25 0 206KB
TP 04 Maple et séries numériques CPGE - Laayoune Essaidi Ali www.mathlaayoune.webs.com MP1-2011-2012 Exercice 01 : Calculer les sommes suivantes : CN
1:
CN
1
>n
n=1
k = 2, 3, 4 ; 2 :
k
5:
>
n=1
CN
>
n C1
n=1
n K1
K1 n
k
CN
1 k = 1, 2, 3 ; 3 : ; 4: n! n=0
>
CN
> n 1n! ;
n=0
1 n C2 ... n C10
. CN
6:
CN
1
CN
1
> n K3 n C2 ; 7 : > arctan n C n C1 ; 8 : >
n=3 CN
2
1
n=0 CN
> n C1 ; 10 : > arctan
n=0
2
n=3
2
in
e
n=1
α=
α
n
k
sin n n
k = 1, 2 ; 9 :
k
1 , 1, 2 2
. O restart; 1 O Sum 2 , n = 1 ..CN n
1
= sum
, n = 1 ..CN ;
2
n N
> n1
n=1
O Sum
1 3
, n = 1 ..CN
= sum
n
= evalf sum
=
2
1 3
1 2 π 6
, n = 1 ..CN ; Sum
n 1 3
, n = 1 ..CN
;
N
n=1
3
1 3
n
n
> n1
(1.1)
=ζ 3
, n = 1 ..CN
N
> n1
1
O Sum
4
, n = 1 ..CN
= 1.202056903
3
n=1
1
= sum
4
n
, n = 1 ..CN ;
n N
> n1
K1
=
4
n=1
O Sum
1 4 π 90
n K1
, n = 1 ..CN
= sum
N
>
K1 n
= ln 2
n
n K1
, n = 1 ..CN
2
= sum
n
>
n K1
, n = 1 ..CN
3
= sum
n
2
, n = 1 ..CN ;
(1.5)
n K1
K1
3
, n = 1 ..CN ;
n n K1
K1
Sum
n K1
K1
K1 n K 1 1 2 = π 2 12 n
n=1
O Sum
(1.4)
n N
K1
, n = 1 ..CN ;
n K1
K1
n=1
K1
(1.3) n K1
n
O Sum
(1.2)
3
, n = 1 ..CN
= evalf sum
3
n N
, n = 1 ..C
n
; N
>
n=1
N
>
n=1
O Sum
n K1
K1
1 , n = 0 ..CN n!
3 K1 n K 1 = ζ 3 3 4 n
K1 n K 1 = 0.9015426772 n3
= sum
(1.6)
1 , n = 0 ..CN ; n! N
> n!1 =e
(1.7)
n=0
O Sum
1 , n = 1 ..CN n$n! ..CN
= evalf sum N
= sum
1 1 , n = 1 ..CN ; Sum , n=1 n$n! n$n!
1 , n = 1 ..CN n$n!
