RE: Re: algebraic solutions
- To: mathgroup@smc.vnet.net
- Subject: [mg11324] RE: [mg11271] Re: [mg11193] algebraic solutions
- From: R Finley <trfin@umsmed.edu>
- Date: Fri, 6 Mar 1998 00:40:39 -0500
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