Dick Hess, Puzzles from around the world, 1997, Problem 116:
  Extension: Torsten Sillke, FRA, Nov. 1997

  Using two positive real c and d (except for (c,d)=(1,1)) you can approximate 
  each 0<x<1 with arbitrary accuracy using +, -, *, /, and sqrt.

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  Let e be Euler's e = exp(1) then we have the approximations:
           ___
  VVVVVVVVV e  - 1    =  1/2**n ( 1  +  1/2**(n+1)  +  O(1/4**n) )
   n sqrts

               _____
  1 - VVVVVVVVV 1/e   =  1/2**n ( 1  -  1/2**(n+1)  +  O(1/4**n) )
       n sqrts

           ___            _____
  VVVVVVVVV e  - VVVVVVVVV 1/e   =  1/2**n ( 1 +  O(1/4**n) )
  n+1 sqrts      n+1 sqrts

           ___            ___
  VVVVVVVVV e  - VVVVVVVVV e     =  1/2**n ( 1  + 3 1/2**(n+1)  +  O(1/4**n) )
  n-1 sqrts       n sqrts

  Only the first two approximations are needed for the general case.
  Without loss of generality we can assume d = 1, as the limit of
  iterating square roots is 1.

 Case: c = exp(1), d=1
	  ___________________
         /           ___
  VVVVVVV    VVVVVVVV e  - 1  
  k sqrts    n sqrts

  can approximate x in (0,1) with arbitrary accuracy.

  So if we fix k, n will be set as

    1/2**(n/q) >= x > 1/2**((n+1)/q)   with q = 2**k.

  Then (k, n) approximates from below.
  Therefore we choose n = round( - 2**k log2(x) ).
  If k increases the approximation become arbitrary close.

  For c > 1 we would get 

    1/2**(n/q) <= x/p < 1/2**((n-1)/q)   with q = 2**k and p = log(c)^(1/q).

  As p = 1 + 1/q log(log(c)) + O(q^2) we can forget c for large k.


 Case: c = exp(-1), d=1
	  ____________________
         /              _____
  VVVVVVV   1 - VVVVVVVV 1/e 
  k sqrts       n sqrts

  can approximate x in (0,1) with arbitrary accuracy.

  So if we fix k, n will be set as

    1/2**(n/q) >= x > 1/2**((n+1)/q)   with q = 2**k.

  For c < 1 we would get 

    1/2**(n/q) <= x/p < 1/2**((n-1)/q)   with q = 2**k and p = log(1/c)^(1/q).

  As p = 1 + 1/q log(log(1/c)) + O(q^2) we can forget c for large k.

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  Torsten Sillke, 17 Nov 1997, FRA