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Question Number 74794 by mathmax by abdo last updated on 30/Nov/19

let A = (((1 2)),((−1 1)) ) 1) calculate A^n 2) find e^A and e^(−A) 3) calculate cosA and sinA 4) calculate ch(A) and sh(A)

$${let}\:{A}\:=\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{A}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{e}^{{A}} \:{and}\:{e}^{−{A}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:{cosA}\:{and}\:{sinA} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\:{ch}\left({A}\right)\:{and}\:{sh}\left({A}\right) \\ $$

Commented by Cmr 237 last updated on 01/Dec/19

1) calcul de A^n on a le polynome caracteristique est P(λ)=det(A−λI_2 ) =λ^2 −2λ+3 deplus on a: P(λ)=0⇒λ_1 =1−i(√2),λ_2 =1+i(√2) on remarque que P(A)=0 donc ∀X∈K[X] on a: X^n =Q(X)P(X)+a_n X+b_n ou Q(X) est un polynome quotient .. λ_1 ^n =Q(λ_1 )P(λ_1 )+a_n λ_1 +b_n λ_2 ^n =Q(λ_2 )P(λ_2 )+a_n λ_2 +b_n { ((a_n λ_1 +b_n =λ_(1 ) ^n )),((a_n λ_2 +b_n =λ_2 ^n )) :} a_n =((λ_2 ^n −λ_1 ^n )/(λ_2 −λ_1 ))=(((1+i(√2))^n −(1−i(√2))^n )/(2i(√2))) b_n =((3(λ_1 ^(n−1) −λ_2 ^(n−1) ))/(2i(√2))) A^n =Q(A)P(A)+a_n A+b_n I_2 =a_n A+b_n I_2 A^n = (((a_n 2a_n )),((−a_n a_n )) )+ (((b_n 0)),((0 b_n )) ) = (((a_n +b_(n ) 2a_n )),((−a_(n ) a_n +b_n )) ) ou a_(n ) et b_n sont les suites donnees en haut 2) calcul de e^A et e^(−A) e^A =Σ_(k=0) ^n (A^k /(k!)) e^(−A) =Σ_(k=0) ^n (A^(−k) /(k!)) 3)calcul de cosA et sinA cosA=Σ_(k=0) ^n (((−1)^k A^(2k) )/((2k)!)) sinA=Σ_(k=0) ^n (((−1)^k A^(2k+1) )/((2k+1)!)) 4) calcul de chA et shA chA=Σ_(k=0) ^n (A^(2k) /((2k)!)) shA=Σ_(k=0) ^n (A^(2k+1) /((2k+1)!)) NB:e^A ,e^(−A) ,cosA,sinA,chA,shA sont sur la forme generale car ∀ n∈N^∗ ,A^n ≠0

