Bounding the extrema of polynomials from $\frac{d^n}{dx^n} \exp(-1/x)$

Question

As laid out on Wikipedia, the function $$f(x):=\begin{cases} \exp(-1/x) & x>0\\ 0 & x\le 0 \end{cases}$$

has the expression for derivatives at $x>0$, $$ f^{(n)}(x) = \frac{p_n(x)}{x^{2n}}f(x) $$ where the polynomials $p_n$ of degree $n-1$ are recursively defined by $p_1=1$, $$ p_{n+1} = (1-2nx)p_{n} + x^2 p_{n}'$$

I am interested in bounding $f^{(n)}$ on sets like $[0,a]$. Since its well known that all derivatives vanish at $x=0$, and $f(x)/x^{2n}$ never vanishes for $x>0$, the local extrema of $f^{(n)}$ occurs at a zero of $p_{n+1}$, so bounding $f^{(n)}(x)$ depends on value of its zeros.

Alternatively, perhaps a crude bound on $p_n$ on these sets $[0,a]$ is obtainable from the recurrence. This sounds easier, but I have yet to make progress here as well.

Question

How do I get a bound on $f^{(n)}$? (Preferrably one that looks like the inequality below)

Motivation

to test my understanding of Gevrey functions, I have decided to check what which space (if any) this function belongs to. This requires getting some bounds on the derivative. My suspicion is that $f$ is Gevrey of order 2, i.e. for any compact set $K$(and in particular any compact set containing $0$) there is a constant $C$ such that $$ \sup_{x\in K} |f^{(n)}(x)| \le C^{n+1} n!^2 $$

Notable "similar question"(currently unanswered)

Bounding $n$-th derivative of $x \mapsto \exp\left({-\frac{1}{x^2}}\right)$

There is a comment there that reaches somewhat the same idea (the associated recursive polynomial is important) but doesn't say any more.

Answer 1

I thought of this after typing the question...

Let $p_n(x) = \sum_{k=0}^{n-1} p_{n,k} x^k$. We know $p_{n,0} = 1$ for all $n$. Also, $$p_{n+1,n} =-2n p_{n,n-1} + (n-1) p_{n,n-1} = -(n+1) p_{n,n-1} $$ More generally, $$ p_{n+1,k} = p_{n,k}-2np_{n,k-1} + (k-1)p_{n,k-1}$$ (with the understanding that for $k>n-1$ or $k<0$, $p_{n,k}=0$.) Thus $$ P_{n+1} := \sup_k |p_{n+1,k}| \le 2(n+1) P_n$$ This means that $$P_{n}\le 2^{n} n! $$ so $$\sup_{x\in [0,a]} |p_n(x)|\le2^n n!\sum_{k=0}^{n-1} a^k =\frac{2^n n!(a^{n}-1)}{a-1}$$ Now, recall that for $x>0$, $e^{x} > \frac{x^j}{j!}$. Rearranging, $$ e^{-x} \le \frac{j!}{x^j}\implies e^{-1/x} \le x^j j! \implies \frac{ e^{-1/x}}{x^{2n} } \le (2n)! \le 4^n n!^2$$ so I'm left with the bound $$ \sup_{x\in [0,a]} |f^{(n)}(x)| \le C_a^{n+1} n!^3$$ So the function is Gevrey of order 3. Don't know if I can improve it...

Answer 2

As in this question, we have $$ f^{(n)}(x)= e^{-1/x} \cdot \sum _{k=0}^{n-1} (-1)^k k! \binom{n}{k} \binom{n-1}{k} x^{-2 n+k} $$ for $x>0$ and $f^{(n)}(x)=0$ for $x=0$.

Replacing $x$ with $x^{-1}$, the maximum of $\left|f^{(n)}(x)\right|$ on $[0,a]$ for $a>0$ is equal to the maximum of $$ \left| e^{-x} \cdot \sum _{k=0}^{n-1} (-1)^k k! \binom{n}{k} \binom{n-1}{k} x^{2 n-k} \right| $$ on $[a^{-1},\infty)$.

Taking out a factor of $x^{n+1}$ and using the triangle inequality, we get a bound of $$ e^{-x}x^{n+1} \cdot \sum _{k=0}^{n-1} k! \binom{n}{k} \binom{n-1}{k} x^{(n-1)-k}. $$ Noting that $k!\binom{n}{k}\le n^k$, we can bound the above using the binomial theorem by: $$ e^{-x}x^{n+1} \cdot (n+x)^{n-1}\le \max\left\{2^{n-1}e^{-x}x^{2n}, 2^{n-1}n^{n-1}e^{-x}x^{n+1}\right\} $$ where the two terms above correspond to the cases $x>n$ and $n>x$, respectively. For any $m>0$, the function $e^{-x}x^m$ for $x\ge 0$ attains its maximum at $x=m$ where its value is $e^{-m}m^m$.

Therefore, in both cases above, we get a bound of the form $A\cdot B^n \cdot n^{2n}$. for $A,B>0$. By Stirling's approximation, $n^n<n! e^n$. Therefore we have that $$ \left| f^{(n)}(x) \right| \le A\cdot C ^n \cdot \left(n!\right)^{2} $$ for $A,C>0$ and so $f(x)$ is Gevrey of order 2.