Different definitions of e (Part 10): Connecting the two definitions

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.

logareagreen line

In yesterday’s post, I proved the following theorem, thus completing a long train of argument that began with the second definition of e and ending with a critical step in the derivation of the continuous compound interest formula. Today, I present an alternate proof of the theorem using L’Hopital’s rule.

Theorem. \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} = P e^{rt}.

Proof #2. Let’s write the left-hand side as

L = \displaystyle \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt}.

Let’s take the natural logarithm of both sides:

\ln L = \displaystyle \ln \left[ \lim_{n \to \infty} P \left( 1 + \frac{r}{n} \right)^{nt} \right]

Since g(x) = \ln x is continuous, we can interchange the function and the limit on the right-hand side:

\ln L = \displaystyle \lim_{n \to \infty} \ln \left[ P \left( 1 + \frac{r}{n} \right)^{nt} \right]

\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + \ln \left( 1 + \frac{r}{n} \right)^{nt} \right]

\ln L = \displaystyle \lim_{n \to \infty} \left[ \ln P + nt \ln \left( 1 + \frac{r}{n} \right)\right]

\ln L = \ln P + \displaystyle \lim_{n \to \infty} \frac{t \displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}

\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \ln \left(1 + \frac{r}{n} \right)}{1/n}

The limit on the right-hand side follows the indeterminate form 0/0, as so we may apply L’Hopital’s Rule. Taking the derivative of both the numerator and denominator with respect to n, we find

\ln L = \ln P + \displaystyle t \lim_{n \to \infty} \frac{\displaystyle \frac{1}{1 + \frac{r}{n}} \cdot \frac{-r}{n^2}}{\displaystyle \frac{-1}{n^2}}

\ln L = \ln P + \displaystyle t \lim_{n \to \infty}\frac{r}{1 + \frac{r}{n}}

\ln L = \ln P + \displaystyle t \frac{r}{1 + 0}

\ln L = rt + \ln P

We now solve for the original limit L:

L = e^{rt + \ln P}

L = e^{rt} e^{\ln P}

L = Pe^{rt}

 

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2 Comments

  1. Different definitions of e: Index | Mean Green Math
  2. Inverse Functions: Index | Mean Green Math

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