Services & ResourcesWolfram ForumsMathGroup Archive

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

Simplification question

• To: mathgroup at smc.vnet.net
• Subject: [mg84726] Simplification question
• From: Yaroslav Bulatov <yaroslavvb at gmail.com>
• Date: Fri, 11 Jan 2008 22:04:22 -0500 (EST)

Is there a way to prove that the following 2 expressions are
equivalent over positives using Mathematica?

Root[a[1, 2] a[2, 2] - 2 Sqrt[a[1, 1] a[1, 2] a[2, 1] a[2, 2]] #1^2 +
a[1, 1] a[2, 1] #1^4 &, 3]
(a[1, 2]^(1/4) a[2, 2]^(1/4))/(a[1, 1]^(1/4) a[2, 1]^(1/4))

------------------
PS: I got the two these two by following the recipe in
http://www.yaroslavvb.com/papers/djokovic-note.pdf for finding
diagonal/doubly-stochastic/diagonal decomposition of a 2x2 symbolic
matrix in two different ways. Simpler solution came from forming
Langrangian and using Solve, more complicated one from the built-in
Minimize function)

f[x_] := Times @@ x;
A = Array[a, {2, 2}];
(* Method 1 *)
posCons = # > 0 & /@ (Flatten[A]~Join~{x1, x2});
min = Minimize[{f[A.{x1, x2}]}~
    Append~(And @@ ({f[{x1, x2}] == 1}~Join~posCons)), {x1, x2}];
Assuming[
 And @@ posCons,
 FullSimplify[min]
 ]

(* Method 2 *)
L[x1_, x2_, \[Lambda]_] :=
  f[A.{x1, x2}] - \[Lambda] (f[{x1, x2}] - 1);
eqs = Table[
   D[L[x1, x2, \[Lambda]], var] == 0, {var, {x1, x2, \[Lambda]}}];
Solve[eqs, {x1, x2}, {\[Lambda]}] // Last

• Follow-Ups:
  • Re: Simplification question
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>