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I am trying to solve the following problem:

Given the following recurrence relation: \begin{equation} a_n = 6a_{n-1} - 8a_{n-2} \ \text{ for } \ n \in \mathbb{Z}_{\ge2} \end{equation}

Find all values of $ r $ so that when $ a_0 = 1 $ and $ a_1 = r $ then $ a_n = r^n \ \ \forall n \in \mathbb{N} $

My attempt:

\begin{align} \text{By distinct roots theorem,} \\ \\ \text{Characteristic equation: } & t^2 - 6t + 8 = 0 \\ & (t-3)^2 + 8 - 9 = 0 \\ & (t-3)^2 = 1 \\ & t = 4 \text{ or } t = 2 \end{align} \begin{align} \text{Let } a_0 = 1 = C(4^0) + D(2^0) \text{ where C, D } \in \mathbb{R} \end{align} \begin{align} 1 = C + D \\ C = 1 - D \end{align}

\begin{align} \text{Let } a_1 = r &= C(4) + D(2) \\ r &= 4C + 2D \\ r &= 4(1-D) + 2D \quad \text{(By substitution)} \\ r &= 4 - 4D + 2D \\ r &= 4 - 2D \end{align}

It seems that I must have gone wrong somewhere as $ r < 0 $ for $ D > 2$. Hence the answer for $ a_n $ would be very strange (i.e. $ a_n < 0 $ when $ r = -4 $). Could anyone please assist me?

Donald
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2 Answers2

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By substituting $a_n$ by $r^n$, you get $r^n = 6 r^{n-1} -8r^{n-2}$ for $n\ge2$.

Thus $r^2 - 6r + 8 =0$ hence $r \in \{2, 4\}$ as you rightly figured out.

The issue with what you did is that you tried to solve the equations once you had $a_n = C 4^n + D 2^n$. You should have figured that either $C = 0$ and $D = 1$ or $D = 0$ and $C = 1$ (or else there exists no $r \in \Bbb R$ such that $\forall n\in \Bbb N, a_n = r^n$)

Astyx
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  • How do you make use of the initial conditions in obtaining the value of r? – Donald Oct 29 '16 at 06:57
  • This is the point : you don't. – Astyx Oct 29 '16 at 07:27
  • Why are the initial conditions irrelevant here? – Donald Oct 29 '16 at 07:27
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    Because you are trying to find a sequence with a global property that does not only rely on the initial conditions rather than expliciting it from the initial conditions. The fact that your sequence is geometric is something that concerns every term of the sequence, not only the first two. – Astyx Oct 29 '16 at 07:36
  • Hi @Astyx, could you elaborate on how was (C = 0, D = 1) and (C = 1, D = 0) obtained? – Keith OYS Oct 31 '16 at 09:03
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    @KeithOYS The first part of my answer ensures that $r\in {2,4}$. Once you have this, simply solve $$ \left{ \begin{aligned} C+D &= 1\ 4C+2D &= 2\ \end{aligned} \right. $$ or $$ \left{ \begin{aligned} C+D &= 1\ 4C+2D &= 4\ \end{aligned} \right. $$ depending on $r$ – Astyx Oct 31 '16 at 09:25
  • Thank you @Astyx. Since the question asks Find all values of r, would the roots r = 2, r = 4 suffice or does it require constant values of C and D as well? – Keith OYS Oct 31 '16 at 10:12
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    @KeithOYS What I would answer is first that if $a_n = r^n$, then $r\in{2,4}$ (see my answer) thus the only solutions are among ${2,4}$. Then if $r\in {2,4}$ and $a_n = r^n$ and $a_{n+1} = r^{n+1}$, we have $a_{n+2} = (6r + 8)r^n = r^{n+2}$ thus $2$ and $4$ are solutions. Therefore the solutions are $r=2$ and $r=4$. No need to solve the general form of sequences verifying this equation. – Astyx Oct 31 '16 at 10:17
  • Hi @LanceHAOH, may I check with you, did you obtain $a_n = 1/2(2^n) + 3/2(4^n)$ for part (b)? – Keith OYS Oct 31 '16 at 16:41
  • @KeithOYS I cannot remember. But since it is a solution to a recurrence, you can check your answer using recurrence solvers like wolframalpha. – Donald Oct 31 '16 at 16:49
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Both roots give solutions: $r=4$ for $a_n = 4^n$ ($C=0, D=1$) and $r=2$ for $a_n = 2^n$ ($C=1, D=0$).

Robert Israel
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