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Fibonacci Numbers and Binet Formula (An Introduction to Number Theory) By: (The Ladies) 2

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Recurrence Sequence each further term of the sequence is defined as a function of the preceding terms (starting seed and rule) Fibonacci sequence (1,1,2,3,5,8,13,21,34,55...) Lucas sequence (2,1,3,4,7,11,18,29,47,76) take 3+8 (1+2)+(3+5) (1+3)+(2+5) (4)+(7) - can be shown to hold in general

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Mathematical induction Fibonacci (1,1,2,3,5,8,13...) 1,1+1,1+1+2,1+1+2+3,1+1+2+3+5,1+1+2+3+5+8... 1,2,4,7,12,20... +1 to each term 2,3,5,8,13,21... because 1+1+2+3+5+8 1+1+1+2+3+5+8 (2+1)+2+3+5+8 (3+2)+3+5+8 (5+3)+5+8 (8+5)+8 (13+8)

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Fibonacci sequence patterns Fibonacci (1,1,2,3,5,8,13...) neither arithmetic nor geometric so write it in a different way 1/1=1 2/1=(1+1)/1 = (1+(1/1)) 3/2=(2+1)/2=1+(1/2)= 1+ 1/(1+(1/1)) 5/3=(3+2)/3=1+(2/3)=1+ 1/(1+ 1/(1+(1/1))) and so on (Golden ratio φ)

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Golden Ratio φ φ=1+1/φ φ 2 =φ+1 quadratic equation φ=(1+sqrt(5))/2 (only the positive answer) φ=1.618033989...

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Golden Ratio and practical application most famous and controversial in history - human aesthetics Converting between km and miles 1 mile= 1.6093 km 13 km = 8 miles Fibonacci (1,1,2,3,5,8,13,21...) OK, using Fibonacci numbers, how many miles are in 50 kilometers?? (show your work)

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Binet Formula A formula to find a term in a Fibonacci numbers without generating previous terms Jacques Binet in 1843 - known to Euler and Bernoulli 100 years before Fibonacci numbers are actually a combo of two geometric progressions Recall φ 2 =φ+1 and τ 2 =τ+1 identities Use them to come up with a formula for the Fibonacci series

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Binet Formula φ 2 =φ+1 and τ 2 =τ+1 identities φ 2 = φ+1 φ 3 =φ(φ 2 )=φ(φ+1)=(φ 2 )+φ=(φ+1)+φ= 2φ+1 φ 4 =φ(φ 3 )=φ(2φ+1)=(2φ 2 )+φ=2(φ+1)+φ= 3φ+2 φ 5 =φ(φ 4 )=φ(3φ+2)=(3φ 2 )+2φ=3(φ+1)+2φ= 5φ+3 φ 6 =φ(φ 5 )=φ(5φ+3)=(5φ 2 )+3φ=5(φ+1)+3φ= 8φ+5 φ 2 =1φ+1 So, φ n =F n φ+F n-1 φ 3 =2φ+1 and φ 4 =3φ+2 τ n =F n τ+F n-1 φ 5 =5φ+3 φ 6 =8φ+5

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Binet Formula φ^n=F n φ+F n-1 τ^n=F n τ+F n-1 φ^n - τ^n=F n φ - F n τ F n = (φ^n - τ^n) / (φ - τ) remember that φ = (1+sqrt(5))/2 and τ = (1+sqrt(5))/2 therefore (φ - τ) = sqrt (5) F n = (φ^n - τ^n) / sqrt(5) F n =(φ^n/sqrt(5)) - (τ^n/sqrt(5)) (two geometric progressions) now for the Fibonacci term 1000 is F 1000 = (φ^(10000) - τ^(10000)) / sqrt(5) = 43466557686937456... (209 digits)

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The Fibonacci Sequence in Nature http://www.youtube.com/watch?v=ahXIMUkSX X0

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Application : The Towers of Hanoi

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The Rules: From here: To here: Without: 1. Moving more than one disk at a time. 2. Placing a larger disk on top of a smaller disk.

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An Example: 12 3 5 76 4

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The Goal: To find the minimum number of moves necessary to complete the puzzle. h n = # of moves required to transfer n disks. Let us find a recurrence rule to predict h n.

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What We Know: h 3 = 7h 5 = 31 h 7 = 123 h 4 = 15h 6 = 63h 8 = 247 h n = small disks + big disk + small disks h n-1 + 1 + h n-1 2h n-1 + 1 Recursive Formula: h n = 2h n-1 + 1

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Closed Formula: 3, 7, 15, 31, 63,... One less than a power of 2? 3 = 2 2 - 1 7 = 2 3 - 1 15 = 2 4 - 1 h n = 2 n -1

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Prediction : End of the World? High on the mountaintops sat a monk who could foretell the end of the world. He had a Tower of Hanoi with 64 gleaming diamond disks and could move one a second. When he stopped the world would end. How long do we have?

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Prediction : Solution Number of moves required 2 64 -1 So... roughly 583,344,214,028 years.

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Prime Numbers: How do we find them? 200 B.C. Eratosthenes invented the sieve.

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Prime Numbers:

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And it stops.

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Why? The number of tests is the # of primes < testing maximum Proof by contradiction: 1. A composite exists in 11-100 2. Thus, it is not a multiple of a P < 10 3. Thus, both factors > 10 4. Therefore, the composite > 100

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Prime Numbers: How many exist? E = P 1 * P 2 * P 3 * P 4... P n now... q = P 1 * P 2 * P 3 * P 4 *... * P n + 1 Following the Composite Theorems (must be factor of unique prime numbers), infinite prime numbers exist.

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Where Aren't the Prime Numbers? 2*3 + 2 = composite2*3 + 3 = composite K = 2 * 3 * 4 *... * (N+1) K+2 = 2 * 3 * 4 *... * (N+1) +2 K+3 = 2 * 3 * 4 *... * (N+1) + 3 K+(N+1) = 2 * 3 * 4 *... * (N+1) + (N+1) K+2, K+3, K+4, K+(N+1) --> all composite, there are runs infinitely long where there are no primes.

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