/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q15E Let \({{\bf{v}}_{\bf{1}}} = \lef... [FREE SOLUTION] | 91影视

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Chapter 8: Q15E (page 437)

Let \({{\bf{v}}_{\bf{1}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{\bf{0}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{2}}} = \left( {\begin{aligned}{{}}{\bf{1}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{3}}} = \left( {\begin{aligned}{{}}{\bf{4}}\\{\bf{2}}\end{aligned}} \right)\), \({{\bf{v}}_{\bf{4}}} = \left( {\begin{aligned}{{}}{\bf{4}}\\{\bf{0}}\end{aligned}} \right)\), and\(p = \left( {\begin{aligned}{{}}{\bf{2}}\\{\bf{1}}\end{aligned}} \right)\). Confirm that

\({\bf{p}} = \frac{{\bf{1}}}{{\bf{3}}}{{\bf{v}}_{\bf{1}}} + \frac{{\bf{1}}}{{\bf{3}}}{{\bf{v}}_{\bf{2}}} + \frac{{\bf{1}}}{{\bf{6}}}{{\bf{v}}_{\bf{3}}} + \frac{{\bf{1}}}{{\bf{6}}}{{\bf{v}}_{\bf{4}}}\)and \({{\bf{v}}_{\bf{1}}} - {{\bf{v}}_{\bf{2}}} + {{\bf{v}}_{\bf{3}}} - {{\bf{v}}_{\bf{4}}} = {\bf{0}}\).

Use the procedure in the proof of Caratheodory鈥檚 Theorem to express p as a convex combination of three of the \({{\bf{v}}_i}\)鈥檚. Do this in two ways.

Short Answer

Expert verified

\({\bf{p}} = \frac{1}{6}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_4}\) and \({\bf{p}} = \frac{1}{2}{{\bf{v}}_1} + \frac{1}{6}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_3}\)

Step by step solution

01

Verify the expression for p

\(\begin{aligned}{}\frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4} &= \frac{1}{3}\left( {\begin{aligned}{{}}1\\0\end{aligned}} \right) + \frac{1}{3}\left( {\begin{aligned}{{}}1\\2\end{aligned}} \right) + \frac{1}{6}\left( {\begin{aligned}{{}}4\\2\end{aligned}} \right) + \frac{1}{6}\left( {\begin{aligned}{{}}4\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}{\frac{1}{3} + \frac{1}{3} + \frac{4}{6} + \frac{4}{6}}\\{0 + \frac{2}{3} + \frac{2}{6} + 0}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}2\\1\end{aligned}} \right)\\ &= {\bf{p}}\end{aligned}\)

02

Verify the equation \({{\bf{v}}_{\bf{1}}} - {{\bf{v}}_{\bf{2}}} + {{\bf{v}}_{\bf{3}}} - {{\bf{v}}_{\bf{4}}} = {\bf{0}}\)

Substitute value in the equation \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\).

\(\begin{aligned}{}{{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} &= \left( {\begin{aligned}{{}}1\\0\end{aligned}} \right) - \left( {\begin{aligned}{{}}1\\2\end{aligned}} \right) + \left( {\begin{aligned}{{}}4\\2\end{aligned}} \right) - \left( {\begin{aligned}{{}}4\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}0\\0\end{aligned}} \right)\\ &= 0\end{aligned}\)

From two steps, the sum of the coefficients is 0, so the combination is an affine dependence relationship. Both the given equations are valid.

03

Consider the coefficients of \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{3}}}\)

From the equations \({\bf{p}} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\) and \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\), the coefficients \({{\bf{v}}_1}\) and \({{\bf{v}}_3}\) are \(\frac{1}{3}\) and \(\frac{1}{6}\).

Subtract \(\frac{1}{6}\) times \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\) from \({\bf{p}} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\).

\(\begin{aligned}{}{\bf{p}} - 0 = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4} - \frac{1}{6}\left( {{{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4}} \right)\\{\bf{p}} = \left( {\frac{1}{3} - \frac{1}{6}} \right){{\bf{v}}_1} + \left( {\frac{1}{3} + \frac{1}{6}} \right){{\bf{v}}_2} + \left( {\frac{1}{6} - \frac{1}{6}} \right){{\bf{v}}_3} + \left( {\frac{1}{6} + \frac{1}{6}} \right){{\bf{v}}_4}\\ = \frac{1}{6}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_4}\end{aligned}\)

Similarly, add \(\frac{1}{6}\) times \({{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4} = 0\) to \({\bf{p}} = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4}\).

