Fixed Point Iterations of F(X) = (S + KX) / (K + X), S > 0

Question

Assume S and K are rational numbers.

So F is a function from the rational numbers to the rational numbers. Assume K > 0 (corresponding results when K < 0) By using google spreadsheets, I came up with following result:

Let p any rational numbers not equal to -K.

Define $p_{k+1} = F(p_{k})$, with $p_{1} = F(p)$

Then {$p_k$} is a Cauchy sequence, and it has a rational number limit if and only if that number is

$\sqrt{S}$

Is there a proof of this without recourse to the real number field?

Alternatively, what mathematical machinery exists to explain this?

Here is the spreadsheet:

[https://docs.google.com/spreadsheets/d/17MkOLqqCmlDWYJUVdlzM8B2bjQz36TWntGJ7NWPtY44/edit?usp=sharing]

Answer 1

Solve

$\displaystyle \frac{S+KX}{K+X} = \frac{S+K(\frac{S+KX}{K+X})}{K+(\frac{S+KX}{K+X})} $

i.e., the fixed point is such that an iteration of your function is equal to the next iteration of your function.

Better yet, solve:

$\displaystyle X = \frac{S+KX}{K+X}$

Answer 2

$$F(X) = \frac{KX+S}{X+K}$$ is a fractional linear transformation. Its iterations are $$ F^{n}(X) = \frac{a_n X + b_n}{c_n X + d_n}$$ where $$ \pmatrix{a_n & b_n\cr c_n & d_n} = \pmatrix{ K & S\cr 1 & K\cr}^n $$ It is useful to normalize by dividing the entries of the matrix $\pmatrix{K & S\cr 1 & K\cr}$ by its largest eigenvalue, which is $K + \sqrt{S}$. Thus we can write $$ F^n(X) = \frac{A_n X + B_n}{C_n X+ D_n}$$ with $$ \pmatrix{A_n & B_n\cr C_n & D_n} = M^n$$ $$M = \pmatrix{K/(K+\sqrt{S}) & S/(K+\sqrt{S})\cr 1/(K+ \sqrt{S}) & K/(K+\sqrt{S})}$$ Using diagonalization, $$M^n \to \pmatrix{ 1/2 & \sqrt{S}/2\cr 1/(2 \sqrt{S}) & 1/2}\ \text{as}\ n \to \infty$$ so that for any $X \ne -\sqrt{S}$ for which all $F^n(X)$ are defined (i.e. the denominator is never $0$), $$F^n(X) \to \frac{X + \sqrt{S}}{X/\sqrt{S} + 1} = \sqrt{S}\ \text{as}\ n \to \infty$$ Note that there are values $X$ other than $-K$ for which the denominator does become $0$ at some point. Thus for $X = -\frac{K}{2} - \frac{S}{2K}$ we have $F(X) = -K$ so that $F^2(X)$ is undefined. Of course, this doesn't happen for $X > 0$.