Least-Squares Polynomials

Because polynomials can be readily manipulated, fitting such functions to data that do not plot linearly is common.

We assume the functional relationship

$\displaystyle y=a_0+a_1x+a_2x^2+\ldots+a_nx^n$

(7.5)

with errors defined by

$\displaystyle e_i=Y_i - y_i=Y_i- (a_0+a_1x+a_2x^2+\ldots+a_nx^n)$

We again use $Y_i$

to represent the observed or experimental value corresponding to $x_i$

, with

free of error.

We minimize the sum of squares;

$\displaystyle S=\sum_{i=1}^N e_i^2=\sum_{i=1}^N (Y_i - a_0 - a_1x - a_2x^2 - \ldots - a_nx^n)^2$

At the minimum, all the partial derivatives $\partial S/\partial a_0,\partial S/\partial a_n$ vanish.

Writing the equations for these gives $n + 1$

equations:

$\begin{indisplay} \begin{array}{l} \frac{\partial S}{\partial a_0}=0=\sum_{i=1}^... ...(Y_i- a_0-a_1x_i-a_2x_i^2-\ldots-a_ix_i^n)(-x_i^n)\\ \end{array}\end{indisplay}$

Dividing each by $-2$

and rearranging gives the $n + 1$

normal equations to be solved simultaneously:

$\begin{indisplay}\begin{array}{rl} a_0N+a_1\sum x_i+a_2 \sum x_i^2+\ldots+a_n\su... ...x_i^{n+2}+\ldots+a_n\sum x_i^{2n}= & \sum x_i^nY_i\\ \end{array}\end{indisplay}$

(7.6)

Putting these equations in matrix form shows the coefficient matrix;

$\begin{indisplay}\left[ \begin{array}{rrrrrl} N & \sum x_i & \sum x_i^2 & \sum x... ... \vdots \\ \sum x_i^n Y_i\\ \end{array}\right] \hspace{-0.5cm}\end{indisplay}$

(7.7)

All the summatins in Eqs. 7.6 and 7.7 run from 1 to $N$

. We will let B stand for the coefficient matrix.

Equation 7.7 represents a linear system.

Degrees higher than 4 are used very infrequently. It is often better to fit a series of lower-degree polynomials to subsets of the data.

Matrix

of Eq. 7.7 is called the normal matrix for the least-squares problem.

There is another matrix that corresponds to this, called the design matrix. It is of the form;

$\begin{indisplay} A=\left[ \begin{array}{rrrrr} 1 & 1 & 1 & 1 & 1 \\ x_1 & x_... ... x_1^n & x_2^n & x_3^n & \ldots & x_N^n\\ \end{array} \right] \end{indisplay}$

is just the coefficient matrix of Eq. 7.7. It is easy to see that $Ay$

, where

is the column vector of $Y$

-values, gives the right-hand side of Eq. 7.7. We can rewrite Eq. 7.7 in matrix form, as

$\displaystyle AA^Ta = Ba= Ay$

so it is to find the solution.

It is illustrated the use of Eq. 7.6 to fit a quadratic to the data of Table 7.7. Figure 7.8 shows a plot of the data.

The data are actually a perturbation of the relation $y=1 -x+0.2 x^2$

**Table 7.7:** Data to illustrate curve fitting.
**Table 7.8:** Figure for the data to illustrate curve fitting.
$\begin{table} % latex2html id marker 6963 \begin{minipage}[h]{0.6\linewidth} To ... ...} \includegraphics[scale=0.5]{images/3.7} \end{center} \end{minipage}\end{table}$

The equations to be solved are:

$\begin{indisplay} \begin{array}{rl} 11a_0+6.01a_1+4.6545a_2& =5.905 \\ 6.01a_... ....1839 \\ 4.6545a_0+4.1150a_1+3.9161a_2& =1.3357 \\ \end{array}\end{indisplay}$