Download - Cálculo Numérico - Prof. Luciano Bittencourt
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5/20/2018 C lculo Num rico - Prof. Luciano Bittencourt
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f(x) =x3
9x+ 3
x
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[a, b]
xk xk+1
f(x)
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x
x
Eax = x x
x
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a) = (11, 4 + 3, 18) + 5, 05 e11, 4 + (3, 18 + 5, 05)
b) = 3, 18.11, 4
5, 05 e
3, 18
5, 05.11, 4
Erx =x x
x
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x
x
f(x)
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f(x)
f(a).f(b) < 0
x =
f(x)
x
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[a, b]
f(x) = x3 9x+ 3
f(x) =x3
9x+ 3
x
f(x)
[5, 3], [0, 1] e [2, 3]
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|xk+1 xk|xk
< , (erro relativo)
xk
x= xk+1 x= xk
f(x)
[a, b]
f(a).f(b)< 0.
f(x)
x [a, b]
[a, b]
xk
xl =a+b
2
f(x) =x.log(x)1
[2, 3]
k a b xk f(xk)
f(a)
f(b)
f(x)
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f(2) = 2.log(2) 1= 0, 3979
f(3)> 0
f(3) = 3.log(3) 1= 0.4314
f(a) < 0
f(b) > 0
a
b
a
b
xk f(xk)
k a() b(+) xk f(xk)
f(xk) xk b
xk f(xk)
k a() b(+) xk f(xk)
= 2, 75 2, 5
2, 5= 0, 1
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k a() b(+) xk f(xk)
k= 5
k a() b(+) xk f(xk)
k log(b a) log()log(2)
f(x) = x3 9x+ 3
x [0, 1]
103
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xk xk+1
xk xk+1
xk+1=xk f(xk)f(xk)
f(x) = 4cos(x)
ex
xk [0, 2] 10
2
xk = 1
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k xk xk+1
f(x)
f(x) =4sen(x) ex
x2 = xk f(x1)f(x1)
x2 = 1 4cos(1) e1
4sen(1) e1x2 = 0, 9084
= |0.9084 1|
1 = 0, 0916
k xk xk+1
x3= 0, 9048
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= 0.004
k xk xk+1
f(x) = 4cos(x) ex
0, 9048
f(x) = x2 +x
6
x0 1, 5
3 2
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xk+1 = xk f(xk)
f(xk)
xk+1 = xkx2 +x 62x+ 1
k= 0
x1 = x0x2 +x
6
2x+ 1
x1 = 1, 5 1, 52 + 1, 5 6
2.1, 5 + 1x1 = 2, 0625
k= 2
k xk xk+1
x
1, 5
102
2x= tg(x)
5x3 +x2 12x+ 4 = 0
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sen(x) ex = 0
x4
8 = 0
f(x)
f(xk) f(xk) f(xk1)xk xk1
xk xk1
xk+1= xk xk xk1f(xk) f(xk1)f(xk)
f(x) =
x5ex
= 102
x
1, 5
x0 1, 5 xk1 x1 1, 7 xk x2
k= 1
x2 = 1, 7 1, 7 1, 5(
1, 7 5e1,7) (1, 7 5e1,7))
1, 7 5e1,7
x2 = 1, 4224
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k xk xk1 xk+1
k= 2
xk = 1, 4224 xk1= 1, 7
xk+1= 1, 4313
k xk xk1 xk+1
102
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f(x) = 2, 7 ln(x)
f(x) =log(x)
cos(x)
f(x) =ex log(x)
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3x + 4y 10z= 3
xy 3z=3ex3 +yz = 0 n
n
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Ax = b n A
A= L+D+R
L
A
D
A
R
A
Ax= b
(L+D+R)x = b
Dx = (L+R)x+bx = D1b D1(L+R)x
xk+1=D1
b D1
(L+R)xk
3x y z = 1
