TP Maple: Séries Numériques - Correction [PDF]

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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)