[PPT] - Ramana Andra Thomas Jackson FTLOMACS OCT 14, 2017 AGENDA PowerPoint Presentation

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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 𝜌 = 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 − 𝑦2𝑒𝑦 = 𝜌 2

∞ −∞

𝜌 4 = 1 − 1 3 + 1 5 − 1 7 + 1 9 − ⋯ = (−1)𝑙 2𝑙 + 1

∞ 𝑙=0

3.1415926535897932384626433832795…….

TRADITIONAL REPRESENTATIONS OF NUMBERS

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𝜒 = 1 + 5 2 = 1.6180339887 … 𝜒 = 1 + 1 + 1 + ⋯ e =

1 𝑜! ∞ 𝑜=0

= 1 +

1 1 + 1 1•2 + 1 1•2•3 + 1 1•2•3•4 + ⋯

e = lim

𝑜→∞(1 + 1 𝑜)𝑜 = 2.718281828459…

2 = 1.41421356237…

TRADITIONAL REPRESENTATIONS OF NUMBERS

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4 𝜌 = 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 + 1 6 + …

Φ = 1 +

1 1 +

1 1 + 1 1 + 1 1 + 1 1 + 1 1 + …

EXAMPLES OF CF’S

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𝜌 2 = 1 - 1 3 −

2•3 1− 1•2 3 − 4•5 1 − 3•4 3 − 6•7 1− 5•6 3 − …

2 = 1 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + 1 2 + ⋯

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 +

𝑐1 𝑏2+

𝑐2 𝑏3 + 𝑐3 𝑏4 + 𝑐4 𝑏5 + 𝑐5 𝑏6…

Simple Continued Fraction (bi = 1) a1 +

1 𝑏2+

1 𝑏3 + 1 𝑏4+ 1 𝑏5 + 1 𝑏6…

= [a1; a2, a3, a4, …] = [a1: a2, a3, a4, …] Convergents

𝑞𝑗 𝑟𝑗= [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

𝑏 𝑐 = 1

𝑐 𝑏

to move forward.

FINITE RATIO EXAMPLE

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47 13 = 3 + 1

13 8

= 3 +

1 1 + 5

8

= 3 +

1 1+ 1

8 5

= = 3 +

1 1+

1 1 + 3 5

47 13 = 3 + 1 1+

1 1 + 1 5 3

= 3 +

1 1+

1 1 + 1 1+2 3

= 3 +

1 1+

1 1 + 1 1+2 3

47 13 = 3 + 1 1+

1 1 + 1 1+1 3 2

= 3 +

1 1+

1 1 + 1 1+ 1 1+1 2

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 𝟐+

1 𝟐 + 1 𝟐+ 1 𝟐+1 𝟑

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 + 1 φ

φ = 1 +

1 1+

1 1 + 1 1+⋯

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 + 1 1

, 3 +

1 1+

1 1 + 1 1+1 1

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 =

𝑞𝑗 𝑟𝑗 represent

the convergents for a given fraction with i ≥ 0.

So c1 = 𝑞1

𝑟1 = a1 = 𝑏1 1

c2 = 𝑞2

𝑟2 = a1 + 1 𝑏2 = 𝑏1𝑏2+1 𝑏2

c3 =

𝑞3 𝑟3 = a1 + 1 𝑏2+ 1

𝑏3

=

𝑏1𝑏2𝑏3+𝑏1+𝑏3 𝑏2𝑏3+1

CONVERGENTS

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c4 =

𝑞4 𝑟4 = a1 + 1 𝑏2+

1 𝑏3+ 1 𝑏4

=

𝑏1𝑏2𝑏3𝑏4+𝑏1𝑏2+𝑏1𝑏4+𝑏3𝑏4+1 𝑏2𝑏3𝑏4+𝑏2+𝑏4

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

𝑢𝑗 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 + 1 2 + ⋯ = [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 +

1

1 2−1

= 1 +

1 2+1

But 2 + 1 = 2 + ( 2 − 1) = 2 +

1

1 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 𝑜2 + 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 [𝑐; 𝑐, 𝑐, 𝑐, 𝑐, 𝑐,… ]

We need to determine the value of B = [b; b, b, b, b, b, …] Just like for A, we can rewrite B = b +

1 [𝑐; 𝑐, 𝑐, 𝑐, 𝑐, 𝑐,… ] 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 =

𝑐+ 𝑐2+4 2

Thus A = a +

2 𝑐+ 𝑐2+4 = a - 𝑐− 𝑐2+4 2

=

2𝑏−𝑐 2

+

𝑐2+4 2

So we have

2𝑏−𝑐 2

+

𝑐2+4 2

= [a; b, b, b, b, …] If b = 2a, then 𝑏2 + 1 = [𝑏; 2𝑏, 2𝑏, 2𝑏, 2𝑏, … ]

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 + 1 2 + ⋯ = [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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