Non-Rigorous Arguments 1: Two Formulas For e

I’m a big fan of non-rigorous arguments, especially in calculus and analysis. I think there should be a book cataloging all the beautiful, morally-true-but-not-actually-true proofs that mathematicians have advanced, but until that time I’ll try to at least catalog a few of them on my blog.

This first one is Euler’s original argument for the equality of two expressions (both of which happen to define e):

\displaystyle{\sum_{n=0}^\infty \frac{1}{n!} = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^n}

I’ll also sketch how this can be made rigorous in non-standard analysis.

The argument is as follows: The limit \lim_{n\to\infty} (1 + 1/n)^n is equal to (1 + 1/N)^N, where N is infinitely large. By the binomial theorem, this is:

\displaystyle{1 + N\frac{1}{N} + {N\choose 2}\frac{1}{N^2} + \cdots + \frac{1}{N^N} = \sum_{i=0}^N {N\choose i}\frac{1}{N^i}}

Since N\choose i is N(N-1)\cdots (N- (i - 1))/ i!, this is the sum as i ranges from {0} to N of:

\displaystyle{\frac{N(N-1)\cdots (N-(i-1))}{N^i}\cdot\frac{1}{i!}}

Now, if i is infinitely large, this term is so small that it may be neglected. On the other hand, if i is finite, then N - j = N for j< i. Therefore

\displaystyle{\frac{N(N-1)\cdots (N-(i-1))}{N^i}=1}

and the whole sum is equal to

\displaystyle {\sum_{i=0}^N \frac{1}{i!} = \sum_{i=0}^\infty \frac{1}{i!}},

as desired.

Now, I’ll sketch how to make this rigorous in non-standard analysis. This is from Higher Trigonometry, Hyperreal Numbers, and Euler’s Analysis of Infinities by Mark McKinzie and Curtis Tuckey, which is the best introductory article on non-standard analysis that I’ve read.

In non-standard analysis, one extends the real numbers \mathbb{R} to a larger field \mathbb{R}^* which contains all the reals, but also a positive \epsilon which is less than every positive real (and hence also a number 1/\epsilon which is greater than every real). For every function f\colon \mathbb{R}^n \to \mathbb{R}, there is a function f^*\colon (\mathbb{R}^*)^n \to \mathbb{R}^*, and the f^*‘s satisfy all the same identities and inequalities formed out of composition that the f‘s do. (For example, \sin^*(2x) = 2\sin^*(x)\cos^*(x) for all hyperreal x.) For that reason, I’ll often omit the {}^*. The range of (x \mapsto \lfloor x \rfloor)^* is called \mathbb{Z}^*, the hyperintegers. Since x - 1\leq \lfloor x \rfloor \leq x, the same is true in the hyperreals and there are therefore infinite hyperintegers.

We call a nonzero hyperreal x\in \mathbb{R}^* infinitesimal if |x| is less than every positive real. We say that x and y are close (written x\simeq y) if x - y is infinitesimal. We say that x is infinite if 1/x is infinitesimal (equivalently, if |x| is greater than every real). We say that x is finite if it’s not infinite (equivalently, if |x| is less than some real).

Let \{s_n\}_{n=0}^\infty be a sequence of hyperreals. We say that it is determinate if \sum_{n=0}^N s_n \simeq \sum_{n=0}^M s_n whenever N and M are infinite

The Summation Theorem can then be proven: If \{s_n\}_{n=0}^\infty and \{t_n\}_{n=0}^\infty are two determinate sequences such that s_n \simeq t_n for all finite n, then \sum_{n=0}^N s_n \simeq \sum_{n=0}^N t_n for all infinite N.

By appropriately replacing “equals” with “is close to”, Euler’s argument above may now be adapted to prove that for all infinite N and M,

\displaystyle{\left(1 + \frac{1}{N}\right)^N \simeq \sum_{n=0}^M \frac{1}{n!}}

(the sequence s_n = 1/n! may be proved determinate by comparison with the geometric sequence, which is easily shown determinate). By a transfer principle, this may in turn be used to prove that \lim_{n\to \infty} (1 + 1/n)^n = \sum_{n=0}^\infty 1/n! (in the regular reals).

2 thoughts on “Non-Rigorous Arguments 1: Two Formulas For e

  1. Try ( 1+ 1/100,000,000 ) ^ 100, 000, 000 on your calculator.

    This is accurate to 7 decimal places. Nothing special. You could use many other series representations for e. Accuracy of 7 digits is more than sufficient for all calculations on planet earth.

    The magnitude e is always used as a rational number even though it is not a number at all.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s