Question 4.6.3: A missile is fired from enemy territory, and its position in......
A missile is fired from enemy territory, and its position in flight is observed by radar tracking devices at the following positions.
Suppose our intelligence sources indicate that enemy missiles are programmed to follow a parabolic flight path—a fact that seems to be consistent with the diagram obtained by plotting the observations on the coordinate system shown in Figure 4.6.4.
Problem: Predict how far down range the missile will land.
Determine the parabola f(t) = α_{0} + α_{1}t + α_{2}t^{2} that best fits the observed data in the least squares sense. Then estimate where the missile will land by determining the roots of f (i.e., determine where the parabola crosses the horizontal axis). As it stands, the problem will involve numbers having relatively large magnitudes in conjunction with relatively small ones. Consequently, it is better to first scale the data by considering one unit to be 1000 miles. If
A = \begin{pmatrix}1 &0& 0\\1 &.25& .0625\\1 &.5& .25\\1 &.75& .5625\\ 1 &1& 1 \end{pmatrix}, x = \begin{pmatrix}α_{0}\\α_{1}\\α_{2}\end{pmatrix}, and b = \begin{pmatrix}0\\.008\\.015\\.019\\.02\end{pmatrix},and if ε = Ax − b, then the object is to find a least squares solution x that minimizes
\sum\limits_{i=1}^{5}ε^{2}_{i} = ε^T ε = (Ax − b)^{T} (Ax − b).We know that such a least squares solution is given by the solution to the system of normal equations A^{T}Ax = A^{T}b, which in this case is
\begin{pmatrix}5 &2.5& 1.875\\2.5 &1.875& 1.5625\\1.875 &1.5625& 1.3828125\end{pmatrix}\begin{pmatrix}α_{0}\\α_{1}\\α_{2}\end{pmatrix} = \begin{pmatrix}.062\\.04375\\.0349375\end{pmatrix}.The solution (rounded to four significant digits) is
x = \begin{pmatrix}−2.286 × 10^{−4}\\3.983 × 10^{−2}\\−1.943 × 10^{−2}\end{pmatrix},
and the least squares parabola is
f(t) = −.0002286 + .03983t − .01943t².
To estimate where the missile will land, determine where this parabola crosses the horizontal axis by applying the quadratic formula to find the roots of f(t) to be t = .005755 and t = 2.044. Therefore, we estimate that the missile will land 2044 miles down range. The sum of the squares of the errors associated with the least squares solution is
\sum\limits_{i=1}^{5}ε^{2}_{i} = ε^{T} ε = (Ax − b)^{T}(Ax − b) = 4.571×10^{−7}.Least Squares vs. Lagrange Interpolation. Instead of using least squares, fit the observations exactly with the fourth-degree Lagrange interpolation polynomial
\ell(t) = \frac{11}{375}t+\frac{17}{750000}t^{2}+\frac{1}{18750000}t^{3}+\frac{1}{46875000000}t^{4}described in Example 4.3.5 on p. 186 (you can verify that \ell(t_{i}) = b_{i} for each observation). As the graph in Figure 4.6.5 indicates, \ell(t) has only one real nonnegative root, so it is worthless for predicting where the missile will land. This is characteristic of Lagrange interpolation.
