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Simplification of absolute value

Steven Kaptein
Posted 2 years ago

Dear Wolfram community, I am trying to simplify the following expression

Simplify[Abs[B Exp[2 z] Cos[x + y]  + A Cos[y - z]] /(  B Exp[2 z] Cos[x + y]  + A Cos[y - z]), 
B Exp[2 z] Cos[x + y]  + A Cos[y - z] > 0]

And it returns

-1 B E^(2 z) Cos[x + y] + A Cos[y - z] < 0
1 True

Why does it not just return 

1

since I specified that

B  Exp[2 z] Cos[x + y] + A Cos[y - z]  > 0
POSTED BY: Steven Kaptein
4 Replies
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One way would be to use Replace, instead of Simplify:

With[{expr = B Exp[2 z] Cos[x + y] + A Cos[y - z]},
 Abs[expr]/expr /. Abs[expr] :> expr]

Perhaps Simplify ignores assumptions that look too complicated, and it seems a reasonable strategy in general.
POSTED BY: Gianluca Gorni
Steven Kaptein
Steven Kaptein
Posted 2 years ago

Yes, it does the job! Thanks a lot.
POSTED BY: Steven Kaptein

This is curious, the result seems to depend on how complicated the expression is:

With[{expr = a b c d e},
 Simplify[Abs[expr]/expr, expr > 0]]
With[{expr = a b c d},
 Simplify[Abs[expr]/expr, expr > 0]]
POSTED BY: Gianluca Gorni
Steven Kaptein
Steven Kaptein
Posted 2 years ago

Indeed, for me it also worked for slightly simpler expressions. I hope there is a way to make it work for the complex ones.
POSTED BY: Steven Kaptein
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