> n 1n! = KγKI π KEi 1, K1
n=1
;
N
> n 1n! = 1.317902151 C0. I
(1.8)
n=1
1 , n = 1 ..CN Product n Ck, k = 1 ..10 1 = sum , n = 1 ..CN ; product n Ck, k = 1 ..10
O Sum
N
1
>
n=1
1 32659200
=
10
? n Ck
(1.9)
k= 1
1
O Sum
, n = 3 ..CN
2
n K3$n C2
1
= sum
, n = 3 ..CN ;
2
n K3$n C2
N
> n K31 n C2 = 1
1
O Sum arctan
(1.10)
2
n=3
, n = 0 ..CN
2
n Cn C1 = evalf sum arctan
1
, n = 0 ..CN
2
;
n Cn C1
N
> arctan
n=0
1 n Cn C1 2
= 1.570796327
(1.11)
sin n sin n , n = 1 ..CN = sum , n = 1 ..CN ; n n sin n sin n Sum , n = 1 ..CN = evalf sum , n = 1 ..CN n n
O Sum
N
> sinn n
=
n=1
1 sin 1 arctan 2 1 Kcos 1
K
;
1 sin 1 arctan 2 K1 Ccos 1
N
> sinn n
= 1.070796327
(1.12)
n=1
O Sum
2
sin n
, n = 1 ..CN
2
= sum
n
2
n N
;
2
2
, n = 1 ..CN ;
n
sin n
Sum
sin n
2
, n = 1 ..CN
= evalf sum
sin n 2
n
2
, n = 1 ..C
N
sin n n=1 n2
>
2
1 1 2 1 polylog 2, e2 I C π K polylog 2, eK2 I 4 12 4
=K
N
sin n n=1 n2
>
1
O Sum
, n = 1 ..CN
2
2
= 1.070796327 C0. I 1
= sum
..CN
, n = 1 ..CN ; Sum
2
n C1
(1.13)
n C1 1
= evalf sum
, n=1
n C1
, n = 1 ..CN
2
1 2
;
n C1
N
> n 1C1 = 12 π coth π
n=1
K
2
1 2
N
> n 1C1 = 1.076674048
exp I$n
O Sum
, n = 1 ..CN
n exp I$n
Sum
(1.14)
2
n=1
= sum
exp I$n
, n = 1 ..CN ;
n , n = 1 ..CN
= evalf sum
exp I$n
n N
, n = 1 ..C
n
; N
>
n=1
N
>
n=1
eI n
eI n n
N
=
eI n
>
n=1
n
= K0.1941089351 C1.043982103 I
(1.15)
n
exp I$n exp I$n , n = 1 ..CN = sum , n = 1 ..CN ; n n exp I$n exp I$n Sum , n = 1 ..CN = evalc sum , n = 1 ..C n n
O Sum
N
; N
eI n = Kln 1 KeI n n=1
>
N
eI n 1 = K ln 2 n=1 n
>
O Sum
exp I$n 2
n
1 Kcos 1
, n = 1 ..CN
2
Csin 1
= sum
2
CI arctan
exp I$n 2
n
sin 1 1 Kcos 1
, n = 1 ..CN ;
(1.16)
Sum
exp I$n
, n = 1 ..CN
2
exp I$n
= evalf sum
2
n N
, n = 1 ..C
n
; N
eI n
>n
2
n=1
= polylog 2, eI
N
eI n = 0.3241377401 C1.013959132 I 2 n=1 n
>
(1.17)
Exercice 02 : Donner un développement asymptotique des sommes : n
1:
CN
1
>
k= 1
>k
; 2:
1 6: ; 7: k= 1 k
>
;
5
k= n
k
n
CN
1
CN
3:
;
k
k= n
>2
4:
k= 1 CN
n
1
>
>
n
k K1
K1
>
k ; 8: ; 2 k = n C1 k k = 1 k C1
9:
>
k= n
1 k K1
n
; 5:
>
k.
k= 1
1 2
k Ck C 1
n
> ln k Cn .
