# Continuous Interest

## Based on infinitely small compounding periods and the number e.

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Continuous Interest

Clever Carol realized that she makes more money when she convinces the bank to give her 12% in two chunks of 6% than only one time at 12%. Carol knew she could convince them to give her 1% at the end of each month for a total of 12% which would be even more than the two chunks of 6%. As Carol makes the intervals smaller and smaller, does she earn more and more money from the bank? Does this extra amount ever stop or does it keep growing forever?

### Continuous Interest

Calculus deals with adding up an infinite number of infinitely small amounts. A result of calculus that is used in finance is the number \begin{align*}e\end{align*} as \begin{align*}k\end{align*} the number of compounding periods, approaches infinity.

\begin{align*}e \approx \left(1+\frac{1}{k}\right)^k \approx 2.71828 \ldots\end{align*} as \begin{align*}k\end{align*} approaches infinity

This means that even when there are an infinite number of infinitely small compounding periods, there will be a limit on the interest earned in a year. The term for infinitely small compounding periods is continuous compoundingA continuously compounding interest rate is the rate of growth proportional to the amount of money in the account at every instantaneous moment in time. It is equivalent to infinitely many but infinitely small compounding periods.

The formula for finding the future value of a present value invested at a continuously compounding interest rate \begin{align*}r\end{align*} for \begin{align*}t\end{align*} years is:

\begin{align*}FV=PV \cdot e^{rt}\end{align*}

Applying this formula, you can determine what the future value of 360 invested for 6 years at a continuously compounding rate of 5% is. \begin{align*}FV=?, \ PV=360, \ r=0.05, \ t=6\end{align*} \begin{align*}FV=PV \cdot e^{rt}=360e^{0.05 \cdot 6}=360e^{0.30} \approx 485.95\end{align*} ### Examples #### Example 1 Earlier, you were asked to compare the amount of money Clever Carol would make using different rates of compounding. Clever Carol could calculate the returns on each of the possible compounding periods for one year. For once per year, \begin{align*}k=1\end{align*}: \begin{align*}FV=PV(1+i)^t=100(1+0.12)^1=112\end{align*} For twice per year, \begin{align*}k=2\end{align*}: \begin{align*}FV=PV(1+i)^t=100 \left(1+\frac{0.12}{2}\right)^2=112.36\end{align*} For twelve times per year, \begin{align*}k=12\end{align*}: \begin{align*}FV=PV(1+i)^t=100 \left(1+\frac{0.12}{12}\right)^{12} \approx 112.68\end{align*} At this point Carol might notice that while she more than doubled the number of compounding periods, she did not more than double the extra pennies. The growth slows down and approaches the continuously compounded growth result. For continuously compounding interest: \begin{align*}FV=PV \cdot e^{rt}=100 \cdot e^{0.12 \cdot 1} \approx 112.75\end{align*} No matter how small Clever Carol might convince her bank to compound the 12%, the most she can earn is around 12.75 in interest. #### Example 2 What is the continuously compounding rate that will grow100 into 250 in just 2 years? \begin{align*}PV=100, \ FV=250, \ r=?, \ t=2\end{align*} \begin{align*}FV &= PV \cdot e^{rt}\\ 250 &= 100 \cdot e^{r2}\\ 2.5 &= e^{r2}\\ \ln 2 &= 2r\\ r &= \frac{\ln 2}{2} \approx 0.3466=34.66 \%\end{align*} #### Example 3 What amount invested at 7% continuously compounding yields1,500 after 8 years?

\begin{align*}PV=? \ FV=1,500, \ t=8, \ r=0.07\end{align*}

\begin{align*}FV &= PV \cdot e^{rt}\\ 1,500 &= PV \cdot e^{0.07 \cdot 8}\\ PV &= \frac{1,500}{e^{0.07 \cdot 8}} \approx \856.81\end{align*}

What is the future value of 500 invested for 8 years at a continuously compounding rate of 9%? \begin{align*}FV=500e^{8 \cdot 0.09} \approx 1027.22\end{align*} #### Example 5 What is the continuously compounding rate which grows27 into 99 in just 4 years? \begin{align*}99=27e^{4r}\end{align*} Solving for \begin{align*}r\end{align*} yields: \begin{align*}r=0.3248=32.48 \%\end{align*} ### Review For problems 1-10, find the missing value in each row using the continuously compounding interest formula.  Problem Number \begin{align*}PV\end{align*} \begin{align*}FV\end{align*} \begin{align*}t\end{align*} \begin{align*}r\end{align*} 1.1,000 7 1.5% 2. $1,575$2,250 5 3. $4,500$5,500 3% 4. $10,000 12 2% 5.$1,670 $3,490 10 6.$17,000 $40,000 25 7.$10,000 $18,000 5% 8.$50,000 30 8% 9. $1,000,000 40 6% 10.$10,000 50 7%

11. How long will it take money to double at 4% continuously compounding interest?

12. How long will it take money to double at 3% continuously compounding interest?

13. Suppose you have $6,000 to invest for 12 years. How much money would you have in 10 years if you earned 3% simple interest? How much money would you have in 10 years if you earned 3% continuously compounding interest? 14. Suppose you invest$2,000 which earns 5% continuously compounding interest for the first 12 years and then 8% continuously compounding interest for the next 8 years. How much money will you have after 20 years?

15. Suppose you invest \$7,000 which earns 1.5% continuously compounding interest for the first 8 years and then 6% continuously compounding interest for the next 7 years. How much money will you have after 15 years?

To see the Review answers, open this PDF file and look for section 13.4.

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### Vocabulary Language: English

TermDefinition
$e$ $e$ is an irrational number that is approximately equal to 2.71828. As $n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e$.
Continuous compounding Continuous compounding refers to a loan or investment with interest that is compounded constantly, rather than on a specific schedule. It is equivalent to infinitely many but infinitely small compounding periods.
Continuously compounding Continuous compounding refers to a loan or investment with interest that is compounded constantly, rather than on a specific schedule. It is equivalent to infinitely many but infinitely small compounding periods.
e $e$ is an irrational number that is approximately equal to 2.71828. As $n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e$.