Among all exponential functions b^t, the one with base e has many special properties.

For any exponential function b^t, the slope of the tangent line is proportional to b^t. The constant of proportionality here is 1 only when the base is the special number e. This can be taken as the defining property of e.

As we shall see, this number e is not a whole number. In fact it's not even a fraction (such as 22/7 or 87/32). If you try to write it as a decimal, you find that the decimal expansion never ends and never repeats; instead it's an example of an irrational number. Its value is approximately:

How to Compute e

Computing the precise value of the number e is a surprisingly mysterious task. We will describe two different ways to do this, even though the full explanation of why the methods work is unfortunately a bit beyond our reach. (For that, you need to know some calculus).

The first method is as follows:

If Java worked here, you'd see something like this. Use the sliding bar to change n in the formula (1+1/n)ⁿ.

Notice how, as n increases, the answers seem to be getting closer and closer to a fixed number. This limiting value, is the magic number e. In mathematical language, this is expressed by saying:

The second method is as follows:

step 1: compute 1+1

step 2: compute 1+1+1/2

step 3: compute 1+1+1/2+1/(2x3)

step 4: compute 1+1+1/2+1/(2x3)+1/(2x3x4)

step 5: compute 1+1+1/2+1/(2x3)+1/(2x3x4)+1/(2x3x4x5)

As you can see, it starts to get a bit cumbersome to write each new term that gets added, even though the pattern should be clear. We can make it a bit easier by introducing a useful piece of notation. Instead of 2x3, we write 3!; instead of 2x3x4 we write 4!, etc. In this notation, n! (pronouced `n factorial') stands for the product of all the integers from 1 up through n. We can thus write step 6 as:

step 6: compute 1+1+1/2!+1/3!+1/4!+1/5!+1/6!

You can probably now guess the pattern:

step n: compute 1+1+1/2!+1/3!+...+1/(n-1)!+1/n!

(The dots mean `continue in this way until you get to'.)

If Java worked here, you'd see something like this. We can write this sum (from step n) more compactly using the greek letter Sigma, as in the applet. Using the slider to change n, and hence the number of terms in the sum.

Mysteriously, as you add more and more terms to this series, you find that the answer gets closer and closer to -- you guessed it -- e. We thus get our second recipe for e:

Although our two methods both give e as a limiting value, the second one converges much more rapidly, as you can see with the applets.

The Function e^t

In fact, both these limiting methods can be used more generally to compute the exponential function e^t. That is, it turns out that the value of e^t can be computed as the limit (as n tends to infinity) of:

or as the limit (as n tends to infinity) of:

Either one of these formulas tells us how to compute exponential functions even when t is not a whole number. Again, the second method, with the summed factorials, converges much faster. For this reason it is the one used in practice by calculators and computers.