tp maple : séries numériques - correction

8/10/2019 Tp Maple : Sries numriques - correction

1/18

(1.1)(1.1)

OO

OO

TP 04Maple et sries numriques

CPGE - Laayoune

Essaidi Aliwww.mathlaayoune.webs.com

MP1-2011-2012

Exercice 01 :

Calculer les sommes suivantes :

1 :>n = 1

CN

1n

k k= 2, 3, 4 ; 2 :

>n = 1

CN

1

nK1

nk

k= 1, 2, 3 ; 3 :>n = 0

CN

1n!

; 4 :>n = 0

CN

1nn!

;

5 : >n = 1

CN

1

nC1 nC2 ... nC10

.

6 :>n = 3

CN

1

n2K3 nC2

; 7 :>n = 0

CN

arctan1

n2CnC1

; 8 :>n = 1

CN

sinkn

nk

k= 1, 2 ; 9 :

>n = 0

CN

1

n2C1

; 10 :>n = 3

CN

arctanein

n

= 12

, 1 , 2

.

restart;

Sum1

n2

, n = 1 . . CN = sum1

n2

, n = 1 . . CN ;

>n = 1

N

1

n2

=1

6

2

Sum1

n3

, n =1 . . CN = sum1

n3

, n =1 . . CN ; Sum1

n3

, n = 1 . . CN

= evalf sum1

n3

, n =1 . . CN ;

>n = 1

N

1

n3

= 3


2/18

(1.5)(1.5)

(1.4)(1.4)

OO

(1.2)(1.2)

(1.7)(1.7)

(1.6)(1.6)

(1.3)(1.3)

>n = 1

N

1

n3

= 1.202056903

Sum1

n4

, n =1 . . CN = sum1

n4

, n =1 . . CN ;

>n = 1

N

1n

4 = 190 4

Sum1

nK1

n

, n =1 . . CN = sum1

nK1

n, n = 1 . . CN ;

>n = 1

N

1 nK1

n= ln 2

Sum1

nK1

n2

, n = 1 . . CN = sum1

nK1

n2

, n =1 . . CN ;

>n = 1

N

1 nK1

n2

=1

12

2

Sum1

nK1

n3

, n = 1 . . CN = sum1

nK1

n3

, n =1 . . CN ;

Sum1

nK1

n3

, n =1 . . CN = evalf sum1

nK1

n3

, n =1 . . C

N ;

>n = 1

N

1 nK1

n3

=3

4 3

>n = 1

N

1 nK1

n3

= 0.9015426772

Sum1

n!, n = 0 . . CN = sum

1

n!, n = 0 . . CN ;

>n = 0

N

1

n! =e

Sum1

n$n!, n =1 . . CN = sum

1

n$n!, n =1 . . CN ; Sum

1

n$n!, n =1

. . CN = evalf sum1

n$n!, n = 1 . . CN ;

>n = 1

N

1

nn!= KIKEi 1, 1


3/18

OO

(1.11)(1.11)

(1.8)(1.8)

(1.12)(1.12)

OO

(1.10)(1.10)

(1.9)(1.9)

>n = 1

N

1

nn!= 1.317902151C0. I

Sum1

Product nCk, k =1 . . 10, n = 1 . . CN

= sum

1

product nCk, k =1 . . 10 , n =1 . .CN

;

>n = 1

N

1

?k= 1

10

nCk

=1

32659200

Sum1

n2K3$nC2

, n = 3 . . CN = sum1

n2K3$nC2

, n =3 . . CN ;

>n = 3

N

1

n2K3 nC2

= 1

Sum ar ct an1

n2CnC1

, n =0 . . CN

= evalf sum ar ct an1

n2CnC1

, n = 0 . . CN ;

>n = 0

N

arctan1

n2CnC1

= 1.570796327

Sumsi n n

n

, n =1 . . CN = sumsi n n

n

, n = 1 . . CN ;

Sumsi n n

n, n = 1 . . CN = evalf sum

si n n

n, n =1 . . CN ;

