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Hi i am trying to understand , consider the matrix :
$M = \begin{pmatrix} 0 &-2 & 4 &-2 \\ 1 &1 &-2 &-1 \\ 0 &0 &0 &0 \\ 1 &-1 &2 &-3 \end{pmatrix} $

$M^2 = \begin{pmatrix} -4 &0 & 0 &8 \\ 0 &0 &0 &0 \\ 0 &0 &0 &0 \\ -4 &0 &0 &8 \end{pmatrix} ,\qquad $ $ M^3 = \begin{pmatrix} 8 &0 & 0 &-16 \\ 0 &0 &0 &0 \\ 0 &0 &0 &0 \\ 8 &0 &0 &-16 \end{pmatrix} $ then $M^n = 2^{n-2}M^2$
how to determine $e^M$ ? i known just that $e^M = \sum_{n=0}^{+\infty}\frac{1}{n!}M^n $ but i don't know determine

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We have :

$$e^M = I + M + M^2 \cdot\sum_{i=2}^{\infty} \frac{2^{i-2}}{i!}$$

Using the fact that : $e^x = \sum_{i=0}^{\infty} \frac{x^i}{i!}$

We get :

$$ \sum_{i=2}^{\infty} \frac{2^{i-2}}{i!} = \frac{1}{4} \cdot \sum_{i=2}^{\infty} \frac{2^i}{i!} = \frac{1}{4} \cdot ( e^2-3)$$

So we have :

$$e^M = I + M + \frac{e^2-3}{4} \cdot M^2$$

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  • $\begingroup$ how to find $e^2 -3$ please? $\endgroup$ – KEVIN DLL Sep 23 '18 at 18:35
  • $\begingroup$ $\sum_{i =2}^{\infty} 2^i/i! = \sum_{i=0}^{\infty} 2^i/i! -1/0!-2/1! = e^2-3$ $\endgroup$ – Thinking Sep 23 '18 at 19:15

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