simultaneous equation solver quadratic 3 unknowns

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SIMULTANEOUS EQUATION SOLVER QUADRATIC 3 UNKNOWNS

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(1)

Usually there will be no set of x_i which exactly satisfies (1). Let us define an error vector e_j by

$\begin{displaymath} \left[ \begin{array} {cccc} a_{11} & a_{12} & \cdots & a_{... ... {c} e_1 \\ e_2 \\ \vdots \\ e_n \\ \end{array} \right]\end{displaymath}$ (2)

It simplifies the development to rewrite this equation as follows (a trick I learned from John P. Berg).

$\begin{displaymath} \left[ \begin{array} {cccc} -c_1 & a_{11} & \cdots & a_{1m... ... {c} e_1 \\ e_2 \\ \vdots \\ e_n \\ \end{array} \right]\end{displaymath}$ (3)

We may abbreviate this equation as

$\begin{displaymath} {\bf Bx} \eq {\bf e}\end{displaymath}$ (4)

where B is the matrix containing c and a. The ith error may be written as a dot product and either vector may be written as the column

$\begin{eqnarraystar} e_i \eq [b_{i1} \quad b_{i2}\quad \cdots] \; \left[ \beg... ...1} \\ b_{i2} \\ \vdots \end{array} \right] \nonumber \\ \end{eqnarraystar}$

Now we will minimize the sum squared error E defined as $\sum {e_i}^2$

$\begin{displaymath} E \eq \sum_i [1 \quad x_1 \quad \cdots] \; \left[ \begin{a... ...\begin{array} {c} 1 \\ x_1 \\ \vdots \end{array} \right]\end{displaymath}$ (5)

The summation may be brought inside the constants

$\begin{displaymath} E \eq [1 \quad x_1 \quad x_2 \quad \cdots] \; \left\{ \sum... ...ay} {c} 1 \\ x_1 \\ x_2 \\ \vdots \end{array} \right]\end{displaymath}$ (6)

The matrix in the center, call it r_ij, is symmetrical. It is a positive (more strictly, nonnegative) definite matrix because you will never be able to find an x for which E is negative, since E is a sum of squared e_i. We find the x with minimum E by requiring $\partial E/\partial x_1 = 0, \ \partial E/\partial x_2 = 0, \, \ldots , \partial E/\partial x_m = 0.$ Notice that this will give us exactly one equation for each unknown. In order to clarify the presentation we will specialize (6) to two unknowns.

$\begin{displaymath} E \eq [1 \quad x_1 \quad x_2] \; \left[ \begin{array} {ccc... ...[ \begin{array} {c} 1 \\ x_1 \\ x_2 \end{array} \right]\end{displaymath}$ (7)

Setting to zero the derivative with respect to x₁, we get

$\begin{displaymath} 0 \eq {\partial E \over \partial x_1} \eq [0 \quad 1 \quad ... ...left[ \begin{array} {c} 0 \\ 1 \\ 0 \end{array} \right]\end{displaymath}$ (8)

Since r_ij = r_ji, both terms on the right are equal. Thus (8) may be written

$\begin{displaymath} 0 \eq {\partial E \over \partial x_1} \eq 2 [r_{10} \quad r... ...[ \begin{array} {c} 1 \\ x_1 \\ x_2 \end{array} \right]\end{displaymath}$ (9)

Likewise, differentiating with respect to x₂ gives

$\begin{displaymath} 0 \eq {\partial E \over \partial x_2} \eq 2 [r_{20} \quad r... ...[ \begin{array} {c} 1 \\ x_1 \\ x_2 \end{array} \right]\end{displaymath}$ (10)

Equations (9) and (10) may be combined

$\begin{displaymath} \left[ \begin{array} {c} 0 \\ 0 \end{array} \right] \e... ...[ \begin{array} {c} 1 \\ x_1 \\ x_2 \end{array} \right]\end{displaymath}$ (11)

This form is two equations in two unknowns. One might write it in the more conventional form

$\begin{displaymath} \left[ \begin{array} {cc} r_{11} & r_{12} \\ r_{21} & r_{... ...ft[ \begin{array} {c} r_{10} \\ r_{20} \end{array} \right]\end{displaymath}$ (12)

The matrix of (11) lacks only a top row to be equal to the matrix of (7). To give it that row, we may augment (11) by

$\begin{displaymath} v \eq r_{00} + r_{01} x_1 + r_{02} x_2\end{displaymath}$ (13)

where (13) may be regarded as a definition of a new variable v. Putting (13) on top of (11) we get

$\begin{displaymath} \left[ \begin{array} {c} v \\ 0 \\ 0 \end{array} \ri... ...[ \begin{array} {c} 1 \\ x_1 \\ x_2 \end{array} \right]\end{displaymath}$ (14)

The solution x of (12) or (14) is that set of x_k for which E is a minimum. To get an interpretation of v, we may multiply both sides by $[1 \quad x_1 \quad x_2]$ , getting

$\begin{displaymath} v \eq [1 \quad x_1 \quad x_2] \; \left[ \begin{array} {c} ... ...[ \begin{array} {c} 1 \\ x_1 \\ x_2 \end{array} \right]\end{displaymath}$ (15)

Comparing (15) with (7), we see that v is the minimum value of E.