>n = 1

N

sin n

n=

1

2arctan

sin 1

1Kcos 1 K

1

2arctan

sin 1

1Ccos 1

>n = 1

N

sin n

n= 1.070796327

Sumsi n n

2

n2

, n = 1 . . CN = sumsi n n

2

n2

, n = 1 . . CN ;

Sumsi n n

2

n2

, n =1 . . CN = evalf sumsi n n

2

n2

, n =1 . . C

N ;


4/18

OO

(1.13)(1.13)

(1.15)(1.15)

(1.16)(1.16)

(1.14)(1.14)

>n = 1

N

sin n2

n2

=1

4polylog 2, e

2 IC

1

12

2K

1

4polylog 2, e

K2 I

>n = 1

N

sin n2

n2

= 1.070796327C0. I

Sum1

n2C1

, n =1 . . CN = sum1

n2C1

, n =1 . . CN ; Sum1

n2C1

, n =1

. . CN = evalf sum1

n2C1

, n = 1 . . CN ;

>n = 1

N

1

n2C1

=1

2coth K

1

2

>n = 1

N

1

n2C1

= 1.076674048

Sum exp I$n

n

, n =1 . . CN = sum exp I$n

n

, n =1 . . CN ;

Sumexp I$n

n

, n =1 . . CN = evalf sumexp I$n

n

, n =1 . . C

N ;

>n = 1

N

eIn

n=>

n = 1

N

eIn

n

>n = 1

N

eIn

n= 0.1941089351C1.043982103 I

Sumexp I$n

n, n =1 . . CN = sum

exp I$n

n, n =1 . . CN ;

Sumexp I$n

n, n =1 . . CN = evalc sum

exp I$n

n, n =1 . . C

N ;

>n = 1

N

e

In

n= ln 1KeI

>n = 1

N

eIn

n=

1

2ln 1Kcos 1

2Csin 1

2CI arctan

sin 1

1Kcos 1

Sumexp I$n

n2

, n =1 . . CN = sumexp I$n

n2

, n =1 . . CN ;


5/18

(2.3)(2.3)

OO

OO

(1.17)(1.17)

(2.1)(2.1)

(2.2)(2.2)

OO

OO

Sumexp I$n

n2

, n =1 . . CN = evalf sumexp I$n

n2

, n =1 . . C

N ;

>n = 1

N

eIn

n2 = polylog 2, e

I

>n = 1

N

eIn

n2

= 0.3241377401C1.013959132 I

Exercice 02 :

Donner un dveloppement asymptotique des sommes :

1 :>k= 1

n

1

k

; 2 : >k= n

CN

1

k5; 3 :>

k= n

CN

1 kK1

k

; 4 :>k= 1

n1

2 k 1

; 5 : >k= 1

n

k.

6 :>k= 1

n

1

k; 7 : >

k= nC1

CN

1

k2

; 8 :>k= 1

nk

kC1; 9 :>

k= n

CN

1

k2CkC1

; 10 : >k= 1

n

ln kCn .

restart;

Sum1

k

, k =1 . . n = asympt sum1

k

, k = 1 . . n , n, 1 ;

>k= 1

n1

k=

2

1

n

C1

2 C

1

2

1

n CO

1

n

Sum1

k5

, k = n . . CN = asympt sum1

k5

, k = n . . CN , n, 5 ;

>k=n

N

1

k5

=1

4 n4CO

1

n5

Sum1

kK1

k, k = n . . CN = asympt sum

1 kK1

k, k = n . . CN , n,

2 ;

>k=n

N

1 kK1

k=

1

2

1 n

n CO

1

n2


6/18


7/18


8/18

(3.3)(3.3)

OO

(3.2)(3.2)

(3.5)(3.5)

(3.4)(3.4)

OO

(3.7)(3.7)

OO

(3.6)(3.6)

(3.1)(3.1)1

2 n3CO

1

n7

ud n/

?k =1

n

nCk

2 nn

; limitu nC1

u n, n =N ; is %!1 ;

u :=n/1

2

?k= 1

n

nCk

nn

4 eK1

false

ud n/l n n

n

n!; limit

u nC1

u n, n =N ; is %!1 ;

u :=n/

ln n n

n!

