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

A moment’s 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 numbers 1. A combinatorial model 2. A geometric model

A combinatorial model for Fibonacci numbers Theorem The number of ways to tile a board of length j with squares and dominoes is f j where f 0 = f 1 = 1 and f j = f j − 1 + f j − 2 . j square domino

f 4 = 5

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

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

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

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

Sums of squares of Fibonacci numbers Theorem f 2 0 + f 2 1 + f 2 2 + · · · + f 2 n = f n f n +1

Proof. Question. How many ways can you tile an n -board and an ( n + 1) -board? n + 1 n

Proof. Question. How many ways can you tile an n -board and an ( n + 1) -board? n + 1 n Answer 1. There are f n and f n +1 tilings of the first and second board, respectively. Therefore there are f n f n +1 tilings of both boards.

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

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.

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 f j ways, so both can be tiled f 2 j ways.

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 f j ways, so both can be tiled f 2 j ways. j =0 f j 2 tilings. Summing over all possible values of j gives � n

A geometric model for Fibonacci numbers Construct a rectangle comprising two adjacent squares of side F 1 = 1 and F 2 = 1 . F 2 F 1

A geometric model For every j ≥ 2 we construct a square of side F j on the larger side of the existing rectangle. F 2 F 1 F 3

A geometric model For every j ≥ 2 we construct a square of side F j on the larger side of the existing rectangle. F 4 F 2 F 1 F 3

A geometric model For every j ≥ 2 we construct a square of side F j on the larger side of the existing rectangle. F 4 F 2 F 1 F 3 F 5

A geometric model For every j ≥ 2 we construct a square of side F j on the larger side of the existing rectangle. F 6 F 4 F 2 F 1 F 3 F 5

A geometric model For every j ≥ 2 we construct a square of side F j on the larger side of the existing rectangle. F 6 F 4 F 2 F 1 F 3 F 5 F 7

A geometric model The j th square has side F j where F 1 = F 2 = 1 and F j = F j − 1 + F j − 2 for n ≥ 2 . F 6 F 4 F 2 F 1 F 3 F 5 F 7

Sums of squares of Fibonacci numbers. Theorem F 2 1 + F 2 2 + · · · + F 2 n = F n F n +1 F 6 F 4 F 2 F 1 F 3 F 5 F 7 Proof. The total area is equal to the sum of its parts.

Sums of cubes of Fibonacci numbers Question. Is there a closed formula for the sum of cubes of Fibonacci numbers?

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

Sums of cubes of Fibonacci numbers Question. Are there combinatorial or geometric methods for determining the sum of cubes of Fibonacci numbers?

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.

Sums of cubes of Fibonacci numbers Theorem n n +2 + ( − 1) n F n − 2 F 3 j = F n +1 F 2 � n F 3 2 j =1 Proof. A. T. Benjamin, B Cloitre and T. A. Carnes, Recounting the Sums of Cubes of Fibonacci Numbers , Proceedings of the Eleventh International Conference on Fibonacci Numbers and their Applications, (2009), 45-51

A preliminary result For the geometric proof we require one preliminary result, n j F j +1 = 1 � F 2 2 F n F n +1 F n +2 . j =1

The centroid of a composite shape � n � n j =1 A j x j j =1 A j y j x = and y = A A

A Fibonacci tiling Example F j F j +1 F j − 1

A Fibonacci tiling Example F j F j +1 F j − 1 1. F 2 j +2 = 4 F j F j +1 + F 2 j − 1

The centroid of the tiling Example F j F j +1 3 4 F j − 1 5 2 1 O F j +2 F j +2 2

The centroid of the tiling Example F j F j +1 3 4 F j − 1 5 2 1 O F j +2 F j +2 2 2 F 2 j − 1 F j + 4 F j F j +1 F j +2 = F 3 j +2 − F 3 j − 1

Theorem n F 3 j + F 3 j +1 + F 3 j +2 − F j − 1 F j F j +1 − 2 � F j F j +1 F j +2 = . 4 j =1

Another Fibonacci tiling Example F j F j +1 F j F j +1

Another Fibonacci tiling Example F j F j +1 F j F j +1 F 2 j +2 = 2 F j F j +1 + F 2 j + F 2 j +1

The centroid of the tiling Example F j F j +1 F j 1 2 F j +1 3 4 O F j +2 F j +2 2 F 3 j + 3 F j F j +1 F j +2 = F 3 j +2 − F 3 j +1

Theorem n 3 F 2 j +1 F j − F 3 j +1 − F 3 j + 1 � F 3 j = . 2 j =1

Sums of cubes Starting point: T n = 1 + 2 + · · · + n = 1 2 n ( n + 1)

Sums of cubes j 1 2 · · · · · · n 1 T n 2 T n

Sums of cubes j T j − 1 T j x j

Sums of cubes j 1 2 j 2

Sums of cubes n n � 2 � j 3 = � � j j =1 j =1

A generalisation The same trick works more generally. Suppose our starting point is n j F j +1 = 1 � F 2 2 F n F n +1 F n +2 . j =1

F 2 1 F 2 F 2 F 2 F 2 2 F 3 3 F 4 n F n +1 · · · 0 1 2 F n F n +1 F n +2 1 4 F n F n +1 F n +2

F 2 j F j +1 0 S j − 1 S j x j

F 2 j F j +1 0 1 2 F j F 2 j +1

n j +1 = 1 � F 3 j F 3 4 F 2 n F 2 n +1 F 2 n +2 j =1

n n � 2 � � � F 3 j F 3 F 2 j +1 = j F j +1 j =1 j =1

Apply the method again n j +1 F 2 j +1 = 1 � F 5 j F 5 8 F 4 n F 4 n +1 F 4 n +2 j =1