Bounding the derivative of a function

Question

Let $f(x) = \frac{\exp(ax)}{1+\exp(ax)}$ for some $a > 0$ be a logistic function. I am looking for an upper bound on the $k$-th derivative of $f$, i.e. $|f^{(k)}(x)|$ on $\mathbb{R}$.

Using Cauchy's integral formula I only obtain a bound of order $k! / (\pi-\epsilon)^{k+1}$ for $a=1$. Is it possible to get below $k!$ and obtain a growth rate that is just exponential?

Thanks!

Answer 1

Given that $y=\dfrac{e^{ax}}{1+e^{ax}}$ we have

\begin{eqnarray}y^\prime&=&\frac{ae^{ax}}{1+e^{ax}}\\ &=&a(1-y)y \end{eqnarray}

from which it follows that

$$ y^{\prime\prime}=a^2(1-2y)(1-y)y $$

etc, for $0< y<1$

So for each $n$ there is a polynomial $P_n(y)$ of degree $n$ such that

$$ y^{(n)}= a^nyP_n(y)\text{ for }y\in(0,1)$$

so the existence of an upper bound is assured.

Addendum: Actually. $(1-y)$ will also be factor of $y^{(n)}$ for $n>0$ so we can make the slightly stronger statement

$$ y^{(n)}= a^ny(1-y)Q_n(y)\text{ for }y\in(0,1) $$

where $Q_n$ is a degree $n-1$ polynomial in $y$.

We have $Q_1(y)=1,\,Q_2(y)=1-2y,\,Q_3(y)=1-6y+6y^2,\cdots$

The sequence of $Q_n(y)$ polynomials are computed recursively by

1. $Q_1(y)=1$

2. $Q_{n+1}(y)=\dfrac{d}{dy}y(1-y)Q_n(y)$

Second Addendum:

\begin{align*} Q_1(y)&=1\\ Q_2(y)&=1-2y\\ Q_3(y)&=1-6y+6y^2\\ Q_4(y)&=1-14y+36y^2-24y^3\\ Q_5(y)&=1-30y+150y^2-240y^3+120y^4\\ Q_6(y)&=1-62y+540y^2-1560y^3+1800y^4-720y^5\\ Q_7(y)&=1-126y+1806y^2-8400y^3+16800y^4-15120y^5+5040y^6\\ Q_8(y)&=1-254y+5796y^2-40834y^3+126000y^4-191520y^5+141120y^6-40320y^7 \end{align*}

In general

\begin{equation} Q_n(y)=\sum_{m=0}^{n-1}a(n,m)y^m \end{equation}

where $\vert a(n,m)\vert$ equals the number of surjections of an $n$-element set onto an $m+1$ element set (according to the Online Encyclopedia of Number Sequences).

Furthermore

\begin{equation} a(n,m)=\sum_{k=1}^{m+1}(-1)^{k+1}\binom{m+1}{k}k^n \end{equation}

Answer 2

Using a CAS to get derivatives and OEIS, I conjecture that the $k$th derivative of $f(x)$ is:

$$f^{(k)}(x) = \frac{a^ke^{ax}}{(e^{ax}+1)^{k+1}}\sum_{m=0}^{k-1} (-1)^{m}A(k, m)e^{axm}$$

Where $A(k, m)$ are the Eulerian numbers. Does this help?