# 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)^napproaches 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...
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.

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.

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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