practica: métodos iterativos para sistemas...

Practica: Métodos iterativos para sistemas

Operaciones con Matrices y vectores

• Definición de matriz

• Definición de vector

• Matriz por vector> a:=matrix(3,3,[1,0,3,1,2,1,3,2,2]);

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 0 31 2 13 2 2

> v:=[1,2,1];

:= v [ ], ,1 2 1> w=evalm(a&*v);

= w [ ], ,4 6 9

• Inversa

• Producto de matrices

• Evaluación float > b:=matrix(3,3,[1,0,1,0,1,1,1,1,0]);

:= b⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 0 10 1 11 1 0

> b1:=evalm(b^(-1));

:= b1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

12

-12

12

-12

12

12

12

12

-12

> c:=b&*b1;

:= c &*b b1> evalm(c);

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 0 00 1 00 0 1

> evalf(b1); #no funciona

b1> b1f:=evalf(evalm(b1));

:= b1f⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.5000000000 -.5000000000 .5000000000-.5000000000 .5000000000 .5000000000.5000000000 .5000000000 -.5000000000

> >

• Producto por escalar

• Suma de matrices> a:=matrix(3,3,[1,1,1,2,1,2,3,4,1]);

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 1 12 1 23 4 1

> c:=3*a;

:= c 3 a> evalm(c);

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3 3 36 3 69 12 3

> b:=matrix(3,3,[1,1,2,1,1,1,0,0,1]);

:= b⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 1 21 1 10 0 1

> d:=a+b;

:= d + a b> evalm(a+b);

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2 2 33 2 33 4 2

• Vectores como matrices columna> a:=matrix(3,3,[1,0,1,1,1,1,2,0,0]);

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 0 11 1 12 0 0

> v:=matrix(3,1,[1,2,1]);

:= v⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

121

> w:=a&*v;

:= w &*a v> evalm(w);

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

242

>

Práctica Métodos iterativos para sistemas de ecuaciones 1

Ejercicio 1Dadas las matrices

A =

⎛⎜⎝ 1 2 31 −1 12 1 −1

⎞⎟⎠ , B =

⎛⎜⎝ −1/5 1/7 01/8 −1/17 1/240 1/27 −1/121

⎞⎟⎠ ,y los vectores

c =

⎛⎜⎝ 1.22.57.1

⎞⎟⎠ , v =

⎛⎜⎝ 123

⎞⎟⎠Calcula de foma exacta y en forma de aproximación decimal(a) D = AB(b) A1 = A

−1

(c) H = 13A+ 2

3B

(d) w = Bv + c(e) Si (

u(0) = v

u(j+1) = Bu(j) + c

calcula una aproximación decimal de u(18).

Solución

(a) D =

⎛⎜⎝ .0 5 . 13632 11951 .0 58539 9449−. 325 . 23871 77093 −4. 99311 2948× 10−2−. 275 . 18985 37193 4. 99311 2948× 10−2

⎞⎟⎠(b) A1 =

⎛⎜⎝ 0 . 33333 33333 . 33333 33333. 2 −. 46666 66667 . 13333 33333. 2 . 2 −. 2

⎞⎟⎠(c) H =

⎛⎜⎝ . 2 . 76190 47619 1.0. 41666 66667 −. 37254 90196 . 36111 11111. 66666 66667 . 35802 46914 −. 33884 29752

⎞⎟⎠(d) w =

⎛⎜⎝ 1. 28571 42862. 63235 29417. 14928 0686

⎞⎟⎠(e) u(18) =

⎛⎜⎝ 1.3332939132. 7996688687. 144644788

⎞⎟⎠


Ejercicio 1

> a:=matrix(3,3,[1,2,3,1,-1,1,2,1,-1]);

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 2 31 -1 12 1 -1

> b:=matrix(3,3,[-1/5,1/7,0,1/8,-1/17,1/24,0,1/27,-1/121]);

:= b

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

-15

17

0

18

-117

124

01

27-1

121> c:=[1.2,2.5,7.1];

:= c [ ], ,1.2 2.5 7.1> v:=[1,2,3];

:= v [ ], ,1 2 3> d:=evalm(a&*b);

:= d

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

120

1461071

851452

-1340

7673213

-1452904

-1140

6103213

1452904

> df:=evalf(evalm(d));

:= df⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.05000000000 .1363211951 .05853994490-.3250000000 .2387177093 -.04993112948-.2750000000 .1898537193 .04993112948

> a1:=evalm(a^(-1));

:= a1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

013

13

15

-715

215

15

15

-15

> a1f:=evalf(evalm(a1));

:= a1f⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0 .3333333333 .3333333333.2000000000 -.4666666667 .1333333333.2000000000 .2000000000 -.2000000000

> h:=1/3*a+2/3*b;

:= h + 13

a23

b

> hf:=evalf(evalm(h));

:= hf⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

.2000000000 .7619047619 1.

