Download Chapter 24 CHAPTER 24: Infinite Series We know that there is an

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Chapter 24
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CHAPTER 24: Infinite Series
We know that there is an infinite, and are ignorant of its nature. As we know it to be false that
numbers are finite, it is therefore true that there is an infinity in number. But we do not know
what it is. It is false that it is even, it is false that it is odd; for the addition of a unit can make no
change in its nature. Yet it is a number, and every number is odd or even (this is certainly true
of every finite number). So we may well know that there is a God without knowing what He
is.120
“There is an infinity in number.” Pascal wrote this over 300 years ago, living in Western Europe.
While this chapter will begin to examine the idea of infinity as a quantity in mathematics as it was
developed in Western European thought, the general idea is not limited to that culture. Consider first the
Hebrew culture of the Old Testament.
The Biblical usage of the concept of the infinite is imprecise by modern mathematical standards. In
the Old Testament, the concept is expressed by two words which have the sense of "not countable".121
Genesis 41:49 shows that this can (and typically does) mean simply "too large to count " (but nonetheless
strictly finite): "Joseph stored up grain in great abundance like the sand of the sea, until he stopped
measuring it, for it was beyond measure (italics mine)." Similarly, Judges 6:5 indicates that the Midian
invaders were "innumerable", and 7:12 indicates that "their camels were without number (italics mine), as
numerous as the sand on the seashore."
When the Old Testament uses similar phrases about God, most of the time the meaning seems
similar to what we have seen in the previous paragraph. For instance, Job 9:10 states that God "does great
things, unfathomable, and wondrous works without number (italics mine)." The point of the passage does
not seem to demand that we go beyond the idea that God does more wondrous deeds than a person could
enumerate in a lifetime. Perhaps Psalm 147:5 is the example that comes closest to a reference to an actual
infinite: "Great is our Lord, and abundant in strength; His understanding is infinite." (On the other hand,
perhaps the same idea is expressed as a mere comparison in Isaiah 55:9, "For as the heavens are higher than
the earth, So are My ways higher than your ways, and My thoughts than your thoughts.") In any case, the
notion of the infinite is only used in reference to God's characteristics or properties. The Bible never uses
the word "infinite" to describe God in His being; He is never referred to as The Infinite.
The anthropologist Jadran Mimica studied the Iqwaye people of Irian Jaya, an island in Indonesia.
In his book, Intimations of Infinity, he writes
The fact that I relate the structures of the Iqwaye counting system to the most sophisticated
concepts of Western mathematics, if anything, attests to the universality of those structures
which we in our culture believe to be exclusively mathematical.122
The "structures" to which he refers are the ideas behind the set theory and number theory of Western
culture , and particularly infinite numbers. The particular details of Western mathematics or Iqwaye
counting in dealing with these ideas is transcended by the universality of the ideas themselves.
Later Mimica claims
120Pascal,
Pensees, p. 83
William Gesenius, A Hebrew and English Lexicon of the Old Testament, Oxford U. P., Oxford, p.708.
122Jadran Mimica, Intimations of Infinity, Oxford: Berg Publishing, 1988, p. 23.
121see
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I am positing [the problem of infinity] as an ontologically real and humanly universal
(emphasis mine) dimension of existence like the reality of the cosmos, that is, the world as a
whole. Its reality is prior to any conscious reflection upon existence. It is a quality present
in the very tissue of human existence, or simply, infinity is an a priori dimension or an a
priori of human existence.123
For our purposes, the importance of these opinions is the suggestion that while Western culture has
analyzed infinity in a mathematical way, perhaps in particular mathematical ways, the underlying concept
of infinity is not an exclusively Western idea. Mimica is claiming that the idea of infinity is one found in all
human cultures. It may be dealt with in different ways in different cultures, but it is there none the less.
Don Richardson, author of Peace Child, has also written a book entitled Eternity in Their Hearts.
The title is taken from Ecclesiastes 3:11, which says, "He has made everything beautiful in its time. He has
also set eternity in the hearts of men; yet they cannot fathom what God has done from beginning to end." It
is Richardson's thesis that "as Yahweh's special revelation ... has continued to reach out into the world
through both the Old and New Testament eras, it has continually found that Yahweh's general revelation ...
is already on the scene..."124 While Richardson does not discuss infinity as an abstract concept, he does see
in the concept of eternity (infinite time?) an idea which points to God. Of course the truths of general
revelation may be mixed with error, or difficult to discover, but the suggestion is that they are there in any
culture, and are the foundation on which the truths of special revelation can be built. And one of these
notions is infinity.
