Question 9.6.4: Use the singular value decomposition technique to determine ...
Use the singular value decomposition technique to determine the least squares polynomial of degree two for the data given in Table 9.5.
Table 9.5
\begin{array}{ccc}\hline i & x_{i} & y_{i} \\\hline 1 & 0 & 1.0000 \\2 & 0.25 & 1.2840 \\3 & 0.50 & 1.6487 \\4 & 0.75 & 2.1170 \\5 & 1.00 & 2.7183 \\\hline\end{array}
This problem was solved using normal equations as Example 2 in Section 8.1.
Here we first need to determine the appropriate form for A, x, and b. In Example 2 in Section 8.1 the problem was described as finding a_{0}, a_{1}, \text { and } a_{2} with
P_{2}(x)=a_{0}+a_{1} x+a_{2} x^{2}
In order to express this in matrix form, we let
x =\left[\begin{array}{l}a_{0} \\a_{1} \\a_{2}\end{array}\right], \quad b =\left[\begin{array}{l}y_{0} \\y_{1} \\y_{2} \\y_{3} \\y_{4}\end{array}\right]=\left[\begin{array}{l}1.0000 \\1.2840 \\1.6487 \\2.1170 \\2.7183\end{array}\right], \quad \text { and }
A=\left[\begin{array}{lll}1 & x_{0} & x_{0}^{2} \\1 & x_{1} & x_{1}^{2} \\1 & x_{2} & x_{2}^{2} \\1 & x_{3} & x_{3}^{2} \\1 & x_{4} & x_{4}^{2}\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\1 & 0.25 & 0.0625 \\1 & 0.5 & 0.25 \\1 & 0.75 & 0.5625 \\1 & 1 & 1\end{array}\right]
The singular value decomposition of A has the form A=U S V^{t} , where
\begin{aligned}&U=\left[\begin{array}{rrrrr}-0.2945 & -0.6327 & 0.6314 & -0.0143 & -0.3378 \\-0.3466 & -0.4550 & -0.2104 & 0.2555 & 0.7505 \\-0.4159 & -0.1942 & -0.5244 & -0.6809 & -0.2250 \\-0.5025 & 0.1497 & -0.3107 & 0.6524 & -0.4505 \\-0.6063 & 0.5767 & 0.4308 & -0.2127 & 0.2628\end{array}\right] \text {, }\\&S=\left[\begin{array}{ccc}2.7117 & 0 & 0 \\0 & 0.9371 & 0 \\0 & 0 & 0.1627 \\0 & 0 & 0 \\0 & 0 & 0\end{array}\right], \quad \text { and } \quad V^{t}=\left[\begin{array}{rrr}-0.7987 & -0.4712 & -0.3742 \\-0.5929 & 0.5102 & 0.6231 \\0.1027 & -0.7195 & 0.6869\end{array}\right] \text {. }\end{aligned}
So
\begin{aligned}c &=U^{t}\left[\begin{array}{l}y_{0} \\y_{1} \\y_{2} \\y_{3} \\y_{4}\end{array}\right]=\left[\begin{array}{rrrrr}-0.2945 & -0.6327 & 0.6314 & -0.0143 & -0.3378 \\-0.3466 & -0.4550 & -0.2104 & 0.2555 & 0.7505 \\-0.4159 & -0.1942 & -0.5244 & -0.6809 & -0.2250 \\-0.5025 & 0.1497 & -0.3107 & 0.6524 & -0.4505 \\-0.6063 & 0.5767 & 0.4308 & -0.2127 & 0.2628\end{array}\right]^{t}\left[\begin{array}{c}1 \\1.284 \\1.6487 \\2.117 \\2.7183\end{array}\right] \\&=\left[\begin{array}{r}-4.1372 \\0.3473 \\0.0099 \\-0.0059 \\0.0155\end{array}\right],\end{aligned}
and the components of z are
\begin{aligned} &z_{1}=\frac{c_{1}}{s_{1}}=\frac{-4.1372}{2.7117}=-1.526, \quad z_{2}=\frac{c_{2}}{s_{2}}=\frac{0.3473}{0.9371}=0.3706, \quad \text { and } \\ &z_{3}=\frac{c_{3}}{s_{3}}=\frac{0.0099}{0.1627}=0.0609 \end{aligned}
This gives the least squares coefficients in P_{2}(x) as
\left[\begin{array}{l}a_{0} \\a_{1} \\a_{2}\end{array}\right]= x =V z =\left[\begin{array}{rrr}-0.7987 & -0.5929 & 0.1027 \\-0.4712 & 0.5102 & -0.7195 \\-0.3742 & 0.6231 & 0.6869\end{array}\right]\left[\begin{array}{c}-1.526 \\0.3706 \\0.0609\end{array}\right]=\left[\begin{array}{l}1.005 \\0.8642 \\0.8437\end{array}\right],
which agrees with the results in Example 2 of Section 8.1. The least squares error using these values uses the last two components of c, and is
\|A x – b \|_{2}=\sqrt{c_{4}^{2}+c_{5}^{2}}=\sqrt{(-0.0059)^{2}+(0.0155)^{2}}=0.0165.