Version 8 of Notes on continued fractions

MS: These are some notes on my playing around with Fraction Math and the reference to the nice notes on Continued Fractions at http://www.inwap.com/pdp10/hbaker/hakmem/cf.html [L1 ].

I have started playing with these things and resisted (for now) looking for other sources. Some of the "errors" I found in the reference may be due to the informal presentation of results there, which might be inaccurate or misunderstood by me: caveat emptor.

DEFINITIONS: a rational approximation p/q to a real number x is "best" iff, for every integer r and s,

  (s <= q) ==> (|x - p/q| <= |x - r/s|)

It is "best on its side" if

  ((s <= q) & (sgn(x-p/q) = sgn(x-r/s))) ==> (|x - p/q| <= |x - r/s|)

i.e, if no other fraction on the sameside of x with a lesser denominator comes closer.

NOTATION: let us identify a (positive) real number x with its regular continued fraction representation

  x = {x[0] x[1] x[3] x[4] ...}

Define the truncation of x after (n+1) terms

  a(x,n) = {x[0] x[1] x[3] x[4] ... x[n]}

and let

  b(x,n,i) = {x[0] x[1] x[3] x[4] ... x[n-1] i}

be a(x,n) with the last element replaced by 0 < i <= x[n]. Note that

  a(x,n) = b(x,n,x[n])

NOTE: [quotient_rep] in Fraction Math computes the highest-order truncation requiring no integers larger than maxint in the rational representation p/q.

Item 101A (3) clearly says (AFAIU, not all claims reproduced here):

The claims are:

   A - a(x,n) is "best"
   B - b(x,n,i) is "best" if i>1
   C - b(x,n,1) is never "best" (note that b(x,n,i) is not defined when x[n] = 1)

Let me provide counterexamples to both B and C, thus showing that they are not true.

The example provided in the text actually contains counterexamples to B. Let x = pi = {3 7 15 1 292 ...}

   . b(x,0,2) = 2/1 = {2} is not best (3/1 is better)
   . b(x,1,2) = 7/2 = {3 2} and b(x,1,3) = 10/3 = {3 3} are not best (3/1 is better) 
   . b(x,2,2) = 47/15 = {3 7 2} is not best (22/7 is better)

For another counterexample to B, easier to follow by hand, consider

  x = 0.51 = {0 1 1 24 2}

The number

  b(x,3,4) = 5/9 = {0 1 1 4} = 0.555...

is not best, as 1/2 = {0 1 1} = {0 2} is better.

For a counterexample to C, consider x = 7/10 = {0 1 2 3}; now

  b(x,2,1) = 1/2 = {0 1 1} = {0 2}

is best.

So, in light of these counterexamples, one proposition I can hope could be true is:

   D - The set of "best on its side" approximations to 0<x<1 coincides with the set of numbers of the form b(x,n,i)

Remark that a simple corollary would be:

   E -if r is a "best" approximation to x, then r = b(x,n,i) for some (n,i)

as a "best" approx has to be "best on its side".

I think (D) can be proved from ITEM 101C in [L2 ] (which may be true AFAIK). I'm working on the details, maybe a restriction (i!=1) will be necessary. After a while I'll stop playing and start reading ...

The example I provided in Fraction Math showing that quotient_rep does not always provide the best approximation is

   quotient_rep 3.1416305 500 --> 355/113

Now, 3.1416305 = {3 7 16 2 ...},and quotient_rep produced

   355/113 = {3 7 16} = a(3.1416305,2)

But the fraction

   377/120 = {3 7 16 1} = b(3.1416305,3,1)

is closer - a new counterexample to C.

Remark that the next truncation involves integers that are too large:

   a(3.1416305,3) = {3 7 16 2} = 732/233

Some code to deal with continued fractions

a lot of this is ripped off from KBK's code in Fraction Math. Some testing of inputs would also be needed.

   #
   # proc cont2frac
   #
   # Returns the fraction corresponding to the list
   # of denominators in a (truncated) regular continued
   # fraction. There is no check for overflow.
   #

   proc cont2frac {lst} {
       foreach {p q p0 q0} {1 0 0 1} break
       foreach a $lst {
           foreach {p p0} [list [expr {$a*$p+$p0}] $p] break 
           foreach {q q0} [list [expr {$a*$q+$q0}] $q] break 
      }
      list $p $q
   }


   # 
   # proc num2cont
   #
   # Returns a list of two lists:
   #   1. the list of denominators in the longest truncated
   #      regular continued fraction expansion of num with both
   #      numerator and denominator <= maxint.
   #   2. the representation of the above list as a fraction
   #
   # Remark that "quotient_rep $num $maxint]" is equivalent to
   # "lindex [num2cont $num $maxint] 1"
   #

 proc num2cont {num { maxint 2147483647 } } {
     foreach {p q p0 q0} {1 0 0 1} break
     set clist {}

     while {1} {
        set a [expr {int($num)}]
        set fract [expr {$num - $a}]

        if {(1.0 * $a * $p + $p0 > $maxint) \
                || (1.0 * $a * $q + $q0 > $maxint)} {
            break
        }
        lappend clist $a
        foreach {p p0} [list [expr {$a * $p + $p0}] $p] break 
        foreach {q q0} [list [expr {$a * $q + $q0}] $q] break 

        if {abs($fract * $maxint) < 1} break
        set num [expr {1.0 / $fract}]
     }
     list $clist [list $p $q]
 }