What are the higher order derivatives of the logistic function?

Question

What are the higher order derivatives of the logistic function

$$\sigma(x) = \frac{1}{1+e^{-x}}$$

and is there a general formula for them?

Answer 1

This holds for $n \in \{0, 1, 2\}$, but a counterexample is $n=3$.

By the statement, we should have

$$F’’’ = F(1-F)(1-2F)(1-3F).$$

However, we have

\begin{align*} F’’’ &= [F(1-F)(1-2F)]’ \\ &= F(1-F)^2(1-2F) - F^2(1-F)(1-2F)-2F^2(1-F)^2 \\ &= F(1-F)(1-2F)[(1-F) - F - \frac{2F(1-F)}{1-2F}] \\ &= F(1-F)(1-2F)[1-3F + F - \frac{2F(1-F)}{1-2F}] \\ &= F(1-F)(1-2F)[1-3F - \frac{F}{1-2F}] \end{align*} which is a contradiction.

Answer 2

For the $n$'th derivative of the logistic function, we can derive formulas that are recursive, a series, or explicit in terms of a special function.

For the recursive formula, use, as in @littleO's hint in the comments, that the logistic function satisfies $\sigma'=\sigma(1-\sigma)$ and apply to that the generalized Leibniz (product rule) formula. Hence, $$\sigma^{(n+1)}=\sigma^{(n)}(1-\sigma)-\sum_{k=1}^{n}\binom{n}{k}\sigma^{\left(n-k\right)} \sigma^{\left(k\right)}$$

For a general formula, we may observe by the geometric series formula that $\displaystyle\sigma(x)=\sum_{i=0}^{\infty}\left(-e^{-x}\right)^{i}$, for positive $x$, which can be termwise differentiated very easily to $\displaystyle\sigma^{(n)}(x)=(-1)^{n}\sum_{i=0}^{\infty}i^n\left(-e^{-x}\right)^{i}$. This series alone provides a good approximation to $\sigma^{(n)}$ for large positive $x$, but also allows writing the derivative explicitly in terms of the special function, the Lerch Transcendent

$$\sigma^{(n)}=(-1)^{n}\Phi(-e^{-x},-n,0)$$

Answer 3

Let $$\sigma^{(n)}=\sigma\sum_{k=0}^{n}L(n,k)\sigma^k$$ and $L(n,k)=0$ when $k\notin [0,n]$. We have the recursion $$L(n+1,k)=(k+1)L(n,k)-kL(n,k-1)$$ First values are easy to find: $L(n,0)=1$, $L(n,1)=1-2^n$. Are they connected with some kind of Stirling numbers?