Talk:Exponential Function is Continuous
| \(\ds \exp(c)\) | \(=\) | \(\ds \sum_{n = 0}^\infty \frac {c^n} {n!}\) | (by definition) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{m \to \infty} \sum_{n = 0}^m \frac {c^n} {n!}\) | (by definition) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{m \to \infty} \sum_{n = 0}^m \lim_{x \to c} \frac {x^n} {n!}\) | (property 1 of limits) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{m \to \infty} \lim_{x \to c} \sum_{n = 0}^m \frac {x^n} {n!}\) | (property 2 of limits) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \to c} \lim_{m \to \infty} \sum_{n = 0}^m \frac {x^n} {n!}\) | (since $\ds \lim_{m \to \infty} \sum_{n = 0}^m \frac {x^n} {n!}$ converges absolutely?) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \to c} \sum_{n = 0}^{\infty} \frac {x^n} {n!}\) | (by definition) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \to c} \ \exp(x)\) | (by definition) |
- KBlott
What is this? If it's a proof, it should be on the page proper. In any event I'm unfamiliar with $\ds \sum_{n=0}^\infty \frac {c^n}{n!}$, 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 $\ds \lim_{m \to \infty}\lim_{x \to c} \sum_{n = 0}^m \frac {x^n} {n!}= \lim_{x \to c} \lim_{m \to \infty} \sum_{n = 0}^m \frac {x^n} {n!}$. The proof that $\ds \sum_{n = 0}^m \frac {c^n} {n!} =\lim_{x \to c} \sum_{n = 0}^m \frac {x^n} {n!}$, on the other hand, is quite well known, though rather longish. --KBlott 21:59, 21 January 2012 (EST)
If one uses the Power Series definition of exp then the result follows because power series are continuous in their radius of convergence (see Power Series is Differentiable on Interval of Convergence). --DominikPeters 07:58, 23 March 2012 (EDT)
- Cool, sounds good. Wanna put up a proof for us, then? Don't worry that you don't know how to make it look good, someone else will clean it up and you can learn from the changes. --GFauxPas 09:01, 23 March 2012 (EDT)
Refactoring
I think it's better to have the real and complex cases on equal footing. Thus there needs to be a subpage for the real exponential like there is for the complex - the two proofs go on there. --Lord_Farin (talk) 08:19, 1 February 2013 (UTC)