Assignment 72

"The universe stands continually open to our gaze, but it cannot be understood unless one first learns to comprehend the language and the characters in which it is written. It is written in the language of mathematics, and its characters are triangles, circles, and other geometric figures." - (Galileo, 1564-1642)


Math History Tidbit:

Luitzen Brouwer (1881-1966: A Dutch mathematician, Brouwer was the leader of the intuitionist mathematicians. The intuitionists believe mathematics is a human activity that originates and thrives in the mind, and that it is independent of the real world. They believe that a logical construction of the system of the real numbers is not acceptable. Their views conflicted with those of the formalist mathematicians led by David Hilbert, and the logistic school of mathematicians led by Alfred North Whitehead and Bertrand Russell.

Herkimer's Corner

Why wouldn't Herkimer let a friend pull a wagonload of sheep over his tray of ice cubes?

Answer: He didn't want anyone pulling the wool over his ice.

Herky wants to know:

Why do those who claim to live by their wits always look so hungry?

Do those who have a desperate urge to paint themselves gold have a gilt complex?


Reading: Review Section 11.3, as necessary.

Written: Assignment sheet provided in class (financial problems).


Items for reflection:

Mathematical word analysis:
COMPUTE: From the Latin computare, a notched tally stick used for calculations by the Romans.

If you make beginning of year deposits of D for n years, and it the deposits earn an annual rate of i, then the accumulation, A, of these deposits at the end of the n-th year is a geometric series:

A = P(1+i) + P(1+i)2 + ... + P(1+i)n = P(1+i)[1 + (1+i) + ... + (1+i)n-1]

= P(1+i)[1(1-(1+i)n)/(1-(1+i))]

= P(1+i)[(1+i)n-1]/i.

Example: Consider 10 beginning-of-year payments of $1,000, accumulating for 20 years at 8%. We will calculate the accumulation of these payment at the end of 20 years different ways.

Using the formula above, we have A = $1000(1.08)[(1.08)20-1)/.08 = $49,422.92.

Using the TI-83 calculator, we have A = sum(seq(1000*1.08^x,x,1,20,1)) = $49,422.92.

Problem: If I make 25 beginning-of-year deposits of $600, what amount will I have at the end of the 25 years if the annual interest rate is 6.75%?

Solution (with communication): Using the accumulation formula, the requested amount is

$600(1.0675)[(1.0675)25-1].0675 = $39,086.07.

Using the TI-83,

sum(seq(600*1.0675^x,x,1,25,1)) = $39,086.07.


Problem: I want to have $50,000 available in ten years. What equal beginning-of-year deposits must I make to accumulate to this amount if the annual interest rate is 7%?

Solution (with communication): If x is the amount of each deposit, then

x(1.07)[(1.07)10 - 1]/.07 = $50,000

==> x = $50,000(.07)/[1.07[(1.07)10 - 1]]

==> x = $3,382.13