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).

3. Compute \(\left\langle {p,q} \right\rangle \), where \(p\left( t \right) = {\bf{4}} + t\), \(q\left( t \right) = {\bf{5}} - {\bf{4}}{t^{\bf{2}}}\).

The values of polynomial \(p\left( t \right) = 4 + t\) are:

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

\(\begin{aligned}p\left( 0 \right) &= 4 + 0\\ &= 4\end{aligned}\)

\(\begin{aligned}p\left( 1 \right) &= 4 + 1\\ &= 5\end{aligned}\)

The values of polynomial \(q\left( t \right) = 5 - 4{t^2}\) are:

\(\begin{aligned}q\left( { - 1} \right) &= 5 - 4{\left( { - 1} \right)^2}\\ &= 5 - 4\\ &= 1\end{aligned}\)

\(\begin{aligned}q\left( 0 \right) &= 5 - 4{\left( 0 \right)^2}\\ &= 5\end{aligned}\)

\(\begin{aligned}q\left( 1 \right) &= 5 - 4{\left( 1 \right)^2}\\ &= 5 - 4\\ &= 1\end{aligned}\)

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 \({t_2} = 1\).

\(\begin{aligned}\left\langle {p,q} \right\rangle &= p\left( { - 1} \right)q\left( { - 1} \right) + p\left( 0 \right)q\left( 0 \right) + p\left( 1 \right)q\left( 1 \right)\\ &= \left( 3 \right)\left( 1 \right) + \left( 4 \right)\left( 5 \right) + \left( 5 \right)\left( 1 \right)\\ &= 3 + 20 + 5\\ &= 28\end{aligned}\)

Thus, the inner product is 28.