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Topic: Positive root of x^n + x = 1 (Read 1362 times) |
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Aryabhatta
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Positive root of x^n + x = 1
« on: Oct 15th, 2007, 3:40pm » |
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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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| « Last Edit: Oct 15th, 2007, 3:40pm by Aryabhatta » |
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iyerkri
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Re: Positive root of x^n + x = 1
« Reply #1 on: Oct 15th, 2007, 8:02pm » |
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| hidden: | 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.
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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.
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~kris
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Aryabhatta
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Re: Positive root of x^n + x = 1
« Reply #2 on: Oct 19th, 2007, 1:52pm » |
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Looks right.
According the Newman, 1 - rn ~ logn/n
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iyerkri
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Re: Positive root of x^n + x = 1
« Reply #3 on: Oct 19th, 2007, 4:50pm » |
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That makes sense.
if we let g(n) = nh(n), then we must have g(n) + log(g(n)) = log(n) , which implies log(n)/2 \leq g(n) \leq log(n).
Sorry for not knowing how to use the math symbols in the forum. I follow latex.
~kris
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