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Title: Positive root of x^n + x = 1 Post by Aryabhatta on Oct 15th, 2007, 3:40pm This is from the book: "A problem Seminar" by Donald J Newman. Show that the equation xn + x = 1 has a unique positive real root rn. Find the limit of rn as n -> oo. How fast is the convergence? |
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Title: Re: Positive root of x^n + x = 1 Post by iyerkri on Oct 15th, 2007, 8:02pm [hideb] For n \geq 0, g(x) = x^n + x -1 is strictly increasing for x \geq 0, with g(0) = -1 and g(1) = 1. Hence the first assertion follows. [/hideb] [hideb] note 0 < r_n < 1. Let r_n = 1 - f(n). suppose, f(n) does not go to zero as n goes to infinity. Then r_n^n goes to zero as n goes to infinity. but that means r_n goes to one, which means f(n) goes to zero, a contradiction. Hence, f(n) goes to zero. which means, r_n goes to 1 as n goes to infinity. Also, r_n^n + r_n =1. thus, (1-f(n))^n = f(n). for large enough n, (1-f(n))^n is of order exp(-nf(n)) Thus, order of f(n), say h(n), should satisfy the equation nh(n) + ln h(n) = 0. I havent been able to find the exact form of h. [/hideb] ~kris |
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Title: Re: Positive root of x^n + x = 1 Post by Aryabhatta on Oct 19th, 2007, 1:52pm Looks right. According the Newman, 1 - rn ~ [hide]logn/n[/hide] |
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Title: Re: Positive root of x^n + x = 1 Post by iyerkri on Oct 19th, 2007, 4:50pm That makes sense. if we let [hide]g(n) = nh(n)[/hide], then we must have [hide] g(n) + log(g(n)) = log(n) [/hide], which implies [hide]log(n)/2 \leq g(n) \leq log(n)[/hide]. Sorry for not knowing how to use the math symbols in the forum. I follow latex. ~kris |
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Title: Re: Positive root of x^n + x = 1 Post by ThudanBlunder on Oct 22nd, 2007, 7:17am on 10/19/07 at 16:50:39, iyerkri wrote:
Welcome to the forum, kris. http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_suggestions;action=display;num=1171644142 |
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