Browse Groups

• ... 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
View Source
--- 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 ---
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.