A recursion function for square roots

OK. This post is NOT a big deal. I posted a previous version on Angelfire several years ago and, after reviewing it last night, decided to polish it up just a tad.

The recursion function

X_n = (X_n-1 + C) + C

is a special case of the continuing fraction

C + 1
    ----------------
    C + 1
        ------------
        C + 1
            --------
            C + ...

which, taken to infinity, equals (C + (C² + 4)^0.5)/2.

The recursive equals that continuing fraction when X₀ = 0.

However, we point out that any real, other than -C, can be used and the result in the infinite limit is a single value.

To see this, we write three iterations of the recursive thus:

((((X₀ + C)^-1 + C) + C)^-1) + C)^-1

And if you write it out in traditional form using 1/z rather than z^-1, it becomes obvious that

lim_n->inf X₀/f(X) = 0, where f(X) is the recursive function.

And note that if X₀ =/= Y₀, then at any step n of the recursive, the values are not equal. There is equality only in the infinite limit.

An example for C = 2, n = 4.

lim_n->inf X_n = 1/2(2 + 8^0.5) = 1 + 2^0.5

X0 = 1

3.0
2.333...
2.428571429
2.411764706
2.414634146

X0 = 1/2

2.5
2.4
2.41666...
2.413793103
2.414285714

X0 = 31

33.0
 2.0333...
 2.492537313
 2.401197605
 2.416458853

X0 = -31

-29.0
  1.965517241
  2.508771930
  2.398601399
  2.416909612

X0 = 1/31

2.032258065
2.492063492
2.401273885
2.416445623
2.413830955

X0 = -1/31

1.967741935
2.508196721
2.398692810
2.416893733
2.413754228