$$\left.\mathrm{1}\right)\:{calcul}\:{de}\:{A}^{{n}} \\ $$$${on}\:{a}\:{le}\:{polynome}\:{caracteristique}\:{est} \\ $$$${P}\left(\lambda\right)={det}\left({A}−\lambda{I}_{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\lambda^{\mathrm{2}} −\mathrm{2}\lambda+\mathrm{3} \\ $$$${deplus}\:{on}\:{a}: \\ $$$${P}\left(\lambda\right)=\mathrm{0}\Rightarrow\lambda_{\mathrm{1}} =\mathrm{1}−{i}\sqrt{\mathrm{2}},\lambda_{\mathrm{2}} =\mathrm{1}+{i}\sqrt{\mathrm{2}} \\ $$$${on}\:{remarque}\:{que}\:{P}\left({A}\right)=\mathrm{0} \\ $$$${donc}\:\forall{X}\in{K}\left[{X}\right]\:{on}\:{a}: \\ $$$${X}^{{n}} ={Q}\left({X}\right){P}\left({X}\right)+{a}_{{n}} {X}+{b}_{{n}} \\ $$$${ou}\:{Q}\left({X}\right)\:\:{est}\:{un}\:{polynome}\:{quotient}\:.. \\ $$$$\lambda_{\mathrm{1}} ^{{n}} ={Q}\left(\lambda_{\mathrm{1}} \right){P}\left(\lambda_{\mathrm{1}} \right)+{a}_{{n}} \lambda_{\mathrm{1}} +{b}_{{n}} \\ $$$$\lambda_{\mathrm{2}} ^{{n}} ={Q}\left(\lambda_{\mathrm{2}} \right){P}\left(\lambda_{\mathrm{2}} \right)+{a}_{{n}} \lambda_{\mathrm{2}} +{b}_{{n}} \\ $$$$\begin{cases}{{a}_{{n}} \lambda_{\mathrm{1}} +{b}_{{n}} =\lambda_{\mathrm{1}\:} ^{{n}} }\\{{a}_{{n}} \lambda_{\mathrm{2}} +{b}_{{n}} =\lambda_{\mathrm{2}} ^{{n}} \:}\end{cases} \\ $$$$\:{a}_{{n}} =\frac{\lambda_{\mathrm{2}} ^{{n}} −\lambda_{\mathrm{1}} ^{{n}} }{\lambda_{\mathrm{2}} −\lambda_{\mathrm{1}} }=\frac{\left(\mathrm{1}+{i}\sqrt{\mathrm{2}}\right)^{{n}} −\left(\mathrm{1}−{i}\sqrt{\mathrm{2}}\right)^{{n}} }{\mathrm{2}{i}\sqrt{\mathrm{2}}} \\ $$$${b}_{{n}} =\frac{\mathrm{3}\left(\lambda_{\mathrm{1}} ^{{n}−\mathrm{1}} −\lambda_{\mathrm{2}} ^{{n}−\mathrm{1}} \right)}{\mathrm{2}{i}\sqrt{\mathrm{2}}} \\ $$$${A}^{{n}} ={Q}\left({A}\right){P}\left({A}\right)+{a}_{{n}} {A}+{b}_{{n}} {I}_{\mathrm{2}} \\ $$$$={a}_{{n}} {A}+{b}_{{n}} {I}_{\mathrm{2}} \\ $$$${A}^{{n}} =\begin{pmatrix}{{a}_{{n}} \:\:\:\mathrm{2}{a}_{{n}} }\\{−{a}_{{n}} \:\:{a}_{{n}} }\end{pmatrix}+\begin{pmatrix}{{b}_{{n}} \:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:{b}_{{n}} }\end{pmatrix} \\ $$$$=\begin{pmatrix}{{a}_{{n}} +{b}_{{n}\:\:\:\:\:\:\:\:\:\:\:\:} \mathrm{2}{a}_{{n}} }\\{−{a}_{{n}\:} \:\:\:\:\:\:\:\:\:{a}_{{n}} +{b}_{{n}} }\end{pmatrix}\:{ou}\:{a}_{{n}\:\:} {et}\:{b}_{{n}} \:{sont}\:{les}\:{suites}\:{donnees}\:{en}\:{haut} \\ $$$$\left.\mathrm{2}\right)\:{calcul}\:{de}\:{e}^{{A}} {et}\:{e}^{−{A}} \\ $$$${e}^{{A}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{A}^{{k}} }{{k}!} \\ $$$${e}^{−{A}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{A}^{−{k}} }{{k}!} \\ $$$$\left.\mathrm{3}\right){calcul}\:{de}\:{cosA}\:{et}\:{sinA} \\ $$$${cosA}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \mathrm{A}^{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!} \\ $$$${sinA}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {A}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$$\left.\mathrm{4}\right)\:{calcul}\:{de}\:{chA}\:{et}\:{shA} \\ $$$${chA}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{A}^{\mathrm{2}{k}} }{\left(\mathrm{2}{k}\right)!} \\ $$$${shA}=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{A}^{\mathrm{2}{k}+\mathrm{1}} }{\left(\mathrm{2}{k}+\mathrm{1}\right)!} \\ $$$${NB}:{e}^{{A}} ,{e}^{−{A}} \:,{cosA},{sinA},{chA},{shA} \\ $$$${sont}\:{sur}\:{la}\:{forme}\:{generale}\:{car}\:\forall\:{n}\in\boldsymbol{{N}}^{\ast} ,{A}^{{n}} \neq\mathrm{0} \\ $$

