Menu Close

let-A-1-1-1-1-1-calculate-A-n-2-find-e-A-e-A-3-find-sinA-and-cosA-4-find-ch-A-and-sh-A-

Tinku Tara 0 Comments
Pin




Question Number 76361 by mathmax by abdo last updated on 26/Dec/19
let A = (((1 1)),((1 1)) ) 1) calculate A^n 2) find e^A ,e^(−A) 3) find sinA and cosA 4) find ch(A) and sh(A)
$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{e}^{{A}} \:\:,{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{sinA}\:{and}\:{cosA} \\ $$$$\left.\mathrm{4}\right)\:{find}\:{ch}\left({A}\right)\:{and}\:{sh}\left({A}\right) \\ $$
Commented by mathmax by abdo last updated on 27/Dec/19
P_c (x) =det(A−xI) = determinant (((1−x 1)),((1 1−x)))=(1−x)^2 −1 =(1−x−1)(1−x+1) =−x(2−x) =x(x−2) so the proper values are λ_1 =0 and λ_2 =2 let divide x^n by P_c (x) ⇒ x^n =Q(x)P_c (x)+u_n x +v_n ⇒0 =v_n 2^n =2u_n ⇒u_n =2^(n−1) we have A^n =u_n A =2^(n−1) (((1 1)),((1 1)) ) ⇒ A^n = (((2^(n−1) 2^(n−1) )),((2^(n−1) 2^(n−1) )) ) 2) e^(A ) =Σ_(n=0) ^∞ (A^n /(n!)) = ((( Σ_(n=0) ^∞ (2^(n−1) /(n!)) Σ_(n=0) ^∞ (2^(n−1) /(n!)))),((Σ_(n=0) ^∞ (2^(n−1) /(n!)) Σ_(n=0) ^∞ (2^(n−1) /(n!)))) ) =(1/2) (((e^2 e^2 )),((e^2 e^2 )) ) e^(−A) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) A^n = (((Σ_(n=0) ^∞ (((−1)^n )/(n!))2^(n−1) ...)),((... ...)) ) = ((((1/2)e^(−2) (1/2)e^(−2) )),(((1/2)e^(−2) (1/2)e^(−2) )) )
$${P}_{{c}} \left({x}\right)\:={det}\left({A}−{xI}\right)\:=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−{x}}\end{vmatrix}=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1} \\ $$$$=\left(\mathrm{1}−{x}−\mathrm{1}\right)\left(\mathrm{1}−{x}+\mathrm{1}\right)\:=−{x}\left(\mathrm{2}−{x}\right)\:={x}\left({x}−\mathrm{2}\right)\:{so}\:{the}\:{proper}\:{values} \\ $$$${are}\:\lambda_{\mathrm{1}} =\mathrm{0}\:\:{and}\:\lambda_{\mathrm{2}} =\mathrm{2}\:\:{let}\:{divide}\:{x}^{{n}} \:{by}\:{P}_{{c}} \left({x}\right)\:\Rightarrow \\ $$$${x}^{{n}} ={Q}\left({x}\right){P}_{{c}} \left({x}\right)+{u}_{{n}} {x}\:+{v}_{{n}} \:\Rightarrow\mathrm{0}\:={v}_{{n}} \\ $$$$\mathrm{2}^{{n}} \:=\mathrm{2}{u}_{{n}} \:\Rightarrow{u}_{{n}} =\mathrm{2}^{{n}−\mathrm{1}} \:\:\:{we}\:{have}\:{A}^{{n}} ={u}_{{n}} \:{A}\:=\mathrm{2}^{{n}−\mathrm{1}} \begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\:{A}^{{n}} \:=\begin{pmatrix}{\mathrm{2}^{{n}−\mathrm{1}} \:\:\:\:\:\:\:\:\:\mathrm{2}^{{n}−\mathrm{1}} }\\{\mathrm{2}^{{n}−\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\mathrm{2}^{{n}−\mathrm{1}} }\end{pmatrix} \\ $$$$\left.\mathrm{2}\right)\:{e}^{{A}\:} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{A}^{{n}} }{{n}!}\:=\begin{pmatrix}{\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}!}\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}!}}\\{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}!}\:\:\:\:\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}^{{n}−\mathrm{1}} }{{n}!}}\end{pmatrix} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{{e}^{\mathrm{2}} \:\:\:\:\:\:\:\:{e}^{\mathrm{2}} }\\{{e}^{\mathrm{2}} \:\:\:\:\:\:\:\:{e}^{\mathrm{2}} }\end{pmatrix}\:\: \\ $$$${e}^{−{A}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{A}^{{n}} \:=\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\mathrm{2}^{{n}−\mathrm{1}} \:\:\:\:\:\:\:\:…}\\{…\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}} }\\{\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}} }\end{pmatrix} \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 27/Dec/19
3) sinA =Σ_(n=0) ^∞ (((−1)^n A^(2n+1) )/((2n+1)!)) = (((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))2^(2n+1−1) ...)),((... ...)) ) = ((((1/2)sin(2) (1/2)sin(2))),(((1/2)sin(2) (1/2)sin(2))) ) cosA =Σ_(n=0) ^∞ (((−1)^n )/((2n)!)) A^(2n) = (((Σ_(n=0) ^∞ (((−1)^n )/((2n)!))2^(2n−1) ...)),((... ...)) ) = ((((1/2)cos(2) (1/2)cos(2))),(((1/2)cos(2) (1/2)cos(2))) )
$$\left.\mathrm{3}\right)\:{sinA}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{A}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:=\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}−\mathrm{1}} \:\:\:\:\:\:…}\\{…\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\right)}\\{\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}\right)}\end{pmatrix} \\ $$$${cosA}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\:{A}^{\mathrm{2}{n}} \:\:=\begin{pmatrix}{\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}\right)!}\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \:\:\:\:\:\:…}\\{…\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…}\end{pmatrix} \\ $$$$=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\right)\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\right)}\\{\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\right)\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{2}\right)}\end{pmatrix} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 27/Dec/19
4) sh(A) =(1/2)(e^A −e^(−A) )=....
$$\left.\mathrm{4}\right)\:{sh}\left({A}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{A}} \:−{e}^{−{A}} \right)=…. \\ $$
Answered by john santu last updated on 27/Dec/19
A^n = (((2^(n−1) 2^(n−1) )),((2^(n−1) 2^(n−1) )) ) =2^(n−1) (((1 1)),((1 1)) )
$${A}^{{n}} =\begin{pmatrix}{\mathrm{2}^{{n}−\mathrm{1}} \:\mathrm{2}^{{n}−\mathrm{1}} }\\{\mathrm{2}^{{n}−\mathrm{1}} \:\mathrm{2}^{{n}−\mathrm{1}} }\end{pmatrix}\:=\mathrm{2}^{{n}−\mathrm{1}} \:\begin{pmatrix}{\mathrm{1}\:\mathrm{1}}\\{\mathrm{1}\:\mathrm{1}}\end{pmatrix} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *