mersenneforum.org Approximation of r by m^(1/n)
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 2021-12-01, 16:37 #1 RomanM   Jun 2021 1011002 Posts 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: ***
2021-12-01, 17:28   #2
Dr Sardonicus

Feb 2017
Nowhere

19·281 Posts

Quote:
 Originally Posted by RomanM $r\pm\epsilon=\sqrt[n]{m}$ 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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