Solving Linear Recurrence: ax_k+1 + bx_k + c

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The discussion focuses on solving the linear recurrence relation defined by x_k+2 = ax_k+1 + bx_k + c, where c is a non-zero constant. Participants clarify that if a + b is not equal to 1, a constant p can be derived such that setting y_k = x_k + p transforms the original recurrence into y_k+2 = ay_k+1 + by_k. The value of p is explicitly calculated as p = c / (a + b - 1), allowing for the determination of the sequence x_k once y_k is established.

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Consider the recurrence x_k+2 = ax_k+1 + bx_k + c where c may not be zero.

If a + b is not equal to 1 show that p can be found such that, if we set y_k = x_k + p, then y_k+2 = ay_k+1 + by_k. [Hence, the sequence x_k can be found provided y_k can be found]


First of all, sorry about the messiness, I don't know how to use LaTeX. Now, this is the question exactly as it is from the question sheet. My problem is, I don't understand the question. And its kind of really hard to start the question without understanding it :mad: . My biggest concern is, what the heck is p and where does it come from? The way I read it, p is just -c.

Thx in advance for any help.
 
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1.You should learn "tex".
2.Hypothetis:[tex]x_{k+2}=ax_{k+1}+bx_{k}+c[/tex] (1)
[tex]y_{k}=x_{k}+p[/tex] (2)
[tex]y_{k+2}=ay_{k+1}+by_{k}[/tex] (3)
3.Question:[tex]p=...?[/tex]

4.From (2) u have:
[tex]y_{k+2}=x_{k+2}+p[/tex] (4)
Combining (1) and (4),u get:
[tex]y_{k+2}=ax_{k+1}+bx_{k}+c+p[/tex] (5)
Equate (5) with (3),make use of (2) and extract 'p':

Answer:[tex]p=\frac{c}{a+b-1}[/tex]

Daniel.
 
Ah, it all makes sense. Can't believe I never saw that. Thx a lot!
 

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