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[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. > > 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[10]:= sols = Reduce[{eq,0<e<1/10,0<v<1,0<ii<1/100}/.consts, p, Reals] > Out[10]= 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

**Follow-Ups**:**Re: Re: an equation containg radicals***From:*Daniel Lichtblau <danl@wolfram.com>

**References**:**an equation containg radicals***From:*dimmechan@yahoo.com