Finding the $18th$ Derivative of a Particular Product at $x = 0$

Question

Let $f_{1}(x) = e^{x^5}$ and $f_{2}(x) = e^{x^3}$. Let $g(x) = f_{1}f_{2}$. Find $g^{(18)}(0)$.

By series expansion at $x = 0$:

$f_{1}(x) = \sum_{k \ge 0} {x^{5k} \over k! }$ and $f_{2}(x) = \sum_{m \ge 0}{x^{3m} \over {m!}}$, then

$$g(x) = \sum_{k, m \ge 0}{x^{5k + 3m} \over {m!k!}}.$$

Substituting $5k + 3m = n$ we get $g(x) = \sum_{n \ge 0} \left( \sum_{5k + 3m = n}{1 \over {m!k!}} \right) x^{n} $.

Solving diophantine equation $5k + 3m = 18$, there are two ordered pairs of non - negative integers $(k, m)$: $(3, 1), (0, 6)$. Thus, $g^{18}(0) = 18! \left[ { {1 \over {3!1!}} + {1 \over {0!6!}}} \right].$

Is there a general method for finding $n^{th}$ derivative of functions $\prod_{1 \le i \le n}f_{i}$? Obviously, if there are no solutions then a derivative of a function at some point will be $0$. But what can be said when there are infinitely many solutions?

UPD: 01.08.2019

Consider function $f(x) = e^{1 \over 1 - x}$. Then by expansion at 0: $$f(x) = e\sum_{n \ge 0} \sum_{x_{1} + 2x_{2} + \cdots = n} {{1} \over {x_{1}!x_{2}!\cdots}} x^{n},$$ which gives an infinite diophantine equation. More general, it can be applied to functions of a form: $f(x)^{g(x)}.$ Referring to my early question, what can be said about a derivative at $x = 0$ of a such function?

Answer 1

This method can be generalized.

For $k,n \in \mathbb{N}$, $k\gt0$ we define $\Omega(k,n):=\{(\omega_1, \ldots, \omega_k) \in \mathbb{N}^k \text{ such that } \sum_{i=1}^k \omega_i = n\}$.

Let $g(x):=\prod\limits_{1 \leq i \leq m}f_i(x)$. Then

$g(x)=\prod\limits_{1 \leq i \leq m}\left(\sum_{k \geq 0} \frac {f_i^{(k)}(0)}{k!} x^{k}\right)$

$g(x)=\sum_{k\in\mathbb{N}^m}\left(\prod\limits_{1 \leq i \leq m} \frac {f_i^{(k_i)}(0)}{k_i!} x^{k_i}\right)$

$g(x)=\sum_{n \geq 0}\left(\sum_{k\in\Omega(m,n)}\left(\prod\limits_{1 \leq i \leq m} \frac {f_i^{(k_i)}(0)}{k_i!}\right)x^n\right)$

$g^{(s)}(x)=\sum_{n \geq 0}\left(\sum_{k\in\Omega(m,n)}\left(\prod\limits_{1 \leq i \leq m} \frac {f_i^{(k_i)}(0)}{k_i!}\right)n^{\underline{s}} \cdot x^{n-s}\right)$

$g^{(s)}(0)=s! \sum_{k\in\Omega(m,s)}\left(\prod\limits_{1 \leq i \leq m} \frac {f_i^{(k_i)}(0)}{k_i!}\right)$

Answer 2

Part I

The Faa di Bruno formula can be described in terms of the Bell polynomials of the second kind $B_{n,k}\bigl(x_1,x_2,\dotsc,x_{n-k+1}\bigr)$ by \begin{equation}\label{Bruno-Bell-Polynomial}\tag{1} \frac{\textrm{d}^n}{\textrm{d}x^n}f\circ h(x)=\sum_{k=1}^nf^{(k)}(h(x)) B_{n,k}\bigl(h'(x),h''(x),\dotsc,h^{(n-k+1)}(x)\bigr), \quad n\ge1. \end{equation} See the site https://math.stackexchange.com/a/4261764/945479.

