Download Module 6 Chapters 10 and 11 Continued Fractions and Fibonacci

Module 6
Chapters 10 and 11
Continued Fractions and Fibonacci numbers.
Continued Fractions
A continued fraction is a way to represent numbers that are improper fractions or, in some cases,
transcendental numbers. A continued fraction takes a whole LOT of room on a page so we
quickly move to an alternate representation.
For example:
87
1
46
 2
 2.12195  2
1
41
87
8
5
Improper fraction (fully reduced); continued fraction; decimal; mixed number
Each of these representations has its uses!
The continued fraction is called “finite and simple” because it ends and each numerator is 1.
There are 2 fraction bars in the continued fraction of the number.
Now, proper fractions do not have a continued fraction representation – only mixed number and
improper fractions.
0.25 =
1
4
no chance for a continued fraction here!
Note that we use the Euclidean Algorithm as a way to get the CF for an improper fraction!
1
87
41
2
41 87
8
5 41
5
15
r :5
r :1
r :0
There is a second way to get CFs
87
5
5
1
1
 2  2  2
 2
41
1
41
41
41
8
5
5
Both ways are similar. You really do need a fully reduced fraction to do this work –
(numerator, denominator) = 1.
Another example:
70
1
 2.12  2 
1
33
8
4
2
70
33
2
33 70
r:4
8
4 33
r :1
4
14
2
r :0
1
8
1
4
Simple finite continued fraction!
Another one:
83
 2
35
2
1
1
1
1
2
1
4
Four fraction bars! Simple finite continued fraction.
3
Popper 11, Question 1
Every fraction can be represented as a continued fraction.
A
True
B
False
4
Continued fractions and Diophantine equations
Here’s a surprising tie in between continued fractions and Diophantine equations!
Here’s an example that is not in the book – it’s the same process, though.
Step 1
Step 2
Step 3
Step 4
Step 5
Step 5
Get the continued fraction
Chop off the last fraction
Reconstitute a new improper fraction
Subtract the new fraction from the original one
Multiply and subtract the numerators getting an equation
Multiply both sides of the equation by the common denominator
Take
9007
29
Notice that I always use a prime number in my improper fractions – it’s a guaranteed way to
make sure the fraction is fully reduced!
The CF work - Step 1:
5
310
29 9007
r :17
1
17 29
1
12 17
r :12
r :5
2
5 12
r:2
2
2 5
r :1
2
12
r :0
1
310 
1
1
1
1
2
1
2
1
2
6
Step 2: chop of the ½:
1
310 
1
1
1
2
1
1
1
1
1
 310 
1
2
1
2
1
2
1
2
Step 3 Reconstitute the new improper fraction:
1
310 
1
2
 310 
1
1

1
1
1
1
5
7
1
 310 
1
1
1
2
 310 
1
1
1
7
5
1
5
2
 310 
1
 310 
1
1
1
2
5
1
5
 310 
12
12
5
3727
12
7
Step 4
Subtract the new fraction from the original one
9007 3727 9007(12)  3727(29)


29
12
29(12)
Step 5
Multiply and subtract the numerators getting an equation
9007 3727 9007(12)  3727(29) 108084  108083
1




