91Ó°ÊÓ

The polynomials at \({t_0} = - 3\), \({t_1} = - 1\), \({t_2} = 1\), and \({t_3} = 3\).

\(\begin{aligned}p\left( {{t_0}} \right) &= p\left( { - 3} \right) = - 27\\{p_0}\left( {{t_0}} \right) &= {p_0}\left( { - 3} \right) = 1\\{p_1}\left( {{t_0}} \right) &= {p_1}\left( { - 3} \right) = - 3\\q\left( {{t_0}} \right) &= q\left( { - 3} \right) = 4\end{aligned}\)

And,

\(\begin{aligned}p\left( {{t_1}} \right) &= p\left( 1 \right) = - 1\\{p_0}\left( {{t_1}} \right) &= {p_0}\left( { - 1} \right) = 1\\{p_1}\left( {{t_1}} \right) &= {p_1}\left( { - 1} \right) = - 1\\q\left( {{t_1}} \right) &= q\left( { - 1} \right) = - 4\end{aligned}\)

And,

\(\begin{aligned}p\left( {{t_2}} \right) &= p\left( 1 \right) = 1\\{p_0}\left( {{t_2}} \right) &= {p_0}\left( 1 \right) = 1\\{p_1}\left( {{t_2}} \right) &= {p_1}\left( 1 \right) = 1\\q\left( {{t_2}} \right) &= q\left( 1 \right) = - 4\end{aligned}\)

And,

\(\begin{aligned}p\left( {{t_3}} \right) &= p\left( 3 \right) = 27\\{p_0}\left( {{t_3}} \right) &= 1\\{p_1}\left( {{t_3}} \right) = {p_1}\left( 3 \right) &= 3\\q\left( {{t_3}} \right) = q\left( 3 \right) &= 4\end{aligned}\)

Find the inner product \(\left\langle {p,{p_0}} \right\rangle \).

\(\begin{aligned}\left\langle {p,{p_0}} \right\rangle &= p\left( { - 3} \right){p_0}\left( { - 3} \right) + p\left( { - 1} \right){p_0}\left( { - 1} \right) + p\left( 1 \right){p_0}\left( 1 \right) + p\left( 3 \right){p_0}\left( 3 \right)\\ &= \left( { - 27} \right)\left( 1 \right) + \left( { - 1} \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( {27} \right)\left( 1 \right)\\ &= 0\end{aligned}\)

Find the inner product \(\left\langle {p,q} \right\rangle \).

\(\begin{aligned}\left\langle {p,q} \right\rangle &= p\left( { - 3} \right)q\left( { - 3} \right) + p\left( { - 1} \right)q\left( { - 1} \right) + p\left( 1 \right)q\left( 1 \right) + p\left( 3 \right)q\left( 3 \right)\\ &= \left( { - 27} \right)\left( 4 \right) + \left( 1 \right)\left( { - 4} \right) + \left( 1 \right)\left( { - 4} \right) + \left( {27} \right)\left( 4 \right)\\ &= 0\end{aligned}\)

Find the inner product \(\left\langle {{p_0},{p_0}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_0},{p_0}} \right\rangle &= {p_0}\left( { - 3} \right){p_0}\left( { - 3} \right) + {p_0}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_0}\left( 1 \right){p_0}\left( 1 \right) + {p_0}\left( 3 \right){p_1}\left( 3 \right)\\ &= 1 + 1 + 1 + 1\\ &= 4\end{aligned}\)

Find the inner product \(\left\langle {{p_1},{p_1}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_1},{p_1}} \right\rangle &= {p_1}\left( { - 3} \right){p_1}\left( { - 3} \right) + {p_1}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_1}\left( 1 \right){p_1}\left( 1 \right) + {p_1}\left( 3 \right){p_1}\left( 3 \right)\\ &= \left( { - 3} \right)\left( { - 3} \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 3 \right)\left( 3 \right)\\ &= 20\end{aligned}\)

Find the inner product \(\left\langle {q,q} \right\rangle \).

\(\begin{aligned}\left\langle {q,q} \right\rangle &= q\left( { - 3} \right)q\left( { - 3} \right) + q\left( { - 1} \right)q\left( { - 1} \right) + q\left( 1 \right)q\left( 1 \right) + q\left( 3 \right)q\left( 3 \right)\\ &= \left( 4 \right)\left( 4 \right) + \left( { - 4} \right)\left( { - 4} \right) + \left( { - 4} \right)\left( { - 4} \right) + \left( 4 \right)\left( 4 \right)\\ &= 64\end{aligned}\)

The best approximation can be done as:

\(\begin{aligned}\widehat {{p_2}} &= \frac{{\left\langle {p,{p_0}} \right\rangle }}{{\left\langle {{p_0},{p_0}} \right\rangle }}{p_0} + \frac{{\left\langle {p,{p_1}} \right\rangle }}{{\left\langle {{p_1},{p_1}} \right\rangle }}{p_1} + \frac{{\left\langle {p,{q_1}} \right\rangle }}{{\left\langle {{p_1},{p_1}} \right\rangle }}{p_1}\\ &= \frac{{20}}{4}\left( 1 \right) + \frac{{164}}{{20}}\left( t \right) + \frac{0}{{64}}\left( {{t^2} - 5} \right)\\ &= \frac{{41}}{5}t\end{aligned}\)

Thus, the best approximation is \(\frac{{41}}{5}t\).