Without getting into a discussion of limits right now, we can get
an idea of what's happening by taking increasingly larger values of n.
We will talk about limits later on in the year. Study the following
table of values and use your calculator to double check the results:
n
(1 + 1/n)n
10
2.593742
100
2.704814
1000
2.716924
10,000
2.718146
100,000
2.718268
1,000,000
2.718280
If you study this chart, you see that the number e approaches
a value of 2.718 . . . The function ex is called the natural exponential function. The graph of ex and e-x are graphed below:
Notice, they fit the pattern of the previous section. The
number e appears in many applications of physics and statistics. We
will take a close look at the number e and how it relates to compound interest.
Compound Interest Formula
Where:
A(t) = amount after time t.
Ao = Initial amount
r = rate in decimal
n = number of times compounded in a year.
t = time in years
Thus, if the interest was paid semiannually, n = 2. Paid
quarterly would make n = 4, Paid monthly, n = 12, etc.
Sample Problems
1)
Find the value of a $1 if it is invested for 1 year at 10% interest compounded
quarterly.
Solution: Initial amount is $1 with r
= .10, n = 4 and t = 1.
A(1) = 1(1 + .10/4)4 = 1.1038 This means that at the end
of a year, each dollar invested in worth 1.1038 or slightly more than $1.10.
The effect of compounding adds another .0038 % to the interest rate.
Thus the effective annual yield is 10.38%. 2) You invest $5000 in
an account paying 6% compounded quarterly for three years. How much
will be in the account at the end of the time period?
Solution: Initial amount is $5000, with
r = .06, n = 4 and t = 3
A(3) = 5000(1 + .06/4)12 = $5978.09. This account pays $978.09 in interest over the
three years. 3) What is the effective
annual yield on $1 invested for one year at 15% interest compounded monthly?
Solution: Initial amount $1, with r =
.15, n = 12 and t = 1
A(1) = 1(1 + .15/12)12 = 1.1608. The effective annual yield is 16.08%. This
is a relatively big increase because of the number of times compounded in
the year. The above problems all had one
thing in common. The number of times compounded was a finite number.
We can also have continuous compounding. That is, compounding basically every second on the
second. This would be rather cumbersome to calculate because the compounding
is extremely large. We can use a similar formula if the compounding
is continuous.
P(t) = Poert
Notice the appearance of the number e. If you look closely
at the compound interest formula, you will see imbedded the definition of
the number e. Only use this formula if you are sure the compounding
is continuous.
Problems
1) $500 is invested in
an account paying 8% interest compounded continuously. They leave it
in the account for 3 years. How much interest is accumulated?
Solution: Initial amount $500, with r
= .08 and t = 3.
P(3) = 500(e.08(3)) = 635.62. This means the interest is $135.62.
2) A population of insects
rapidly increases so that the population after t days from now is given by
A(t) = 5000e.02t.
Answer the following questions: a)
What is the initial population? b)
How many will there be after a week? c)
How many will there be after a month? (30 days)
Solutions:
a) The initial population is 5000 from the formula.
b) A(7) = 5000e.14 = 5751
c) A(30) = 5000e.6 = 9111
On to
log functions:
Back to
the previous section:
Current quizaroo # 5a
1) (3-2 +3-3)-1 =
a)
36
b) 1/36
c) 27/4
d) 4/27
e) 35
2) 92 = 273x, find x
a) 4/9
b) 9/4
c) 1/3
d) 2/3
e) 0
3) The half-life of an isotope is 6 days. If 4.5 kg
are present now, how much will be present after 12 days?
a) 2.25
kg
b) 1.125
kg
c) 3
kg
d) 9
kg
e) .565
kg
4) Find the amount of
interest earned if $1000 is invested for 3 years at 7% compounded quarterly.
a) $1231.44
b) $210
c) $1210
d) $100
e) $231.44
5)
Find the amount of money you will have if you invest $2000 at 11% compounded
continuously for 6 years.
Definition of the irrational number e 