.4166666667 -.3725490196 .3611111111

.6666666667 .3580246914 -.3388429752> w:=evalm(b&*v+c);

:= w [ ], ,1.285714286 2.632352941 7.149280686> u0:=v;

n:=17; for i from 0 to n do u.(i+1):=evalf(evalm(b&*u.i+c)); od;

:= u0 [ ], ,1 2 3 := n 17Page 1

:= u1 [ ], ,1.285714286 2.632352941 7.149280686 := u2 [ ], ,1.318907563 2.803756691 7.138409589 := u3 [ ], ,1.336755158 2.797370314 7.144847720 := u4 [ ], ,1.332273299 2.800245188 7.144557980 := u5 [ ], ,1.333580367 2.799503773 7.144666851 := u6 [ ], ,1.333213037 2.799715306 7.144638491 := u7 [ ], ,1.333316722 2.799655765 7.144646560 := u8 [ ], ,1.333287479 2.799672564 7.144644289 := u9 [ ], ,1.333295728 2.799667826 7.144644929 := u10 [ ], ,1.333293401 2.799669162 7.144644749 := u11 [ ], ,1.333294057 2.799668785 7.144644800 := u12 [ ], ,1.333293872 2.799668891 7.144644785 := u13 [ ], ,1.333293924 2.799668861 7.144644789 := u14 [ ], ,1.333293910 2.799668870 7.144644788 := u15 [ ], ,1.333293914 2.799668868 7.144644788 := u16 [ ], ,1.333293913 2.799668868 7.144644788 := u17 [ ], ,1.333293913 2.799668868 7.144644788 := u18 [ ], ,1.333293913 2.799668868 7.144644788

>

Page 2


Libreria de álgebra lineal linalg

> with(linalg);Warning, new definition for normWarning, new definition for trace

BlockDiagonal GramSchmidt JordanBlock LUdecomp QRdecomp Wronskian addcol addrow adj adjoint angle, , , , , , , , , , ,[augment backsub band basis bezout blockmatrix charmat charpoly cholesky col coldim colspace colspan companion, , , , , , , , , , , , , ,concat cond copyinto crossprod curl definite delcols delrows det diag diverge dotprod eigenvals eigenvalues, , , , , , , , , , , , , ,eigenvectors eigenvects entermatrix equal exponential extend ffgausselim fibonacci forwardsub frobenius gausselim, , , , , , , , , , ,gaussjord geneqns genmatrix grad hadamard hermite hessian hilbert htranspose ihermite indexfunc innerprod, , , , , , , , , , , ,intbasis inverse ismith issimilar iszero jacobian jordan kernel laplacian leastsqrs linsolve matadd matrix minor, , , , , , , , , , , , , ,minpoly mulcol mulrow multiply norm normalize nullspace orthog permanent pivot potential randmatrix randvector, , , , , , , , , , , , ,rank ratform row rowdim rowspace rowspan rref scalarmul singularvals smith stack submatrix subvector sumbasis, , , , , , , , , , , , , ,swapcol swaprow sylvester toeplitz trace transpose vandermonde vecpotent vectdim vector wronskian, , , , , , , , , , ]

• determinante, inversa> a:=matrix(2,2,[1,2,4,5]);

:= a ⎡⎣⎢

⎤⎦⎥

1 24 5

> a1:=inverse(a);

:= a1

⎡

⎣

⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥

-53

23

43

-13

> d:=det(a);

:= d -3> evalm(d*a1);

⎡⎣⎢

⎤⎦⎥

5 -2-4 1

> b:=matrix(3,3,[1,2,x,1,x,2,0,1,0]);

:= b⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 2 x1 x 20 1 0

> det(b);

− + 2 x

• resolucion de sistemas lineales> a:=matrix(3,3,[1,2,1,2,-1,2,1,3,3]);

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 2 12 -1 21 3 3

> b:=[1,0,1];

:= b [ ], ,1 0 1> s:=linsolve(a,b);

:= s⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

25

25

-15

> evalm(a&*s);

[ ], ,1 0 1

• Función vectorial, jacobiano> f:=(x,y,z)->[x+y^2,x*y-z,z^2+y^2];

