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Fraction finder

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  • purushaz
    ... Here s a useful operation to find a rational fraction from the continued fraction form [a,b,c....]. (applications will be given later). Find [1,2,3,4].
    Message 1 of 1 , Sep 27, 2002
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      --- In Guruevents@y..., "purushaz" <purushaz@y...> wrote:
      Here's a useful operation to find a rational fraction from the
      continued fraction form [a,b,c....]. (applications will be given
      later). Find [1,2,3,4].

      1. Start at left, let's label the above [a,b,c,d] and under a, write
      1/a = 1/1 in this case.

      2. Bring down b into numerator position below. For denominator,
      multiply b by previous denominator (a 1) and add previous numerator
      (a 1). So we write 2/3 under the 2.

      3. For all successive c,d...and more (a finite number...an infinite
      number of repeating, or cyclical terms will have a different
      procedure i.e. bar[a,b,c,d] =[a b c d a b c d a b c d...]; at any
      rate, with our simpler example the procedure is: for all successive
      numerators going to the right, multiply nth heading term such as "c"
      by previous (n-1th) numerator and add n-2th numerator. So under the 3
      we write 7. Analogous procedure for denominators, so we get a 10 for
      the denominator under the c position. Similarly, under the d
      position, our numerator is 30 and denominator is 43. That's our

      answer: 30/43 = [1,2,3,4]; with the terms 1/1, 2/3 and 7/10 being

      partial quotients, note neighbors have unit +or- modulus or
      determinant, a property of "convergents", a string of which in
      physics may be called a "concatenation". Thus, we can arbitrarily
      form a concatenation between any of our partial quotients by taking
      mediants (add numerators and denominators separately:. e.g. some
      mediants between 2/3 and 7/10 could be: 2/3 11/16, 9/13 & our 7/10.

      Alternative procedure: The following is the reverse of the Euclidean
      algorithm used to get [1,3,3,4]. Since, if we start with 30/43 ,
      using the E.A. procedure: 43/30 = 1 & 13/30, while 30/13 = 2 and
      4/13, while 13/4 = 3 and 1/4, and 4/1 = 4; but the fractions 13/30,
      4/13, and 1/4 are not in the previous set of partial quotients. To
      recover the same fractions gotten in the E.A. procedure, we do the
      following:
      1. write out our 1 2 3 4 and starting from the
      RIGHT, place inverse of "d" (a 4) under the "c", getting 1/4 under
      the 3. For all successive terms, going to the left, do the following;

      2. Move current denominator into next (going to the left numerator
      position). Next (going to left) denominator is current heading term
      (a 3) times current denominator (a 4), PLUS current numerator (a 1).
      So denominator under the 2 is a 13. Then move the 13 into numerator
      position under the 1, and denominator under the 1 is a 30 = (2 times
      13, plus 4).

      3. Do the above procedure one more time getting a fraction to the
      LEFT of "a", which is our 30/43. This gives the exact reverse of the
      E.A. operation we used to get [1,2,3,4]. The terms here are
      remainders, not convergents; and they will (all) not have the unit
      modulus unless the fractions are of the form [N,N,N...] i.e.
      convergents to positive Silver Mean constants. In that case,
      fractions under a,b,c...d will be identical using both the first
      procedure and the 2nd procedure.. Have fun! ....Gary Adamson
      --- End forwarded message ---
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