The Fibonacci Sequence and Recursion


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1 The Fibonacci Sequence and Recursion (Handout April 6, 0) The Fibonacci sequence,,,3,,8,3,,34,,89,44,... is defined recursively by {, for n {0,}; F n = F n + F n, for n. Note that the first two terms in the sequence are defined explicitly; thereafter, each term is the sum of the previous two terms. This is an example of a recursive definition. In order to evaluate F 00 using this definition, we would first have to evaluate the previous hundred terms in the sequence. We will soon find a closed form expression for F n which does not require evaluation of all the previous terms in the sequence in order to find a given term. Both recursion and induction may be described as chain reactions; the difference is that mathematical induction is used to prove an infinite sequence of statements, whereas recursion is used to define an infinite sequence of quantities. The connection is a close one; and in order to prove facts about recursively defined sequences, it is natural to try to use induction. The following is an example of a fact about Fibonacci numbers, which we prove by induction. Theorem. For every n, we have Fn = F n+ F n + ( ) n. Proof. The result holds for n = since F = = = = F F 0 + ( ). Now assume that the result holds for n = k, i.e. Fk = F k+f k + ( ) k where k is some positive integer. Then Fk+ = (F k + F k )F k+ (using the recursive formula for F n ) = F k+ F k + F k+ F k = F k+ F k + Fk ( ) k (by the inductive hypothesis) = (F k+ + F k )F k + ( ) k+ = F k+ F k + ( ) k+ (again using the recursive formula for F n )
2 so that the result holds for n = k + whenever it holds for n = k. By induction, the identity holds for all n as required. We now derive a closed formula for F n, i.e. an explicit formula not requiring recursion. The entire sequence is embodied in the series F n x n = + x + x + 3x 3 + x 4 + 8x + 3x 6 + x 7 +. The idea is to find a simple way of evaluating f(x), from which we can identify all coefficients in the series for f(x). In order to make use of the recursion formula for F n, we write which can be solved for f(x): + x + = + x + = + x + = + x + x F n+ x n+ (F n+ + F n )x n+ F n+ x n+ + F n x n+ F n+ x n+ + x F n x n = + x + x ( f(x) ) + x f(x) x x. Thus f(x) is a rational function in x, i.e. a quotient of two polynomials in x. If we can compute the coefficients in the Taylor series of this function, then we will have a formula for F n. However, it is not necessary to know anything about Taylor series to solve this problem; all we need is a little algebra skill, and one very basic series, the geometric series x = + x + x + x 3 + x 4 +. Even if you haven t seen (or don t remember) this series, you can quickly verify it by crossmultiplying: ( x)( + x + x + x 3 + ) = x + x x + x x 3 + x 3 + =.
3 This suggests possibly writing (x + x ) = + (x+x ) + (x+x ) + (x+x ) 3 + (x+x ) 4 + = + (x+x ) + (x +x 3 +x 4 ) + (x 3 +3x 4 +3x +x 6 ) + (x 4 + ) + = + x + x + 3x 3 + x 4 + which is encouraging since we see the Fibonacci numbers appearing as coefficients; however, this doesn t seem to lead to an explicit formula for the Fibonacci numbers. A better idea is to find a partial fraction decomposition of f(x). We first factor the denominator as where x x = ( αx)( βx) α = , β = We may describe α and β as the reciprocal roots of the quadratic, rather than the roots themselves; the reason why they are easier to use will soon become apparent. We try to find constants A and B such that x x = ( αx)( βx) = A αx + B βx. You may have seen this technique used in Calculus II; if not, keep in mind the basic idea, which is as follows. The right hand side of the latter expression can be combined using a common denominator to obtain a rational function whose denominator is ( αx)( βx). Instead what we want is the reverse process: to split up the rational function into two terms, each having a single linear factor in the denominator. Multiplying both sides by ( αx)( βx) in order to clear denominators, we obtain the identity = ( βx)a + ( αx)b. This is a polynomial identity! so equality must hold if x is replaced by any constant. In particular we evaluate at x = α since this causes the B term to vanish, thereby allowing us to solve for A: ( = β ) A = α β α α A = α A, so A = α. Similarly, evaluating at x = β causes the A term to vanish, allowing us to solve for B: = ( α ) B = β α β β B = 3 β B, so B = β.
