Re: Number of roots from Solve?

*To*: mathgroup at smc.vnet.net*Subject*: [mg48443] Re: [mg48423] Number of roots from Solve?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 30 May 2004 06:12:04 -0400 (EDT)*References*: <200405290706.DAA20383@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I think having inexact coefficients does not make any difference here since I am pretty sure Solve first applies Rationalize to everything. In fact I don&t think it would not make any sense to do otherwise, given that Solve uses only algebraic and not numerical techniques. Since the equations are non-polynomial I can see nothing at all strange about them having very different numbers of roots. Andrzej Kozlowski Since the equation is non-polynomial I do not see an On 29 May 2004, at 16:06, Goyder Dr HGD wrote: > I am working with the expression > > a =.; b =.; e = (1/z)((2 + z^2 (b - 1))/(b + 1))^((1 + b)/(2(b - 1))) > > and was not surprised to find a number of complex and real roots (6) > for > the particular case > > a = 10.; b = 1.4; s = Solve[a == e]; {Length[s], s} > > I was surprised that the number of roots changed to 41 for the case > > a = 10.; b = 1.05; s = Solve[a == e]; {Length[s], s} > > With > > a = 10.; b = 1.03; s = Solve[a == e]; {Length[s], s} > > I never got an answer (or the calculation takes longer than my > patience). > > Is it possible to predict the number of roots that an expression like e > above will produce? > Does Mathematica find all the roots, particularly as we are dealing > with > approximate numbers? > > Many thanks > > Hugh Goyder > > -- > This message has been scanned for viruses and > dangerous content by the Cranfield MailScanner, and is > believed to be clean. > >

**References**:**Number of roots from Solve?***From:*Goyder Dr HGD <h.g.d.goyder@cranfield.ac.uk>