Continued Fractions: Invisible Patterns Ramana Andra ......These continued fractions are not...
Transcript of Continued Fractions: Invisible Patterns Ramana Andra ......These continued fractions are not...
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Continued Fractions: Invisible Patterns
Ramana Andra
Thomas Jackson
FTLOMACS
OCT 14, 2017
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• Representations on Numbers
• Continued Fraction Notation
• Finite Ratio Example
• Convergents
• Radicals
• Algebraic Numbers
• π
• Java Programs
• Invisible Connections
AGENDA
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Infinite Series, Periodic and Nonperiodic Decimal Expansions,
Integrals
𝜋 = 1 +1
2 + 1
3+1
4−1
5+1
6+1
7+1
8+1
9+ −
1
10+1
11+1
12−1
13…
1 − 𝑥2𝑑𝑥 =𝜋
2
∞
−∞
𝜋
4= 1 −
1
3+1
5−1
7+1
9− ⋯ =
(−1)𝑘
2𝑘 + 1
∞
𝑘=0
3.1415926535897932384626433832795…….
TRADITIONAL
REPRESENTATIONS OF NUMBERS
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𝜑 =1 + 5
2= 1.6180339887 …
𝜑 = 1 + 1 + 1 + ⋯
e = 1
𝑛!∞𝑛=0 = 1 +
1
1+1
1•2+1
1•2•3+
1
1•2•3•4+⋯
e = lim𝑛→∞(1 +
1
𝑛)𝑛 = 2.718281828459…
2 = 1.41421356237…
TRADITIONAL
REPRESENTATIONS OF NUMBERS
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𝜋 = 1 +
12
2 + 32
2 + 52
2 + 72
2 + 92
2 + 112
2 + …
EXAMPLES OF CF’S
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e = 2 +1
1 + 1
2 + 1
1 + 1
1 + 1
4 + 1
1 + 1
1 + 16 + …
Φ = 1 +1
1 + 1
1 + 1
1 + 1
1 + 1
1 + 11 + …
EXAMPLES OF CF’S
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𝜋
2 = 1 -
1
3 − 2•3
1− 1•2
3 − 4•5
1 − 3•4
3 − 6•7
1− 5•63 − …
2 = 1 +1
2 +1
2 +1
2 +1
2 +1
2 +12 + ⋯
EXAMPLES OF CF’S
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CONTINUED FRACTIONS TYPES
Continued Fractions
Simple
Finite Infinite
Periodic Non-
Periodic
General
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General Continued Fraction
a1 + 𝑏1
𝑎2+
𝑏2
𝑎3 +
𝑏3
𝑎4 +
𝑏4
𝑎5 + 𝑏5𝑎6…
Simple Continued Fraction (bi = 1)
a1 + 1
𝑎2+
1
𝑎3 +
1
𝑎4+
1
𝑎5 + 1 𝑎6…
= [a1; a2, a3, a4, …] = [a1: a2, a3, a4, …]
Convergents
𝑝𝑖
𝑞𝑖
= [a1; a2 a3, a4, a5,…an]
CONTINUED FRACTIONS FORMS
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π = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, ...]
e = [2; 1, 2, 1, 1, 4, 1, 1, 6, …]
φ = [1; 1, 1, 1, 1, 1, 1, … ]
FAMOUS CONTINUED FRACTIONS
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Consider the fraction 47
13. Can we find the continued fraction (cf.) for this
finite ratio?
We should consider breaking up the fraction into mixed form.
So 47
13 = 3 +
8
13
We need the numerator to be one for our cf. form so we can apply the
following algebraic identity 𝑎
𝑏=1𝑏
𝑎
to move forward.
