The formulas (5.63) apply only when the sample times are symmetric around 0 . When the sample points $t_1, \ldots, t_n$ are equally spaced, so $t_{i+1}-t_i=h$ for all $i=1, \ldots, n-1$, then there is a simple trick to convert the least squares problem into a symmetric form.
(a) Show that the translated sample points $s_i=t_i-\bar{t}$, where $\bar{t}=\frac{1}{n} \sum_{i=1}^n t_i$ is the average, are symmetric around 0 . (b) Suppose $q(s)$ is the least squares polynomial for the data points $\left(s_i, y_i\right)$. Prove that $p(t)=q(t-\bar{t})$ is the least squares polynomial for the original data $\left(t_i, y_i\right)$. (c) Apply this method to find the least squares polynomials of degrees 1 and 2 for the following data:
\begin{tabular}{c|cccccc}
$t_i$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline$y_i$ & -8 & -6 & -4 & -1 & 1 & 3
\end{tabular}