Derivatives At 0

If we write out the Taylor Polynomial, we find that

$f(x)=0+\displaystyle \sum_{n=1}^{\infty}\frac{n x^n}{n!}=\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{(n-1)!}$ .

Using the convergence test, we see that $f$ converges everywhere, so we just need to calculate $f(1)$ as above:

$f(1)=\displaystyle \sum_{n=1}^{\infty}\frac{1}{(n-1)!}$ .

This sum converges (by definition) to $\boxed{e}$ .

To solve the generalization, let $k$ be a defined positive integer. We start with the Taylor Polynomial once again:

$f(x)=\displaystyle \sum_{n=1}^{\infty}\frac{n^k x^n}{n!}$ .

We proceed by induction, and claim that for the given integer $k$ , $f(1)=B_ke$ , where $B_k$ is the $k$ th Bell number. The case of $k=1$ is covered above. Assuming that $f(1)=B_ke$ for $1\leq k\leq j$ for some positive integer $j$ , the Taylor polynomial for $j+1$ then becomes:

$f(1)=\displaystyle \sum_{n=1}^{\infty}\frac{n^{j+1}}{n!}=0+\displaystyle \sum_{n=1}^{\infty}\frac{(n+1)^{j+1}}{(n+1)!}=\displaystyle \sum_{n=1}^{\infty}\frac{(n+1)^j}{n!}=\displaystyle \sum_{n=1}^{\infty}\displaystyle \sum_{i=0}^{j}\frac{{j \choose i}n^i}{n!}$ $=\displaystyle \sum_{i=0}^{j}\displaystyle \sum_{n=1}^{\infty}\frac{{j \choose i}n^i}{n!}=\displaystyle \sum_{i=0}^{j}{j \choose i}\displaystyle \sum_{n=1}^{\infty}\frac{n^i}{n!}=\displaystyle \sum_{i=0}^{j}{j \choose i}B_ie$ .

Here, we note the recurrence formula for Bell numbers, $B_{j+1}=\displaystyle \sum_{i=0}^{j}{j \choose i}B_i$ . With that in mind, we proceed and find that $f(1)=B_{j+1}e$ , completing the proof. Thus $\boxed{f(1)=B_ke.}$