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Question Number 61895 by maxmathsup by imad last updated on 10/Jun/19

$$letA=\left(\begin{array}{c}1-1\\ 01\end{array}\right)$$$$1)calculate{A}^n$$$$2)find{e}^A,e^{-A}$$$$3)determine{e}^{iA}thencosAandsinA.$$

Commented by maxmathsup by imad last updated on 11/Jun/19

$$wehaveA=\left(\begin{array}{c}10\\ 01\end{array}\right)+\left(\begin{array}{c}0-1\\ 00\end{array}\right)=I+J$$$$wehave{J}^2=\left(\begin{array}{c}0-1\\ 00\end{array}\right).\left(\begin{array}{c}0-1\\ 00\end{array}\right)=\left(\begin{array}{c}00\\ 00\end{array}\right)$$$$A^n={(I+J)}^n=\sum _{k=0}^n{C}_n^k{J}^{k}I{}^{n-k}=I^n+nJ{I}^{n-1}+0=I+nJ$$$$=\left(\begin{array}{c}10\\ 01\end{array}\right)+\left(\begin{array}{c}0-n\\ 00\end{array}\right)=\left(\begin{array}{c}1-n\\ 01\end{array}\right)forn=1theresultistrue$$$$letsuppise{A}^n=\left(\begin{array}{c}1-n\\ 01\end{array}\right)\Rightarrow A^{n+1}=A.A^n=\left(\begin{array}{c}1-1\\ 01\end{array}\right).\left(\begin{array}{c}1-n\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}1-n-1\\ 01\end{array}\right)=\left(\begin{array}{c}1-(n+1)\\ 01\end{array}\right)theresultistrueatterm(n+1)$$$$2)wehave{e}^A=\sum _{n=0}^{\mathrm{\infty }}\frac{A^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\left(\begin{array}{c}1-n\\ 01\end{array}\right)=\left(\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}-\sum _{n=0}^{\mathrm{\infty }}\frac{n}{n!}\\ 0\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\end{array}\right)$$$$wehave{e}^x=\sum _{n=0}^{\mathrm{\infty }}\frac{x^n}{n!}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=e$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{n}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{(n-1)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}=e\Rightarrow e^A=\left(\begin{array}{c}e-e\\ 0e\end{array}\right)$$$$wehavealso{e}^{-A}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-A)}^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{A^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}\left(\begin{array}{c}1-n\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}-\sum _{n=0}^{\mathrm{\infty }}n\frac{{(-1)}^n}{n!}\\ 0\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}\end{array}\right)$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}=e^{-1}and\sum _{n=0}^{\mathrm{\infty }}\frac{n{(-1)}^n}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(n-1)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^{n+1}}{n!}$$$$=-e^{-1}\Rightarrow e^{-A}=\left(\begin{array}{c}\frac{1}{e}\frac{1}{e}\\ 0\frac{1}{e}\end{array}\right)$$

Commented by maxmathsup by imad last updated on 11/Jun/19

$$3)wehave{e}^{iA}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(iA\right)}^n}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{i^n}{n!}\left(\begin{array}{c}1-n\\ 01\end{array}\right)=\left(\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}\frac{i^n}{n!}-\sum _{n=0}^{\mathrm{\infty }}\frac{{ni}^n}{n!}\\ 0\sum _{n=0}^{\mathrm{\infty }}\frac{i^n}{n!}\end{array}\right)$$$$weknowthat{e}^z=\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{n!}\Rightarrow \sum _{n=0}^{\mathrm{\infty }}\frac{i^n}{n!}=e^i$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{{ni}^n}{n!}=\sum _{n=1}^{\mathrm{\infty }}\frac{i^n}{(n-1)!}=\sum _{n=0}^{\mathrm{\infty }}\frac{i^{n+1}}{n!}=i{e}^i\Rightarrow$$$$e^{iA}=\left(\begin{array}{c}{e}^i-i{e}^i\\ 0e^i\end{array}\right)$$$$cosA=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n{A}^{2n}}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}\left(\begin{array}{c}1-2n\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n!}-2\sum _{n=0}^{\mathrm{\infty }}\frac{n{(-1)}^n}{n!}\\ 0\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n:}\end{array}\right)=.....$$

Commented by maxmathsup by imad last updated on 11/Jun/19

$$forgivecosA=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}{A}^{2n}=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}\left(\begin{array}{c}1-2n\\ 01\end{array}\right)$$$$=\left(\begin{array}{c}\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}-2\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^{n}n}{\left(2n\right)!}\\ 0\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}\end{array}\right)$$$$but\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{\left(2n\right)!}=cos\left(1\right),\sum _{n=0}^{\mathrm{\infty }}\frac{n{(-1)}^n}{\left(2n\right)!}=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n}2n}{\left(2n\right)!}$$$$=\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^n}{(2n-1)!}{=}_{n=p+1}\frac{1}{2}\sum _{p=0}^{\mathrm{\infty }}\frac{{(-1)}^{p+1}}{(2p+1)!}=-\frac{1}{2}sin\left(1\right)\Rightarrow$$$$cosA=\left(\begin{array}{c}cos\left(1\right)sin\left(1\right)\\ 0cos\left(1\right)\end{array}\right)$$

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