Let \({{\bf{P}}_{\bf{3}}}\) have the inner product given by evaluation at \( - {\bf{3}}\), \( - {\bf{1}}\),1, and 3. Let \({p_{\bf{0}}}\left( t \right) = {\bf{1}}\), \({p_{\bf{1}}}\left( t \right) = t\), and \({p_{\bf{2}}}\left( t \right) = {t^{\bf{2}}}\).

1. Compute the orthogonal projection of \({p_{\bf{2}}}\) onto the sub-spaced spanned by , and \({p_{\bf{1}}}\).
2. Find the polynomial q that is orthogonal to \({p_{\bf{0}}}\) and , such that \(\left\{ {{p_{\bf{0}}},{p_{\bf{1}}},q} \right\}\) is an orthogonal basis for . Scale the polynomial q so that its vector of values at \(\left( { - {\bf{3}}, - {\bf{1}},{\bf{1}},{\bf{3}}} \right)\) is \(\left( {{\bf{1}}, - {\bf{1}}, - {\bf{1}},{\bf{1}}} \right)\).

The value of \({p_0}\left( t \right)\) is 1 for all values of \(t\).

The values of \({p_1}\left( t \right) = t\) are:

\(\begin{aligned}{p_1}\left( { - 3} \right) = - 3\\{p_1}\left( { - 1} \right) = - 1\\{p_1}\left( 1 \right) = 1\\{p_1}\left( 3 \right) = 3\end{aligned}\)

The values of \({p_2}\left( t \right) = {t^2}\) are:

\(\begin{aligned}{p_2}\left( { - 3} \right) = 9\\{p_2}\left( { - 1} \right) = 1\\{p_2}\left( 1 \right) = 1\\{p_2}\left( 3 \right) = 9\end{aligned}\)

Find the inner product \(\left\langle {{p_2},{p_0}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_2},{p_0}} \right\rangle &= {p_2}\left( { - 3} \right){p_0}\left( { - 3} \right) + {p_2}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_2}\left( 1 \right){p_0}\left( 1 \right) + {p_2}\left( 3 \right){p_0}\left( 3 \right)\\ &= \left( 9 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 9 \right)\left( 1 \right)\\ &= 20\end{aligned}\)

Find the inner product \(\left\langle {{p_2},{p_1}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_2},{p_1}} \right\rangle &= {p_2}\left( { - 3} \right){p_1}\left( { - 3} \right) + {p_2}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_2}\left( 1 \right){p_1}\left( 1 \right) + {p_2}\left( 3 \right){p_1}\left( 3 \right)\\ &= \left( 9 \right)\left( { - 3} \right) + \left( 1 \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 9 \right)\left( 3 \right)\\ &= 0\end{aligned}\)

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 \(\left\langle {{p_1},{p_1}} \right\rangle \).

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