Analyse-Numérique-et-Algorithmique-examens-02

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LU A
A=
2 1 1
4 6 1
27 3
.
Ax =b b =
3
5
1
f(x) = x2+x1
xexacte =51
20.61803
a0= 0.5b0= 1
x0= 1
f(1,1) (0,1)
(1,1) (2,5)
I=Z1
0
1
1 + x2dx
[0,1] I103
I
x0= 0 x1=1
2x2= 1
L1=
1 0 0
4
21 0
2
20 1
, A(1) =L1A=
2 1 1
041
08 4
,
L2=
1 0 0
0 1 0
0 2 1
, A(2) =L2A(1) =L2L1A=
2 1 1
0 4 1
0 0 6
.
L=
1 0 0
2 1 0
12 1
, U =
2 1 1
0 4 1
0 0 6
A=
2 1 1
4 6 1
27 3
=
1 0 0
2 1 0
12 1
2 1 1
0 4 1
0 0 6
=LU
Ax =b Ly =b U x =y
1 0 0
2 1 0
12 1
y1
y2
y3
=
3
5
1
y1= 3
2y1+y2= 5
y12y2+y3=1
y1= 3
y2=1
y3=6
2 1 1
0 4 1
0 0 6
x1
x2
x3
=
3
1
6
2x1+x2x3= 3
4x2+x3=1
6x3=6
x1= 1
x2= 0
x3=1
2 1 1
4 6 1
27 3
1
0
1
=
3
5
1
f(x) = x2+x1
a0= 0.5b0= 1 f(a0) =
f(0.5) = 0.25 <0f(b0) = f(1) = 1 >0f(a0+b0
2) = f(0.75) = (0.75)2+
0.75 1=0.3125 >0a1=a0= 0.5b1=a0+b0
2= 0.75
0.5< xexacte <0.75 0.75 0.5 = 0.25
f(a1+b1
2) = f(0.625) = (0.625)2+
0.625 1=0.015625 >0a2=a1= 0.5b2=a0+b0
2= 0.625
0.5< xexacte <0.625 0.625 0.5 = 0.125
a0= 0.5b0= 1 c0=a0
b0a0
f(b0)f(a0)f(a0)c0= 0.510.5
f(1)f(0.5) f(0.5) = 0.50.5
10.25 (0.25) = 2
3f(c0) =
f(2
3) = 1
9>0a1=a0= 0.5b1=c0=2
3
1
2< xexacte <2
3
2
31
2=1
60.166666
c1=a1b1a1
f(b1)f(a1)f(a1) = 1
2
2
31
2
f(2
3)f(1
2)f(1
2) =
1
2
1
6
13
36
(1
4) = 8
13 f(8
13 ) = 1
169 <0a2=c1=8
13 b2=b1=2
3
8
13 < xexacte <2
3
2
38
13 =2
39 0.051282
xn+1 =xnf(xn)
f0(xn)x0= 1
x1=x0f(x0)
f0(x0)= 1 1
3=2
3x1=2
3
e1=|x1xexacte|= 0.0486326
x2x1=x1f(x1)
f0(x1)=2
31/9
7/3=2
31
21 =13
21 0.619047
e2=|x2xexacte|= 0.00101
n
n+ 1 (xi, yi)i=n
i=0
Pn(x) =
i=n
X
i=0 yi
j=n
Y
j=0,j6=i
xxj
xixj!
n= 3
L1(x) = x(x1)(x2)
(10)(11)(12) =x33x2+ 2x
6
L2(x) = (x+ 1)(x1)(x2)
(0 + 1)(0 1)(0 2) =x32x2x+ 2
2
L3(x) = x(x+ 1)(x2)
(1 + 1)(1 0)(1 2)) =x3x22x
2
L4(x) = x(x+ 1)(x1)
(2 + 1)(2 0)(2 1) =x3x
6
P3(x) = L1(x)L2(x)L3(x)+5L2(x) = x3x1
P3(x) = a0+a1(x+ 1) + a2(x+ 1)(x0) + a3(x+ 1)(x0)(x1)
P3(x) = 1 + x(x+ 1)(x1) = x3x1
x f(x)
1 = a0
011(1)
0(1) = = a1
111(1)
10= 0 = 00
1(1) = = a2
2 5 5(1)
21= 6 60
20= 3 30
2(1) = = a3
f: [a, b]RnNn2h=ba
n
x0=a xn=b xk=x0+kh xk=xk1+xk
21knRb
af(x)dx
Ic
P M (f) = h
n
X
k=1
f(xk),
f C2
Zb
a
f(x)dx Ic
P M (f)
(ba)h2
24 max
x[a,b]|f00(x)|
Ic
T(f) = h
2
n
X
k=1
f(xk1+f(xk)),
f C2
Zb
a
f(x)dx Ic
T(f)
(ba)h2
12 max
x[a,b]|f00(x)|
Ic
S(f) = h
6
n
X
k=1
[f(xk1+ 4f(xk) + f(xk)] ,
f C4
Zb
a
f(x)dx Ic
S(f)
h4
2880 max
x[a,b]|f(4)(x)|
I103
103
(ba) = 1 f(4)(x) = 24
(1+x2)3+144x+48
(1+x2)496
(1+x2)5maxx[0,1] |f(4)(x)| ≤
312
h42880
312 ×103h0.31
h=1
44
I
x0= 0 x1=1
2x2= 1 h=1
2
Ic
S(f) = h
6f(0) + f(1) + 2f(1
2)+4f(1
4) + f(3
4)
Ic
S(f) = 1
12 1 + 1
2+ 2 ×5
4+ 4 16
17 +16
25= 0.86039
Iπ
4|IIc
S(f)|=|π
40.86039| ≤ 0.075
1 / 5 100%

Analyse-Numérique-et-Algorithmique-examens-02

Telechargé par YOUSSEF MOUMMOU
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