Commented by mathmax by abdo last updated on 01/Dec/19

1)we have P_c (x)=det(A−xI)= determinant (((1−x 2)),((−1 1−x))) =(1−x)^2 +2 =x^2 −2x +3 Δ^′ =(−1)^2 −3 =1−3=−2 ⇒λ_1 =1+i(√2) and λ_2 =1−i(√2) cayley hamilton give P_c (A)=0 let divide x^n by x^2 −2x+3 ⇒ x^n =q(x)P_c (x)+a_n x +b_n ⇒λ_1 ^n =a_n λ_1 +b_n λ_2 ^n =a_n λ_2 +b_n so we get the systeme { ((λ_1 a_n +b_n =λ_1 ^n )),((λ_2 a_n +b_n =λ_2 ^n )) :} ⇒(λ_1 −λ_2 )a_n =λ_1 ^n −λ_2 ^n ⇒a_n =((λ_1 ^n −(λ_1 ^− )^n )/(λ_1 −λ_2 )) =((2i Im(λ_1 ^n ))/(2i(√2))) =(1/(√2)) Im(λ_1 ^n ) we have ∣λ_1 ∣ =(√(1+2))=(√3) ⇒λ_1 =(√3)e^(iarctan((√2))) ⇒ a_n =((((√3))^n )/(√2)) sin(narctan((√2)) and b_n =λ_1 ^n −λ_1 a_n =λ_1 ^n −λ_1 ((λ_1 ^n −λ_2 ^n )/(λ_1 −λ_2 )) =((λ_1 ^(n+1) −λ_2 λ_1 ^n −λ_1 ^(n+1) +λ_1 λ_2 ^n )/(λ_1 −λ_2 )) =((λ1λ_2 )/(λ_1 −λ_2 )){ λ_1 ^(n−1) −(λ_1 ^− )^(n−1) } =(3/(2i (√2))) (2i ((√3))^(n−1) sin((n−1)arctan((√2))) =((((√3))^(n+1) )/(√2)) sin{(n−1)arctan((√2))) we have A^n =a_n A +b_n I =((((√3))^n )/(√2))sin(narctan((√2))) (((1 2)),((−1 1)) ) +((((√3))^(n+1) )/(√2))sin{(n−1)arctan((√2))) (((1 0)),((0 1)) )

