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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