# Different definitions of e (Part 5): Continuous compound interest

In this series of posts, I consider how two different definitions of the number $e$ are related to each other. The number $e$ is usually introduced at two different places in the mathematics curriculum:

1. Algebra II/Precalculus: If $P$ dollars are invested at interest rate $r$ for $t$ years with continuous compound interest, then the amount of money after $t$ years is $A = Pe^{rt}$.
2. Calculus: The number $e$ is defined to be the number so that the area under the curve $y = 1/x$ from $x = 1$ to $x = e$ is equal to $1$, so that

$\displaystyle \int_1^e \frac{dx}{x} = 1$.

These two definitions appear to be very, very different. One deals with making money. The other deals with the area under a hyperbola. Amazingly, these two definitions are related to each other. In this series of posts, I’ll discuss the connection between the two.

I should say at the outset that the second definition is usually considered the true definition of $e$. However, compound interest usually appears earlier in the mathematics curriculum than definite integrals, and so an informal definition of $e$ is given at that stage of the curriculum.

At this point in the exposition, I have justified the formula $A = \displaystyle P \left(1 + \frac{r}{n} \right)^{nt}$ for computing the value of an investment when interest is compounded $n$ times a year. We have also made the informal definition

$\displaystyle \left(1 + \frac{1}{n} \right) \to e \qquad$ as $\qquad n \to \infty$.

We are now in position to give an informal derivation of the continuous compound interest formula. Though this derivation is informal, I have found it to be very convincing for my Precalculus students (as well as to my class of future high school teachers).

The basic idea is to rewrite the discrete compound interest formula so that it contains a term like $\displaystyle \frac{1}{\hbox{something}}$ instead of $\displaystyle \frac{r}{\hbox{something}}$. In this way, we can think like an MIT freshman and reduce to previous work.

To this end, let $n = mr$. Then the discrete compound interest formula becomes

$A = P\displaystyle \left( 1 + \frac{r}{n} \right)^{nt}$

$A = P\displaystyle \left( 1 + \frac{r}{rm} \right)^{rmt}$

$A = P\displaystyle \left( 1 + \frac{1}{m} \right)^{mrt}$

$A = P\displaystyle \left[ \left( 1 + \frac{1}{m} \right)^m \right]^{rt}$

Inside of the brackets is our familiar friend $\displaystyle \left( 1 + \frac{1}{m} \right)^m$, except that the name of the variable has changed from $n$ to $m$. But that’s no big deal: as $n$ tends to infinity, then $m$ does as well since $m = n/r$ and both $n$ and $r$ are positive. Therefore, as interest is compounded more frequently, we have replace the thing inside the brackets with the number $e$. This leads us to the formula for continuous compound interest:

$A =P e^{rt}$

Again, my experience is that college students have no conceptual understanding of this formula or even a memory of seeing it derived once upon a time. They remember is it as coming out of nowhere, as a number in a formula or as a button on a calculator. It really shouldn’t be this way. The above calculation is perhaps a harder sell to high school students that the other calculations that I’ve posted in this series, but I firmly believe that this explanation is within the grasp of good students at the time that they take Algebra II and Precalculus.
Of course, the above derivation is highly informal. For starters, it rests upon the limit

$\displaystyle \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n = e$,

which cannot be formally proven using only the tools of Algebra II and Precalculus. Second, the above computation rests upon the continuity of the function $A(x) = P x^{rt}$, so that we can simply replace $\displaystyle \left( 1 + \frac{1}{m} \right)^m$ with its limit $e$. My experience is that students are completely comfortable making this substitution, even though professional mathematicians realize that interchanging limits requires continuity.

So, mathematically speaking, the above argument should not be considered a proper derivation of the continuous compound interest formula. Still, I have found that the above argument to be quite convincing to Algebra II and Precalculus students, appropriate to their current level of mathematical development.

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