An iteration formula I found (please don't jump at me if it's already been discovered)

Question

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Where the modulus-like symbol actually denotes iteration of a radical function. Sorry for the messy work everyone- I am new to this stuff, and I literally just found this iteration formula like 15 minutes ago. $X[i]$ starts as being an arbitrary number greater than $n!/(n-2)!$ which I should have mentioned earlier but didn't have space to (apologies). So the business is: you start with a number e.g. 1, then if you want to find say $50$, you compute $sqrt(1 + 50!/48!)$ then replace $1$ on your calculator with the result you just got a.k.a. $Ans$ etc. and keep iterating till you get (at a surprisingly fast convergence rate!) the very number you started with, n.

I appreciate this might not be the most interesting insight or $n!/(n-2)!$ is probably just me not realising that it could also be presented as $n(n-1) = n^2-n$. But still, please don't mass-downvote it all, remember it's a 15-year old person who figured this out...

Answer 1

You can do this formally by just looking at $$\sqrt{n!/(n-2)!+\sqrt{n!/(n-2)! + \dots}}$$ and so on to infinity. Suppose this has a limit. Then we'd expect that $$L=\sqrt{n!/(n-2)! + L}$$ This gives $$L^2-L=n!/(n-2)!=n^2-n$$ which has the obvious solution of $L=n$.

Analytically, this could be done with some sort of fixpoint theorem (such as the Banach fixed-point theorem), setting $$f(x)=\sqrt{n!/(n-2)! + x}$$ and using it to show that $\lim_{m\to \infty}\underbrace{f(\cdots f(x)\cdots)}_{m\text{ times}}$ is the solution to the fixpoint equation $$f(L)=L.$$

Answer 2

So you're saying that the sequence $$X_0=1$$ $$X_{k+1}=\sqrt{X_k+\frac{n!}{(n-2)!}}$$ tends to towards $n$. Firstly notice that $$\frac{n!}{(n-2)!}=n(n-1)=n^2-n$$ Hence your sequence is equivalent to $$X_{k+1}=\sqrt{X_k+n^2-n}$$ Then note that for $0\le X_k\lt n$ we have $$X_k\lt\sqrt{X_k+n^2-n}\lt n$$ So your sequence is strictly increasing and bounded above. Also note that $$\sqrt{n+n^2-n}=\sqrt{n^2}=n$$ Hence the sequence has a stationary point at $X_k=n$. This is sufficient to prove that for a given initial value $0\le X_0\le n$ the sequence tends to $n$. In other words $$\lim_{k\to\infty}X_k=n$$ In fact one can prove that this sequence tends to $n$ for any initial value by also using the fact that $\sqrt{X_k+n^2-n}\lt X_k$ when $X_k\gt n$.