; 10 :
k= 1
O restart; 1 O Sum , k = 1 ..n k n
>
k= 1
O Sum
1
1
= asympt sum
, k = 1 ..n , n, 1 ;
k 2
=
1 n
k
1
, k = n ..CN 5 k
Cζ
1 2
1 2
C
1
= asympt sum
5
1 1 CO n n
, k = n ..CN , n, 5 ;
k N
> k1
k= n
5
=
1 1 CO 5 4 4n n
k K1
O Sum
K1 k
(2.1)
(2.2) k K1
, k = n ..CN
= asympt sum
K1 k
, k = n ..CN , n,
2 ; N
>
k= n
kK1
K1 k
1 =K 2
K1 n
n
CO
1 2 n
(2.3)
1
O Sum
, k = 1 ..n
1
= asympt sum
k K1
2$
, k = 1 ..n , n,
2$ k K1
1 ; n
1
>2
k K1
k= 1
O Sum
k , k = 1 ..n n
>
k= 1
O Sum
2 1 n
k = 3
1 , k = 1 ..n k
1
C
3/2
k= 1
2
, k = n C1 ..N
(2.4)
k , k = 1 ..n , n, 1 ; 1 2
= asympt sum CγC
CO 1
1 n
Cζ K
1 n
2
> 1k = ln n
1
1
= asympt sum
n
O Sum
=
C
1 24
1 1 CO n n
(2.5)
1 , k = 1 ..n , n, 5 ; k
1 1 1 1 K C CO 5 2 4 2n 12 n 120 n n
= asympt sum
k
1
(2.6)
, k = 1 ..n , n, 5 ;
2
k N
1 1 2 1 1 1 1 = π K C K CO 5 2 2 3 6 n k= n C1 k 2n 6n n
>
O Sum
k , k = 1 ..n k C1
k = n C1 Kln n > k C1
KγK
k= 1
1
O Sum
, k = n ..N
2
k , k = 1 ..n , n, 5 ; k C1
= asympt sum
n
(2.7)
3 13 1 119 1 C K 3 C CO 5 2 4 2n 12 n n 120 n n 1
= asympt sum
, k = n ..CN ,
2
k Ck C1
(2.8)
k Ck C1
n, 5 ; N
> k Ck1 C1 = 1n C n1
k= n
2
2
C
1 I 2
3
K
1 I 2
3
C
1 12
1 CI 3
C
1 I 3
K
1 I 3 3
C
3
1 1 K I 6 4
1 1 K I 3 3 3 n 2
1 1 C I 3 2 2
1 1 K I 2 2
1 1 1 C I 3 C 6 4 2 3 C
1 2
1 1 C I 3 2 2
K
1 1 C I 2 2
1 2
K
C
1 I 3
3
2
3 1 2
K
C
1 2
3
K
(2.9)
1 2
1 1 C I 3 12 12
1 1 K I 2 2
2
3
C
1 12
K
1 I 12
C
1 n4
1 I 3
K
1 12
K
K
1 I 3
3
K
1 I 2
3
3 C
1 12
K1 CI 1 8
3
2
2
K
3
1 CI 1 32
1 32
C 3
2
K1 CI
3
3
3 3
K
1 24
1 1 K I 2 2 1 24
K
1 1 C I 3 2 2
C
1 CI 1 120
K1 CI 3
1 120
C 2
3
2
3
K
1 12
1 2
K
1 n5
CO
= asympt sum ln k Cn , k = 1 ..n , n, 5 ;
n
>ln k Cn
3
1 1 K I 3 2 2
1 1 C I 2 2
1 CI 3
O Sum ln k Cn , k = 1 ..n
1 1 K I 2 2 2
K1 CI 3
1 1 C I 2 2 1 8
2
3
= ln n K1 C2 ln 2
n Cln
k= 1
π K1 K ln 2
e
e
π
2
Kln
K1
e
e
K
1 24 n
(2.10)
1 n3
CO
Exercice 03 : Etudier la convergence des série suivantes : 1:
;
> >
6
6
n C3 n Kn ; 2 :
K1
>
n C1
n C2 ... n Cn
;
n
2n
3:
>
n
ln n ; n!
4:
n K1 n ln nC1
n
n! 3 ; 3n !
> 6 : > sin
5:
2
2
n Cn C1 n C1
π ; 7:
3n 4 nK1
>
2 n C1
; 8:
2 n
> a C1 a C1 ... a C1 ; 9 : > ln ch 1 sin 1 n n ; 10 : > sin nπ C 1 C 1 ; 11 : > cos π n ln n n n K1 n a
2
n
2
2
.
O restart; O asympt
6
6
2
n C3 n Kn, n ; (3 1)
1 1 CO 7 3 2n n
(3.1)
n
O u d n/
?
n Ck
k=1
u n C1 , n = N ; is % ! 1 ; u n
; limit
n
2 n
n
? n Ck
1 u := n/ 2
k= 1
nn
4 eK1 false O u d n/
n
ln n
; limit
u n C1
n!
