Multidimensional Newton-Raphson method

From Rosetta Code
Multidimensional Newton-Raphson method is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Create a program that finds and outputs the root of a system of nonlinear equations using Newton-Raphson method.

C#[edit]

For matrix inversion and matrix and vector definitions - see C# source from Gaussian elimination

 
using System;
 
namespace Rosetta
{
internal interface IFun
{
double F(int index, Vector x);
double df(int index, int derivative, Vector x);
double[] weights();
}
 
class Newton
{
internal Vector Do(int size, IFun fun, Vector start)
{
Vector X = start.Clone();
Vector F = new Vector(size);
Matrix J = new Matrix(size, size);
Vector D;
do
{
for (int i = 0; i < size; i++)
F[i] = fun.F(i, X);
for (int i = 0; i < size; i++)
for (int j = 0; j < size; j++)
J[i, j] = fun.df(i, j, X);
J.ElimPartial(F);
X -= F;
//need weight vector because different coordinates can diffs by order of magnitudes
} while (F.norm(fun.weights()) > 1e-12);
return X;
}
}
}
 
 
using System;
 
//example from https://eti.pg.edu.pl/documents/176593/26763380/Wykl_AlgorOblicz_7.pdf
namespace Rosetta
{
class Program
{
class Fun: IFun
{
private double[] w = new double[] { 1,1 };
 
public double F(int index, Vector x)
{
switch (index)
{
case 0: return Math.Atan(x[0]) - x[1] * x[1] * x[1];
case 1: return 4 * x[0] * x[0] + 9 * x[1] * x[1] - 36;
}
throw new Exception("bad index");
}
 
public double df(int index, int derivative, Vector x)
{
switch (index)
{
case 0:
switch (derivative)
{
case 0: return 1 / (1 + x[0] * x[0]);
case 1: return -3*x[1]*x[1];
}
break;
case 1:
switch (derivative)
{
case 0: return 8 * x[0];
case 1: return 18 * x[1];
}
break;
}
throw new Exception("bad index");
}
public double[] weights() { return w; }
}
 
static void Main(string[] args)
{
Fun fun = new Fun();
Newton newton = new Newton();
Vector start = new Vector(new double[] { 2.75, 1.25 });
Vector X = newton.Do(2, fun, start);
X.print();
}
}
}
 
Output:

2.54258545959024 1.06149981539336

Go[edit]

Translation of: Kotlin


We follow the Kotlin example of coding our own matrix methods rather than using a third party library.

package main
 
import (
"fmt"
"math"
)
 
type vector = []float64
type matrix []vector
type fun = func(vector) float64
type funs = []fun
type jacobian = []funs
 
func (m1 matrix) mul(m2 matrix) matrix {
rows1, cols1 := len(m1), len(m1[0])
rows2, cols2 := len(m2), len(m2[0])
if cols1 != rows2 {
panic("Matrices cannot be multiplied.")
}
result := make(matrix, rows1)
for i := 0; i < rows1; i++ {
result[i] = make(vector, cols2)
for j := 0; j < cols2; j++ {
for k := 0; k < rows2; k++ {
result[i][j] += m1[i][k] * m2[k][j]
}
}
}
return result
}
 
func (m1 matrix) sub(m2 matrix) matrix {
rows, cols := len(m1), len(m1[0])
if rows != len(m2) || cols != len(m2[0]) {
panic("Matrices cannot be subtracted.")
}
result := make(matrix, rows)
for i := 0; i < rows; i++ {
result[i] = make(vector, cols)
for j := 0; j < cols; j++ {
result[i][j] = m1[i][j] - m2[i][j]
}
}
return result
}
 
func (m matrix) transpose() matrix {
rows, cols := len(m), len(m[0])
trans := make(matrix, cols)
for i := 0; i < cols; i++ {
trans[i] = make(vector, rows)
for j := 0; j < rows; j++ {
trans[i][j] = m[j][i]
}
}
return trans
}
 
func (m matrix) inverse() matrix {
le := len(m)
for _, v := range m {
if len(v) != le {
panic("Not a square matrix")
}
}
aug := make(matrix, le)
for i := 0; i < le; i++ {
aug[i] = make(vector, 2*le)
copy(aug[i], m[i])
// augment by identity matrix to right
aug[i][i+le] = 1
}
aug.toReducedRowEchelonForm()
inv := make(matrix, le)
// remove identity matrix to left
for i := 0; i < le; i++ {
inv[i] = make(vector, le)
copy(inv[i], aug[i][le:])
}
return inv
}
 
