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

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