The Bell polynomials of the second kind satisfy \begin{equation}\label{Bell(n-k)}\tag{2} B_{n,k}\bigl(abx_1,ab^2x_2,\dotsc,ab^{n-k+1}x_{n-k+1}\bigr) =a^kb^nB_{n,k}(x_1,x_2,\dotsc,x_{n-k+1}) \end{equation} for $n\ge k\ge0$. See the site https://math.stackexchange.com/a/4262657/945479.

For $n\ge1$, by the formulas \eqref{Bruno-Bell-Polynomial} and \eqref{Bell(n-k)}, we have \begin{align} \bigl[e^{1/(1-x)}\bigr]^{(n)} &=\sum_{k=1}^ne^{1/(1-x)}B_{n,k}\biggl(\frac{1!}{(1-x)^2}, \frac{2!}{(1-x)^3}, \dotsc,\frac{(n-k+1)!}{(1-x)^{n-k+2}}\biggr)\\ &=e^{1/(1-x)}\sum_{k=1}^n\frac{1}{(1-x)^{n+k}}B_{n,k}(1!, 2!, \dotsc,(n-k+1)!)\\ &=e^{1/(1-x)}\sum_{k=1}^n\frac{1}{(1-x)^{n+k}}\binom{n}{k}\binom{n-1}{k-1}(n-k)!\\ &\to e\sum_{k=1}^n\binom{n}{k}\binom{n-1}{k-1}(n-k)!, \quad x\to0, \end{align} where we used the formula \begin{equation}\label{Bell-factorial-eq}\tag{1} B_{n,k}(1!,2!,\dotsc,(n-k+1)!)=\binom{n}{k}\binom{n-1}{k-1}(n-k)!, \quad n\ge k\ge0, \end{equation} which was reviewed at https://math.stackexchange.com/a/4262340/945479. Consequently, the series expansion of $e^{1/(1-x)}$ around $x=0$ is \begin{equation} e^{1/(1-x)}=e\sum_{n=0}^\infty \Biggl[\sum_{k=1}^n\binom{n}{k}\binom{n-1}{k-1}(n-k)!\Biggr]\frac{x^n}{n!} =e\sum_{n=0}^\infty \Biggl[\sum_{k=1}^n\frac{1}{k!}\binom{n-1}{k-1}\Biggr]x^n, \quad |x|<1. \end{equation}

Part II

Let $f_1(x)=e^{x^3}$ and $f_2(x)=e^{x^5}$. Then \begin{align} [f_1(x)f_2(x)]^{(18)} &=\sum_{n=0}^{18}\binom{18}{n}f_1^{(n)}(x)f_2^{(18-n)}(x)\\ &=\sum_{n=0}^{18}\binom{18}{n}\Biggl[e^{x^3}\sum_{k=0}^n x^{3k-n} \sum _{j=k}^n 3^js(n,j)S(j,k)\Biggr]\\ &\quad\times\Biggl[e^{x^5}\sum_{k=0}^{18-n} x^{5k-(18-n)} \sum _{j=k}^{18-n}5^j s(18-n,j)S(j,k)\Biggr]\\ &\to\sum_{n=0}^{18}\binom{18}{n}\lim_{x\to0}\Biggl(\frac1{x^{18}}\Biggl[\sum_{k=0}^n x^{3k} \sum _{j=k}^n 3^js(n,j)S(j,k)\Biggr] \\ &\quad\times\Biggl[\sum_{k=0}^{18-n} x^{5k} \sum _{j=k}^{18-n}5^j s(18-n,j)S(j,k)\Biggr]\Biggr), \quad x\to0, \end{align} where we used the formula \begin{equation*} \bigl(e^{x^\alpha}\bigr)^{(n)} =e^{x^\alpha}\sum_{k=0}^n x^{k\alpha-n} \sum _{j=k}^n s(n,j)\alpha^jS(j,k), \quad n\ge0 \end{equation*} at the site https://math.stackexchange.com/a/4262657/945479. The rest argument is left to readers and the proposer of this question.