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Inductive Proofs and Recurrence Relation Solutions in Discrete Mathematics

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ophrah
13 diapositives · 292 Views · Aug 14, 2014

This document explores the methodologies for solving recurrence relations using induction techniques. It details how to verify candidate solutions through basis steps and inductive hypotheses for sequences defined by specific recurrence formulas. Key examples include solving homogeneous and non-homogeneous linear recurrence relations, along with the characteristic equations leading to general solutions. The explanations are structured to guide students through various assignments and practice problems, enhancing their understanding of mathematical induction and the behavior of sequences.

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Inductive Proofs and Recurrence Relation Solutions in Discrete Mathematics
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  1. CS 427 Practice Problems for Assignment # 2. • Appendix B: #1 b,c; 12 a,c; 20 (1a)

  2. # 1.b. Use induction to verify the candidate solution t n = t n – 1 + 5 for n > 1 t 1 = 2 The candidate solution is t n = 5n – 3. Basis Step: let n = 1 then t 1 = 5(1) – 3  t 1 = 2 Inductive Hypothesis: Assume that t n = 5n – 3 for n > 1 Inductive Step: We need to show that t n+1 = 5(n+1) – 3 Now use the recurrence equation for t n+1 t n+1 = t n + 5 Now substitute the candidate solution for t n

  3. Inductive Step (continued): Now substitute the candidate solution for t n t n+1 = t n + 5  t n+1 = (5n – 3) + 5 t n+1 = (5n – 3) + 5  t n+1 = 5n – 3 + 5 t n+1 = 5n + 5 – 3  t n+1 = 5(n+1) – 3 Which is what we needed to show.

  4. # 1.c. Use induction to verify the candidate solution t n = t n – 1 + n for n > 1 t 1 = 1 The candidate solution is t n = Basis Step: let n = 1 then t 1 =  2/2  1 as defined Inductive Hypothesis: Assume that t n = for n > 1 Inductive Step: Show that t n+1 = n(n+1) 2 1(1+1) 2 n(n+1) 2 (n+1)(n+2) 2

  5. Inductive Step (continued): Use the recurrence equation for t n+1 t n+1 = t n + (n + 1) next substitute the characteristic equation for t n t n+1 = + (n + 1) + t n+1 =   This is what we wanted to show. n(n+1) 2 n2 + n 2 2n +2 2 n2 + 3n + 2 2 (n+1)(n+2) 2

  6. # 12.a. Solve the recurrence equations using the c.e. This is a homogeneous linear recurrence ( see p.469) t n = 4t n – 1 - 3t n – 2 for n > 1 t 0 = 0, t 1 = 1 Obtain the characteristic equation. t n - 4t n – 1 + 3t n – 2 = 0 t n = r n then we get r n - 4r n-1 + 3r n-2 = 0 then factor out r n-2 to getr n-2 (r 2 - 4r + 3) = 0 r = 0 is a root, but we need to find the nonzero roots. (r 2 - 4r + 3) = 0  (r – 3)(r – 1) = 0 so we have nonzero roots at 3 and 1. Roots = [3 1] Using these roots we see our general solution is t n = c1 3 n + c 2 1 n  c1 3 n + c 2

  7. Now we use our initial conditions to find the values of c1 and c 2 t n = c1 3 n + c 2 t 0 = 0  c1 3 0 + c 2 = 0 t 1 = 1  c1 3 1 + c 2 = 1 Now we have two equations and two unknowns. c1 + c 2 = 0  c 2 = - c 1 3c1 + c 2 = 1 next substitute 3c1 - c1 = 1  2c 1 = 1  c 1 = 1/2 c 2 = - ½ So the general solution with the correct coefficients is: t n = (1/2)3 n – 1/2

  8. # 12.c. Solve the recurrence equations using the c.e. This is a non-homogeneous linear recurrence (p 477. Theorem B.3 on p 479) t n = 5t n – 1 - 6t n – 2 + 5 n for n > 1 t 0 = 0, t 1 = 1 First reorganize the recurrence equation. t n - 5t n – 1 + 6t n – 2 = 5n (1) Now use theorem B.3 to find the characteristic equation b = 5, p(n) = 1 so d = 0, k = 2 ( r k – 5r k-1 + 6r k-2 )(r – 5) 1 = 0 r k-2 ( r 2 – 5r + 6)(r – 5) = 0 Remember, we are looking for nonzero roots so… ( r 2 – 5r + 6)(r – 5) = 0 now factor ( r – 3 )( r – 2 )( r – 5) = 0 so the roots are [3 2 5]

  9. The roots are [3 2 5] so we apply theorem B.1 to find the general solution: t n = c12 n + c23 n + c35 n Now we use the initial conditions to find the coefficients. t 0 = 0 = c12 0 + c23 0 + c35 0 t 1 = 1 = c12 1 + c231 + c35 1 Now there are two equations, but three unknowns so we need another initial condition. Use the recurrence to find t 2 t 2 = 5t 1 - 6t0 + 52 t 2 = 5(1) – 6(0) + 25  t 2 = 30 Now there are three equation and three unknowns. Solve. t 0 = 0 = c1 20 + c2 3 0 + c3 5 0 c1 + c2 + c3 = 0 t 1 = 1 = c1 21 + c2 31 + c3 5 1  2c1 + 3c2 + 5c3 = 1 t 2 = 30 = c1 22 + c2 32 + c3 52  4c1 + 9c2 + 25c3 = 30

  10. You can solve for the unknowns c 1, c 2, and c 3 however you like. c 1 c 2 c 3 sum • 1 1 0 • 3 5 1 • 4 9 25 30 • 1 1 0 • 0 1 3 1 • 4 9 25 30 • 1 1 0 • 0 1 3 1 • 0 5 21 30 -2*row1 + row2 -4*row1 + row3 • 1 1 0 • 0 1 3 1 • 0 0 1 25/6 • 1 1 0 • 0 1 0 -23/2 • 0 0 1 25/6 • 1 1 0 • 0 1 3 1 • 0 0 6 25 -3*row3 + row2 -5*row2 + row3 (row3)/6 • 0 1 23/2 • 0 1 0 -23/2 • 0 0 1 25/6 • 0 0 22/3 • 0 1 0 -23/2 • 0 0 1 25/6 c1 = 22/3 c2 = -23/2 c3 = 25/6 -1*row2 + row1 -1*row3 + row1

  11. c1 = 22/3 c2 = -23/2 c3 = 25/6 Remember the general solution with the unknowns: t n = c12 n + c23 n + c35 n Now replace the unknowns with the values we found: t n = (22/3)2 n + (-23/2)3 n + (25/6)5 n Now we are finished.

  12. # 20 - 1.a. Solve the recurrence equations using the substitution method The substitution method is described on pages 486-487 t n = 4t n – 1 for n > 1 t 1 = 3 Use the recurrence equation to generate values from t n to t 1. t n = 4 t n-1 t n-1 = 4 t n-2 t n-2 = 4 t n-3 …... t 2 = 4 t 1 t 1 = 3

  13. Next substitute each equation into the previous one. t n = 4 t n-1 = 41 t n-1 t n = 4 ( 4 t n-2)= 42 t n-2 t n = 4 ( 4 ( 4 t n-3))= 43 t n-3 t n = 4 (4 ( 4 ( 4 t n-4)))= 44 t n-4 …….. …….. t n = 4(4(4(4(….4 t2))))…)))= 4 n-2 t 2 t n = 4(4(4(4(4(….4 t1))))…)))) = 4 n-1 t 1 t 1 = 3 So… The general solution is: t n = 3( 4 n-1 )