FINITE RATIO EXAMPLE
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47
13 = 3 +
113
8
= 3 + 1
1 + 5
8
= 3 + 1
1+ 185
= = 3 + 1
1+ 1
1 + 35
47
13 = 3 +
1
1+ 1
1 + 153
= 3 + 1
1+ 1
1 + 1
1+23
= 3 + 1
1+ 1
1 + 1
1+23
47
13 = 3 +
1
1+ 1
1 + 1
1+132
= 3 + 1
1+ 1
1 + 1
1+1
1+12
FINITE RATIO EXAMPLE
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This procedure is an example of an algorithm
called the Euclidean Algorithm which was
developed by Euclid in his famous book The
Elements. We can also look at the problem from
geometrical and coding perspectives.
FINITE RATIO EXAMPLE
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GEOMETRIC CONSTRUCTION
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Geometric Construction
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Let’s consider the calculations already performed for the fraction 47
13 .
47 = 3(13) + 8
13 = 1(8) + 5
8 = 1(5) + 3
5 = 1(3) + 2
3 = 1(2) + 1
2 = 2(1) + 0
EUCLIDEAN ALGORITHM
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We continue this operation on the two numbers and it will always stop when
the length of the shortest rectangle is one and the remainder after the last
triangle has been divided is zero.
We can find the cf. form from the leading numbers from the list:
47
13= 3 +
1
𝟏+ 1
𝟏 + 1
𝟏+1
𝟏+1𝟐
EUCLIDEAN ALGORITHM
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We all know that φ is derived from the roots of the quadratic equation
φ 2 - φ - 1 = 0
φ = 1 + 1
φ
φ = 1 + 1
1 + 1
φ
φ = 1 + 1
1+ 1
1 + 1φ
φ = 1 + 1
1+ 1
1 + 11+⋯
φ = [1; 1, 1, 1, 1, 1, 1, 1, …]
PHI DERIVATION
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We can look at the partial cfs for the fraction 47
13 to find better approximations
to the exact value of this fraction. These values are called convergents.
3, 3 + 1
1 , 3 +
1
1+ 1
1
, 3 + 1
1+ 1
1 + 11
, 3 + 1
1+ 1
1 + 1
1+11
3, 4, 7
2, 11
3, 18
5, 47
13 = 3, 4, 3.5, 3.666.., 3.6, 3.65…
CONVERGENTS
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We can generalize this procedure. Let ci =𝑝𝑖
𝑞𝑖 represent
the convergents for a given fraction with i ≥ 0.
So c1 =𝑝1
𝑞1
= a1 = 𝑎1
1
c2 = 𝑝2
𝑞2
= a1 + 1
𝑎2
= 𝑎1𝑎2+1
𝑎2
c3 = 𝑝3
𝑞3 = a1 +
1
𝑎2+1
𝑎3
= 𝑎1𝑎2𝑎3+𝑎1+𝑎3
𝑎2𝑎3+1
CONVERGENTS
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c4 = 𝑝4
𝑞4 = a1 +
1
𝑎2+1
𝑎3+1𝑎4
= 𝑎1𝑎2𝑎3𝑎4+𝑎1𝑎2+𝑎1𝑎4+𝑎3𝑎4+1
𝑎2𝑎3𝑎4+𝑎2+𝑎4
So we can “see” the following recursions for the
convergent terms if we look carefully:
pi = aipi-1 + pi-2
qi = aiqi-1 + qi-2
CONVERGENTS
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Using these recursion formulas, the convergents for the golden ratio φ are
1
1, 2
1,3
2, 5
3, 8
5, 13
18, …..
We can see that the numbers in the numerators and denominators are terms from the Fibonacci sequence!
We should notice that
pi = aipi-1 + pi-2 = (1)pi-1 + pi-2 = pi-1 + pi-2
qi = aiqi-1 + qi-2 = (1)qi-1 + qi-2 = qi-1 + qi-2
and ci = ti+1
𝑡𝑖
where ti are the terms of the Fibonacci sequence.
CONVERGENTS
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2 = 1 +1
2 +1
2 +1
2 +1
2 +1
2 +12 + ⋯
= [1; 2, 2, 2, 2, 2, 2, … ]
SQUARE ROOT 2
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Let’s see how we can derive this cf representation.
