Lecture Notes The Fibonacci Sequence age De nition: The Fibonacci sequence starts with and and for all other terms in the sequence, we must add the last two terms. F F and for all n, + + + So the rst few terms of the Fibonacci sequence are ; ; ; 3; ; 8; 3; ; 3; ; ; ; : : : The de nition shown above is a recursive one. If we are needed to comute the 00th term of the sequence, we would be forced to comute rst the rst 99 terms in the sequence. So we are naturally interested in nding a formula that enables us to comute the 00th element directly. Such a formula is callled elicit. Like so many things about this sequence, the elicit formula for its nth term is fascinating and surrising. We will derive this formula later. The Fibonacci sequence is named after Leonardo Fibonacci and has very strange and beautiful roerties. of these roerties are connected to the golden mean, ' + : 68 033 988 79 9. Consider now another sequence, fq n g that is formed by taking the quotients of consecutive term in the Fibonacci sequence. That is, ; ; 3 ; 3 ; 8 ; 3 8 ;... q n + for all natural number n. The decimal resentations of the terms in this sequence show an interesting attern. 3 : 3 :66667 8 :6 3 8 :6 3 :6 38 6 3 : 69 07 6 A lot The quotients oscillate back and forth and seem to be closer and closer to each other. Amazingly, there is only one number that is inside all of the "swirls" shown on the icture above. These ratios aroach a single number. We call this number the limit of this sequence and we comute its eact value in the samle roblems. De nition: A Fibonacci-tye of a sequence starts with any two real numbers and the rest of the sequence is generated the same way the Fibonacci sequence is. f ; f R and for all n, f n + f n+ f n+ Suose we start with f 3 and f. The rst few terms of this Fibonacci-tye sequence are 3; ; 7; ; ; 6; ; 67; 08; 7; : : : We can de ne oerations on Fibonacci-tye sequences. Consider fa n g and fb n g de ned as follows: fa n g : 3; ; 7; ; 8; 9; 7; 76; 3; 99; : : : fb n g : ; 7; 9; 6; ; ; 66; 07; 73; 80; : : : c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age We can multily a sequence by a number by multilying each term by that number: fa n g : 6; 8; ; ; 30; ; 8; 3; 6; 30; : : : and the resulting sequence is still Fibonacci-tye. We can also add two sequences by adding them term by term: c n a n + b n c n fa n + b n g : ; ; 6; 7; 3; 70; 3; 83; : : : and the sum is again Fibonacci-tye. determined by its rst two terms. Also, it is very easy to see that every Fibonacci-tye sequence is uniquely These roerties are used when we derive the elicit formula for the nth term of the Fibonacci sequence. c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 3 Samle Problems. Solve the equation +.. De ne ' + and. Prove each of the following. a) ' b) ' 3. Find the limit of of the sequence formed from consecutive terms in the Fibonacci sequence. In short, + comute lim. n!. De nition: Two ositive integers are relatively rime if their greatest common divisor is. Prove that any two consecutive terms of the Fibonacci sequence are relatively rime.. Consider a Fibonacci-tye of a sequence with rst term and second term. Is there a value of for which all terms of the sequence fall between 00 and 00? 6. De nition: A geometric sequence is de ned as a; ar; ar ; ar 3 ; ar ; ::::::. The number r is called the common ratio of the sequence because if r 6 0, then r a n+ for all n N. It is clear that a geometric sequence a n is determined by its rst element and common ratio. One great advantage of a geometric sequence over a Fibonacci-tye of a sequence is that there is a very easy elicit formula for the nth term of the sequence: a n ar n : Is there a Fibonacci-tye sequence that is also a geometric series? 7. Consider the geometric sequence de ned by rst element and common ratio r +. Comute the eact value of the 9th term in the sequence. 8. Use results form the revious roblems to nd the elicit formula for the nth term of the Fibonacci sequence. c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age Solutions - Samle Problems. Solve the equation + Solution. Solution. Using the quadratic formula Comleting the square 0 ; ( ) 0 + 0 {z } 0! 