0
$\begingroup$

Given the sequence:

$x_{n+1} + x_{n} - 6x_{n-1} = 0$

I need to give a necessary and sufficient condition on $x_{0}$ and $x_{1}$ such that the sequence is bounded below.

I am confused about "necessary and sufficient". Is it enough to say that for the sequence to be bounded below by the first term it has to be increasing, so $x_{1} > x_{0}$?

  • 0
    Welcome to math stack exchange! If the $6$ would be $-6$, the solution would be much nicer. Please check, whether the given recurrence is actually true. – 2017-02-26
  • 0
    You are right! Sorry for that, I edited the question. – 2017-02-26
  • 0
    You are supposed to solve this, that is, to express every $x_n$ in terms of $(x_0,x_1)$, and then to inspect the formula. Which step is a problem? – 2017-02-26
  • 0
    I found the general formula for xn, but I don't know how to express it in terms of (x0, x1) - shouldn't I know their values to do that? Thanks. – 2017-02-26
  • 0
    I am actually starting to think it can not be bounded below. – 2017-02-26
  • 0
    Of course it can, for example if $x_0=x_1=0$... My impression is that you need first and foremost to review (or, to view?) the general method of resolution of such linear recurrences. – 2017-02-27

1 Answers 1

1

Such recurrence relations can be solved by solving the characteristic quadratic equation, in this case

$$x^2+x-6=0$$

The roots are $-3$ and $2$ and the general solution is $$x_n=c_1\cdot (-3)^n+c_2\cdot 2^n$$

This is bounded from below if and only if $c_1=0$ and $c_2\ge 0$, so we have the general solution $$x_n=c_2\cdot 2^n$$

Inserting $n=0$ and $n=1$ , we have $x_0=c_2$ and $x_1=2c_2$

Hence the condition is $x_1=2x_0\ge 0$

  • 0
    This method always works assuming the roots of the characteristic equation are distinct (They can even be complex). With induction, you can prove the general solution to satisfy the recurrence relation. – 2017-02-26