I've watched the khan academy pre-calculus playlist about compound interest and constant e on youtube Khan Academy. First he said that you can compute the final payment like this:

Let P = Principal, let r = interest rate in decimal, let t = time period, let F = final payment, then the equation would be like this:

$P(1 + r)^t = F$

For example if I borrowed \$50 for 1 year for 15%, then after 20 years I would need to repay \$818.

But then he says that this equations equals to this:


But this is not completely equal to the other equation. Can you explain me what did he mean by this last equation, is this equation even right?

  • 1
    $\begingroup$ The $e$ appears in the case when there are infinite compound periods. It is called continuos compound. Instead of a finite number of periods $n$ you have to evaluate a limit and that limit gives an exponential function. $\endgroup$ Jul 17 '11 at 18:59
  • $\begingroup$ @AmericoTavares Can you provide an example? $\endgroup$
    – anonymous
    Jul 17 '11 at 19:19
  • $\begingroup$ Do you know what would be awesome. For banks to give your continuos compound interest on your savings. $\endgroup$
    – PyRulez
    Feb 16 '13 at 17:30

If you were to compound the interest only once per year, you would have the equation $$ P(1 + r)^t = F. $$ If you compound twice per year, you would accrue half of your annual interest every six months (half a year). This gives a slightly higher final payment and follows the formula $$ P\left(1 + \frac{r}{2} \right)^{2t} = F. $$ Following this pattern, if you compounded $n$ times every year, the equation is $$ P\left(1 + \frac{r}{n} \right)^{nt} = F. $$ It turns out that the bigger the value of $n$ you choose, the higher your final payment. However, the increase in final payment from, say, $n = 1$ to $n = 2$ is much more significant than the increase from $n = 100$ to $n = 101$. This phenomenon allows the final payment to have a limiting value. That is, if you let $n$ "equal" infinity (in other words, let $n$ grow as large as you like), the term $$ \left(1 + \frac{r}{n} \right)^{n} $$ doesn't approach infinity itself, but rather approaches the constant $e^r$. This is called continuous compounding. It is as though you are compounding every moment of every day; an "infinite number" of compoundings each year. Continuous compounding gives the highest possible final payment.

  • 1
    $\begingroup$ corrected the exponents $\endgroup$
    – GEdgar
    May 20 '15 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.