Services & Resources / Wolfram Forums / MathGroup Archive
-----

MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: wrong answer or no answer?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg82796] Re: [mg82710] wrong answer or no answer?
  • From: DrMajorBob <drmajorbob at bigfoot.com>
  • Date: Wed, 31 Oct 2007 06:11:40 -0500 (EST)
  • References: <21140268.1193712179571.JavaMail.root@m35>
  • Reply-to: drmajorbob at bigfoot.com

No solution exists. Here's a demonstration that's pretty convincing, I  
think:

m2={{-1,x},{x,1}};
r={{a,b},{c,d}};
r.r-m2//Flatten;
Thread[%==0];
soln=Solve[%,{a,b,c,d}]//Simplify
r.r==m2/.soln//Simplify
soln/.x->I//Quiet

{{a->-((1+x^2+Sqrt[-1-x^2])/(Sqrt[2]  
(-1-x^2)^(3/4))),d->(-1-x^2+Sqrt[-1-x^2])/(Sqrt[2]  
(-1-x^2)^(3/4)),b->x/(Sqrt[2] (-1-x^2)^(1/4)),c->x/(Sqrt[2]  
(-1-x^2)^(1/4))},{a->-((\[ImaginaryI] (-1-x^2+Sqrt[-1-x^2]))/(Sqrt[2]  
(-1-x^2)^(3/4))),d->(\[ImaginaryI] (1+x^2+Sqrt[-1-x^2]))/(Sqrt[2]  
(-1-x^2)^(3/4)),b->(\[ImaginaryI] x)/(Sqrt[2]  
(-1-x^2)^(1/4)),c->(\[ImaginaryI] x)/(Sqrt[2]  
(-1-x^2)^(1/4))},{a->(\[ImaginaryI] (-1-x^2+Sqrt[-1-x^2]))/(Sqrt[2]  
(-1-x^2)^(3/4)),d->-((\[ImaginaryI] (1+x^2+Sqrt[-1-x^2]))/(Sqrt[2]  
(-1-x^2)^(3/4))),b->-((\[ImaginaryI] x)/(Sqrt[2]  
(-1-x^2)^(1/4))),c->-((\[ImaginaryI] x)/(Sqrt[2]  
(-1-x^2)^(1/4)))},{a->(1+x^2+Sqrt[-1-x^2])/(Sqrt[2]  
(-1-x^2)^(3/4)),d->(1+x^2-Sqrt[-1-x^2])/(Sqrt[2]  
(-1-x^2)^(3/4)),b->-(x/(Sqrt[2] (-1-x^2)^(1/4))),c->-(x/(Sqrt[2]  
(-1-x^2)^(1/4)))}}

{True,True,True,True}

{{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity},{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity},{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity},{a->Indeterminate,d->Indeterminate,b->ComplexInfinity,c->ComplexInfinity}}

Visually examining the poles in Solve's output (as x tends to I), one can  
see that the Indeterminate results are actually infinite, like the others.

Bobby

On Mon, 29 Oct 2007 05:54:37 -0500, Roger Bagula <rlbagula at sbcglobal.net>  
wrote:

> M = {{-1, I}, {I, 1}}
> MatrixPower[M, 1/2]
> gives
> {{0, 0}, {0, 0}}
>
> So try it as {{a,b},{c,d}} squared:
> c = b; a = I*Sqrt[1 + b^2]; d = Sqrt[1 - b^2
> FullSimplify[{a^2 + b c + 1 == 0, a b + b d - I == 0, a c +
>          c d - I == 0, b c + d^2 - 1 == 0}]
> {True, -I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 +
> b^2)]== 0, -I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 +
> b^2)]== 0, True}
> Solve[-I + b*Sqrt[1 - b^2] +I* b*Sqrt[1 +
> b^2)]== 0,b]
> {}
> which says there is no solution.
>
> This problem comes from the graph of SU(2) and of U(1)*SU(2)
> as a two vertex 3 directed connections (i,j,k) and a 2 vertex with 4
> directed connections (1,i,j,k}.
> Basically there is either a solution or there is none.
> Mathematica gives zero and the null set from two different approaches.
> {{0,i+k},{j,0}} and {{0,i+k},{j+IdentityMatrix[2],0}}
>
> It is pretty much a break down of mathematical definitions.
> The matrix M does appear to have not one, but four solutions
> the way I do it:
>
> M2={{+/-I*Sqrt[1 + b^2], b}, {b, +/-Sqrt[1 + b^2]}}
> I really may be doing it all wrong.
> b=+/-Sqrt[+/-1/2+I/2]
> which gives the stange answers from this code:
> Clear[b]
> M2 = {{I*Sqrt[1 + b^2], b}, {b, Sqrt[1 + b^2]}}
> Det[M2]
> Solve[Det[M2] == 0, b]
> b0 = b /. Solve[Det[M2] == 0, b][[2]]
> M20 = {{-I*Sqrt[1 + b0^2], b0}, {b0, Sqrt[1 + b0^2]}}
> FullSimplify[M20]
> FullSimplify[M20.M20]
>
> All this leaves me really puzzled.
> Usually Mathematica takes away my doubts,
> but here it isn't any help at all.
> Maybe it is a paradox?
> Roger Bagula
>
>



-- 

DrMajorBob at bigfoot.com


  • Prev by Date: Re: wrong answer or no answer?
  • Next by Date: Re: Re: Selecting Rows Where All Columns Satisfy a Condition-Plot of Code speeds
  • Previous by thread: Re: wrong answer or no answer?
  • Next by thread: MultivariateDistribution mathematica 6