Solving Continued FractionsDate: 08/08/98 at 18:43:53 From: kristy Subject: Solving continued fractions I've read the archives and determined that the problem in my review packet is in fact a continued fraction. I also know, from reading the archives, that it is equal to sqrt(2). My question is how in the world do you get sqrt(2) from: 1 ---------------- 2 + 1 ------------ 2 + 1 -------- 2 + .... Any help on your part would be greatly appreciated. Date: 08/09/98 at 20:15:05 From: Doctor Pete Subject: Re: Solving continued fractions Hi, The key in solving continued fractions is to use a bit of algebra. Say the above continued fraction is x. Then notice that: x = 1/(2 + x) This is because the continued fraction contains a copy of itself. So solving this equation for x, we find: x(2+x) = 1 or x^2 + 2x - 1 = 0. The quadratic formula gives: x = (-2 + Sqrt[8])/2 or (-2 - Sqrt[8])/2 Thus: x = Sqrt[2] - 1 or -Sqrt[2] - 1 Since the second solution is negative, and it is obvious that the continued fraction is positive, it follows that x = Sqrt[2] - 1. Another example: 1 1 + --------------- 1 2 + ----------- 1 1 + ------- 1 2 + --- 1 ... Here, the 1's and 2's alternate. Then we see that: 1 x = 1 + ------- 1 2 + --- x Again solving for x, we find: x = (3x+1)/(2x+1) 2x^2 - 2x - 1 = 0 Thus: x = (1 + Sqrt[3])/2 or (1 - Sqrt[3])/2 and discarding the negative root gives x = (1 + Sqrt[3])/2. - Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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