Recurrence relations. Warm-up: The Fibonacci sequence

Transcription

1 1 Recurrence relations Warm-up: The Fibonacci sequence The Fibonacci sequence is the sequence f n satisfying f 0 = 0, f 1 = 1 f n+2 = f n + f n+1 0,1,1,2,3,,8,13,21,34,,... Such an expression for a term in a sequence as a function of previous terms is called a recurrence relation. Other examples: h 0 = 1 h n+1 = h n + 3 h 0 = 1 h n+1 = 3h n In these cases, we can easily find an expression for h n in terms of n. Can we do this for f n? To do, we should consider the more general problem where we vary the initial values f 0 and f 1, and just consider sequences f n satisfying the recurrence relation f n+2 = f n + f n+1. If f n and f n are two such sequences, then is c 1 f n + c 2 f n for any c 1, c 2 (i.e. the lutions form a vector space). So if we can find me lutions to the Fibonacci recurrence relation, we can easily generate more - perhaps including the Fibonacci sequence itself. (In fact, if you recall your linear algebra, you should be able to see that we only need to find two linearly independent sequences to generate all of them) Let s loo for lutions of the particularly simple form f n = q n with q 0. Then the recurrence relation becomes q n+2 = q n + q n+1 q n (q 2 q 1) = 0 (q 2 q 1) = 0 (since q 0) This is a quadratic equation, it has two lutions. They are φ = 1+ 2, φ = 1 2. (φ is nown as the Golden Ratio; it is the unique positive real satisfying 1+φ φ = φ)

2 2 So for any c 1 and c 2, c 1 φ n + c 2 φ n satisfies the Fibonacci recurrence relation. Let s find such a sequence which satisfies the initial conditions of the Fibonacci sequence, f 0 = f 1 = 1; then it must be the Fibonacci sequence. c 1 + c 2 = 0 c 1 φ + c 2 φ = 1 We can lve this system of simultaneous equations So we obtain ( c1 c 2 ( ) ( ) ( ) 1 1 c1 0 φ φ = c 2 1 ) = 1 ( ) ( ) φ 1 0 = 1 ( ) 1 φ φ φ Theorem: The Fibonacci numbers are f n = φn φ n where φ = 1+, φ = Note that φ = 1 φ, we could al write this as f n = φn (1 φ) n. Homogeneous Linear Recurrence Relations with Constant Coefficients A homogeneous linear recurrence relation with constant coefficients is an equation h n+ = a 0 h n + a 1 h n a 1 h n+ 1, with a i complex numbers. is the order of the recurrence relation. A number sequence h n satisfying the recurrence relation is called a lution to the recurrence relation. If we add initial conditions h 0 = c 0,..., h 1 = c 1, this clearly uniquely determines a lution. Examples: (i) The Fibonacci sequence. (ii) Geometric sequences, h n+1 = ah n.

3 3 (iii) The life-cycle of inventioni exemplicus is as follows: a new hatchling remains in the larval stage until its first summer, then spends a year maturing, and in the subsequent summer lays a clutch of 7 eggs (which quicly hatch into larvae), then in the summer after lays a second clutch of 6 eggs, then dies. All excemplicus are female (they reproduce parthenogenetically). Suppose no exemplicus die except at the end of their life cycle. If 100 exemplicus hatchlings are introduced one summer, how many exemplicus larvae will there be at the end of the nth summer thereafter? Solution: At the end of the nth summer, there are 7 larvae born from each 2- year-old exemplicus, and 6 from each 3-year-old. So h n = 6h n 3 + 7h n 2 for n 3, i.e. h n+3 = 6h n + 7h n+1 for n 0. We al have the initial conditions h 0 = 100, h 1 = 0, h 2 = 700. So we get h 3 = 600, h 4 = 4900, h = 8400, h 6 = 37900,... We proceed to generalise the lution to the Fibonacci recurrence relation to lve general homogeneous linear recurrence relation with constant coefficients. Given a recurrence relation h n+ = a 0 h n + a 1 h n a 1 h n+ 1, i.e. h n+ = 1 j=0 a jh n+j, we again loo for lutions h n = q n. Clearly h n = q n is a lution iff q = a 0 + a 1 q a 1 q 1, i.e. q a 1 q 1... a 1 q a 0 = 0. The polynomial x a 1 x 1... a 1 x a 0 is called the characteristic polynomial of the recurrence relation. It is a degree polynomial, has roots in the complex numbers (counting multiplicities). Suppose that it has distinct roots, q 1,..., q. (See the Bonus section for what happens when we have repeated roots.) Claim: the vectors ((q1, 0..., q1 1 ),..., (q 0,..., q 1 )) are linearly independent in C, Proof: Otherwise, considering the columns of the -by- matrix whose rows are these vectors,

