Solve Difference Equation: Find β for x_k=x_{k-1}+x_{k-2}

[jameswill1am]

Homework Statement

In analogy with differential equations, the difference equation

[tex]x_{k}=x_{k-1}+x_{k-2}[/tex]

has two solutions [tex]x_{k}=\beta^{k}[/tex] for some [tex]\beta\neq0[/tex]. Determine the two possible values of [tex]\beta[/tex].

Homework Equations

[tex]x_{k}=x_{k-1}+x_{k-2}[/tex]
[tex]x_{k}=\beta^{k}[/tex]
[tex]\beta\neq0[/tex]

The Attempt at a Solution

So I've read that for equation in the form [tex]a_{n}=Aa_{n-1}+Ba_{n-2}[/tex] the roots are found from [tex]S^{2}-As-B=0[/tex] so applying that to [tex]x_{k}=x_{k-1}+x_{k-2}[/tex] with A=1 and B=1 i get something like [tex]\beta^{2}-\beta-1=0[/tex] but I'm sure this is not correct.

I'm wondering if the clue is in the phrasing "in analogy with differential equations" or if I'm just miles off.

Here is the rest of the next question to sort of show you what we are going towards;

Any solution of the equation [tex]x_{k}=x_{k-1}+x_{k-2}[/tex] can be written as [tex]x_{k}=\alpha_{1}\beta^{k}_{1}+\alpha_{2}\beta^{k}_{2}[/tex] where [tex]\beta_{1}[/tex] [tex]\beta_{2}[/tex] were found by you in the previous step and [tex]\alpha_{1}[/tex] and [tex]\alpha_{2}[/tex] are determined by [tex]x_{0}[/tex] and [tex]x_{1}[/tex]. Using these facts, determine [tex]p_{k}[/tex] and [tex]q_{k}[/tex] as functions of k.

So from this i get the impression it doesn't want me to know [tex]x_{k}=\alpha_{1}\beta^{k}_{1}+\alpha_{2}\beta^{k}_{2}[/tex] yet. So any help to nudge me in the right direction for finding [tex]\beta[/tex] would be much appreciated.

 

[tiny-tim]

Hi jameswill1am!

Yes, basically you've got it right …

the method is the same as for the differential equation y'' = y' + y …

you find the roots ß₁ and ß₂ of the characteristic equation (x² = x + 1), and then the general solution is any linear combination of solutions of a_n+1 = ßa_n, which of course is a_n = Cßⁿ.

See the PF library on https://www.physicsforums.com/library.php?do=view_item&itemid=158" for more details.