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Given: $a < b < 0$ and
$y_1 = a$
$y_2 = b$
$y_n = \frac{1}{3}y_{n-1} + \frac{2}{3}y_{n-2}$, for $n > 2$

I was able to show that this sequence was contractive and now I'm asked to find the limit. I had a problem similar to this one where I was able to find a geometric series out of the recursion, but this one has not been as obvious.

After looking the first few terms of the sequence, we get that
$y_3 = \frac{1}{3}b + \frac{2}{3}a$
$y_4 = \frac{7}{9}b + \frac{2}{9}a$
$y_5 = \frac{13}{27}b + \frac{14}{27}a$

The only observation I see so far is that the coefficients of each term in the sequence are adding up to 1.

I also know that since I was able to show it converges, then every subsequence of $y_n$ must also converge to this limit. Therefore, I need to find some "nice" subsequence that I can see what its limit is. Any help as to where I go about this?

  • this is linear homogeneous recurrence. there exists standard method to do these – Lost1 Apr 14 '13 at 21:49
  • I wasn't taught this. I'll find some material about it and read it. Thanks. –  Apr 14 '13 at 21:52
  • do you know how to solve say $4y'' + 3y'+2y=0$? this is kind of like the discrete analogue of this ODE. the method is looking for an eigenvalue, just like in the ODE thing. that is why you would have solutions looking like $q^n$ in the same way for the ODE you get $e^ax$ – Lost1 Apr 14 '13 at 21:53
  • Yes. So I can think of the subscript of $y_n$ as the degree of a term in a polynomial and finding the limit of the sequence is equivalent to say solving: $r^n - \frac{1}{3}r^{n-1} - \frac{2}{3} r^{n-2} = 0$? –  Apr 14 '13 at 21:58
  • I think it is something like that yes. Check a reliable source. you put in your boundary condition as $y(1)=a$ and $y(2)=b$, then bingo. – Lost1 Apr 14 '13 at 22:04
  • Here is the technique for solving a linear homogenous recurrence relation. – Mhenni Benghorbal Apr 14 '13 at 22:47
  • Was able to find the limit, Thanks! This is really quite nice. –  Apr 14 '13 at 23:02

1 Answers1

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If you can find a closed formula of the form $y_n=c\cdot q^n$, then you can conclude that $q^2=\frac13q+\frac23$. Actually, you should find two different kinds of solution this way. It is possible to make combine these to match the conditions $y_1=a, y_2=b$.

Hagen von Eitzen
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  • Limit is $\frac{3}{5}b + \frac{2}{5}a$ Thanks for the tip, Professor never covered this technique. –  Apr 14 '13 at 23:02