Approximation of r by m^(1/n)

RomanM · 2021-12-01, 16:37

$r\pm\epsilon=\sqrt[n]{m}$
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

Dr Sardonicus · 2021-12-01, 17:28

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,... x_n 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 k^th root of the k^th Lucas number has limiting value equal to the root r > 1 of P(x) = 0.

2021-12-01, 16:37	#1
RomanM Jun 2021 47₁₀ 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: ***

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