K th -order linear difference equation (Normal Form)

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1 The standard form of the general K th -order difference equation is x n + 1 = F(x n, x n - 1, x n - 2,..., x n - K + 1, P) where P is a vector of parameters. However, linear equations are normally expressed with the indices shifted so that the lowest index is n and the highest is n + K. When F is linear, this difference equation is written in the normal form as a sum of sequence terms, between X n and X n + K : K th -order linear difference equation (Normal Form) X n + K + A 1 X n + K A 2 X n + K A 3 X n + K A K - 1 X n A K X n = γ where γ does not depend on any X i. If γ = 0 the equation is homogeneous.

2 For K = 2, the normal form is usually written with coefficients a and b (rather than using A 1 and A 2 ): X n a X n b X n = γ We will assume that the coefficients are constant. Homogeneous Linear Second-order Difference Equation X n+2 + a X n+1 + b X n = 0 H-DE

3 If the constant b = 0 this equation reduces to a first-order homogeneous equation, x n ax n + 1 = 0 or, shifting the indices down by one, in the standard form x n + 1 = -a x n. Thus, we would expect that the solution of the second-order equation should be similar to and an extension of the solution to this first-order equation, which is x n = The second-order equation solution is assumed to be similar, a term raised to the power n. However, not simply the coefficient -a.

4 An unknown parameter is introduced: λ (Lambda) Consider the sequence defined by X n = C λ n where C and λ are constants to be determined. Our objective is to find values of the constants C and λ that will make the sequence {X n } a solution of the difference equation H-DE. According to this formula, the (n + 1) st and (n + 2) nd terms of the sequence are: X n+1 = and X n+2 =

5 Substituting these into the difference equation X n+2 + ax n+1 + bx n = 0 gives C λ n+2 + C λ n+1 + C λ n = 0 If C = 0 or λ = 0 the solution is said to be trivial. For X n = C λ n to be a non-trivial solution of the equation λ must be a root of the Auxiliary equation: λ 2 + λ + b = 0 The two roots of this quadratic equation are and λ 1 = λ 2 = For each root the sequence λ n } is a solution of the H-DE.

6 The most general solution is a linear combination of these two solutions, which can be written as X n = C 1 λ 1 n + C 2 λ 2 n where C 1 and C 2 are arbitrary constants.

7 However, there are actually three different forms of the general solution, corresponding to the three types of roots that a quadratic equation can have. Remember that the quadratic λ 2 + λ + b = 0 will have either (i) two distinct real roots, (ii) or, two identical or repeated roots, (iii) or, two complex roots and no real roots. λ = 0.5( - a ± a 2 &4b ) The character of the roots will depend on the value of the discriminant D = a 2-4b

8 The roots, which are called eigen values, are: λ = 0.5( ) - a ± a 2 &4b (i) both real and distinct if D = a 2-4 b > 0; (ii) both real and equal if D = a 2-4 b = 0; (iii) complex conjugates if D = a 2-4 b < 0.

9 Case 1: When D = a 2-4b is positive, the roots are real and distinct and the general solution is the sum of C 1 λ n 1 and C 2 λ n 2 : X n = C 1 λ 1 n + C 2 λ 2 n Case 2: When D = a 2-4b= 0, The repeated roots are λ 1 = -a/2. The general solution is then X n = (C 1 + C 2 ) λ 1 n Case 3: When D = a 2-4b < 0, The roots are complex numbers of the form λ = A ± B i i = &1 A = a B = 0.5 4b&a 2

10 y (i) B r O-- - A x A complex numbers A + Bi has an alternative representation, r e ±iθ, in terms of its distance from the origin, its complex modulus or radius r, and its angle measured from the x-axis, θ.

11 Thus, the complex roots, can be expressed as λ = r e ±iθ The modulus r is simply b, while the angle θ must be found by solving & a 2 cos(θ) = for θ = Arccos( & a ) r 2r You will not need knowledge of complex numbers to work the exercises. To evaluate the Arccos function, on most calculators, you can uses the button followed by the Cos button. Inverse As we allow no calculators in exams, you will not need to evaluate Arccos( ).

12 If then λ = r e ±iθ λ n = (r e ±iθ ) n = r n (e ±iθ ) n = r n e ±i@n@θ The dual representation of complex numbers states that hence becomes e ±nθ@i = cos(±nθ) + i sin(±nθ) X n = C 1 λ 1 n + C 2 λ 2 n X n = r n [C 1 e nθ@i + C 2 e -nθ@i ] X n = r n [ C 1 {cos(nθ) + i sin(nθ)} + C 2 {cos(-nθ) + i sin(-nθ)} ] Does this mean that the solution is a complex number?