// note: this mutates the matrix in place
func (m matrix) toReducedRowEchelonForm() {
lead := 0
rowCount, colCount := len(m), len(m[0])
for r := 0; r < rowCount; r++ {
if colCount <= lead {
return
}
i := r
 
for m[i][lead] == 0 {
i++
if rowCount == i {
i = r
lead++
if colCount == lead {
return
}
}
}
 
m[i], m[r] = m[r], m[i]
if div := m[r][lead]; div != 0 {
for j := 0; j < colCount; j++ {
m[r][j] /= div
}
}
 
for k := 0; k < rowCount; k++ {
if k != r {
mult := m[k][lead]
for j := 0; j < colCount; j++ {
m[k][j] -= m[r][j] * mult
}
}
}
lead++
}
}
 
func solve(fs funs, jacob jacobian, guesses vector) vector {
size := len(fs)
var gu1 vector
gu2 := make(vector, len(guesses))
copy(gu2, guesses)
jac := make(matrix, size)
for i := 0; i < size; i++ {
jac[i] = make(vector, size)
}
tol := 1e-8
maxIter := 12
iter := 0
for {
gu1 = gu2
g := matrix{gu1}.transpose()
t := make(vector, size)
for i := 0; i < size; i++ {
t[i] = fs[i](gu1)
}
f := matrix{t}.transpose()
for i := 0; i < size; i++ {
for j := 0; j < size; j++ {
jac[i][j] = jacob[i][j](gu1)
}
}
g1 := g.sub(jac.inverse().mul(f))
gu2 = make(vector, size)
for i := 0; i < size; i++ {
gu2[i] = g1[i][0]
}
iter++
any := false
for i, v := range gu2 {
if math.Abs(v)-gu1[i] > tol {
any = true
break
}
}
if !any || iter >= maxIter {
break
}
}
return gu2
}
 
func main() {
/*
solve the two non-linear equations:
y = -x^2 + x + 0.5
y + 5xy = x^2
given initial guesses of x = y = 1.2
 
Example taken from:
http://www.fixoncloud.com/Home/LoginValidate/OneProblemComplete_Detailed.php?problemid=286
 
Expected results: x = 1.23332, y = 0.2122
*/

f1 := func(x vector) float64 { return -x[0]*x[0] + x[0] + 0.5 - x[1] }
f2 := func(x vector) float64 { return x[1] + 5*x[0]*x[1] - x[0]*x[0] }
fs := funs{f1, f2}
jacob := jacobian{
funs{
func(x vector) float64 { return -2*x[0] + 1 },
func(x vector) float64 { return -1 },
},
funs{
func(x vector) float64 { return 5*x[1] - 2*x[0] },
func(x vector) float64 { return 1 + 5*x[0] },
},
}
guesses := vector{1.2, 1.2}
sol := solve(fs, jacob, guesses)
fmt.Printf("Approximate solutions are x = %.7f, y = %.7f\n", sol[0], sol[1])
 
/*
solve the three non-linear equations:
9x^2 + 36y^2 + 4z^2 - 36 = 0
x^2 - 2y^2 - 20z = 0
x^2 - y^2 + z^2 = 0
given initial guesses of x = y = 1.0 and z = 0.0
 
Example taken from:
http://mathfaculty.fullerton.edu/mathews/n2003/FixPointNewtonMod.html (exercise 5)
 
Expected results: x = 0.893628, y = 0.894527, z = -0.0400893
*/

 
fmt.Println()
f3 := func(x vector) float64 { return 9*x[0]*x[0] + 36*x[1]*x[1] + 4*x[2]*x[2] - 36 }
f4 := func(x vector) float64 { return x[0]*x[0] - 2*x[1]*x[1] - 20*x[2] }
f5 := func(x vector) float64 { return x[0]*x[0] - x[1]*x[1] + x[2]*x[2] }
fs = funs{f3, f4, f5}
jacob = jacobian{
funs{
func(x vector) float64 { return 18 * x[0] },
func(x vector) float64 { return 72 * x[1] },
func(x vector) float64 { return 8 * x[2] },
},
funs{
func(x vector) float64 { return 2 * x[0] },
func(x vector) float64 { return -4 * x[1] },
func(x vector) float64 { return -20 },
},
funs{
func(x vector) float64 { return 2 * x[0] },
func(x vector) float64 { return -2 * x[1] },
func(x vector) float64 { return 2 * x[2] },
},
}
guesses = vector{1, 1, 0}
sol = solve(fs, jacob, guesses)
fmt.Printf("Approximate solutions are x = %.7f, y = %.7f, z = %.7f\n", sol[0], sol[1], sol[2])
}
Output:
Approximate solutions are x = 1.2333178,  y = 0.2122450

Approximate solutions are x = 0.8936282,  y = 0.8945270,  z = -0.0400893


Julia[edit]

NLsolve is a Julia package for nonlinear systems of equations, with the Newton-Raphson method one of the choices for solvers.

