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Continued fractions
Name: existin
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: Around 1995
Question:
I am very interested in continued fractions. I was wondering if there was a
way to multiply them easily without changing them back to a simple fraction
and then multiplying.
Replies:
The book is CONTINUED FRACTIONS (big surprise) by H. S. Wall, published by
Chelsea Publishing Co., 1973. It is pretty high level stuff: I am still
perusing Chapter 1 to see if I can give a quick answer to your question.
You might have to go to a university library to find that book.
hawley
Another interesting question that I would never have thought to ask! I
looked in several books in Iowa State University's library, and none of them
mention multiplication (or addition) of continued fractions (CF's). This,
and consideration of some simple cases, lead me to guess that there is no
easy way (a method might exist, but using it may be a lot more laborious
than just converting the CF's to fractions, multiplying, and converting the
result to its CF representation). For example, consider multiplication of a
CF by integers. Let us use:
12 1
-- = 2 + ------- or, in the notation I have
5 2 + 1
-
2
[2,2,2]. The CF of an integer N is just [N].
We get
[2]*[2,2,2] = [4,1,4], [3]*[2,2,2] = [7,5],
[4]*[2,2,2] = [9,1,1,2], [5]*[2,2,2] = [12],
[6]*[2,2,2] = [14,2,2], etc.
I would be amazed to find out that there is a reasonable rule, even for a
simple case like this (let alone the general case), for finding the length
and entries of such CF products. By the way, numerical "experiments" with
CF's are a lot easier if you have access to a symbolic algebra package like
Mathematica or Macsyma (or Maple?). For example, in Macsyma I can enter
"CF([3,2,1]*[1,2,3]);" and it returns [4,1,3,5].
rcwinther
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Update: June 2012
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