0

true

asympt 1 n$ n $l n

nK1

nC 1, n, 2 ;

2 1 n

1

n CO

1

n

3/2

ud n/n!

3

3$ n !

; limitu nC1

u n

, n =N ; is %!1 ;

u :=n/n!

3

3 n !

1

27

true

Sd asymptn

2CnC1

nC1$, n, 5 ;

S:=nC

n

K

n2C

n3K

n4CO

1

n5

asympt 1 n$ si n SKn$ , n, 2 ;

1 n

n CO

1

n2

ud n/3$n

4$nK1

2$nC1

; limitu nC1

u n, n =N ; is %!1 ;


9/18

(3.11)(3.11)

(3.13)(3.13)

(3.9)(3.9)

(3.10)(3.10)

(3.12)(3.12)

OO

(3.8)(3.8)

OO

(3.14)(3.14)

u :=n/3 n

4 nK1

2 nC1

9

16

true

ud n/a

n2

product aC1 k

, k =1 . . n; simplify u nC1

u n;

u :=n/a

n2

?k= 1

n

aC1 k

a2 nC1

aC1 KnK1

Sd solve a2!abs aC1 , a ;

S:=RealRange Open

1

2 K

1

2 5 ,Open

1

2 C

1

2 5

asympt l n cosh1

n$si n

1

n, n ;

1

2 n4CO

1

n6

asympt 1 n$si n

1

nC

1

n2

, n, 2 ;

1 n

n

CO1

n2

Td asympt $n2$l n

n

nK1, n, 4 ;

T:=nC1

2C

1

3

n CO

1

n2

asympt 1 n$cos TKn$ , n, 4 ;

1

3

1 n

n CO

1

n2

Exercice 04 :

Etudier la srie sin nkCnC1

kpour k2{ 2, 3, 4, 5, 6, 7, 8, 9, 10 }.

restart;

kd 2; Ad asympt nkCnC1

k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ;


10/18

(4.4)(4.4)

(4.2)(4.2)

(4.3)(4.3)

(4.1)(4.1)

(4.5)(4.5)

k:= 2

A :=nC1

2C

3

8

n CO

1

n2

1 nK

9

128

1 n

2

n2 CO

1

n3


k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ;k:= 3

A :=nC1

3

n CO

1

n2

1

3

1 n

n CO

1

n2

kd 4; Ad asympt nk

CnC1

k

$, n, 2 ; asympt 1

n

$si n AKn$ ,n ;

k:= 4

A :=nCO1

n2

O1

n2


k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ; k:= 5

A :=nCO1

n3

O1

n3


k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ;k:= 6

A :=nCO 1

n4

O1

n4


k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ;k:= 7


11/18

(4.6)(4.6)

(4.9)(4.9)

(4.7)(4.7)

(5.2)(5.2)

(4.8)(4.8)

(5.1)(5.1)

OO

A :=nCO1

n5

O1

n5


k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ;k:= 8

A :=nCO1

n6

O1

n6


k$, n, 2 ; asympt 1

n$si n AKn$ ,

n ;

k:= 9

A :=nCO1

n7

O1

n6


k$, n, 2 ; asympt 1

n$si n AKn

$ , n ;k:= 10

A :=nCO1

n8

O1

n6

Exercice 05 :

Dterminer les rels et pour que la srie> n C nC1 C nC2converge. Calculer, dans ce cas, sa somme.

restart;

ud n C$ nC1 C $ nC2 ;

u := n C nC1 C nC2

asympt u, n, 1 ;


12/18

(6.4)(6.4)

(6.3)(6.3)

(6.5)(6.5)

OO

(5.2)(5.2)

(5.5)(5.5)

(6.1)(6.1)

(5.4)(5.4)

(6.6)(6.6)

(6.2)(6.2)

(5.3)(5.3)

1CC

1

n

C1

2C

1

n CO

1

n

3/2

Sd solve 1CC,1

2$C , , ;

S:= = 2, = 1

assign S ;

u;

n K2 nC1 C nC2

evalf sum u, n =0 . .N ;1.000000000

Exercice 06 :

Dterminer les rels aet bpour que la srie>cos n3Can2Cbn3 soit

convergente.