# from the NLSolve documentation: to solve
# (x, y) -> ((x+3)*(y^3-7)+18, sin(y*exp(x)-1))
using NLsolve
 
function f!(F, x)
F[1] = (x[1]+3)*(x[2]^3-7)+18
F[2] = sin(x[2]*exp(x[1])-1)
end
 
function j!(J, x)
J[1, 1] = x[2]^3-7
J[1, 2] = 3*x[2]^2*(x[1]+3)
u = exp(x[1])*cos(x[2]*exp(x[1])-1)
J[2, 1] = x[2]*u
J[2, 2] = u
end
 
println(nlsolve(f!, j!, [ 0.1; 1.2], method = :newton))
 
Output:
Results of Nonlinear Solver Algorithm
 * Algorithm: Newton with line-search
 * Starting Point: [0.1, 1.2]
 * Zero: [-3.7818e-16, 1.0]
 * Inf-norm of residuals: 0.000000
 * Iterations: 4
 * Convergence: true
   * |x - x'| < 0.0e+00: false
   * |f(x)| < 1.0e-08: true
 * Function Calls (f): 5
 * Jacobian Calls (df/dx): 4


Kotlin[edit]

A straightforward approach multiplying by the inverse of the Jacobian, rather than dividing by f'(x) as one would do in the single dimensional case, which is quick enough here.

As neither the JDK nor the Kotlin Standard Library have matrix functions built in, most of the functions used have been taken from other tasks.

// Version 1.2.31
 
import kotlin.math.abs
 
typealias Vector = DoubleArray
typealias Matrix = Array<Vector>
typealias Func = (Vector) -> Double
typealias Funcs = List<Func>
typealias Jacobian = List<Funcs>
 
operator fun Matrix.times(other: Matrix): Matrix {
val rows1 = this.size
val cols1 = this[0].size
val rows2 = other.size
val cols2 = other[0].size
require(cols1 == rows2)
val result = Matrix(rows1) { Vector(cols2) }
for (i in 0 until rows1) {
for (j in 0 until cols2) {
for (k in 0 until rows2) {
result[i][j] += this[i][k] * other[k][j]
}
}
}
return result
}
 
operator fun Matrix.minus(other: Matrix): Matrix {
val rows = this.size
val cols = this[0].size
require(rows == other.size && cols == other[0].size)
val result = Matrix(rows) { Vector(cols) }
for (i in 0 until rows) {
for (j in 0 until cols) {
result[i][j] = this[i][j] - other[i][j]
}
}
return result
}
 
fun Matrix.transpose(): Matrix {
val rows = this.size
val cols = this[0].size
val trans = Matrix(cols) { Vector(rows) }
for (i in 0 until cols) {
for (j in 0 until rows) trans[i][j] = this[j][i]
}
return trans
}
 
fun Matrix.inverse(): Matrix {
val len = this.size
require(this.all { it.size == len }) { "Not a square matrix" }
val aug = Array(len) { DoubleArray(2 * len) }
for (i in 0 until len) {
for (j in 0 until len) aug[i][j] = this[i][j]
// augment by identity matrix to right
aug[i][i + len] = 1.0
}
aug.toReducedRowEchelonForm()
val inv = Array(len) { DoubleArray(len) }
// remove identity matrix to left
for (i in 0 until len) {
for (j in len until 2 * len) inv[i][j - len] = aug[i][j]
}
return inv
}
 
fun Matrix.toReducedRowEchelonForm() {
var lead = 0
val rowCount = this.size
val colCount = this[0].size
for (r in 0 until rowCount) {
if (colCount <= lead) return
var i = r
 
while (this[i][lead] == 0.0) {
i++
if (rowCount == i) {
i = r
lead++
if (colCount == lead) return
}
}
 
val temp = this[i]
this[i] = this[r]
this[r] = temp
 
if (this[r][lead] != 0.0) {
val div = this[r][lead]
for (j in 0 until colCount) this[r][j] /= div
}
 
