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CHAPTER TWO RECURRENCE RELATION

CHAPTER TWO RECURRENCE RELATION. BY Dwee. Background. Computer programming Given a function to find faktorial n! in Pascal We can make this function in another way, which is defined by recalling itself (recursive function). f unction faktorial (n:integer): longint ; Var i:integer;

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CHAPTER TWO RECURRENCE RELATION

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  1. CHAPTER TWORECURRENCE RELATION BY Dwee

  2. Background • Computer programming • Given a function to find faktorial n! in Pascal • We can make this function in another way, which is defined by recalling itself (recursive function) function faktorial(n:integer):longint; Var i:integer; Begin faktorial:=1; for i:=1 to n do faktorial:=faktorial*i; End;

  3. Background • Recursive function • This function is more simple but sometimes it use more memory than the function before • Some time we can solve a problem easily by recursive function than using iteration function faktorial(n:integer):longint; Begin if n=1 then faktorial:=1 else faktorial:=n*faktorial(n-1); End;

  4. Background • After this course, you should be able to • Solve a some simple kinds of recurrrence relation • Use forward and backward subtition to find the solution of recurrence relation • Make recurrence relation model from a problem • Solve 1st and 2nd order homogenous linear recurrence relation with constant coeficient • Solve 1st and 2nd order non-homogenous linear recurrence relation with constant coeficient

  5. INTRODUCTION • Let a geometric series: 5, 15, 45, 135, …. • If we write the terms as a1, a2, a3, …, an we have a2/a1=a3/a2=a4/a3=….=an/an-1=3 or we can write the relation between the terms in recurrence relation as an = 3 an-1, n  0 (*) • But recurrence relation (*) doesn’t define a unique geometric series, because 1,3,9, 27 ,…. also satisfy the relation. • To pinpoint the particular sequence described by an = 3 an-1, n  0, we need to know one of the terms of that sequence

  6. INTRODUCTION • Hence:, an = 3 an-1, n  0, a0=5 uniquely define sequence 5, 15, 45, ….

  7. Recurrence Relation • Recurence Relation : a relation that describe a function a(n), written as an, n 0, where an is depend on the the prior terms an-1, an-2, an-3, …, a1, a0. • Example: Fibonacci Series: 1, 1, 2, 3, 5, 8, …. By recurence relation it is given as a0= 1, a1=1, an=an-1+an-2 (n2)

  8. Recurrence Relation • Recurrence Relation consists of two parts • A set of initial or boundary condition • A rule or recurrence part • Example: Fibonacci series an=an-1+an-2 (recurrence part) with a0= 1, a1=1,(n2) (initial condition)

  9. Recurrence Relation • Solution of recurrence relation • General Solution • Iterative procedure computing • Forward subtitution • Backward subtitution • Example: • Find the general solution of recurrence relation a. an= 3an-1, a1= 2, n2 b. an = an-1 + n, a0=0, n1 c. an = 2an-1 + 1, a0=0, n1

  10. Recurrence Relation • Example: • Find a8 from this recurrence relation using Forward and Backward subtitution a0= 1, a1=1, an=an-1+an-2 (n2) • Find a6 from this recurrence relation using forward and backward subtitution with a1 =1, a2=1, n n3

  11. Modelling Problems to Recurrence Relation • A Bank pay 6% annual interest in savings, compounding the interest monthly. If Bonnie deposits $1000 on the first day on May, how much will this deposit b e worth a year later? Express the problem in the recurrence relation first. • Find a recurrence relation for the number of regions into which the plane is divided by n straight lines if every pair of lines intersect, but no three lines, meet the common point • Find a Recurrence Relation for the number of n digits binary sequences with no consecutive ones.

  12. Modelling Problems to Recurrence Relation • Recurrence Relation with more than one variable • Examples: Pascal Identity C(n,r) = C(n-1, r-1) + C(n-1,r) with boundary condition C(n,0)=1 and C(n,n)= 1. Use backward subtition to find C(5,4) by this recurrence relation! • Example: Find Recurrence relation for P(n,r) the number of r-permutation from {x1, x2, …, xn}!

  13. Linear Recurrence Relation with Constant Coefficient • General linear recurrence relation of degree k an + h1(n) an-1+ h2(n)an-2 + …+ hr(n)an-k=f(n) where h and f are functions and hr 0. If f(n) = 0 then the relation is called homogeneous, otherwise the relation is inhomogenous. And f is called inhomogenous part. If h’s are constant functions then the relation is called linear recurrence relation with constant coefficient.

