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FREE SIMULTANEOUS QUADRATIC EQUATION SOLVER
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Thank you for visiting our site! You landed on this page because you entered a search term similar to this: free simultaneous quadratic equation solver. We have an extensive database of resources on free simultaneous quadratic equation solver. Below is one of them. If you need further help, please take a look at our software "Algebrator", a software program that can solve any algebra problem you enter!

  (28)
Now the weighted sum-squared error will be  
 \begin{displaymath} E \eq \sum_i w_i {e_i}^2\end{displaymath} (29)
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 which satisfies the simultaneous equations  
 \begin{displaymath} \left\{ \sum_{i} w_i \; \left[ \begin{array} {c} -c_i \\ ... ...{array} {c} v \\ 0 \\ 0 \\ \vdots \end{array} \right]\end{displaymath} (30)
Choice of a set of weights is often a rather subjective matter. However, if data are of uneven quality, it cannot be avoided. Omitting w is equivalent 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, in principle, 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 $w \rightarrow \infty$.

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  

 \begin{displaymath} \left[ \begin{array} {rcrccc} -M & 1 & 1 & 1 & 1 & \cdots \... ...t] \eq \left[ \begin{array} {c} 0 \\ 0 \end{array} \right]\end{displaymath} (31)
We write a general set of k constraint equations as  
 \begin{displaymath} {\bf G} \left[ \begin{array} {c} 1 \\ x_1 \\ x_2 \\ ... ...eq \left[ \begin{array} {c} 0 \\ \vdots \end{array} \right]\end{displaymath} (32)
Minimizing the error as $w \rightarrow \infty$ of the equations
\begin{eqnarraystar} \sqrt{w} {\bf Gx} &\approx & {\bf 0} \nonumber \\ {\bf Bx} &\approx & {\bf 0} \end{eqnarraystar}
is algebraically similar to minimizing the error of Bx $\approx$ 0. The rows of $\sqrt{w} {\bf G}$ are just like some extra rows for B. The resulting equation for x is  
 \begin{displaymath} \left\{ \sum^n_{i = 1} \left[ \begin{array} {c} -c_i \\ ... ... \begin{array} {c} v \\ 0 \\ \vdots \end{array} \right]\end{displaymath} (33)
Now we will take all the wi to equal $1/\varepsilon$ and we will let $\varepsilon$tend to zero. Also let
      \begin{eqnarray} {\bf x} &= & {\bf x}^{(0)} + \varepsilon {\bf x}^{(1)} + \var... ... \varepsilon {\bf v}^{(1)} + \varepsilon^2 {\bf v}^{(2)} + \cdots\end{eqnarray} (34)
(35)
With this, (33) may be written  
 \begin{displaymath} \left( {\bf B}^T {\bf B} + {1 \over {\varepsilon}} {\bf G}^T... ...\cdots) \eq {\bf v}^{(0)} + {\bf v}^{(1)} \varepsilon + \cdots\end{displaymath} (36)
Identify coefficients of powers of $\varepsilon$
      \begin{eqnarray} \varepsilon^{-1} & : & {\bf G}^T {\bf Gx}^{(0)} = {\bf 0} \\ \... ... \varepsilon^1, \varepsilon^2 & : & \hbox{not required} \nonumber\end{eqnarray} (37)
(38)
Equation (37) is m equations in m unknowns. It will automatically be satisfied if the k equations in (32) are satisfied. Equation (38) appears to involve the m unknowns in ${\bf x}^{(0)}$ plus m more unknowns in ${\bf x}^{(1)}$. In fact, we do not need ${\bf x}^{(1)}$;the k unknowns  
 \begin{displaymath} \lambda \eq {\bf Gx}^{(1)}\end{displaymath} (39)
will suffice.

Arranging (38) and (32) together and dropping superscripts, we get a square matrix in m + k unknowns.  

 \begin{displaymath} \left[ \begin{array} {c} {\bf B}^T {\bf B} \\ \vbox{\h... ...t[ \begin{array} {c} v \\ 0 \\ \vdots \end{array} \right]\end{displaymath} (40)

Equation (40) is now a simultaneous set for the unknowns ${\bf x}$and $\lambda$. It might also be thought of as the solution to the problem of minimizing the quadratic form

\begin{eqnarraystar} E &= & [{\bf x}^T \quad \lambda^T]\left[ \begin{array} {cc}... ... \lambda^T {\bf Gx} + {\bf x}^T {\bf G}^T \lambda \nonumber \end{eqnarraystar}
and since we can always transpose a scalar,  
 \begin{displaymath} E \eq {\bf x}^T {\bf B}^T {\bf Bx} + 2\lambda^T {\bf Gx}\end{displaymath} (41)

 

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