for (k in 0 until rowCount) {
if (k != r) {
val mult = this[k][lead]
for (j in 0 until colCount) this[k][j] -= this[r][j] * mult
}
}
 
lead++
}
}
 
fun solve(funcs: Funcs, jacobian: Jacobian, guesses: Vector): Vector {
val size = funcs.size
var gu1: Vector
var gu2 = guesses.copyOf()
val jac = Matrix(size) { Vector(size) }
val tol = 1.0e-8
val maxIter = 12
var iter = 0
do {
gu1 = gu2
val g = arrayOf(gu1).transpose()
val f = arrayOf(Vector(size) { funcs[it](gu1) }).transpose()
for (i in 0 until size) {
for (j in 0 until size) {
jac[i][j] = jacobian[i][j](gu1)
}
}
val g1 = g - jac.inverse() * f
gu2 = Vector(size) { g1[it][0] }
iter++
}
while (gu2.withIndex().any { iv -> abs(iv.value - gu1[iv.index]) > tol } && iter < maxIter)
return gu2
}
 
fun main(args: Array<String>) {
/* solve the two non-linear equations:
y = -x^2 + x + 0.5
y + 5xy = x^2
given initial guesses of x = y = 1.2
 
Example taken from:
http://www.fixoncloud.com/Home/LoginValidate/OneProblemComplete_Detailed.php?problemid=286
 
Expected results: x = 1.23332, y = 0.2122
*/

 
val f1: Func = { x -> -x[0] * x[0] + x[0] + 0.5 - x[1] }
val f2: Func = { x -> x[1] + 5 * x[0] * x[1] - x[0] * x[0] }
val funcs = listOf(f1, f2)
val jacobian = listOf(
listOf<Func>({ x -> - 2.0 * x[0] + 1.0 }, { _ -> -1.0 }),
listOf<Func>({ x -> 5.0 * x[1] - 2.0 * x[0] }, { x -> 1.0 + 5.0 * x[0] })
)
val guesses = doubleArrayOf(1.2, 1.2)
val (xx, yy) = solve(funcs, jacobian, guesses)
System.out.printf("Approximate solutions are x = %.7f, y = %.7f\n", xx, yy)
 
/* solve the three non-linear equations:
9x^2 + 36y^2 + 4z^2 - 36 = 0
x^2 - 2y^2 - 20z = 0
x^2 - y^2 + z^2 = 0
given initial guesses of x = y = 1.0 and z = 0.0
 
Example taken from:
http://mathfaculty.fullerton.edu/mathews/n2003/FixPointNewtonMod.html (exercise 5)
 
Expected results: x = 0.893628, y = 0.894527, z = -0.0400893
*/

 
println()
val f3: Func = { x -> 9.0 * x[0] * x[0] + 36.0 * x[1] * x[1] + 4.0 * x[2] * x[2] - 36.0 }
val f4: Func = { x -> x[0] * x[0] - 2.0 * x[1] * x[1] - 20.0 * x[2] }
val f5: Func = { x -> x[0] * x[0] - x[1] * x[1] + x[2] * x[2] }
val funcs2 = listOf(f3, f4, f5)
val jacobian2 = listOf(
listOf<Func>({ x -> 18.0 * x[0] }, { x -> 72.0 * x[1] }, { x -> 8.0 * x[2] }),
listOf<Func>({ x -> 2.0 * x[0] }, { x -> -4.0 * x[1] }, { _ -> -20.0 }),
listOf<Func>({ x -> 2.0 * x[0] }, { x -> -2.0 * x[1] }, { x -> 2.0 * x[2] })
)
val guesses2 = doubleArrayOf(1.0, 1.0, 0.0)
val (xx2, yy2, zz2) = solve(funcs2, jacobian2, guesses2)
System.out.printf("Approximate solutions are x = %.7f, y = %.7f, z = %.7f\n", xx2, yy2, zz2)
}
Output:
Approximate solutions are x = 1.2333178,  y = 0.2122450

Approximate solutions are x = 0.8936282,  y = 0.8945270,  z = -0.0400893

Perl 6[edit]