:= f → ( ), ,x y z [ ], , + x y2 − x y z + z2 y2

> f(1,2,3);

[ ], ,5 -1 13> jf:=jacobian(f(x,y,z),[x,y,z]);

Page 1

:= jf⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 2 y 0y x -10 2 y 2 z

• Sustitución en el jacobiano> v:=[1,2,3];

:= v [ ], ,1 2 3> jfv:=subs({x=v[1],y=v[2],z=v[3]},evalm(jf));

:= jfv⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 4 02 1 -10 4 6

>

Page 2


Ejercicio 2: Libreria linalg

(a) Resuelve el sistema, calcula una aproximación decimal de la solución.⎧⎪⎨⎪⎩10x+ y + z = 12−x+ 25y − 2z = −232x− 3y + 24z = 16

(b) Define la función vectorial

F

⎛⎜⎝ xyz

⎞⎟⎠ =⎛⎜⎝ x3 − 2xzx− 2xy + z2x2 − yz

⎞⎟⎠

calcula la imágen del vector v =

⎛⎜⎝ 1−10

⎞⎟⎠ .(c) Calcula el jacobiano de F y su valor en v.(d) Resuelve el sistema, calcula una aproximación decimal de la solución.⎧⎪⎨⎪⎩

10x+ 7y + z = −12−5x+ 25y − 2z = −1232x− 13y + 24z = 160

Solución

(a) x = 1. 23744 2922, y = −0. 83375 61306, z = 0. 45932 69068

(b) F

⎛⎜⎝ 1−10

⎞⎟⎠ =⎛⎜⎝ 131

⎞⎟⎠(c) JF =

⎛⎜⎝ 3x2 − 2z 0 −2x1− 2y −2x 2z2x −z −y

⎞⎟⎠, JF

⎛⎜⎝ 1−10

⎞⎟⎠ =⎛⎜⎝ 3 0 −23 −2 02 0 1

⎞⎟⎠(d) x = 1. 39104 614, y = −4. 30424 8515, z = 4. 21927 8209


Ejercicio 2

*********** (a) *********************> with(linalg):

a:=matrix(3,3,[10,1,1,-1,25,-2,2,-3,24]); b:=[12,-23,16];

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 1 1-1 25 -22 -3 24

:= b [ ], ,12 -23 16> s:=linsolve(a,b);

:= s⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

271219

-49305913

27165913

> sf:=evalf(evalm(s));

:= sf [ ], ,1.237442922 -.8337561306 .4593269068************ (b) **********************> f:=(x,y,z)->[x^3-2*x*z,x-2*x*y+z^2,x^2-y*z];

:= f → ( ), ,x y z [ ], , − x3 2 z x − + x 2 x y z2 − x2 y z> f(1,-1,0);

[ ], ,1 3 1************ (c) **********************> jf:=jacobian(f(x,y,z),[x,y,z]);

:= jf⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

− 3 x2 2 z 0 −2 x − 1 2 y −2 x 2 z2 x −z −y

> jfv:=subs({x=1,y=-1,z=0},evalm(jf));

:= jfv⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3 0 -23 -2 02 0 1

*********** (d) *************************> a:=matrix(3,3,[10,7,1,-5,25,-2,2,-13,24]);

b:=[-12,-123,160]; s:=linsolve(a,b); sf:=evalf(evalm(s));

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 7 1-5 25 -22 -13 24

:= b [ ], ,-12 -123 160

:= s⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

30452189

-94222189

92362189

:= sf [ ], ,1.391046140 -4.304248515 4.219278209>

Page 1


Ejercicio 3: Método iterativo lineal

Consideramos el sistema⎧⎪⎨⎪⎩10x+ y + z = 1−x+ 25y − 2z = −132x− y + 24z = 10

(a) Define las matrices A y b de la forma matricial Ax = b(b) Define la matriz N correspondiente al método de jacobi(c) Calcula P = N−A(d) Calcula M = N−1P(e) Calcula c = N−1b(f) Estudia la convergencia(g) Determina el número de iteraciones necesarias para obtener 6 decimalesexactos, tomando x(0) = ~0(h) Calcula el valor aproximado(i) Calcula la solución exacta y verifica el resultado(j) Repite todos los pasos emplando ahora el método de Gauss-Seidel.

