A moments thought: physical derivations of Fibonacci summations - - PowerPoint PPT Presentation

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A moments thought: physical derivations of Fibonacci summations - - PowerPoint PPT Presentation

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A moments thought: physical derivations of Fibonacci summations David Treeby Two pretty formulas n n 2 j 3 = j j =1 j =1 n n 2 F 3 j F 3 F 2 j +1 = j F j +1 j =1 j =1 Two models of the Fibonacci

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A moment’s thought: physical derivations of Fibonacci summations

David Treeby

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Two pretty formulas

n

  • j=1

j3 =

  • n
  • j=1

j 2

n

  • j=1

F 3

j F 3 j+1 =

  • n
  • j=1

F 2

j Fj+1

2

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Two models of the Fibonacci numbers

  • 1. A combinatorial model
  • 2. A geometric model
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A combinatorial model for Fibonacci numbers

Theorem

The number of ways to tile a board of length j with squares and dominoes is fj where f0 = f1 = 1 and fj = fj−1 + fj−2. j square domino

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f4 = 5

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Proof.

Consider a j-board. Suppose that this can be tiled in fj ways. j

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Proof.

Consider a j-board. Suppose that this can be tiled in fj ways. j Case 1. If the first tile is a square then there are fj−1 ways to tile the remaining (j − 1)-board. j − 1

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Proof.

Consider a j-board. Suppose that this can be tiled in fj ways. j Case 1. If the first tile is a square then there are fj−1 ways to tile the remaining (j − 1)-board. j − 1 Case 2. If the first tile is a domino then there are fj−2 ways to tile the remaining (j − 2)-board. j − 2

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Proof.

Consider a j-board. Suppose that this can be tiled in fj ways. j Case 1. If the first tile is a square then there are fj−1 ways to tile the remaining (j − 1)-board. j − 1 Case 2. If the first tile is a domino then there are fj−2 ways to tile the remaining (j − 2)-board. j − 2 Therefore fj = fj−1 + fj−2.

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Sums of squares of Fibonacci numbers

Theorem

f2

0 + f2 1 + f2 2 + · · · + f2 n = fnfn+1

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Proof.

  • Question. How many ways can you tile an n-board and an

(n + 1)-board? n + 1 n

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Proof.

  • Question. How many ways can you tile an n-board and an

(n + 1)-board? n + 1 n Answer 1. There are fn and fn+1 tilings of the first and second board, respectively. Therefore there are fnfn+1 tilings of both boards.

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Answer 2. Condition on the position j corresponding to the last common edge of each tiling. j

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Answer 2. Condition on the position j corresponding to the last common edge of each tiling. j To avoid future common edges, there is exactly one way to finish the tiling.

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Answer 2. Condition on the position j corresponding to the last common edge of each tiling. j To avoid future common edges, there is exactly one way to finish the tiling. Prior to this, the first and second board can each be tiled fj ways, so both can be tiled f2

j ways.

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Answer 2. Condition on the position j corresponding to the last common edge of each tiling. j To avoid future common edges, there is exactly one way to finish the tiling. Prior to this, the first and second board can each be tiled fj ways, so both can be tiled f2

j ways.

Summing over all possible values of j gives n

j=0 fj2 tilings.

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A geometric model for Fibonacci numbers

Construct a rectangle comprising two adjacent squares of side F1 = 1 and F2 = 1.

F1 F2

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A geometric model

For every j ≥ 2 we construct a square of side Fj on the larger side

  • f the existing rectangle.

F1F3 F2

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A geometric model

For every j ≥ 2 we construct a square of side Fj on the larger side

  • f the existing rectangle.

F1F3 F2 F4

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A geometric model

For every j ≥ 2 we construct a square of side Fj on the larger side

  • f the existing rectangle.

F1F3 F5 F2 F4

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A geometric model

For every j ≥ 2 we construct a square of side Fj on the larger side

  • f the existing rectangle.

F1F3 F5 F2 F4 F6

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A geometric model

For every j ≥ 2 we construct a square of side Fj on the larger side

  • f the existing rectangle.

F1F3 F5 F7 F2 F4 F6

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A geometric model

The jth square has side Fj where F1 = F2 = 1 and Fj = Fj−1 + Fj−2 for n ≥ 2.

F1F3 F5 F7 F2 F4 F6

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Sums of squares of Fibonacci numbers.

