5

Let $f(1) = f(2) = 1$ and for $n>2:$

$$f(n) = \frac{f(n-1)^2 (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3}$$

So we get

  • $f(3) = \frac{1}{2} = 0.5$
  • $f(4) = \frac{9}{128} = 0.0703125$
  • $f(5) = \frac{56307}{16777216} = 0.00335615...$
  • $f(6) = \frac{38821347033039473}{295147905179352825856} = 0.000131531...$

and the sequence is going to $0$ in the limit.

Now we are getting to zero at an exponential rate

$$ \lim_{n \to \infty} \frac{f(n+1)}{f(n)} = C$$

Where $C$ is a constant

$$C = 0.003879..$$

What is the closed form for $C$ ??

C is probably algebraic !!

see the analogue simpler limits that give algebraic numbers :

$ a_n = \frac{a_{n-1}(a_{n-1} + 1)}{a_{n-2}}.$ and $ T = 3.73205080..$?

some toughts

I was thinking about using Stolz-Cesàro and substituting the defining identity maybe, but it is not clear to me how exactly that would work.

( https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem )

$$ \lim_{n \to \infty} \frac{f(n)}{f(n-1)} = \lim_{n \to \infty} \frac{f(n) - f(n-1)}{f(n-1) - f(n-2)} = \lim_{n \to \infty} \frac{\frac{f(n-1)^2 (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3} - f(n-1)}{f(n-1) - f(n-2)} = C$$

or

$$ \lim_{n \to \infty} \frac{f(n)}{f(n-1)} = \frac{f(n-1) (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3}$$

But that seems trivial because the limit of $\frac{(f(n-1)+1)^2}{(f(n-2)+1)^3} = 1$. (*)

Because of * it MIGHT BE TRUE that (based on the above)

$$\lim_{n \to \infty} \frac{\frac{f(n-1)^2}{f(n-2)} - f(n-1)}{f(n-1) - f(n-2)} = C$$

??

what may or may not be a valid step and may or may not be helpful.

I think I made a mistake already somewhere...

edit

As noted in the comment the limit depends on the intial value so using Stolz-Cesàro alone is certainly insufficient.

See also the related question with a similar recursion and identical initial values.

Period $T$ such that $f(n+T) = f(n)$ and $f(n) = \frac{(f(n-1)+1)^2 (f(n-1)+2)^2}{(f(n-2)+1)(f(n-2)+2)^3}$

mick
  • 17,886
  • 1
    No S-C cannot work here and the reason is simple: the limit you get depends on the initial values $f(1)$ and $f(2)$ and not just the tail of the sequence. This means you can't get much useful info about $C$ out of any purely asymptotic analysis. My naive guess is that you will need something close to knowing the full solution $f(n)$ to be able to deduce this limit. – Winther Aug 08 '24 at 22:51
  • @Winther good point +1. Indeed I did get different values for different starting values, so indeed it does matter ! I forgot about that somehow ... I do not think there is a closed form for this proces unless maybe something fibonacci like ... – mick Aug 08 '24 at 23:08
  • Let me clarify, I think some set of initial values have a closed form, but I assume they are not dense or not continu on real intervals. – mick Aug 08 '24 at 23:13
  • 1
    FWIW, by calculating a bunch more terms by computer, it seems that $C \approx 0.038793915349577$. But I can't figure out what the exact closed form would be. – Dan Aug 09 '24 at 02:22
  • Maybe the analogue differential equation gives insight ? But im not sure if that one can be solved in closed form, so just a thought. – mick Aug 10 '24 at 00:29
  • If I did my numerical computation correctly, $f(n) \approx 38603.78521955633 \times 0.03879391534957667^n$ for sufficiently large $n$. – Dan Aug 12 '24 at 16:23
  • 1
    And in case this information is at all helpful in finding a closed-form solution:

    $$f(0) = -1 + \frac{1}{2} \left(- \frac{1}{\sqrt[3]{27 - 3 \sqrt{78}}} - \frac{\sqrt[3]{9 - \sqrt{78}}}{3^{2/3}}\right) \approx -1.7606898534022837$$

     – Dan Aug 15 '24 at 02:40
  • @Dan any ideas about that $38603.78..$ number ? – mick Aug 18 '24 at 23:23
  • 1
    @mick: If you call that number $A$, then $AC^{3.25}$ is pretty close to $1$. I'm not sure what to make of that. – Dan Aug 19 '24 at 16:52
  • Should I post to MO ... – mick Sep 03 '24 at 22:53

1 Answers1

2

Partial answer

Let $g(n) = \frac{f(n+1)}{f(n)}$. Then $g(1) = \frac{f(2)}{f(1)} = 1$.

Substituting $f(n) = f(n-1)g(n-1)$ in the recurrence relation gives:

$$f(n-1)g(n-1) = \frac{f(n-1)^2 (f(n-1)+1)^2}{f(n-2)(f(n-2)+1)^3}$$ $$g(n-1)f(n-2)(f(n-2)+1)^3 = f(n-1)(f(n-1)+1)^2$$

Substituting $f(n-1) = f(n-2)g(n-2)$ gives:

$$g(n-1)f(n-2)(f(n-2)+1)^3 = f(n-2)g(n-2)(f(n-2)g(n-2)+1)^2$$ $$g(n-1)(f(n-2)+1)^3 = g(n-2)(f(n-2)g(n-2)+1)^2$$

Substituting $f(n-2) = f(n-3)g(n-3)$ gives:

$$g(n-1)(f(n-3)g(n-3)+1)^3 = g(n-2)(f(n-3)g(n-3)g(n-2)+1)^2$$

Substituting $f(n-3) = f(n-4)g(n-4)$ gives:

$$g(n-1)(f(n-4)g(n-4)g(n-3)+1)^3 = g(n-2)(f(n-4)g(n-4)g(n-3)g(n-2)+1)^2$$

Substituting $f(n-4) = f(n-5)g(n-5)$ gives:

$$g(n-1)(f(n-5)g(n-5)g(n-4)g(n-3)+1)^3 = g(n-2)(f(n-5)g(n-5)g(n-4)g(n-3)g(n-2)+1)^2$$

Continuing the pattern to $k$ steps gives:

$$g(n-1)(f(n-k)g(n-k)g(n-k+1)\dots g(n-3)+1)^3 = g(n-2)(f(n-k)g(n-k)g(n-k+1)\dots g(n-2)+1)^2$$

Let $k=n-2$.

$$g(n-1)(f(2)g(2)g(3)\dots g(n-3)+1)^3 = g(n-2)(f(2)g(2)g(3)\dots g(n-2)+1)^2$$

But $f(2) = 1$, so:

$$g(n-1)(g(2)g(3)\dots g(n-3)+1)^3 = g(n-2)(g(2)g(3)\dots g(n-2)+1)^2$$

This eliminates $f$ from the equation, and just leaves $g$. For convenience, let $h(n) = \prod_{k=1}^n g(k)$.

$$g(n-1)(h(n-3)+1)^3 = g(n-2)(h(n-2)+1)^2$$

To be continued: Solve the recurrence for $g$...

Dan
  • 18,262
  • 1
    Your last equation says it converges pretty fast. Afterall $g(n-1)$ and $g(n-2)$ converge to the same constant. And $h(n)$ must go to zero fast. In fact $h$ can be expressed in $f$ by the telescoping connection between $f,g,h$. Even stronger I think the last equation is just a restatement of the first since $g$ and $h$ can be restated by $f$. Therefore I am not sure this helps – mick Aug 10 '24 at 00:27