x + 3y + z = 3
2x + y + 4z = 7
Ax= b
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3 1 1
1 3 12 1 4
x
yz
=
1
37
D=
3 0 0
0 3 0
0 0 4
D1
D1 =
1/3 0 0
0 1/3 0
0 0 1/4
L=
0 0 01 0 02 1 0
R=
0 1 10 0 1
0 0 0
L+R=
0 1 11 0 12 1 0
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D1b
D1b=
1/317/4
(L+R)D1
(L+R)D1 =
0 1/3 1/3
1/3 0 1/3
1/2 1/4 0
xk+1
yk+1
zk+1
=
1/3
1
7/4
0 1/3 1/31/3 0 1/31/2 1/4 0
xk
yk
zk
xk+1
yk+1
zk+1
= 1/3
1
7/4
yk/3
zk/3
xk/3 +zk/3xk/2 +yk/4
xk+1
yk+1
zk+1
=
1/3 +yk/3 +zk/3
1 +xk/3 zk/37/4 xk/2 yk/4
xk+1, yk+1e zk+1 k= 0
x0= y0=z0= 0
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x1
y1z1
=
0, 3333
17/4
x2 y2 z2
x2
y2
z2
=
1, 25
0, 5278
1, 3333
x15
y15
z15
=
1
1
1
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Ax= b
k = (n
j=1,j=k
|akj |)/|akk|,
= max(alphak)< 1 xk
x0
10 2 1
1 5 1
2 3 10
1=2 + 1
10 = 0, 3
2=1 + 1
5 = 0, 4
3=2 + 3
10
= 0, 5
max(k) = 0, 5 < 1
102
x0= 0, 7 y0=1, 6 z0= 0, 6
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10x + 2y + z = 7
x + 5y + z = 82x + 3y + 10z = 6
A
x + 0, 2y + 0, 1z = 0, 7
0, 2x + y + 0, 2z = 1, 60, 2x + 0, 3y + z = 0, 6
x
y
z
k
k+ 1
x = 0, 2y 0, 1z + 0, 7y = 0, 2x 0, 2z 1, 6z = 0, 2x 0, 3y + 0, 6
xk+1 = 0, 2yk 0, 1zk + 0, 7yk+1 = 0, 2xk 0, 2zk 1, 6zk+1 =
0, 2xk
0, 3yk + 0, 6
x0 y0 z0
k= 4
102
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k xk yk zk
x
y
z
=
0, 9979
1, 99960, 9968
10x + y z = 10x + 10y + z = 12
2x y + 10z = 11
10x + y z = 102x + 10y + 8z = 20
7x + y + 10z = 30
Ax= b
(L +I+ R)x= b
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(L +I)x=Rx+b
x=(L +I)1Rx+ (L +I)1b,
xk+1=
(L +I)Rxk+ (L
+I)1b
xk+1
(L =I)xk+1=Rxk+b,
xk+1=Lxk+1 Rxk+b,
xk+1
(L + I)1
xk+1
xk+1
103
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5x + y + z = 5
3x + 4y + z = 63x + 3y + 6z = 0
A
x + 0, 2y + 0, 2z = 1
0, 75x + y + 0, 25z = 1, 50, 5x + 0, 5y + z = 0
x
y
z
xk+1 = 0, 2yk 0, 2zk + 1yk+1 = 0, 75xk+1 0, 25zk + 1, 5zk+1 = 0, 5xk+1 0, 5yk+1
x0= y0=z0= 0
k xk yk zk
x4 y4 z4 k= 4
x = 1, 0016
y = 0, 9987
z = 1.0002
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max1jn
i< 1
i =i1j=1
|aij| j+n
j=i+1
|aij|
5x + y + z = 5
3x + 4y + z = 6
3x + 3y + 6z = 0
x + 0, 2y + 0, 2z = 1
0, 75x + y + 0, 25z = 1, 5
0, 5x + 0, 5y + z = 0
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1=|0, 2| + |0, 2|= 0, 42=|0, 75|(0, 4) + |0, 25|= 0, 3 + 0, 25 = 0, 553=|0, 5|(0, 4) + |0, 5|(0, 55) = 0, 2 + 0, 275 = 0, 475
max1jn
i= 0, 55< 1
A
102
4x + 2y + 6z = 1
4x 1y + 3z = 2x + 5y + 3z = 3
5x + 2y + z = 7
x + 4y + 2z = 32x 3y + 10z = 1
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f(x)
g(x)
g(x)
f(x)
x0, x1,...,xn (n+ 1) yi= f(xi), i= 0,...,n
pn(x) n f x0,...,xn
pn(x) pn(x) = y0L0(x) + y1L1(x),...,ynLn(x)
Lk(x) n i pn(xi) =yi
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pn(xi) =y0L0(xi) +y1L1(xi) +...+ynLn(xi) =yi.
