6.2 Compound Interest

Robert Foth

6.2 Compound Interest

Learning Objectives

Upon completion of this section, you should be able to

Calculate the future value of an account in applications for which interest is compounded continuously.
Calculate the principal amount needed in applications for which interest is compounded continuously.

Compound Interest Continuously

Before we get into what we mean by compounding continuously lets look back at an example of compounding
discretely. What we want to investigate is what happens as we compound more often during the same time frame.

Does compounding more frequently earn us more money?

The answer to this question is Yes, but there is some limiting factor. As we increase the number of times we
compound we will earn more money, but there is an upper bound associated with the amount of money we can earn
with the same interest rate compounded more often in the same time interval.

Let us look at what happens when we start with a principal of $3,200 in an account with an annual interest rate
of 5.45% invested for 1.5 years and we vary the compounding period.

Table of values for different compounding periods for the same amount of time.
Compounding Rule	Number of times a year	Value in account after 1.5 years with an annual rate 5.45%
Yearly	n = 1	$P = 3200 {(1 + \frac{.0545}{1})}^{(1) (1.5)} = 3465.132568$
Quarterly	n = 4	$P = 3200 {(1 + \frac{.0545}{4})}^{(4) (1.5)} = 3470.674292$
Monthly	n = 12	$P = 3200 {(1 + \frac{.0545}{12})}^{(12) (1.5)} = 3471.947686$
Daily	n = 365	$P = 3200 {(1 + \frac{.0545}{365})}^{(365) (1.5)} = 3472.569143$
Hourly	n = 8760	$P = 3200 {(1 + \frac{.0545}{8760})}^{(8760) (1.5)} = 3472.589452$
Every Minute	n = 525600	$P = 3200 {(1 + \frac{.0545}{525600})}^{(525600) (1.5)} = 3472.590321$

Now what you should notice is that when we first started to increase the compounding period the gains in the
increase in compounding were fairly noticeable, but look what happened when we went from monthly to daily. We
are compounding by about 20 times more frequently, yet the gains are shrinking. This is even more noticeable as
we go from hourly to every second – this is compounding 60 times more frequently and the gain isn’t even
measurable when we round to the nearest penny.

If we wish to compound every fraction of a second and then every fraction of that fraction of a second we are
on our way to compounding continuously. To find a way to calculate what this limit is for compounding more often
we will take a look at a slightly different expression before examining our formula for compounding discretely
as n goes to infinity.

Look at the following table – it shows what happens to the expression
${(1 + \frac{1}{m})}^{m}$
as m goes to infinity (gets very large) and introduces us to the number “e“.

m	${(1 + \frac{1}{m})}^{m}$
1	2
5	2.48832
10	2.59374246
50	2.691588029
100	2.704813829
500	2.715568521
1000	2.716923932
5000	2.71801005
10000	2.718145927
50000	2.718254646
100000	2.718268237
500000	2.71827911
1000000	2.718280469
5000000	2.718281555

As you can see the values of m are getting really large (what we mean when we say going towards
infinity) that the values of the expression
${(1 + \frac{1}{m})}^{m}$
is approaching some fixed constant. We can see that the decimal places are starting to converge, but do not
follow any fixed pattern (so it doesn’t appear to be a rational number, but rather an irrational number like
π). We call this number “e” (unlike π it gets a regular English character to represent itself
while many other numbers represented by a character use greek letters). We sometime refer to this as
“Euler’s Number” – read
more about Euler’s Number in Nature.

More formally we have this result:
$\lim_{m \to \infty} {(1 + \frac{1}{m})}^{m} = e ≅ 2.71828…$

Let’s see how this result works with the discrete compounding formula as k goes to infinity.

$\begin{array}{l} FV & = P {(1 + \frac{r}{n})}^{t \cdot n}, let m = \frac{n}{r} \\ = P {(1 + \frac{1}{m})}^{m \cdot r \cdot t}, we substituted for n : n = m r \\ = P {[{(1 + \frac{1}{m})}^{m}]}^{r \cdot t}, now look at middle term \\ = P {(e)}^{r \cdot t}, as n goes to infinity, so does m \end{array}$

Compounding Interest Continuously

$F V = P \cdot e^{(r \cdot t)}$

FV is the balance in the account after t years also known as the future value.
P is the starting balance of the account (also called initial deposit, or principal).
r is the annual interest rate (in decimal form).

The most important thing to remember about using this formula is that it assumes that we put money in the
account once and let it sit there earning interest while compounding continuously.

To finish up our example with depositing $3200 into an account with an annual rate of 5.45% for 1.5 years. We
say that as we compounded more often in the year we approached some limiting value close to $3472.59. If we were
to compound continuously we would have the following:

$\begin{array}{l} F V = P e^{r t} \\ F V = 3200 e^{(0.0545) (1.5)} \\ F V = 3200 (1.085184) \\ F V = 3472.590335 \\ F V \approx $ 3472.59 \end{array}$

The actual value for compounding continuously compared to compounding every second was off by $0.00014 or less
than a penny.

Evaluating exponents with e on the calculator

When we need to calculate something like e³ it would be unwise to use an estimate like
2.718 to do the calculations if we can avoid it. As we increase the value on that exponent the estimation
would become worse and worse.

Many scientific calculators have a button for e or e to an exponent. It is typically
either labeled like
$\begin{matrix} e \end{matrix}$ or
$\begin{matrix} e^{x} \end{matrix}$

To evaluate e ³ we’d type
$\begin{matrix} e \end{matrix}$

$\begin{matrix} \land \end{matrix}$
3, or
$\begin{matrix} e^{x} \end{matrix}$
3. Try it out – you should get something around 20.08553692.

If you don’t have the constant e on your calculator use the approximation 2.718 for e for now.

