1
$\begingroup$

In the documentation for FullSimplify, under Options > Assumptions it is stated that:

When assumptions are given as an argument, $Assumptions is used as well:

Assuming[x > 0, FullSimplify[E^(LogGamma[x] + LogGamma[y]), y > 0]]
(* Gamma[x] Gamma[y] *)

Specifying assumptions as an option value prevents FullSimplify from using $Assumptions:

Assuming[x > 0, FullSimplify[E^(LogGamma[x] + LogGamma[y]), Assumptions -> y > 0]]
(* Gamma[x] Gamma[y] *)

The resuls are just the same. What is the point?

$Version
(* 10.1.0  for Microsoft Windows (64-bit) (March 24, 2015) *)
Improve this question
$\endgroup$
5
  • $\begingroup$ Try FullSimplify[E^(LogGamma[x] + LogGamma[y])] without assumptions. $\endgroup$
    – Michael E2
    Commented Jun 12, 2015 at 17:14
  • $\begingroup$ I get the same result. $\endgroup$
    – Giovanni F.
    Commented Jun 12, 2015 at 17:17
  • 1
    $\begingroup$ So the assumptions are irrelevant, since the simplification is valid more generally. $\endgroup$
    – Michael E2
    Commented Jun 12, 2015 at 17:20
  • $\begingroup$ This documentation example just adds nothing then. $\endgroup$
    – Giovanni F.
    Commented Jun 12, 2015 at 17:23
  • $\begingroup$ Yes, it's strange. I wonder if it has changed since some version. $\endgroup$
    – Michael E2
    Commented Jun 12, 2015 at 17:26

1 Answer 1

Reset to default
1
$\begingroup$

It's a badly written example. What it is trying to show is something like the following behavior, where the local Assumptions option overrides the global $Assumptions.

$Assumptions = x == -1;
FullSimplify[E^(LogGamma[x] + LogGamma[y])]

FullSimplify::infd : "Expression LogGamma[x] + LogGamma[y] simplified to ∞. 

∞

FullSimplify[E^(LogGamma[x] + LogGamma[y]), Assumptions -> x == 1]

Gamma[y]

However, Assumming only adds to $Assumptions, so

Assuming[x == 1, FullSimplify[E^(LogGamma[x] + LogGamma[y])]]

produces

$Assumptions::cas : Warning : contradictory assumption (s) x == 1 && x == -1 encountered. >>

1
Improve this answer
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.