Re:
- To: mathgroup@smc.vnet.net
- Subject: [mg11319] Re: [mg11271]
- From: "Fred Simons" <wsgbfs@win.tue.nl>
- Date: Fri, 6 Mar 1998 00:40:30 -0500
> 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