       Re: an equation containg radicals

• To: mathgroup at smc.vnet.net
• Subject: [mg69719] Re: an equation containg radicals
• From: dimmechan at yahoo.com
• Date: Thu, 21 Sep 2006 07:31:09 -0400 (EDT)
• References: <200609190945.FAA28444@smc.vnet.net><eeqq46\$ohv\$1@smc.vnet.net>

```Dear Daniel,
Thanks for the response.

Reduce did help me a lot.
>From application of the argument principle I know that there are no
solutions
in the complex domain for the restrictions of parameters I have in the
problem.
I just have one question.

Can the following command be a verification that mine equation does not
exhibit
solutions in the complex domain under the restrictions of parameters?

K[p_] := 1 - 4*(1 - v)*ii^2*p^2*(1 - Sqrt[e^2 - p^2]/Sqrt[a^2 - p^2])
a = 1/ii;
eq = K[p] == 0;
consts = {e -> 1/1000, v -> 3/10, ii -> 10^(-5)};

Reduce[{eq, 0 < e < 1/10, 0 < v < 1, 0 < ii < 1/100} /. consts, p,
Reals]
False

Reduce[{eq, 0 < e < 1/10, 0 < v < 1, 0 < ii < 1/100} /. consts, p,
Complexes]
False

BTW, why we must mention Real as the third argument?
Reduce search by deafult in the Real domain as far as I know.
Am I missing something?

Regards
Dimitris

Î?/Î? Daniel Lichtblau Î­Î³Ï?Î±Ï?Îµ:
> dimmechan at yahoo.com wrote:
> > Hello to all.
> >
> > In a crack problem appeared the following function.
> >
> > K[p_] := 1 - 4*(1 - v)*Î»^2*p^2*(1 - Sqrt[e^2 - p^2]/Sqrt[a^2 - p^2])
> > a = 1/Î»;
> >
> > Here are some typical values for the involving constants
> >
> > consts = {e -> 1/1000, v -> 3/10, Î» -> 10^(-5)};
> >
> > Here is the solution obtained with Solve
> >
> > sols = FullSimplify[Solve[eq = K[p] == 0, p]]
> > {{p -> (-Sqrt)*Sqrt[-(1/(Î»^2*(-9 + 8*v + Sqrt[-15 + 16*v +
> > 64*e^2*(-1 + v)^2*Î»^2])))]},
> >   {p -> Sqrt*Sqrt[-(1/(Î»^2*(-9 + 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1
> > + v)^2*Î»^2])))]},
> >   {p -> (-Sqrt)*Sqrt[1/(Î»^2*(9 - 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1
> > + v)^2*Î»^2]))]},
> >   {p -> Sqrt*Sqrt[1/(Î»^2*(9 - 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1 +
> > v)^2*Î»^2]))]}}
> >
> > What I need now is to see which roots (or if all roots) are extreneous
> > (i.e. they do not satisfy the intial equation K[p]=0).
> > This is a difficult task for Mathematica.
> >
> > TimeConstrained[FullSimplify[eq /. sols], 300, "Failed"]
> > "Failed"
> >
> > However replacing the values for the constants it is verified that all
> > solutions (for this typical values of the constants) are extreneous.
> >
> > eq /. sols /. consts
> > {False, False, False, False}
> >
> > This can be verified also by the following command
> >
> > Solve[eq /. consts, p]
> > {}
> >
> > My question is if it a way to "help" a bit Mathematica in the above
> > verification with the symbolic parameters.
> >
> > Thanks in advance for any help.
>
> I had some trouble with the fonts so I'm using ii for the last variable
> in your 'consts' replacement list.
>
> K[p_] := 1 - 4*(1 - v)*ii^2*p^2*(1 - Sqrt[e^2 - p^2]/Sqrt[a^2 - p^2])
> a = 1/ii;
> eq = K[p] == 0;
> consts = {e -> 1/1000, v -> 3/10, ii -> 10^(-5)};
>
> If you use Reduce and place some restrictions on teh parameters you will
> see that there are no real-valued solutions in a neighborhood of the
> consts values in parameter space.
>
> In:= sols = Reduce[{eq,0<e<1/10,0<v<1,0<ii<1/100}/.consts, p, Reals]
> Out= False
>
> I'm not sure this tells you what you need but it may give some ideas for
>   using Reduce on this equation.
>
> If you want to consider complex valued solutions, another possibility
> might be to use Solve as you have it above, then use FullSimplify as
> above but with assumptions on the parameters. Again, this might just
> hang, but giving assumptions will probably improve the chances that it
> does not, and that it gives a meaningful result.
>
>
> Daniel Lichtblau
> Wolfram Research

```

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