Different definitions of e (Part 6): 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.

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In yesterday’s post, I presented an informal derivation of the continuous compound interest formula A = Pe^{rt} from the discrete compound interest formula A = \displaystyle P \left( 1 + \frac{r}{n} \right)^{nt}. In today’s post, I’d like to give the more formal derivation using calculus.

What does it mean for something to compound continuously? In a nutshell, the rate at which the money increases should be proportional to the amount currently present. In other words, \$10,000 should earn ten times as much interest as \$1,000. Since A'(t) is the rate at which the money increases and A(t) is the current amount, that means

A'(t) = r A(t)

for some constant of proportionality r. This is a differential equation which can be solved using standard techniques. We divide both sides by A(t) and then integrate:

\displaystyle \frac{A'(t)}{A(t)} = r

\displaystyle \int \frac{A'(t)}{A(t)} dt = \int r dt

\ln |A(t)| = r t + C

|A(t)| = e^{rt+C} = e^rt e^C = C_1 e^{rt}

A(t) = \pm C_1 e^{rt}

A(t) = C_2 e^{rt}

(Technically, a better solution would use an integrating factor [see also MathWorld], but I find that the above derivation is much more convincing to students who are a few semesters removed from a formal course in differential equations.) When presenting this in class, I’ll sometimes lazily write C in place of C_1, with the understanding that e to an arbitrary constant is just an arbitrary positive constant. Also, on the last line, plus or minus an arbitrary constant is just an arbitrary constant (which I’ll usually write as C instead of C_2).

To solve for the missing constant C_2, we use the initial condition A(0) = P:

A(0) = C_2 e^{r\cdot 0}

P = C_2 \cdot 1

P = C_2

Replacing C_2 by P, we have arrived at the continuous compound interest formula A(t) = Pe^{rt}.

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2 Comments

  1. 2048 and algebra: Index | Mean Green Math
  2. Different definitions of e: Index | Mean Green Math

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