\(\begin{aligned}{}{\bf{p}} - 0 = \frac{1}{3}{{\bf{v}}_1} + \frac{1}{3}{{\bf{v}}_2} + \frac{1}{6}{{\bf{v}}_3} + \frac{1}{6}{{\bf{v}}_4} + \frac{1}{6}\left( {{{\bf{v}}_1} - {{\bf{v}}_2} + {{\bf{v}}_3} - {{\bf{v}}_4}} \right)\\{\bf{p}} = \left( {\frac{1}{3} + \frac{1}{6}} \right){{\bf{v}}_1} + \left( {\frac{1}{3} - \frac{1}{6}} \right){{\bf{v}}_2} + \left( {\frac{1}{6} + \frac{1}{6}} \right){{\bf{v}}_3} + \left( {\frac{1}{6} - \frac{1}{6}} \right){{\bf{v}}_4}\\ = \frac{1}{2}{{\bf{v}}_1} + \frac{1}{6}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_3}\end{aligned}\)

So, the expressions for p are\({\bf{p}} = \frac{1}{6}{{\bf{v}}_1} + \frac{1}{2}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_4}\) and \({\bf{p}} = \frac{1}{2}{{\bf{v}}_1} + \frac{1}{6}{{\bf{v}}_2} + \frac{1}{3}{{\bf{v}}_3}\).

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Most popular questions from this chapter

Let \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) be cubic B茅zier curves with control points \(\left\{ {{{\bf{p}}_{\bf{o}}}{\bf{,}}{{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}} \right\}\)and \(\left\{ {{{\bf{p}}_{\bf{3}}}{\bf{,}}{{\bf{p}}_{\bf{4}}}{\bf{,}}{{\bf{p}}_{\bf{5}}}{\bf{,}}{{\bf{p}}_{\bf{6}}}} \right\}\) respectively, so that \({\bf{x}}\left( t \right)\) and \({\bf{y}}\left( t \right)\) are joined at \({{\bf{p}}_3}\) . The following questions refer to the curve consisting of \({\bf{x}}\left( t \right)\) followed by \(y\left( t \right)\). For simplicity, assume that the curve is in \({\mathbb{R}^2}\).

a. What condition on the control points will guarantee that the curve has \({C^1}\) continuity at \({{\bf{p}}_3}\) ? Justify your answer.

b. What happens when \({\bf{x'}}\left( 1 \right)\) and \({\bf{y'}}\left( 1 \right)\) are both the zero vector?

In Exercises 1-6, determine if the set of points is affinely dependent. (See Practice Problem 2.) If so, construct an affine dependence relation for the points.

1.\(\left( {\begin{aligned}{{}}3\\{ - 3}\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\6\end{aligned}} \right),\left( {\begin{aligned}{{}}2\\0\end{aligned}} \right)\)

Question: 14. Show that if \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\), then aff \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is the plane through \({{\rm{v}}_{\rm{1}}}{\rm{, }}{{\rm{v}}_{\rm{2}}}\) and \({{\rm{v}}_{\rm{3}}}\).

Questions: Let \({F_{\bf{1}}}\) and \({F_{\bf{2}}}\) be 4-dimensional flats in \({\mathbb{R}^{\bf{6}}}\), and suppose that \({F_{\bf{1}}} \cap {F_{\bf{2}}} \ne \phi \). What are the possible dimension of \({F_{\bf{1}}} \cap {F_{\bf{2}}}\)?

In Exercises 1-6, determine if the set of points is affinely dependent. (See Practice Problem 2.) If so, construct an affine dependence relation for the points.

5.\(\left( {\begin{aligned}{{}}1\\0\\{ - 2}\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\1\\1\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 1}\\5\\1\end{aligned}} \right),\left( {\begin{aligned}{{}}0\\5\\{ - 3}\end{aligned}} \right)\)
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