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Re:



> Date:          Wed, 4 Mar 1998 01:39:27 -0500
> From:          Daniel Lichtblau <danl@wolfram.com>
To: mathgroup@smc.vnet.net
> To:            mathgroup@smc.vnet.net
> Subject: [mg11319]       [mg11271] Re: [mg11193] algebraic solutions

> Daniel Teitelbaum wrote:
> > 
> > Hi all,
> > 
> > I'm a fairly novice Mathematica user, and I'm having a problem.  I asked
> > a more experienced user and he could solve it, either.  I hope there is
> > a solution and that you all can help.
> > 
> > I want to find the roots of the following equation:
> > 
> >     z^5  + 2z^3 - p + 1 = 0
> > 
> > I want to solve for z in terms of p.  Now, if I pick some random number
> > for p, I can get mathematica to solve for z, but I cant get a solution
> > in terms of p.  Alternatively, I would like to be able to plot this
> > function with p included as part of the vertical axis.
> > 
> > Thanks in advance for your help,
> > 
> > Daniel
> 
> 
> You have five functions of p, not one. Here is one way to plot one of
> them. It relies on the fact that the first root of an odd-degree
> algebraic function in Mathematica is always real-valued.
> 
> In[3]:= algfuns = Solve[z^5  + 2z^3 - p + 1 == 0, z];
> 
> In[4]:= zp = z /. %[[1]]
>                           3     5
> Out[4]= Root[-1 + p - 2 #1  - #1  & , 1]
> 
> (* I assume you want p to be the independent variable, that is, along
> the horizontal axis. *)
> 
> In[5]:= Plot[%, {p,0,3}]
> Out[5]= -Graphics-
> 
> Will only work for other root functions in ranges where they are
> real-valued.
> 
> Alternative methods using FindRoot or NSolve could also be coded without
> too much trouble.
> 
> 
> Daniel Lichtblau
> Wolfram Research
> 

A simpler approach for the graphs of the real-valued roots  is to make a
ContourPlot of  z^5  + 2z^3 - p + 1  with level 0.

Fred Simons
Eindhoven University of Technology
Fred Simons
Eindhoven University of Technology



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