The Natural Base e

Compounding faster and faster

You already have the discrete compounding formula. An amount $P$, growing at annual rate $r$ and compounded $n$ times a year for $t$ years, becomes

Strip it down to the simplest possible case so nothing distracts from the one question we care about. Take one dollar of principal, $P = 1$; take an absurdly generous rate of $100\%$, so $r = 1$; and take one year, $t = 1$. Everything collapses:

Only $n$ is left, the number of times a year the interest is added. Each extra compounding cuts the interest into smaller pieces but puts every piece to work sooner, so it should be worth something. The question is how much. Compute the balance for a few values of $n$:

Two things jump out. Every row does beat the row above it, so compounding more often really does pay more. But look at the size of the gain. Going from yearly to twice a year adds $0.25$ dollars, a quarter on the dollar. Going from daily to hourly adds only $0.003559$ dollars, about a third of a cent. Going from hourly to every single second adds $0.000155$ dollars, under two hundredths of a cent, even though the number of compoundings grew by a factor of over three thousand. The column is not marching upward without end. It is pressing against a ceiling somewhere just past $2.718$.

The number the column is pressing against

Two facts decide the story.

The values increase. Every row of the table beats the row above it, and that continues for every $n$, not just the seven values shown.

The values never reach 3. However large you make $n$, the balance stays below $3$ dollars.

A column of numbers that keeps rising and never gets past $3$ cannot run off to infinity, and it cannot jump around. Such a column has to close in on one single value, and it has to approach that value from below. That value is the number we want.

Now be clear about what has and has not been shown, because this is the honest part. The table is evidence, not proof. The first fact is at least believable, since every piece of interest starts earning sooner when you compound more often, so finer compounding cannot pay less. The second fact is not obvious at all, and neither one is proved here, because neither can be proved with the algebra you have. Even the claim before this paragraph is a loan: that a rising, bounded column of numbers must close in on something is a theorem about the real numbers themselves. That theorem is the first serious result of a calculus course. We are borrowing all three, out in the open. What follows is the definition they earn us.

The number that $\left(1 + \frac{1}{n}\right)^{n}$ closes in on, as $n$ grows without end, is called $e$:

Like $\pi$, it is irrational: it is not a ratio of whole numbers, and its decimal expansion never ends and never repeats. So $2.71828$ is a rounding, not the number. (The block $1828$ appearing twice in a row is a coincidence that makes $e$ easy to memorize; the pattern breaks apart immediately afterward at $459045$.) That $e$ is irrational was proved in the 1730s, and that proof, like the convergence itself, waits for calculus.

The punchline of the table is worth stating on its own: compounding more often does not pay unboundedly more. Split the year into a thousand pieces, or a billion, and your dollar still comes back as $2.71828\ldots$ dollars. The infinitely greedy bank and the merely hourly bank owe you almost exactly the same money.

Where e enters the general formula

The table used a $100\%$ rate over one year because it made the algebra vanish. Real rates are not $100\%$, so we have to put $r$ and $t$ back. The surprise is that no new constant appears: the same $e$ does all the work.

The formula $A = Pe^{rt}$ is called continuous compounding, or, away from money, the continuous growth model. There is no $n$ in it at all: the interest is credited so often that the question “how often” has stopped mattering. Two notes on using it. The rate $r$ goes in as a decimal ($6\%$ means $r = 0.06$, never $6$). Also, $r$ and $t$ must use the same unit of time, so a yearly rate demands $t$ in years.

When the quantity shrinks instead of grows, the same formula is written $N = N_{0}e^{kt}$ with $k$ negative. Be precise about what that is. The proof above assumed $r > 0$, since it divides by $r$ to build $m$, so it does not derive the decay case at all. There, $e^{kt}$ with $k < 0$ is adopted as the model of a quantity that continuously loses a fixed fraction of itself, rather than derived from compounding. Every method in the rest of this lesson still applies to it unchanged.

The natural logarithm

Every exponential function has an inverse, and you already know what it is called: a logarithm. Apply that to base $e$ and you get the logarithm this chapter has been building toward.

For $x > 0$, the natural logarithm of $x$ is the logarithm with base $e$:

It is read “ell en of $x$”. Unpacking the definition of a logarithm at this base gives the statement you will use constantly, for $x > 0$:

Both directions hold, and both are just the definition of $\log_e$ read aloud. If $\ln x = y$, then $y$ is the exponent that turns $e$ into $x$, so $e^{y} = x$. And if $e^{y} = x$, then $y$ is that same exponent, so $\ln x = y$. Feeding one statement into the other gives the two undoing rules:

Now the important structural point, and the reason this section is short: $\ln$ is not a new kind of object. It is the same logarithm you met two lessons ago, with the base pinned at $e$. Nothing about the properties depended on which base was used, so every one of them survives the substitution untouched. For positive $x$ and $y$ and any real $k$:

The third of those is the change-of-base rule with $\ln$ as the common base. That rule holds for any valid base $b$, meaning $b > 0$ and $b \neq 1$, so that $\ln b$ is defined and is not zero. It is the reason a calculator needs only two logarithm keys to compute a logarithm in any base at all.

Why “natural”? For now, because it is the logarithm that the growth constant hands you. Any question about continuous growth reaches into the exponent of $e^{rt}$, and $\ln$ is the tool that reaches there. There is a deeper reason the name is deserved, and you will meet it in calculus.

How big is the gap, really?

Continuous compounding sounds like a different world from the discrete kind. Put the two models side by side on the same money and see. Deposit $1000$ dollars for $10$ years at $6\%$:

Daily compounding and continuous compounding differ by nine cents on a thousand dollars across a decade. The continuous figure is the ceiling: no compounding schedule, however fine, reaches it, and none exceeds it. That is exactly the statement the table at the top of this lesson made, now written with real numbers instead of a bare dollar. The reason to use $e^{rt}$ is not that it pays noticeably more. It is that $e^{rt}$ is a cleaner object to work with, since it has no $n$ in it. It is also that most real-world growth (populations, radioactive decay, cooling, charging capacitors) does not happen in scheduled jumps at all.

Common mistakes