Solución

(a) A =

⎛⎜⎝ 10 1 1−1 25 −22 −1 24

⎞⎟⎠ , b =⎛⎜⎝ 1−1310

⎞⎟⎠ (b) N =

⎛⎜⎝ 10 0 00 25 00 0 24

⎞⎟⎠(c)P =

⎛⎜⎝ 0 −1 −11 0 2−2 1 0

⎞⎟⎠ (d)M =

⎛⎜⎝ 0 − 110− 110

125

0 225

− 112

124

0

⎞⎟⎠ (e) c =

⎛⎜⎝110

−1325512

⎞⎟⎠(f) kMk∞ = 1

5< 1, el método es convergente

(g) (0.2)j

0.8× 13

25≤ 0.5× 10−6 =⇒ j ≥ 8.747, necesitamos 9 iteraciones

(h) Calculamos con 14 decimalesx = 0.10972945715932, y = −0.48462443285167, z = 0.38732986060353(i) x = 0. 10972 945723408, y = −0. 48462 443286843, z = 0. 38732 986052764°°°α− x(9)°°° = −0.7589× 10−10(j) x(9) = [.10972945723411,−.48462443286843, .38732986052764]


Ejercicio 3: Método iterativo lineal

> with(linalg):Warning, new definition for normWarning, new definition for trace> a:=matrix(3,3,[10,1,1,-1,25,-2,2,-1,24]);

b:=[1,-13,10]; n:=matrix(3,3,[a[1,1],0,0,0,a[2,2],0,0,0,a[3,3]]); n1:=inverse(n); p:=evalm(n-a); m:=evalm(n1&*p); c:=evalm(n1&*b);

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 1 1-1 25 -22 -1 24

:= b [ ], ,1 -13 10

:= n⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 0 00 25 00 0 24

:= n1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

110

0 0

0125

0

0 01

24

:= p⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0 -1 -11 0 2

-2 1 0

:= m

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

0-110

-110

125

0225

-112

124

0

:= c⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

110

-1325

512

Convergencia> nm:=norm(m,infinity);

nc:=norm(c,infinity);

:= nm15

:= nc1325

||M||<1, el método es convergente> ineq:=nm^j/(1-nm)*nc<0.5*10^(-6);

:= ineq < 1320

⎛

⎝⎜⎜

⎞

⎠⎟⎟

15

j

.5000000000 10-6

> plot(ineq,j=5..10,thickness=3);

Page 1

j1098765

1

0.8

0.6

0.4

0.2

0

gráficamente, j>8.7> Digits:=14;

a:=matrix(3,3,[10,1,1,-1,25,-2,2,-1,24]); b:=[1,-13,10]; n:=matrix(3,3,[a[1,1],0,0,0,a[2,2],0,0,0,a[3,3]]); n1:=inverse(n); p:=evalm(n-a); m:=evalm(n1&*p); c:=evalm(n1&*b); nmax:=8; x0:=[0,0,0]; for i from 0 to nmax do x.(i+1):=evalf(evalm(m&*x.i+c)); od;

>

:= Digits 14

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 1 1-1 25 -22 -1 24

:= b [ ], ,1 -13 10

:= n⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 0 00 25 00 0 24

:= n1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

110

0 0

0125

0

0 01

24

:= p⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0 -1 -11 0 2

-2 1 0

:= m

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

0-110

-110

125

0225

-112

124

0

:= c⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

110

-1325

512

:= nmax 8Page 2

:= x0 [ ], ,0 0 0 := x1 [ ], ,.10000000000000 -.52000000000000 .41666666666667 := x2 [ ], ,.11033333333333 -.48266666666667 .38666666666667 := x3 [ ], ,.10960000000000 -.48465333333333 .38736111111112 := x4 [ ], ,.10972922222222 -.48462711111111 .38733944444445 := x5 [ ], ,.10972876666667 -.48462367555556 .38732976851852 := x6 [ ], ,.10972939070370 -.48462446785185 .38732994962963 := x7 [ ], ,.10972945182222 -.48462442840148 .38732986461420 := x8 [ ], ,.10972945637873 -.48462443275798 .38732986116476 := x9 [ ], ,.10972945715932 -.48462443285167 .38732986060353

> s:=linsolve(a,b);

:= s⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

6535951

-28845951

23055951

> sf:=evalf(evalm(s));

:= sf [ ], ,.10972945723408 -.48462443286843 .38732986052764> er9:=evalm(sf-x9);

