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How to solve Non linear Equation solving
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Now the weighted sum-squared error will be
Following the method of the last section,
it is easy to show that the x
which minimizes the weighted error E
of (29) is the x
satisfies the simultaneous equations
Choice of a set of weights
is often a rather subjective matter.
if data are of uneven quality,
it cannot be avoided.
to choosing it equal to unity.
A case of common interest
is where some equations should be solved exactly.
Such equations are called constraint equations.
Constraint equations often
arise out of theoretical considerations so they may,
not have any error.
The rest of the equations often involve some measurement.
Since the measurement can often be made many times,
it is easy to get a lot more
equations than unknowns.
Since measurement always involves error,
we then use the method of least squares
to minimize the average error.
In order to be certain that the constraint equations are solved exactly,
one could use the trick of applying very large weight factors
to the constraint equations.
A problem is that
``very large'' is not well defined.
A weight equal 1010
might not be large enough
to guarantee the constraint equation is satisfied
with sufficient accuracy.
On the other hand, 1010 might lead to
disastrous round-off when solving
the simultaneous equations in a computer
with eight-digit accuracy.
The best approach is to analyze the situation
theoretically for .
An example of a constraint equation
is that the sum of the xi equals M.
Another constraint would be x1 = x2.
Arranged in a matrix,
these two constraint equations are
We write a general set of k
constraint equations as
Minimizing the error as
of the equations
is algebraically similar to
minimizing the error of Bx 0
The rows of
are just like some extra rows for B
The resulting equation for x
Now we will take all the wi
and we will let
tend to zero. Also let
With this, (33) may be written
Identify coefficients of powers of
Equation (37) is m
equations in m
It will automatically be satisfied
if the k
equations in (32) are satisfied.
Equation (38) appears to involve
more unknowns in
we do not need
Arranging (38) and (32)
together and dropping superscripts,
we get a square matrix in m + k unknowns.
Equation (40) is now a
simultaneous set for the unknowns and .
It might also be thought of as the solution to the problem
of minimizing the quadratic form
and since we can always transpose a scalar,