Re: [Q] Solve?
- To: mathgroup at smc.vnet.net
- Subject: [mg23410] Re: [mg23408] [Q] Solve?
- From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
- Date: Sun, 7 May 2000 21:17:57 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
on 00.5.7 8:14 AM, James at research at proton.csl.uiuc.edu wrote: > > > Hi! > > I come across a problem that I don't understand. > I generate systems of equations > and try to solve them using mathematica. > For example, > > --------------------------------------------------------------- > In[1]= Solve[((4/2)*( 1.50*z+ 2.00)+(4-4/2)* > Sqrt[( 1.50*z+ 0.00)^2+ 4.00]) + > ((4/2)*( 1.50*z+ 2.00)+(4-4/2)* > Sqrt[( 1.50*z+ 0.00)^2+ 4.00])- 8.10*z==0, z] > > Out[1]= {} > --------------------------------------------------------------- > > But mathematica can't solve it, simply returing empty. > This equation can be solved, though, as you see. > I restarted mathematica to run again, but it doesn't help. > > > The weird thing is some equations are solved, like the below. > --------------------------------------------------------------- > In[2]= Solve[((4/2)*( 1.50*z+ 2.00)+(2-4/2)* > Sqrt[( 1.50*z+ 0.00)^2+ 4.00]) + > ((4/2)*( 1.50*z+ 2.00)+(3-4/2)* > Sqrt[( 1.50*z+ 0.00)^2+ 4.00])- 8.10*z==0, z] > > Out[2]= {{z->13.4976}} > --------------------------------------------------------------- > > What's wrong with it and how can I solve it? > I need to use mathematica, because the number of the systems > of equations are large. > (Please disregard any redundunt terms in the equations.. > they are generated by a program.) > With that, when does mathematica return '{}' as an answer? > Thank you. > > > --- James, newbie in mathematica. > > But Mathematica is completely right: there are no solutions. If you do not believe it, let's do this "by hand". After simplification your equation becomes: In[1]:= eq = ((4/2)*(1.50*z + 2.00) + (4 - 4/2)* Sqrt[(1.50*z + 0.00)^2 + 4.00]) + ((4/2)*(1.50*z + 2.00) + (4 - 4/2)*Sqrt[(1.50*z + 0.00)^2 + 4.00]) - 8.10*z == 0 // FullSimplify Out[1]= 2 8. + 4 Sqrt[4. + 2.25 z ] == 2.1 z Let's move the 8 to the RHS: In[2]:= % /. Equal[a_ + b_?NumericQ, c_] -> Equal[a, c - b] Out[2]= 2 4 Sqrt[4. + 2.25 z ] == -8. + 2.1 z Now square both sides and expand: In[3]:= Thread[#^2 &[%], Equal] // ExpandAll Out[3]= 2 2 64. + 36. z == 64. - 33.6 z + 4.41 z Now solve the equation we got: In[4]:= Solve[%, z] Out[4]= {{z -> -1.06363}, {z -> 0.}} Now check if the solutions satisfy the original equation: In[5]:= eq /. % Out[5]= {False, False} In other words we only got "parasite" solutions. There are no genuine solutions at all! If you wanted to see the parasite solutions Mathematica could have given them in one go: In[6]:= Solve[eq, VerifySolutions -> False] Out[6]= {{z -> -1.06363}, {z -> 0.}} Your other equation of course does have a solution and Mathematica correctly finds it. -- Andrzej Kozlowski Toyama International University Toyama, Japan http://sigma.tuins.ac.jp/