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FullSimplify does not get rid of the z factor in the following expression:

FullSimplify[(E^(-I x) y z + (1 + E^(I y)) (x + y) z)/z, z != 0]

Upon removing a single piece of the expression, such as the first y factor or the 1+ in the brackets, the simplification suddenly succeeds. As far as I can tell it is only this special combination of variables that fails.

Is there any obvious problem with simplifying this?
Is there a way to tweak the FullSimplify function to make it succeed? (this is just a part of a longer expression - and I want to avoid looking for such problems manually).

I'm using Mathematica version 9.0.0.0 on Win7 64 bit.

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    $\begingroup$ How about Simplify[(E^(-I x) y z + (1 + E^(I y)) (x + y) z)/z // Factor] ? $\endgroup$ Commented Jan 27, 2013 at 16:04
  • $\begingroup$ Maybe FullSimplify[Expand@(E^(-I x) y z + (1 + E^(I y)) (x + y) z)/z, z != 0]? $\endgroup$
    – user1066
    Commented Jan 27, 2013 at 16:37
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    $\begingroup$ Or FullSimplify[Cancel[(E^(-I x) y z + (1 + E^(I y)) (x + y) z)/z], z != 0]. What I find interesting is that if you use e.g. (x+a) instead of (x+y) in your original expression, it seems to work directly... $\endgroup$ Commented Jan 27, 2013 at 16:42
  • $\begingroup$ All these suggestions work. I'm not sure how to judge this, but the one involving Cancel seems like the most direct. As for an explanation for this strange behavior, I guess I shouldn't hold my breath. @PinguinDirk - do you want to turn your comment into an answer so I can accept it? $\endgroup$
    – Joe
    Commented Jan 28, 2013 at 7:23

1 Answer 1

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As in the comment above and requested by the OP as an answer, I can only offer a workaround:

FullSimplify[Cancel[(E^(-I x) y z + (1 + E^(I y)) (x + y) z)/z], z != 0]

Also, I'd like to note again that using e.g. (x+a) instead of (x+y) in the original expression seems to make it work, out of the box (without Cancel)

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    $\begingroup$ (+1) Interesting observation about a. That seems to indicate that the culprit is the same as in this question $\endgroup$
    – Jens
    Commented Jan 28, 2013 at 7:43
  • $\begingroup$ @Jens: ah, I wasn't aware of the question - interesting indeed. I did try various expressions instead of (x+y) there and couldn't figure out any pattern in Mathematica's behaviour. Interesting indeed. $\endgroup$ Commented Jan 28, 2013 at 7:47

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