As one example of this, Richardson quotes several hymns of the Karen people of Burma. "By means
of these hymns, awe and reverence for Y'wa, the true God, were kept alive in the hearts of the Karen people
so that they would not lapse into Buddhism with its idolatry."125 Here are the hymns:
Y'wa is eternal, his life is long.
One aeon -- he dies not!
Two aeons -- he dies not!
He is perfect in meritorious attributes.
Aeons follow aeons -- he dies not!
Who created the world in the beginning?
Y'wa created the world in the beginning!
Y'wa appointed everything.
Y'wa is unsearchable.126
Putting the ideas of these two authors together, I would make the following suggestion. The God of
the Bible, the One True God, is the Creator of humankind. He has revealed truth about Himself which is
generally available to all human beings. This makes sense because it is entirely reasonable that the Creator
would leave traces of Himself in both the world around us and in our own minds. This is Paul's point in
Romans 1: 20, which J. B. Phillips translates this way, "For since the beginning of the world the invisible
attributes of God, e.g. his eternal power and deity, have been plainly discernible through things which he
has made and which are commonly seen and known..."127 So the notion of infinity points us beyond itself
to God. Keep Him in mind as we look at the details of the current Western mathematical analysis of the
subject of infinity.
123Mimica,
p. 96.
Richardson, Eternity in Their Hearts, Ventura, CA : Regal Books, 1981.
125Richardson, p. 76.
126Richardson, p. 76.
127J. B. Phillips, The New Testament in Modern English, Revised Edition, New York: Macmillan, 1984.
124Don
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Zeno's Paradox
Zeno was a Greek philosopher who lived around 450 B.C., which means he lived after Pythagoras
but before Euclid. He argued philosophically against the reality of motion. Philosophically, he believed
that real things never change. Any change which we believe we perceive is not real, hence motion, which is
change in position, is not real. His arguments involve paradoxes, apparent contradictions which arise
when what we perceive is subjected to critical analysis.
The paradox which we wish to consider involves a race between the Greek hero Achilles and a
Tortoise. Achilles provides the lowly and very slow Tortoise with a head start. For the purposes of the
paradox, the amount of the head start really doesn't matter. Zeno claims that it is impossible for Achilles to
catch the Tortoise (let alone pass him); this claim seems ridiculous in light of our experience in which
runners who are behind in a race often end up winning.
Here is Zeno's argument. For the sake of working with specific numbers, assume the Tortoise has a
1 meter head start, and Achilles is 10 times as fast as the Tortoise. The first thing Achilles must do is run the
1 m to where the Tortoise was at the start. However, while Achilles is doing this, the Tortoise moves ahead
a distance of 0.1 meters. Now Achilles must run the 0.1 meters to the Tortoise's second position. But in the
mean time, the Tortoise has advanced again, 0.01 meters. So Achilles will need to run that distance, but the
Tortoise will again have advanced. Zeno concludes that the process has an infinite number of stages,
continuing like this forever, with Achilles always trying to catch the Tortoise, but never able to accomplish
that goal.
Why does Zeno's way of looking at this race seem to lead to a surprising conclusion? At issue
seems to be the question whether a person can "do" an infinite number of "things", with Zeno suggesting
that the answer is no. (We will discuss the idea of "infinity" as an actual number in the following chapter.
In this chapter we will never really work with an actual infinity.) There are two possible approaches to
altering the sense of Zeno's "paradox".
One approach is to recognize that dealing with mathematical points is dealing with abstract
concepts, not real "things". For instance, a meter stick may have 100 markings, one for each centimeter.
From a mathematical perspective, however, the meter stick (thought of as a line segment) contains infinitely
many points. To move from one end of the stick to the other end is to traverse infinitely many points (at
least in mathematical terms). Hence the mathematical analysis of motion always involves traversing an
infinite number of points. So what? Mathematical points are abstract, not real (some would say), and hence
there is no real problem.
Another approach is to recognize that in the story of Achilles and the Tortoise, the distances run by
Achilles (and the Tortoise) are getting shorter and shorter, and hence are traversed in less and less time.
