Question

Find a polynomial \({p_{\bf{3}}}\) such that \(\left\{ {{p_{\bf{0}}},{p_{\bf{1}}},{p_{\bf{2}}},{p_{\bf{3}}}} \right\}\) (see Exercise 11) is an orthogonal basis for the subspace \({{\bf{P}}_{\bf{3}}}\) of \({{\bf{P}}_{\bf{4}}}\). Scale the polynomials \({p_{\bf{3}}}\) so that vector of values is \(\left( { - {\bf{1}},{\bf{2}},{\bf{0}}, - {\bf{2}},{\bf{1}}} \right)\).

Short answer

The vector \({p_3}\) is \(\frac{5}{6}\left( {{t^3} - \frac{{17}}{5}t} \right)\).

Step-by-step solution

Find the vector \({p_{\bf{3}}}\)

Let the subspace W is defined as:

\(W = {\rm{Span}}\left\{ {{p_0},{p_1},{p_2}} \right\}\)

The vector \({p_3}\) is:

\(\begin{aligned}{p_3} &= p - {\rm{pro}}{{\rm{j}}_W}p\\ &= {t^3} - \frac{{17}}{5}t\end{aligned}\)

Thus, \({p_3}\) makes \(\left\{ {{p_0},{p_1},{p_2},{p_3}} \right\}\) and orthogonal basis for the subspace \({{\bf{P}}_3}\) and \({{\bf{P}}_4}\).

Find the values of \({p_{\bf{3}}}\) 

The values of \({p_3}\) are:

\(\begin{aligned}{p_3}\left( { - 2} \right) &= - \frac{6}{5}\\{p_3}\left( { - 1} \right) &= \frac{{12}}{5}\\{p_3}\left( 0 \right) &= 0\\{p_3}\left( 1 \right) &= - \frac{{12}}{5}\\{p_3}\left( 2 \right) &= \frac{6}{5}\end{aligned}\)

So, scaling the vector by \(\frac{5}{6}\), the vector \({p_3}\) can be expressed as \({p_3} = \frac{5}{6}\left( {{t^3} - \frac{{17}}{5}t} \right)\).