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

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SUMMARY

The difference equation x_{k}=x_{k-1}+x_{k-2} has solutions of the form x_{k}=\beta^{k} for non-zero values of β. The characteristic equation derived from this difference equation is β^{2}-β-1=0, which can be solved to find the two roots β_{1} and β_{2}. These roots are essential for expressing the general solution as a linear combination of the form x_{k}=\alpha_{1}\beta_{1}^{k}+\alpha_{2}\beta_{2}^{k}, where α_{1} and α_{2} are determined by initial conditions x_{0} and x_{1}. The method parallels the approach used in solving differential equations.

PREREQUISITES
  • Understanding of difference equations
  • Familiarity with characteristic equations
  • Knowledge of solving quadratic equations
  • Basic concepts of linear combinations in solutions
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  • Study the roots of the characteristic equation β^{2}-β-1=0
  • Learn about linear combinations of solutions in difference equations
  • Explore the relationship between difference equations and differential equations
  • Investigate initial condition applications in solving difference equations
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Students and educators in mathematics, particularly those focusing on difference equations, linear algebra, and differential equations. This discussion is beneficial for anyone seeking to deepen their understanding of solving recurrence relations.

jameswill1am
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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.
 
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Hi jameswill1am! :smile:

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 an+1 = ßan, which of course is an = Cßn.

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