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I have the following recurrence relation: $$T(i) = \sqrt{T(i-1) \left(T(i+1) + k\right)},$$ with $k \geq 0$, a fixed constant.

I know that when $k=0$, we have: $$T(i) = \sqrt{T(i-1) T(i+1)},$$ which solution is $T(i) = e^{ai+b}$ (geometric progression), but I do not know if there is a solution for the general case.

  • If there is no closed-form solution to it, as it seems to be the case, is there an easy way to compute $T(i),\;1 < i < n$ given $T(1) = A$ and $T(n) = B$ ?

  • More generally, is there a systematic, elegant way to (numerically) compute recurrences when the initial conditions are not "contiguous" ?

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And a bit of notation: Is "$[\cdots]$" supposed to be the integer part of a number? E.g. $[4.2] = 4, [-4.2] = -4$? Or is it just a bracket? –  user2468 Sep 4 '12 at 14:36
    
Just a regular bracket. I edited to avoid confusion. –  doc Sep 4 '12 at 14:50
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Asymptotically, it will be given by the same geometric progression (because eventually $T \gg k$). For small times there might be quite some difference. I doubt however that there is a closed form solution. –  Fabian Sep 4 '12 at 15:28
    
'...when the initial conditions are not "contiguous"...' I'm not sure if I can connect that description with the problem statement. Didn't you, by chance, mean to write T(i+1+k) instead of T(i+1)+k , no? –  leonbloy Jan 26 '13 at 21:06
    
@leonbloy no, I meant the initial conditions ($T(1) = A$ and $T(n) = B$). –  doc Jan 31 '13 at 15:49

1 Answer 1

Remind that $T(i)=\sqrt{T(i-1)T(i+1)}$ can find its general solution is just a luck:

$T(i)=\sqrt{T(i-1)T(i+1)}$

$(T(i))^2=T(i-1)T(i+1)$

Let $T(i)=e^{U(i)}$ ,

Then $T(i-1)=e^{U(i-1)}$ , $T(i+1)=e^{U(i+1)}$

$\therefore(e^{U(i)})^2=e^{U(i-1)}e^{U(i+1)}$

$e^{2U(i)}=e^{U(i-1)+U(i+1)}$

$2U(i)=U(i-1)+U(i+1)$

$U(i+1)-2U(i)+U(i-1)=0$

$U(i)=\Theta_1(i)i+\Theta_2(i)$, where $\Theta_1(i)$ and $\Theta_2(i)$ are arbitrary periodic functions with unit period

$\therefore T(i)=e^{\Theta_1(i)i+\Theta_2(i)}$, where $\Theta_1(i)$ and $\Theta_2(i)$ are arbitrary periodic functions with unit period

It does not mean that $T(i)=\sqrt{T(i-1)\left(T(i+1)+k\right)}$ is as lucky as $T(i)=\sqrt{T(i-1)T(i+1)}$ .

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