Department of Mathematics, Statistics
and Computer Science
Wim Ruitenburg's Fall 2012 MATH 1300-101
Last updated: September 2012
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.
We spent some time on continuously compounded interest.
- There is a so-called Euler number e = 2.71828182845999988... with
special properties described below.
- For big values of n we have (1 + 1/n)^n is approximately equal to e.
For example, (1 + 1/365)^365 = e approximately.
- Better still, for big n and for values y not too big we have (1 + y/n)^n
is approximately equal to e^y.
Another way to say this is that when a is modest in size and b is huge, then
(1+a)^b is approximately equal to e^(a*b).
- Example (1 + 0.2/365)^365 = e^0.2 approximately (and equals about
1.2214).
- Another example: (1 + 0.0002)^30000 = e^(0.0002 * 30000) = e^6
approximately (and equals about 403.42879).
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.
- In class (a) student(s) observed that the number d above is very close
to 0.05 / 365 = 0.000136986, or 0.0136986 percent, which is close to
correct.
In many situations such a nice simple alternative formula is good enough.
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 interst 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, 23 (give the correct formulas only)
- Which of the following is the best approximation of (1 + 0.0007)^3000?
- e^2.1
- 2.21
- e^1.2
- 1.21
- 0
- Recommended problems from Chapter 10 of the book:
31, 33 (give the correct formulas only)
- 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