Above I said that Zeno concludes that the process continues forever. However, it may not be axiomatic
(self-evident, obvious to any rational person) that traversing an infinite number of points requires an
infinite amount of time. Perhaps there is a sense in which doing an infinite number of things in a finite
amount of time makes sense if you do successive things more rapidly.
In any case, let's consider the "arithmetic" of the problem. The sum of the distances run by Achilles
in trying to catch the Tortoise is
1 + 0.1 + 0.01 + 0.001 + 0.0001 + .........
It would seem clear that the sum of this "infinitely long" addition problem is simply 1.1111... in our decimal
notation (an advantage the Greeks did not have); this number is less than 1.2 meters, so if you believe
Achilles can run 1.2 meters, you must believe he could pass the Tortoise (sometime earlier) and be ahead
when he has run 1.2 meters. If it seems strange that we would be able to "add" infinitely many numbers,
this example links such a problem to rational numbers that require infinitely many digits to write, and most
people are willing to accept that idea. 1.1111... is a rational number since it is a repeating decimal. But
what fraction equals 1.1111…?
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Repeating Decimals
Repeating decimals represent ratonal numbers, i.e., fractions. We will illustrate the technique for
finding this fraction by using 1.111… Represent 1.1111... by the letter A (for Achilles). If we multiply A
by 10, we will simply shift the digits one decimal place, and the part of 10A which is after the decimal point
will look just like the part of A after the decimal point. A subtraction and division will complete the
solution, as you can see.
Example 1: Write 1.1111..... as a fraction, a ratio of whole numbers.
Solution: Let A = 1.1111.....
So
Then
10A
–A
9A
=
=
=
A
=
11.1111........
1.1111........
10.
10
9 .
Perhaps several more examples of the above process would be helpful. Consider the following.
Example 2: Add
0.3 + 0.03 + 0.003 + 0.0003 + .....
Solution: Let A = 0.3 + 0.03 + 0.003 + 0.0003 + ..... = 0.3333...... Then
10A
–A
9A
So
=
=
=
3
A = 9
3.3333......
0.3333......
3
1
= 3
.
This is not surprising since
1
0.3 + 0.03 + 0.003 + 0.0003 + ..... = 0.3333..... = 3
Example 3: Add 0.27 + 0.0027 + 0.000027 + ......
Solution : Let A = 0.27 + 0.0027 + 0.000027 + ...... = 0.272727........ Then
100A =
–A =
99A = 27
27.272727.....
0.272727.....
27
3
3
So A = 99 = 11 . Check out 11 on your calculator. Note that in this example
we multiplied A by 100. This was because the repeating pattern had 2 digits.
The above technique (or trick) works fine if the "addition" leads to a repeating decimal. But it
doesn't do so well if the problem looks like this:
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1
1 + 2
1
+ 4
1
+ 8
1
+ 16
187
+ ... .
This is another type of example which occurs in other versions of Zeno's paradox. Here's a sample
story. Is it possible to walk 2 miles? According to Zeno, the walk could be analyzed this way: the next
stage in the process of walking is to walk half of the remaining distance to the end. So the first stage is to
1
walk half the distance, 1 mile. The second stage is to walk half the distance remaining, 2 of 1 mile. The
1
1
1
third stage is to walk 2 of 2 mile, i.e., 4 mile. There will always be a next stage since each stage only
covers half of the distance remaining.
Converting to decimals seems to be of little help:
1 + 0.5 + 0.25 + 0.125 + 0.0625 + ........
I can see at least two problems. First, when looking at the decimal form I'm not sure what the next term
would be in the sum. There is a pattern, but it is not as easy to see as in the earlier examples. In the fraction
form, it is much easier to notice that each term is half of the previous term; that is, you get the next term by
multiplying by 0.5. A second problem is that it is not at all clear how you could write this sum as a single
decimal, repeating or not. What, for instance, would be the first digit after the decimal point? The three
numbers in that position, 5, 2 and 1, add to 8. But in the second decimal place, we have 5, 2 and 6, which
add to 13, so we will "carry" a 1. So the first digit is at least 9. If we included more terms, maybe we would
need to carry a 2. Who knows?? We seem to be unable to apply the above process to this problem.
Geometric Series
Consider the problem in this form (which could utilize fractions as easily as decimals)
1 + 0.5 + 0.25 + 0.125 + 0.0625 + ..... = 1 + 0.5 + (0.5)2 + (0.5)3 + (0.5)4 + ......