restart;

ud $ n3Ca$n

2Cb$n

3;

u := n3Can

2Cbn

1/3

vd asympt u, n, 2 ;

v :=nC

1

3 aC

1

3bK

1

9a

2

n C

O

1

n2

wd asympt 1 n$cos vK$n , n, 2 ;

w := 1 n

cos1

3a K

1 n

sin1

3a

1

3bK

1

9a

2

n CO

1

n2

Sd solve cos1

3$$a , a ;

S:= a =3

2

assign S ;

ad aC3$k;

a :=3

2 C3 k

solve 1

3bK

1

9a

2, b ;

b =3

4 1C2 k

2


13/18

(8.2)(8.2)

(7.1)(7.1)

OO

(7.2)(7.2)

(8.3)(8.3)

(7.3)(7.3)

OO

(8.1)(8.1)

Exercice 07 :

Dterminer les rels aet bpour que la srie> nC1nK2

n

- a 1Cb

nsoit

convergente.

restart;

udnC1

nK2

n

Ka$ 1Cb

n;

u := nC1

nK2

n

Ka 1Cb

n

vd asympt u, n, 3 ;

v := e3KaC

3

2e

3Kab

n CO

1

n2

Sd solve e3Ka, 32 e3Ka$b , a, b ;

S:= a = e3,b =

3

2

Exercice 08 :

Dterminer les rels aet bpour que la srie> cos 1n

-n

2Ca

n2Cb

converge le plus

rapidement possible.

restart;

ud cos1

nK

n2Ca

n2Cb

;

u := cos1

n K

n2Ca

n2Cb

vd asympt u, n, 7 ;

v :=

aCbK1

2

n2 C

aCb bC1

24

n4 C

aKb b2K

1

720

n6 CO

1

n7

Sd solve 1

2KaCb,

1

24K aCb $b , a, b ;

S:= a =5

12,b =

1

12

assign S ;

asympt u, n, 7 ;


14/18

(9.4)(9.4)

(9.2)(9.2)

(9.1)(9.1)

(9.3)(9.3)

(8.4)(8.4)

OO

1

480 n6CO

1

n7

Exercice 09 :

Dterminer les rels a,b,c,det epour que la srie> sin1

n -a$n

4Cb$n

2Cc

n5Cd$n

3Ce$n

converge le plus rapidement possible.

restart;

ud si n1

nK

a$n4Cb$n

2Cc

n5Cd$n

3Ce$n

;

u := sin1

n K

an4Cbn

2Cc

n5Cdn

3Cen

vd

asympt u, n, 10 ;

v :=1Ka

n C

1

6 KbCad

n3

C

cCaeK bCad dC1

120

n5

C

bCad eK cCaeCdbKad2

dK1

5040

n7

C

1

362880 K cCaeCdbKad

2eK ebK2 eadCdcKd

2bCad

3d

n9

CO1

n10

pd convert v, polynom ;

p :=1Ka

n C

1

6 KbCad

n3

C

cCaeK bCad dC1

120

n5

C

bCad eK cCaeCdbKad2

dK1

5040

n7

C

1362880

K cCaeCdbKad2 eK ebK2 eadCdcKd2bCad3 d

n9

Ed coeffs p, n ;

E:= 1Ka,1

6 KbCad,

1

362880 K cCaeCdbKad

2eK ebK2 eadCdc

Kd2bCad

3d, bCad eK cCaeCdbKad

2dK

1

5040, cCaeK


15/18

(10.4)(10.4)

(9.6)(9.6)

(10.1)(10.1)

2.2.

(10.5)(10.5)

(10.2)(10.2)

1.1.

OO

(10.3)(10.3)

(9.5)(9.5)

bCad dC1

120

Sd solve E, a, b, c, d, e ;

S:= a = 1,b =53

396,c =

551

166320,d=

13

396,e =

5

11088

assign S ;

asympt u, n, 12 ;11

457228800 n11 CO

1

n12

Exercice 10 :

Soient a,b,c,d 2=et on considre la srie> sin anCb

Ksin c

nCd.