#!/usr/bin/env perl6
 
# Reference:
# https://github.com/pierre-vigier/Perl6-Math-Matrix
# Mastering Algorithms with Perl
# By Jarkko Hietaniemi, John Macdonald, Jon Orwant
# Publisher: O'Reilly Media, ISBN-10: 1565923987
# https://resources.oreilly.com/examples/9781565923980/blob/master/ch16/solve
 
use v6;
 
sub solve_funcs ($funcs, @guesses, $iterations, $epsilon) {
my ($error_value, @values, @delta, @jacobian); my= $epsilon;
for 1 .. $iterations {
for ^+$funcs { @values[$^i] = $funcs[$^i](|@guesses); }
$error_value = 0;
for ^+$funcs { $error_value += @values[$^i].abs }
return @guesses if $error_value ≤ ε;
for ^+$funcs { @delta[$^i] = [email protected]values[$^i] }
@jacobian = jacobian $funcs, @guesses, ε;
@delta = solve_matrix @jacobian, @delta;
loop (my $j = 0, $error_value = 0; $j < +$funcs; $j++) {
$error_value += @delta[$j].abs ;
@guesses[$j] += @delta[$j];
}
return @guesses if $error_value ≤ ε;
}
return @guesses;
}
 
sub jacobian ($funcs is copy, @points is copy, $epsilon is copy) {
my (, @P, @M, @plusΔ, @minusΔ);
my Array @jacobian; my= $epsilon;
for ^+@points -> $i {
@plusΔ = @minusΔ = @points;
= (ε * @points[$i].abs) || ε;
@plusΔ[$i] = @points[$i] +;
@minusΔ[$i] = @points[$i] -;
for ^+$funcs { @P[$^k] = $funcs[$^k](|@plusΔ); }
for ^+$funcs { @M[$^k] = $funcs[$^k](|@minusΔ); }
for ^+$funcs -> $j {
@jacobian[$j][$i] = (@P[$j] - @M[$j]) / (2 *);
}
}
return @jacobian;
}
 
sub solve_matrix (@matrix_array is copy, @delta is copy) {
# https://github.com/pierre-vigier/Perl6-Math-Matrix/issues/56
{ use Math::Matrix;
my $matrix = Math::Matrix.new(@matrix_array);
my $vector = Math::Matrix.new(@delta.map({.list}));
die "Matrix is not invertible" unless $matrix.is-invertible;
my @result = ( $matrix.inverted dot $vector ).transposed;
return @result.split(" ");
}
}
 
my $funcs = [
{ 9*$^x² + 36*$^y² + 4*$^z² - 36 }
{ $^x² - 2*$^y² - 20*$^z }
{ $^x² - $^y² + $^z² }
];
 
my @guesses = (1,1,0);
 
my @solution = solve_funcs $funcs, @guesses, 20, 1e-8;
 
say "Solution: ", @solution;
 
Output:
Solution: [0.8936282344764825 0.8945270103905782 -0.04008928615915281]

Phix[edit]

Translation of: Go

Uses code from Reduced_row_echelon_form#Phix, Gauss-Jordan_matrix_inversion#Phix, Matrix_transposition#Phix, and Matrix_multiplication#Phix
See std distro for a complete runnable version.

-- demo\rosetta\Multidimensional_Newton-Raphson_method.exw
function solve(sequence fs, jacob, guesses)
integer size := length(fs),
maxIter := 12,
iter := 0
sequence gu1, g, t, f, g1,
gu2 := guesses,
jac := repeat(repeat(0,size),size)
atom tol := 1e-8
while true do
gu1 := gu2
g := matrix_transpose({gu1})
t := repeat(0, size)
for i=1 to size do
t[i] := call_func(fs[i],{gu1})
end for
f := matrix_transpose({t})
for i=1 to size do
for j=1 to size do
jac[i][j] := call_func(jacob[i][j],{gu1})
end for
end for
g1 := sq_sub(g,matrix_mul(inverse(jac),f))
gu2 := vslice(g1,1)
iter += 1
bool any := find(true,sq_gt(sq_sub(sq_abs(gu2),gu1),tol))!=0
if not any or iter >= maxIter then exit end if
end while
return gu2
end function
 
function f1(sequence v) atom {x,y} = v return -x*x+x+0.5-y end function
function f2(sequence v) atom {x,y} = v return y+5*x*y-x*x end function
function f3(sequence v) atom {x,y,z} = v return 9*x*x+36*y*y+4*z*z-36 end function
function f4(sequence v) atom {x,y,z} = v return x*x-2*y*y-20*z end function
function f5(sequence v) atom {x,y,z} = v return x*x-y*y+z*z end function
 
