- **Miscellaneous Math**
(*https://www.mersenneforum.org/forumdisplay.php?f=56*)

- - **Approximation of r by m^(1/n)**
(*https://www.mersenneforum.org/showthread.php?t=27369*)

Approximation of r by m^(1/n)[TEX]r\pm\epsilon=\sqrt[n]{m}[/TEX]
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 |

[QUOTE=RomanM;594283][TEX]r\pm\epsilon=\sqrt[n]{m}[/TEX]
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[/QUOTE]It depends on what you're given first. If r = x[sub]1[/sub] is a Pisot number (an algebraic integer > 1 whose algebraic conjugates x[sub]2[/sub],... x[sub]n[/sub] all have absolute value less than 1) then for positive integer k, the sums [tex]S_{k}\;=\;\sum_{i=1}^{n}x_{i}^{k}[/tex] are all rational integers, and all the terms except the first tend to 0 as k increases without bound. Thus [tex]r\;\approx\;\(S_{k}\)^{\frac{1}{k}}[/tex] 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[sup]th[/sup] root of the k[sup]th[/sup] Lucas number has limiting value equal to the root r > 1 of P(x) = 0. |

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