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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 3 7, Wednesday, December 3

MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 3 7, Wednesday, December 3. Theorem. Let a n = ca n/k + f(n) be recurrence relation with positive constant c and the positive function f(n).

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MATH 310, FALL 2003 (Combinatorial Problem Solving) Lecture 3 7, Wednesday, December 3

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  1. MATH 310, FALL 2003(Combinatorial Problem Solving)Lecture 37, Wednesday, December 3

  2. Theorem • Let an = can/k + f(n) be recurrence relation with positive constant c and the positive function f(n). • (a) If for large n, f(n) grows proportional to n logk c, then an grows proportional to nlogk c log2 c. • (b) If for large n f(n) · pnq, where p is a positive constant and q < logk c, then an grows at most at rate proportional to nlogk c.

  3. 7.3. Solution of Linear Recurrence Relations • Homework (MATH 310#11W): • Read 7.4 • Do 7.3: all odd numbered problems • Turn in 7.3: 2,4,6,8,10

  4. Homogeneous Linear Recurrence • an = c1an-1 + c2an-2 + ... cran-r. • ar - c1ar-1- c2ar-2 -... - cr= 0 is called the charateristic equation. • General solution: • an = A1a1n + A2a2n + ... + Ararn = • The constants are determined by solving the linear system • A1a1k + A2a2k + ... + Arark = ak,0 · k · r – 1. • If theroot a has multiplicity m, then the individual solutions an, nan, ..., nm-1an should be used.

  5. Example 1: Doubling Rabbit Population • an = 2an-1. a0 = 6. • Solution: an = 6 £ 2n.

  6. Example 2: Second-Order Recurrence Relation • Solve the recurrence relation an = 2an-1 + 3an-2 with a0 = a1 = 1. • Solution: • an = A13n + A2(-1)n where A1 = A2 = ½.

  7. Example 3: Fibonacci Relation • an = an-1 + an-2, a0 = a1 = 1. • Solution: • an = (½ + ½ sqrt(5))n+1/sqrt(5) + (½ - ½ sqrt(5))n+1/sqrt(5)

  8. Example 4: Complex and Multiple Roots • an = -2an-2 – an-4, • a0 = 0, a1 = 1, a2 = 2, a3 = 3. • Note that the solution from the book reduces to: • a4k = -4k • a4k+1 = 1 – 8k • a4k+2 = 4k+2 • a4k+3 = 8k + 3

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