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| Syntax |
snewton(function f, function df, fval, xini, mitr) |
| See Also | brent , dnlsq |
x with the same type and
dimension as fval, such that
f(x ) = fval for i = 1, ... , M, and j = 1, ... , N
i,j i,j
M and N are the
row and column dimension of fval respectively.
The matrix xini has the same type and dimension as fval.
The (i,j)-th element of xini is the starting
point when Newton's method is applied to solve the
corresponding equation above.
snewton returns the current estimate.
If you suspect this method might fail,
you should check that the equations are solved by the return value.
(Note the equations may be solved for a subset of the indices.)
(x)
The (i,j)-th element of the return value of (x)
is the value f(x(i, j)).
The argument x and the return value
f(x) have the same type and dimension as fval.
(x)
The (i,j)-th element of the return value of (x)
is the derivative of f evaluated at
the point x(i, j).
The argument x and the return value
df(x) have the same type and dimension as fval.
snewton to compute the
square root of the integers between 1 and 4
(though we would normally use the sqrt
function for this task).
The initial starting points are random numbers between
zero and one.
function f(x) begin
return x^2.
end
function df(x) begin
return 2. * x
end
fval = real(seq(4))
xini = rand(4, 1)
mitr = 10
snewton(function f, function df, fval, xini, mitr)
returns
{
1
1.41421
1.73205
2
}
which are the square roots of 1, 2, 3, and 4, respectively.