Continued FractionsDate: 07/02/97 at 09:47:23 From: Julia MacDonell Subject: Continued fraction What exactly IS a continued fraction? Thanks! Jules Date: 07/02/97 at 09:56:06 From: Doctor Rob Subject: Re: Continued fraction Jules, That is a very interesting question. I'm glad you asked. A continued fraction is an expression of the form: b0 a0 + ----------------------- b1 a1 + ------------------ b2 a2 + ------------- b3 a3 + -------- a4 + ... Here all the ai's and bi's are integers, i >= 0. It may terminate after a finite number of steps. In that case, when you evaluate it, you will get a rational number (quotient of two integers). It may be infinite, in the same way that an infinite series is, or an infinite decimal expansion is. There is a well-developed theory of continued fractions, especially "simple" ones (bi = 1 for all i, and ai > 0 for all i > 1). They have many uses. They can be helpful in approximating irrational numbers, solving Diophantine equations, finding units in quadratic number fields, and other places. If you want more information, you can write again, or consult the following reference: C. D. Olds, _Continued Fractions_, New Math Library, vol. 9 (1963: New York, Random House) -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 07/03/97 at 20:00:25 From: Julia MacDonell Subject: expressing continued fractions How would you express (sqrt A) - B as a continued fraction? Thank you! Date: 07/04/97 at 05:49:35 From: Doctor Anthony Subject: Re: expressing continued fractions The general expression: sqrt(N) + b ----------- where N, b, r are integers (N > 0) r can be expressed as a simple recurring continued fraction. We can write the stages as follows: sqrt(N)+b1 sqrt(N)-b2 r2 ---------- = a1 + ----------- = a1 + ---------- r1 r1 sqrt(N)+b2 1 = a1 + ------------ sqrt(N)+b2 ----------- r2 a1 is the integral part of [sqrt(N)+b1]/r1 and: b2 = a1*r1 - b1 N - b2^2 r2 = ---------- r1 Apply these three expressions for calculating a(n), b(n+1), r(n+1) to the expression of: sqrt(37)+8 ---------- as a continued fraction. 9 Here a1 = integral part of expression = 1, also b1 = 8, r1 = 9. Then b2 = 1 x 9 - 8 = 1 r2 = [37-1]/9 = 4 So the continued fraction at this stage is: 1 1 + ---------- sqrt(37)+1 ---------- 4 It is better to lay out the working in tabular form as follows, using the expressions for a(n), b(n+1), r(n+1) given above. n 1 2 3 4 5 6 ------------------------------------------- b 8 1 3 4 5 3 r 9 4 7 3 4 7 a 1 1 1 3 2 1 This gives: sqrt(37)+8 1 1 1 1 1 ---------- = 1 + ---- ----- ----- ---- ----- 9 1+ 1+ 3+ 2+ 1+ The example you gave [sqrt(a)-b]/1 is done exactly as shown above, with r1 = 1. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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