Talk:Exponential Dominates Polynomial

$\ds \sum_{m \ge 0} \frac{(\alpha n)^m}{m!} > \frac{(\alpha n)^{k+1}}{(k+1)!}$

Why? --GFauxPas (talk) 19:48, 9 July 2013 (UTC)

Because the the latter is a summand of the former when $m=k+1$ --Ybab321 (talk) 00:05, 18 October 2013 (UTC)

Real polynomials

It seems the proof would be identical for the growth of polynomials in $\R$, am I wrong? If so, should I copy - paste - find-and-replace this theorem into Exponential Dominates Real Polynomial ? --GFauxPas (talk) 15:14, 15 November 2016 (EST)

Meh. I'd rather this (appallingly carelessly imprecisely stated) proof was made to apply to all reals than another proof written. Using Archimedean principle, for any $x \in \R$ there is an $n \in \N$ such that $n > x$ and the job is done.

But the answer is always: go to your source works rather than trying to craft stuff out of whole cloth -- at this level of mathematics there is no research left to be done, and a "new" proof or a "new" result will either already be out there somewhere or (worse) just not be any good. --prime mover (talk) 15:47, 15 November 2016 (EST)

Well my source has this as an exercise, so I was looking for someone to do my homework for me, so to speak. --GFauxPas (talk) 16:39, 15 November 2016 (EST)

One doesn't even need the Archimedean principle. Just allow $n$ to be a real number in the proof. In the theorem text, there is no mention of $n$ having to be a natural number, so the theorem text does not need to be changed. The variable name $n$ should be changed to $x$, though.

By the way, my change of $n \in \N$ to $n \in \N_{>0}$ was wrong. I forgot the convention $0^0 = 1$, sorry. --Ivar Sand (talk) 07:10, 16 November 2016 (EST)