Solving%20Recurrence%20Relations%20by%20Iteration

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Solving Recurrence Relations by Iteration

• Lecture 41
• Section 8.2
• Fri, Apr 13, 2007

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Solving Recurrence Relations

• Our method will involve two steps.
• Guess the answer.
• Verify the guess, using mathematical induction.

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Guessing the Answer

• Write out the first several terms, as many as
  necessary.
• Look for a pattern.
• Two strategies
• Do the arithmetic.
• Spot the pattern in the resulting numbers.
• Postpone the arithmetic.
• Spot the pattern in the algebraic formulas.

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Example Do the Arithmetic

• Define an by
• a1 2,
• an 2an 1 1, for all n ? 2.
• Find a formula for an.
• First few terms 2, 5, 11, 23, 47, 95, 191.

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Example Do the Arithmetic

• Define an by
• a1 2,
• an 2an 1 1, for all n ? 2.
• Find a formula for an.
• First few terms 2, 5, 11, 23, 47, 95, 191.
• Compare to 1, 2, 4, 8, 16, 32, 64.

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Example Do the Arithmetic

• Define an by
• a1 2,
• an 2an 1 1, for all n ? 2.
• Find a formula for an.
• First few terms 2, 5, 11, 23, 47, 95, 191.
• Compare to 1, 2, 4, 8, 16, 32, 64.
• Guess that an 3?2n 1 1.

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Example Postpone the Arithmetic

• Define an by
• a1 1,
• an 2an 1 5, for all n ? 2.
• Find a formula for an.
• First few terms 1, 7, 19, 43, 91.
• What is an?

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Example Postpone the Arithmetic

• Calculate a few terms
• a1 1.
• a2 2 ? 1 5.
• a3 22 ? 1 2 ? 5 5.
• a4 23 ? 1 22 ? 5 2 ? 5 5.
• a5 24 ? 1 23 ? 5 22 ? 5 2 ? 5 5.
• It appears that, in general,
• an 2n 1 (2n 2 2n 3 1) ? 5.

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Lemma Geometric Series

• Lemma Let r ? 1. Then

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Example Postpone the Arithmetic

• an 2n 1 (2n 2 2n 3 1) ? 5
• 2n 1 (2n 1 1)/(2 1) ? 5
• 2n 1 (2n 1 1) ? 5
• 2n 1 5 ? 2n 1 5
• 6 ? 2n 1 5
• 3 ? 2n 5.

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Example

• Define an by
• a0 a,
• an ran 1 b, for all n ? 1.
• Find a formula for an.
• a1 ra b.
• a2 r(ra b) b r2a (rb b).
• a3 r(r2a (rb b)) b
• r3a (r2b rb b).

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Example Future Value of an Annuity

• It appears that, in general,

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Verifying the Answer

• Use mathematical induction to verify the guess.

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Solving First-Order Linear Recurrence Relations

• A first-order linear recurrence relation with
  constant coefficients is a recurrence relation of
  the form
• an san 1 t, n ? 1,
• with initial condition
• a0 u,
• where s, t, and u are real numbers.

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Solving First-Order Linear Recurrence Relations

• Theorem Depending on the value of s, the
  recurrence relation will have one of the
  following solutions
• If s 0, the solution is a0 u, an t, for all
  n ? 1.
• If s 1, the solution is an u nt, for all n
  ? 0.

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Solving First-Order Linear Recurrence Relations

• If s ? 0 and s ? 1, then the solution is of the
  form
• an Asn B, for all n ? 0,
• for some real numbers A and B.

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Solving First-Order Linear Recurrence Relations

• To solve for A and B in the general case,
  substitute the values of a0 and a1 and solve the
  system for A and B.
• a0 A B u
• a1 As B su t

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Example

• Solve the recurrence relation
• a0 1,
• an 2an 1 1, n ? 1.
• Solve the recurrence relation
• a0 1,
• an 2an 1 5, n ? 1.