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I am trying to isolate out the real and complex terms in a fairly large expression. During the process, I have made several assumptions that last throughout the entire process.

$Assumptions = Element[{x,y},Reals] && {x,y}>0 

When I try separating the real and complex parts, the Re and Im functions have trouble with terms where variables appear in the denominator. For example

Im[1/(x^2 + 4 y^2)^(9/2)] //Simplify = Im[1/(x^2 + 4 y^2)^(9/2)]

Even though all the terms are real and positive, meaning there should be no complex component, and the expression should be zero

I have checked, and if the denominator does not involve two or more terms, Im does work successfully. For example

Im[1/(x^2)^(9/2) ] //Simplify = 0
Im[1/(4 y^2)^(9/2)] //Simplify = 0

Can someone please comment on if this is a bug in the version I have, or am I doing something incorrect here? A mwe is provided below. For reference, I am using version 10.0.0.0

-Thanks

In[421]:=
$Assumptions = Element[{x, y}, Reals] && {x, y} > 0 ;
Simplify[Im[1/(x^2 + 4*y^2)^(9/2)]]
Simplify[Im[1/(x^2)^(9/2)]]
Simplify[Im[1/(y^2)^(9/2)]]

Out[424]= Im[1/(x^2+4 y^2)^(9/2)]
Out[425]= 0
Out[426]= 0

Edit

If I modify the assumptions to define each one separately, the code runs fine. e.g

$Assumptions = Element[x, Reals] && Element[y, Reals] && x > 0 && y > 0 ;

In my main program, I have a large number of variables that are both real and positive, is there a way to assign these all at once, or do I need to do each one separately?

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  • 2
    $\begingroup$ To the edit: If you have those variables in a list varlist then $Assumptions = And @@ (Element[#, Reals] && # > 0 & /@ varlist) will do the trick. $\endgroup$ – Marius Ladegård Meyer Nov 10 '15 at 7:07
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    $\begingroup$ Try ComplexExpand[Im[...]]. That should work. Same for the real part $\endgroup$ – Lukas Nov 10 '15 at 7:08
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{x,y} > 0 is not doing what you think it does. In contrast to Element which, as its documentation states, accepts arbitrary patterns and in which first argument, being a List of more than one elements, evaluates to Alternatives:

$Assumptions = {x, y} ∈ Reals
(* (x | y) ∈ Reals *)
x ∈ Reals // Refine
(* True *)
y ∈ Reals // Refine
(* True *)

{x,y} > 0 in assumptions is just stating that literal list expression {x, y} is greater than zero and not that individual elements of list are greater than zero:

$Assumptions = {x, y} > 0
(* {x, y} > 0 *)
{x, y} > 0 // Refine
(* True *)
x > 0 // Refine
(* x > 0 *)
y > 0 // Refine
(* y > 0 *)

Using expression in inequality in assumptions implies that this expression is real:

$Assumptions = x > 0;
x ∈ Reals // Refine
(* True *)

For some reason, although {x, y} > 0 assumption, assumes that only literal list is greater than zero, it also results in individual elements being assumed real:

$Assumptions = {x, y} > 0;
{x, y} ∈ Reals // Refine
(* True *)
x ∈ Reals // Refine
(* True *)
y ∈ Reals // Refine
(* True *)

Going back to expression from question it's enough to assume that individual symbols are greater than zero, this will automatically imply that they are real.

$Assumptions = And @@ Thread[{x, y} > 0]
(* x > 0 && y > 0 *)
Im[1/(x^2 + 4*y^2)^(9/2)] // Refine
(* 0 *)
Im[1/(x^2)^(9/2)] // Refine
(* 0 *)
Im[1/(y^2)^(9/2)] // Refine
(* 0 *)
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  • $\begingroup$ That works exactly like what I was looking for. Thanks for the clear answer $\endgroup$ – jpdomann Nov 10 '15 at 15:03
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$Assumptions = And @@ Thread[Greater[{x, y, d, e}, 0]] && Element[{x, y, d, e}, Reals]
x > 0 && y > 0 && d > 0 && e > 0 && (x | y | d | e) ∈ Reals
ComplexExpand[Im[1/(y^2)^(9/2)]]
0 
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  • $\begingroup$ By the way, Greater[{x, y, d, e}, 0] is just {x, y, d, e} > 0 even inside Thread $\endgroup$ – ybeltukov Nov 10 '15 at 12:25
  • $\begingroup$ @ybeltukov, I don't think so. Greater[{x, y, d, e}, 0] === Thread[Greater[{x, y, d, e}, 0]] evaluates to flase False $\endgroup$ – rhermans Nov 11 '15 at 14:06
  • $\begingroup$ I mean a more obvious thing: Thread[{x, y, d, e} > 0] === Thread[Greater[{x, y, d, e}, 0]] $\endgroup$ – ybeltukov Nov 11 '15 at 14:16

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