I'm reading "Enumerative Combinatorics Vol 1." by Stanley who states " Throughout we will be making heavy use of the principle that formal power series can be treated as if they were functions."

Suppose $$F(x) = \sum_{n \geq 0}\frac{a_n}{n!}x^n $$is an EGF. The formal derivative $F'(x)$ of $F(x)$ is, $$ F'(x) = \sum_{n \geq 0}\frac{a_{n+1}}{n!}x^n. $$

QUESTION: What is the relationship between the formal derivative of aa formal sum and the derivative of the function that that formal sum represents?

For example (p20 of Enumerative Combinatorics Vol 1.), suppose $f(i)$ is a counting function (in the sense of Stanley) and $$G(x) = \sum_{n \geq 0}\frac{f(n)}{n!}x^n = \exp\left(x +\frac{x^2}{2} \right).$$ Does it follow that $$G'(x) = \frac{d}{dx}\left(\exp\left(x +\frac{x^2}{2}\right)\right),$$and $$ \sum_{n \geq 1}\frac{f(n-1)}{n!}x^n = \int_0^x\exp\left(y +\frac{y^2}{2} \right)dy$$ And if the above equalities do hold, are there particular functions that correspond to exponential generating functions where this breaks down?


If $F(x)$ does not converge, the question is moot.

If the series obtained by treating the formal power series for $F(x)$ as a Taylor series converges absolutely, then the derivative of the function $f(x)$ represented by that power series is always the function whose Taylor series is the formal derivative $F'(x)$.

If the formal power series converges, but does not converge absolutely, as is the case for $F(x) = \sum\frac{(-1)^{n-1}(n-1)!}{n!}x^n$ at $x=1$ then the situation is more subtle because the sum could be re-arranged to yield any value you please. However, you can replace the definition of the generating function by the limit, as $n$ goes to infinity, of the $n$-th partial sum. With this caveat, yes, the formal derivative always matches the function derivative.


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