Hello Jaroslav
I have found an easy way
Algorithmus:
E = expm(A)
Situation: A ∈ R4×4
tol := 0.001
Identity Matrix E, I∈ R4×4
k = 1
repeat
I := A∗ I/k
E := E + I
k := k + 1
until E < tol
Return: E
Von: jaroslavmencl [email removed]
Gesendet: Mittwoch, 14. Januar 2015 22:42
An: [email removed]
Betreff: Re: matrix exponential [mathnetnumerics:571946]
From: jaroslavmencl
Hello,
I'm working on the same. Fitting and exponential.
I still cannot found any solution (for free). Did you make any progress?
Regards,
Jaroslav