A Monster of a Continued FractionDate: 11/09/96 at 00:36:11 From: Patrick Beebe Subject: How to solve an infinitely repetitious problem I'm curious about solving a specific problem and others similar to it. Given the forever repeating fraction shown below, what is the solution? 1 + ... 1 + -------- 3 + ... 1 + ------------- 5 + ... 3 + -------- 1 + ... 1 + ------------------ 1 + ... 5 + -------- 3 + ... 3 + ------------- 5 + ... 1 + -------- 1 + ... 1 + ----------------------- 1 + ... 1 + -------- 3 + ... 5 + ------------- 5 + ... 3 + -------- 1 + ... 3 + ------------------ 1 + ... 5 + -------- 3 + ... 1 + ------------- 5 + ... 1 + -------- 1 + ... It is an absolute monster! As I see it, there are two distinct infinite patterns within it: the 1 + 1/3... pattern, and the 1 + 5/1... pattern. I have tried substituting these patterns with x and y, and then attempting to solve from there, but to no avail. I am still stuck. Is there another way to solve these sorts of problems? Thanks! Date: 11/17/96 at 23:15:01 From: Doctor Sam Subject: Re: How to solve an infinitely repititious problem Patrick, You are correct...this IS a monster. I have seen continued fractions before, but never such a combination. I have a possible solution but I would like someone else to verify it so that we could both be confident that it is correct! I liked your idea of substitution. Here are the ones I made: 1 + ... <<. <<<<<<< . 1 + -------- . . 3 + ... . . 1 + ------------- A . 5 + ... . . 3 + -------- . . 1 + ... <<. . 1 + ------------------ A 1 + ... <<. . 5 + -------- . . 3 + ... . . 3 + ------------- B . 5 + ... . . 1 + -------- . . 1 + ... <<. <<<<<<<< . 1 + ----------------------- 1 + ... <<. <<<<<<<< . 1 + -------- . . 3 + ... . . 5 + ------------- X . 5 + ... . . 3 + -------- . . 1 + ... << . . 3 + ------------------ B 1 + ... <<. . 5 + -------- . . 3 + ... . . 1 + ------------- Y . 5 + ... . . 1 + -------- . . 1 + ... <<. <<<<<<<< . So I see the numerator of this expression two ways. It is both A and 1 + A/B. This gives one condition on A and B. The trick is to see the denominator of this monster also in terms of A and B. The entire denominator is just B. But what are X and Y? I can't be sure of what all of your "..." stand for but I *think* that X is 4 + A. (Instead of thinking of the numerator of X as 5 + ... I am thinking of it as 4 + 1 + ... and that makes it 4+A.) Similarly, I think the denominator, Y, of the monster is B - 2. (Think of the 3 + ... as -2 + 5 + ... ). If I am correct in this interpretation, then the denomintor is both B and 3 + (4+A)/(B-2). This gives a second condition on A and B. We now have to solve the simultaneous equations: A = 1 + A/B and B = 3 + (4+A)/(B-2) I get a cubic equation in B with a single real solution of B approximately equal to 4.84732. Since A = 1 + A/B, A = B/(B-1) = 1.25992 And A is also the value of the entire monster. I think. I hope that helps! -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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