Chapter 10.8, Problem 51WE

(a)

To determine

Show that if interest is compounded daily , that is , n is very large, the Compound Interest Formula is $A\approx P{e}^{rt}$

Explanation of Solution

Formula Used:

  $\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n=e$

Calculation:

Compound Interest Formula is :

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

where

P= Principal

r = annual interest rate in decimal

t = time in years

A = Final amount after t years

n = number of times the interest is compounded in a year

Since n is very large , apply limit on each side:

  $\begin{array}{l}\underset{n\to \infty }{\lim }A=\underset{n\to \infty }{\lim }P{\left(1+\frac{r}{n}\right)}^{nt}\\ \Rightarrow A=P\underset{n\to \infty }{\lim }{\left(1+\frac{1}{\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{nt}\mathrm{..............}[\underset{n\to \infty }{\lim }c=c;c\text{ is a constant]}\\ \Rightarrow A=P\underset{\frac{n}{r}\to \infty }{\lim }{\left(1+\frac{1}{\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$r$}\right.}\right)}^{nt}\mathrm{...................}[n\to \infty \Rightarrow \frac{n}{r}\to \infty \text{ since r is the rate in decimal , positive]}\\ \Rightarrow A=P\underset{m\to \infty }{\lim }{\left(1+\frac{1}{m}\right)}^{mrt}\mathrm{................}[\frac{n}{r}=m\Rightarrow n=mr]\\ \Rightarrow A=P\underset{m\to \infty }{\lim }{\left[{\left(1+\frac{1}{m}\right)}^m\right]}^{rt}\\ \Rightarrow A\approx P{e}^{rt}\mathrm{...............................}[\underset{n\to \infty }{\lim }{\left(1+\frac{1}{n}\right)}^n\approx e]\end{array}$

Hence proved.

(b)

To determine

Find the amount of money that you would have after 1 year if you invest $1000 at 6% interest compounded daily.

Answer to Problem 51WE

$1,061.8

Explanation of Solution

Calculation:

From part(a) , we have the Compound Interest Formula , if the interest is compounded daily:

  $A\approx P{e}^{rt}$

where

P= Principal = 1000

r = annual interest rate in decimal = 6% = 0.06

t = time in years = 1 year

A = Final amount after t years

Substitute the values and solve for A :

  $\begin{array}{l}A\approx P{e}^{rt}\\ \Rightarrow A\approx 1000e^{(0.06)1}\\ \Rightarrow A\approx 1000(1.0618)\\ \Rightarrow A\approx 1061.8\end{array}$

So, the amount after 1 year is about $1,061.8