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Old 2021-12-01, 16:37   #1
RomanM
 
Jun 2021

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Default Approximation of r by m^(1/n)

r\pm\epsilon=\sqrt[n]{m}
where r - real root of polynomial P(r), order >=6, and m, n - integers, and P(r+/-eps)<1
P.S. I'm suspect that the simpler the question look like, the less likely it is to get an answer

Last fiddled with by RomanM on 2021-12-01 at 16:42 Reason: ***
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Old 2021-12-01, 17:28   #2
Dr Sardonicus
 
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Quote:
Originally Posted by RomanM View Post
r\pm\epsilon=\sqrt[n]{m}
where r - real root of polynomial P(r), order >=6, and m, n - integers, and P(r+/-eps)<1
P.S. I'm suspect that the simpler the question look like, the less likely it is to get an answer
It depends on what you're given first.

If r = x1 is a Pisot number (an algebraic integer > 1 whose algebraic conjugates x2,... xn all have absolute value less than 1) then for positive integer k, the sums

S_{k}\;=\;\sum_{i=1}^{n}x_{i}^{k}

are all rational integers, and all the terms except the first tend to 0 as k increases without bound. Thus

r\;\approx\;\(S_{k}\)^{\frac{1}{k}}

becomes an increasingly good approximation as k increases.

The simplest case is with the polynomial P(x) = x^2 - x - 1. The sums are the Lucas numbers.

So the kth root of the kth Lucas number has limiting value equal to the root r > 1 of P(x) = 0.
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