# Faa di Bruno's formula and alternating functions

Suppose you have a function $$f(x)$$ such that $${\rm sgn}\Big(\frac{d^k}{dx^k}\big(f(x)\Big) = (-1)^k$$ and a function $$g(x)$$ such that $${\rm sgn} \Big(\frac{d^k}{dx^k}g(x)\Big) = (-1)^{(k+1)}$$, $$\forall k \in \mathbb{Z}^+, x \in \mathbb{R}^+$$.

$${\rm sgn(x)}$$ is the sign (or signum) function.

I want to prove that $${\rm sgn} \Big(\frac{d^k}{dx^k} f(g(x))\Big) = (-1)^k$$.

After trying to find patterns in the expression that arose after differentiation, I realized that this problem could be approached using the Faa di Bruno's rule and Bell's polynomials: $$\frac{d^n}{dx^n}f(g(x)) = \sum_{k=1}^n f^{(k)}(g(x))\cdot B_{n,k}(g^{(1)},g^{(2)},...,g^{(n-k+1)})$$

where $$g^{(m)}(x)$$ represents the $$m$$-th derivative of $$g$$ wrt $$x$$.

However, I am unable to generalize this and find an inductive or, a somewhat out of my depth, combinatorial proof for this. How should I go about it? Any advice would be appreciated.

eg: $$f(x) = e^{-x}$$ and $$g(x) = \sqrt{x}$$.

Rewriting Faa di Bruno's polynomial as $$\frac{d^n}{dx^n}f(g(x)) = \sum_{k=1}^n \frac{n!}{\prod_{i=1}^n m_i!(i!)^{m_i}} f^{(\sum_{i=1}^n m_i)}(g(x))\cdot \prod_{j=1}^n(g^{(j)}(x))^{m_j}$$ subject to $$\sum_{i=1}^n im_i = n$$.
What we want to prove now is the following: if $$n$$ is even, $$(\sum m_i),(\sum m_{2j})$$ are both even or are both odd.
• I assume the $k$ is not fixed, right? Also, not convinced that it's true; IMHO $f = x$ and $g = -x$ should be a counterexample. – darij grinberg May 11 at 0:55
• What does ${\rm sgn}\Big(\frac{d^k}{dx^k}f(x)\Big) = (-1)^k$ mean? Is it shorthand for $\forall x_0: {\rm sgn}\Big(\frac{d^k}{dx^k}f(x)\Big|_{x=x_0}\Big) = (-1)^k$? – Peter Taylor May 14 at 7:11