Compound Interest Rate

This article is part of the MathHelp Tutoring Wiki

This article is part of the EconHelp Tutoring Wiki

General Formula

Given an annual interest rate r, we can use the number following formula to compute the increase in the principle when the interest is compound annually.

$A(t)=P(1+r)^{t}$

If the interest is compounded in n times a year (in our example n=12), the amount after t years is,

$A(t)=P({\frac {1+r}{n}})^{tn}$

If we take a limit as n goes to infinity, we say that interest is compounded continuously.

As we know, $e=\lim _{n\to \infty }(1+{\frac {1}{n}})^{n}$

Let's use the substitution m=n/r, so n=mr

${\begin{aligned}A(t)&=\lim _{m\to \infty }P(1+{\frac {1}{m}})^{tmr}\\&=P(\lim _{m\to \infty }(1+{\frac {1}{m}})^{tmr}\\&=P(\lim _{m\to \infty }((1+{\frac {1}{m}})^{m})^{tr}\\&=Pe^{tr}\end{aligned}}$

Practice Question

Q: say P=$10,000 with r=0.06 and t=3, how much will you have if interest is compounded.

(1) 6 times a year

(2) 24 times a year

(3) continuously

A:

(1) $A=P(1+r/n)^{tn}=10000(1+0.06/6)^{3\times 6}=11961.47$

(2) $A=10000(1+0.06/24)^{3\times 24}=11969.48$

(3) $A=Pe^{rt}=10000e^{0.06\times 3}=11972.17$

• Back to Differential Calculus
• Back to MathHelp