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)


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
Question Number 74794 by mathmax by abdo last updated on 30/Nov/19

$l e t A = (\begin{matrix} 1 2 \\ - 1 1 \end{matrix})$ $1) c a l c u l a t e A^{n}$ $2) f i n d e^{A} a n d e^{- A}$ $3) c a l c u l a t e c o s A a n d s i n A$ $4) c a l c u l a t e c h (A) a n d s h (A)$
Commented by Cmr 237 last updated on 01/Dec/19
$1) calcul de A^n on a le polynome caracteristique est P(λ)=det(A−λI_2 ) =λ^2 −2λ+3 deplus on a: P(λ)=0⇒λ_1 =1−i(√2),λ_2 =1+i(√2) on remarque que P(A)=0 donc ∀X∈K[X] on a: X^n =Q(X)P(X)+a_n X+b_n ou Q(X) est un polynome quotient .. λ_1 ^n =Q(λ_1 )P(λ_1 )+a_n λ_1 +b_n λ_2 ^n =Q(λ_2 )P(λ_2 )+a_n λ_2 +b_n { ((a_n λ_1 +b_n =λ_(1 ) ^n )),((a_n λ_2 +b_n =λ_2 ^n )) :} a_n =((λ_2 ^n −λ_1 ^n )/(λ_2 −λ_1 ))=(((1+i(√2))^n −(1−i(√2))^n )/(2i(√2))) b_n =((3(λ_1 ^(n−1) −λ_2 ^(n−1) ))/(2i(√2))) A^n =Q(A)P(A)+a_n A+b_n I_2 =a_n A+b_n I_2 A^n = (((a_n 2a_n )),((−a_n a_n )) )+ (((b_n 0)),((0 b_n )) ) = (((a_n +b_(n ) 2a_n )),((−a_(n ) a_n +b_n )) ) ou a_(n ) et b_n sont les suites donnees en haut 2) calcul de e^A et e^(−A) e^A =Σ_(k=0) ^n (A^k /(k!)) e^(−A) =Σ_(k=0) ^n (A^(−k) /(k!)) 3)calcul de cosA et sinA cosA=Σ_(k=0) ^n (((−1)^k A^(2k) )/((2k)!)) sinA=Σ_(k=0) ^n (((−1)^k A^(2k+1) )/((2k+1)!)) 4) calcul de chA et shA chA=Σ_(k=0) ^n (A^(2k) /((2k)!)) shA=Σ_(k=0) ^n (A^(2k+1) /((2k+1)!)) NB:e^A ,e^(−A) ,cosA,sinA,chA,shA sont sur la forme generale car ∀ n∈N^∗ ,A^n ≠0$
$1) c a l c u l d e A^{n}$ $o n a l e p o l y n o m e c a r a c t e r i s t i q u e e s t$ $P (λ) = d e t (A - λ I_{2})$ $= λ^{2} - 2 λ + 3$ $d e p l u s o n a :$ $P (λ) = 0 \Rightarrow λ_{1} = 1 - i \sqrt{2}, λ_{2} = 1 + i \sqrt{2}$ $o n r e m a r q u e q u e P (A) = 0$ $d o n c \forall X \in K [X] o n a :$ $X^{n} = Q (X) P (X) + a_{n} X + b_{n}$ $o u Q (X) e s t u n p o l y n o m e q u o t i e n t . .$ $λ_{1}^{n} = Q (λ_{1}) P (λ_{1}) + a_{n} λ_{1} + b_{n}$ $λ_{2}^{n} = Q (λ_{2}) P (λ_{2}) + a_{n} λ_{2} + b_{n}$ ${\begin{cases} a_{n} λ_{1} + b_{n} = λ_{1}^{n} \\ a_{n} λ_{2} + b_{n} = λ_{2}^{n} \end{cases}$ $a_{n} = \frac{λ_{2}^{n} - λ_{1}^{n}}{λ_{2} - λ_{1}} = \frac{{(1 + i \sqrt{2})}^{n} - {(1 - i \sqrt{2})}^{n}}{2 i \sqrt{2}}$ $b_{n} = \frac{3 (λ_{1}^{n - 1} - λ_{2}^{n - 1})}{2 i \sqrt{2}}$ $A^{n} = Q (A) P (A) + a_{n} A + b_{n} I_{2}$ $= a_{n} A + b_{n} I_{2}$ $A^{n} = (\begin{matrix} a_{n} 2 a_{n} \\ - a_{n} a_{n} \end{matrix}) + (\begin{matrix} b_{n} 0 \\ 0 b_{n} \end{matrix})$ $= (\begin{matrix} a_{n} + b_{n} 2 a_{n} \\ - a_{n} a_{n} + b_{n} \end{matrix}) o u a_{n} e t b_{n} s o n t l e s s u i t e s d o n n e e s e n h a u t$ $2) c a l c u l d e e^{A} e t e^{- A}$ $e^{A} = \underset{k = 0}{\sum^{n}} \frac{A^{k}}{k!}$ $e^{- A} = \underset{k = 0}{\sum^{n}} \frac{A^{- k}}{k!}$ $3) c a l c u l d e c o s A e t s i n A$ $c o s A = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} A^{2 k}}{(2 k)!}$ $s i n A = \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} A^{2 k + 1}}{(2 k + 1)!}$ $4) c a l c u l d e c h A e t s h A$ $c h A = \underset{k = 0}{\sum^{n}} \frac{A^{2 k}}{(2 k)!}$ $s h A = \underset{k = 0}{\sum^{n}} \frac{A^{2 k + 1}}{(2 k + 1)!}$ $N B : e^{A}, e^{- A}, c o s A, s i n A, c h A, s h A$ $s o n t s u r l a f o r m e g e n e r a l e c a r \forall n \in N^{*}, A^{n} \neq 0$
