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Re: an equation containg radicals
*To*: mathgroup at smc.vnet.net
*Subject*: [mg69665] Re: an equation containg radicals
*From*: "Scout" <Scout at nodomain.com>
*Date*: Wed, 20 Sep 2006 02:44:17 -0400 (EDT)
*References*: <eeofde$2mo$1@smc.vnet.net>
Solve[] gives generic solutions only. You might use Reduce[]
In[1]:= k = 1-4*(1-v)*b^2*p^2*(1-Sqrt[e^2-p^2]/Sqrt[(1/b)^2-p^2]);
In[2]:= sol = Reduce[k==0,p,Reals];
In[3]:= sol2 = LogicalExpand[sol];
In[4]:= consts = {e->10^-3, v->3/10, b->10^-5};
In[5]:= FindInstance[sol2/.consts,{p}]
Out[5]= {}
That means that your values of the parameters don't match any solution of
the above equation k.
Let's find just one solution, assigning a value for 'e' and solve for the
other (v,b,p):
In[6]:= inst = FindInstance[sol2/.{e->10^-3},{v,b,p}]
Out[6]={{v->-81, b->-250*Sqrt[2/41], p->-1/1000}}
In[7]:= Simplify[k==0 /.{e->10^-3} /. inst]
Out[7]= {True}
Regards,
~Scout~
> 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[2])*Sqrt[-(1/(Î»^2*(-9 + 8*v + Sqrt[-15 + 16*v +
> 64*e^2*(-1 + v)^2*Î»^2])))]},
> {p -> Sqrt[2]*Sqrt[-(1/(Î»^2*(-9 + 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1
> + v)^2*Î»^2])))]},
> {p -> (-Sqrt[2])*Sqrt[1/(Î»^2*(9 - 8*v + Sqrt[-15 + 16*v + 64*e^2*(-1
> + v)^2*Î»^2]))]},
> {p -> Sqrt[2]*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.
>
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