Solving Linear Recurrence Relations

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SUMMARY

This discussion focuses on solving linear recurrence relations, specifically addressing three examples. The first relation is defined by the equation t(n) - 5t(n-1) + 6t(n-2) = 0 with initial conditions t(1) = 1 and t(2) = 4. The second relation, a(n) = 4a(n-1) - 4a(n-2), has initial conditions a(0) = 4 and a(1) = 12. The third relation is t(n) + 2t(n-1) + t(n-2) = 0 with initial conditions t(1) = 3 and t(2) = 3, leading to the characteristic equation t_{n + 2} - 5t_{n + 1} + 6t_n = 0.

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  • Understanding of linear recurrence relations
  • Familiarity with characteristic equations
  • Knowledge of initial conditions in recurrence relations
  • Basic algebra skills for solving equations
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  • Study the method of solving linear recurrence relations with constant coefficients
  • Learn about the characteristic polynomial and its roots
  • Explore the general solution forms for different types of roots
  • Investigate applications of recurrence relations in algorithm analysis
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Mathematicians, computer scientists, and students studying algorithms or discrete mathematics will benefit from this discussion, particularly those interested in recurrence relations and their solutions.

taya
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Solve each of the following linear recurrence relations:
1. t(1)=1 t(2)=4
t(n) - 5t(n-1) + 6t (n-2)= 0 for n>1

2. a(n)=4a(n-1) - 4a (n-2)
with initial conditions a(0) = 4 and a(1)=12

3. t(1)=3 t(2)=3
t(n) + 2t (n-1) + t(n-2) = 0
 
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taya said:
1. t(1)=1 t(2)=4
t(n) - 5t(n-1) + 6t (n-2)= 0 for n>1
Rewrite this as [math]t_{n + 2} - 5t_{n + 1} + 6 t_n = 0[/math]. What is your characteristic equation?

-Dan
 
Hello and welcome to MHB, taya! :D

For future reference, we ask that no more than 2 questions be asked in the initial post of a thread.

In case you find repeated characteristic roots, what form will your general solution take?
 

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