Donner une condition ncessaire et suffisante de convergence.Dans ce cas, dterminer a, b, cet dpour que la srie converge le plus vite

possible.

restart;

ud si n a

nCb

Kt an c

nCd

;

u := sin a

nCbKtan

c

nCd

vd asympt u, n, 2 ;

v := aKc 1

n C

1

2abK

1

6a

3C

1

2cdK

1

3c

3

1

n

3/2CO

1

n2

assign a =c ;

asympt u, n, 3 ;

1

2cbK

1

2c

3C

1

2cd

1

n

3/2

C1

4c

3bC

3

8cb

2K

1

8c

5K

3

8cd

2

C1

2c

3d

1

n

5/2

CO1

n3

pd conver t %, pol ynom ;

p :=1

2cbK

1

2c

3C

1

2cd

1

n

3/2

C1

4c

3bC

3

8cb

2K

1

8c

5K

3

8cd

2

C1

2c

3d

1

n

5/2

Ed seq coef f p, 1/ n ^ 2* kC1 / 2 , k =1 . . 2 ;

E:=1

2cbK

1

2c

3C

1

2cd,

1

4c

3bC

3

8cb

2K

1

8c

5K

3

8cd

2C

1

2c

3d


16/18


17/18

(12.3)(12.3)

(11.5)(11.5)

(12.1)(12.1)

(13.1)(13.1)

(13.2)(13.2)

(12.2)(12.2)

OO

zd subs eK1nsin

1

2a

= 1 ,z ;

z :=1

8

cos1

2a 4 bCa

2 1

n

n CO

1

n2

Exercice 11 :

Soit a > 0. Etudier la nature de la srie> arccos 32 C

1 n

na

6.

r estar t; assume xO0 ;

ud cosK1 3

2 Cx

6;

u := arccos1

2 3 Cx~ K

1

6

vd series u, x=0, 3 ;v := 2x~K2 3 x~

2CO x~

3

subs x=1

n

na

, v ;

2 1 n

na K

2 3 1 n 2

na 2

CO1

n 3

na 3

Exercice 12 :

Calculer l'exponentiel de chaqu'une des matrices suivantes et vrifier que det exp(A) =exp( tr(A)) :

Ad

0 2 1

3 2 0

2 2 1

;Bd

1 0 0

1 3 0

8 2 4

; Cd

0 1 1

0 0 1

0 1 0

;

restart;with LinearAlgebra :

AdMatrix 0 , 2 , 1 , 3 , 2 , 0 , 2 , 2 , 1 ;

A :=

0 2 1

3 2 0

2 2 1

expAdMatrixExponential A ;


18/18

(13.7)(13.7)

(13.3)(13.3)

(13.4)(13.4)

(13.5)(13.5)

(13.6)(13.6)

(13.9)(13.9)

(13.8)(13.8)

(13.2)(13.2)expA :=

1

3 1C2 e

6eK4 2

5 e

5K1 e

K4 1

15 10 e

6K9 e

5K1 e

K4

1

2 1Ce

6eK4 1

5 3C2 e

5eK4 1

10 5 e

6K6 e

5C1 e

K4

1

3

1Ce6

eK4 2

5

e5K1 e

K4 1

15

1C5 e6C9 e

5eK4

simplify Determinant expA ; exp Trace A ;

eK1

eK1

BdMatrix 1 , 0 , 0 , 1 , 3 , 0 , 8 , 2 , 4 ;

B :=

1 0 0

1 3 0

8 2 4

expBdMatrixExponential B ;

expB :=

e 0 0

1

2e

3C

1

2e e

30

e3K3 eC2 e

42 e

3C2 e

4e

4

simplify Determinant expB ; exp Trace B ;

e8

e8

Cd

Matrix 0, 1, 1 , 0, 0, 1 , 0, 1, 0 ;

C:=

0 1 1

0 0 1

0 1 0

expCdMatrixExponential C ;

expC:=

1 sin 1 Kcos 1 C1 cos 1 C1Ksin 1

0 cos 1 sin 1

0 sin 1 cos 1

simplify Determinant expC ; exp Trace C ;1

1

tp maple : séries numériques - correction

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