1
$\begingroup$

I have the following recurrence relation: $$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\hspace{1cm}(*)$$

If we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that $r=\frac{a+\sqrt{a^2-4}}{2}$ or $r=\frac{a-\sqrt{a^2-4}}{2}$.

By using the initial values $x_1=1, x_2=a$, I found that $$x_n=\frac{1}{\sqrt{a^2-4}}\left(\frac{a+\sqrt{a^2-4}}{2}\right)^n-\frac{1}{\sqrt{a^2-4}}\left(\frac{a-\sqrt{a^2-4}}{2}\right)^n$$ is a solution for the recurrence relation (*).

Question: How do we know whether this is the only solution for the recurrence relation $(*)$? Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions will have this form.

$\endgroup$
  • $\begingroup$ Use method of difference on the recursive definition $ x_{n+2}-x_{n+1}=(a-2)x_{n+1}+x_{n+1}-x_{n}$ and add these equations for all values of n. $\endgroup$ – Pi_die_die Jul 8 '18 at 20:57
  • $\begingroup$ In other words, prove that it satisfies condition $(*)$. $\endgroup$ – steven gregory Jul 8 '18 at 20:59
  • $\begingroup$ Yes you would get the sum which would be the type of sum you would expect from a gp $\endgroup$ – Pi_die_die Jul 8 '18 at 21:01
3
$\begingroup$

You have specific initial values $x_0$ and $x_1$. The recurrence relation fully determines all other values. Thus, there's exactly one solution. If you've found it, that's it, there can't be any others, no matter what approach you took in order to find it.

If you want to show that all solutions of $x_{n+2}=ax_{n+1}-x_n$ are combinations of the geometric series you found, you can argue as follows:

This is a linear second-order recurrence. Its solutions form a two-dimensional vector space. (A vector space because of the linearity of the recurrence, and a two-dimensional one because the solution space is spanned by the two solutions for the initial conditions $x_1=1$, $x_2=0$ and $x_1=0$, $x_2=1$.) You've found two linearly independent solutions; hence they span the entire solution space.

$\endgroup$
  • $\begingroup$ Hi @joriki I would appreciate some clarifications before I accept your answer. I understand that the set of solutions is a subspace of the vector space of sequences $\{f:\mathbb{N}\to\mathbb{C}: f\mbox{ is a function}\}$. I also understand that I found two solutions that are linearly independent. I do not know why the dimension of the solution space is 2 and not greater than 2. Also, what do you mean by $a_0$ and $a_1$? $\endgroup$ – Chilote Jul 8 '18 at 21:23
  • $\begingroup$ @Chilote: I'm sorry, I meant $x_1$ and $x_2$; I've fixed that. The solution for given initial conditions $x_1=\hat x_1$ and $x_2=\hat x_2$ is $\hat x_1s_1+\hat x_2 s_2$, where $s_1$ is the solution for $x_1=1$ and $x_2=0$ and $s_2$ is the solution for $x_1=0$ and $x_2=1$; so $s_1$ and $s_2$ span the solution space; hence it's at most two-dimensional (in fact exactly two-dimensional, since $s_1$ and $s_2$ are linearly independent). $\endgroup$ – joriki Jul 8 '18 at 21:31
  • 1
    $\begingroup$ Got it, the solution space of $x_{n+2}=ax_{n+1}-x_n$ is $S=\{(x_n)_n:x_{n+2}=ax_{n+1}-x_n\}$ which is a vector space. Any vector in this space is of the form $x=(x_1,x_2,ax_2-x_1,a(ax_2-x_1)-x_2,\dots)=x_1 s_1+x_2 s_2$ where $s_1$ and $s_2$ are two particular solutions that are linearly independent. $\endgroup$ – Chilote Jul 8 '18 at 21:49
  • 1
    $\begingroup$ You've found two linearly independent solutions Just to nitpick, that's iff $a \ne \pm 2$. $\endgroup$ – dxiv Jul 8 '18 at 23:29
  • $\begingroup$ @Chilote: This was actually overkill, since you only asked about the uniqueness of one particular solution, not about all solutions being combinations of the geometric series. I edited the answer accordingly. $\endgroup$ – joriki Jul 9 '18 at 5:22
2
$\begingroup$

A bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy $$ x_n^2 - a \, x_n \, x_{n+1} + x_{n+1}^2 = 1 $$

Try consecutive values in $$ 1, \; \; a, \; \; a^2 - 1, \; \; a^3 - 2a, \; \ldots $$

This comes from the matrix $$ \left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) $$ which has determinant $1$ and trace $a.$ It also gives an automorphism of the quadratic form $x^2 - a xy+y^2.$ An automorphism matrix $P$ for a quadratic form that has Hessian matrix $H$ satisfies $P^T HP = H$

In this case $$ \left( \begin{array}{cc} 0 & -1 \\ 1 & a \end{array} \right) \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) = \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right) $$

The explicit relation with the sequence is

$$ \left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) \left( \begin{array}{c} x_n \\ x_{n+1} \end{array} \right) = \left( \begin{array}{c} x_{n+1} \\ x_{n+2} \end{array} \right) $$

$\endgroup$
  • $\begingroup$ How can I get the first equality? $\endgroup$ – Chilote Jul 8 '18 at 21:52
  • 1
    $\begingroup$ @Chilote the automorphism group for an (indefinite) binary quadratic form, along with the Pell type equation $\tau^2 - \Delta \sigma^2 = 4,$ is discussed in many number theory and quadratic forms books. $\endgroup$ – Will Jagy Jul 8 '18 at 23:24
  • $\begingroup$ I think the quadratic form corresponding to the matrix you wrote above is $ay^2$ $\endgroup$ – Chilote Jul 8 '18 at 23:59
  • 1
    $\begingroup$ @Chilote I put in a little about the automorphism. The matrix that gives the quadratic form is the symmetric one; it is just the Hessian matrix of second partial derivatives $\endgroup$ – Will Jagy Jul 9 '18 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.