Solving Recurrence Relations - Discrete Mathematics and its Applications - Lecture Slides, Slides of Discrete Mathematics

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Lecture

section 6.

Solving Recurrence Relations

Recurrence Relations

Recurrence Relations can take many forms, and most forms are hard, if not impossible to solve. There are however a certain subset that can be solved explicitly. They are of the form:

n n k n k k

a c a c a c

This is a linear, homogeneous recurrence relation of degree k with constant coefficients -linear because we don’t have terms: with F(.) a nonlinear function. -homogeneous because we don’t have terms:. -degree k, because it depends on k terms in the past a(n-1) ... a(n-k).

• Note that the equation: is also of degree k (some c’s are zero). -constant coefficients because c(k) does not depend on n.

F a ( n −k)

G n ( ) n k n k a c a − =

Solving Linear RR (degree 1)

The trick in many of these cases is to try out a parameterized form of a solution and solve for the remaining parameters (educated guess, ansatz). In the case of linear equation we have already seen the solution to the bank problem: B(n) = c B(n-1) = c^2 B(n-2) = c^3B(n-3) Now we can guess a solution: B(n) = d r^n  stick into the equation: d r^n=d c r^(n-1)  r=c B(n) = d c^n This is a solution for all values of `d’. Now impose initial value: B(0) = 1 d c^0 = 1 d=1  final solution B(n) = c^n (unique)

Solving RR (degree 2)

-For higher degree RR the idea is exactly the same: try a solution of the d r^n. Let’s try second degree: Fibonacci RR: F(n) = F(n-1) + F(n-2). d r^n = d r^(n-1) + d r^(n-2) multiplicative constants can never be determined by the RR (they will be determined by the initial conditions). r^n = r^(n-1) + r^(n-2)  r^2 = r + 1  r^2-r-1=0  Two solutions! In fact any linear combination will be a solution: F(n) = d1 r1^n + d2 r2^n (try it by inserting this solution in the RR). d1 and d2 are free, but should be determined by the initial values: F(0)=0  d1 + d2 = 0; F(1) = 1  d1 r1 + d2 r2 = 1. This represents 2 linear equations with 2 unknowns: solve to find: d1 = 1/sqrt(5) d2 = -1/sqrt(5). Now the solution is unique. 1,

r = ±

Overview first and second degree

1. determine if the RR is homogeneous, linear, const. coeff. and find it’s degree.
2. Insert a(n) = r^n into the equation and find the roots of the resulting polynomial equation degree 1 degree 2 (2 different roots) degree 2 (2 equal roots)
3. Determine the coefficients d, d1, d2 from the initial conditions. ( ) n a n = d r ( )^1 1 2 2 n n a n = d r + d r 1 2 ( ) ( ) n a n = d +d n r

Higher Degree RR

Theorem: c1,...,ck real numbers with ck NOT 0. Suppose the characteristic equation: has k distinct roots r1,...,rk. Then the sequence {an} is a solution of the recurrence relation: if and only if for n=0,1,2,3,... and arbitrary constants d1,...,dk we have: if we have t distinct roots, each with multiplicity m1,...mt then the sequence {an} will be a solution iff 1 1 ...^0 k k r c r ck − − − − = an = c a 1 n (^) − 1 + ...+c ak n −k 1 1 ... n n n k k a = d r + +d r 1 2 1 1 1,0 1,1 1, 1 1 2 1 2,0 2,1 2, 1 2 1 ,0 ,1 , 1 ( ... ) ( ... ) ..... ( ... ) t m n n m m n m mt n t t t m t a d d n d n r d d n d n r d d n d n r − − − − − − = + + +

For every distinct root we have an arbitrary polynomial of degree m(t)-1 in front.

Examples

a[n]=-3a[n-1]-3a[n-2]-a[n-3] a[0]=1, a[1]=-2, a[2]=- Insert r^n into RR: r^3 + 3r^2 + 3r + 1 = 0 (r+1)^3= General solution: a[n] = (d1+d2 n + d3 n^2) (-1)^n Initial conditions: d1 = 1 (d1+d2+d3) x (-1) = - (d1+4d2+9d3) x 1 = -  d1=1, d2=3,d3=- Final solution: a[n]=(1+3n-2n^2) x (-1)^n

Inhomogeneous Terms

1 1

n n k n k

a c a c a F n

− −

Linear, inhomogeneous RR of degree k with constant coefficients Again, in general this is a hard problem, but for certain cases we can guess a particular solution to the full equation. Once we have one solution, we can immediately write down the general solution according to: Theorem: If {bn} is a particular solution to the inhomogeneous RR, and {an} is the general solution to the associated homogeneous RR, then the general solution to the inhomogeneous RR is given by: {gn} with gn = an+bn. Proof: Show that gn-bn must be a solution to the homogeneous RR, which we know: in full generality {an}. Thus gn = an+bn

Finding Particular Solutions

Example: a[n]=5a[n-1]-6a[n-2] + 7^n Now try bn = d 7^n as a solution: Insert in the equation: d 7^n = 5 d 7^(n-1) – 6 d 7^(n-2) + 7^n  7^2 d – 5x7d + 6d - 7^2 = 0 d = 49/ General form for solution: gn = d1 r1^n + d2 r2^n + 49/20 7^n Now solve homogeneous equation to get: r1 = 3, r2=2, d1,d2 arbitrary because we didn’t specify initial conditions.

Theorem: If we have a inhomogeneous RR of the form: where: then a) if s is not a root of the characteristic equation of the associated homogeneous equation then there exists a particular solution of the form: b) If s is a root with multiplicity m then the following solution exists:

General Case

0 1 ( ) ( ... ) t n t F n = b + b n + + b n ×s 1 1

n n k n k

a c a c a F n

− −

0 1 ( ) t n n t b = p + p n + p n s 0 1 ( ) m t n n t b = n p + p n + p n s

One more example

a[n]=6a[n-1]-9a[n-2] + (n^2+1) 3^n a[0]=1, a[1]=2;

1. linear inhomogeneous RR with constant coefficients of degree k=2.
2. The inhomogeneous term fits the special form with: t=2, b0=1, b1=0, b2=1, s=3.
3. r^2-6r+9=(r-3)^2=0  r=2, m(r) = 2.
4. Insert:  divide by 3^(n-2), match all powers of n, solve p0,p1,p
5.  impose initial conditions, solve for d0, d1. 2 2 ( 0 1 2 ) 3 n bn = n p + p n + p n ( 0 1 ) 3 n an = d + d n +bn