Re: Solve Question - 2 Non zero values

*To*: mathgroup at smc.vnet.net*Subject*: [mg112973] Re: Solve Question - 2 Non zero values*From*: Peter Pein <petsie at dordos.net>*Date*: Fri, 8 Oct 2010 04:50:49 -0400 (EDT)*References*: <i7seni$pph$1@smc.vnet.net>

Am Tue, 28 Sep 2010 10:06:42 +0000 (UTC) schrieb "Lea Rebanks" <lrebanks at netvigator.com>: > Dear All, > > > > Enclosed are the workings to Solve for BOTH values of V1 and V2. > > I have already shown below the Solve for just one value either V1 or > V2 and there is no problem. > > The correct values required for V1 and V2 are shown below when I > worked them out individually. > > However, when I setup a Solve function to evaluate BOTH V1 and V2 in > the same Solve function then I run into problems. > > > > Question 1:-In the instance shown below the result shows V1 and V2 to > equal' zero'. > > Therefore to avoid this could someone show me how to set Solve, (or > Reduce, or whatever) up to provide a result greater than zero.(I am > sure this is easy, but I don' t know how to do it.) > > > > Secondly, I suspect there is a deeper or more involved mathematical > method to extract these 2 required values. Any ideas anyone? > > Many thanks for everyones' help & attention. > > Best regards, > > Lea... > ... Hi Lea, your posting has been a while ago but let's see... It is almost ever a good idea to simplify expressions before feeding them into Solve or Reduce. In[1]:= M=Simplify[45-ArcTan[Sqrt[(4-Pi)/Pi]]/Degree//FunctionExpand]//Apart radiusD[x_]=Tan[M*Degree]*x//FunctionExpand//ExpandAll//TrigExpand//Simplify Out[1]= 45-(180 ArcTan[Sqrt[-1+4/\[Pi]]])/\[Pi] Out[2]= ((-Sqrt[4-\[Pi]]+Sqrt[\[Pi]]) x)/(Sqrt[4-\[Pi]]+Sqrt[\[Pi]]) so we've got a simple radiusD :-) With the same simplifications and replacement of floating point numbers by their exact values, the first equation reduces to 18 == V1^2 and the second equation to V2^2 == 8. So we get V1 == 3 Sqrt[2] and V2 == 2 Sqrt[2]. The list of equations reduces to two times the same equation: Simplify[TrigExpand[ExpandAll[FunctionExpand[ {(((15*360)* Sqrt[radiusD[-(((15*Sqrt[2])*V1)/(Tan[135*Degree] - Tan[M*Degree]))]^2 + (((15*Sqrt[2])*V1)/(Tan[135*Degree] - Tan[M*Degree]))^2])/(360*15))^2*Pi - (((45*360)* Sqrt[radiusD[-((((45/2)*Sqrt[2])*V2)/(Tan[135*Degree] - Tan[M*Degree]))]^2 + ((((45/2)* Sqrt[2])*V2)/(Tan[135*Degree] - Tan[M*Degree]))^2])/(360*45))^2* Pi == 0, (((13*360)* Sqrt[radiusD[-(((13*Sqrt[2])*V1)/(Tan[135*Degree] - Tan[M*Degree]))]^2 + (((13*Sqrt[2])*V1)/(Tan[135*Degree] - Tan[M*Degree]))^2])/(360*13))^2*Pi - (((39*360)* Sqrt[radiusD[-((((39/2)*Sqrt[2])*V2)/(Tan[135*Degree] - Tan[M*Degree]))]^2 + ((((39/2)* Sqrt[2])*V2)/(Tan[135*Degree] - Tan[M*Degree]))^2])/(360*39))^2* Pi == 0}]]]] gives: {4*V1^2 == 9*V2^2, 4*V1^2 == 9*V2^2} Using Reduce[4*V1^2 == 9*V2^2 && V1 > 0 < V2, {V1, V2}] yields V1 > 0 && V2 == (2*V1)/3, which confirms the relation between V1 and V2 found above. Regards, Peter