Examen d'Analyse Numérique - Décomposition LU, Interpolation

LU A
A=
2 1 −1
4 6 −1
2−7 3 
.
Ax =b b =
3
5
−1
f(x) = x2+x−1
xexacte =√5−1
2≈0.61803
a0= 0.5b0= 1
x0= 1
f(−1,−1) (0,−1)
(1,−1) (2,5)
I=Z1
0
1
1 + x2dx
[0,1] I10−3
I
x0= 0 x1=1
2x2= 1

L1=
1 0 0
−4
21 0
−2
20 1 
, A(1) =L1A=
2 1 −1
041
0−8 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
1−2 1 
, U =
2 1 −1
0 4 1
0 0 6 
A=
2 1 −1
4 6 −1
2−7 3 
=
1 0 0
2 1 0
1−2 1 
2 1 −1
0 4 1
0 0 6 
=LU
Ax =b Ly =b U x =y
1 0 0
2 1 0
1−2 1 
y1
y2
y3
=
3
5
−1
⇒
y1= 3
2y1+y2= 5
y1−2y2+y3=−1
⇒
y1= 3
y2=−1
y3=−6
2 1 −1
0 4 1
0 0 6 
x1
x2
x3
=
3
−1
−6
⇒
2x1+x2−x3= 3
4x2+x3=−1
6x3=−6
⇒
x1= 1
x2= 0
x3=−1
2 1 −1
4 6 −1
2−7 3 
1
0
−1
=
3
5
−1
f(x) = x2+x−1
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−
b0−a0
f(b0)−f(a0)f(a0)c0= 0.5−1−0.5 f(1)−f(0.5) f(0.5) = 0.5−0.5 1−0.25 (−0.25) = 2 3f(c0) =
f(2
3) = 1 9>0a1=a0= 0.5b1=c0=2 3
1
2< xexacte <2 3
2
3−1 2=1 6≈0.166666
c1=a1−b1−a1
f(b1)−f(a1)f(a1) = 1 2−
2
3−1 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
3−8 13 =2 39 ≈0.051282
xn+1 =xn−f(xn)
f0(xn)x0= 1
x1=x0−f(x0)
f0(x0)= 1 −1 3=2 3x1=2 3
e1=|x1−xexacte|= 0.0486326
x2x1=x1−f(x1)
f0(x1)=2 3−1/9 7/3=2 3−1 21 =13 21 ≈0.619047
e2=|x2−xexacte|= 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
x−xj
xi−xj!
n= 3
L1(x) = x(x−1)(x−2)
(−1−0)(−1−1)(−1−2) =−x3−3x2+ 2x 6
L2(x) = (x+ 1)(x−1)(x−2)
(0 + 1)(0 −1)(0 −2) =x3−2x2−x+ 2 2
L3(x) = x(x+ 1)(x−2)
(1 + 1)(1 −0)(1 −2)) =−x3−x2−2x 2
L4(x) = x(x+ 1)(x−1)
(2 + 1)(2 −0)(2 −1) =x3−x 6
P3(x) = −L1(x)−L2(x)−L3(x)+5L2(x) = x3−x−1
P3(x) = a0+a1(x+ 1) + a2(x+ 1)(x−0) + a3(x+ 1)(x−0)(x−1)
P3(x) = −1 + x(x+ 1)(x−1) = x3−x−1

x f(x)
−1 = a0
0−1−1−(−1)
0−(−1) = = a1
1−1−1−(−1)
1−0= 0 = 0−0 1−(−1) = = a2
2 5 5−(−1)
2−1= 6 6−0 2−0= 3 3−0 2−(−1) = = a3
f: [a, b]→Rn∈Nn≥2h=b−a
n
x0=a xn=b xk=x0+kh xk=xk−1+xk
21≤k≤nRb
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)
≤(b−a)h2
24 max
x∈[a,b]|f00(x)|
Ic
T(f) = h
2
n
X
k=1
f(xk−1+f(xk)),
f C2
Zb
a
f(x)dx −Ic
T(f)
≤(b−a)h2
12 max
x∈[a,b]|f00(x)|
Ic
S(f) = h
6
n
X
k=1
[f(xk−1+ 4f(xk) + f(xk)] ,
f C4
Zb
a
f(x)dx −Ic
S(f)
≤h4
2880 max
x∈[a,b]|f(4)(x)|
I10−3
10−3

(b−a) = 1 f(4)(x) = −24
(1+x2)3+144x+48 (1+x2)4−96 (1+x2)5maxx∈[0,1] |f(4)(x)| ≤
312
h4≤2880
312 ×10−3⇒h≤0.31
h=1
44
I
x0= 0 x1=1
2x2= 1 h=1 2
Ic
S(f) = h
6f(0) + f(1) + 2f(1 2)+4f(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|I−Ic
S(f)|=|π
4−0.86039| ≤ 0.075