MHB Solving Linear Recurrence Relations

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The discussion focuses on solving linear recurrence relations with specific initial conditions. The first relation involves finding a solution for t(n) defined by t(n) - 5t(n-1) + 6t(n-2) = 0, with initial values t(1)=1 and t(2)=4. The second relation, a(n)=4a(n-1) - 4a(n-2), is solved with initial conditions a(0)=4 and a(1)=12. The third relation, t(n) + 2t(n-1) + t(n-2) = 0, is also analyzed with initial values t(1)=3 and t(2)=3. The discussion emphasizes the importance of identifying the characteristic equation and the general solution form, especially in cases of repeated roots.
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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