Matrix Exponential and series idenfication

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SUMMARY

The discussion focuses on calculating the matrix exponential \(e^{\mathbf{A}t}\) for the matrix \(\mathbf{A} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -4 & -5 & -4 \end{bmatrix}\). The user initially attempted to identify the series expansion but faced challenges due to irrational and complex eigenvalues. A diagonalization approach was suggested, utilizing the formula \(e^{At} = U^*D^{At}U\), where \(U\) is unitary and \(D\) contains the eigenvalues. The user later discovered a correction in their calculations, which resolved the issue.

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  • Understanding of matrix exponentials and series expansions
  • Familiarity with eigenvalues and eigenvectors
  • Knowledge of diagonalization of matrices
  • Experience with numerical methods for matrix computations
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  • Learn about numerical methods for approximating matrix diagonalization
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  • Review the document outlining "19 dubious ways" to compute matrix exponentials
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Dustinsfl
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Let
\[
\mathbf{A} =
\begin{bmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
-4 & -5 & -4
\end{bmatrix}
\]
Then I want to find \(e^{\mathbf{A}t}\).
\[
\mathbf{I} + \mathbf{A}t +\frac{\mathbf{A}^2t^2}{2!} + \frac{\mathbf{A}^3t^3}{3!} + \cdots
\]
I have up to the 6th term but I can't identify the series.
\[
\begin{bmatrix}
1 - \frac{2t^3}{3} + \frac{2t^4}{3} - \frac{11t^5}{30} + \frac{7t^6}{45} &
t - \frac{5t^3}{6} + \frac{2t^4}{3} - \frac{13t^5}{40} + \frac{2t^6}{15} &
\frac{t^2}{2} - \frac{2t^3}{3} + \frac{11t^4}{24} - \frac{7t^5}{30} + \frac{73t^6}{720}\\
-2t^2 + \frac{8t^3}{3} - \frac{11t^4}{6} + \frac{14t^5}{15} - \frac{73t^6}{180} &
1 - \frac{5t^2}{2} + \frac{8t^3}{3} - \frac{13t^4}{8} + \frac{4t^5}{5} - \frac{253t^6}{720} &
-\frac{49 t^6}{180} + \frac{73 t^5}{120} - \frac{7 t^4}{6} + \frac{11 t^3}{6} - 2t^2 + t\\
\frac{49 t^6}{45} - \frac{73 t^5}{30} + \frac{14 t^4}{3} - \frac{22 t^3}{3} + 8t^2 - 4t &
\frac{43 t^6}{45} - \frac{253 t^5}{120} + 4t^4 - \frac{13 t^3}{2} + 8t^2 - 5t &
\frac{59 t^6}{80} - \frac{49 t^5}{30} + \frac{73 t^4}{24} - \frac{14t^3}{3} + \frac{11 t^2}{2} - 4t + 1
\end{bmatrix}
\]
 
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In principle you can diagonalise $A$, $A = U^*DU$, where $U$ is unitary and $D$ is diagonal (with the eigenvalues of $A$ as its diagonal elements). Then $e^{At} = U^*D^{At}U$, and you can easily evaluate $D^{At}$ by taking the exponentials of the eigenvalues. Unfortunately in this case the eigenvalues are irrational (and two of them are complex). Maybe you can use some numerical method to get an approximate diagonalisation?
 
Opalg said:
In principle you can diagonalise $A$, $A = U^*DU$, where $U$ is unitary and $D$ is diagonal (with the eigenvalues of $A$ as its diagonal elements). Then $e^{At} = U^*D^{At}U$, and you can easily evaluate $D^{At}$ by taking the exponentials of the eigenvalues. Unfortunately in this case the eigenvalues are irrational (and two of them are complex). Maybe you can use some numerical method to get an approximate diagonalisation?

I tried that method first that is why I moved on to this form.
 
I can't pretend to be an expert, but http://www.cs.cornell.edu/cv/researchpdf/19ways+.pdf outlining 19 "dubious" ways to compute a matrix exponential. Use at your own risk!
 
Ackbach said:
I can't pretend to be an expert, but http://www.cs.cornell.edu/cv/researchpdf/19ways+.pdf outlining 19 "dubious" ways to compute a matrix exponential. Use at your own risk!

I have read that plus the numerous other links that come up from Googling the topic.
 
dwsmith said:
I have read that plus the numerous other links that come up from Googling the topic.

Well, then you know more than I do about matrix exponentials! I hope you find a do-able method for your problem. The command

Code:
N[MatrixExp[{{0, 1, 0}, {0, 0, 1}, {-4, -5, -4}}t]] // MatrixForm

gives a result in Mathematica. It's pretty long and ugly - I don't know how exact a solution you need.
 
Found the problem, I should have had a -6 not -5 and everything works.
 

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