pentic root
in [Math]
|
From: Robert Israel on 9 Dec 2009 19:15 hagman <google(a)von-eitzen.de> writes: > On 6 Dez., 03:20, "Jon" <jon8...(a)peoplepc.com> wrote: > > The root to ax^5+bx+c=3D0 is approximately, > > > > x =3D { -c/({a^2+b^2}^(1/2)) }^(1/5) > > > > Using this formula, the roots to, > > > > x^5+x-34=3D0 =A0x=3D1.888 =A0should be 2 > > 32x^5+4x-3=3D0 x=3D0.623 should be 1/2 > > x^5+x-0.10001 x=3D0.5887 should be 1/10 > > That approximation is exact if either b=3D0 or c=3D0, but as your examples > show of doubtful quality in the general case Here's a better approximation, for when b is small: x = 1/a*(-c*a^4)^(1/5) + (-1/5*a^2/(-c*a^4)^(3/5))*b +(-1/25*a^5/(-c*a^4)^(7/5))*b^2 + (-1/125*a^8/(-c*a^4)^(11/5))*b^3 +(-21/15625*a^2/(-c*a^4)^(4/5)/c^3)*b^5 + O(b^6) On the other hand, when c is small, the root near 0 is approximately (-1/b)*c + a/b^6*c^5 + (-5*a^2/b^11)*c^9 + O(c^13) -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |