Re: Number of roots from Solve?

*To*: mathgroup at smc.vnet.net*Subject*: [mg48444] Re: [mg48423] Number of roots from Solve?*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 30 May 2004 06:12:05 -0400 (EDT)*References*: <200405290706.DAA20383@smc.vnet.net> <DCCD3B24-B159-11D8-9808-000A95B4967A@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I should have added that you can indeed predict the number of roots. In[1]:= e := (1/z)*((2 + z^2*(b - 1))/(b + 1))^ ((1 + b)/(2*(b - 1))) In[2]:= a = 10; b = Rationalize[1.4]; In[3]:= Numerator[Together[e - 10]] Out[3]= z^6 + 15*z^4 + 75*z^2 - 2160*z + 125 Hence you will get 6 roots. In[4]:= a = 10; b = Rationalize[1.05]; In[5]:= Limit[Numerator[Together[e - 10]]/z^41, z -> Infinity] Out[5]= Sqrt[41] So this time you will indeed get 41 roots. Andrzej On 29 May 2004, at 19:20, Andrzej Kozlowski wrote: > 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>