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

Expert verified

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

Step by step solution

01

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 \({{\bf{P}}_4}\).

02

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

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Most popular questions from this chapter

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

In Exercises 9-12, find a unit vector in the direction of the given vector.

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Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

5. \(\left( {\frac{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}{{{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} }}} \right){\mathop{\rm v}\nolimits} \)

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)

Let \(\overline x = \frac{1}{n}\left( {{x_1} + \cdots + {x_n}} \right)\), and \(\overline y = \frac{1}{n}\left( {{y_1} + \cdots + {y_n}} \right)\). Show that the least-squares line for the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) must pass through \(\left( {\overline x ,\overline y } \right)\). That is, show that \(\overline x \) and \(\overline y \) satisfies the linear equation \(\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x \). (Hint: Derive this equation from the vector equation \({\bf{y}} = X{\bf{\hat \beta }} + \in \). Denote the first column of \(X\) by 1. Use the fact that the residual vector \( \in \) is orthogonal to the column space of \(X\) and hence is orthogonal to 1.)

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