13 Because cos(-x) = cos(x) sin(-x) = -sin(x) X n = r n [ C 1 {cos(nθ) + i sin(nθ)} + C 2 {cos(nθ) - i sin(nθ)} ] by choosing C 1 = 0.5 (c 1 - c 2 i) and C 2 = 0.5 (c 1 + c 2 i) the imaginary part of this formulas disappears to give X n = [ c 1 cos (nθ) + c 2 sin (nθ)] r n

14 The general solution of the constant coefficient second-order difference equation X n a X n b X n = 0 depends on the roots λ 1 and λ 2 of the Auxiliary Equation λ 2 + a@λ + b = 0 If λ 1 = 0.5( - a + a 2 &4b ) and λ 2 = 0.5( - a - a 2 &4b ) are Case 1: real and unequal, then X n = C 1 λ 1 n + C 2 λ 2 n Case 2: If λ 2 = λ 1, then X n = (C 1 + n@c 2 )λ 1 n

15 Case 3: If λ 1 and λ 2 are complex, then X n = [ C 1 cos (nθ) + C 2 sin (nθ)] r n where b &a r = and cos(θ) = = 2 b &a 2r

16 Determine the (general) solution of x n x n x n = 0 What is a particular solution satisfying x 0 = 1 and x 1 = 5

17 Determine the (general) solution of x n x n x n = 0 What is a particular solution satisfying x 0 = 0 and x 1 = -1

18 Determine the solution of x n x n x n = 0 What is a particular solution satisfying x 0 = 2 and x 1 = 4

19 Modeling multi-generation populations. Basic Change Model of a population {X n } x n = Births - Deaths Consider a short term model of a population of long living animals (so that the death rate is 0, for practical purposes) that do not reproduce in their first generation and have one offspring each generation after that. At generation n, the animals at least two generations old are those counted in the n-1st generation. Thus, the model is Births - Deaths x n = -

20 Leonardo Fibonacci c Fibonacci, or more correctly Leonardo da Pisa, was born in Pisa in 1175AD. He was the son of a Pisan merchant who also served as a customs officer in North Africa. He traveled widely in Barbary (Algeria) and was later sent on business trips to Egypt, Syria, Greece, Sicily and Provence. In 1200 he returned to Pisa and used the knowledge he had gained on his travels to write Liber abaci, in which he introduced the Latin-speaking world to the Arabic decimal number system, i.e, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to replace the Roman system using I, V, X, L, D, M e.g., 1746 instead of MDCCXLVI. The Fibonacci Sequence {f n } satisfies the equation f n+1 = f n + f n-1 See

21 f n+1 = f n + f n-1 If f 0 and f 1 are integers, for n > 1 what type of number will f n be? If f 0 = 1 and f 1 = 1, what is f 6? Then, what is f 600?

22 Assume that a population s size at generation n is given by y n and that the each individual produces two offspring each generation, after which two individuals die for each member that is two generations old. Model this population: Births = Deaths = y n = y n = Births - Deaths Standard form of this equation:

23 Auxiliary equation: Roots: λ 1 = λ 2 = Solution: y n =

24 Assume that a population s size at generation n is given by y n and that the each individual produces three offspring each generation, after which four individuals die for each member that is two generations old. Model this population: Births = Deaths = y n = y n = Births - Deaths Standard form of this equation:

25 Auxiliary equation: Roots: λ 1 = λ 2 = Solution: y n =

26 Assume that a population s size at generation n, y n, is described by the solution of the equation 6 y n y n y n = 0 (1) If y 1 = 2 and y 2 = 1 what is the general solution? (2) What happens to the size of this population as n 6 4?

27 Assume that a population s size at generation n is given by y n and that the each individual produces one offspring each generation, after which four individuals die for each member that is two generations old. Model this population: Births = Deaths = y n = y n = Births - Deaths Standard form of this equation:

28 Auxiliary equation: Roots: λ 1 = λ 2 = Solution: y n =

29 Assume that a population s size at generation n, y n, is described by the solution of the equation 6 y n y n y n = 0 (1) If y 1 = 2 and y 2 = 1 what is the general solution? (2) What happens to the size of this population as n 6 4?

30 If a solution of X n a X n b X n = 0 is X n = 3 n - 2@5 n what are the parameters a and b?

31 lim λ n = n64 lim n@λ n = n64

32 Problem. A quantity X n varies so that the change in the quantity between two successive indices is always one half the change between the previous two indices. Model this system and determine its steady state level. x n = 0.5 x n-1

33 What is the limiting size of a population satisfying the Diff. Eq. X n a X n b X n = 0 lim X n = n64

34 Case 1. lim n64 X n = lim n64 C 1 λ 1 n + C 2 λ 2 n = Case 2. lim X n = lim (C 1 + n@c 2 )λ n 1 = n64 n64

35 Case 3. When r = b and cos(θ) = &a 2 b lim X n = lim [ C 1 cos (nθ) + C 2 sin (nθ)] r n = n64 n64