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Lecture15

Lecture15. section 6.2 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:.

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Lecture15

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  1. Lecture15 section 6.2 Solving Recurrence Relations

  2. 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: 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.

  3. RR Examples: non-linear no constant coefficients non-homogeneous Reminder:There can be more than one solution to a recurrence relation! Only when the initial conditions are specified is the solution unique. For a recurrence relation of degree k, you need k contiguous initial conditions.

  4. 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)

  5. 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.

  6. Degenerate roots What happens if we have a quadratic equation with 2 equal roots? The solution we have a different form: a(n) = (d1 + d2 n) r^n This extra n factor is different! Example: a(n) = 6 a(n-1) – 9 a(n-2) a0 = 1, a1 = 6. Try r^n  r^n = 6 r^(n-1) – 9 r^(n-2)  r^2 – 6 r + 9 = 0  (r-3)^2=0  r=3 (2x). So now the general form of the solution is: a(n) = (d1 + d2 n) 3^n d1,d2 must follow from the initial conditions: d1 = 1, (d1+d2) 3 = 6  d2 = 1.  a(n) = (1 + n) 3^n.

  7. 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.

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

  9. Examples a[n] = 6a[n-1]-11a[n-2]+6a[n-3] a[0]=2, a[1]=5, a[2]=15. Insert r^n in the equation:  r^3 – 6r^2 + 11r – 6 = 0 = (r-1)(r-2)(r-3)=0 General form of solution: a[n]=d1 1^n + d2 2^n + d3 3^n = d1 + d2 2^n + d3 3^n . Initial condition give 3 linear equations to solve d1,d2,d3: d1+d2+d3=2 d1+2d2+3d3=5 d1+4d2+9d3=15 d1 = 1, d2=-1, d3=2. a[n]=1-2^n+2x3^n

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

  11. Inhomogeneous Terms 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

  12. How to find a particular solution? This problem can be solved in the case: n’th power of a constant s polynomial Example: g[n]=3g[n-1]+2n. g1=3 Try a solution of the bn = c n+d c n+d = 3c(n-1)+3d+2n  c n = 3 c n + 2n d = 3d-3c  c=-1 & d=-3/2 General solution: gn = -n-3/2 + d 3^n the coefficients before every power of n must be equal. Initial conditions: d=11/6 solution inhom. RR solution hom. RR

  13. 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 = 49/20 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.

  14. General Case 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:

  15. Flow Diagram 1) Determine of RR is linear with constant coefficients and determine degree. 2) Determine if inhomogeneous part fits the special form. 3) Solve associated homogeneous RR (determine roots by inserting r^n), but leave the parameters d1,...dk undetermined. 4) Insert an trial particular solution and determine the parameters by inserting it in the RR. (s is root with multiplicity m) (s is not a root) This term is also there when s=1 is a root with mult. m. 5) Add particular solution and homogeneous part and determine d1,...dk from initial conditions (if specified). m1, m2,... are multiplicities roots r1, r2,... d1,0, d1,1,...,dt,0, dt,1,... are determined by imposing the initial conditions on this beast (don’t forget to include bn) +

  16. 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=3, m(r) = 2. 4) Insert:  divide by 3^(n-2), match all powers of n, solve p0,p1,p2 5)  impose initial conditions, solve for d0, d1.

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