My textbook gives the formula for compound interest as:

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

Where: P = The principal, r=the annual rate of interest, n= the frequency of compounding, t=Time in years and A is the total interest accrued over time.

It then goes onto show how if we compound £1 continuously at a rate of 100% for 1 year, for greater and greater values of $n$ we get:

$\left( 1+\dfrac {1}{n}\right) ^{n}\rightarrow e$

And then uses this to derive the formula for continuously compounded interest:

$A\left( t\right) =Pe^{rt}$

The book says it uses "a little calculus and the definition of e" to derive this, but how exactly does it do this?


Let's see. The limit claim is pretty widely discussed on MSE, so I'm assuming you're willing to believe that the limit does approach $e$. Once you have that, you can look at the formula $$ A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt} $$ and do a little fiddling. Let $m = \dfrac{n}{r}$, so that $n = rm$. Then rewrite: $$ A_n(t) = P \left(1+ \frac{r}{n}\right)^{nt} = P \left(1 + \frac{1}{m}\right)^{mrt}= P \left(\left(1 + \frac{1}{m}\right)^m \right)^{rt} $$

Now as $n \to \infty$, the thing inside the large parentheses approaches $e$, so you get

$$ A(t) = Pe^{rt}. $$

As for the main limit, the usual approach is to say you want to find $$ L = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n $$ but instead, you compute \begin{align} \ln L &= \lim_{n \to \infty} \ln \left(1 + \frac{1}{n} \right)^n \\ &= \lim_{n \to \infty} n \ln \left(1 + \frac{1}{n} \right) \\ &= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n}, \end{align} which you can evaluate with L'hopital's rule (take derivative of top and bottom, since both go towards 0): \begin{align} \ln L &= \lim_{n \to \infty} \frac{\ln \left(1 + \frac{1}{n} \right)}{1/n} \\ &= \lim_{n \to \infty} \frac{\frac{1}{1 + \frac{1}{n}}\left(\frac{-1}{n^2}\right)} {\frac{-1}{n^2}}\\ &= \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}}\\ &= 1. \end{align} Since the natural log of your limit is $1$, the limit itself must be $e$.$$$$$$

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    $\begingroup$ What is the purpose of the m=n/r substitution? $\endgroup$ – seeker Dec 21 '13 at 15:52
  • $\begingroup$ I did that so that I'd get a limit that looked like the one that the authors had given $(1 + \frac{1}{n})^n$. In my second equation, you can see how the thing inside the large parens is of this form, and therefore we can use the authors' statement to jump right to the limit. If I'd known at the outset that I was going to expand on how to find the limit in the second part, I might have skipped that. $\endgroup$ – John Hughes Dec 21 '13 at 16:15

There is an easier way to derive (without Calculus), using $(1 + \frac 1 m)^m = {\rm e }$.

If $A = P (1 + \frac r n)^{nt}$, then $A = P \left( 1 + \frac 1 {\frac n r} \right)^{nt}$ (we basically just took the reciprocal of $\frac r n$ and put a $1$ on top again to make it equivalent.)

Now set $\frac n r = m$ and substitute: $A = P (1 + \frac 1 m)^{nt}$. It seems pretty close now, doesn't it!

Let's reform the $\frac n r = m$ statement, shall we? We can change that into $n = mr$; substitute that back into the $nt$ exponent part for $n$, and you will get $A = P (1 + \frac 1 m)^{mrt}$ (hey, that looks kind of familiar, doesn't it!).

Substitute back in from the beginning formula $(1 + \frac 1 m)^m = {\rm e}$ and you get $A = P({\rm e})^{rt}$.

  • $\begingroup$ Well, $(1 + 1/m)^m \ne e$ - that statement only holds in the limit as $m \to \infty$. And this uses calculus. (Also, is there a reason your entire post is in bold font?) $\endgroup$ – user296602 Jan 18 '16 at 16:47

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