Is it legal to use factorials of negative integers when summing Taylor series of different derivatives?

Question

My question is about the use of the factorial function appearing in the Taylor series, in the case that expansions of the derivatives of several order, of the same function, around the same point, are linearly combined together.

Consider the classical Taylor expansion of $f(h)$ around $x=0$, $$f_h = \sum_{k=0}^{+\infty} f_0^{(k)} \frac{h^k}{k!}, \tag{1}$$ as well as that of its derivative in the same point $f'(h)$, $$f'_h = \sum_{q=0}^{+\infty} f_0^{(q+1)} \frac{h^q}{q!}$$ which can be conveniently rewritten by the change of variable $q=k-1$, $$f'_h = \sum_{k=1}^{+\infty} f_0^{(k)} \frac{h^{k-1}}{(k-1)!} \tag{2}$$

Now consider the sum of expressions (1) and (2), the latter multiplied by $h$, $$f_h + hf'_h = \sum_{k=0}^{+\infty} f_0^{(k)} \frac{h^k}{k!} + \sum_{k=1}^{+\infty} f_0^{(k)} \frac{h^{k}}{(k-1)!}$$ It is clear that I can collect the terms form $k=1$ on in one summation, thus obtaining $$f_h + hf'_h = f_0^{(0)} \frac{h^0}{0!} + \sum_{k=1}^{+\infty} f_0^{(k)}h^k \left(\frac{1}{k!} + \frac{1}{(k-1)!}\right) \tag{3}$$

I don't think I made errors up to (3).

My question is: Can I include the first addend at the right hand side in the summation by changing the lower bound of the summation to $k=0$?

Doing so would lead to the more concise, but maybe wrong, form $$f_h + hf'_h = \sum_{k=0}^{+\infty} f_0^{(k)}h^k \left(\frac{1}{k!} + \frac{1}{(k-1)!}\right)$$ Here I know that the second addend in the parenthesis is not defined for $k=0$, since it leads to a term $(-1)!$, but

• the second addend in the parenthesis should nothing more than absent when $k=0$ (as it is absent in (3));
• the generalization of the factorial to the whole complex plain, i.e. the Gamma function, does tend to $+\infty$ as its argument approaches negative integers, so I guess it could make sense to consider $1/(k-1)! \to 0$ as $k \to 0$ (or, more generally, as $k \to -n$ with $n \in \mathbb N$);
• there are other cases where an abuse of notation makes little threat and works well, e.g. the Dirac delta "function" used as if it was a function (instead of a distribution).

Answer 1

No! Sometimes you just have to accept that you can't express things as compactly as you'd like because it would include undefined terms.

Or, of course you can describe that you're going to do so, but most of your readers will fall in one of two categories:

1. Those that doesn't understand that it's just for brevity, and ends up manipulating your formula in ways it wasn't meant to and are also undefined ending up in expressions that are just wrong.

2. Those who know better and thinks it's a stupid idea to try to save a little typing by defining that you can use undefined terms in your formulas.