Lk(xi) =
0 se k = i1 se k = i
Lk(x)
Lk(x) =
(x
x0)(x
x1)...(x
xk1)(x
xk+1)...(x
xn)
(xkx0)(xkx1)...(xkxk1)(xkxk+1)...(xkxn)
pn(x) =n
k=0
ykLk(x)
Lk(x) =
nj=0,j=k
(x xj)
nj=0,j=k
(xk xj)
pn(x)
x
f(x)
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pn(x) =y0L0(x) +y1L1(x) +y2L2(x),
Ln(x)
L0(x) = (x x1)(x x2)(x0 x1)(x0 x2)=
(x 0)(x 2)(1 0)(1 2)=
x2 2x3
L1(x) = (x x0)(x x2)(x1 x0)(x1 x2)=
(x+ 1)(x 2)(0 + 1)(0 2) =
x2 x 22
L2(x) = (x x0)(x x1)(x2 x0)(x2 x1)=
(x+ 1)(x 0)(2 + 1)(2 0) =
x2 +x6
pn(x) = 4
(x2 2x
3
+ 1
(x2 x 2
2
+ (1)(
x2 +x
6
pn
(x) = 1
7
3x+
2
3x2
f(x)
pn(x) x
pn(x)
f(3, 5)
x
f(x)
n= 4
L0(x) =(x 3)(x 4)(x 5)(1 3)(1 4)(1 5) =
x3 12x2 + 47x 6024
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L1(x) =(x 1)(x 4)(x 5)(3
1)(3
4)(3
5)
=x3 10x2 + 29x 20
4
L2(x) =(x 1)(x 3)(x 5)(4 1)(4 3)(4 5) =
x3 9x2 + 23x 153
L3(x) =(x 1)(x 3)(x 4)(5 1)(5 3)(5 4) =
x3 8x2 + 19x 128
pn(x)
pn(x) =3x3 30x2 + 87x 60
2
8x3 + 72x2 184x+ 1201
+15x3 120x2 + 285x 180
2
pn(x) =2x3 6x2 + 4x
2
pn(x) =x3
3x2 + 2x
f(3, 5)
pn(3, )
pn(3, 5) = 3, 53 3.3, 52 + 2.3, 5 = 13, 125.
Pn(x)
f(x) =ln(x)
ln(3, 7)
f(x)
Pn(x)
x
f(x)
f(x) =ex +x 1
Pn(x) f(x) Pn(0, 7)
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I=
/2
0
dx1 x2sen2(x)
I
e3,1
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y = f(x)
[a, b]
Pn(x) f(x)
f(x)
(x0, f(x0)), (x1, f(x1)), (x2, f(x2)) x0 = a xn = b
f(x)
f(x)
f(x)
t0
s
(t3/2 s3/2)2/3ds
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[a, b]
f(x)
f(x)
x0 x1
f(x)
Pn(x)
ba
P1(x)dx=h
2[f(x0) +f(x1)]
h= b a
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I= 10
1
1 +x2dx
h= 1 0 = 1f(0) = 1
1+0= 1
f(1) = 11+1
= 0, 5
I=
1
0
1
1 +x2dx=
1
2[1 + 0, 5] = 0, 75
0, 75
[1, b]
N
h= ban
x0= a xN=b
[xj, xj+1] j = 0, 1,...,N1
xNx0
f(x)dx=h
2{f(x0) + 2[f(x1) +f(x2) +...f(xN1)] +f(xN)}
f(x)
2
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h= 0.2
1,2
0
excos(x)
f(x)
0
1, 2
h
x
excos(x)
1,2
0
excos(x) =0, 2
2[1 + 2(1, 197 + 1, 374 + 1, 503 + 1, 552 + 1, 468) + 1, 202]
1,2
0
excos(x) = 1, 639
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2
f(x)
3
[xj, xj+2]
j = 0, 2,..., 2N 2
x0Nx0
f(x)dx=h
3[f(x0) + 4f(x1) +f(x2)]
h
x0N
x0
f(x)dx=h
3{f(x0)+4[f(x1) + f(x3) + ...]+2[f(x2) + f(x4)...] + f(xN)
}
h/3
f(x0) f(xN)
f(x)
4
f(x)
2
1,2
0
excos(x) =h
3{f(x0)+4[f(x1)+f(x3)+f(x5)]+2[f(x2)+f(x4)+f(x5)+f(xN)}
1,2
0
