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I need to prove that there exists $a> 0$ and $n_0\in\mathbb N$ such that $$\forall n> n_0,\quad \sup\limits_{|x|\leq a} |f^{(n)}(x)| \geq (n!)^{\frac 32}.$$ Where $f$ is defined by $f(x)=\exp(-\frac 1{x^2})$. I know that $f^{(n)}(x)=\frac{P_n(x)}{x^{3n}}\exp(-1/x^2)$ where the polynomials $P_n$ verify the relation $P_{n+1}=X^3 P'_n+(2-3nX^2)P_n$

thank you in advance for your support

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  • $\begingroup$ You want such a precise estimate? Maybe there is some exponential correction but the growth is still captured by the factorial to $3/2$. $\endgroup$
    – Giorgio Metafune
    Commented Mar 29, 2021 at 11:13
  • $\begingroup$ Yes I am trying to have precisely this lower bound $\endgroup$
    – Paul
    Commented Mar 29, 2021 at 19:50

1 Answer 1

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Math experiment done with Mathematica

f[n_] := NMaximize[{Evaluate[RealAbs[D[Exp[-1/x^2], {x, n}]]]/
n!^(3/2),   RealAbs[x] <= 1 && n >= 1 && n \[Element] PositiveIntegers}, {x}, 
Method -> {"DifferentialEvolution", "ScalingFactor" -> 1}, 
AccuracyGoal -> 5, PrecisionGoal -> 5, WorkingPrecision -> 15]
Table[f[k], {k, 1, 20}]

results in $$\left( \begin{array}{cc} 0.819832557883720 & \{x\to -0.816496580926071\} \\ 1.05309210878418 & \{x\to -0.520851654621382\} \\ 1.43213203276870 & \{x\to -0.403266928496221\} \\ 2.02929498643239 & \{x\to -0.487430660529263\} \\ 3.99625841701811 & \{x\to -0.397684505142533\} \\ 7.32790921886967 & \{x\to -0.341047324550614\} \\ 12.8456602395717 & \{x\to -0.301641954804728\} \\ 20.1279218513637 & \{x\to 0.333941189526955\} \\ 40.8839394652186 & \{x\to 0.299174580083694\} \\ 79.3007302990450 & \{x\to -0.272480728040435\} \\ 148.468049756260 & \{x\to 0.251254191655359\} \\ 270.267689843771 & \{x\to 0.233912856932535\} \\ 491.924121098429 & \{x\to -0.249799234272746\} \\ 976.765593945151 & \{x\to -0.233559061606742\} \\ 1885.77886324325 & \{x\to -0.219827966408129\} \\ 3555.54552839170 & \{x\to 0.208043856897294\} \\ 6569.30249774506 & \{x\to 0.197802964493932\} \\ 11926.3007981150 & \{x\to 0.188807540307310\} \\ 11115.5329809164 & \{x\to 0.165430061926195\} \\ 48517.0571216129 & \{x\to -0.189175462374212\} \\ \end{array} \right) ,$$ suggesting $a=1$ and $n_0=2$.

Addition. A numerical proof with Mathematica can be done as follows.

f[n_?NumericQ] := NMaximize[{RealAbs[Evaluate[D[Exp[-1/x^2], {x, n}]]]/n!^(3/2), 
RealAbs[x] <= 1 && n >= 1 && 
 n \[Element] PositiveIntegers}, {x},][[1]]
NMinimize[{Evaluate[f[n]], n >= 2 && n \[Element] Integers}, n, 
Method -> {"DifferentialEvolution", "ScalingFactor" -> 1}, 
 AccuracyGoal -> 5, PrecisionGoal -> 5, WorkingPrecision -> 15]

$$\{1.05309210878418,\{n\to 2\}\} $$

It should be noticed that the execution of the above code takes a lot of time.

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  • $\begingroup$ Thank you for this verification but how to show it $\endgroup$
    – Paul
    Commented Mar 23, 2021 at 16:09

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