FIBONACCI NUMBERS AND THE GOLDEN RATIO

1 FIBONACCI NUMBERS AND THE GOLDEN RATIOROBERT this expository paper written to commemorate FIBONACCI Day 2016, wediscuss famous relations involving the FIBONACCI sequence, the GOLDEN RATIO , continuedfractions and nested radicals, and show how these fit into a more general frameworkstemming from the quadratic numbersTheFibonacci numbersare an interesting sequence of integers discovered by the promi-nent medieval mathematician Leonardo FIBONACCI , related to the shapes of flower petalsand tree branches, the birth rates of rabbits, and other natural phenomena, that showsup in many places in mathematics you can even find the sequence in Pascal s triangle[2]. Thenth FIBONACCI numberFnis defined forn= 0 andn= 1 byF0=F1= 1,and forn 2 by(1)Fn=Fn 1+Fn example, we haveF2=F1+F0= 1 + 1 = 2F3=F2+F1= 2 + 1 = 3F4=F3+F2= 3 + 2 = 5,and so on. When you define a sequence of NUMBERS like this, building up thenth termfrom the previous terms, it is called arecursive sequence, and an equation like (1) thatproduces the sequence is called arecursion relations can lead to all sorts of lovely and intricate patterns, by pluggingthe terms back into themselves in creative ways.

2 For example, the reader might like totry taking the definition (1) above, and repeatedly using the observationFn=Fn 1+Fn 2=Fn 1+ (Fn 3+Fn 4) =Fn 1+Fn 3+ (Fn 5+Fn 6)and so on, to show the following an odd number thenFN=F0+F2+F4+F6+..+FN Fodd can be written as the sum of all the smaller FIBONACCI NUMBERS of the form Feven . For instance,F5= 8 can be writtenF5=F0+F2+F4= 1 + 2 + about ifNis anevennumber above? The reader is encouraged to experimentwith (1) to answer this question, and to discover other FIBONACCI relations for GOLDEN ratioIt is well known that the FIBONACCI NUMBERS are connected to another famous number,thegolden RATIO , which is related to the shapes of pineapples, seashells, and other objectsin nature, as well as to fractals and other self-similar mathematical objects [2]. The goldenratio studied since the time of Euclid is equal to the larger of the two roots1of thepolynomial(2)x2 x quadratic formula gives the exact value2 =1 + 52= beautiful fact is that is related to its own reciprocal by a very simple GOLDEN RATIO is equal to its own reciprocal plus1: = 1 +1 To see this, notice from expression (2) that 2 1 = 0 implies 2= 1 +.

3 Now divide both sides of this equation by , to finish the 2 gives us another kind of recursion relation. Observe that is present onboth the left-hand side, and in the denominator on the right. So we can plug the entireright-hand side of Proposition 2 (which is equal to ) back into itself in place of the inthe denominator: = 1 +11 +1 There still is in the denominator on the right-hand side, so we can repeat this substi-tution, and repeat it again, any number of times: = 1 +11 +11+1 = 1 +11 +11+11+1 = 1 +11 +11+11+11+1 =..These stacked fractions-within-fractions are calledcontinued fractions, and the numbersmaking up the fraction are calledcoefficients. Notice that at every stage, the left-handside does not change: the continued fraction is still equal to . Proceeding in this mannerforever, we get a very famous can write the GOLDEN RATIO as an infinite continued fraction with allthe coefficients equal to 1: = 1 +11 +11+11+11+.

4 1 The quadratic formula says the roots, or zeros , ofAx2+Bx+Care given by B B2 , the other root1 52is equal to 1/ . FIBONACCI NUMBERS AND THE GOLDEN RATIO3In the next section we will use this formula to show how is related to the Fibonaccisequence. The interested reader is referred to Hardy and Wright [1] for more aboutcontinued FIBONACCI ratiosThe connection between the FIBONACCI numbersFnand the GOLDEN RATIO is ratioFnFn 1approaches the very beginning of the FIBONACCI sequence we see the ratioFn/Fn 1oscillatesaround = , getting closer and closer to the GOLDEN RATIO :F1/F0= 1F2/F1= 2F3/F2= 3/2 = 5/3 = 8/5 = (which is getting pretty close to already)Skipping up the sequence just a few terms, we haveF10/F9= 89/55 = ,which is indeed a very close approximation to (and it keeps getting better).It is easy to see why this is true, taking a recursive approach. Using the definition (1)of thenth FIBONACCI number, we can rewrite this RATIO as(3)FnFn 1=Fn 1+Fn 2Fn 1= 1 +Fn 2Fn 1= 1 +1Fn 1/Fn exactly the same principle, we can rewriteFn 1Fn 2= 1 +1Fn 2/Fn 3,Fn 2Fn 3= 1 +1Fn 3/Fn 4,and so on, and substitute these one after another forFn 1/Fn 2in the right-hand side ofequation (3):FnFn 1= 1 +1Fn 1/Fn 2= 1 +11 +1Fn 2/Fn 3= 1 +11 +11+1Fn 3/Fn 4=.

