Calculating the Value of e

There are several ways to calculate the value of e. Let's look at the historical development.

Using a Binomial Expansion

If n is very large (approaches infinity) the value of `(1+1/n)^n`approaches e.

This is not an efficient way to find `e`. Even if we go out to n = 100,000, our value is only correct to the 4th decimal place.

`e~~[(1+1/n)^n]_(n=100000)` `=2.718268237`

Here's the graph demonstrating this expansion:

5 10 15 20 25 -5 -10 -15 -20 -25 1 2 3 4 5 6 7 n y
e = 2.71828...
`y=(1+1/n)^n`

Graph of `y=(1+1/n)^n`, showing the limit as `n->+-oo` is e.

Another Expansion

As n becomes very small, `(1+n)^(1"/"n)` approaches the value of e.

We can obtain reasonable accuracy with a very small value of n.

`e~~[(1+n)^(1"/"n)]_(n=0.000000001)` `=2.718281827`

Let's see the graph of the situation.

1 2 3 4 5 -1 -2 1 2 3 4 5 6 n y
e = 2.71828...
`y=(1+n)^(1//n)`

Graph of `y=(1+n)^(1"/"n)`, showing the y-intecept (the limit as `x->0`) is e.

(There is actually a "hole" at n = 0. Can you understand why?)

Newton's Series Expansion for e

The series expansion for e is

`e^x=1+x+1/2x^2+1/6x^3+` `1/24x^4+...`

Replacing x with 1, we have:

`e=1+1+1/2(1)^2+` `1/6(1)^3+` `1/24(1)^4+...`

We can write this as:

`e=sum_(n=0)^oo(1/(n!))`

This series converges to give us the answer correct to 9 decimal places using 12 steps:

`e~~sum_(n=0)^12(1/(n!))=2.718281828`

Continues below

Brothers' Formulae

Recently, new formulae have been developed by Brothers (2004) which make the calculation of e very efficient.

`e=sum_(n=0)^oo(2n+2)/((2n+1)!`

We only need 6 steps for 9 decimal place accuracy:

`e=sum_(n=0)^6(2n+2)/((2n+1)!)=` `2.718281828`

Graphical Demonstration of e

The area under the curve `y=1/x` between 1 and e is equal to `1` unit2.

1 2 3 4 5 -1 -2 1 2 3 4 -1 -2 -3 -4 x y
e

Area under the curve `y=1/x` between `1` and `e`.

Reference

Brothers, H.J. 2004. Improving the convergence of Newton's series approximation for e. College Mathematics Journal 35(January):34-39..