To discuss this problem, we introduce a new phrase. A partial sum will be the sum you get by
adding just a finite number of the terms. For instance, the fourth partial sum would be the sum of the first
four terms, in this case, 1 + 0.5 + 0.25 + 0.125 = 1.875. The sixth partial sum would be
1 + 0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 = 1.96875.
First we observe that any partial sum will have a value less than 2. In fact, the partial sum would
be 2 minus the last term added. (In the story, the runner always has half of the remaining distance left to
run.) So in general, the partial sum would have the formula 2 – (0.5)n. Now imagine what this would
look like if n was very, very large. The larger n gets, the closer (0.5)n approaches to 0. In calculus
language, 0 is the limit, the value which (0.5)n approaches "as n approaches infinity". Thus, 1 – (0.5)n
approaches 1 – 0 = 1. So we say
1 + 0.5 + 0.25 + 0.125 + 0.0625 + ........ = 2.
The way we were able to reason to a formula for the partial sum in the above problem was very
special. A more general approach using some algebra works in a limited number of other cases. These
special cases are called geometric series. Consider the following form for a problem (which includes all the
problems we have examined so far), where r is a number to be determined by the particular problem we
want to solve:
1 + r + r2 + r3 + r4 + ........
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Let's try to get a formula for a partial sum. The following trick bears some resemblance to what we
did above to write a repeating decimal as a fraction. Multiply the partial sum of the first n terms by 1 - r.
By the rules of algebra, this means multiplying each term in the partial sum by 1, multiplying each term in
the partial sum by r, and subtracting the results:
( 1 + r + r2 + ..... + rn ) (1 – r)
=
( 1 + r + r2 + ..... + rn ) –
=
1 + r + r2 + ..... + rn
=
1 + r – r + r2 – r2 + .... + rn –
=
1
–
–
( r + r2 + ..... + rn+1)
r
–
r2
–
....
rn –
–
rn –
rn+1
rn+1
rn+1
By dividing both sides of this equation by 1 – r , we get a formula for the partial sum of the first n
terms:
1 + r + r2 + ..... + rn =
1
– rn+1
1 – r
.
If r is a number between – 1 and 1, the number rn+1 will approach 0 "as n approaches infinity".
So we have
Theorem: If r is between – 1 and 1, then
1 + r + r2 + r3 + r4 + ........
=
1
1 – r
.
Example 4: The problem at the beginning of this section is
1 + 0.5 + (0.5)2 + ( 0.5)3 + ( 0.5)4 + ......
This appears to match the form
1 + r + r2 + r3 + r4 + ........
with r = .5 .
So the formula we derived yields to result
1 + 0.5 + (0.5)2 + ( 0.5)3 + ( 0.5)4 + ...... =
Example 5: Add 1 + .2 + (.2)
2
+ (.2)
3
1
1 – .5
=
1
.5
=
2.
+ …
Solution: This matches the form of a geometric series with r = .2
2
3
1 + .2 + (.2) + (.2) + …
1
1
1
10 5
=
= =
=
=
1 − .2 .8 810 8 4
Here's how we can use this formula to do all our previous examples. Example 1 is straightforward.
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Chapter 24
Example 1 (again): Write 1.1111.... as a fraction of integers.
1.1111…
= 1 + 0.1 + 0.01 + 0.001 + 0.0001 + .......
=
1 + 0.1 + 0.12 + 0.13 + 0.14 + .......
=
1
1 - .1
=
1
.9
=
10
9
[ r = .1]
Example 2 and 3 requires a bit more creativity. A geometric series requires a number "r " which
appears repeatedly to higher and higher powers. In order to get an "r" in this example, we will "factor out"
a number which turns out to be the "r". This is actually a sophisticated operation, a generalization of an
axiom we discussed earlier: a ( b + c) = a b + a c . While it doesn't always work, in all the cases we will
consider, we can generalize to
a ( b + c + d + ...... ) = a b + a c + ad + ..... .
In our particular setting, we will use this in the form
a + ar + ar2 + ar3 + ........ =
a ( 1 + r + r2 + r3 + ........ )
Example 2(again):
Add 0.3 + 0.03 + 0.003 + 0.0003 + .....
Solution:
0.3 + 0.03 + 0.003 + 0.0003 + .....
