Question: 29. Prove that the open ball \(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x:}}\left\| {{\rm{x - p}}} \right\| < \delta } \right\}\)is a convex set. (Hint: Use the Triangle Inequality).

Short Answer

Expert verified

It is shown that the open ball\(B\left( {{\rm{p}},\delta } \right) = \left\{ {{\rm{x}}:\left\| {{\rm{x}} - {\rm{p}}} \right\| < \delta } \right\}\)is a convex set.

Step by step solution

01

Assume some vectors in an open ball set

Assume \({\rm{x,y}} \in B\left( {{\rm{p}},\delta } \right)\) and for \(0 \le t \le 1\), the vector \(z\) satisfies \(z = \left( {1 - t} \right)x + ty\).

02

Evaluate \(\left\| {z - p} \right\|\)

\(\begin{array}{c}\left\| {z - {\rm{p}}} \right\| = \left\| {\left( {\left( {1 - t} \right){\rm{x}} + t{\rm{y}}} \right) - {\rm{p}}} \right\|\\ = \left\| {\left( {\left( {1 - t} \right)\left( {{\rm{x}} - {\rm{p}}} \right) + t\left( {y - {\rm{p}}} \right)} \right)} \right\|\end{array}\)

03

Use Triangle inequality and \(x,y \in B\left( {p,\delta } \right)\)

According to triangle inequality \(\left( {1 - t} \right)\left\| {\left( {{\rm{x}} - {\rm{p}}} \right)} \right\| + t\left\| {\left( {{\rm{y}} - {\rm{p}}} \right)} \right\| < \left( {1 - t} \right)\delta + t\delta \) and as \({\rm{x,y}} \in B\left( {{\rm{p}},\delta } \right)\).

So, \(\left\| {{\rm{z}} - {\rm{p}}} \right\| < \left( {1 - t} \right)\delta + t\delta \).

04

Draw a conclusion

From \(\left\| {z - p} \right\| < \left( {1 - t} \right)\delta + t\delta \), it can be concluded that \(z \in B\left( {{\rm{p}},\delta } \right)\) and \(B\left( {{\rm{p}},\delta } \right)\) is a convex set of the ball.

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Show that a set\(\left\{ {{{\bf{v}}_{\bf{1}}},...,{{\bf{v}}_p}} \right\}\)in\({\mathbb{R}^{\bf{n}}}\)is affinely dependent when \(p \ge n + 2\).

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Explain why a cubic Bezier curve is completely determined by \({\mathop{\rm x}\nolimits} \left( 0 \right)\), \(x'\left( 0 \right)\), \({\mathop{\rm x}\nolimits} \left( 1 \right)\), and \(x'\left( 1 \right)\).

Question: Let \({{\bf{a}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{ - {\bf{1}}}\\{\bf{5}}\end{array}} \right)\), \({{\bf{a}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{\bf{3}}\end{array}} \right)\), \({{\bf{a}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{1}}}\\{\bf{6}}\\{\bf{0}}\end{array}} \right)\), \({{\bf{b}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{5}}\\{ - {\bf{1}}}\end{array}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{3}}}\\{ - {\bf{2}}}\end{array}} \right)\),\({{\bf{b}}_{\bf{3}}} = \left( {\begin{array}{*{20}{c}}{\bf{2}}\\{\bf{2}}\\{\bf{1}}\end{array}} \right)\) and \({\bf{n}} = \left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\\{ - {\bf{2}}}\end{array}} \right)\), and let \(A = \left\{ {{{\bf{a}}_{\bf{1}}},{{\bf{a}}_{\bf{2}}},{{\bf{a}}_{\bf{3}}}} \right\}\) and \(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\). Find a hyperplane H with normal n that separates A and B. Is there a hyperplane parallel to H that strictly separates A and B?

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