, n = N ; is % ! 1 ;
u n u := n/
(3.2)
ln n n!
n
0 true O asympt
n
K1 $ n $ln
n K1 , n, 2 ; nC 1
K2 K1 O u d n/
n!
(3.3)
3
; limit
3$ n !
1 CO n
n
u n C1 u n
1 n
3/2
(3.4)
, n = N ; is % ! 1 ;
n!3 u := n/ 3n ! 1 27 true
(3.5)
2
n Cn C1 O S d asympt $π, n, 5 ; n C1 π π π π 1 S := π n C K 2 C 3 K 4 CO 5 n n n n n O asympt
(3.6)
n
K1 $ sin S Kn$π , n, 2 ; K1 n π 1 CO 2 n n
3$n O u d n/ 4$n K1
2$n C 1
; limit
u n C1 u n
(3.7)
, n = N ; is % ! 1 ;
u := n/
3n 4 n K1 9 16 true
2 n C1
(3.8)
n2
a
O u d n/ product
; simplify
k
a C1 , k = 1 ..n
u n C1 u n
;
2
u := n/
an n
? a C1
k
k= 1 2 n C1
a
a C1
Kn K 1
(3.9)
2
O S d solve a ! abs a C1 , a ; 1 1 S := RealRange Open K 2 2 1 1 $sin n n
O asympt ln cosh
5 , Open
1 1 C 2 2
5
(3.10)
,n ;
1 1 CO 6 4 2n n O asympt
(3.11)
1 1 C 2 , n, 2 ; n n K1 n 1 CO 2 n n
n
K1 $sin
(3.12)
n , n, 4 ; n K1 1 π 1 1 CO 2 T := π n C π C 3 n 2 n
2
O T d asympt π$n $ln
O asympt
(3.13)
n
K1 $cos T Kn$π , n, 4 ; n 1 K1 π 1 K CO 2 3 n n
(3.14)
Exercice 04 : Etudier la série ∑ sin
k
k
n Cn C1 pour k 2 { 2, 3, 4, 5, 6, 7, 8, 9, 10 }.
O restart; O k d 2; A d asympt n ;
k
k
n Cn C1 $π, n, 2 ; asympt
n
K1 $sin A Kn$π ,
k := 2 A := n π C K1 O k d 3; A d asympt
k
n
1 3 π 1 πC CO 2 2 8 n n n 2
π
K1
9 K 128
2
CO
n
1 n3
k
n Cn C1 $π, n, 2 ; asympt
(4.1) n
K1 $sin A Kn$π ,
n ; k := 3 1 π 1 A := n π C CO 2 3 n n 1 3 O k d 4; A d asympt
k
K1 n π 1 CO 2 n n
k
n Cn C1 $π, n, 2 ; asympt
(4.2) n
K1 $sin A Kn$π ,
n ; k := 4 A := n π CO O O k d 5; A d asympt
k
1 n2
1 n2
(4.3)
k
n Cn C1 $π, n, 2 ; asympt
n
K1 $sin A Kn$π ,
n ; k := 5 A := n π CO O O k d 6; A d asympt
k
1 n3
1 n3
(4.4)
k
n Cn C1 $π, n, 2 ; asympt
n
K1 $sin A Kn$π ,
n ; k := 6 A := n π CO O O k d 7; A d asympt
k
1 n4
1 4 n
k
n Cn C1 $π, n, 2 ; asympt
n ; k := 7
(4.5) n
K1 $sin A Kn$π ,
A := n π CO O O k d 8; A d asympt
k
1 n5
1 n5
(4.6)
k
n Cn C1 $π, n, 2 ; asympt
n
K1 $sin A Kn$π ,
n ; k := 8 A := n π CO O O k d 9; A d asympt
k
1 n6
1 n6
(4.7)
k
n Cn C1 $π, n, 2 ; asympt
n
K1 $sin A Kn$π ,
n ; k := 9 A := n π CO O O k d 10; A d asympt
k
1 n7
1 n6
(4.8)
k
n Cn C1 $π, n, 2 ; asympt
n
K1 $sin A Kn
$π , n ; k := 10 A := n π CO O
1 n8
1 6 n
(4.9)
Exercice 05 : Déterminer les réels α et β pour que la série converge. Calculer, dans ce cas, sa somme.
