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let-A-1-1-0-1-1-calculate-A-n-2-find-e-A-e-A-3-determine-e-iA-then-cosA-and-sinA-

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Question Number 61895 by maxmathsup by imad last updated on 10/Jun/19
let A = (((1 −1)),((0 1)) ) 1) calculate A^n 2) find e^A ,e^(−A) 3) determine e^(iA) then cosA and sinA .
$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{e}^{{A}} \:\:,{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{e}^{{iA}} \:\:\:{then}\:\:{cosA}\:\:{and}\:{sinA}\:. \\ $$
Commented by maxmathsup by imad last updated on 11/Jun/19
we have A = (((1 0)),((0 1)) ) + (((0 −1)),((0 0)) ) =I +J we have J^2 = (((0 −1)),((0 0)) ) . (((0 −1)),((0 0)) ) = (((0 0)),((0 0)) ) A^n =(I +J)^n =Σ_(k=0) ^n C_n ^k J^k I^(n−k) = I^n + n J I^(n−1) +0 =I +nJ = (((1 0)),((0 1)) ) + (((0 −n)),((0 0)) ) = (((1 −n)),((0 1)) ) for n=1 the result is true let suppise A^n = (((1 −n)),((0 1)) ) ⇒ A^(n+1) = A.A^n = (((1 −1)),((0 1)) ) . (((1 −n)),((0 1)) ) = ((( 1 −n−1)),((0 1)) ) = (((1 −(n+1))),((0 1)) ) the result is true at term (n+1) 2) we have e^A =Σ_(n=0) ^∞ (A^n /(n!)) =Σ_(n=0) ^∞ (1/(n!)) (((1 −n)),((0 1)) ) = (((Σ_(n=0) ^∞ (1/(n!)) −Σ_(n=0) ^∞ (n/(n!)))),((0 Σ_(n=0) ^∞ (1/(n!)))) ) we have e^x =Σ_(n=0) ^∞ (x^n /(n!)) ⇒Σ_(n=0) ^∞ (1/(n!)) =e Σ_(n=0) ^∞ (n/(n!)) =Σ_(n=1) ^∞ (1/((n−1)!)) =Σ_(n=0) ^∞ (1/(n!)) =e ⇒ e^A = (((e −e)),((0 e)) ) we have also e^(−A) =Σ_(n=0) ^∞ (((−A)^n )/(n!)) =Σ_(n=0) ^∞ (−1)^n (A^n /(n!)) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) (((1 −n)),((0 1)) ) = (((Σ_(n=0) ^∞ (((−1)^n )/(n!)) −Σ_(n=0) ^∞ n(((−1)^n )/(n!)))),((0 Σ_(n=0) ^∞ (((−1)^n )/(n!)))) ) Σ_(n=0) ^∞ (((−1)^n )/(n!)) =e^(−1) and Σ_(n=0) ^∞ ((n(−1)^n )/(n!)) =Σ_(n=1) ^∞ (((−1)^n )/((n−1)!)) =Σ_(n=0) ^∞ (((−1)^(n+1) )/(n!)) =−e^(−1) ⇒ e^(−A) = ((((1/e) (1/e))),((0 (1/e))) )
$${we}\:{have}\:{A}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:={I}\:+{J} \\ $$$${we}\:{have}\:{J}^{\mathrm{2}} \:\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$${A}^{{n}} \:=\left({I}\:+{J}\right)^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{J}^{{k}} \:{I}\:^{{n}−{k}} \:=\:{I}^{{n}} \:+\:{n}\:{J}\:{I}^{{n}−\mathrm{1}} \:+\mathrm{0}\:={I}\:+{nJ} \\ $$$$=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:+\begin{pmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:{for}\:{n}=\mathrm{1}\:\:{the}\:{result}\:{is}\:{true} \\ $$$${let}\:{suppise}\:{A}^{{n}} \:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\Rightarrow\:{A}^{{n}+\mathrm{1}} \:=\:{A}.{A}^{{n}} \:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:.\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{n}−\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−\left({n}+\mathrm{1}\right)}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:\:{the}\:{result}\:{is}\:{true}\:{at}\:{term}\:\left({n}+\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{we}\:{have}\:\:{e}^{{A}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{A}^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:=\:\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{n}!}\:\:\:\:\:\:\:\:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}}{{n}!}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}}\end{pmatrix} \\ $$$${we}\:{have}\:{e}^{{x}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}} }{{n}!}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:={e}\: \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}}{{n}!}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{n}!}\:={e}\:\Rightarrow\:{e}^{{A}} \:=\:\begin{pmatrix}{{e}\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{e}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}}\end{pmatrix} \\ $$$${we}\:{have}\:{also}\:{e}^{−{A}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−{A}\right)^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:\frac{{A}^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\:\:\:\:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:{n}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}}\end{pmatrix} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:={e}^{−\mathrm{1}} \:\:\:\:\:\:{and}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}!} \\ $$$$=−{e}^{−\mathrm{1}} \:\Rightarrow\:{e}^{−{A}} \:=\:\begin{pmatrix}{\frac{\mathrm{1}}{{e}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{e}}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{e}}}\end{pmatrix} \\ $$