4 4 the vectors ((q1, 0..., q 0 ),..., (q 1 1,..., q 1 )) are linearly dependent. i.e. there are b 0,..., b 1 C not all 0, such that the polynomial b 0 + b 1 x b 1 x 1 has roots q 1,..., q. But a degree 1 polynomial can t have distinct roots; contradiction. So any given initial conditions h 0 = c 0,..., h 1 = c 1 can be written as a linear combination h n = b 1 q n b q n = i=1 b iq n i. Taing this as a definition of h n for all n, we see that not only does it satisfy the initial conditions by choice of b i, but it satisfies the recurrence relation; indeed h n+ = i=1 b iq n+ i = i=1 b iqi n qi = i=1 b iqi n ( 1 j=0 a jq j i ) = 1 j=0 i=1 b iqi n a j q j i = 1 j=0 a j i=1 b iq n+j i = 1 j=0 a jh n+j Example: Let s lve the exemplicus example. h n+3 = 6h n + 7h n+1, h 0 = 100, h 1 = 0, h 2 = 700. The characteristic polynomial is x 3 7x 6 = (x 3)(x + 2)(x + 1) the lutions are of the form h n = b 1 3 n + b 2 ( 2) n + b 3 ( 1) n. Solving 100 = h 0 = b 1 + b 2 + b 3 0 = h 1 = 3b 1 2b 2 b = h 2 = 9b 1 + 4b 2 + b 3 gives b 1 = 4, b 2 = 80, b 3 = 2. So the lution is h n = 4 3 n + 80 ( 2) n 2 ( 1) n.

5 Bonus: Solving recurrence relations with generating functions Generating functions provide a convenient device for lving recurrence relations (although in theoretical terms, they only provide a different way to pacage the same linear algebra). If g(x) is the generating function for the sequence h n, i.e. the coefficient of x n in g(x) is h n, then the coefficient of x n+1 in xg(x) is h n. So if h n satisfy a recurrence relation h n+ = a 0 h n + a 1 h n a 1 h n+ 1 then in g(x) a 0 x g(x) a 1 x 1 g(x)... a 1 xg(x), x n+ has coefficient 0 for n 0, i.e. this is a polynomial of order 1. Using initial conditions, we can find this polynomial, and express g(x) as a rational function. For example, consider the Fibonacci relations f n+2 = f n + f n+1, f 0 = 0, f 1 = 1. If g(x) is the generating function, then g(x) xg(x) x 2 g(x) = f 0 +f 1 x+f 2 x 2 +f 3 x f 0 x f 1 x 2 f 2 x 3... f 0 x 2 f 1 x 3... = f 0 + (f 1 f 0 )x + (f 2 f 1 f 0 )x 2 + (f 3 f 2 f 1 )x = f 0 + (f 1 f 0 )x = 0 + (1 0)x = x, (1 x x 2 )g(x) = x g(x) = x. 1 x x 2 Furthermore, by factoring the denominator and finding partial fractions, we can expand this as a power series and lve the recurrence equations. In this case, the lutions to 1 x x 2 = 0 are the reciprocals of the lutions φ, φ to x 2 x 1, g(x) = x 1 x x 2 x = = (x φ 1 )(x φ 1 ) φφ x (1 φx)(1 φ x) b 1 φ x = a 1 φx + where 0 = a + b φφ = φ a φb => b = a => φφ = a(φ φ ) => a = 1 (using the definitions of φ, φ ) => b = 1

6 6 g(x) = 1 ((1 φx) 1 (1 φ x) 1 ) = 1 (( n=0 φn ) ( n=0 φ n )) = 1 n=0 (φn φ n ) we reclaim the formula we found before, f n = φn φ n, and we had to do the same algebra to get there. Bonus: abstract reformulation, and handling repeated roots Consider the (infinite dimensional) complex vector space of complex number sequences h = h 0, h 1,... Let σ be the downshift operator, which from a sequence h obtains the new sequence (σh) n = h n+1. Note that this is a linear map. Then a linear homogeneous constant coefficient recurrence relation, which we can write as i=0 a ih n+i = 0, with a 0, can be rewritten as ( i=0 a iσ i )h = 0. The subspace of lutions to this is then the ernel of the linear operator i=0 a iσ i. This is a finite dimensional vector space. The lution method described above is a matter of finding a basis of eigenvectors of σ on this space. Note that eigenvectors are precisely geometric series cq n. In general of course, the eigenvectors of σ won t span the space. But its generalised eigenvectors will. One can chec inductively that the th generalised q-eigenspace of σ, i.e. the ernel of (σ q), is the space of sequences f(n)q n where f is a polynomial of degree at most 1. So if the characteristic polynomial factors as i=0 a iσ i = Π(σ q i ) i, the space of lutions has a basis of eigensequences n j qi n where j < i, any lution can be expressed as a linear combination of these. We might as well note that this abstract formulation al applies to homogeneous linear differential equations over the constants: replace the space of sequences with, say, the space of complex analytic functions in one variable, and replace σ with the differentiation operator. The th generalised λ-eigenspace consists of f(x)e λx for f(x) a polynomial of degree at most 1.