  14. Linear Recurrence Relation with Constant Coefficient • A well defined linear recurrence relation of degree k consist of a recurrence relation part and k initial condition for k consecutive values  define one and only one function (solution) • Example : an – 5 an-1 + 6 an-2 = 0 is satisfied by an = C12n + C23n for any coonstant C1 and C2. Let initial condition are a0 = 2 and a1=5, we need a0 = 2 = C120 + C230  2 = C1 + C2 a1 = 5 = C121 + C231  5 = 2C1 + 3C2 this system has the solution C1=1 and C2 = 1. So the solution of the recurrence relation is an = 2n + 3n

  15. Linear Recurrence Relation with Constant Coefficient • Given a recurrence part of degree k, the strategy is to find a solution with k arbitrary constant C1, C2, …, Ck such that we can satisfy any set of k consecutive initial condition by solving a system of k simultaneous equation (one for each initial condition) in k unknowns (Ci) • Such a solution is called a general solution

  16. Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root • Superposition Principle: If g1(n) is a solution of an + c1 an-1+ c2an-2 + …+ ckan-k=f1(n) and if g2(n) is a solution of an + c1 an-1+ c2an-2 + …+ ckan-k=f2(n) then C1g1(n) +C2 g2(n) is a solution of an + c1 an-1+ c2an-2 + …+ ckan-k=C1 f1(n)+ C2 f2(n) for any constant C1 and C2

  17. Its follow immediately if g1 (n), g2(n), …, gk(n) are solutions of an + c1 an-1+ c2an-2 + …+ ckan-k=0 then so is C1g1 (n)+ C2g2(n)+ …+Ckgk(n) • How do we find different solution gi(n)? One of the choice is to look for solution in form rn for some number r

  18. Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root • Given Recurrence relation: an + c1 an-1+ c2an-2 + …+ ckan-k=0 • Let an = rn , subtituting to the recurrence relation yields rn + c1rn-1+ c2rn-2 + …+ ckrn-k=0 rn-k(rk + c1rk-1+ c2rk-2 + …+ ck)=0 rk + c1rk-1+ c2rk-2 + …+ ck=0 • The last equation is called by characteristic equation. Its roots are called characteristic roots. • Base on the characteristic roots, we have two case of the solutions • Case 1 : All the roots are distinct • Case 2 :There are some multiplicity roots

  19. Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root • Case 1: All of characteristic roots are distinct If r1, r2, …, rk is distinct characteristic root, so by using superposition principle, the general solution of the recurrence relation is an = C1r1n + C2r2n+…+Ckrkn • Example: Solve the recurrence relation 1) an + an-1 – 6 an-2 = 0, a0=1, a1=2, n2 2) an = an-1 + an-2, a0=0, a1=1, n2 (Fibonacci series) 3) an=2an-1+an-2-2an-3 , a0 =0, a1=1, a2=1, n3

  20. Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root • If the charactristic roots are complex numbers • Recall De Moivre’s Theorem: (cos + i sin )n = cos n+ i sin (n). If z = x + iy, we can write z = r (cos + i sin ) where r = sqrt(x2+ y2) and (y/x) = tan  then zn = (r (cos + i sin ) )n= rn(cos n+ i sin n). • Example: Solve the recurrence relation an = 2an-1-2an-2 , a0=1, a1=2, n2

  21. Homogenous Linear Recurrence Relation with Constant Coefficient : The Method of Characteristic Root • Case 2: The characteristic equation has multiplicity roots • Example: Solve the recurrence relation an = 4an-1-4an-2 , a0 =1, a1=3, n2

  22. If r is a characteristic root of multiplicity m, then it contributes m solution: rn, nrn, n2rn, …, nm-1rn. • Examples: Solve The recurrence relation a0=1, a1=1, a2=2, an = 4an-1-5an-2+2an-3

  23. Inhomogeneous Recurrence Relation • Towers of Hanoi. Consider n circular disks (having different diameters) with holes in their centers. These disks can be stacked on any of the pegs shown in Fig. 10.11. In the figure, n = 5 and the disks are stacked on peg 1 with no disk resting upon a smaller one. The objective is to transfer the disks one at a time so that we end up with the original stack on peg 3. Each of pegs 1, 2, and 3 may be used as a temporary location for any disk(s), but at no time are we allowed to have a larger disk on top of a smaller one on any peg. What is the minimum number of moves needed to do this for n disks?

