# 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:

*e*= 2.71828...

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.

*e*= 2.71828...

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`

## 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` unit^{2}.

*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..

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