:= er9 [ ], ,.7476 10-10 -.1676 10-10 -.7589 10-10

Método de Gauss-Seidel> Digits:=16;

a:=matrix(3,3,[10,1,1,-1,25,-2,2,-1,24]); b:=[1,-13,10]; n:=matrix(3,3,[a[1,1],0,0,a[2,1],a[2,2],0,a[3,1],a[3,2],a[3,3]]); n1:=inverse(n); p:=evalm(n-a); m:=evalm(n1&*p); c:=evalm(n1&*b); nmax:=8; x0:=[0,0,0]; for i from 0 to nmax do x.(i+1):=evalf(evalm(m&*x.i+c)); od;

:= Digits 16

:= a⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 1 1-1 25 -22 -1 24

:= b [ ], ,1 -13 10

:= n⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

10 0 0-1 25 02 -1 24

:= n1

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

110

0 0

1250

125

0

-496000

1600

124

:= p⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0 -1 -10 0 20 0 0

:= m

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

0-110

-110

0-1

25019250

049

600023

2000

:= c⎡

⎣⎢⎢

⎤

⎦⎥⎥, ,

110

-129250

23216000

Page 3

:= nmax 8 := x0 [ ], ,0 0 0

:= x1 [ ], ,.1000000000000000 -.5160000000000000 .3868333333333333 := x2 [ ], ,.1129166666666667 -.4845366666666667 .3870679166666666 := x3 [ ], ,.1097468750000000 -.4846446916666667 .3873275649305555 := x4 [ ], ,.1097317126736111 -.4846245262986111 .3873296686814236 := x5 [ ], ,.1097294857617188 -.4846244470750174 .3873298575583977 := x6 [ ], ,.1097294589516620 -.4846244330372617 .3873298603774756 := x7 [ ], ,.1097294572659786 -.4846244328791628 .3873298605245366 := x8 [ ], ,.1097294572354626 -.4846244328686186 .3873298605275190 := x9 [ ], ,.1097294572341100 -.4846244328684341 .3873298605276394

> er9:=evalm(sf-x9);

:= er9 [ ], ,-.300 10-13 .41 10-14 .6 10-15

>

Page 4


Ejercicio 4: Método Newton-Raphson

Consideramos el sistema (x2 + y2 = 4y = (x− 1)2

(a) Representa conjutamente las curvas y calcula una aproximación de lospuntos de corte(b) Plantea el sistema como una ecación vectorial F(x) = 0(c) Define la función vectorial F(x)(d) Calcula el jacobiano(e) Escribe un programa que permita aplicar el método de Newton-Raphson,

toma como valor inicial x(0) =

Ã1.80.7

!


Ejercicio 4: Método de Newton-Raphson

> with(plots):> eq1:=x^2+y^2-4;

eq2:=y-(x-1)^2;

:= eq1 + − x2 y2 4

:= eq2 − y ( ) − x 1 2

> implicitplot({eq1=0,eq2=0},x=-1..4,y=-1..4);

x321-1

y

4

3

2

1

0

-1

Solución de primer cuadrante x=1.8, y=0.7;> f:=unapply([eq1,eq2],x,y);

:= f → ( ),x y [ ], + − x2 y2 4 − y ( ) − x 1 2

> with(linalg):> jf:=jacobian(f(x,y),[x,y]);

:= jf ⎡⎣⎢

⎤⎦⎥

2 x 2 y− + 2 x 2 1

> x0:=[1.8,0.7]; n:=3; for i from 0 to n do `************`,i+1,`*************`; jfx:=subs(x=x.i[1],y=x.i[2],evalm(jf)); x.(i+1):=evalm(x.i-inverse(jfx)&*f(x.i[1],x.i[2])); od;

:= x0 [ ],1.8 .7 := n 3

, ,************ 1 *************

:= jfx ⎡⎣⎢

⎤⎦⎥

3.6 1.4-1.6 1

:= x1 [ ],1.8606164383562 .73698630136986, ,************ 2 *************

:= jfx ⎡⎣⎢

⎤⎦⎥

3.7212328767124 1.4739726027397-1.7212328767124 1

:= x2 [ ],1.8589453355656 .73778429690542, ,************ 3 *************

:= jfx ⎡⎣⎢

⎤⎦⎥

3.7178906711312 1.4755685938108-1.7178906711312 1

:= x3 [ ],1.8589441280931 .73778501518413Page 1

, ,************ 4 *************

:= jfx ⎡⎣⎢

⎤⎦⎥

3.7178882561862 1.4755700303683-1.7178882561862 1

:= x4 [ ],1.8589441280924 .73778501518444>

Page 2

practica: métodos iterativos para sistemas...

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