>
n Cα n C1 C β n C2
O restart; O ud
n Cα$ n C1 C β$ n C2 ; u := n Cα n C1 Cβ
n C2
(5.1)
O asympt u, n, 1 ; (5.2)
1 Cα Cβ 1 n O S d solve
1 Cα Cβ,
1 α Cβ 2
C
1 CO n
3/2
1 n
(5.2)
1 $α Cβ , α, β ; 2 S := α = K2, β = 1
(5.3)
n K2
(5.4)
O assign S ; O u; O evalf sum u, n = 0 ..N
n C1 C n C2
; K1.000000000
(5.5)
Exercice 06 : Déterminer les réels a et b pour que la série convergente.
>cos
π
3
3
2
n Can Cbn
soit
O restart; O u d π$
3
3
2
n Ca$n Cb$n ; u := π n3 Ca n2 Cb n
1/3
(6.1)
O v d asympt u, n, 2 ; 1 v := π n C πaC 3 O w d asympt
π
1 1 2 bK a 3 9 n
1 $π$ a , a 3
(6.2)
O assign S ; O a d a C3$ k;
1 1 2 bK a, b 3 9
1 1 b K a2 3 9
CO
1 2 n
(6.3)
;
S := a =
O solve
1 n2
n
K1 $cos v Kπ$n , n, 2 ; 1 K1 n sin πa π 1 3 n w := K1 cos πa K 3 n
O S d solve cos
CO
3 2
a :=
3 C3 k 2
3 4
1 C2 k
(6.4)
(6.5)
; b=
2
(6.6)
Exercice 07 : Déterminer les réels a et b pour que la série
>
n C1 n K2
n
- a 1C
b n
soit
convergente. O restart; O ud
n C1 n K2
n
Ka$ 1 C
b ; n
u :=
n
n C1 n K2
Ka 1 C
b n
(7.1)
O v d asympt u, n, 3 ;
O S d solve
3 3 e Ka b 2 1 3 v := e Ka C CO 2 n n 3 3 e3 Ka, e Ka $b , a, b ; 2 3 S := a = e3, b = 2
(7.2)
(7.3)
Exercice 08 : Déterminer les réels a et b pour que la série
>
2
1 n Ca cos n n2 Cb
converge le plus
rapidement possible. O restart; O u d cos
1 n
2
n Ca
K
2
;
n Cb u := cos
1 n
K
n2 Ca n2 Cb
(8.1)
O v d asympt u, n, 7 ; 1 1 1 Ka Cb K K Ka Cb b C K a Kb b2 K 2 24 720 1 v := C C CO 7 2 4 6 n n n n O S d solve
1 1 Ka Cb, K Ka Cb $b , a, b 24 2 5 1 S := a = K , b = 12 12
K
(8.2)
; (8.3)
O assign S ; O asympt u, n, 7 ; (8 4)
1 1 CO 7 6 480 n n
(8.4)
Exercice 09 : Déterminer les réels a,b,c,d et e pour que la série
>
4
2
1 a$n Cb$n C c sin n n5 Cd$n3 C e$n
converge le plus rapidement possible. O restart; O u d sin
1 n
4
K
2
a$n Cb$n Cc 5
3
n Cd$n Ce$n 1 u := sin n
; K
a n4 Cb n2 Cc n5 Cd n3 Ce n
