A DIFFERENCE EQUATION INVOLVING FIBONACCI NUMBERS. Ezio Marchi. IMA Preprint Series #2471. (March 2016)

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A DIFFERENCE EQUATION INVOLVING FIBONACCI NUMBERS By Ezio Marchi IMA Preprint Series #2471 (March 2016) INSTITUTE FOR MATHEMATICS AND ITS APPLICATIONS

1 A DIFFERENCE EQUATION INVOLVING FIBONACCI NUMBERS By Ezio Marchi IMA Preprint Series #2471 (March 2016) INSTITUTE FOR MATHEMATICS AND ITS APPLICATIONS UNIVERSITY OF MINNESOTA 400 Lind Hall 207 Church Street S.E. Minneapolis, Minnesota Phone: Fax: URL:

2 1 A DIFFERENCE EQUATION INVOLVING FIBONACCI NUMBERS by Ezio Marchi *) Abstract: In this short note we solve a non linear difference equation which becomes related to Fibonacci numbers. *) Director del Instituto de Matemática Aplicada San Luis. CONICET. Universidad Nacional de San Luis. Ejercito de los Andes 950 (5700) San Luis, Argentina.

3 2 1- A NON LINEAR DIFFERENCE EQUATION Consider the difference equation given by: x = α n x n x + β n x n = 1, 2, where x n is obtained from the previous values in a non linear way in the form expressed in it. The problem to be solved is to find x n in terms of the firsts values of it. We will propose a general form for the values of x n called quadrature and generating function procedure. The values of β n will not be arbitrary but in the solution they will depend upon other parameters. Simply the proposal is as follows (1) x n = α n c(n) + β n c(n) (2) then x n x = α n α c(n)c(n 1) + β n c(n) α c(n) c(n 1) + α n β c(n 1) + β n β c(n)c(n 1) (3) Introducing this expression in (1) and x, then after identifying terms we obtain the following recursive of difference equations α n+1 c(n + 1) = α n α n α c(n)c(n 1) β n+1 c(n + 1) = β n β α n c(n)c(n 1) (4) (5) On the other hand, c(n) α n β c(n 1) = β α α α c(n 2) (6) then α n β α α + β n = 0 (7) In a similar way 1 β n α c(n 1) β α n α c(n 2) β α = β n α α c(n) (8) from which it follows α nα n β n + β n β α α = 0 (9)

4 3 or β n = α n α n β n β α α (10) which is a condition to be fulfilled by the coefficient β n as we already mentioned. From the equality β n = V α n β α α = α n α n β n β α α (11) one derives α n β n = α 2 β α β α (12) This last difference equation is the same as that equation obtained by multiplying (4) and (5). Now the equation (12) has the general form of ɤ n = ε ɤ ɤ (13) Developing the first terms of (13) we have ɤ n = ε ε ɤ 2 ɤ n 3 = ε ε ε 2 n 3 ɤ 3 n 3 ɤ n 4 (14) = ε ε ε 2 n 3 ε 3 n 4 ɤ 5 n 4 ɤ n 5 = ε ε ε n 3 ε n 4 ε n 5 ɤ n 5 ɤ n = ε ε ε n 3 ε n 4 ε n 5 ε n 6 ɤ 13 8 n 6 ɤ n 7 Where yon immediately see that Fibonacci numbers appear in the developing of ɤ n. We define (15) p(0) p(1) p(2) p(3) p(4) p(5) p(6) p(7) p(8) as the Fibonacci numbers which are recursively given by with p(0) = 0 and p(1) = 1. p(n + 1) = p(n) + p(n 1)

5 4 In order to find the solution of (13) let us consider some equalities. First we well prove that ε s n = ε s (16) In order to see that this equality hold true consider the equality ε s p(s 1) = (17) which is obtained just by a change of variables s + 1 = s. Using this last equality then since p(0) = 0 and p(1) + p(0) = p(2) then p(s 1) p(s+1) = p(1) ε = εn s p(s+1) p(2) p(s+1) ε = εn s (18) Now with the change of variable s 1 = s we have n ε n+1 s p(s+1) = (19) and therefore the equality (16) is valid. Now we will claim that ɤ n = ɤ 1 p(n) ɤ0 p() (20) is the solution of the difference equation (13). We will prove it by induction. For n=2 we have 1 ɤ 2 = ε 2 s ɤ 1 p(2) ɤ0 p(1) = ε1 ɤ 1 ɤ 0 (21) Now assume that the formula is true for j n 1 then we have to prove that it is valid for n. This is equivalent to prove that the next equality is true ɤ n = p(n) p() ε n s ɤ 1 ɤ0 (22)

6 5 We remember that ɤ n = ε ɤ ɤ (13) p(n) p() ɤ 1 ɤ0 = ε ε s n 3 p() p() ɤ 1 ɤ0 ε s ɤ 1 p() ɤ0 p(n 3) (23) or equivalently = ε ε s n 3 ε s (24) Make the change of variable s + 2 = s in the first product of the right hand of (24) and s + 1 = s in the second product of the same hand, we have = ε p(s 2) p(1) p(s 2)+p(s 1) = ε ε εn s p(1) = ε ε εn s (25) and since p(1) = p(2) p(1) p(2) = ε ε εn s and this the equality given by (24) or (25) is true. Then applying the formula (22) to the recurrence equation (12) one gets the solution α n β n = α n s 2 p(n) p(n) p() p() α 1 β1 α0 β0 (26) On the other hand the equation (4) has the solution α n c(n) = α n s p(n) p(n) p() p() α 1 c1 α0 c0 (27) and equation (5) (28) β n c(n) = α n s p(n) p() β 1 β0 p(n) p() c 1 c0

7 6 Thus the general solution (2) is given by x n = α n s p(n) p() p(n) p(n) p() p() β [α 1 c1 α0 c0 + 1 β0 p(n) p() c ] 1 c0 (29) The equality (11) which is a restriction on the coefficients β n can be expressed as β n = α nα α n s 2 p() p() p() p() α 1 β1 α0 β0 (30) This giving the values α 1, α 0, β 0, β 1 and c 1, c 0 we have solved in a suitable way the problem of finding a general solution for equation (1) by quadrature.

FURTHER REMARKS FOR DIFFERENCE EQUATIONS. Ezio Marchi. IMA Preprint Series #2472. (March 2016)

FURTHER REMARKS FOR DIFFERENCE EQUATIONS By Ezio Marchi IMA Preprint Series #47 (March 16) INSTITUTE FOR MATHEMATICS AND ITS APPLICATIONS UNIVERSITY OF MINNESOTA 4 Lind Hall 7 Church Street S.E. Minneapolis,