** e**, the base of all continuously-growing exponential functions. You will need to have a little background in probability to get the most out of this section. You can skip it and still be able to work continuous-growth problems ... but it is very interesting.

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

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*.

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

$$ \begin{align} (a + b)^0 &= 1\\[3pt] (a + b)^1 &= a + b\\[3pt] (a + b)^2 &= a^2 + 2 ab + b^2 \\[3pt] (a + b)^3 &= a^3 + 3 a^2b + 3 ab^2 + b^3 \\[3pt] (a + b)^4 &= a^4 + 4 a^3b + 6 a^2b^2 + 4 ab^3 + b^4 \: \dots \end{align}$$

... 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**.

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:

$$(a + b)^n ~ \sum_{k = 0}^n \; a^{n - k}b^k$$

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 = ).

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:

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.

$$ \begin{align} &\;1 \\[4pt] 1 &\phantom{00} 1 \\[4pt] 1 \phantom{00} &2 \phantom{00} 1 \\[4pt] 1 \phantom{00} 3 &\phantom{00} 3 \phantom{00} 1 \\[4pt] 1 \phantom{00} 4 \phantom{00} &6 \phantom{00} 4 \phantom{00} 1 \\[4pt] 1 \phantom{00} 5 \phantom{00} 10 &\phantom{00} 10 \phantom{00} 5 \phantom{00} 1 \end{align}$$

and so on ...

$$ \begin{align} &\color{#E90F89}{\longrightarrow} (a + b)^0 = 1 \\[3pt] &\color{#E90F89}{\longrightarrow} (a + b)^1 = a + b\\[3pt] &\color{#E90F89}{\longrightarrow} (a + b)^2 = a^2 + 2ab + b^2\\[3pt] &\color{#E90F89}{\longrightarrow} (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \\[3pt] &\color{#E90F89}{\longrightarrow} (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\\[3pt] &\color{#E90F89}{\longrightarrow} (a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5 \end{align}$$

and so on ...

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 n^{th} row, and so on. We'd like to be able to calculate the coefficients of the n^{th} 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:

$$ \begin{align} &(a + b)^4 = \\[4pt] &a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 = \\[4pt] &\phantom{000} 1\cdot aaaa + 4 \cdot aaab + 6 \cdot aabb \, + \\[4pt] &\phantom{000} 4 \cdot abbb + 1 \cdot bbbb \end{align}$$

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 | 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

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.

$$\binom{n}{k} = \frac{n!}{k! \, (n - k)!}$$

The term on the left, notation for the binomial coefficient, is sometimes read, "n-choose k." On the right is the number of permutations of n objects, taken k at a time if order of those objects doesn't matter.

Now that we have our coefficients, we can write a general expression for the **n ^{th}** order binomial expansion

**Narrow screens**: Many of the equations below are wide, but you can scroll each left-to-right if you're working with a narrow device.

$$(a + b)^n = \sum_{k = 0}^n \frac{n!}{k! \, (n - k)!} a^{n - k} b^k$$

Terms of the binomial expansion

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

$$n! = 1 \cdot 2 \cdot 3 \cdot \; \dots \; (n - k) \cdot (n - k + 1) \cdot \; \dots \; \cdot n \; \;$$

$$\color{#E90F89}{\text{ and }} \; \; (n - k)! = 1 \cdot 2 \cdot 3 \cdot \; \dots \; \cdot (n - k)$$

We can then write the factorial part of the binomial expansion as:

$$\frac{n!}{(n - k)!} = (n - k + 1) \cdot \; \dots \; \cdot n$$

This gives us a new way to express the binomial coefficients:

$$\binom{n}{k} = \frac{1}{k!} \frac{n!}{(n - k)!} = \frac{(n - k + 1) \cdot \; \dots \; \cdot (n - 2)(n - 1) n}{k!}$$

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

$$(a + b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^2 + \binom{n}{3} a^{n - 3} b^3 + \dots + \binom{n}{n} a^0 b^n$$

Now replace the binomial coefficients with our factorial expressions:

$$(a + b)^n = 1 a^n b^0 + n a^{n - 1} b + \frac{n(n - 1)}{2!} a^{n - 2} b^2 + \frac{n(n - 1)(n - 2)}{3!} a^{n - 3} b^3 + \dots + 1 a^0 b^n$$

Now plug in **1** for **a** and **1/n** for **b**:

$$\left(1 + \frac{1}{n}\right)^n = 1^n \left( \frac{1}{n} \right)^0 + n 1^{n - 1} \left( \frac{1}{n} \right)^1 + \frac{n(n - 1)}{2!} 1^{n - 2} \left( \frac{1}{n} \right)^2 + \frac{n(n - 1)(n - 2)}{3!} 1^{n - 3} \left( \frac{1}{n} \right)^3 + \dots$$

... and simplify:

$$\left(1 + \frac{1}{n}\right)^n = \left( \frac{1}{n} \right)^0 + n\left( \frac{1}{n} \right)^1 + \frac{n(n - 1)}{2!}\left( \frac{1}{n} \right)^2 + \frac{n(n - 1)(n - 2)}{3!} \left( \frac{1}{n} \right)^3 + \dots$$

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

$$\left(1 + \frac{1}{n}\right)^n = 1 + 1 + \frac{\left( 1 - \frac{1}{n} \right)}{2!} + \frac{\left( 1 - \frac{1}{n} \right)\left( 1 - \frac{1}{2n} \right)}{3!} + \frac{\left( 1 - \frac{1}{n} \right)\left( 1 - \frac{1}{2n} \right)\left( 1 - \frac{1}{3n} \right)}{4!} + \dots$$

Now it's easy to see that as $n \rightarrow \infty,$ each term containing n tends toward zero, leaving:

$$\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots$$

which we can write as a tidy sum:

$$\lim_{n\to\infty} \left( 1 + \frac{1}{n} \right)^n = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = \sum_{n = 0}^{\infty} \; \frac{1}{n!}$$

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. The magenta numbers in the growing sum show the digits that are "locked in" and won't change further as more terms are added to the sum.

n | n! | 1/n! | sum of terms |
---|---|---|---|

0 | 1 | 1.000000000000000 | 1.0000000000 |

1 | 1 | 1.000000000000000 | |

2 | 2 | 0.500000000000000 | |

3 | 6 | 0.166666666666667 | |

4 | 24 | 0.041666666666667 | |

5 | 120 | 0.008333333333333 | |

6 | 720 | 0.001388888888889 | |

7 | 5040 | 0.00019841268413 | |

8 | 40320 | 0.000024801587302 | |

9 | 362880 | 0.000002755731922 | |

10 | 3628800 | 0.000000025052108 | |

11 | 39916800 | 0.0000000275573192 | |

12 | 479001600 | 0.000000002087676 | |

13 | 6227020800 | 0.000000000160590 | |

14 | 87178291200 | 0.000000000011471 | |

15 | 1307674368000 | 0.000000000000765 |

We began this long process by asking about the limit of the function $f(t) = A_o (1 + 1/n)^{nt}$ as **n** becomes infinitely large. We found that $f(t) = (1 + 1/n)^n$ approaches the transcendental number * e* as

If we had left the ** A _{o}**,

All continuously-growing exponential processes proceed according to

where A_{o} is the initial value of the function, r is the (decimal) growth rate, and *e* = 2.718281 ...

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