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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)

letA=(1211)1)calculateAn2)findeAandeA3)calculatecosAandsinA4)calculatech(A)andsh(A)

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

1)calculdeAnonalepolynomecaracteristiqueestP(λ)=det(AλI2)=λ22λ+3deplusona:P(λ)=0λ1=1i2,λ2=1+i2onremarquequeP(A)=0doncXK[X]ona:Xn=Q(X)P(X)+anX+bnouQ(X)estunpolynomequotient..λ1n=Q(λ1)P(λ1)+anλ1+bnλ2n=Q(λ2)P(λ2)+anλ2+bn{anλ1+bn=λ1nanλ2+bn=λ2nan=λ2nλ1nλ2λ1=(1+i2)n(1i2)n2i2bn=3(λ1n1λ2n1)2i2An=Q(A)P(A)+anA+bnI2=anA+bnI2An=(an2ananan)+(bn00bn)=(an+bn2ananan+bn)ouanetbnsontlessuitesdonneesenhaut2)calculdeeAeteAeA=nk=0Akk!eA=nk=0Akk!3)calculdecosAetsinAcosA=nk=0(1)kA2k(2k)!sinA=nk=0(1)kA2k+1(2k+1)!4)calculdechAetshAchA=nk=0A2k(2k)!shA=nk=0A2k+1(2k+1)!NB:eA,eA,cosA,sinA,chA,shAsontsurlaformegeneralecarnN,An0

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)) )

1)wehavePc(x)=det(AxI)=|1x211x|=(1x)2+2=x22x+3Δ=(1)23=13=2λ1=1+i2andλ2=1i2cayleyhamiltongivePc(A)=0letdividexnbyx22x+3xn=q(x)Pc(x)+anx+bnλ1n=anλ1+bnλ2n=anλ2+bnsowegetthesysteme{λ1an+bn=λ1nλ2an+bn=λ2n(λ1λ2)an=λ1nλ2nan=λ1n(λ1)nλ1λ2=2iIm(λ1n)2i2=12Im(λ1n)wehaveλ1=1+2=3λ1=3eiarctan(2)an=(3)n2sin(narctan(2)andbn=λ1nλ1an=λ1nλ1λ1nλ2nλ1λ2=λ1n+1λ2λ1nλ1n+1+λ1λ2nλ1λ2=λ1λ2λ1λ2{λ1n1(λ1)n1}=32i2(2i(3)n1sin((n1)arctan(2))=(3)n+12sin{(n1)arctan(2))wehaveAn=anA+bnI=(3)n2sin(narctan(2))(1211)+(3)n+12sin{(n1)arctan(2))(1001)

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