Chapter Contents

• Exponential & Logarithmic Functions
• 1. Definitions: Exponential and Logarithmic Functions
• 2. Graphs of Exponential and Logarithmic Equations
• 3. Logarithm Laws
• 4. Logarithms to Base 10
• 5. Natural Logarithms (base e)
• Dow Jones Industrial Average

• Calculating the value of e

• 6. Exponential and Logarithmic Equations
• World Population Live
• 7. Graphs on Logarithmic and Semilogarithmic Paper
• Exponential & Logarithmic Functions Problem Solver

• Comments, Questions?

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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.

The largest that Scientific Notebook can handle is about n = 100,000 and this is only correct to the 4th decimal place.

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

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`

The graph of `y=(1+n)^(1//n)` is as follows:

(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.