Example 1

Suppose you deposit $3000 in a CD paying 6% interest, compounded continuously. How much will you have
in the account after 20 years? How did this compare to compounding Monthly?

Solution

In this example,

$\begin{array}{l} P & = & $ 3000 \\ \begin{array}{l} r \\ t \end{array} & \begin{array}{l} = \\ = \end{array} & \begin{array}{l} 0.06 \\ 20 \end{array} \end{array} \begin{array}{l} the initial deposit \\ \begin{array}{l} 6% annual rate \\ since we’re looking for how much we’ll have after 20 years \end{array} \end{array}$

So,
$F V = 3000 e^{(0.06) (20)} \approx $ 9, 960.35$
(round your answer to the nearest penny).

Compare this now to compounding monthly:
$F V = 3000 {(1 + \frac{0.06}{12})}^{(20 \cdot 12)} \approx $ 9, 930.61$

What we see is that by compounding continuously we have an additional gain of roughly $30 over that same time
period (or roughly a 0.3% increase in value for compounding continuously).

Try it Now 1

If $3500 is invested at 9% compounded continuously, what will the value be in four years?

Hint 1

The key word to look for in this problem is that the interest is being compounded continuousyly. This means
we will be using
$FV = P e^{r t}$
to calculate the amount in the account from an initial deposit. When using the formula make sure to always
convert the percentage to a decimal before putting it into the formula.

Answer

Use the formula for compounding continuously with the given information:

$\begin{array}{l} FV = $ 3500 \\ t = 4 \\ r = .09 \\ FV = P e^{r t} \\ FV = 3500 e^{.09 \times 4} = $ 5016.65 \end{array}$

Example 2

You know that you will need $40,000 for your child’s education in 18 years. If your account earns
4% compounded continuously, how much would you need to deposit now to reach your goal?

Solution

In this example (as we saw in the last section), we’re looking for P.

$\begin{array}{l} r & = & 0.04 & 4 % \\ t & = & 18 & Since we know the balance in 18 years \\ FV & = & $ 40, 000 & The amount we have in 18 years \end{array}$

Use the equation for continuous compounding to set up an equation to solve for P₀.

$\begin{array}{l} 40000 = P e^{0.04 \times 18} \\ 40000 = P (2.0544) \\ P = \frac{40000}{2.0544} = $ 19470.40 \end{array}$

You would need to deposit approximately $19,470.40 now to have $40,000 in 18 years.

As a side note you will have noticed we rounded
$e^{0.04 \times 18}$
at the second step to 2.0544. This rounding introduced a small amount of error in our result when we go to
divide 40,000 by 2.0544. If we had left the exact expression in and divided the total would have been
$19.470.09 instead (see below):

$P = \frac{40000}{e^{0.04 \times 18}} = $ 19470.09$

Try it Now 2

The Tran family is expecing a new baby in a few days. They want to put away money for college and figure they
will need about $100,000 in 18 years. If they make a single lump sum investment now into an account that
compounds interest continuously with an annual rate of 3.4%, then how much would that be?

Hint 1

In this problem we are given the future value in the account and will be looking for the principal (the
lump sum payment into the accout at the start of the 18 years).

Answer

Find P in the continuous compounding interest formula with the following from the question
statement

$\begin{array}{l} F V = $ 100, 000 \\ r = 0.034 \\ t = 18 \end{array}$

Substitute the values into the formula
$F V = P \cdot e^{(r \cdot t)}$
and solve for P.

$\begin{array}{l} F V = P \cdot e^{(r \cdot t)} \\ 100, 000 = P e^{(0.034 \cdot 18)} \\ \frac{100, 000}{e^{(0.034 \cdot 18)}} = P \\ P \approx $ 54, 226.53 \end{array}$

The Tran family needs to make a single lump sum deposit of $54,226.54 in the account.

Exercises

If $8,000 is invested at an annual rate of 9.2% compounded continuously, what will the final amount be in 4
years?
What is the value of $750 invested into an account with an annual rate of 3.25% compounded continuously for
4 years and 3 months?
Find the value of a deposit of $8,000 earning an annual rate of 9.2% compounded continuously for N years.
How much interest is earned after 2 years? How much interest is earned after 4 years? Were you surprised
that it was not twice as much? Why or why not.
Sam’s grandparents was planning to give him $8,000 in 10 years. Sam has convinced his grandparents to pay
him $4,000 now, instead. If Sam invests this $4,000 at an annual rate of 5.7% compounded continuously, how
much money will he have in 10 years?
How much should be deposited into an account that has an annual rate of 4.15% compounded continuously so
that in 15 years there is $10,000?
Rosa knows she needs $2350 for tuition in 9 months from now. How much should she deposit into an account
that compounds continuously at an annual rate of 5.2% to achieve that goal?
Brandon plans on buying a car when he graduates in three years from now. He has some inheritance he can set
away for this purchase. If he assumes the car will cost him $23,000 how much of that money should he set
aside in an account that has an annual rate of 7.8% compounded continuously?
As an experiment you plan on depositing $5 for your future great great grandchild to inherit 100 years from
now. If the account compounds interest continuously with an annual rate of 5.5%, then how much will they
inherit? What if you want to leave it for someone 200 years from now?
Javier invests $500 into an account with the bank that compounds continuously with an annual rate of 9.0%
and Maria invests $500 into a different account that compounds annually with an annual rate of 9.03%. Find
the value in each account after 10 and 30 years. What did you notice? Should you compare more than just the
annual rate to determine which account to invest in? Why or why not.

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Learning Objectives

Compound Interest Continuously

Compounding Interest Continuously

Evaluating exponents with e on the calculator

Example 1

Try it Now 1

Hint 1

Answer

Example 2

Try it Now 2

Hint 1

Answer

Exercises

License

Share This Book