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RE: Re: algebraic solutions



Daniel,
There have been a number of correct answers concerning the inutility of
 a closed form solution of z as a function of p.  However, you
originally  asked for a way to visualize the solution and the form of
your equation  gives a particularly simple way of doing so.  Since the
equation is  linear in p, rather than thinking of z as a function of
p, turn it  around and think of p as a function of z and graph it as
such:  p[z]:>5 + 2 z^3 + 1      This will give what you asked for,
p on the  vertical axis and you will simply have to think a bit about
what the  graphs are telling you about the solution.  For any real z
there is a  single unique real p that is a root of the original
equation.  Plot[ p[z], {z, -5,5} ]    Depending on your application,
this may be  all you need.  If you want to explore the entire solution
space for p  and/or z complex you will have to look at the real and
imaginary plots  of p plotted at real and complex values of z. For
example: Plot3D[ Re[ p[x + I y] ], {x,-5,5},{y,-5,5} ]  and similarly
for  Im[p[z]]
Hope that helps...
RF

-----Original Message-----
From:	Daniel Lichtblau [SMTP:danl@wolfram.com] To:
mathgroup@smc.vnet.net
Sent:	Wednesday, March 04, 1998 12:39 AM To:	mathgroup@smc.vnet.net
Subject:	[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 
>
> 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]:
lgfuns olve[z^5  + 2z^3 - p + 1 Ð0, z];

In[4]:p  /. %[[1]]
                          3     5
Out[4]oot[-1 + p - 2 #1  - #1  & , 1]

(* I assume you want p to be the independent variable, that is, along
the horizontal axis. *)

In[5]:	lot[%, {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




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