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Chapter 6: Q9E (page 331)

Let \({{\bf{P}}_{\bf{3}}}\) have the inner product given by evaluation at \( - {\bf{3}}\), \( - {\bf{1}}\),1, and 3. Let \({p_{\bf{0}}}\left( t \right) = {\bf{1}}\), \({p_{\bf{1}}}\left( t \right) = t\), and \({p_{\bf{2}}}\left( t \right) = {t^{\bf{2}}}\).

  1. Compute the orthogonal projection of \({p_{\bf{2}}}\) onto the sub-spaced spanned by , and \({p_{\bf{1}}}\).
  2. Find the polynomial q that is orthogonal to \({p_{\bf{0}}}\) and , such that \(\left\{ {{p_{\bf{0}}},{p_{\bf{1}}},q} \right\}\) is an orthogonal basis for . Scale the polynomial q so that its vector of values at \(\left( { - {\bf{3}}, - {\bf{1}},{\bf{1}},{\bf{3}}} \right)\) is \(\left( {{\bf{1}}, - {\bf{1}}, - {\bf{1}},{\bf{1}}} \right)\).

Short Answer

Expert verified

a. 5

b. \(\frac{1}{4}\left( {{t^2} - 5} \right)\)

Step by step solution

01

Find the values of polynomials

The value of \({p_0}\left( t \right)\) is 1 for all values of \(t\).

The values of \({p_1}\left( t \right) = t\) are:

\(\begin{aligned}{p_1}\left( { - 3} \right) = - 3\\{p_1}\left( { - 1} \right) = - 1\\{p_1}\left( 1 \right) = 1\\{p_1}\left( 3 \right) = 3\end{aligned}\)

The values of \({p_2}\left( t \right) = {t^2}\) are:

\(\begin{aligned}{p_2}\left( { - 3} \right) = 9\\{p_2}\left( { - 1} \right) = 1\\{p_2}\left( 1 \right) = 1\\{p_2}\left( 3 \right) = 9\end{aligned}\)

02

Find the inner products

Find the inner product \(\left\langle {{p_2},{p_0}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_2},{p_0}} \right\rangle &= {p_2}\left( { - 3} \right){p_0}\left( { - 3} \right) + {p_2}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_2}\left( 1 \right){p_0}\left( 1 \right) + {p_2}\left( 3 \right){p_0}\left( 3 \right)\\ &= \left( 9 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 9 \right)\left( 1 \right)\\ &= 20\end{aligned}\)

Find the inner product \(\left\langle {{p_2},{p_1}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_2},{p_1}} \right\rangle &= {p_2}\left( { - 3} \right){p_1}\left( { - 3} \right) + {p_2}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_2}\left( 1 \right){p_1}\left( 1 \right) + {p_2}\left( 3 \right){p_1}\left( 3 \right)\\ &= \left( 9 \right)\left( { - 3} \right) + \left( 1 \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 9 \right)\left( 3 \right)\\ &= 0\end{aligned}\)

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

\(\begin{aligned}\left\langle {{p_0},{p_0}} \right\rangle &= {p_0}\left( { - 3} \right){p_0}\left( { - 3} \right) + {p_0}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_0}\left( 1 \right){p_0}\left( 1 \right) + {p_0}\left( 3 \right){p_1}\left( 3 \right)\\ &= 1 + 1 + 1 + 1\\ &= 4\end{aligned}\)

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

\(\begin{aligned}\left\langle {{p_1},{p_1}} \right\rangle &= {p_1}\left( { - 3} \right){p_1}\left( { - 3} \right) + {p_1}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_1}\left( 1 \right){p_1}\left( 1 \right) + {p_1}\left( 3 \right){p_1}\left( 3 \right)\\ &= \left( { - 3} \right)\left( { - 3} \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 3 \right)\left( 3 \right)\\ &= 20\end{aligned}\)

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Most popular questions from this chapter

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Question: In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

11. \(A = \left( {\begin{aligned}{{}{}}{\bf{4}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{ - {\bf{5}}}&{\bf{1}}\\{\bf{6}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{5}}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{9}}\\{\bf{0}}\\{\bf{0}}\\{\bf{0}}\end{aligned}} \right)\)

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

5. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{1}}\\{\bf{3}}\\{\bf{8}}\\{\bf{2}}\end{aligned}} \right)\)

A certain experiment produces the data \(\left( {1,1.8} \right),\left( {2,2.7} \right),\left( {3,3.4} \right),\left( {4,3.8} \right),\left( {5,3.9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = {\beta _1}x + {\beta _2}{x^2}\)

Such a function might arise, for example, as the revenue from the sale of \(x\) units of a product, when the amount offered for sale affects the price to be set for the product.

a. Give the design matrix, the observation vector, and the unknown parameter vector.

b. Find the associated least-squares curve for the data.

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

5. \(\left( {\begin{aligned}{{}{}}1\\{ - 4}\\0\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}7\\{ - 7}\\{ - 4}\\1\end{aligned}} \right)\)
See all solutions

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