Backward from infinity?
The following question has been raised and answered lately:
Please take a look at the Reverse method part of the
answer
given by this author.
What's happening there is that we start with some "infinitely" large number $n=$LARGE,
some accompanying value of $1-1/(n+1) < x_{n+1} < 1$ - the precise value of $x_{n+1}$ doesn't matter much
- and then iterate down to $n=1$ where the desired value $x_1$ is found. I've learned that this is called
Backward Recursion .
The LARGEr the number $n$ we start with, the more accurate the value of $x_1$ will be;
that accuracy is improving with an amount of approximately $\times 1/2$ with every single
increase LARGE := LARGE $+ 1$ which is one significant bit at a time.
This is the basic Backward Recursive formula:
$$
x_n := \frac{x_{n+1}}{\frac{1}{2n}+\sqrt{\left(\frac{1}{2n}\right)^2+x_{n+1}}}
$$
Our excercise has been limited by 64 bits double precision of standard Delphi Pascal.
Note that LARGE is the only variable in the programs that is responsible for
the limitations imposed.
But suppose that we have hardware and software at our disposal which is free of any
limitations - I've always thought that such an ideal machinery may be simply called
mathematics . In a truly mathematical machine we can make LARGE as large as we
want. Now the following wild idea comes into mind. Suppose that we start with $n=\infty$,
hence $x_{\infty+1}=1$ and iterate back from infinity. Then informally we have (hope you get the idea,
which is sort of limit LARGE $\to \infty$) :
$$
x_{\infty} := \frac{x_{\infty+1}}{\frac{1}{2\infty}+\sqrt{\left(\frac{1}{2\infty}\right)^2+x_{\infty+1}}}
\quad \Longrightarrow \quad x_\infty := \sqrt{x_{\infty+1}} = 1
$$
Therefore the next iterand, hence all iterands remain $1$ and finally $x_1 = 1$, which is wrong. So we are getting nowhere.
(Surely that has been a long way back to go :-)
Question: Is there any way to yet depart from infinity, get around this "stuck with $1$" and
arrive at the correct value of $x_1$?