Together witha1=1=F2anda2=2=F3, we obtainan=Fn+1.2.Again consider the set of different possible strings. This set may be divided into two nonoverlappingsubsets: those strings for which the one andnare not interchanged, and those strings for which theyare interchanged. For the former, the number of different strings is given byan=Fn+1. For the latter,the number of different strings is given byan-2=Fn-1. We therefore havebn=Fn+1+Fn-1.From (1.3), the relation satisfied bybnis the same as that satisfied by thenth Lucas number, so thatbn=Ln.
90APPENDIX C. PROBLEM AND PRACTICE QUIZ SOLUTIONSSolutions to thePractice quiz: The Fibonacci numbers1.c. See alsoLecture 1. The relevant table for the rabbit pair population ismonthJFMAMJJASONDJjuvenile101123581321345589adult01123581321345589144total1123581321345589144233The number of adult rabbit pairs in the fifth month is 3.2.c. See alsoLecture 1, Problem 1. The Fibonacci numbers can be extended to negative indices usingthe recursion relation. The result isF-n= (-1)n+1Fn. Whenn=5, we haveF-5=F5=5.3.c.See alsoLecture 1, Problem 4.If we line up the Fibonacci numbers and the Lucas numbersproperly, we haveFibonacci011235813Lucas13471118and it can be observed (without proof) thatLn=Fn-1+Fn+1. We can also use the Fibonacci recursionrelation to lower the index ofFn+1(andFn). We haveFn+1=Fn+Fn-1= (Fn-1+Fn-2) +Fn-1=2Fn-1+Fn-2.Therefore,Ln=Fn-1+Fn+1=3Fn-1+Fn-2.Both A and B are true.
91Solutions to theProblems for Lecture 31.(a)Φ-1=√5+12-1=√5-12=φ,(b)1Φ=21+√5×1-√51-√5=21-√5-4=√5-12=φ.(c)Φ2=√5+12!2=5+2√5+14=√5+32=Φ+1.(d)φ2=√5-12!2=5-2√5+14=-√5+32=-φ+1.2.We multiply1Φ=Φ-1byΦand rearrange to obtainΦ2=Φ+1.
92APPENDIX C. PROBLEM AND PRACTICE QUIZ SOLUTIONSMultiplying both sides byΦn-1yields the desired result:Φn+1=Φn+Φn-1.3.We substituteφ=1/Φinto1Φ=Φ-1to obtainφ=1φ-1.Multiplying both sides byφnand rearranging terms yields the desired result:φn-1=φn+φn+1.
93Solutions to theProblems for Lecture 41.WriteFk+nFk=Fk+nFk+n-1×Fk+n-1Fk+n-2× · · · ×Fk+1Fk.Then taking limk→∞, and usinglimj→∞FjFj-1=Φ,one obtains directlylimk→∞Fk+nFk=Φn.2.We prove (4.2) by mathematical induction.Base case:Forn=1, the relation (4.2) becomesΦ=Φ, which is true.Induction step:Suppose that (4.2) is true for positive integern=k. Then we haveΦk+1=ΦΦk=Φ(FkΦ+Fk-1)(from induction hypothesis)=FkΦ2+Fk-1Φ=Fk(Φ+1) +Fk-1Φ(fromΦ2=Φ+1)= (Fk+Fk-1)Φ+Fk=Fk+1Φ+Fk(from recursion relation)so that (4.2) is true forn=k+1. By the principle of induction, (4.2) is therefore true for all positiveintegers.
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- Jamie Watson
- Fibonacci number, Golden ratio