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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 $g(x) = \sqrt{x}$.

Headway I have made:

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.

... this problem seems to be harder than I thought it would be.

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  • $\begingroup$ 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. $\endgroup$ – darij grinberg May 11 at 0:55
  • $\begingroup$ apologies, i have edited the question $\endgroup$ – user207526 May 11 at 0:59
  • $\begingroup$ What does ${\rm sgn}\Big(\frac{d^k}{dx^k}f(x)\Big) = (-1)^k$ mean? Is it shorthand for $\forall x_0: {\rm sgn}\Big(\frac{d^k}{dx^k}f(x)\Big|_{x=x_0}\Big) = (-1)^k$? $\endgroup$ – Peter Taylor May 14 at 7:11

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