O v d asympt u, n, 10 ; 1 1 K Kb Ca d Kc Ca e K Kb Ca d d C 1 Ka 6 120 v := C C 3 5 n n n 1 K Kb Ca d e K Kc Ca e Cd b Ka d2 d K 5040 C 7 n 1 K Kc Ca e Cd b Ka d2 e K e b K2 e a d Cd c Kd2 b Ca d3 d 362880 C 9 n 1 CO 10 n O p d convert v, polynom ; 1 1 K Kb Ca d Kc Ca e K Kb Ca d d C 1 Ka 6 120 p := C C 3 5 n n n 1 K Kb Ca d e K Kc Ca e Cd b Ka d2 d K 5040 C 7 n 1 K Kc Ca e Cd b Ka d2 e K e b K2 e a d Cd c Kd2 b Ca d3 d 362880 C n9 O E d coeffs p, n ; 1 1 E := 1 Ka, K Kb Ca d, K Kc Ca e Cd b Ka d2 e K e b K2 e a d Cd c 6 362880 1 Kd2 b Ca d3 d, K Kb Ca d e K Kc Ca e Cd b Ka d2 d K , Kc Ca e K 5040
(9.1)
(9.2)
(9.3)
(9.4)
1 120
Kb Ca d d C
O S d solve E, a, b, c, d, e ; 53 551 13 5 S := a = 1, b = K ,c= ,d= ,e= 396 166320 396 11088
(9.5)
O assign S ; O asympt u, n, 12 ; 11 1 K CO 12 11 457228800 n n
(9.6)
Exercice 10 : Soient a,b,c,d 2 = et on considère la série
>
a
sin
c
Ksin
n Cb n Cd 1. Donner une condition nécessaire et suffisante de convergence. 2. Dans ce cas, déterminer a, b, c et d pour que la série converge le plus vite possible.
.
O restart; O u d sin
a
c
Ktan
n Cb
;
n Cd a
u := sin
n Cb
Ktan
c
O v d asympt u, n, 2 ; 1 1 1 1 1 3 v := a Kc C K a b K a3 C c dK c n 2 6 2 3 O assign a = c ; O asympt u, n, 3 ; 1 1 3 1 K c bK c C cd 2 2 2 C
1 3 c d 2
1 n
5/2
CO
1 n
3/2
1 3 c d 2
1 n
1 n
3/2
CO
1 2 n
1 3 3 1 5 3 c bC c b2 K c K c d2 4 8 8 8
(10.2)
(10.3)
1 3 n
O p d convert %, polynom ; 1 1 3 1 1 p := K c b K c C cd 2 2 2 n C
C
(10.1)
n Cd
3/2
C
1 3 3 1 5 3 2 2 c bC cb K c K cd 4 8 8 8
(10.4)
5/2
O E d seq coeff p, 1 / n ^ 2 * k C1 / 2 , k = 1 ..2 ; 1 1 3 1 1 3 3 1 5 3 1 3 2 2 E := K c b K c C c d, c bC cb K c K cd C c d 2 2 2 4 8 8 8 2
(10.5)
O s d solve
E , b, c, d ; s := b = b, c = 0, d = d , b = Kc2 Cd, c = c, d = d
O assign s 2 ; O asympt u, n, 4 ; simplify % ; 1 1 3 1 2 1 1 cdK c Cc c K d 2 2 2 2 n 1 3 K c 2 K
1 2 1 3 c K d K c 2 2 4
3/2
5 1 2 1 c c K d 8 2 2 1 CO 4 n
Kc2 Cd
1 30
3 1 3 1 c d2 C c d K c5 8 2 8
C K
1 2 1 c K d 2 2
5 3 2 1 5 13 7 1 3 c d C c dK c K c 8 3 240 2
C
2
c7
K
1 n
2
Kc Cd
1 2 1 c K d 2 2
3 3 c 8
C
(10.6)
2
C
1 2 1 c K d 2 2
1 1 K30 O 4 n n 3 n
1 5 c 24
5/2
5 c d3 16
C
1 2 1 c K d 2 2 1 n
Kc2 Cd
7/2
n3 (10.7)
Exercice 10 : Soient a,b 2 =. Etudier la nature de la série
>
2 sin π n Can Cb
e
K1 .