= 0.3 ( 1 + 0.1 + 0.01 + 0.001 + ..... )
= 0.3 ( 1 + 0.1 + 0.12 + 0.13 + ....... )
1
= 0.3 ( 1 - .1
1
= 0.3 ( .9
)
)
1
= 3
Example 3 (again):
Add 0.27 + 0.0027 + 0.000027 + ......
= 0.27 ( 1 + 0.01 + 0.0001
..... )
[ r = 0.1]
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Chapter 24
= 0.27 ( 1 + 0.01 + 0.012 ....... )
=
1
.27 ( 1 - .01 )
=
1
.27 ( .99 )
=
100
.27 ( 99
=
27
99
=
[ r = 0.01]
)
3
11 .
There are two methods by which problems can be solved using the formula for a geometric series
even if the "1" at the beginning is missing. In this technique, a "1" is placed in the beginning to make the
problem match the formula and then "1" is subtracted so that the value is not changed. Let me illustrate:
0.4 + (0.4)2 + (0.4)3 + ......
Example 6:
Add
Solution:
0.4 + (0.4)2 + (0.4)3 +
2
3
= 1 + .4 + (4) + (.4) ..... − 1
1
−1
1 − .4
10 6 4 2
1
− = =
= −1 =
.6
6 6 6 3
=
Example 7:
Solution:
1
Add 13 + 19 +
+ .....
27
1 + 1 + 1 + .....
3
9 27
1 1 2 1 3
= 1 + + ( ) + ( ) + ........− 1
3
3
3
1
3
1
=
− 1 = − 1= .
1
2
2
1− 3
The previous two problems could also be done by using the formula a(b+c+….) = ab + ac + … , in
a way that is similar to what we did earlier (but not identical).
Example 6 (again):
Solution:
Add
0.4 + (0.4)2 + (0.4)3 + ......
0.4 + (0.4)2 + (0.4)3 + ……
= 0.4 • (l + 0.4 + (0.4)2 + . . .
 1 
 1
 = .4  = 2
= 0.4 • 
3
 1 − .4 
 .6 
Example 7 (again):
1
Add 13 + 19 +
+ .....
27
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Chapter 24
Solution:
1 + 1 + 1 + .....
3
9 27
 1 1

= 13 • 1 + + + ...


3 9
 1 12

= 13 • 1 + +   + ...
3 3


=
1
1
1 1
1
•
= •
= .
3 1− 1 3 23 2
3
Recall that earlier I asserted that 0.9999..... = 1. This is now justifiable by the two techniques
studied in this chapter.
Let A = 0.9999...... Then
10A
–A
9A
=
=
=
9
A = 9
So
9.9999......
0.9999......
9
= 1 .
So 0.9999.... = 1.
Alternately,
0.9999....
= 0.9 + 0.09 + 0.009 + 0.0009 + .....
= 0.9 ( 1 + 0.1 + 0.01 + 0.001 + ..... )
= 0.9 ( 1 + 0.1 + 0.12 + 0.13 + ....... )
1
= 0.9 ( 1 - .1
1
= .9 ( .9
[ r = 0.1]
)
)
= 1.
Finally, let's consider a different story involving Zeno's paradox which all of us see happen
everyday. Consider a clock with an hour hand and a minute hand (digital clocks won't work for this story).
Suppose it is 6 o'clock (AM or PM, it doesn't matter). The minute hand is pointed straight up at the 12, and
the hour hand is pointed straight down at the 6. By 7 o'clock, the minute hand will have moved past the
hour hand. When does that happen? Or, if you believe Zeno, does it really happen??
Here's Zeno's version. The minute hand must first move to where the hour hand was, i.e., to
straight down at the 6. But during the 30 minutes it takes to do that, the hour hand will have moved to halfway between the 6 and the 7. Then the minute hand will need to move to that position, while the hour hand
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moves further on. The process goes on forever, hence the minute hand will never catch the hour hand.
What's interesting about this version of Zeno's paradox is how clearly time is involved in the problem.
To analyze this problem, first notice that the minute hand makes one complete revolution in an
1
hour, whereas the hour hand only goes 12 of the way around in an hour. So the hour hand is moving at a
1
speed which is 12 times the speed of the minute hand. Now let's add the times involved in the stages of
the movement of the hands of the clock.
