# [SOLVED]Matrix Exponential and series idenfication

#### dwsmith

##### Well-known member
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.
$\begin{bmatrix} 1 - \frac{2t^3}{3} + \frac{2t^4}{3} - \frac{11t^5}{30} + \frac{7t^6}{45} & t - \frac{5t^3}{6} + \frac{2t^4}{3} - \frac{13t^5}{40} + \frac{2t^6}{15} & \frac{t^2}{2} - \frac{2t^3}{3} + \frac{11t^4}{24} - \frac{7t^5}{30} + \frac{73t^6}{720}\\ -2t^2 + \frac{8t^3}{3} - \frac{11t^4}{6} + \frac{14t^5}{15} - \frac{73t^6}{180} & 1 - \frac{5t^2}{2} + \frac{8t^3}{3} - \frac{13t^4}{8} + \frac{4t^5}{5} - \frac{253t^6}{720} & -\frac{49 t^6}{180} + \frac{73 t^5}{120} - \frac{7 t^4}{6} + \frac{11 t^3}{6} - 2t^2 + t\\ \frac{49 t^6}{45} - \frac{73 t^5}{30} + \frac{14 t^4}{3} - \frac{22 t^3}{3} + 8t^2 - 4t & \frac{43 t^6}{45} - \frac{253 t^5}{120} + 4t^4 - \frac{13 t^3}{2} + 8t^2 - 5t & \frac{59 t^6}{80} - \frac{49 t^5}{30} + \frac{73 t^4}{24} - \frac{14t^3}{3} + \frac{11 t^2}{2} - 4t + 1 \end{bmatrix}$

#### Opalg

##### MHB Oldtimer
Staff member
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?

#### dwsmith

##### Well-known member
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.

#### Ackbach

##### Indicium Physicus
Staff member
I can't pretend to be an expert, but here's a paper outlining 19 "dubious" ways to compute a matrix exponential. Use at your own risk!

#### dwsmith

##### Well-known member
I can't pretend to be an expert, but here's a paper 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.

#### Ackbach

##### Indicium Physicus
Staff member
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.

#### dwsmith

##### Well-known member
Found the problem, I should have had a -6 not -5 and everything works.