Efficient way of calculation matrix exponential in matlab
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    Hayao
 el 12 de Nov. de 2015
  
    
    
    
    
    Comentada: Walter Roberson
      
      
 el 4 de Ag. de 2016
            I am currently trying to solve a matrix exponential of a matrix that has the form of
[a 0 b 0]
[0 c 0 d]
[e 0 f g]
[0 h i j]
This is obviously not symmetry, and I am very sure this is hard to solve.
1)Is there a way to solve this more time efficiently on Matlab?
2)How complicated of a matrix can Matlab solve for matrix exponential? (I know "complicated" is a very qualitative word, but I really have no other way of expressing this)
3)How exactly do MATLAB try to solve matrix exponential?
1 comentario
  Pawel Tokarczuk
 el 4 de Ag. de 2016
				You shouldn't use "i" or "j" as symbols because those are interpreted as imaginary units - just try typing one or the other at the command line to see the point.
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  John D'Errico
      
      
 el 12 de Nov. de 2015
        If your matrix is symbolic, expect it to be difficult, and incredibly nasty looking even if you could obtain a solution. You might think that this should be easy, but to no surprise, it is not a trivial thing to compute. Even the simple case of the matrix exponential of a general 2x2 matrix is a bit of a mess. A 3x3 matrix exponential is right now completely bogging down one of the cores of my cpu in a test case.
If your matrix is purely numeric, so you have values for those elements, then it is trivial. Use expm.
Just wanting code to be more efficient is not sufficient. There is no magic to be found, unless your name is Harry Potter.
11 comentarios
  Walter Roberson
      
      
 el 24 de Nov. de 2015
				185 hours so far and there is no sign that it is getting closer to finding the solution. You need to confirm soon that you need me to keep running it and you need to tell me which output format you need the answer it.
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  Star Strider
      
      
 el 12 de Nov. de 2015
        2 comentarios
  Star Strider
      
      
 el 12 de Nov. de 2015
				I know I can’t write any more efficient code than expm, so I can’t offer any further help.
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