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 e^{x} is called the natural exponential function. The graph of e^{x}
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.

A_{o} = 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) = P_{o}e^{rt}

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)
3^{5}

2) 9^{2} = 27^{3x}, 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.