Occasionally, it is more convenient to have the essential equations in partitioned matrix form. In partitioned matrix form, we have for the error (6)

$\begin{displaymath} E \eq [1 \ \vdots \ {\bf x}]^T \left[ \begin{array} {c} -... ...1 \\ \hbox to 0.25in{\dotfill} \\ {\bf x} \end{array} \right]\end{displaymath}$ (16)

The final equation (14) splits into

$\begin{eqnarray} V &= & {\bf c}^T{\bf c} - {\bf c}^T {\bf Ax} \\ {\bf 0} &= & -{\bf A}^T {\bf c} + {\bf A}^T {\bf Ax}\end{eqnarray}$ (17)

(18)

where (18) represents simultaneous equations to be solved for x. Equation (18) is what you have to set up in a computer. It is easily remembered by a quick and dirty (very dirty) derivation. That is, we began with the overdetermined equations ${\bf Ax} \approx {\bf c}$ ;premultiplying by ${\bf A}^T$ gives $({\bf A}^T {\bf A}) {\bf x} = {\bf A}^T {\bf c}$ which is (18).

In physical science applications, the variable z_j is frequently a complex variable, say z_j = x_j + iy_j. It is always possible to go through the foregoing analyses, treating the problem as though x_i and y_i were real independent variables. There is a considerable gain in simplicity and a saving in computational effort by treating z_j as a single complex variable. The error E may be regarded as a function of either x_j and y_j or z_j and $\overline{z}_j$ . In general $j = 1,\, 2, \, \ldots , \, N,$ but we will treat the case N = 1 here and leave the general case for the Exercises. The minimum is found where

$\begin{eqnarray} 0 &= & {dE \over dx} \eq {\partial E \over \partial z} {dz \ov... ...er \partial z} - {\partial E \over \partial \overline{z}} \right)\end{eqnarray}$ (19)

(20)

Multiplying (20) by i and adding and subtracting these equations, we may express the minimum condition more simply as

$\begin{eqnarray} 0 &= & {\partial E \over \partial z} \\ 0 &= & {\partial E \over \partial \overline{z}}\end{eqnarray}$ (21)

(22)

However, the usual case is that E is a positive real quadratic function of z and $\overline{z}$ and that $\partial E/\partial z$ is merely the complex conjugate of $\partial E/\partial \overline{z}$ . Then the two conditions (21) and (22) may be replaced by either one of them. Usually, when working with complex variables we are minimizing a positive quadratic form like

$\begin{displaymath} E(z^*, z) \eq \vert{\bf Az} - {\bf c}\vert^2 \eq ({\bf z}^* {\bf A}^* - {\bf c}^*) ({\bf Az} - {\bf c})\end{displaymath}$ (23)

where ^* denotes complex-conjugate transpose. Now (22) gives

$\begin{displaymath} 0 \eq {\partial E \over \partial z^*} \eq {\bf A}^* ({\bf Az} - {\bf c})\end{displaymath}$ (24)

which is just the complex form of (18).

Let us consider an example. Suppose a set of wave arrival times t_i is measured at sensors located on the x axis at points x_i. Suppose the wavefront is to be fitted to a parabola $t_i \approx a + bx_i + {cx_i}^2$ .Here, the x_i are knowns and a, b, and c are unknowns. For each sensor i we have an equation

$\begin{displaymath}[-t_i\quad 1 \quad x_i \quad {x_i}^2] \; \left[ \begin{array}... ...\ a \\ b \\ c \end{array} \right] \quad\approx\quad 0\end{displaymath}$ (25)

When i has greater range than 3 we have more equations than unknowns. In this example, (14) takes the form

$\begin{displaymath} \left\{ \sum^n_{i = 1} \left[ \begin{array} {c} -t_i \\ ... ...begin{array} {c} v \\ 0 \\ 0 \\ 0 \end{array} \right]\end{displaymath}$ (26)

This may be solved by standard methods for a, b, and c.

The last three rows of (26) may be written

$\begin{displaymath} \sum^n_{i = 1} \left[ \begin{array} {c} 1 \\ x_i \\ ... ...left[ \begin{array} {c} 0 \\ 0 \\ 0 \end{array} \right]\end{displaymath}$ (27)

$74.99