The first stage (the minute hand gets to where the hour hand started) takes 30 minutes. During this
1
time, the hour hand will have moved 12 (30) "minutes" = 2.5 "minutes" of distance, i.e., the hour hand will
be pointed at 32.5 minutes on the face of the clock, half way between the 6 and the 7. The second stage then
takes 2.5 minutes, while the minute hand moves from the 6 to where the hour hand was, pointed toward
32.5 minutes. But during this time, the hour hand will have moved
1
1
1
(12 ) (12 ) (30) = (12 ) 2 (30) "minutes" of distance, i.e., the hour hand will be just a little farther
along. The process continues, leading to
1
1
1
1
30 + 12 (30) + (12 )2 (30) + (12 ) 3 (30) + (12 ) 4 (30)+ ........
1
1
1
1
= 30 [ 1 + 12 + (12 )2 + (12 ) 3 + (12 ) 4 ....... ]
= 30 (
1
1
1- 12
)
=
1
30 ( 11
12
) =
12
30 ( 11 )
=
1
(r = 12 )
360
11
= 32.72727272........
That is, the minute hand will catch the hour hand at 32.727272.... minutes past 6 o'clock, when they will both
be pointing at 32.727272..... minutes.
By the way, we could also derive this time algebraically. We are looking for the time t (in minutes
after 6 o'clock) when the minute hand and the hour hand will be at the same position. Given the "head
start" which the hour hand has, the two hands will move to equal positions. So we can get an algebra
equation this way:
movement of minute hand
t
=
=
11
( 12 ) t
"head start" + movement of hour hand
1
30
+
( 12 ) t
=
t
12
= 30 (11 )
30
=
360
11
Fibonacci Sequence
Can you determine the next term in the sequence 1, 1, 2, 3, 5, ... ? You are correct if you said the next term is
8. The terms are obtained by adding the two previous terms. 8 is the sum of 3 and 5.
Continuing this process we have 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,... . All sequences formed in this
manner are called Fibonacci type sequences, but the one given here is the Fibonacci Sequence. Fibonacci
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193
was a mathematician who lived in Wesstern Europe in the 1200’s. He played a significant role in the
introduction of Hindu-Arabic numerals to a society in whick Roman numerals were deeply entrenched.
Fibonacci formed this sequence as an answer to a problem given to him by a friend having to do with the
number of rabbits one would have at any given time given a reproduction rate.While this sequence is of
some interest in its own right, what makes this sequence so special is its occurrence in a variety of places in
nature. If one looks at the center of a sunflower, for example, we see that the seeds of the sunflower are
arranged in two sets of spirals, one set clockwise and one counterclockwise. If we were to count the spirals,
we would find that the number of spirals clockwise and the number counterclockwise are consecutive terms
in the Fibonacci Sequence. The same is true of pine cones, pine apples, and daisies. An octave on the piano
keyboard contains 5 black keys and 8 white keys which are consecutive terms in the Fibonacci Sequence.
The Sequence also appears in the spiral of a Sea Nautilus as well as in a number of other places.
I once read that a rectangle with sides whose measures are consecutive terms in the Fibonacci Sequence is
more pleasant to look at than other rectangles. Such a rectangle is called a Golden Rectangle. This appeared
to me to be a rather bold statement since I am of the opinion that I deccided for myself what is most
pleasant to look at and believed everyone else did the same. To show that the statement was wrong, I
determined to test my students. Over the course of several semesters I gave each of my students a piece of
paper containing four rectangles one of which was a Golden Rectangle. They were instructed to place an
“X” in the rectangle that was most pleasant to look at. Every semester, the Golden Rectangle was selected
most often. Apparently, the Golden Rectangle is more pleasant to look at. These findings have implications
for artists, photographers, architects, etc.
We read in Romans 1:20 that God’s invisible qualities are clearly seen through His creation. I would suggest
that, in cases such as the above, the language of mathematics gives us the surest way of communicating
what we have observed.
A Puzzle
Another interesting sequence arises from the following question. Is it possible to stack blocks of wood of
equal measure on a table in such a way that one block will eventually be completely beyond the edge of the
table? Pictorially, is this kind of thing possible?
The basic principle is that there must be (at least) as much weight of wood “on” the table as there is “off”
the table.
1
1
the block will be on the table and
will be off
2
2
the table. If two blocks are to be balanced, then the weight equivalent to one block must be on the table and
3
1
one block off the table. In other words, we will have
of the weight of one block on the table and
off
4
4
1
3
while the other block will be
on the table and
off the table. To understand what is going on, it might
4
4
With one block of wood balanced on the edge of the table,
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Chapter 24
194
help to draw a chart.