4 This gives us the desired partial fraction decomposition of f(x): x x = A αx + B βx = ( α αx β ). βx This is just what we need to expand f(x) as a power series: we expand both terms as geometric series to obtain ] [α( + αx + α x + α 3 x 3 + ) β( + βx + β x + β 3 x 3 + ) = (α n+ β n+ )x n. Reading off the coefficient of x n gives F n = αn+ β n+ for all n 0. All of this is demonstrated in the attached Maple worksheet. Recall that the set of polynomials in x with real coefficients, forms the polynomial ring R[x]. The set of rational functions in x with real coefficients is the field of rational functions R(x) = { f(x) g(x) } : f(x),g(x) R[x], g(x) 0. This is a field, not just a ring: it is closed under division. Here we see yet another analogy between the ring Z of integers and the ring R[x] of polynomials: both rings can be extended to fields by formally introducing quotients. We call Q the fraction field of Z; and R(x) is the fraction field of R[x]. Note that R[x] R(x). We also consider the set of all power series in x with real coefficients: R[[x]] = { a 0 +a x+a x +a 3 x 3 + : a 0,a,a,a 3,... R }. This is not a field; it is a ring with R[x] R[[x]]. The latter inclusion follows from the fact that every polynomial may be regarded as a power series where most of the coefficients (all but finitely many) are zero. Elements of R[x], or of R(x), or R[[x]], are first and foremost regarded as formal objects. In some (but not all) cases, they may represent actual functions. You are quite accustomed to using elements of R[x] and R(x) to represent functions; but we have tried to also highlight applications where they are used in other ways without serving as functions. An example is x x R(x) used above to find a closed formula for the Fibonacci numbers. Never did we evaluate f(a) for any number a R; so the popular term rational function is misleading. 4
5 This point is even more important when studying power series. In order to study an arbitrary sequence a 0,a,a,a 3,..., it is often useful to study instead the associated power series g(x) = a n x n = a 0 + a x + a x + a 3 x 3 +, popularly known as the generating function of the original sequence. In many cases this does not represent a function at all; so in general, the name generating function is quite a misnomer. For example the factorial sequence gives rise to a power series p(x) = n!x n = + x + x + 6x 3 + 4x 4 + 0x +. The values p(a) are undefined for any nonzero real number a, although the series itself is a perfectly useful and interesting object. An important part of Calculus II is devoted to the study of when a given power series converges, and so represents an actual function; but none of this is relevant to our discussion! In the case of the series for f(x) defined above, it turns out that the series does converge on a suitable open interval centered at 0; but we don t care about this. Some sources may refer to power series in our setting as formal power series to emphasize the role of power series as purely formal objects without using them to represent functions. However, the term formal is redundant, and used for emphasis only: a power series is, without any added considerations, a purely formal object. Some additional examples of infinite sequences and their generating functions appear below: The sequence,,,,,... has generating function x n = + x + x + x 3 + x 4 + = x. The sequence,,4,8,6,3,... generated recursively by a 0 = ; a n+ = a n for all n 0 has generating function n x n = + x + 4x + 8x 3 + 6x 4 + = x. The sequence,, 3, 4,,... has generating function (n + )x n = + x + 3x + 4x 3 + x 4 + = The sequence, 0,, 0,, 0,... has generating function + x + x 4 + x 6 + x 8 + = x. The sequence,3,6,0,,,8,... defined by a n = ( n+ n 0 has generating function ( n + ) x n = + 3x + 6x + 0x 3 + x 4 + = ( x). ) = ( n+ )( n+ ) for all ( x) 3.
6 We first demonstrate how to list the Fibonacci numbers through recursively: > F:=array(0..00): > F[0]:=: F[]:=: > for n from to 00 do F[n]:=F[n]+F[n]: od: > seq(f[n],..00); () Next, we expand in a power series and observe the appearance of the Fibonacci numbers as its sequence of coefficients: > f:=/(xx^); () > series(f,x=0,0); (3)
7 Finally, we demonstrate how the formula for the n th Fibonacci number is used to find the 00 th Fibonacci number: > alpha:=(+sqrt())/; beta:=(sqrt())/; (4) > Fib:=n>simplify((alpha^(n+)beta^(n+))/sqrt()); () > Fib(00); (6)
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