Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Spring 2013 MATH 1300-101
Last updated: March 2013
Comments and suggestions: Email wimr@mscs.mu.edu
Money from chapter 10
Despite all the money talk, chapter 10 is really about exponential growth, with
examples from money management.
From the chapter we can learn:
- Increasing an amount like 100 by 5 percent is essentially the same as
multiplying 100 by (1 + 5/100) or by 105/100 or by 1.05, and the result equals
105.
Decreasing an amount like 100 by 5 percent is essentially the same as
multiplying 100 by (1 - 5/100) or by 95/100 or by 0.95, and the result equals
95.
- Compound interest is very powerful.
When we start with just 100 (dollars, say), and increase the amount by 5
percent each year for 10 years, the new amount equals 100 * 1.05 * 1.05 * 1.05
* 1.05 * 1.05 * 1.05 * 1.05 * 1.05 * 1.05 * 1.05, also written 100 * 1.05^10,
and which equals approximately 162.88946 (a bank may round this off to
162.89).
- Increasing an amount like 100 by 5 percent and then reducing the new
amount by 5 percent is NOT neutral.
For first we multiply 100 by 1.05 to get 105, and then we multiply 105 by
0.95 and get as final result only 99.75.
- In the following example we ignore other factors like insurance and
taxes and other overhead costs.
We pretend that money devalues at a fixed 5 percent each year in the sense
for example, 100 dollars last year has the same purchasing power as 105 dollars
today.
So 1000 dollars in year 2000 (let us call them Y2K dollars) has the same
purchasing power as 1000 * 1.05^13 = 1885.65 in year 2013.
We may say that 1885.65 dollars in 2013 is worth 1000 Y2K dollars.
Now 3000 dollars in year 2018 is worth 3000 / 1.05^18 = 1246.56 dollars in year
2000, that is, 1246.56 Y2K dollars.
Combined,
- 1885.65 dollars in 2013 equals 1000 Y2K dollars, and
- 3000 dollars in 2018 equals 1246.56 Y2K dollars.
So 3000 dollars in 2018 is worth more than 1885.65 dollars in 2013.
We spent some time on geometric sequences.
- We verified the formula on page 378 of the book.
- We used the formula in an example compatible with the Red Mustang
example of page 384.
We said a few things about logarithms.
- To find the daily interest rate d such that we end up with annual
interest 0.05 or 5 percent, we must solve the equation (1+d)^365 = 1 + 0.05 =
1.05.
Here is a sketch using natural logarithms (usually called ln on a
calculator).
Before doing anything with the calculator, get the correct equation:
Start with (1+d)^365 = 1.05.
Write in the logarithms and get ln( (1+d)^365 ) = ln( 1.05 ).
With the first important logarithm property we get 365 * ln( 1+d ) =
ln( 1.05 ).
With ordinary division we get ln( 1+d ) = ln( 1.05 ) / 365.
With the second important logarithm property we get 1+d = e^( ln( 1.05
) / 365 ).
With the calculator we get 1+d = e^( 0.048790164 / 365 ) approximately.
More calculator work: 1+d = e^( 0.00013367168 ) approximately.
Yet more calculator work: 1+d = 1.0001336806 approximately.
Thus d = 0.0001336806 approximately, or 0.01336806 percent.
We spent some time on continuously compounded interest.
- There is a so-called Euler number e = 2.71828182845999988... with
special properties described below.
- When a is small in size and b is big and a*b is modest in size, then
(1+a)^b is approximately equal to e^(a*b).
For example, (1 + 0.01)^100 = e^1 = e approximately.
Another example: (1 + 0.0002)^30000 = e^(0.0002 * 30000) = e^6
approximately (and equals about 403.42879).
- Recall that above we try to find the daily interest rate d such that we
end up with annual interest 0.05 or 5 percent.
Here is a sketch by which we get a close enough approximation of d rather than
the exactly correct value for d.
Before doing anything with the calculator, get the correct equation:
Start with (1+d)^365 = 1.05.
Rewrite using our approximation formula and get e^(d*365) = 1.05.
Write in the logarithms and get ln( e^(d*365) ) = ln( 1.05 ).
With the first important logarithm property we get d * 365 = ln( 1.05 ).
With ordinary division we get d = ln( 1.05 ) / 365.
More calculator work: d = 0.00013367168 approximately.
Note how very close this is to the correct approximation above.
- Since d * 365 is quite small, the following formula gives a decent
approximation.
Simply set
d = 0.05 / 365 = 0.000136986.
Fairly close to the correct approximation.
Example Problem(s)
- Recommended problems from Chapter 10 of the book:
1, 3, 9
- Suppose we put an amount X in the bank at an interest rate of 3 percent
annual interest.
After 10 years we have 2000 dollars.
Which of the following is the correct formula for X?
- X = 2000 * 1.03^10
- X = 2000 / 1.03^10
- X = 1.03^10 / 2000
- X = 2000 * 10^1.03
- X = 2000 / 10^1.03
- Recommended problems from Chapter 10 of the book:
21, 22, 31, 32 (give the correct formulas only)
- Suppose a fixed inflation rate as explained earlier on this page, of 4
percent each year.
Suppose we have 750 dollars in 2014.
Which of the following is the correct formula for this amount in Y2K dollars?
- 750 * 1.04^14
- 1.04^14 / 750
- 750^14 * 1.04
- 750 / 1.04^14
- 750^14 / 1.04
- Let d be the daily compounded interest rate such that the annual
interest rate equals 8 percent.
So (1+d)^365 = 1.08.
Which of the following formulas is the correct one for d?
- (1+d)^108 = 3.65
- 1+d = 1.08 / 365
- 1+d = e^( ln( 1.08 ) / 365 )
- 1+d = ln( 1.08 ) / 365
- 1+d = 108 / 365
- Which of the following is the best approximation of (1 + 0.007)^300?
- e^2.1
- 2.21
- e^1.2
- 1.21
- 0