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).

5. Compute \(\left\| p \right\|\] and \[\left\| q \right\|\), for p and q in Exercise 3.

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

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

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

And,

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

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

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

The value of \[\left\| p \right\|\] can be calculated as follows:

\(\begin{array}\left\| p \right\| &= \sqrt {\left\langle {p,p} \right\rangle } \\ &= \sqrt {p\left( { - 1} \right)p\left( { - 1} \right) + p\left( 0 \right)p\left( 0 \right) + p\left( 1 \right)p\left( 1 \right)} \\ &= \sqrt {\left( 3 \right)\left( 3 \right) + \left( 4 \right)\left( 4 \right) + \left( 5 \right)\left( 5 \right)} \\ &= \sqrt {50} \end{array}\)

Thus, the value of \(\left\| p \right\|\) is \(\sqrt {50} \).

The value of \[\left\| q \right\|\] can be calculated as follows:

\(\begin{array}\left\| q \right\| &= \sqrt {\left\langle {q,q} \right\rangle } \\ &= \sqrt {q\left( { - 1} \right)q\left( { - 1} \right) + q\left( 0 \right)q\left( 0 \right) + q\left( 1 \right)q\left( 1 \right)} \\ &= \sqrt {\left( 1 \right)\left( 1 \right) + \left( 5 \right)\left( 5 \right) + \left( 1 \right)\left( 1 \right)} \\ &= \sqrt {27} \end{array}\)

Thus, the value of \(\left\| q \right\|\) is \(\sqrt {27} \).