Commented by maxmathsup by imad last updated on 11/Jun/19
3) we have e^(iA) =Σ_(n=0) ^∞ (((iA)^n )/(n!)) =Σ_(n=0) ^∞ (i^n /(n!)) (((1 −n)),((0 1)) ) = (((Σ_(n=0) ^∞ (i^n /(n!)) −Σ_(n=0) ^∞ ((ni^n )/(n!)))),((0 Σ_(n=0) ^∞ (i^n /(n!)))) ) we know that e^z =Σ_(n=0) ^∞ (z^n /(n!)) ⇒Σ_(n=0) ^∞ (i^n /(n!)) =e^i Σ_(n=0) ^∞ ((ni^n )/(n!)) =Σ_(n=1) ^∞ (i^n /((n−1)!)) =Σ_(n=0) ^∞ (i^(n+1) /(n!)) =i e^i ⇒ e^(iA) = ((( e^i −i e^i )),((0 e^i )) ) cosA =Σ_(n=0) ^∞ (((−1)^n A^(2n) )/(n!)) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) (((1 −2n)),((0 1)) ) = (((Σ_(n=0) ^∞ (((−1)^n )/(n!)) −2 Σ_(n=0) ^∞ ((n(−1)^n )/(n!)))),((0 Σ_(n=0) ^∞ (((−1)^n )/(n:)))) ) =.....
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{e}^{{iA}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left({iA}\right)^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{i}^{{n}} }{{n}!}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:−{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:=\:\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{i}^{{n}} }{{n}!}\:\:\:\:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{ni}^{{n}} }{{n}!}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{i}^{{n}} }{{n}!}}\end{pmatrix} \\ $$$${we}\:{know}\:{that}\:{e}^{{z}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{z}^{{n}} }{{n}!}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{i}^{{n}} }{{n}!}\:={e}^{{i}} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{ni}^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{i}^{{n}} }{\left({n}−\mathrm{1}\right)!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{i}^{{n}+\mathrm{1}} }{{n}!}\:={i}\:{e}^{{i}} \:\Rightarrow \\ $$$${e}^{{iA}} \:=\:\begin{pmatrix}{\:\:\:{e}^{{i}} \:\:\:\:\:\:\:\:\:\:\:−{i}\:{e}^{{i}} }\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{e}^{{i}} }\end{pmatrix} \\ $$$${cosA}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{A}^{\mathrm{2}{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{2}{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}\left(−\mathrm{1}\right)^{{n}} }{{n}!}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}:}}\end{pmatrix}\:=….. \\ $$
Commented by maxmathsup by imad last updated on 11/Jun/19
forgive cosA =Σ_(n=0) ^∞ (((−1)^n )/((2n)!)) A^(2n) =Σ_(n=0) ^∞ (((−1)^n )/((2n)!)) (((1 −2n)),((0 1)) ) = (((Σ_(n=0) ^∞ (((−1)^n )/((2n)!)) −2 Σ_(n=0) ^∞ (((−1)^n n)/((2n)!)))),((0 Σ_(n=0) ^∞ (((−1)^n )/((2n)!)))) ) but Σ_(n=0) ^∞ (((−1)^n )/((2n)!)) =cos(1) , Σ_(n=0) ^∞ ((n(−1)^n )/((2n)!)) =(1/2)Σ_(n=1) ^∞ (((−1)^n 2n)/((2n)!)) =(1/2) Σ_(n=1) ^∞ (((−1)^n )/((2n−1)!)) =_(n=p+1) (1/2) Σ_(p=0) ^∞ (((−1)^(p+1) )/((2p+1)!)) =−(1/2)sin(1) ⇒ cosA = (((cos(1) sin(1))),((0 cos(1))) )
$${forgive}\:\:{cosA}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:{A}^{\mathrm{2}{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{2}{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$=\:\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {n}}{\left(\mathrm{2}{n}\right)!}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}}\end{pmatrix} \\ $$$${but}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:={cos}\left(\mathrm{1}\right)\:\:\:\:,\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{n}\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \mathrm{2}{n}}{\left(\mathrm{2}{n}\right)!} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)!}\:=_{{n}={p}+\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{p}+\mathrm{1}} }{\left(\mathrm{2}{p}+\mathrm{1}\right)!}\:=−\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${cosA}\:=\:\begin{pmatrix}{{cos}\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:{sin}\left(\mathrm{1}\right)}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\left(\mathrm{1}\right)}\end{pmatrix} \\ $$

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