5 Eventually we run out of FIBONACCI NUMBERS to put in the denominator, and end up with(4)Fn/Fn 1= 1 +11 +11+11+..+1F1/F0= 1 +11 +11+11+..+1,where the final coefficient on the right is 1 becauseF1/F0= 1. Noting that the continuedfraction on the right side gets longer and longer asnincreases (the number of coefficientsincreases proportionally), then asnapproaches the continued fraction in (4) gets closerand closer to the infinite continued fraction in Proposition 3, which equals .That it turns out to be the exact number approached by ratios of consecutive Fibonaccinumbers, which are interesting in their own right, is another impressive property of .4 ROBERT radicalsWe want to point out one other beautiful (and exotic-looking) property of the goldenratio involving square roots, , radicals . The equation 2 1 = 0 implies 2= 1+ ;taking the square root of both sides leads to the following GOLDEN RATIO is equal to the square root of itself plus1: = 1 + Viewing this as a recursion relation, just as with the continued fractions previously, wecan now repeatedly substitute the entire right-hand side of Proposition 5 for the underthe radical on the right: = 1 + 1 + = 1 + 1 + 1 + = 1 + 1 + 1 + 1 + =.

6 These roots-within-roots are callednested radicals, the more obscure cousins of continuedfractions. Continuing in this fashion forever, then can be written as an infinite nestedradical containing only 1 s, like the continued fraction representation in Proposition that we have the expressions all in one place, we collect this observation and ourprevious formulas for in the following amazing GOLDEN RATIO is equal to =1 + 52= 1 +11 +11+11+11+..= 1 + 1 + 1 + 1 +..In the next section, we will write a very similar system of equations for the roots ofanyquadratic polynomial with rational quadratic formula and beyondHere we will see there is a more general framework containing the previous results aboutthe GOLDEN RATIO , FIBONACCI NUMBERS , continued fractions and nested radicals all goingback to the quadratic , the sign in the quadratic formula yields two roots, a plus root and a minus root.

7 For rational numbersaandb, let (a, b) denote the plus root of the polynomial(5)x2 ax the quadratic formula gives the exact value3for this number: (a, b) =a+ a2+ 4b2 The GOLDEN RATIO is the special case = (1,1). In fact, these NUMBERS (a, b) possessmany of the nice properties enjoyed by . For instance, it follows from (5) that (a, b)2 a (a, b) b= can use the minus roota a2+4b2=a (a, b) = b/ (a, b) for similar results to those thatfollow, but with some sign changes. The reader is encouraged to work this case out, NUMBERS AND THE GOLDEN RATIO5 Then similar steps to those we applied to yield a pair of familiar-looking can write (a, b)in the following ways: (a, b) =a+b (a, b)(so long as (a, b)6= 0in the denominator)= b+a (a, b)Notice how the casea=b= 1 reduces to Propositions 2 and 5. Using these twoidentities, and following the exact steps that proved the corresponding equations for previously, we can generalize Proposition 6 as follows.

8 We omit the proofs, however, asthe reader might enjoy working out the details number (a, b)is equal to (a, b) =a+ a2+ 4b2=a+ba+ba+ba+ba+..= b+a b+a b+a b+..We can also define a generalization of the FIBONACCI sequence that connects with (a, b).Let us defineFn(a, b) byF0(a, b) = 1, F1(a, b) =a,and forn 2 by(6)Fn(a, b) =aFn 1(a, b) +bFn 2(a, b).The usual FIBONACCI NUMBERS are the special caseFn=Fn(1,1). It turns out theseFn(a, b) behave similarly toFn, as we see in the following ratioFn(a,b)Fn 1(a,b)approaches (a, b) proposition follows from the definition (6) with the observation(7)Fn(a, b)Fn 1(a, b)=aFn 1(a, b) +bFn 2(a, b)Fn 1(a, b)=a+bFn 1(a, b)/Fn 2(a, b),exactly like the proof of Proposition 4, so that after repeatedly using (7) we end up withFn(a, b)Fn 1(a, b)=a+ba+ba+ba+..+bF1(a,b)/F0(a,b)=a+ba+b a+ba+..+ , the far right side of this equation looks more and more like the infinitecontinued fraction in Proposition 8, that is, more and more like (a, b).

9 What if we defineF0(a, b) to be a different number from 1? What if we changeF1(a, b)too? Does this affect Proposition 9? And what if we use more than two terms in therecursion relation (6), for instance, if we defineFn(a1, a2, a3, .., ak) usingkterms? Is therea special number (a1, a2, a3, .., ak) associated with such a sequence, as (a, b) is withFn(a, b)? The reader is encouraged to experiment with FIBONACCI -like sequences, and alsoto check out important variations on the FIBONACCI NUMBERS , such as Lucas NUMBERS (see[1]), that produce other interesting SCHNEIDERA cknowledgmentsThis work is dedicated to my son, Max Schneider, a talented young mathematician,musician, programmer and thinker, who happily worked through these proofs with me asa child over doughnuts on Sunday [1] G. H. Hardy and E. M. Wright,An introduction to the theory of NUMBERS , Oxford UniversityPress, 1979.

10 [2] M. Livio,The GOLDEN RATIO : The story of phi, the world s most astonishing number, BroadwayBooks, of Mathematics and Computer ScienceEmory University400 Dowman Dr., W401 Atlanta, Georgia 30322E-mail