I've been manipulating the series definitions of the Taylor (Maclaurin) and higher-order derivatives.

The definitions I'm working with are as follows:

\begin{align*} T(x) &\equiv \sum_{n=0}^\infty\frac{D^nf(0)}{n!}x^n,\\ D^nf(x)&\equiv \lim_{h\to0}\sum _{k=0}^n \binom nk \frac{(-1)^k}{h^n}f(x+(n-k)h).\end{align*}

Meaning that \begin{align*}f(x) &= \lim_{h \to 0} \sum_{n=0}^\infty\frac{x^n}{n!}\sum _{k=0}^n\binom nk \frac{(-1)^k}{h^n}f((n-k)h)\\ &= \lim_{h \to 0} \sum_{n=0}^\infty\sum _{k=0}^n\frac{(-1)^k x^n}{k!(n-k)!h^n}f((n-k)h)\\ &= \lim_{h \to 0} \sum_{n=0}^\infty\sum _{k=0}^n\frac{(-1)^k x^k}{k!h^k} \frac{x^{n-k}f(h(n-k))}{(n-k)!h^{n-k}},\end{align*} from which I reasoned that with Cauchy's formula, $ \sum a_n \sum b_n = \sum_{n=0}^\infty\sum_{k=0}^na_kb_{n-k} $: \begin{align*}f(x) &= \lim_{h \to 0} \sum_{k=0}^\infty\frac{(-1)^k x^k}{k!h^k}\sum _{n=0}^\infty\frac{x^nf(hn)}{n!h^{n}}\\&= \lim_{h \to 0} e^{-x/h}\sum _{n=0}^\infty\frac{x^nf(hn)}{n!h^{n}}.\end{align*}

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    $\begingroup$ What is the question? Is it about whether that last equation is correct for nice enough $f$? $\endgroup$
    – Ian
    Mar 19 at 16:44
  • $\begingroup$ The first thing I would try before posting is to check this numerically for a few examples. $\endgroup$
    – Kurt G.
    Mar 19 at 16:45
  • $\begingroup$ Another good place to start is to check each monomial: for $x^k$ you have $e^{-x/h} \sum x^n h^k n^k/(n! h^n)=e^{-x/h} \sum x^n n^k/(n! h^{n-k})$ and now you factor out an $x^k$ and try to show that what is left goes to $1$. The part that demands an approximation is the fact that $n^k/n!$ is not $1/(n-k)!$. $\endgroup$
    – Ian
    Mar 19 at 16:56
  • $\begingroup$ Being more direct, this is probably true for at least any analytic function provided that $\lim_{h \to 0} e^{-x/h} \sum_{n=0}^\infty \frac{n^k x^{n-k}}{n! h^{n-k}}=1$. It's definitely not true if this isn't the case. $\endgroup$
    – Ian
    Mar 19 at 17:23


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