$$\left.\mathrm{1}\right){we}\:{have}\:{P}_{{c}} \left({x}\right)={det}\left({A}−{xI}\right)=\begin{vmatrix}{\mathrm{1}−{x}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{1}−{x}}\end{vmatrix} \\ $$$$=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} +\mathrm{2}\:={x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{3} \\ $$$$\Delta^{'} =\left(−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\:=\mathrm{1}−\mathrm{3}=−\mathrm{2}\:\Rightarrow\lambda_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{2}}\:{and}\:\lambda_{\mathrm{2}} =\mathrm{1}−{i}\sqrt{\mathrm{2}} \\ $$$${cayley}\:{hamilton}\:{give}\:\:{P}_{{c}} \left({A}\right)=\mathrm{0}\:{let}\:{divide}\:{x}^{{n}} \:{by}\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\:\Rightarrow \\ $$$${x}^{{n}} ={q}\left({x}\right){P}_{{c}} \left({x}\right)+{a}_{{n}} {x}\:+{b}_{{n}} \:\:\Rightarrow\lambda_{\mathrm{1}} ^{{n}} ={a}_{{n}} \lambda_{\mathrm{1}} +{b}_{{n}} \\ $$$$\lambda_{\mathrm{2}} ^{{n}} \:={a}_{{n}} \lambda_{\mathrm{2}} \:+{b}_{{n}} \:\:{so}\:{we}\:{get}\:{the}\:{systeme}\:\:\:\begin{cases}{\lambda_{\mathrm{1}} {a}_{{n}} +{b}_{{n}} =\lambda_{\mathrm{1}} ^{{n}} }\\{\lambda_{\mathrm{2}} {a}_{{n}} +{b}_{{n}} =\lambda_{\mathrm{2}} ^{{n}} }\end{cases} \\ $$$$\Rightarrow\left(\lambda_{\mathrm{1}} −\lambda_{\mathrm{2}} \right){a}_{{n}} =\lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{2}} ^{{n}} \:\Rightarrow{a}_{{n}} =\frac{\lambda_{\mathrm{1}} ^{{n}} \:−\left(\overset{−} {\lambda}_{\mathrm{1}} \right)^{{n}} }{\lambda_{\mathrm{1}} −\lambda_{\mathrm{2}} }\:=\frac{\mathrm{2}{i}\:{Im}\left(\lambda_{\mathrm{1}} ^{{n}} \right)}{\mathrm{2}{i}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{Im}\left(\lambda_{\mathrm{1}} ^{{n}} \right)\:\:{we}\:{have}\:\mid\lambda_{\mathrm{1}} \mid\:=\sqrt{\mathrm{1}+\mathrm{2}}=\sqrt{\mathrm{3}}\:\Rightarrow\lambda_{\mathrm{1}} =\sqrt{\mathrm{3}}{e}^{{iarctan}\left(\sqrt{\mathrm{2}}\right)} \:\Rightarrow \\ $$$${a}_{{n}} =\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} }{\sqrt{\mathrm{2}}}\:{sin}\left({narctan}\left(\sqrt{\mathrm{2}}\right)\:{and}\:{b}_{{n}} =\lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{1}} {a}_{{n}} \right. \\ $$$$=\lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{1}} \:\frac{\lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{2}} ^{{n}} }{\lambda_{\mathrm{1}} −\lambda_{\mathrm{2}} }\:=\frac{\lambda_{\mathrm{1}} ^{{n}+\mathrm{1}} −\lambda_{\mathrm{2}} \lambda_{\mathrm{1}} ^{{n}} −\lambda_{\mathrm{1}} ^{{n}+\mathrm{1}} +\lambda_{\mathrm{1}} \lambda_{\mathrm{2}} ^{{n}} }{\lambda_{\mathrm{1}} −\lambda_{\mathrm{2}} } \\ $$$$=\frac{\lambda\mathrm{1}\lambda_{\mathrm{2}} }{\lambda_{\mathrm{1}} −\lambda_{\mathrm{2}} }\left\{\:\lambda_{\mathrm{1}} ^{{n}−\mathrm{1}} −\left(\overset{−} {\lambda}_{\mathrm{1}} \right)^{{n}−\mathrm{1}} \right\}\:=\frac{\mathrm{3}}{\mathrm{2}{i}\:\sqrt{\mathrm{2}}}\:\left(\mathrm{2}{i}\:\left(\sqrt{\mathrm{3}}\right)^{{n}−\mathrm{1}} \:{sin}\left(\left({n}−\mathrm{1}\right){arctan}\left(\sqrt{\mathrm{2}}\right)\right)\right. \\ $$$$=\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}+\mathrm{1}} }{\sqrt{\mathrm{2}}}\:\:{sin}\left\{\left({n}−\mathrm{1}\right){arctan}\left(\sqrt{\mathrm{2}}\right)\right) \\ $$$${we}\:{have}\:{A}^{{n}} ={a}_{{n}} {A}\:+{b}_{{n}} \:{I} \\ $$$$=\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}} }{\sqrt{\mathrm{2}}}{sin}\left({narctan}\left(\sqrt{\mathrm{2}}\right)\right)\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\:\:\mathrm{1}}\end{pmatrix}\:\:+\frac{\left(\sqrt{\mathrm{3}}\right)^{{n}+\mathrm{1}} }{\sqrt{\mathrm{2}}}{sin}\left\{\left({n}−\mathrm{1}\right){arctan}\left(\sqrt{\mathrm{2}}\right)\right)\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\:\:\:\mathrm{1}}\end{pmatrix} \\ $$

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