O restart; O u d π$ n2 Ca$n Cb ; n2 Ca n Cb
u := π
(11.1)
O v d asympt u, n, 2 ; v := π n C O w d asympt
1 πaC 2
π
1 1 2 bK a 2 8 n
CO
1 n2
(11.2)
n
K1 $sin v Kπ$n , n, 2 ; 1 K1 n cos πa π 1 2 n w := K1 sin πa C 2 n
1 1 b K a2 2 8
1 n2
CO
(11.3)
O z d asympt exp w K1, n, 2 ;
z := e
1 K1 n sin πa 2
CO
1 n2
K1 K
1 8
e
1 K1 n sin πa 2
cos
1 π a π K4 b Ca2 2 n
K1
n
(11.4)
O z d subs
e
1 K1 n sin πa 2
=1 , z ; 1 cos π a π K4 b Ca2 1 2 z := K 8 n
n
K1
CO
1 n2
C
K1 na
(11.5)
Exercice 11 :
>
Soit a > 0. Etudier la nature de la série
3
arccos
2
π 6
n
K
.
O restart; assume x O 0 ; O u d cosK1
π 3 Cx K ; 2 6 u := arccos
1 2
3 Cx~ K
O v d series u, x = 0, 3 ; v := K2 x~K2 O subs x =
K1 na
1 π 6
(12.1)
3 x~2 CO x~3
(12.2)
n
,v ; 2 K1 na
K
n
K
2
3
K1
n 2
K1
CO
a 2
n 3
(12.3)
a 3
n
n
Exercice 12 : Calculer l'exponentiel de chaqu'une des matrices suivantes et vérifier que det exp(A) = exp( tr(A)) : 1 0 0 0 1 K1 0 2 K1 Ad
3 K2 0 K2 2
1
; B d K1 3 0 ; C d 0 0
;
0 K1 0
8 2 4
O restart; with LinearAlgebra : O A d Matrix 0 , 2 , K1 , 3 , K2 , 0 ,
K2 , 2 , 1 0 2 K1
A :=
1
3 K2 K2
2
0
; (13.1)
1
O expA d MatrixExponential A ;
(13.2)
expA :=
1 3
1 C2 e6 eK4
1 2
K1 Ce6 eK4
1 3
K
2 5
e5 K1 eK4
1 5
10 e6 K9 e5 K1 eK4
1 10
5 e6 K6 e5 C1 eK4
3 C2 e5 eK4 K
2 5
K1 Ce6 eK4
1 15
K
1 15
e5 K1 eK4
(13.2)
1 C5 e6 C9 e5 eK4
O simplify Determinant expA ; exp Trace A ; eK1 1,0,0 ,
O B d Matrix
eK1 K1 , 3 , 0 , 8 , 2 , 4 1 0 0 B :=
(13.3) ;
K1 3 0
(13.4)
8 2 4 O expB d MatrixExponential B ; e expB :=
1 3 1 e C e 2 2
K
0
0
e3
0
(13.5)
e3 K3 e C2 e4 K2 e3 C2 e4 e4 O simplify Determinant expB ; exp Trace B ; e8 O C d Matrix
e8 0, 1,K1 , 0, 0, 1 , 0,K1, 0 ; 0 1 K1 C :=
0
0
1
0 K1
0
(13.6)
(13.7)
O expC d MatrixExponential C ; 1 sin 1 Kcos 1 C1 Kcos 1 C1 Ksin 1 expC :=
0
cos 1
sin 1
0
Ksin 1
cos 1
O simplify Determinant expC ; exp Trace C ; 1 1 O
(13.8)
(13.9)