0!! + 0 + + +. De ne ' + a) ' Solution: b) ' Solution: ' and. Prove each of the following. + + We rationalize the radical eression ' + + + ' using its conjugate. + + c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 3. Find the limit of of the sequence formed from consecutive terms in the Fibonacci sequence. In short, + comute lim. n! Solution : Let us assume rst that the consecutive terms ; ; 3 ; 3 ; 8 ; 3 8 ; 3 ; 3 ; 3 ; ; ; : : : do aroach a single number. Imagine we are much further into the sequence, after millions and millions of terms. Then these numbers are very close to and thus also very close to each other. Sort of like we are in the Fibonacci sequence: + + + + + Using more general notation, we arrive to the same conclusion. For very large values of n, + + we are in the Fibonacci sequence: + + + + + + + + + + + + + + + + + We solve the equation + + multily by + 0 ; We rule out because it is negative, and the sequence clearly aroaches a number above 0:6. other solution, +, the golden mean is the limit. The c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 6 Solution : This is the same comutation but this time it is resented with calculus notation. + Let us denote lim by. n! + lim lim n! n! + lim lim n! n! + lim lim n! n! + lim lim n! n! + + + + F n+ Fn + F n+ + + Fn + + + lim lim + n! n! + + + 0 ; We rule out because it is negative, and the sequence clearly aroaches a number above 0:6. + other solution, the golden mean is the limit. lim +. n!. De nition: Two ositive integers are relatively rime if their greatest common divisor is. Prove that any two consecutive terms of the Fibonacci sequence are relatively rime. Solution: this is an interesting alication of roofs by contradiction. Suose for a contradiction that there eist two consecutive terms F k and F k+ (for some natural number k) that are not relatively rime. Then there eists a ositive integer d > such that d is a divisor of both F k and F k+. Then there eist and ositive integers such that F k d and F k+ d. We claim that then d is also a divisor of F k. F k + F k F k+ F k F k+ F k d d d ( ) Thus d also divides F k. Net we similarly rove that d is then also a divisor of F k and F k 3 ; and so on, all the way till F. Thus d is a divisor of F. This is imossible because d > and F. This is a contradiction comleting our roof.. Consider a Fibonacci-tye of a sequence with rst term and second term. Is there a value of for which all terms of the sequence fall between 00 and 00? Solution: The Fibonacci sequence vey quickly becomes very large. The question is: how can we ensure that the terms of the sequence do not become large? Consider a Fibonacci-tye sequence with rst term and second term. ; ; + ; + ; 3 + ; + 3; 8 + ; 3 + 8; + 3; : : : : eventually the terms are +. If is ositive, even if tiny, the other art alone, will ensure that the nth term is very large. Thus, if we want the terms to stay small, we need to be negative. This idea generalizes. We want every second term ositive and every other term negative, because two consecutive terms with the same sign guarantee that the terms after that get very large. Suose that a and b are two consecutive terms with the same sign. Then from then on, we have that a; b; a + b; a + b; a + 3b; 3a + b; a + 8b; : : : ; a + + b ; : : : : So we want alternating sings in the sequence That is: ; ; + ; + ; 3 + ; + 3; 8 + ; 3 + 8; + 3; : : : : The c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 7 ositive, negative, + ositive, + negative, 3 + ositive, + 3 negative, 8 + ositive, 3 + 8 negative, + 3 ositive, and so on. We solve all these inequalities: < 0 + > 0 ) > + < 0 ) < 3 + > 0 ) > 3 + 3 < 0 ) < 3 8 + > 0 ) > 8 3 + 8 < 0 ) < 8 3 + 3 > 0 ) > 3 It looks like must be between the values de ned by consecutive terms of the Fibonacci sequence. These rations dislay a strange behavior, they siral around over a smaller and smaller interval (see roblem 3). The only di erence here is that we are looking at instead of +. The ratios in this roblem aroach + the negative recirocal of the golden mean. We rationalize and obtain. This is the only + number that will work for. This sequence will have terms with alternating signs and thus each term will have a smaller absolute value than the revious term. It is an amazing thought that for any other values of, the sequence will reach huge numbers and outgrow any bound. We can resent a bit more formal comutation: denote the sequence by a n : fa n g : ; ; + ; + ; 3 + ; + 3; 8 + ; 3 + 8; + 3; : : : : Notice that for all n 3 where f g is the Fibonacci sequence. a n + f g : ; ; ; 3; ; 8; 3; ; 3; ; ; We want a, a 3, a,... ositive and a ; a ; a 6 ;... negative. For all n, we need a n+ > 0 and a n < 0. a n + and a n+ + + > 0 and + < 0 > < < and < for all n for all n Since these quotiens oscillate around and enclose only a single number, must be that number. Thus lim n! F m+ lim m! F m + c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 8 6. De nition: A geometric sequence is de ned as a; ar; ar ; ar 3 ; ar ; ::::::. The number r is called the common ratio of the sequence because if r 6 0, then r a n+ for all n N. It is clear that a geometric sequence a n is determined by its rst element and common ratio. One great advantage of a geometric sequence over a Fibonacci-tye of a sequence is that there is a very easy elicit formula for the nth term of the sequence: a n ar n : Is there a Fibonacci-tye sequence that is also a geometric series? Solution: Let fa n g be a Fibonacci-tye geometric sequence with rst element a and comon ratio r. Then the rst three elements (since geometric) are a; ar; ar Let us assume that a 6 0. (The constant zero sequence is both Fibonacci-tye and geometric, but not very interesting.) The sequence is also Fibonacci-tye and so a + ar ar 0 ar ar a factor out a 0 a r r divide by a 0 r r r ; ( ) At this oint, we are not surrised that we again bumed into the golden mean. will be very useful later on. Suose that a. Then one sequence is Both solutions work, which ; + ; 3 + ; + ; 7 + 3 ; : : : is an increasing sequence that grows unbounded. The other sequence is 3 7 3 ; ; ; ; ; : : : is a sequence with alternating sings and thus small terms, and the two sequences aear to be conjugates of each other, term by term. The comutation above shows that only r will result in a non-zero Fibonacci-tye sequence. On the other hand, all other such sequences are just constant muliles of these two. All Fibonacci-tye sequences with rst term a are of the form +! 3 +! a; a ; a ; a + 7 + 3! ; a ; : : : and!! 3 7 3! a; a ; a ; a ; a ; : : : What makes these sequences secial is that their nth term can be so easily determined because they are geometric sequences as well. 7. Consider the geometric sequence de ned by rst element and common ratio r +. Comute the eact value of the 9th term in the sequence. Solution: a 9 ar 8 +! 8 +! 8 c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 9 We start with +! +! + 6 + 3 + We square this number: +! +! 3 3 +! 3 + + 6 7 + 3 We square again: +! 8 +! 3! 7 + 3 7 + 3 9 + + 9 + + 9 + 7 + and so a 9 7 +. This might seem laborous but if n is large, it is still much better than having to comute all revious terms. 8. Use results form the revious roblems to nd the elicit formula for the nth term of the Fibonacci sequence. Solution: We will eress the Fibonacci sequence as the sum of two Fibonacci-tye geometric sequences. First, we need to verify that the constant mutlies and sums of Fibonacci-tye sequences are still Fibonaccitye. Second, we will use the fact that the rst two elements uniquely determine any Fibonaccy-tye of sequence. De ne fa n g and fb n g geometric sequences as follows. a and r a + Thus a n +! n and b n! n and b n and r b. These sequences are also Fibonacci-tye (see roblem 6). Thus, any constant multiles and sums formed from these sequences will still be Fibonacci-tye. Could we use fa n g and fb n g to "concoct" the Fibonacci sequence? Let and y be real numbers such that for all n c n a n a n and c and c If we could nd such and y, we would be done because a Fibonacci-tye sequence that begins with and is THE bonacci-sequence. c a a ) +!! c a a ) +!! c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age 0 8 >< +!! The system >: +!! it. We simlify both equations. Since is a linear system in and y so we should be able to solve +! + 6 + 3 + the system is 8>< >: +! 3 +! 3!! Let us multily by + We solve for y in the rst equation: 3 + 3 y + and substitute into the second equation and solve for : 3 + + + 3 3 + + + 3 + 3 + 3 + 3 + + 6 + + + multily by add subtract 6 factor out divide by We substitute this into the eression eressing y. y + and c Hidegkuti, 03 Last revised: August, 03
Lecture Notes The Fibonacci Sequence age y +! + 0 So and y. + + + + Consequently, the nth term in the Fibonacci sequence is 0! + + 0 +! n! n! This formula seems very unlikely to roduce integers. Let us see the rst few elements generated by the formula. F +!!! + + 0 F @ +!! A 3 + 3! 3 + 3 + Before comuting F 3, let us comute +! 3. +! 3 +! +! +! 3 +! + 3 + 8 + + We similarly obtain the eact value of 0 F 3 @ +! 3! 3 and then we are ready to comute F 3.! 3 A + For more documents like this, visit our age at htt://www.teaching.martahidegkuti.com and click on Lecture Notes. Our goal is to emower students to learn and enjoy mathematics free of charge. If you have any questions or comments, e-mail to mhidegkuti@ccc.edu. c Hidegkuti, 03 Last revised: August, 03