Re: radical question again

*To*: mathgroup at smc.vnet.net*Subject*: [mg72123] Re: radical question again*From*: dh <dh at metrohm.ch>*Date*: Wed, 13 Dec 2006 06:39:04 -0500 (EST)*Organization*: hispeed.ch*References*: <elgm4j$ha9$1@smc.vnet.net>

Hi Dimitris, I can not fully answer your question, but here is a hint: In Root notation we can indicate the number of the root. The first root must, by definition, be real for a 3.order poly. In Radical notation we have no such notation. As long as we have a symbolic parameter, radical notation may not be able to indicate the real root. This may explain why Cubics->True does not work as soon as you restrict the domain to real. It may also prevent radical notation if you want to pick out the real root, having a symbolic parameter. Daniel dimitris wrote: > Before proceeding to the post let me thank Andrzej and Daniel > for their responses to my previous question appeared here > > http://groups.google.com/group/comp.soft-sys.math.mathematica/browse_thread/thread/78d57749388d7384/e072ef4f6aa4f9c0?hl=en#e072ef4f6aa4f9c0 > > So consider again the following equation > > req = (2 - x)^2 - 4*Sqrt[1 - x]*Sqrt[1 - c*x] == 0; > > with 0<x<1 and 0<c<1. > > A quick solution is provided by > > Reduce[{req, 0 < x < 1, 0 < c < 1}, {x}, Reals] > 0 < c < 1 && x == Root[-16 + 16*c + (24 - 16*c)*#1 - 8*#1^2 + #1^3 & , > 1] > > Here comes my first question: > > Why the following setting returns the same output? > > Reduce[{req,0<x<1,0<c<1},{x},Reals,Cubics->True(*the default is > False*)] > 0 < c < 1 && x == Root[-16 + 16*c + (24 - 16*c)*#1 - 8*#1^2 + #1^3 & , > 1] > > The following plots verify the obtained solution by Reduce > > f[c_] = x /. ToRules[Last[%]]; > Plot[f[c], {c, 0, 1}] > Plot[Chop[First[req] /. x -> f[c]], {c, 0, 1}] > > Let > > g[c_] = ToRadicals[f[c]] > 8/3 - (8 - 48*c)/(6*(-17 + 45*c + 3*Sqrt[3]*Sqrt[11 - 62*c + 107*c^2 - > 64*c^3])^(1/3)) + > (2/3)*(-17 + 45*c + 3*Sqrt[3]*Sqrt[11 - 62*c + 107*c^2 - > 64*c^3])^(1/3) > > with > > g[c] == x /. DeleteCases[Solve[req, x], {x -> 0}][[1]] > True > > Now > > Block[{Message}, Plot[g[c], {c, 0, 1}, PlotPoints -> 100]] > Show[GraphicsArray[Block[{$DisplayFunction = Identity}, > (Plot[Chop[#1[g[c]]], {c, 0, 1}, Frame -> True, Axes -> False, > PlotPoints -> 100] & ) /@ {Re, Im}]], ImageSize -> 500] > > The graphs can be justified due the choise of principal cube root by > Mathematica. > In the case of negative numbers, it turns out that it is not a real > number. > (see relevant descussion here: > http://groups.google.com/group/comp.soft-sys.math.mathematica/browse_thread/thread/403b579399a8b61/3d24f97466c328ed?hl=en#3d24f97466c328ed > ) > > And here comes my second question: > > With the Root object we obtain a closed form root which is valid for > 0<x,c<1. > Converting it to radical we lost that and we must consider in more > detail the range > of c. > > Is it possible to get a radical expression which is equivalent to the > Root object? > If yes how? > > Thanks > Dimitris >