Following is the formula to calculate continuous compounding
A = P e^(RT)
Continuous Compound Interest Formula where, P = principal amount (initial investment)
r = annual interest rate (as a decimal)
t = number of years
A = amount after time tThe above is specific to continuous compounding. The general compounding formula is $$A=P\left(1+\frac{r}{n}\right)^{nt}$$
I want to understand how continuous compounding formula is derived from general compounding formula, given t=1, n=INFINITY.
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$\begingroup$One of the more common definitions of the constant $e$ is that: $$ e = \lim_{m \to \infty} \left(1 + \frac{1}{m}\right)^m $$ Thus we have, with a change of variables $n = mr$, that $$ \lim_{n \to \infty} P\left(1 + \frac{r}{n}\right)^{nt}\\ = \lim_{m \to \infty} P\left(1 + \frac{1}{m}\right)^{mrt}\\ = P\left(\lim_{m \to \infty}\left(1 + \frac{1}{m}\right)^m\right)^{rt}\\ = Pe^{rt} $$ and you have your continuous compounding formula.
$\endgroup$ 0 $\begingroup$Let's start with the general compounding formula and take the limit as $N \to \infty$.
$$\frac{A}{P}=\lim_{n \to \infty}{\left( 1+\frac{r}{n}\right)^{nt}}$$We start by taking logs:$$\log\left(\frac{A}{P}\right)=\log\lim_{n \to \infty}{\left( 1+\frac{r}{n}\right)^{nt}}$$ And because $log(x)$ is continuous at $x\neq0$, we can swap the limit and the log:$$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{\log\left( 1+\frac{r}{n}\right)^{nt}}$$ $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{nt \log\left( 1+\frac{r}{n}\right)}$$ Next let's take the Taylor series of the log:$$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{nt \left( \frac{r}{n} - \frac{r^2}{2n^2} + \frac{r^3}{3n^3} - \frac{r^4}{4n^4} + \dots\right)}$$ $$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{ \left( rt - \frac{tr^2}{2n} + \frac{tr^3}{3n^2} - \frac{tr^4}{4n^3} + \dots\right)}$$
Finally let's distribute the limit operator (it's linear) and take the limits:$$\log\left(\frac{A}{P}\right)=\lim_{n \to \infty}{rt} - \lim_{n \to \infty}{ \frac{tr^2}{2n} + \lim_{n \to \infty}\frac{tr^3}{3n^2} - \lim_{n \to \infty}\frac{tr^4}{4n^3} + \dots}$$ $$\log\left(\frac{A}{P}\right)= rt $$ $$\frac{A}{P}= e^{rt}$$ $$A=Pe^{rt}$$
Which was what we wanted.
$\endgroup$ $\begingroup$Rewrite your formula as $$ A=P \left[ \left( 1+ \frac{r}{n} \right)^{\frac{n}{r}} \right]^{rt} $$ and let $ w=\frac{n}{r} $ then
$$ A= P\left[ \left( 1+\frac{1}{w} \right)^w \right]^{rt} $$
As the frequency of compounding $n$ is increased the newly created $w$ will increase as well; thus as $ n \rightarrow \infty $ , $w \rightarrow \infty $ as well and the bracketed expression tends to the number $e$.
Consequently we find that as $ n \rightarrow \infty $
$$ A = Pe^{rt} $$
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