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. >