29
12
29(12)
29(12)
29(12
Step 5
Multiply both sides of the equation by the common denominator
108084 108083  1
Which is:
9007(12)  3727(29)  1
Now look at it hard:
( x0 , y0 )  (12, 29)
9007 x  3727 y  1
There’s a Diophantine equation! With the initial solution showing!
Popper 11, Question 2
Diophantine Equations have continued fractions as solutions.
A
True
B
False
8
Now for infinite repeating continued fractions
All irrationals of the form number have infinite repeating continued fractions.
We’ll definitely need new notation for this!
10 is a nonrepeating, nonterminating decimal…no place to put a repeat
superscripted bar…
Example
1
10  3 
1
6
6
1
6  ...
10  [3, 6 ]
How cool is that? A nice compact and REPEATING representation?
The square root of 10 has period 1.
Example
1
3  1
 [1,1, 2 ]
1
1
1
2
1
1
2  ...
Period 2
Other common notation:
1,[1, 2]
1,1, 2
9
57  [7,1,1, 4,1,1,14]
Example:
period 6
Popper 11, Question 3
[1,1, 2,3]
A
3
B
4
has period
We’ll explore only one theorem on how to find these. There has been a LOT of work done and
lots is known.
Theorem on page 127
If an irrational number has the form
fraction is [a, 2a].
a 2  1 where a is a natural number, then its continued
Let’s illustrate this theorem!
12  1  2  [1, 4]
a=1
1
2  1.414168...  1 
1
4
4
 [1, 4]
1
4  ...
10
Will this work for 11?
a=3
10
a=4
17
nope!
Let’s check the theorem:
5 = 22  1
From a calculator:
5  2.2360679...
From the theorem:
5  [2, 4]
2 + ¼ = 2.25
2
1
1
4
4
= 2.2352941
…
2 + 6 fraction bars = 2.2360648
Take a limit to inifinty…sure!
Popper 11, Question 3
26  [5,10]
A
True
B
False
11
Now let’s take a minute and look at continued fractions from a set perspective
Reals
Algebraic:
Rationals (finite simple cf) and some irrationals (infinite repeating cf)
Transcendental:
Some have cf but they’re definitely not simple!
Example for a transcendental number:

4

1
12
1
32
2
52
2
72
2
2  ...
The nice thing is that you can see the pattern. The problem is that it’s not a repeating pattern so
there’s nowhere for a repeat bar.
This is an infinite complicated continued fraction!
12
Example
e, the Euler number
e is irrational and approximately 3.7, non-terminating and non-repeating…
It shows up in lots of natural phenomena and in continuously compounded interest in man-made
phenomena.
It has an infinite complicated continued fraction with a visible pattern:
1
e  2
1
1
1
2
1
1
4  ...
Again, no place for a repeat bar…
But for the first time there’s a pattern to see with some of these irrational numbers!
Popper 11, Question 5
Only algebraic numbers have continued fraction representations.
A
True
B
False
Homework hint: Google Continued Fraction for homework help!
End video 1
13
Chapter 11: Fibonacci Numbers
Pages 133 – 144
Fibonacci was an Italian mathematician. He discovered a sequence of natural numbers created
by addition that shows up frequently in natural phenomena. If you google the topic, you’ll get
millions of hits. There is even a Fibonacci Society!
The sequence:
first
f1
1
second
f2
1
third
f3
2
fourth
f4
3
fifth
f5
5
sixth
f6
8
seventh ...
f n  f n1  f n2
n3
Why 3?
Well if you try it with n = 1, you get f1  f 0  f 1 and that’s not ok.
Side note:
on the final you’ll have a calculator, a list of primes, and a list of
Fibonacci numbers on links. If you don’t see the links, get some help
from the proctors!
Now let’s look at n = 10
f10  f9  f8 so we really need a list of these to do things!
Popper 12, Question 1
Given f 58
A
f 50
B
f 55
C
f 60
D
f 56
what is f n2 ?
14
Now let’s work with our formula a little bit:
f n  f n 1  f n  2
f n  f n 1  f n  2
The second equation is used almost as much as the first equation so be comfy with either, please.
Popper 12, Question 2
f11  89
f13  233
f12  _____
A
322
B
122
C
233
89
D
89 2  1
E
12
Let’s take a really nice continued fraction and see where it leads us:
it doesn’t get simpler than this one…
[1,1]
1
1
1
1
1
1
1  ...
15
Does this one converge to a number?
1
1/1
f 2 f1
1
1+1
2/1
f3 f 2
2
3/2
f3 f 4
1.5
5/3
f5 f 4
1.6
1
1
11
1
1
1
1
11
…
1.625
…
1.618
…
non-terminating and non-repeating decimal

