Re: Solution of this equation

• To: mathgroup at smc.vnet.net
• Subject: [mg20654] Re: Solution of this equation
• From: John Doty <jpd at w-d.org>
• Date: Sun, 7 Nov 1999 02:09:56 -0500
• Organization: The Internet Access Company, Inc.
• References: <7vrc3p\$2nd@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Dave Richardson wrote:
>
> Can anyone offer insight here?
>
> This Mathematica expression gives 3 solutions to the equation.
>
> Solve[Pr == (8*Tr)/(3*vr - 1) - 3/vr^2, vr]
>
> The problem is that there are 3 Real solutions, and Mathematica is giving
> solutions with (granted a small) imaginary component.
>
> And hitting it with  a full simplify is just not a good idea...

This is life with cubic equations. Solutions in terms of radicals
generally involve complex numbers, even when the roots are real. This
was a *major* puzzle of 16th century mathematics. See Paul Nahin's
wonderful book, "An Imaginary Tale: the Story of Sqrt[-1]" (title given
in OutputForm :-).

FullSimplify cannot generally cancel the imaginary components of the
solutions, although it can in exact calculations of cases where the
roots are all real:

In[1]:=
8 Tr      3
s = Solve[Pr == -------- - ---, vr];
3 vr - 1     2
vr

In[2]:=
s /. {Tr -> -2, Pr -> 1}

Out[2]=
5                16
{{vr -> -(-) + --------------------------- +
3                            1/3
(-1188 + 324 I Sqrt[15])

1                         1/3
- (-1188 + 324 I Sqrt[15])   },
9

5         8 (1 + I Sqrt[3])
{vr -> -(-) - --------------------------- -
3                            1/3
(-1188 + 324 I Sqrt[15])

1                                          1/3
-- (1 - I Sqrt[3]) (-1188 + 324 I Sqrt[15])   },
18

5         8 (1 - I Sqrt[3])
{vr -> -(-) - --------------------------- -
3                            1/3
(-1188 + 324 I Sqrt[15])

1                                          1/3
-- (1 + I Sqrt[3]) (-1188 + 324 I Sqrt[15])   }}
18

In[3]:=
FullSimplify[%]

Out[3]=
{{vr -> -2 + Sqrt[5]}, {vr -> -2 - Sqrt[5]}, {vr -> -1}}

In inexact numerical calculations, the cancellation of imaginary
components will generally not be exact.