Exercise 3-8 refer to \({{\bf{P}}_{\bf{2}}}\) with the inner product given by evaluation at \( - {\bf{1}}\), 0, and 1. (See Example 2).

4. Compute \(\left\langle {p,q} \right\rangle \), where \(p\left( t \right) = {\bf{3}}t - {t^{\bf{2}}}\), \(q\left( t \right) = {\bf{3}} + {\bf{2}}{t^{\bf{2}}}\).

The values of polynomial \(p\left( t \right) = 3t - {t^2}\) are:

\(\begin{aligned}p\left( { - 1} \right) &= 3\left( { - 1} \right) - {\left( { - 1} \right)^2}\\ &= - 3 - 1\\ &= - 4\end{aligned}\)

\(\begin{aligned}p\left( 0 \right) &= 3\left( 0 \right) - {\left( 0 \right)^2}\\ &= 0\end{aligned}\)

\(\begin{aligned}p\left( 1 \right) &= 3\left( 1 \right) - {\left( 1 \right)^2}\\ &= 3 - 1\\ &= 2\end{aligned}\)

The values of polynomial \(q\left( t \right) = 3 + 2{t^2}\) are:

\(\begin{aligned}q\left( { - 1} \right) &= 3 + 2{\left( { - 1} \right)^2}\\ &= 5\end{aligned}\)

\(\begin{aligned}q\left( 0 \right) &= 3 + 2{\left( 0 \right)^2}\\ &= 3\end{aligned}\)

\(\begin{aligned}q\left( 1 \right) &= 3 + 2{\left( 1 \right)^2}\\ &= 5\end{aligned}\)

The inner product for \(\left\langle {p,q} \right\rangle \) is defined as:

\(\left\langle {p,q} \right\rangle = p\left( {{t_0}} \right)q\left( {{t_0}} \right) + p\left( {{t_1}} \right)q\left( {{t_1}} \right) + p\left( {{t_2}} \right)q\left( {{t_2}} \right)\)

Substitute \({t_0} = - 1\), \({t_1} = 0\) and \({t_2} = 1\).

\(\begin{aligned}\left\langle {p,q} \right\rangle &= p\left( { - 1} \right)q\left( { - 1} \right) + p\left( 0 \right)q\left( 0 \right) + p\left( 1 \right)q\left( 1 \right)\\ &= \left( { - 4} \right)\left( 5 \right) + \left( 0 \right)\left( 3 \right) + \left( 2 \right)\left( 5 \right)\\ &= - 20 + 0 + 10\\ &= - 10\end{aligned}\)

Thus, the inner product is \( - 10\).