Talk:Exponential Function is Continuous

- KBlott

What is this? If it's a proof, it should be on the page proper. In any event I'm unfamiliar with $\displaystyle \sum_{n=0}^\infty \frac {c^n}{n!}$, if that's another approach to $\exp$ can you put it up here? --GFauxPas 00:59, 19 January 2012 (EST)
 * In any event, the "properties of limits" you use only work if the function is continuous, am I wrong? So aren't you begging the question? --GFauxPas 01:31, 19 January 2012 (EST)
 * Property 1 above $does$ follow from the fact that $f_n(x) = \frac {x^n} {n!}$ is continuous. If you remind me again later (on my talk page) I can dig up the proof.  Property 2 is easily proven by induction.  The next step after that one is a bit out of my league.  I suppose one could also prove it by induction but I’m not certain. --KBlott 05:15, 19 January 2012 (EST)

Let me quote (rather, paraphrase, as I have to translate in the process) from my lecture notes on sequences, series and convergence:

'Let $V \subseteq \C$, and let for $k \in \N, k\ge l$, $f_k:V\to \C$ be given. Suppose there exists a sequence $M_k\ge0$, such that:
 * $x \in V \implies |f_k(x)|\le M_k$
 * The sequence $S_l:= \sum_{k=p}^l M_k$ is bounded.

Then there is a unique $s:V\to \C$ such that $\lim_{l\to\infty}\sum_{k=p}^lf_k(x) = s(x)$, where the convergence is uniform on $V$. If all the $f_k$ are continuous, then so is $s$.'

This appears to be applicable. --Lord_Farin 08:30, 19 January 2012 (EST)
 * I’m afraid that I couldn’t give a satisfactory proof that $\displaystyle\lim_{m \to \infty}\lim_{x \to c} \sum_{n = 0}^m  \frac {x^n} {n!}= \lim_{x \to c} \lim_{m \to \infty} \sum_{n = 0}^m  \frac {x^n} {n!}$.  The proof that $\displaystyle\lim_{m \to \infty} \sum_{n = 0}^m \frac {c^n} {n!} = \lim_{m \to \infty} \lim_{x \to c} \sum_{n = 0}^m  \frac {x^n} {n!}$, on the other hand, is quite well known, though rather longish. --KBlott 21:59, 21 January 2012 (EST)