  24. Inhomogeneous Linear Recurrence Relation with Constant Coefficient • Recurrence relation: an + c1 an-1+ c2an-2 + …+ ckan-k= f(n) with r consecutive initial conditions • Our strategy is the same as that used in the homogeneous case. • Find the general solution to the recurrene part, and use the initial condition to set up a system of simultaneous equation • By superposition principle: If anh is the solution of an + c1 an-1+ c2an-2 + …+ ckan-k= 0 and if anp is the solution of an + c1 an-1+ c2an-2 + …+ ckan-k= f(n) then anh + anp is also the solution of the recurence relation

  25. Inhomogeneous Linear Recurrence Relation with Constant Coefficient • anh is homogeneous part solution and anp is particular solution • How to find anh? Using characteristic equation • How to find anp? No general method, but some techniques are available for certain special case (method of undetermined coefficient). • Example: • Solve the recurrence relation: an- 3an-1=5 (7)n, a0=2, n1 • Solve the recurrence relation: an+3an-1=4n2-2n, a1=-4, n2 • Solve the recurrence relation: an- 4an-1+4an-2= 2n, a0=0 a1=1, n2

  26. Inhomogeneous Linear Recurrence Relation with Constant Coefficient • Solve the recurrence relation :an- 3an-1=5 (7)n, a0=2, n1 • Answer: • Homogenous part: an- 3an-1=0 Characteristic equation: r-3=0, charactristic root :r=3 Homogenous part solution: G(n)=c3n • Inhomogenous part: an- 3an-1=5(7)n We are “guessing” that the particular solution p(n)=A(7n) Subtituting p(n) into RR: A(7n)-3 A(7n-1)=5 (7n) 7n-1 (7A-3A)=7n-1(5.7) 4A = 35 A=35/4 • General solution: an = G(n)+p(n) = c 3n + 35/4 (7)n with a0 = 2, we get c= … so the solution is an = …..

  27. Inhomogeneous Linear Recurrence Relation with Constant Coefficient • Solve the recurrence relation: an+3an-1=4n2-2n, a1=-4, n2 • Answer: • Homogenous part: an+3an-1=0 Homogenous part solution: G(n)=c1(-3)n • Inhomogenous part: an+3an-1=4n2-2n Particular solution is polunomial of degree 2 or higher Guess that p(n) is polynomial of degree 2, p(n) = An2+Bn+C. Subtituting p(n) the RR: An2+Bn+C + 3(A(n-1)2+B(n-1)+C)= 4n2-2n 4An2+(4B-6A)n+(3A-3B+4C)=4n2-2n Solving the equation ,we have: A=1, B=1, C=0. The particular solution is p(n) = n2+n • General solution is an=G(n)+p(n)= c1(-3)n + n2+n Find the solution based on the initial condition

  28. Solve the recurrence relation an-an-1=n, a0=1, n1 • Answer: • Homogenous part: an-an-1= 0 Homogenous part splution part: anh = c(1)n=c • Inhomogenous part: an-an-1=n Because f(n) =n so the particular solution is polynomial of degree 1 or higher. Find the particular solution using guess • p(n) = An +B • p(n)= An2 +Bn +C which one give the solution? Why?

  29. Inhomogeneous Linear Recurrence Relation with Constant Coefficient • How to determine the degree of the polynomial guess? • The degree of anp is the degree of f(n) plus the multilicity of 1 in the characteristic equation • If 1 is not a root of characteristic equation, the degree of anp is the degree of f • Similarly if f(n) is of the form rn, then the particular solution will be Ankrn, where k is the multiplicity of r in the characteristic equation

  30. Inhomogeneous Linear Recurrence Relation with Constant Coefficient • Solve the recurrence relation an- 4an-1+4an-2= 2n, a0=0 a1=1, n2 • Answer: • Homogenous part: an- 4an-1+4an-2= 0 Characteristic equation: r2 – 4r + 4 = 0 Characteristic root: r = 2 (multiplicity of degree 2) Homogenous part solution: G(n) = …… • Inhomogeneous part: an- 4an-1+4an-2= 2n Particular solution is p(n)= … • The general solution is an = ….

  31. Given a linear nonhomogeneous recurrence relation (with constant coefficients) of the form C0an + C1an-1 + C2an-2 + • • • + Ckan-k = f(n) • Let Gn denote the homogeneous part of the solution an.

  32. EXERCISE. • a0=1, an-an-1= n2 • a0=1,a1=2, an-5an-1+6an-2 = 2n+1, n>=2 • a0=4, an-an-1=2n2-n-1, n>=1 • a0=1, a1=0, an-2an-1+an-2=2, n>=2 • a0=1, an+3an-1 =2n, n>=1 • a0=1, an+2an-1=2n-n2 , n>=1

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