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Does anyone know the answer to this question regarding recurrence relations?

To determine the general solution of the following recursive relation with the initial conditions. Write down the roots in ascending order.

$$a_0=0,a_1=8:a_n=7a_{n-2}+6a_{n-1},n\ge2$$

Since this question was from an exam I had that I failed, I will roughly translate it.

Edit: Here's the empty fields that need requireing, I will put x as a placeholder for the answers

$$a_n=x * x^n + x * x^n $$

Jean Marie
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bakerino
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Another User Jun 10 '22 at 10:31
  • Will do, i had an exam that i failed and this was one of the questions. – bakerino Jun 10 '22 at 10:32
  • There is a derivation of such linear recurrences in answer to How to construct a closed form formula for a recursive sequence?. Does this answer your question? You'll have to plug in some numbers, though. – emacs drives me nuts Jun 10 '22 at 10:39
  • It does give me a start where to learn, honestly though they give 0 proper details for these exams so we can't honestly have a clue on how to solve them. – bakerino Jun 10 '22 at 10:42

1 Answers1

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That hint is almost the complete solutions, as it tells you that the explicit formula for $a_n$ has the form

$$a_n = \alpha x^n + \beta y^n$$

Plugging in $n=0$ yields $a_0 = 0 = \alpha x^0 + \beta y^0 = \alpha + \beta$, hence $\beta = -\alpha$ and the form of the equation has been narrowed down to

$$a_n = \alpha (x^n - y^n)$$

Then plug in $n=1$, $n=2$, $n=3$ to get

$$\begin{align} a_1 = 8 &= \alpha(x-y) \tag1 \\ a_2 = 6\cdot 8 &= \alpha(x^2-y^2) \tag2 \\ a_3 = \cdots &= \alpha(x^3-y^3) \tag3 \\ \end{align}$$ respectively. Dividing $(2)$ by $(1)$ gives you the value of $x+y$, and dividing $(3)$ by $(1)$ gives you the value of $x^2+xy+y^2$. This results in a quadratic equation for $x$ resp. $y$.

You should be able to proceed from here.

emacs drives me nuts
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  • Thank you very much kind sir, you are a life saver :D – bakerino Jun 10 '22 at 11:10
  • I agree with the form $\alpha x_1^n + \beta x_2^n$ but I don't agree with the treatment you do. The easiest way one can advise in general is to compute first $x_1,x_2$ which are the roots of $x^2-7-6x=0$ giving $x_1=7$ and $x_2=-1$. Knowing $x_1,x_2$ it remains to find $\alpha,\beta$. – Jean Marie Jun 10 '22 at 12:40
  • @Jean Marie: If you don't agree, then my treatment is wrong? I tries to answer on a level such that the OP can use it. I don't know if the OP already knows (roots of) characteristic equations and how linear algebra and eigenvalues relate to linear recurrence; presumably not. – emacs drives me nuts Jun 10 '22 at 13:05
  • No, by no way it is wrong ; don't take it as a big critic. In the case at hand where $\alpha = - \beta$, you just have to manage $3$ unknowns, but in the general cas, you are dealing with 4 simultaneous equations with 4 unknowns, whereas computing first the roots of the characteristic polynomials make the computation simpler. – Jean Marie Jun 10 '22 at 15:38