Fibonacci Numbers By: Sara Miller Advisor: Dr. Mihai Caragiu
Abstract We will investigate various ways of proving identities involving Fibonacci Numbers, such as, induction, linear algebra (matrices), and combinatorics (0-1 sequences). We will also look at one educational activity.
Summary 1. Introduction. Mathematical Induction 3. Linear Algebra (matrices) 4. Combinatorics (0-1 sequences) 5. Educational Activity
1. Introduction Leonardo Pisano Born around 1175 in Pisa, Italy His nickname was Fibonacci He traveled extensively with his father Wrote book: Liber Abbaci in 10 Presented a numerical series now referred to as the Fibonacci numbers
Fibonacci Numbers 1, 1,, 3, 5, 8, 13, 1, 34, 55, 89, 144, 33,
Rabbit Problem A certain man put a pair of rabbits in a place surrounded by a wall. How many pairs of rabbits can be produced from that pair in a year if it is supposed that every month each pair begets a new pair from which the second month on becomes productive? (Liber Abbaci, chapter 1, p. 83-4)
Chart of Rabbit Problem Month Adult Pairs Baby Pairs Total Pairs January 1 0 1 February 1 1 March 1 3 April 3 5 May 5 3 8 June 8 5 13 July 13 8 1 August 1 13 34 September 34 1 55 October 55 34 89 November 89 55 144 December 144 89 33 January 33 144 377
The Fibonacci Linear Recurrence Initial Conditions: f = 1 and f = 1 1 Recurrence relation: fn = f + f n 1 n
n f n 1 3 4 5 6 7 1 1 3 5 8 13
. Mathematical Induction.1 Sum of Squares of Fibonacci Numbers: f + f +... + f = 1 n f n f n+ 1, n 1 I will initially carry out the proof of this identity by induction. Then I will provide a visual proof.
Proof by induction: (a) For n= 1 the formula takes the form f = f f 1 = (1)(1) thus 1= 1 1 1 Thus the first part of the induction (basis step) is finished.
(b) Assuming the formula holds true for n, we will prove it for n+ 1. Therefore, f +... 1 f + + f n = f n f n+ 1 And we will prove f+ f+... + f f f f 1 n + = n+ 1 n+ 1 n+
Indeed, f+ f+... + f f 1 n + n+ 1 = f f + f n n+ 1 n+ 1 By inductive hypothesis f = n 1 f n+ f + n+ 1 = f f n+ 1 n+ By f + 1 f n = n+ f n+ Thus the induction is complete, and I have proved that the formula holds for all n. QED.
Visual Proof f +... 1 f + + f n = f n f n+ 1
. Sum of Odd Fibonacci Numbers: f..., 1 1 + f 3 + f 5 + + f 1 = f n n n
.3 Sum of Even Fibonacci Numbers: f +... 1, 0 0 f + f + + = 4 f n f n+ 1 n
.4 Sum of Products of Consecutive Fibonacci Numbers: f f + f f + f f +... + f f = f, n 1 1 3 34 n 1n n Proof: (a) For n= 1 the formula takes the form f f = f 1 1 =1 thus 1= 1 1 Thus the first part of the induction (basis step) is finished.
(b) Assuming the formula holds true for n, we will prove it for n+ 1. For the inductive step, we will assume f f + f f + f f +... + f f = f 1 3 34 n 1n n And we will prove f f + f f + f f +... + f f + f f + f f = f 1 3 34 n 1n n n+ 1 n+ 1n+ n+
Indeed, f f + f f + f f +... + f f + f f + f f 1 3 34 n 1n n n+ 1 n+ 1n+ = f + f f + f f n n n+ 1 n+ 1 n+ By inductive Hypothesis = f f + f + f f n n n+ 1 n+ 1 n+
= f f + f + f f n n n+ 1 n+ 1 n+ = f f + f f n n+ n+ 1 n+ = f f + f n+ n n+ 1 = f n+ By f + f = f n n+ 1 n+ Thus the induction is complete, and I have proved that the formula holds for all n. QED
3. Linear Algebra (Matrices) 3.1 Cassini s Identity: f f f 1 n 1 n 1 n = + n First we will introduce the transition matrix It is easy to see that this matrix satisfies f f n 1 T n + = for all n 1 f f n n 1 1 1 T = 1 0
It is easy to prove (by induction) that Tn = f n+ 1 fn f n f n 1 for any n=1,,3,
Let s take the determinants of both sides: det n T = f f f n+ 1 n 1 n But, n n det n T = dett = 1 Therefore, n f f f 1 n 1 n 1 n = + Thus we have proven the Cassini s Identity. QED
4. Combinatorics 4.1 0-1 Sequences One important fact is that the number of 0 1 sequences of length n without consecutive 1s is f for every n 1. n+ Let s prove this!
First denote by A n the number of 0-1 sequences of length n without consecutive 1s. Here is an example of a string of length 8 without consecutive 1s: 0 100 10 10
For n= 1, we have a single, so we have two possibilities a 0 or a 1, Thus A =, 1 f = 1 f = + 3 holds true for n= 1. therefore An = f n+ For n=, we have two, so we have three possibilities 00, 01, 10 Thus A = 3, 3 f = f = + 4 for n=. and An = f holds true n+ Thus the first part of the induction (basis step) is finished.
I shall prove the induction step in the following way: Assume A = f is true for all. k k k < + n We want An = f n+
n 1 cells 01001010 A n 1 n cells Ending with a 0 01001010 n cells 01001001 A n Ending with a 1 Therefore, A n = A n 1 + A n = f n 1+ + f n + = f n+ 1 + f n = f n+ QED
4. An Example of Combinatorial Proof: We consider the identity f n + f = f n+ 1 n+ 1 n 1 n 1 n 1 n 1 n n 0 010 central cell central cell is 0 central cell is 1 n 1 + 1= n 1 f f = f n+ 1 n+ 1 n+ 1 f f f n n= n Therefore, f = f + f n+ 1 n+ 1 n.
5. Educational Activity: Many activities can be used in the classroom to generate and investigate Fibonacci sequences. One is to have students place 1 and cent stamps across the top of a postcard (facing with correct side up) in different arrangements to make up certain postage amounts. The number of different arrangements will be a Fibonacci number.
END :O)~