Assignment 70

"Mathematicians are like lovers...Grant a mathematician the least principle, and he will draw from it a consequence which you must also grant him, and from this consequence another." - (Fontenelle, 1657-1757)


Math History Tidbit:

Zero: The number we call 0 has a long and fascinating history. While the concept of 0 was used by the Chinese in the fourth century B.C., it did not surface in the Western world until the 1200's. Leonardo of Pisa, better known as Fibonacci (1180-1250) advocated strongly for the use of Hindu-Arabic numerals (which included 0) in Western Europe. For a variety of reasons, the West resisted the use of the efficient Hindu-Arabic numerals. As late as 1298, the City Council of Florence (Italy) outlawed any system but Roman numerals. (How would you like to do operations like multiplication and division with Roman numerals?)

I guarantee you that research on the number 0 will yield some extremely interesting historical information.

Herkimer's Corner

Why did Herkimer insist on placing lamps with his autograph collection that was stored in his attic?

Answer: He wanted to have his lights up in names.

Herky wants to know:

If a pig loses its voice, is it disgruntled?

If lawyers are disbarred and clergymen are defrocked, can electricians be delighted, musicians denoted, cowboys deranged, tree surgeons debarked, and dry cleaners depressed?


Reading: Section 11.3, pages 666-670.

Written: Page 670/33-44


Items for reflection:

Mathematical word analysis:
CENTURY: This was originally the Latin term for a collection of one hundred items. In the Roman army, a company of one hundred men was called a centurion.


A geometric series has the form a, ar, ar2, ar3, ...

The n-th term of a geometric series is arn-1.

The sum of the first n terms of a geometric series is a(1-rn)/(1-r). (See derivation below.)

Consider the series 2,6,18,54,... . It is geometric with a = 2 and r = 3. The 6th term is 2(36-1) = 2(35) = 486. The sum of the first six terms is 2(1-36)/(1-3) = 728.

We can define the sequence explicitly by writing f(n) = 2(3n-1) for n = 1,2,3,... Note that we can get the sum of the first six terms on our calculator: sum(seq(2*3^(x-1),x,1,6,1)) = 728.

Derivation of formula for sum of a geometric series (with assumption that r is not equal to 1).

(A): S = a + ar + ar2 + ... + arn-1 (There are n terms here.)

(B): rS = ar + ar2 + ... + arn-1 + arn

(C) = (A)-(B): S - rS = a - arn

==> S(1-r) = a - arn

==> S = (a - arn)/(1-r) = a(1-rn)/(1-r).

Problem: What is the 10th term of a geometric sequence whose first term is 3 and whose common ratio is 2? And, what is the sum of the first ten terms of this sequence?

Solution (with communication):

We have a = 3 and r = 2. Hence the 10th term is

3(210-1) = 3(29) = 1,536

and the sum of the first ten terms is

3(1 - 210)/(1-2) = 3,069.

Problem: If the second term of a geometric sequence of positive numbers is 15 and the fourth term is 135, what is the sixth term?

Solution (with communication):

We have ar = 15 and ar3 = 135. Hence

ar3 = ar(r2) = 15r2 = 135

==> r2 = 9 ==> r = -3 or r = 3. Since the sequence consists of positive numbers, we must have r = 3. Hence ar = 15 ==> a = 5.

The sixth term is ar5 = 5(35) = 1,215.