Which, after some algebra turns out to be

1 5
2
Let’s look at that algebra
1
x
1
1
1
x  1
1
x
(page 133)
If I drop below the first fraction bar, notice that I’ve got x again
1
1  ...
incredible but true
16
Now, multiply both sides by x
And then subtract to get a quadratic = 0
which we know is NOT zero!
x2  x  1
x 2 x 1  0
Now let’s use the quadratic formula with A = 1, B = −1, and C = −1.
b  b 2  4ac
x
2a
x
1  1  4( 1)
2
x
1 5
2
And we ignore the negative answer because it’s a negative number and not the result of adding
positives.
Now, in the homework you need to look up PHI, this number. It’s used often in art!
This is an amazing result because the algebra agrees with our work on the continued fraction.
And the trick to getting the algebra to work was so clever! Can you see how we’d do this to any
repeating continued fraction?
Popper 12, Question 3
f n 1
 1.6
fn
A
True
B
False
n  100
17
Properties of Fibonacci Numbers
Theorem:
n
f
i 1
i
 fn2  1 .
The big capital sigma means add from the bottom index to the top.
So this says, the sum of n Fibonacci numbers is f n 2  1 .
Let’s illustrate this theorem. I’ll go up to n = 8.
n
f
i 1
8
i
 f n  2  1   fi  f10  1
i 1
Adding the first 8 Fibonacci numbers:
1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 = 54
f10  55
Checking the list
55  1  54
Perfect! It works.
Another illustration:
20
f
i 1
i
 f 22  1
n  20
17, 710  17, 711  1
18
On the left, well, I added them but didn’t want to type them!
Now there’s a nifty illustration of WHY this works on page 136. It’s a process called
“telescoping”. And it has lots and lots of cancelling in it. It’s actually fun to check out.
Popper 12, Question 4
147
f
i 1
i
 f149  1
A
True
B
False
Another property and theorem:
f n 12  f n  f n  2  (1) n
Let’s illustrate this using n = 5:
f n 12  f n  f n  2  (1) n 
f 6 2  f 5  f 7  (1)5
82  5(13)  1
64  65  1
And perhaps most astonishing of all:
The Fibonacci numbers can be found in Pascal’s Triangle on diagonals!
The
Page 137
” rising diagonals”
19
1
1
1
1
2
1
3
1
1
3
4
6
1
4
1
And on page 138, we have the Fibonacci numbers and Areas
5
8
2
3
1
1
20
And from this we find another theorem to illustrate:
n
f
i 1
2
i
 f n  f n 1
Doesn’t that look awful?
Let’s do it with n = 6 – we’ve got the first 6 Fibonacci numbers in the area above.
The height is 8 ( f 6 ) and the length is 13 ( f 7 ). Yep!
5
8
2
3
1
1
21
Homework hint:
We’ll go back to geometry and learn about the “geometric mean”
If we have “b is the geometric mean of a and c” for 3 natural numbers we mean the result of this
proportionality:
a b
  ac  b 2  b  ac
b c
we only want the positive number
Now we switch subjects for the very last time this semester at the bottom of page 139.
Let’s take a line of length 1 and split it into 2 carefully chosen pieces: x and 1 – x in a way that
the ancient Greeks thought was particularly pleasing:
1
0
x
1-x
Now we want
1
x

x 1 x
1  x  x2
x2  x 1  0
x
1  5
2
22
We only want the positive number so we ignore the minus!
Notice how close this is to Phi!
Now take the reciprocal of both sides:
1
2

x 1  5
In the book are the steps that show you that for the reciprocal of x, we do have Phi!
The ancient Greeks called this way of breaking up a line: The Golden Section.
It turns out that the rectangles built with the Fibonacci numbers become closer and closer to
length Phi with a height of 1 and a break at   1 as you make them larger and larger by adding
on to the right and top. A Golden Rectangle has sides  :1 and is supposed to be perfectly
pleasing to the eye.
This is pretty cool really. The Golden Rectangle, as it is known, is used often by architects and
artists.
The Pentagram also uses an arithmetic mean. And I leave that to you for homework!
Meanwhile, it has been a pleasure getting to know you over the semester. Please do remind me
to publish Hints for the Final as soon as possible!
Thanks!
Ms. Leigh
Popper 12, Question 5
The final is in CASA Testing by appointment.
A
True
B
False
23
24