excos(x) =0, 2
3[1 + 4(1, 197 + 1, 503 + 1, 468) + 2(1, 374 + 1, 552) + 1, 202]
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1,2
0
excos(x) = 1, 639
1,3
1
xdx
h= 0, 05
0,8
0
cos(x)dx
h= 0, 1
1,6
1,2
senxdx
h= 0, 2
0,8
0
xexdx
h= 0, 1
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y =f(x, y)
f
x
y
y
x
y = f(x, y)
y =
y(x)
x [a, b]
y(x) = f(x, y(x))
y(x) =C ex
C
y = y
y(x)
y(x0
) = y0
y = f(x, y)
y(x0) = y0
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xn+1= xn+h
yn+1= yn+h.f(xn, yn)
h
h2
dy
dx= 1 x+ 4y
y(0) = 1
h= 0, 5
y(0) = 1
y0 = 1 x0 = 0
n
xn yn f(x, y h.f(x, y)
n = 0
f(xn, yn) = 1 xn+ 4yn
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n xn yn f(xn, yn) h.f(xn, yn)0 0 1 5 2, 5
f(xn, yn) = 10 + 4.1 = 5 n= 1
xn+1= xn + h x2=x1 + h= 0 + 0, 5 = 0, 5
y2 yn+1= yn+h.f(xn, yn) = 1 + 2, 5 =
3, 5
n xn yn f(xn, yn) h.f(xn, yn)0 0 1 5 2, 51 0, 5 3, 5 14, 5 7, 3
y
96, 5
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n xn yn f(xn, yn) h.f(xn, yn)0 0 1 5 2, 51 0, 5 3, 5 14, 5 7, 32 1 10, 7 43 21, 53 1, 5 32, 3 128, 5 64, 3
4 2 96, 5 385 192, 5
yn
yn+1=yn+h
6(k1+ 2k2+ 2k3+k4)
k1 = f(xn, yn)
k2 = f(xn+h
2, yn+
h
2k1)
k3 = f(xn+h
2, yn+
h
2k2)
k4 = f(xn+h, yn+hk3)
y(1, 5)
y = 2xy y(1) = 1
h= 0, 1
k1 k2 k3 k4
n= 0
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k1 = f(xn, yn)= f(x0, y0)
= 2.x0.y0
= 2.1.1
= 2
k2 = f(xn+h
2, yn+
h
2k1)
= f(x0+h
2, y0+
h
2k1)
= f(1 +0, 1
2 , 1 +
0, 1
2 .2)
= f((1, 05), (1, 1))
= 2.(1, 05).(1, 1)
= 2, 31
k3 = f(xn+h
2, yn+
h
2k2)
= f(x0+h
2, y0+
h
2k2)
= f(1 +0, 12
, 1 +0, 12
(2, 31))
= f((1, 05), (1, 1155))
= 2.(1, 05).(1, 1155)
= 2, 3425
k4 = f(xn+h, yn+hk3)
= f(x0+h, y0+k3)
= f((1 + 0, 1), (1 + 0, 1.(2, 3425)))
= f((1, 1), (1, 2342))
= 2.(1, 1).(1, 2342)
= 2, 7154
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yn+1 = yn+
h
6 (k1+ 2k2+ 2k3+k4)
y1 = y0+h
6(k1+ 2k2+ 2k3+k4)
= 1 +0, 1
6(2 + 2.(2, 31) + 2.(2, 3425) + 2, 7154)
= 1, 2337
x = 1, 5
y(1, 5)
n xn yn0 1 11 1, 1 2, 23372 1, 2 1, 55273 1, 3 1, 99374 1, 4 2, 6117
5 1, 5 3, 4904
y =
0, 3679ex2
y(1, 5 ) = 3, 4905
=|3, 4905 3, 4904|= 0, 0001
y = y+x+ 2y(0) = 20x0, 3; h= 0, 1
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y = y1/3
y(0) = 0
x [0;0, 3]; h= 0, 1
xy = x yy(2) = 2
y(2, 1)
h= 0, 1
y = 2y
x+1+ (x+ 1)3
y(0) = 3
x [0;0, 2]; h= 0, 1
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