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Fibonacci Numbers, Polynomial Coefficients, and Vector Programs.

Fibonacci Numbers, Polynomial Coefficients, and Vector Programs. Leonardo Fibonacci. In 1202, Fibonacci proposed a problem about the growth of rabbit populations. Inductive Definition or Recurrence Relation for the Fibonacci Numbers. Stage 0, Initial Condition, or Base Case:

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Fibonacci Numbers, Polynomial Coefficients, and Vector Programs.

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  1. Fibonacci Numbers, Polynomial Coefficients, and Vector Programs.

  2. Leonardo Fibonacci • In 1202, Fibonacci proposed a problem about the growth of rabbit populations.

  3. Inductive Definition or Recurrence Relation for theFibonacci Numbers Stage 0, Initial Condition, or Base Case: Fib(0) = 0; Fib (1) = 1 Inductive Rule For n>1, Fib(n) = Fib(n-1) + Fib(n-2)

  4. Sneezwort (Achilleaptarmica) Each time the plant starts a new shoot it takes two months before it is strong enough to support branching.

  5. Counting Petals • 5 petals: buttercup, wild rose, larkspur, columbine (aquilegia) 8 petals: delphiniums 13 petals: ragwort, corn marigold, cineraria, some daisies • 21 petals: aster, black-eyed susan, chicory 34 petals: plantain, pyrethrum 55, 89 petals: michaelmas daisies, the asteraceae family.

  6. Pineapple whorls • Church and Turing were both interested in the number of whorls in each ring of the spiral. The ratio of consecutive ring lengths approaches the Golden Ratio.

  7. Definition of  (Euclid) • Ratio obtained when you divide a line segment into two unequal parts such that the ratio of the whole to the larger part is the same as the ratio of the larger to the smaller. A B C

  8. Expanding Recursively

  9. Continued Fraction Representation

  10. Continued Fraction Representation

  11. 1,1,2,3,5,8,13,21,34,55,…. • 2/1 = 2 • 3/2 = 1.5 • 5/3 = 1.666… • 8/5 = 1.6 • 13/8 = 1.625 • 21/13 = 1.6153846… • 34/21 = 1.61904… •  = 1.6180339887498948482045

  12. 1 + X + X2 1 – X 1 -(1 – X) X2 X3 X -(X – X2) -(X2 – X3) 1 + X + X2 + X3 + X4+ X5 + X6 + X7 + … = How to divide polynomials? 1 ? 1 – X …

  13. 1 1 + X1 + X2 + X3 + … + Xn + ….. = 1 - X The Infinite Geometric Series

  14. 1 1 + X1 + X2 + X3 + … + Xn + ….. = 1 - X • (1-X) ( 1 + X1 + X2 + X 3 + … + Xn + … ) • = 1 + X1 + X2 + X 3 + … + Xn + Xn+1 + …. • - X1 - X2 - X 3 - … - Xn-1 – Xn - Xn+1 - … • = 1

  15. 1 1 + X1 + X2 + X3 + … + Xn + ….. = 1 - X 1 + X 1 – X 1 -(1 – X) X2 X3 X -(X – X2) + X2 + … -(X2 – X3) …

  16. X + X2 + 2X3 + 3X4 + 5X5 + 8X6 1 – X – X2 X -(X – X2 – X3) X2 + X3 -(X2 – X3 – X4) 2X3 + X4 -(2X3 – 2X4 – 2X5) 3X4 + 2X5 -(3X4 – 3X5 – 3X6) 5X5 + 3X6 8X6 + 5X7 -(5X5 – 5X6 – 5X7) -(8X6 – 8X7 – 8X8) Something a bit more complicated X 1 – X – X2

  17. Hence X • = F0 1 + F1 X1 + F2 X2 +F3 X3 + F4 X4 + F5 X5 + F6 X6 + … 1 – X – X2 = 01 + 1 X1 + 1 X2 + 2X3 + 3X4 + 5X5 + 8X6 + …

  18. Going the Other Way F0 = 0, F1 = 1 • (1 - X- X2)  ( F0 1 + F1 X1 + F2 X2 + … + Fn-2 Xn-2 + Fn-1 Xn-1 + Fn Xn + …

  19. Going the Other Way • (1 - X- X2)  ( F0 1 + F1 X1 + F2 X2 + … + Fn-2 Xn-2 + Fn-1 Xn-1 + Fn Xn + … • = ( F0 1 + F1 X1 + F2 X2 + … + Fn-2 Xn-2 + Fn-1 Xn-1 + Fn Xn + … • - F0 X1 - F1 X2 - … - Fn-3 Xn-2 - Fn-2 Xn-1 - Fn-1 Xn - … • - F0 X2 - … - Fn-4 Xn-2 - Fn-3 Xn-1 - Fn-2 Xn - … • = F0 1 + ( F1 – F0 ) X1 F0 = 0, F1 = 1 = X

