Computing $\frac{d^n}{dx^n} ( f(x)^m) \Big|_{x=0}$ if $\frac{d^n}{dx^n} f(x) \Big|_{x=0}=0$ for $n$ odd

Question

Let $f$ be continuous and infinitely integrable function.

I am interested in finding

\begin{align} \frac{d^n}{dx^n} ( f(x)^m) \Big|_{x=0} \end{align}

Where $m$ and $n$ are positive integers. Here we use notation $f(x)^m$ to mean to the power of $m$ and $f^{(m)}(x)$ to mean $m$-th derivative of $f$.

Moreover, we have the following condition on $f$:

\begin{align} \frac{d^n}{dx^n} f(x) \Big|_{x=0}&=c_n \text{ for } n=\text{even},\\ \frac{d^n}{dx^n} f(x) \Big|_{x=0}&=0 \text{ for } n=\text{odd}. \end{align}

I tried using the generalized Leibniz Rule \begin{align} \frac{d^n}{dx^n}\left( \prod_{i=1}^m f \right)=\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}f^{(k_{t})}, \end{align}

Since we have that $\frac{d^n}{dx^n} f(x) \Big|_{x=0}=0$ for $n$ odd, a lot of terms will cancel.

However, I am stuck with figuring out what terms will remain in the above expression. Please help.

Thank you.

Answer 1

It immediately follows from the properties of the derivatives that if any $k_t$ is odd then

$$ \prod_{1\le t\le m}f^{(k_{t})} = 0$$

and so the only non-zero terms in the sum will be those for which all $k_t$'s are even

\begin{align} \frac{d^n}{dx^n}\left( \prod_{i=1}^m f \right)=\sum_{2k_1+2k_2+\cdots+2k_m=n} {n \choose 2k_1, 2k_2, \ldots, 2k_m} \prod_{1\le t\le m}c_{2k_t}, \end{align}

The condition for the sum gives $2(k_1+k_2+\cdots+k_m)=n$ from which it follows that if $n$ is odd, no combination of $k_t$'s will satisfy the condition and hence \begin{align} \frac{d^{2n+1}}{dx^{2n+1}}\left( \prod_{i=1}^m f \right)=0 \end{align}

In the other case, the expression simplifies slightly

\begin{align} \frac{d^{2n}}{dx^{2n}}\left( \prod_{i=1}^m f \right)=\sum_{k_1+k_2+\cdots+k_m=n} {2n \choose 2k_1, 2k_2, \ldots, 2k_m} \prod_{1\le t\le m}c_{2k_t}, \end{align}