue as in your graphs (an explicit bound $O(x^{-2})$ shouldn't be hard to find. edit: just integrate by parts twice as suggested in comments by Fedor Petrov!).

Also $\|f^{(n)}\|_\infty =o(1)$ as $n\to\infty$ because for all $x$ $$|f^{(n)}(x)|\le \int_0^1(1-t)t^n dt= \frac1{(n+1)(n+2)}$$

Answer 2

Lemma (implicit in Dunkell's solution to problem 339 in the AMM): Let $g\in C^{n+1}(\mathbb{R})$. If $\lim_{|x| \to \infty} g^{(n)}(x)=0$ then $$\|g^{(n)}\|_{\infty} \le \frac{1}{n+1}\|(xg(x))^{(n+1)}\|_{\infty}.$$

Corollary: $\|f^{(n)}\|_{\infty} \le \frac{1}{(n+1)(n+2)}$. Proof: Apply Lemma with $g = f$ and $g=xf$ to find $\|f^{(n)}\|_{\infty} \le \frac{1}{n+1}\|(xf(x))^{(n+1)}\|_{\infty} \le \frac{1}{(n+1)(n+2)}\|(x^2f(x))^{(n+2)}\|_{\infty}$. To conclude, observe $|(x^2 f(x))^{(n+2)}|\in \{|\cos x |,|\sin x|\}$.

Proof of Lemma: We first show $\sup g^{(n)} \le\frac{1}{n+1}\|(xg(x))^{(n+1)}\|_{\infty}$. If $\sup g^{(n)}=0$ we are done. Otherwise $g^{(n)}$ attains its global maximum at a point $c$ with $g^{(n+1)}(c)=0$. We differentiate $n+1$ times the relation $x \cdot g(x) = xg(x)$ to obtain $(n+1) g^{(n)}(x) + x g^{(n+1)}(x) = (xg(x))^{(n+1)}$; at $x=c$ this gives the required bound. A matching lower bound on $\inf g^{(n)}$ is proved similarly.

Answer 3

Your function is essentially $g(x)=\sin(x/2)^2/(x/2)^2$ up to dilation. Now, consider that $$\|f^{(k)}(x)\|_{L^\infty} \lesssim \int_{\mathbb{R}} |\xi|^k |\hat{f}(\xi)|d\xi.$$ Now, write the Fourier transform of $g$ as a convolution of a square function with itself, which is up to constants essentially $1_{[-1,1]}(\xi) (1-|\xi|)$. You are left with estimating $$\int_{0}^1 \xi^k (1-\xi)d\xi=\frac{1}{k} - \frac{1}{k+1}=O(1/k^2).$$