function j1(sequence v) atom {x} = v return -2*x+1 end function
function j2(sequence /*v*/) return -1 end function
function j3(sequence v) atom {x,y} = v return 5*y-2*x end function
function j4(sequence v) atom {x} = v return 1+5*x end function
function j11(sequence v) atom {x} = v return 18*x end function
function j12(sequence v) atom {?,y} = v return 72*y end function
function j13(sequence v) atom {?,?,z} = v return 8*z end function
function j21(sequence v) atom {x} = v return 2*x end function
function j22(sequence v) atom {?,y} = v return -4*y end function
function j23(sequence /*v*/) return -20 end function
function j31(sequence v) atom {x} = v return 2*x end function
function j32(sequence v) atom {?,y} = v return -2*y end function
function j33(sequence v) atom {?,?,z} = v return 2*z end function
 
procedure main()
sequence fs, jacob, guesses
/*
solve the two non-linear equations:
y = -x^2 + x + 0.5
y + 5xy = x^2
given initial guesses of x = y = 1.2
 
Example taken from:
http://www.fixoncloud.com/Home/LoginValidate/OneProblemComplete_Detailed.php?problemid=286
 
Expected results: x = 1.23332, y = 0.2122
*/
fs = {routine_id("f1"),routine_id("f2")}
jacob = {{routine_id("j1"),routine_id("j2")},
{routine_id("j3"),routine_id("j4")}}
guesses := {1.2, 1.2}
printf(1,"Approximate solutions are x = %.7f, y = %.7f\n\n", solve(fs, jacob, guesses))
 
/*
solve the three non-linear equations:
9x^2 + 36y^2 + 4z^2 - 36 = 0
x^2 - 2y^2 - 20z = 0
x^2 - y^2 + z^2 = 0
given initial guesses of x = y = 1.0 and z = 0.0
 
Example taken from:
http://mathfaculty.fullerton.edu/mathews/n2003/FixPointNewtonMod.html (exercise 5)
 
Expected results: x = 0.893628, y = 0.894527, z = -0.0400893
*/
 
fs = {routine_id("f3"), routine_id("f4"), routine_id("f5")}
jacob = {{routine_id("j11"),routine_id("j12"),routine_id("j13")},
{routine_id("j21"),routine_id("j22"),routine_id("j23")},
{routine_id("j31"),routine_id("j32"),routine_id("j33")}}
guesses = {1, 1, 0}
printf(1,"Approximate solutions are x = %.7f, y = %.7f, z = %.7f\n", solve(fs, jacob, guesses))
 
end procedure
main()
Output:
Approximate solutions are x = 1.2333178,  y = 0.2122450

Approximate solutions are x = 0.8936282,  y = 0.8945270,  z = -0.04008929

zkl[edit]

This doesn't use Newton-Raphson (with derivatives) but a hybrid algorithm.

var [const] GSL=Import.lib("zklGSL");    // libGSL (GNU Scientific Library)
 
// two functions of two variables: f(x,y)=0
fs:=T(fcn(x,y){ x.atan() - y*y*y }, fcn(x,y){ 4.0*x*x + 9*y*y - 36 });
v=GSL.VectorFromData(2.75, 1.25); // an initial guess at the solution
GSL.multiroot_fsolver(fs,v);
v.format(11,8).println(); // answer overwrites initial guess
 
fs.run(True,v.toList().xplode()).println(); // deltas from zero
Output:
 2.59807621, 1.06365371
L(2.13651e-09,2.94321e-10)

A condensed solver (for a different set of functions):

v:=GSL.VectorFromData(-10.0, -15.0);
GSL.multiroot_fsolver(T( fcn(x,y){ 1.0 - x }, fcn(x,y){ 10.0*(y - x*x) }),v)
.format().println(); // --> (1,1)
Output:
1.00,1.00

Another example:

v:=GSL.VectorFromData(1.0, 1.0, 0.0);	// initial guess
fxyzs:=T(
fcn(x,y,z){ x*x*9 + y*y*36 + z*z*4 - 36 }, // 9x^2 + 36y^2 + 4z^2 - 36 = 0
fcn(x,y,z){ x*x - y*y*2 - z*20 }, // x^2 - 2y^2 - 20z = 0
fcn(x,y,z){ x*x - y*y + z*z }); // x^2 - y^2 + z^2 = 0
(v=GSL.multiroot_fsolver(fxyzs,v)).format(12,8).println();
 
fxyzs.run(True,v.toList().xplode()).println(); // deltas from zero
Output:
  0.89362824,  0.89452701, -0.04008929
L(6.00672e-08,1.0472e-08,9.84017e-09)