Theorem

F 2

1 + F 2 2 + · · · + F 2 n = FnFn+1

F1F3 F5 F7 F2 F4 F6

Proof.

The total area is equal to the sum of its parts.

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Sums of cubes of Fibonacci numbers

  • Question. Is there a closed formula for the sum of cubes of

Fibonacci numbers?

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Sums of cubes of Fibonacci numbers

  • Question. Is there a closed formula for the sum of cubes of

Fibonacci numbers?

  • Answer. Yes, by Binet’s formula for the nth Fibonacci

number there has to be. However, the answer is not expressible as the product of Fibonacci numbers.

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Sums of cubes of Fibonacci numbers

  • Question. Are there combinatorial or geometric methods for

determining the sum of cubes of Fibonacci numbers?

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Sums of cubes of Fibonacci numbers

  • Question. Are there combinatorial or geometric methods for

determining the sum of cubes of Fibonacci numbers?

  • Answer. Yes and yes.
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Sums of cubes of Fibonacci numbers

Theorem

n

  • j=1

F 3

j = Fn+1F 2 n+2 + (−1)nFn − 2F 3 n

2

Proof.

  • A. T. Benjamin, B Cloitre and T. A. Carnes, Recounting the Sums
  • f Cubes of Fibonacci Numbers, Proceedings of the Eleventh

International Conference on Fibonacci Numbers and their Applications, (2009), 45-51

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A preliminary result

For the geometric proof we require one preliminary result,

n

  • j=1

F 2

j Fj+1 = 1

2FnFn+1Fn+2.

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The centroid of a composite shape

x = n

j=1 Ajxj

A and y = n

j=1 Ajyj

A

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A Fibonacci tiling

Example

Fj Fj+1 Fj−1

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A Fibonacci tiling

Example

Fj Fj+1 Fj−1 1. F 2

j+2 = 4FjFj+1 + F 2 j−1

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The centroid of the tiling

Example

1 4 2 3 Fj Fj+1 Fj−1 5

Fj+2 2

O Fj+2

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The centroid of the tiling

Example

1 4 2 3 Fj Fj+1 Fj−1 5

Fj+2 2

O Fj+2 2F 2

j−1Fj + 4FjFj+1Fj+2 = F 3 j+2 − F 3 j−1

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Theorem

n

  • j=1

FjFj+1Fj+2 = F 3

j + F 3 j+1 + F 3 j+2 − Fj−1FjFj+1 − 2

4 .

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Another Fibonacci tiling

Example

Fj Fj+1 Fj+1 Fj

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Another Fibonacci tiling

Example

Fj Fj+1 Fj+1 Fj F 2

j+2 = 2FjFj+1 + F 2 j + F 2 j+1

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The centroid of the tiling

Example

1 2 3 4 Fj Fj+1 Fj Fj+1 O

Fj+2 2

Fj+2 F 3

j + 3FjFj+1Fj+2 = F 3 j+2 − F 3 j+1

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Theorem

n

  • j=1

F 3

j =

3F 2

j+1Fj − F 3 j+1 − F 3 j + 1

2 .

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Sums of cubes

Starting point: Tn = 1 + 2 + · · · + n = 1 2n(n + 1)

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Sums of cubes

1

1 2Tn

Tn 2 · · · j · · · n

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Sums of cubes

j xj Tj−1 Tj

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Sums of cubes

j

1 2j2

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Sums of cubes

n

  • j=1

j3 =

  • n
  • j=1

j 2

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A generalisation

The same trick works more generally. Suppose our starting point is

n

  • j=1

F 2

j Fj+1 = 1

2FnFn+1Fn+2.

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F 2

1 F2F 2 2 F3

F 2

3 F4

· · · F 2

nFn+1 1 4FnFn+1Fn+2 1 2FnFn+1Fn+2

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F 2

j Fj+1

xj Sj−1 Sj

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F 2

j Fj+1 1 2FjF 2 j+1

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n

  • j=1

F 3

j F 3 j+1 = 1

4F 2

nF 2 n+1F 2 n+2

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n

  • j=1

F 3

j F 3 j+1 =

  • n
  • j=1

F 2

j Fj+1

2

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Apply the method again

n

  • j=1

F 5

j F 5 j+1F2j+1 = 1

8F 4

nF 4 n+1F 4 n+2