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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:

$Pe^{rt}=F$

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?

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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. –  Américo Tavares Jul 17 '11 at 18:59
    
@AmericoTavares Can you provide an example? –  anonymous Jul 17 '11 at 19:19
    
Do you know what would be awesome. For banks to give your continuos compound interest on your savings. –  PyRulez Feb 16 '13 at 17:30

1 Answer 1

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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 $$ P\left(1 + \frac{r}{2} \right)^{\frac{t}{2}} = F. $$ Following this pattern, if you compounded $n$ times every year, the equation is $$ P\left(1 + \frac{r}{n} \right)^{\frac{t}{n}} = F. $$ 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 $$ \left(1 + \frac{r}{n} \right)^{\frac{1}{n}} $$ 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.

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