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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
8/10/2019 Tp Maple : Sries numriques - correction
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
8/10/2019 Tp Maple : Sries numriques - correction
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 ;
8/10/2019 Tp Maple : Sries numriques - correction
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 ;
8/10/2019 Tp Maple : Sries numriques - correction
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
8/10/2019 Tp Maple : Sries numriques - correction
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8/10/2019 Tp Maple : Sries numriques - correction
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8/10/2019 Tp Maple : Sries numriques - correction
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$nC1
; limitu nC1
u n, n =N ; is %!1 ;
8/10/2019 Tp Maple : Sries numriques - correction
9/18
(3.11)(3.11)
(3.13)(3.13)
(3.9)(3.9)
(3.10)(3.10)
(3.12)(3.12)
OO
(3.8)(3.8)
OO
(3.14)(3.14)
u :=n/3 n
4 nK1
2 nC1
9
16
true
ud n/a
n2
product aC1 k
, k =1 . . n; simplify u nC1
u n;
u :=n/a
n2
?k= 1
n
aC1 k
a2 nC1
aC1 KnK1
Sd solve a2!abs aC1 , a ;
S:=RealRange Open
1
2 K
1
2 5 ,Open
1
2 C
1
2 5
asympt l n cosh1
n$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 ;
8/10/2019 Tp Maple : Sries numriques - correction
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
kd 3; Ad asympt nkCnC1
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
kd 5; Ad asympt nkCnC1
k$, n, 2 ; asympt 1
n$si n AKn$ ,
n ; k:= 5
A :=nCO1
n3
O1
n3
kd 6; Ad asympt nkCnC1
k$, n, 2 ; asympt 1
n$si n AKn$ ,
n ;k:= 6
A :=nCO 1
n4
O1
n4
kd 7; Ad asympt nkCnC1
k$, n, 2 ; asympt 1
n$si n AKn$ ,
n ;k:= 7
8/10/2019 Tp Maple : Sries numriques - correction
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
kd 8; Ad asympt nkCnC1
k$, n, 2 ; asympt 1
n$si n AKn$ ,
n ;k:= 8
A :=nCO1
n6
O1
n6
kd 9; Ad asympt nkCnC1
k$, n, 2 ; asympt 1
n$si n AKn$ ,
n ;
k:= 9
A :=nCO1
n7
O1
n6
kd 10; Ad asympt nkCnC1
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 ;
8/10/2019 Tp Maple : Sries numriques - correction
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$k;
a :=3
2 C3 k
solve 1
3bK
1
9a
2, b ;
b =3
4 1C2 k
2
8/10/2019 Tp Maple : Sries numriques - correction
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(8.2)(8.2)
(7.1)(7.1)
OO
(7.2)(7.2)
(8.3)(8.3)
(7.3)(7.3)
OO
(8.1)(8.1)
Exercice 07 :
Dterminer les rels aet bpour que la srie> nC1nK2
n
- a 1Cb
nsoit
convergente.
restart;
udnC1
nK2
n
Ka$ 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 $b , a, b ;
S:= a =5
12,b =
1
12
assign S ;
asympt u, n, 7 ;
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(9.4)(9.4)
(9.2)(9.2)
(9.1)(9.1)
(9.3)(9.3)
(8.4)(8.4)
OO
1
480 n6CO
1
n7
Exercice 09 :
Dterminer les rels a,b,c,det epour que la srie> sin1
n -a$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
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(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
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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 ;
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(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
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