Math Core

Compound interest more and more often (monthly, daily, every second) and the balance grows, but not without limit. It creeps up to a ceiling set by one special number, $e \approx 2.718$ . That number turns out to be the most natural base for exponential functions, which is why scientists and calculators use it everywhere.

Compounding more and more often

Imagine a bank that pays a generous $100\%$ interest per year on $1. The compound interest formula $A = P\left(1 + \tfrac{r}{n}\right)^{nt}$ with $P = 1$ , $r = 1$ and $t = 1$ gives

A = \left(1 + \frac{1}{n}\right)^{n},

where $n$ is the number of times per year the interest is compounded. Here is what happens as $n$ grows:

Compounding	$n$	$\left(1 + \tfrac{1}{n}\right)^n$
yearly	$1$	$2$
twice a year	$2$	$2.25$
quarterly	$4$	$2.441406\ldots$
monthly	$12$	$2.613035\ldots$
daily	$365$	$2.714567\ldots$
a million times	$1{,}000{,}000$	$2.718280\ldots$

More frequent compounding helps, but the gains shrink. Going from daily to a million times a year adds less than half a cent. The values are closing in on a fixed number.

Definition

The number e

The number $e$ is the value that $\left(1 + \dfrac{1}{n}\right)^n$ approaches as $n$ grows without bound:

e \approx 2.718281828\ldots

Like $\pi$ , $e$ is irrational: its decimal never ends or repeats.

The number is named after the mathematician Leonhard Euler, who studied it extensively. On your calculator, $e^x$ is usually the second function of the $\ln$ key.

Continuous compounding

If interest were compounded infinitely often, every instant, the $1 at $100\%$ would grow to exactly $e$ dollars in a year. For a general principal $P$ , rate $r$ and time $t$ , the same limit gives a remarkably clean formula.

Continuously compounded interest

A = Pe^{rt}

$P$ is the principal, $r$ the annual interest rate as a decimal, and $t$ the time in years.

Worked example: Continuous vs. monthly

You invest $5,000 at $4\%$ annual interest for $10$ years. Find the balance if interest is compounded (a) continuously and (b) monthly.

Solution.

(a) $A = 5000e^{0.04 \cdot 10} = 5000e^{0.4} \approx 5000(1.491825) \approx 7459.12$ . The balance is about $7,459.12.

(b) $A = 5000\left(1 + \tfrac{0.04}{12}\right)^{120} \approx 7454.16$ . The balance is about $7,454.16.

Continuous compounding earns only about $5 more. It's the upper limit for compounding at a given rate, not a dramatic bonus.

The natural base in general

Any exponential function can be written with base $e$ :

y = ae^{kx}.

This is called the natural exponential form. If $k > 0$ it's growth; if $k < 0$ it's decay. Scientists write models this way because $k$ has a direct meaning: it's the continuous rate. A population with $k = 0.02$ is growing at $2\%$ per year continuously.

To find the ordinary growth factor per unit, notice that $ae^{kx} = a\left(e^k\right)^x$ . So the base is $b = e^k$ .

y = e^x lies between y = 2^x and y = 3^x, since 2 < e < 3.Open in grapher →

The graph of $y = e^x$ has all the features you already know: domain all real numbers, range $y > 0$ , $y$ -intercept $(0, 1)$ , and horizontal asymptote $y = 0$ . Because $2 < e < 3$ , it sits between $y = 2^x$ and $y = 3^x$ . It also has a special property you'll meet in calculus: at every point, its slope equals its height.

Worked example: A natural decay model

A drug's concentration in the blood, in mg/L, is modeled by $C(t) = 30e^{-0.2t}$ , where $t$ is in hours.

What is the initial concentration?
What is the concentration after $5$ hours?
By what percent does the concentration drop each hour?

Solution.

$C(0) = 30e^0 = 30$ mg/L.
$C(5) = 30e^{-1} \approx 30(0.367879) \approx 11.04$ mg/L.
The hourly factor is $e^{-0.2} \approx 0.8187$ . Since $0.8187 = 1 - 0.1813$ , the concentration drops about $18.13\%$ each hour.

Notice that the continuous rate $0.2$ ( $20\%$ ) is larger than the actual hourly drop ( $18.13\%$ ). The decrease happens a little at a time, always to a slightly smaller amount.

Common mistake

In $y = ae^{kx}$ , the number $e^k$ is the growth factor, not $1 + k$ . A continuous rate of $5\%$ gives an annual factor of $e^{0.05} \approx 1.0513$ , an effective annual rate of about $5.13\%$ , not exactly $5\%$ .

Tip

Type $e^{0.4}$ with the $e^x$ key, not as $2.718^{0.4}$ . Rounding $e$ early introduces error that grows with the exponent.