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. hawleyAnother 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 Click here to return to the Mathematics Archives

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