# MandysNotes

Wednesday, 11 May 2011 23:16

## The Exponential Function

By  Gideon
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(Note: this is a quick and dirty introduction to exponentials. I will not bother with any epsilons or deltas, but for those of you who are concerned about such things, be assured that all convergence is uniform, all sequences are Cauchy, etc. The series that defines the exponential is about as well behaved as an infinite series can be.)

The function $e^{x}$is defined to be the limit as $$n$$ approaches infinity of $( 1 + \frac{x}{n})^{n}:$

exp(x) = $e^{x} = \lim_{n \to \infty} (1 + \frac{x}{n})^{n} .$

Using the binomial theorem:

$(x + h)^{n} = \sum_{i=0}^{n}\binom{n}{i}x^{n-i}h^{i} ,$

we have:

$e^{x} = \lim_{n \to \infty} \sum_{k=0}^{\infty} \binom{n}{k} \frac{x}{n}^{k}.$

In order to make sense of this, we have to find what

$\binom{n}{k} \left(\frac{1}{n}\right)^{k} = \frac{n!}{(n-k)!k!n^{k}}$

becomes when $$n$$ tends to infinity.

$% \lim_{n \to \infty} \frac{n!}{(n-k)!n^{k}k!} = \frac{1}{k!}$

Note that:

$\frac{n!}{(n-k)!n^{k}} = \frac{n(n-1)...(n – k + 1)}{n(n)...(n)}.$

This has k terms on top and k terms on the bottom.

Therefore:

$\lim_{n \to \infty} \frac{n!}{(n-k)!n^{k}} = \lim_{n \to \infty} \frac{n(n-1)...(n – k + 1)}{n(n)...(n)} = 1.$

So that:

$\lim_{n \to \infty} \frac{n!}{(n-k)!n^{k}k!} = \frac{1}{k!}.$

Putting this all together, we find:

$e^{x} = \lim_{n \to \infty} (1 + \frac{x}{n})^{n}$

$= \sum_{k=0}^{\infty} \frac{x^{k}}{k!}.$