One block:
1
off the table
2
Two blocks:
the second block will be
Three blocks:
the third block will be
Four blocks:
the fourth block will be
3 1 1
1
= + off the table and the first will be .
4 2 4
4
1 1 1 11
+ + =
off the table.
2 4 6 12
1 1 1 1 25
+ + + =
beyond the edge of the table.
2 4 6 8 24
This means that one block of wood is now completely beyond the edge of the table! Can we two blocks
beyond the edge? Three?
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Chapter 24
The infinite series formed is
195
1 1 1 1 1
1
+ + + + + + ... . If we add all these terms together, then
2 4 6 8 10 12
1 1 1 1 1
1
1
1
1
1
1
1
+ + + + + +
+ + + ... +
+ ... + ... =
2 4 6 8 10 12 14 16 18
32 34
64
1  1 1
1 + +  +
2  2 3
1  1 1 1 1 1
1 
 +  + + +  +  + ... +  + ... ≥
4  5 6 7 8 9
18  
1  1  1 1  1 1 1 1  1
1 
1 + +  +  +  + + +  +  + ... +  + ... =
2  2  4 4   8 8 8 8   16
16  

1 1 1 1 1 1
 + + + + + ... = =
2 2 2 2 2 2

1
[1 + 1 + 1 + 1 + 1 + ...] = ∞
2
Mathematically speaking, we can get as many blocks of wood as we want beyond the edge of the table.
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Chapter 24
Chapter 24
196
Homework
1. Use the first technique presented in this chapter to write the given repeating decimal as a fraction. Check
your answer using your calculator. Write your answer as a reduced fraction.
a.
b.
c.
d.
1.23232323.....
0.27272727......
0.55555555.....
2.34234234234...
2. Use the first technique presented in this chapter to write the given repeating decimal as a fraction. Check
your answer using your calculator. Write your answer as a reduced fraction.
a.
b.
c.
d.
3.111111.....
0.252525......
0.357357357.....
0.833333...
3. Find the sums of the following infinite series. Write your answer as a reduced fraction.
a. 1 + 0.36 + 0.0036 + 0.000036 + .....
b. 0.57 + 0.0057 + 0.000057 + ......
c. 0.46 + 0.0046 + 0.000046 + ......
4. Find the sums of the following infinite series. Write your answer as a reduced fraction.
a. 1 + 0.63 + 0.0063 + 0.000063 + .....
b. 0.75 + 0.0075 + 0.000075 + ......
c. 0.64 + 0.0064 + 0.000064 + ......
5. Find the sums of the following infinite series. Write your answer as a reduced fraction.
a.
b.
c.
d.
e.
1 + 0.2 + 0.22 + 0.23 + .......
1 + 0.15 + 0.152 + 0.153 + .......
1 + 0.36 + 0.362 + 0.363 + .......
0.16 + 0.162 + 0.163 + .......
0.58 + 0.582 + 0.583 + .......
6. Find the sums of the following infinite series. Write your answer as a reduced fraction.
a.
b.
c.
d.
e.
1 + 0.6 + 0.62 + 0.63 + .......
1 + 0.75 + 0.752 + 0.753 + .......
1 + 0.9 + 0.92 + 0.93 + .......
0.76 + 0.762 + 0.763 + .......
0.55 + 0.552 + 0.553 + .......
7. EXTRA CREDIT: Suppose you drop a ball from 1 meter above the floor. Each time the ball hits the floor
after falling from a height of h meters, it rebounds to a height of 0.8h. (So you drop it from 1 meter, it hits
the floor, it rebounds to 0.8 meters, falls to the floor, rebounds to (0.8) (0.8) = 0.82 meters, etc.) Find the total
distance the ball travels up and down.
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Chapter 24
Answers:
122
1. a. 99
b.
3
11
28
2. a. 9
25
b. 99
135
3. a. 99
b.
19
33
4. a.
18
11
b.
25
33
5. a.
5
4
b.
20
17
6. a.
5
2
b. 4
c. 10
7. Hint: Start with 1 + 0.8 + .8 + (0.8)2 + (0.8)2 + …
then add like terms, factor out the 2, factor out the 0.8, use formula.
197