2 = 1 + ( 2 - 1) = 1 +11
2−1
= 1 + 1
2+1
But 2 + 1 = 2 + ( 2 − 1) = 2 + 11
2−1
= 2 +1
2+1
Thus 2 = 1 + 1
2+1
2+1
If we use this rationalization method again repeatedly, we find the
cf. of 2 consists of a single 1 followed entirely by 2’s.
Thus 2 = [1; 2, 2, 2, 2, 2, … ]
SQUARE ROOT 2
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Is there a pattern in the representation?
Can we generalize numbers of the form 𝑛2 + 1?
Let’s give it a go!
SQUARE ROOT 17
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Let’s consider a simple case and suppose
A = [a; b, b, b, b, b,…]
We can rewrite A in the form A = a + 1
[𝑏; 𝑏, 𝑏, 𝑏, 𝑏, 𝑏,… ]
We need to determine the value of B = [b; b, b, b, b, b, …]
Just like for A, we can rewrite B = b + 1
[𝑏; 𝑏, 𝑏, 𝑏, 𝑏, 𝑏,… ]
PERIODIC CF’S
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So we have B = b + 1
B which can be rewritten as B2 – bB -1 = 0
Using the Quadratic Formula, we have B = 𝑏+ 𝑏2+4
2
Thus A = a + 2
𝑏+ 𝑏2+4 = a -
𝑏− 𝑏2+4
2 = 2𝑎−𝑏
2+𝑏2+4
2
So we have 2𝑎−𝑏
2+𝑏2+4
2 = [a; b, b, b, b, …]
If b = 2a, then 𝑎2+ 1 = [𝑎; 2𝑎, 2𝑎, 2𝑎, 2𝑎, … ]
PERIODIC CF’S
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Let’s try to derive the continued fraction for
3 on the board together!
SQUARE ROOT 3
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3 = 1 +1
1 +1
2 +1
1 +1
2 +1
1 +12 + ⋯
= [1; 1, 2, 1, 2, 1, 2, … ] .
SQUARE ROOT 3
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PERIODIC CF’S
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What about cube roots?
These continued fractions are not periodic.
We will find their cf. representation
from their decimal first.
From the decimal part,
we will repeatedly invert the fractional part.
CUBE ROOTS
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CUBE ROOTS
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INFINITE CONTINUED
FRACTIONS
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Π 1/4
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Π
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CUBE ROOTS
We can find the cube roots using the bisection method.
Easy for high school students to understand.
Can be easily coded in a programming language.
We have some simple Java programs to demonstrate.
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Rational Numbers
Square Root
Cube Root
π
JAVA PROGRAMS
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How do we evaluate Pi using a formula?
We use an identity that came up recently in our Math club
meeting.
π = 16 arctan 1
5 − 4 arctan
1
239
FORMULA FOR Π
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Ramanujan proved two connections between π, e and φ:
RAMANUJAN
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A CONTINUED FRACTION
APPROXIMATION OF THE GAMMA
FUNCTION
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You Tube video: Continued Fractions - Professor John Barrow
https://www.youtube.com/watch?v=zCFF1l7NzVQ&t=670s
The Topsy-Turvy World of Continued Fractions:
https://www.math.brown.edu/~jhs/frintonlinechapters.pdf
Cube Root of a Number:
http://www.geeksforgeeks.org/find-cubic-root-of-a-number/
Gamma Function:
http://www.sciencedirect.com/science/article/pii/S0022247X12009274
A Continued Fraction Approximation of the Gamma Function:
http://www.sciencedirect.com/science/article/pii/S0022247X12009274
https://ac.els-cdn.com/S0022247X12009274/1-s2.0-S0022247X12009274-main.pdf?_tid=4981f0e4-b07a-11e7-bb6a-
00000aab0f26&acdnat=1507942672_ff44d4811825271fcf37c06314338fe1
Java Code:
https://dansesacrale.wordpress.com/2010/07/04/continued-fractions-sqrt-steps/
Continued Fractions:
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/cfCALC.html
REFERENCES
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