Question

Let \({p_{\bf{0}}}\), \({p_{\bf{1}}}\), and \({p_{\bf{2}}}\)be the orthogonal polynomials described in Example 5, where the inner product on \({{\bf{P}}_{\bf{4}}}\) is given by evaluation at \( - {\bf{2}}\), \( - {\bf{1}}\), 0, 1, and 2. Find the orthogonal projection of \({t^{\bf{3}}}\) onto \({\bf{Span}}\left\{ {{p_{\bf{0}}},{p_{\bf{1}}},{p_{\bf{2}}}} \right\}\).

Short answer

The orthogonal projection is \(\frac{{17}}{5}t\).

Step-by-step solution

Write the polynomials in vector form

The polynomials can be written as:

\({p_0} = \left( {\begin{aligned}{*{20}}1\\1\\1\\1\\1\end{aligned}} \right)\), \({p_1} = \left( {\begin{aligned}{*{20}}{ - 2}\\{ - 1}\\0\\1\\2\end{aligned}} \right)\), \({p_2} = \left( {\begin{aligned}{*{20}}4\\1\\0\\1\\4\end{aligned}} \right)\)

Find the inner products

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

\(\begin{aligned}\left\langle {p,{p_1}} \right\rangle & = \left\langle {{t^3},t} \right\rangle \\ &= p\left( { - 2} \right){p_1}\left( { - 2} \right) + p\left( { - 1} \right){p_1}\left( { - 1} \right) + p\left( 0 \right){p_1}\left( 0 \right) + p\left( 1 \right){p_1}\left( 1 \right) + p\left( 2 \right){p_1}\left( 2 \right)\\ &= {\left( { - 2} \right)^3}\left( { - 2} \right) + {\left( { - 1} \right)^3}\left( { - 1} \right) + 0 + {1^3}\left( 1 \right) + {2^3}\left( 2 \right)\\ &= 34\end{aligned}\)

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

\(\begin{aligned}\left\langle {p,{p_0}} \right\rangle &= \left\langle {{t^3},1} \right\rangle \\ &= p\left( { - 2} \right){p_0}\left( { - 2} \right) + p\left( { - 1} \right){p_0}\left( { - 1} \right) + p\left( 0 \right){p_0}\left( 0 \right) + p\left( 1 \right){p_0}\left( 1 \right) + p\left( 2 \right){p_0}\left( 2 \right)\\ &= {\left( { - 2} \right)^3}\left( 1 \right) + {\left( { - 1} \right)^3}\left( 1 \right) + {\left( 0 \right)^3}\left( 1 \right) + {1^3}\left( 1 \right) + {2^3}\left( 1 \right)\\ &= 0\end{aligned}\)

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

\(\begin{aligned}\left\langle {p,{p_2}} \right\rangle &= \left\langle {{t^3},{t^2}} \right\rangle \\ &= p\left( { - 2} \right){p_2}\left( { - 2} \right) + p\left( { - 1} \right){p_2}\left( { - 1} \right) + p\left( 0 \right){p_2}\left( 0 \right) + p\left( 1 \right){p_2}\left( 1 \right) + p\left( 2 \right){p_2}\left( 2 \right)\\ &= {\left( { - 2} \right)^3}{\left( { - 2} \right)^2} + {\left( { - 1} \right)^3}{\left( { - 1} \right)^2} + {\left( 0 \right)^3}{\left( 0 \right)^2} + {\left( 1 \right)^3}{\left( 1 \right)^2} + {\left( 2 \right)^3}{\left( 2 \right)^2}\\ &= 0\end{aligned}\)

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

\(\begin{aligned}\left\langle {{p_0},{p_0}} \right\rangle &= \left\langle {1,1} \right\rangle \\ &= {p_0}\left( { - 2} \right){p_0}\left( { - 2} \right) + {p_0}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_0}\left( 0 \right){p_0}\left( 0 \right) + {p_0}\left( 1 \right){p_0}\left( 1 \right) + {p_0}\left( 2 \right){p_0}\left( 2 \right)\\ &= \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right)\\ &= 5\end{aligned}\)

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

\(\begin{aligned}\left\langle {{p_1},{p_1}} \right\rangle &= \left\langle {t,t} \right\rangle \\ &= {p_1}\left( { - 2} \right){p_1}\left( { - 2} \right) + {p_1}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_1}\left( 0 \right){p_1}\left( 0 \right) + {p_1}\left( 1 \right){p_1}\left( 1 \right) + {p_1}\left( 2 \right){p_1}\left( 2 \right)\\ &= \left( { - 2} \right)\left( { - 2} \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 0 \right)\left( 0 \right) + \left( 1 \right)\left( 1 \right) + \left( 2 \right)\left( 2 \right)\\ &= 10\end{aligned}\)

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

\(\begin{aligned}\left\langle {{p_2},{p_2}} \right\rangle &= \left\langle {{t^2},{t^2}} \right\rangle \\ &= {p_2}\left( { - 2} \right){p_2}\left( { - 2} \right) + {p_2}\left( { - 1} \right){p_2}\left( { - 1} \right) + {p_2}\left( 0 \right){p_2}\left( 0 \right) + {p_2}\left( 1 \right){p_2}\left( 1 \right) + {p_2}\left( 2 \right){p_2}\left( 2 \right)\\ &= {\left( { - 2} \right)^2}{\left( { - 2} \right)^2} + {\left( { - 1} \right)^2}{\left( { - 1} \right)^2} + {\left( 0 \right)^2}{\left( 0 \right)^2} + {\left( 1 \right)^2}{\left( 1 \right)^2} + {\left( 2 \right)^2}{\left( 2 \right)^2}\\ &= 34\end{aligned}\)

Find the orthogonal projection

The orthogonal projection can be calculated as follows:

\(\begin{aligned}\hat p &= \frac{{\left\langle {p,{p_0}} \right\rangle }}{{\left\langle {{p_0},{p_0}} \right\rangle }}{p_0} + \frac{{\left\langle {p,{p_1}} \right\rangle }}{{\left\langle {{p_1},{p_1}} \right\rangle }}{p_1} + \frac{{\left\langle {p,{p_2}} \right\rangle }}{{\left\langle {{p_2},{p_2}} \right\rangle }}{p_2}\\ &= \frac{0}{5}\left( 1 \right) + \frac{{34}}{{10}}t + \frac{0}{{34}}{t^2}\\ &= \frac{{34}}{{10}}t\\ &= \frac{{17}}{5}t\end{aligned}\)

Thus, the orthogonal projection is \(\frac{{17}}{5}t\).