Linear Algebra

402 Chapter Five. SimilarityIntroducing the vectors and taking the n-th power, we have( ) ( ) n ( )f(n + 1) 1 1 f(1)=f(n) 1 0 f(0)( √1+ 5 1− √ ) ( 5 1+ √ n) ( )51√5= 2 220− 1−√ 5 (f(1) )2 √ 5n1 1f(0)01− √ 52√−11+ √ 55 2 √ 5We can compute f(n) from the second component of that equation.[( ) n ( ) n ]f(n) = √ 1− 51 + √ 521 − √ 52Notice that f is dominated by its first term because (1 − √ 5)/2 is less thanone, so its powers go to zero. Although we have extended the elementary modelof population growth by adding a delay period before the onset of fertility, wenonetheless still get an (asmyptotically) exponential function.In general, a linear recurrence relation has the formf(n + 1) = a n f(n) + a n−1 f(n − 1) + · · · + a n−k f(n − k)(it is also called a difference equation). This recurrence relation is homogeneousbecause there is no constant term; i.e, it can be put into the form 0 = −f(n+1)+a n f(n)+a n−1 f(n−1)+· · ·+a n−k f(n−k). This is said to be a relation of orderk. The relation, along with the initial conditions f(0), . . . , f(k) completelydetermine a sequence. For instance, the Fibonacci relation is of order 2 andit, along with the two initial conditions f(0) = 1 and f(1) = 1, determines theFibonacci sequence simply because we can compute any f(n) by first computingf(2), f(3), etc. In this Topic, we shall see how linear algebra can be used tosolve linear recurrence relations.First, we define the vector space in which we are working. Let V be the setof functions f from the natural numbers N = {0, 1, 2, . . .} to the real numbers.(Below we shall have functions with domain {1, 2, . . .}, that is, without 0, butit is not an important distinction.)Putting the initial conditions aside for a moment, for any recurrence, we canconsider the subset S of V of solutions. For example, without initial conditions,in addition to the function f given above, the Fibonacci relation is also solved bythe function g whose first few values are g(0) = 1, g(1) = 1, g(2) = 3, g(3) = 4,and g(4) = 7.The subset S is a subspace of V . It is nonempty because the zero functionis a solution. It is closed under addition since if f 1 and f 2 are solutions, thena n+1 (f 1 + f 2 )(n + 1) + · · · + a n−k (f 1 + f 2 )(n − k)= (a n+1 f 1 (n + 1) + · · · + a n−k f 1 (n − k))= 0.+ (a n+1 f 2 (n + 1) + · · · + a n−k f 2 (n − k))