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Proof of Formula for Continuous Compounding

*Copyright 1995 by Campbell R. Harvey and Stephen Gray.
All rights reserved. No part of this lecture may be reproduced without
the permission of the author.*

Latest Revision: August 1996.

When there are *n* compounding periods per year, we saw that the
*effective annual interest rate* is equal to *(1+R/n)*^{n}
- 1

We wish to show that if interest compounds continuously, then the *effective
annual interest rate* is equal to *e*^{R} - 1

We can prove this, if we can show that as there are more and more compounding
periods per year, then the *effective annual interest rate* moves
closer and closer to *e*^{R} - 1

In the language of Calculus, this is the same as showing that:

This is equivalent to:

In the special case where *R* is zero, both sides of this equation
are equal, so it is true. We will examine what happens when *R* is
nonzero.

If we take the natural logarithm of the left hand side, we get:

You may recall, from Calculus, that the definition of a derivative is:

The expression on the right is called a *difference quotient*.
The trick in our proof is to make our equation look like a difference quotient.
In order to do this, we need two more tricks.

The first is to multiply and divide the right hand side of the equation
by *R*. With some basic algebra, the equation becomes:

The second trick is to realize that *ln*(1) is equal to zero, so
that if we subtract it from the top of the fraction, it will not change
the value.

Our equation now looks like:

Note that if we let *h = (R/n)* and *x = 1*
then the bracketed expression on the right is just a difference quotient.
Our equation becomes:

As *n* gets large, then *h = R/n* will go to zero.

This last equality holds, because *(1/x)* is the derivative
of *ln(x)* and we set *x* = 1.

We are almost done. We have shown that as n gets large, the natural
logarithm of *(1+R/n)*^{n} goes to *R*. Since *e*^{ln(x)}
= x, we can conclude that: