# Talk:Exponential Dominates Polynomial

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

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)

- 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.