let give A=(_(2 2) ^(1 2) ) find A^n and e^A and e^(tA) . we remind that e^A = Σ


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Question Number 27343 by abdo imad last updated on 05/Jan/18

$l e t g i v e A = (_{2 2}^{1 2}) f i n d A^{n} a n d e^{A}$ $a n d e^{t A} . w e r e m i n d t h a t e^{A} = \sum_{} \frac{A^{n}}{n!}$
Commented by Rasheed.Sindhi last updated on 05/Jan/18

${What}^{'} s n in e^{A} = \sum_{} \frac{A^{n}}{n!} ?$
Commented by abdo imad last updated on 05/Jan/18

$n i n t e g e r .$
Commented by Rasheed.Sindhi last updated on 06/Jan/18

$My question is actually this that$ $why n is involved in e^{A} ?$
Commented by Rasheed.Sindhi last updated on 07/Jan/18

$t h a n k s s i r . I m i s u n d e r s t o o d .$
Commented by prakash jain last updated on 06/Jan/18

$e^{x} = \underset{i = 0}{\sum^{\infty}} \frac{x^{i}}{i!}$
Commented by Rasheed.Sindhi last updated on 06/Jan/18

$Is e^{A} = \sum_{} \frac{A^{n}}{n!} correct ?$
Commented by abdo imad last updated on 06/Jan/18

$e^{A} i s a l s o a m a t r i s e d e f i n e d b y$ $e^{A} = \sum_{n ⩾ 0} \frac{A^{n}}{n!} l i k e e^{x} = Σ \frac{x^{n}}{n!}$
	Answered by prakash jain last updated on 07/Jan/18

	$A^{n} = (\begin{matrix} a_{n - 1} & b_{n - 1} \\ c_{n - 1} & d_{n - 1} \end{matrix}) \cdot (\begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix})$ $a_{n} = a_{n - 1} + 2 b_{n - 1}$ $b_{n} = 2 a_{n - 1} + 2 b_{n - 1}$ $c_{n} = 2 c_{n - 1} + 2 d_{n - 1}$ $c_{n} = 2 c_{n - 1} + 2 d_{n - 1}$ $b_{n - 1} = \frac{a_{n} - a_{n - 1}}{2}$ $b_{n} = 2 a_{n - 1} + 2 b_{n - 1} = a_{n} + a_{n - 1}$ $a_{n} = a_{n - 1} + 2 b_{n - 1} = a_{n - 1} + 2 a_{n - 1} + 2 a_{n - 2}$ $a_{n} - 3 a_{n - 1} - 2 a_{n - 2} = 0$ $x^{2} - 3 x - 2 = 0 \Rightarrow x = \frac{3 \pm \sqrt{17}}{2}$ $a_{n} = c_{1} {(\frac{3 + \sqrt{17}}{2})}^{n} + c_{2} {(\frac{3 - \sqrt{17}}{2})}^{n}$ $a_{1} = 1 \Rightarrow c_{1} (3 + \sqrt{17}) + c_{2} (3 - \sqrt{17}) = 1$ $a_{2} = 5 \Rightarrow c_{1} {(3 + \sqrt{17})}^{2} + c_{2} {(3 - \sqrt{17})}^{2} = 20$ $c_{1} = \frac{20 - (3 - \sqrt{17})}{{(3 + \sqrt{17})}^{2} - (3 + \sqrt{17}) (3 - \sqrt{17})}$ $= \frac{17 + \sqrt{17}}{9 + 17 + 6 \sqrt{17} - 9 + 17} = \frac{17 + \sqrt{17}}{34 + 6 \sqrt{17}}$ $= \frac{1 + \sqrt{17}}{2 (\sqrt{17} + 3)}$ $c_{2} = \frac{20 - (3 + \sqrt{17})}{{(3 - \sqrt{17})}^{2} - (3 + \sqrt{17}) (3 - \sqrt{17})} = \frac{17 - \sqrt{17}}{34 - 6 \sqrt{17}}$ $c_{2} = \frac{1 - \sqrt{17}}{2 (3 - \sqrt{17})}$ $Similarly we can find b_{n}, c_{n} and d_{n} .$
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