Commented by mathmax by abdo last updated on 01/Dec/19
$1)we have P_c (x)=det(A−xI)= determinant (((1−x 2)),((−1 1−x))) =(1−x)^2 +2 =x^2 −2x +3 Δ^′ =(−1)^2 −3 =1−3=−2 ⇒λ_1 =1+i(√2) and λ_2 =1−i(√2) cayley hamilton give P_c (A)=0 let divide x^n by x^2 −2x+3 ⇒ x^n =q(x)P_c (x)+a_n x +b_n ⇒λ_1 ^n =a_n λ_1 +b_n λ_2 ^n =a_n λ_2 +b_n so we get the systeme { ((λ_1 a_n +b_n =λ_1 ^n )),((λ_2 a_n +b_n =λ_2 ^n )) :} ⇒(λ_1 −λ_2 )a_n =λ_1 ^n −λ_2 ^n ⇒a_n =((λ_1 ^n −(λ_1 ^− )^n )/(λ_1 −λ_2 )) =((2i Im(λ_1 ^n ))/(2i(√2))) =(1/(√2)) Im(λ_1 ^n ) we have ∣λ_1 ∣ =(√(1+2))=(√3) ⇒λ_1 =(√3)e^(iarctan((√2))) ⇒ a_n =((((√3))^n )/(√2)) sin(narctan((√2)) and b_n =λ_1 ^n −λ_1 a_n =λ_1 ^n −λ_1 ((λ_1 ^n −λ_2 ^n )/(λ_1 −λ_2 )) =((λ_1 ^(n+1) −λ_2 λ_1 ^n −λ_1 ^(n+1) +λ_1 λ_2 ^n )/(λ_1 −λ_2 )) =((λ1λ_2 )/(λ_1 −λ_2 )){ λ_1 ^(n−1) −(λ_1 ^− )^(n−1) } =(3/(2i (√2))) (2i ((√3))^(n−1) sin((n−1)arctan((√2))) =((((√3))^(n+1) )/(√2)) sin{(n−1)arctan((√2))) we have A^n =a_n A +b_n I =((((√3))^n )/(√2))sin(narctan((√2))) (((1 2)),((−1 1)) ) +((((√3))^(n+1) )/(√2))sin{(n−1)arctan((√2))) (((1 0)),((0 1)) )$
$1) w e h a v e P_{c} (x) = d e t (A - x I) = \| \begin{matrix} 1 - x 2 \\ - 1 1 - x \end{matrix} \|$ $= {(1 - x)}^{2} + 2 = x^{2} - 2 x + 3$ $Δ^{^{'}} = {(- 1)}^{2} - 3 = 1 - 3 = - 2 \Rightarrow λ_{1} = 1 + i \sqrt{2} a n d λ_{2} = 1 - i \sqrt{2}$ $c a y l e y h a m i l t o n g i v e P_{c} (A) = 0 l e t d i v i d e x^{n} b y x^{2} - 2 x + 3 \Rightarrow$ $x^{n} = q (x) P_{c} (x) + a_{n} x + b_{n} \Rightarrow λ_{1}^{n} = a_{n} λ_{1} + b_{n}$ $λ_{2}^{n} = a_{n} λ_{2} + b_{n} s o w e g e t t h e s y s t e m e {\begin{cases} λ_{1} a_{n} + b_{n} = λ_{1}^{n} \\ λ_{2} a_{n} + b_{n} = λ_{2}^{n} \end{cases}$ $\Rightarrow (λ_{1} - λ_{2}) a_{n} = λ_{1}^{n} - λ_{2}^{n} \Rightarrow a_{n} = \frac{λ_{1}^{n} - {({\bar{λ}}_{1})}^{n}}{λ_{1} - λ_{2}} = \frac{2 i I m (λ_{1}^{n})}{2 i \sqrt{2}}$ $= \frac{1}{\sqrt{2}} I m (λ_{1}^{n}) w e h a v e ∣ λ_{1} ∣ = \sqrt{1 + 2} = \sqrt{3} \Rightarrow λ_{1} = \sqrt{3} e^{i a r c t a n (\sqrt{2})} \Rightarrow$ $a_{n} = \frac{{(\sqrt{3})}^{n}}{\sqrt{2}} s i n (n a r c t a n (\sqrt{2}) a n d b_{n} = λ_{1}^{n} - λ_{1} a_{n}$ $= λ_{1}^{n} - λ_{1} \frac{λ_{1}^{n} - λ_{2}^{n}}{λ_{1} - λ_{2}} = \frac{λ_{1}^{n + 1} - λ_{2} λ_{1}^{n} - λ_{1}^{n + 1} + λ_{1} λ_{2}^{n}}{λ_{1} - λ_{2}}$ $= \frac{λ 1 λ_{2}}{λ_{1} - λ_{2}} {λ_{1}^{n - 1} - {({\bar{λ}}_{1})}^{n - 1}} = \frac{3}{2 i \sqrt{2}} (2 i {(\sqrt{3})}^{n - 1} s i n ((n - 1) a r c t a n (\sqrt{2}))$ $= \frac{{(\sqrt{3})}^{n + 1}}{\sqrt{2}} s i n {(n - 1) a r c t a n (\sqrt{2}))$ $w e h a v e A^{n} = a_{n} A + b_{n} I$ $= \frac{{(\sqrt{3})}^{n}}{\sqrt{2}} s i n (n a r c t a n (\sqrt{2})) (\begin{matrix} 1 2 \\ - 1 1 \end{matrix}) + \frac{{(\sqrt{3})}^{n + 1}}{\sqrt{2}} s i n {(n - 1) a r c t a n (\sqrt{2})) (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum