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You are not logged in. #1 20121019 19:39:44
Continued fractionHow do i find a number's convergent,could you help me find sqrt(3)'s convergents #2 20121019 19:55:54
Re: Continued fractionHi Ronald; The convergents are: In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20121019 21:24:31
Re: Continued fractionYes,I know how to get the representation,I just want to know what is the method of finding convergents #4 20121019 21:48:12
Re: Continued fractionYou use two recurrences. Call the representation above a1,a2,a3,a4,a5... You start with the numerators of the a's and use this recurrence. with the initial conditions of The first few numerators are The denominators are done in the same way. with the initial conditions of The first few denominators are; For example the fifth convergent is 19 / 11. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #5 20121019 22:17:50
Re: Continued fractionAre the h1,h2,k1,k2 defined for every representation,or are they depending on the representation? #6 20121019 22:29:23
Re: Continued fractionThey are the numerators and the denominators of the first two of the cf representation. See the 1,1 at the beginning? That can be expressed as 1 / 1 and 1 / 1. So h[1] = 1 and h[2] = 1. ( the numerators ) k[1] = 1 and k[2] = 1. ( the denominators ) Now isn't that the neatest thing you ever saw! Computational mathematics! In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #8 20121019 22:57:25
Re: Continued fractionIs that .9? In some books you will see them start at a0 and h[0] rather than a1 and h[1]. Just shift all indices down by one in that case. I used the higher indices because programming languages start at 1 not 0 for an array. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 