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From: SourceForge.net <noreply@so...>  20061011 08:54:59

Bugs item #1575120, was opened at 20061011 01:54 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core  Simplification Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Some laws are still missing Initial Comment: is(equal((x/y)^z,(x^z/y^z))); is(equal((x*y)^z,(x^z*y^z))); is(equal((x^y)^z,(x^(y*z)))); of course they are equal! Those are laws! Mario/Mexico  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 
From: SourceForge.net <noreply@so...>  20061011 10:09:28

Bugs item #1575120, was opened at 20061011 03:54 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core  Simplification Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Some laws are still missing Initial Comment: is(equal((x/y)^z,(x^z/y^z))); is(equal((x*y)^z,(x^z*y^z))); is(equal((x^y)^z,(x^(y*z)))); of course they are equal! Those are laws! Mario/Mexico  >Comment By: Barton Willis (willisbl) Date: 20061011 05:09 Message: Logged In: YES user_id=895922 For real x,y,z, the equation (x*y)^z = x^z*y^z isn't an identity. To see this, let x > 1, y > 1, and z > 1/2. If Maxima did is(equal((x*y)^z,(x^z*y^z))) > true that would be a bug. Similarly, all your other laws are not valid for all real numbers. (1) We're working on improving the function equal; it has many known problems. (2) The function 'radcan' does (%i16) radcan((x*y)^z); (%o16) x^z*y^z Maybe you would like to use it.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 
From: SourceForge.net <noreply@so...>  20061021 20:26:12

Bugs item #1575120, was opened at 20061011 02:54 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core  Simplification Group: None >Status: Pending >Resolution: Rejected Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Some laws are still missing Initial Comment: is(equal((x/y)^z,(x^z/y^z))); is(equal((x*y)^z,(x^z*y^z))); is(equal((x^y)^z,(x^(y*z)))); of course they are equal! Those are laws! Mario/Mexico  >Comment By: Robert Dodier (robert_dodier) Date: 20061021 14:26 Message: Logged In: YES user_id=501686 As mentioned by Barton, Maxima's default behavior is correct (since there are examples for which those equations fail to hold). I find that assuming x > 0 and y > 0, Maxima does evaluate those to true; I believe that is correct. assume(x>0,y>0); is(equal((x/y)^z,(x^z/y^z))); => true is(equal((x*y)^z,(x^z*y^z))); => true is(equal((x^y)^z,(x^(y*z)))); => true A possible enhancement (very far away at this point) would be for is(equal(...)) to return a result with one or more guard clauses specifying the applicability of various particular results. I won't try to spec that here. Marking this report "rejected" and "pending" (so that it will be closed automatically in 2 weeks, in case the original poster comes back).  Comment By: Barton Willis (willisbl) Date: 20061011 04:09 Message: Logged In: YES user_id=895922 For real x,y,z, the equation (x*y)^z = x^z*y^z isn't an identity. To see this, let x > 1, y > 1, and z > 1/2. If Maxima did is(equal((x*y)^z,(x^z*y^z))) > true that would be a bug. Similarly, all your other laws are not valid for all real numbers. (1) We're working on improving the function equal; it has many known problems. (2) The function 'radcan' does (%i16) radcan((x*y)^z); (%o16) x^z*y^z Maybe you would like to use it.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 
From: SourceForge.net <noreply@so...>  20061105 03:20:15

Bugs item #1575120, was opened at 20061011 01:54 Message generated for change (Comment added) made by sfrobot You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. Category: Lisp Core  Simplification Group: None >Status: Closed Resolution: Rejected Priority: 5 Private: No Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: Some laws are still missing Initial Comment: is(equal((x/y)^z,(x^z/y^z))); is(equal((x*y)^z,(x^z*y^z))); is(equal((x^y)^z,(x^(y*z)))); of course they are equal! Those are laws! Mario/Mexico  >Comment By: SourceForge Robot (sfrobot) Date: 20061104 19:20 Message: Logged In: YES user_id=1312539 This Tracker item was closed automatically by the system. It was previously set to a Pending status, and the original submitter did not respond within 14 days (the time period specified by the administrator of this Tracker).  Comment By: Robert Dodier (robert_dodier) Date: 20061021 13:26 Message: Logged In: YES user_id=501686 As mentioned by Barton, Maxima's default behavior is correct (since there are examples for which those equations fail to hold). I find that assuming x > 0 and y > 0, Maxima does evaluate those to true; I believe that is correct. assume(x>0,y>0); is(equal((x/y)^z,(x^z/y^z))); => true is(equal((x*y)^z,(x^z*y^z))); => true is(equal((x^y)^z,(x^(y*z)))); => true A possible enhancement (very far away at this point) would be for is(equal(...)) to return a result with one or more guard clauses specifying the applicability of various particular results. I won't try to spec that here. Marking this report "rejected" and "pending" (so that it will be closed automatically in 2 weeks, in case the original poster comes back).  Comment By: Barton Willis (willisbl) Date: 20061011 03:09 Message: Logged In: YES user_id=895922 For real x,y,z, the equation (x*y)^z = x^z*y^z isn't an identity. To see this, let x > 1, y > 1, and z > 1/2. If Maxima did is(equal((x*y)^z,(x^z*y^z))) > true that would be a bug. Similarly, all your other laws are not valid for all real numbers. (1) We're working on improving the function equal; it has many known problems. (2) The function 'radcan' does (%i16) radcan((x*y)^z); (%o16) x^z*y^z Maybe you would like to use it.  You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1575120&group_id=4933 
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