Skip to main content
Search IntMath
Close

450+ Math Lessons written by Math Professors and Teachers

5 Million+ Students Helped Each Year

1200+ Articles Written by Math Educators and Enthusiasts

Simplifying and Teaching Math for Over 23 Years

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.

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`

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..

Problem Solver


This tool combines the power of mathematical computation engine that excels at solving mathematical formulas with the power of GPT large language models to parse and generate natural language. This creates math problem solver thats more accurate than ChatGPT, more flexible than a calculator, and faster answers than a human tutor.

Tips, tricks, lessons, and tutoring to help reduce test anxiety and move to the top of the class.