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BUG? f(x):=x*x(-1) (x-scalar!, x-vector?)
  Rank: Administration Groups: Registered, Advanced Member Joined: 23/06/2009(UTC) Posts: 1,740   Was thanked: 318 time(s) in 268 post(s)
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Helo Andrey, Just a reminder, in the case you've forgotten. f(x)←x*x^{-1}f(5)=1??? f(mat(1;0;0;0;1;0;0;0;1;3;3))=1??? mat(1;0;0;0;1;0;0;0;1;3;3)*mat(1;0;0;0;1;0;0;0;1;3;3)^{-1}—mat(1;0;0;0;1;0;0;0;1;3;3)Regards, Radovan Edited by user 22 September 2009 13:56:05(UTC)
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When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!" |
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Rank: Administration Groups: Registered, Advanced Member Joined: 23/06/2009(UTC) Posts: 1,740 Was thanked: 318 time(s) in 268 post(s)
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Hello Andrey, Actually, regarding the recent comments about "eval" it flashed to my mind and I found the way about this one (it took me a long time to figure out  ). We just have to use "eval" in the case of x*x^(-1) in order to work either with scalar or square matrix. The result with a scalar will be unity and with a square matrix wil be the identity matrix - as it should be. Here it is: f(x)←eval(x)*eval(x^{-1})f(5)=1f(mat(1;0;0;5;-2;0;0;0;1;3;3))=mat(1;0;0;0;1;0;0;0;1;3;3)Regards, Radovan Edited by user 15 March 2010 03:12:45(UTC)
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When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!" |
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Rank: Administration Groups: Registered, Advanced Member Joined: 23/06/2009(UTC) Posts: 1,740 Was thanked: 318 time(s) in 268 post(s)
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Hello Andrey, I just wonder how the Disabling optimization will work in this case. Do we still need to use "eval" or not? Regards, Radovan PS. What do you think to have Disabling, Enabling optimization globally - as an option, applied on the entire workshet? Edited by user 02 April 2010 12:46:43(UTC)
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When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!" |
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BUG? f(x):=x*x(-1) (x-scalar!, x-vector?)
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