Whence e?

An interesting thing happens if we examine the formula for compound exponential growth in the limit as n approaches infinity—that is, we divide the growth rate into infinitely small chunks and compound it an infinite number of times. Recall that the compound growth function is

For simplicity, we'll first let r = 1 and t = 1. We'll relax those constraints later. So we're interested in finding this limit:

If you aren't familiar with limit statements, the question it's asking is, "what does (1 + 1/n)n equal when n gets infinitely large?"

It turns out that there are two ways to look at this limit, both of which are incorrect. Nevertheless, it's helpful to think them through.

  • First, take the limit of 1/n as n → ∞. That's zero because as n gets very large (in the positive direction), 1/n gets very small (but is never negative). So inside the parenthesis, we have 1 + 0 = 1, implying that the limit is 1. That's incorrect.
  • Then again, if 1/n never quite reaches zero, then the number in parenthesis is always a little bit bigger than one. And if we take a number a little larger than one to an infinitely large power, we get infinity. Also incorrect.

The trick with such limits is understanding that the n in the denominator of 1/n and the exponent n are both approaching infinity together – they compete in a way, and there is a possibility of a finite limit. In this case, there is; it's the very important number, e.


Finding e (one of many ways)

Below is the rather lengthy process of finding e from the limit. I like to go through this with my advanced classes because it involves a wide variety of mathematical thinking, but doesn't require any tricks from calculus. Here we go:

First, we note that (1 + 1/n)n is a binomial that is being raised to the powers 1, 2, 3, ... So we'll treat the limit like a binomial expansion. Here are a few powers of the binomial (a + b):

(a + b)0 = 1
(a + b)1 = a + b
(a + b)2 = a2 + 2 ab + b2
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(a + b)4 = a4 + 4 a3b + 6 a2b2 + 4 ab3 + b4 … and so on.

Each term of each power of (a + b)

consists of a constant multiplied by a product of powers of the coefficients a and b.

The a's and b's

Note that the powers of a begin with the exponent of the binomial, n, and are reduced by one as we go across, from left-to-right, to zero at the end. Likewise, the powers of b begin at zero on the left and count up to n, in a regular, predictable pattern. We can write that pattern like this:

In this summation, the variable k is introduced to count from 0 up to n. Try writing a few terms of this sum and you'll see that it captures our binomial expansion, except for the numerical coefficients (that's why I wrote ~ instead of = ).

The coefficients

Convince yourself that the above represents most of the expansions of the powers of (a + b). What it leaves out it the coefficients. For example, the coefficients (left to right) of the terms of the expansion (that means we multiply it completely through) of (a + b)4 are: 1, 4, 6, 4, 1. For our shorthand formula, we need a way to get the coefficients correct, too. Here's how.

If we write down the coefficients of the terms containing a and b in our binomial power expansions, they form Pascal's Triangle:

Pascal's triangle is formed by writing a 1, then 1   1 on the next line, then each

successive line starts and ends with 1, with the numbers in the middle being the sum of the two numbers diagonally above. Here's a diagram:

The figure below shows the correspondence between the numbers in the triangle and the coefficients of several binomial expansions.

 

 

It's like magic, isn't it? but the trouble with Pascal's triangle is that it's necessary to know its (n-1)th row to get the nth row, and so on. We'd like to be able to calculate the coefficients of the nth row directly, for very large n —without writing the whole triangle.

To do this, take a look at (a + b)4, a term big enough to see some patterns:

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

= 1·aaaa + 4·aaab + 6·aabb + 4·abbb + 1·bbbb

Now take a look at this table. The number of ways of rearranging those a's and b's (the number of permutations of each group) is exactly the numerical coefficient that precedes it.

Permutations of four objects of two types
Term Coefficient Number of permutations
aaaa 1 aaaa
aaab 4 aaab, aaba, abaa, baaa
aabb 6 aabb, abba, bbaa, abab, baba, baab
abbb 4 abbb, babb, bbab, bbba
bbbb 1 bbbb

Those coefficients are just the binomial coefficients, the number of ways of grouping k objects from a set of n, when order doesn't matter ( → )

In our final equation for the binomial expansions, k will count from 0 to n (the power of the binomial) as we move from left to right across the terms.

n-choose-k

Now that we have our coefficients, we can write a general expression for the nth order binomial expansion (a + b)n . You should write a few terms for yourself and confirm that it works. Let's pause a moment here and remind ourselves, in this long-ish process, that we're trying to put our exponential expression (1+1/n)n into a form where we can properly evaluate the limit as n→∞.

Now just a short detour: It will be convenient to express factorial expressions in a slightly different way. Because 0 ≤ k ≤ n, and n! = 1 · 2 · 3 · . . . · n, we have

derivation equation 1

We can then write the factorial part of the binomial expansion (see blue box above) as

derivation equation 2.

This gives us an alternate expression for the binomial coefficients:

derivation equation 3

Now we're in a position to write down an expression for (a + b)n for any n. Here it is:

derivation equation 4

Now replace the binomial coefficients with our factorial expressions:

derivation equation 5

Plug in   1   for   a   and   1/n   for   b:

derivation equation 6

... and simplify:

derivation equation 7

Now we're almost ready to evaluate this series as n approaches infinity. A little bit of creative factoring gives:

derivation equation 8

Now it's easy to see that as n→∞, each term containing n tends toward zero, leaving:

derivation equation 9,

which we can write as a sum:

derivation equation 10

Remember that 0! = 1. Notice that the terms of this sum get smaller and smaller. That is, the sum converges. Values of the sum are shown in the table, and you can see that it converges to the number 2.718 ..., which we call e, the base of all continuously-growing exponential functions.

convergence of e table

We began this long process by asking about the limit of the function f(t) = Ao(1 + 1/n)nt as n becomes infinitely large. We found that f(t) = (1 + 1/n)n approaches the transcendental number e as n gets large. If we had left the Ao, r and t in as we went through the same procedure, we'd have arrived at the expression for continuous growth below.

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