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Hessian - Сообщения
#1 Опубликовано: 02.10.2012 01:13:52
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Radovan Omorjan 02.10.2012 03:04:00
#2 Опубликовано: 02.10.2012 02:27:02
:d

Russia ☭ forever, Viacheslav N. Mezentsev
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Radovan Omorjan 02.10.2012 03:04:00
#3 Опубликовано: 02.10.2012 03:16:45
#4 Опубликовано: 02.10.2012 03:36:06
#5 Опубликовано: 02.10.2012 04:35:18
#6 Опубликовано: 06.10.2012 03:17:29
I think that this is the better way to work with the Jacobian.
Alvaro.
Jacobian.sm (22 КиБ) скачан 95 раз(а).
Alvaro.
Jacobian.sm (22 КиБ) скачан 95 раз(а).
#7 Опубликовано: 06.10.2012 07:24:02
Hello Alvaro,
But what to do if the vector function is defined this way?
[MATH=eng]Φ(X,t):-5*t*mat(el(X,1),el(X,2),2,1)+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1)^2,2,1)[/MATH]
It will not work until t is a constant (scalar). Take a look at the pictures please.
Regards,
Radovan

But what to do if the vector function is defined this way?
[MATH=eng]Φ(X,t):-5*t*mat(el(X,1),el(X,2),2,1)+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1)^2,2,1)[/MATH]
It will not work until t is a constant (scalar). Take a look at the pictures please.
Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
#8 Опубликовано: 06.10.2012 15:26:35
#9 Опубликовано: 07.10.2012 02:40:21
Hello Alvaro,
Actually, that is the point. I think that's the problem of the symbolic engine.
The first one gives the symbolic result
[MATH=eng](-5*t)*mat(el(X,1),el(X,2),2,1)+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1),2,1)=mat(-5*el(X,1),-5*el(X,2),2,1)*t+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1),2,1)[/MATH]
and the result is not a vector, therefore length() does not work.
The second one actually give you the vector symbolically
[MATH=eng](-5)*mat(el(X,1)*t,el(X,2)*t,2,1)+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1),2,1)=mat(el(X,1)*(-5*t+el(X,2)*el(X,1)),el(X,2)*(-5*t+el(X,2)*el(X,1)),2,1)[/MATH]
Maybe the solution would be to say in advance what the variable t was - either scalar, vector or matrix. Something like "assume". If SMath assumed that t was a scalar, it should behave in accordance to that and give you the second correct result.
Regards,
Radovan
Actually, that is the point. I think that's the problem of the symbolic engine.
The first one gives the symbolic result
[MATH=eng](-5*t)*mat(el(X,1),el(X,2),2,1)+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1),2,1)=mat(-5*el(X,1),-5*el(X,2),2,1)*t+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1),2,1)[/MATH]
and the result is not a vector, therefore length() does not work.
The second one actually give you the vector symbolically
[MATH=eng](-5)*mat(el(X,1)*t,el(X,2)*t,2,1)+mat(el(X,2)*el(X,1)^2,el(X,2)^2*el(X,1),2,1)=mat(el(X,1)*(-5*t+el(X,2)*el(X,1)),el(X,2)*(-5*t+el(X,2)*el(X,1)),2,1)[/MATH]
Maybe the solution would be to say in advance what the variable t was - either scalar, vector or matrix. Something like "assume". If SMath assumed that t was a scalar, it should behave in accordance to that and give you the second correct result.
Regards,
Radovan
When Sisyphus climbed to the top of a hill, they said: "Wrong boulder!"
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