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The exponential grows faster than polynomials

April 26, 2012

Earlier this week, I was trying to correctly write up why the gamma function \Gamma(s) converges for s > 0 and I realised that I did not know how to show that the exponential function grows faster than any polynomial function. It’s only just a simple application of the Taylor expansion of the exponential but I’ll write it up since I’m sure a lot of people never saw the proof.

First, what I mean by that is that for any n \in \textbf{N}, there exists an N \in \textbf{N} such that

x \geq N \Rightarrow e^x > \displaystyle\sum_{i=0}^{n} |a_ix^i|

for any a_i \in \textbf{R}, i = 1, \ldots, n. To show this, we will show that

\displaystyle\lim_{x \to \infty} \frac{e^x}{\sum |a_ix^i|} = \infty.

This suffices because then for x large enough this ratio will be greater than 1 and therefore we will have e^x > \sum |a_ix^i|. In fact, we could only show that this limit is greater than 1.

First, note that

\displaystyle\frac{\sum_{i=0}^n a_ix^i}{a_nx^n} = \left( 1 + \displaystyle\frac{a_{n-1}}{a_nx} + \cdots + \frac{a_0}{a_nx^n} \right) \to 1

as x \to \infty. This is often written \sum a_ix^i \sim_{x \to \infty} a_nx^n. Thus, for any \epsilon > 0, when x is large enough we have

\left|\displaystyle\sum_{i=0}^n a_ix^i - a_nx^n\right| < \epsilon.

Therefore, we only need to show that

\displaystyle\lim_{x \to \infty} \frac{e^x}{|a_n|x^n} = \infty.

Since this limit goes to infinity iff \frac{e^x}{x^n} goes, we will drop the |a_n| in the denomiator.

Expanding e^x in its Taylor series, we get

\displaystyle\frac{e^x}{x^n} = \displaystyle\sum_{k=0}^{\infty} \frac{x^k}{x^nk!}

= \displaystyle\frac{1 + x + x^2/2! + x^3/3! + \cdots + x^n/n! + x^{n+1}/(n+1)! + \cdots}{x^n}

= \displaystyle\frac{1}{x^n} + \frac{1}{x^{n-1}} + \frac{1}{x^{n-2}2!} + \cdots + \frac{1}{n!} + \frac{x}{(n+1)!} + \cdots

> \displaystyle\frac{x}{(n+1)!}

which tends to \infty as x \to \infty.

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  1. Convergence of the gamma function « arbourj's blog

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