  20. Thus • F0 1 + F1 X1 + F2 X2 + … + Fn-1 Xn-1 + Fn Xn + … X = 1 – X – X2 = X/(1- X)(1 – (-)-1 X) -1= 1, --1=1

  21. Linear factors on the bottom X (1 – X)(1- (-)-1X) ? n=0..∞ Xn =

  22. an+1 – bn+1 a- b Geometric Series (Quadratic Form) (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 = (1 – aX)(1-bX) n=0..∞ Xn =

  23. n+1 – (--1)n+1 √5 Geometric Series (Quadratic Form) 1 (1 – X)(1- (--1X)) n=0.. ∞ Xn =

  24. n+1 – (--1)n+1 √5 Power Series Expansion of F X (1 – X)(1- (--1X) n=0.. ∞ Xn+1 =

  25. Leonhard Euler (1765) J. P. M. Binet (1843)A de Moivre (1730) The ith Fibonacci number is:

  26. an+1 – bn+1 a- b Let’s Derive This (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 = (1 – aX)(1-bX) n=0..∞ Xn =

  27. Substituting Y = aX … 1 1 + Y1 + Y2 + Y 3 + … + Yn + ….. = 1 - Y

  28. Geometric Series (Linear Form) 1 1 + aX1 + a2X2 + a3X 3 + … + anXn + ….. = 1 - aX

  29. Geometric Series (Quadratic Form) (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 (1 – aX)(1-bX)

  30. Suppose we multiply this out to get a single, infinite polynomial.What is an expression for Cn? (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 + c1X1 + .. + ck Xk + …

  31. cn =a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0 (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 + c1X1 + .. + ck Xk + …

  32. If a = b then cn = (n+1)(an)a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0 (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 + c1X1 + .. + ck Xk + …

  33. an+1 – bn+1 a- b a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0 = • (a-b) (a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0) • = a1bn +… ai+1bn-i… + anb1 + an+1b0 • - a0bn+1 – a1bn… ai+1bn-i… - an-1b2 - anb1 • = - bn+1 + an+1 • = an+1 – bn+1

  34. an+1 – bn+1 a- b if a  b then cn =a0bn + a1bn-1 +… aibn-i… + an-1b1 + anb0 (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 + c1X1 + .. + ck Xk + …

  35. an+1 – bn+1 a- b Geometric Series (Quadratic Form) (1 + aX1 + a2X2 + … + anXn + …..) (1 + bX1 + b2X2 + … + bnXn + …..) = 1 = (1 – aX)(1-bX) n=0.. Xn = n=0.. or Xn (n+1)an when a=b

  36. Sequences That Sum To n • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. • Example: f5 = 5

  37. Sequences That Sum To n • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. • Example: f5 = 5 • 4 = 2 + 2 • 2 + 1 + 1 • 1 + 2 + 1 • 1 + 1 + 2 • 1 + 1 + 1 + 1

  38. Sequences That Sum To n • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. f1 f3 f2

  39. Sequences That Sum To n • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. f1 = 1 0 = the empty sum f3 = 2 2 = 1 + 1 2 f2 = 1 1 = 1

  40. Sequences That Sum To n • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. fn+1 = fn + fn-1

  41. Sequences That Sum To n • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. fn+1 = fn + fn-1 # of sequences beginning with a 2 # of sequences beginning with a 1

  42. Fibonacci Numbers Again • Let fn+1 be the number of different sequences of 1’s and 2’s that sum to n. fn+1 = fn + fn-1 f1 = 1 f2 = 1

  43. Visual Representation: Tiling • Let fn+1 be the number of different ways to tile a 1 × n strip with squares and dominoes.

  44. Visual Representation: Tiling • Let fn+1 be the number of different ways to tile a 1 × n strip with squares and dominoes.

  45. Visual Representation: Tiling • 1 way to tile a strip of length 0 1 way to tile a strip of length 1: 2 ways to tile a strip of length 2:

  46. fn+1 = fn + fn-1 • fn+1 is number of ways to tile length n. fn tilings that start with a square. fn-1 tilings that start with a domino.

  47. Let’s use this visual representation to prove a couple of Fibonacci identities.

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