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I have a seemingly straightforward homogeneous second degree recurrence relationship.

$$Z(a+2) = 3 Z(a+1) - 2 Z(a) $$

If given boundary condition of $a$ at 0 and 1, I could solve it using standard methods. I won't bore people with detail here.

However, if boundary condition is given not at 0 and 1, but rather at 0 and $N$, is this still solvable? $Z(a)$ would be undefined for $a < 0 $ or $a > N$

Essentially I'm given the following. $$Z(a) = 0 \space \space \space if \space a = 0.$$ $$Z(a) = 1 \space \space \space if \space a = N$$ $$Z(a) = 3 Z(a-1) - 2 Z(a-2) \space \space \space if \space 0 < a < N $$

user3240688
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1 Answers1

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We have $$Z(a+2)-2Z(a+1)=Z(a+1)-2Z(a)$$ so, $$Z(a+1)-2Z(a)=Z(a)-2Z(a-1)=\cdots =Z(1)-2Z(0)=Z(1)\tag1$$

On the other hand, we have $$Z(a+2)-Z(a+1)=2(Z(a+1)-Z(a))$$ so, $$Z(a+1)-Z(a)=2(Z(a)-Z(a-1))=\cdots =2^a(Z(1)-Z(0))=2^aZ(1)\tag2$$

Now $(1)-(2)$ gives $$-Z(a)=Z(1)-2^aZ(1),$$ i.e. $$Z(a)=(2^a-1)Z(1)$$

So, if $Z(N)=1$, then you have $$Z(N)=(2^N-1)Z(1)=1.$$

So, you can set $Z(1)=\frac{1}{2^N-1}$ for a given $N\gt 1$.

mathlove
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  • I don't think that's right. Say you're given the recurrence, and I guess a value of Z(1). Using that, I can define Z(a) for all a between 0 and N. I'll know whether my original guess is too big or too small when I see if Z(n) = 1. There exists a real number Z(1) such that Z(n) would be 1. I want to find it. – user3240688 Oct 17 '15 at 19:09
  • @user3240688: I understand what you mean, but the body of your question does not say what you want to say at all... Please edit it. – mathlove Oct 17 '15 at 19:36
  • @user3240688: I rewrote an answer. I hope this answers your question written in the comment. – mathlove Oct 17 '15 at 19:51
  • this is beautiful. thank you so much! – user3240688 Oct 19 '15 at 04:03