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Mathematics Wolfram Language Symbolic Computations

The Simplify function is not completely simplified

Zhenyu Zeng

Posted 1 month ago

Hello, In Mathematica, FullSimplify[(Sqrt[1/a^2] a G \[Pi] u Sqrt[a^2/(a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], Assumptions -> a > 0] will produce (a G \[Pi] u Sqrt[1/(a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)] But FullSimplify[(Sqrt[1/a^2] a G \[Pi] u Sqrt[a^2/(a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], Assumptions -> a > 0 && Subscript[R, 1] > 0] will produce the right result a G \[Pi] u (-(1/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)]) + 1/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]) What is the reason?

Hello,

In Mathematica,

FullSimplify[(Sqrt[1/a^2]  a  G  \[Pi]  u  Sqrt[a^2/(a^2 + 

\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\))]  (Sqrt[a^2 + 

\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + 

\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + 

\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], Assumptions -> a > 0]

will produce

(a G \[Pi] u Sqrt[1/(a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]

But

FullSimplify[(Sqrt[1/a^2] a G \[Pi] u Sqrt[a^2/(a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], 
 Assumptions -> a > 0 && Subscript[R, 1] > 0]

will produce the right result

a G \[Pi] u (-(1/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)]) + 1/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)])

What is the reason?

enter image description here

POSTED BY: Zhenyu Zeng

8 Replies

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Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 1 month ago

The condition `r1!=Sqrt[-1]a` is not enough to allow the simplification; for example, `r1 = 2 I a` is also problematic: expr0 = (Sqrt[1/a^2] a G \[Pi] u Sqrt[ a^2/(a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)] expr2 = FullSimplify[expr0, Assumptions -> a > 0 && Subscript[R, 1] > 0] {expr0, expr2} /. Subscript[R, 1] -> 2 I a /. a -> 1 // Simplify

The condition r1!=Sqrt[-1]a is not enough to allow the simplification; for example, r1 = 2 I a is also problematic:

expr0 = (Sqrt[1/a^2]    a    G    \[Pi]    u    Sqrt[
           a^2/(a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\))]   
    (Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]
expr2 = FullSimplify[expr0, 
    Assumptions -> a > 0 && Subscript[R, 1] > 0]
{expr0, expr2} /. Subscript[R, 1] -> 2 I  a /. a -> 1 // Simplify

POSTED BY: Gianluca Gorni

Zhenyu Zeng

Posted 1 month ago

Yes! I know now.

POSTED BY: Zhenyu Zeng

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 1 month ago

The two expressions are not equivalent if you allow `R_1` to be complex: expr1 = FullSimplify[(Sqrt[1/a^2] a G \[Pi] u Sqrt[ a^2/(a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], Assumptions -> a > 0] expr2 = FullSimplify[(Sqrt[1/a^2] a G \[Pi] u Sqrt[ a^2/(a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\))] (Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + \!\(\SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], Assumptions -> a > 0 && Subscript[R, 1] > 0] {expr1, expr2} /. {a -> 1, Subscript[R, 1] -> 2 I, Subscript[R, 2] -> 1, G -> 1, u -> 1} // N Here is a much simpler situation exhibiting the same behaviour: expr1 = FullSimplify[Sqrt[1/r^2] Sqrt[r^2]] expr2 = FullSimplify[Sqrt[1/r^2] Sqrt[r^2] , Assumptions -> r > 0] {expr1, expr2} /. r -> I

The two expressions are not equivalent if you allow R_1 to be complex:

expr1 = FullSimplify[(Sqrt[1/a^2]   a   G   \[Pi]   u   Sqrt[
      a^2/(a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\))]   (Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], 
  Assumptions -> a > 0]
expr2 = FullSimplify[(Sqrt[1/a^2]   a   G   \[Pi]   u   Sqrt[
      a^2/(a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\))]   (Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(1\), \(2\)]\)] - Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)]))/Sqrt[a^2 + 
\!\(\*SubsuperscriptBox[\(R\), \(2\), \(2\)]\)], 
  Assumptions -> a > 0 && Subscript[R, 1] > 0]
{expr1, expr2} /. {a -> 1, Subscript[R, 1] -> 2 I, 
   Subscript[R, 2] -> 1, G -> 1, u -> 1} // N

Here is a much simpler situation exhibiting the same behaviour:

expr1 = FullSimplify[Sqrt[1/r^2]  Sqrt[r^2]]
expr2 = FullSimplify[Sqrt[1/r^2]  Sqrt[r^2] , Assumptions -> r > 0]
{expr1, expr2} /. r -> I

POSTED BY: Gianluca Gorni

Zhenyu Zeng

Posted 1 month ago

Thanks a lot.

POSTED BY: Zhenyu Zeng

Bill Nelson

Posted 1 month ago

If you do not tell Mathematica that variables are positive or real then it will often assume that variables may be complex. For your problem that may mean that you have zeros in denominators and multiplying by zero may give solutions that you will not accept. Telling it `r1>0` or `Element[r1,Reals]` is enough to know that you do not have zero denominators. Telling it that `r1!=Sqrt[-1]a` is not enough for it to understand that.