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?
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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