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Nonlinear Least Squares
Syntax nlsq(function fx0scalemaxitlevel)
See Also nlsqbox , dnlsq

Description
Uses the Gauss-Newton method to solve the problem
                    2
     minimize |f(x)|  with respect to x

The function call f(x) returns a column vector with the same type as x0 provided that x has the same type and dimension as x0. The real or double-precision column vector x0 specifies the point at which to start the Gauss-Newton method. The vector scale has the same type and dimension as x0; scale(i) is the maximum absolute change in x(i) per iteration, and .0001 scale(i) is the central difference step size used to approximate the partial derivative of f with respect to x(i). The integer scalar maxit specifies the maximum number of iterations to try before giving up on convergence. The integer scalar level specifies the amount of tracing within this function. If level > 1, the value of |f(x)|^2 is printed at each iteration.

The vector scale also specifies convergence criteria. Convergence is accepted when the absolute change in x(i) between iterations is less than .0001 scale(i) for all i.

The return value is a matrix with i-th column equal to the value of x at the beginning of the i-th iteration. The return value has the same type and row dimension as x0. The last column of the return value is the approximate solution.

Example
The following program minimizes the function
                                2                       2
     f(x) =  [exp(x )  - exp(1)]   + [exp(x )  - exp(2)]
                   1                       2
with respect to x. The resulting value of x is (1, 2).

clear
#
function f(x) begin
     return {exp(x(1)) - exp(1.), exp(x(2)) - exp(2.)}
end
level  = 0
maxit  = 20
x0     = {0., 0.}
scale  = {1., 1.}
result = nlsq(function f, x0, scale, maxit, level)
xout   = result.col(coldim(result))
print xout