> restart:
> # NEWTON'S METHOD FOR SYSTEMS ALGORITHM 10.1
> #
> # To approximate the solution of the nonlinear system F(X)=0 given
> # an initial approximation X:
> #
> # INPUT:   Number n of equations and unknowns: initial approximation
> #          X=(X(1),...,X(n)): tolerance TOL: maximum number of
> #          iterations N.
> #
> # OUTPUT:  Approximate solution X=(X(1),...,X(n)) or a message
> #          that the number of iterations was exceeded.
> LINSYS := proc(N,OK,A,Y) local K, I1, Z, IR, IA, J, C, L, JA:
> K := N-1:
> OK := TRUE:
> I1 := 1:
> while OK = TRUE and I1 <= K do
> Z := abs(A[I1-1,I1-1]):
> IR := I1:
> IA := I1+1:
> for J from IA to N do
> if abs(A[J-1,I1-1]) > Z then
> IR := J:
> Z := abs(A[J-1,I1-1]):
> fi:
> od:
> if Z <= 1.0e-20 then
> OK := FALSE:
> else
> if IR <> I1 then
> for J from I1 to N+1 do
> C := A[I1-1,J-1]:
> A[I1-1,J-1] := A[IR-1,J-1]:
> A[IR-1,J-1] := C:
> od:
> fi:
> for J from IA to N do
> C :=A[J-1,I1-1]/A[I1-1,I1-1]:
> if abs(C) <= 1.0e-20 then
> C := 0:
> fi:
> for L from I1 to N+1 do
> A[J-1,L-1] := A[J-1,L-1]-C*A[I1-1,L-1]:
> od:
> od:
> fi:
> I1 := I1+1:
> od:
> if OK = TRUE
> then if abs(A[N-1,N-1]) <= 1.0e-20 then
> OK := FALSE:
> else
> Y[N-1] := A[N-1,N]/A[N-1,N-1]:
> for I1 from 1 to K do
> J := N-I1:
> JA := J+1:
> C := A[J-1,N]:
> for L from JA to N do
> C := C-A[J-1,L-1]*Y[L-1]:
> od:
> Y[J-1] := C/A[J-1,J-1]:
> od:
> fi:
> fi:
> if OK = FALSE then
> print(`Linear system is singular`):
> fi:
> end:
> print(`This is the Newton Method for Nonlinear Systems.`):
> OK := FALSE:
> while OK = FALSE do
>    print(`Input the number n of equations.`):
>    N := scanf(`%d`)[1]: print(`n = `):print(N):
>    if N >= 2 then
>       OK := TRUE:
>    else
>       print(`n must be an integer greater than 1.`):
>    fi:
> od:
> for I1 from 1 to N do
> print(`Input the function f[i](x1,x2, ... xn) in terms of x1, x2, ... for i = `):print(I1):
> print(`For example: x1-x2^2+1`):
> F[I1] := scanf(`%a`)[1]:print(`function is `):print(F[I1]):
> od:
> for I1 from 1 to N do
> for J from 1 to N do
> P[I1,J] := diff(F[I1],evaln(x || J)):print(`partial derivative i,j `):print(I1,J):print(P[I1,J]):
> P[I1,J] := unapply(P[I1,J],evaln(x || (1..N))):
> od:
> od:
> for I1 from 1 to N do
> F[I1] := unapply(F[I1],evaln(x || (1..N))):
> od:
> OK := FALSE:
> while OK = FALSE do
> print(`Input the Tolerance.`):
> TOL := scanf(`%f`)[1]:  print(`Tolerance = `): print(TOL):
> if TOL > 0 then
> OK := TRUE:
> else
> print(`Tolerance must be positive.`):
> fi:
> od:
> OK := FALSE:
> while OK = FALSE do
> print(`Input the maximum number of iterations.`):
> NN := scanf(`%d`)[1]: print(`Maximum number of iterations - `): print(NN):
> if NN > 0 then
> OK := TRUE:
> else
> print(`Must be a positive integer.`):
> fi:
> od:
> for I1 from 1 to N do
> print(`Input initial approximation X(I), I = `, I1):
> X[I1-1] := scanf(`%f`)[1]: print(`Input is `): print(X[I1-1]):
> od:
> if OK = TRUE then
> print(`Select output destination`):
> print(`1. Screen`):
> print(`2. Text file`):
> print(`Enter 1 or 2`):
> FLAG := scanf(`%d`)[1]: print(`Response is `):print(FLAG):
> if FLAG = 2 then
> print(`Input the file name in the form - drive\\:name.ext`):
> print(`for example   A:\\OUTPUT.DTA`):
> NAME := scanf(`%s`)[1]: print(`Output file is `): print(NAME):
> OUP := fopen(NAME,WRITE,TEXT):
> else
> OUP := default:
> fi:
> print(`Select amount of output\n`):
> print(`1. Answer only\n`):
> print(`2. All intermeditate approximations\n`):
> print(`Enter 1 or 2\n`):
> FLAG := scanf(`%d`)[1]: print(`Input is `):print(FLAG):
> fprintf(OUP, `NEWTONS METHOD FOR NONLINEAR SYSTEMS\n\n`):
> if FLAG = 2 then
> fprintf(OUP, `Iteration, Approximation, Error\n`):
> fi:
> # Step 1
> K := 1:
> # Step 2
> while OK = TRUE and K <= NN do
> # Step 3
> for I1 from 1 to N do
> for J from 1 to N do
> A[I1-1,J-1] := evalf(P[I1,J](seq(X[i],i=0..N-1))):
> od:
> A[I1-1,N] := evalf(-F[I1](seq(X[i],i=0..N-1))):
> od:
> # Step 4
> LINSYS(N,OK,A,Y):
> if OK = TRUE then
> # Step 5
> R := 0:
> for I1 from 1 to N do
> if abs(Y[I1-1]) > R then
> R := abs(Y[I1-1]):
> fi:
> X[I1-1] := X[I1-1]+Y[I1-1]:
> od:
> if FLAG = 2 then
> fprintf(OUP, ` %2d`, K):
> for I1 from 1 to N do
> fprintf(OUP, ` %11.8f`, X[I1-1]):
> od:
> fprintf(OUP, `\n%12.6e\n`, R):
> fi:
> # Step 6
> if R < TOL then
> OK := FALSE:
> fprintf(OUP, `Iteration %d gives solution:\n\n`, K):
> for I1 from 1 to N do
> fprintf(OUP, ` %11.8f`, X[I1-1]):
> od:
> fprintf(OUP, `\n\nto within tolerance %.10e\n`, TOL):
> # Step 7
> else
> K := K+1:
> fi:
> fi:
> od:
> if K > NN then
> # Step 8
> fprintf(OUP, `Procedure does not converge in %d iterations\n`, NN):
> fi:
> if OUP <> default then
> fclose(OUP):
> print(`Output file `,NAME,` created sucessfully`):
> fi:
> fi: