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Chapter 8: Q21E (page 437)

Question: If A and B are convex sets, prove that \(A + B\) is convex.

Short Answer

Expert verified

\(A + B\)is convex.

Step by step solution

01

Make assumptions for x and y \(\left( {{\bf{x}},{\bf{y}} \in A + B} \right)\) 

Let \({\bf{a}}\), \({\bf{c}}\) are in A, and \({\bf{c}}\), \({\bf{d}}\) are in B, such that \({\bf{x}} = {\bf{a}} + {\bf{b}}\), \({\bf{y}} = {\bf{c}} + {\bf{d}}\).

02

Write w as the combination of x and y

For any \(t\), let \(w = \left( {1 - t} \right){\bf{x}} + t{\bf{y}}\).

So, it can be written as shown below:

\(\begin{array}{c}w = \left( {1 - t} \right){\bf{x}} + t{\bf{y}}\\ = \left( {1 - t} \right)\left( {{\bf{a}} + {\bf{b}}} \right) + t\left( {{\bf{c}} + {\bf{d}}} \right)\\ = \left( {\left( {1 - t} \right){\bf{a}} + t{\bf{c}}} \right) + \left( {\left( {1 - t} \right){\bf{b}} + t{\bf{d}}} \right)\end{array}\)

As A is convex, so \(\left( {1 - t} \right){\bf{a}} + t{\bf{c}} \in A\). Similarly, B is also convex\(\left( {1 - t} \right){\bf{b}} + t{\bf{d}} \in B\). So, \(w\) is in \(A + B\).

So, \(A + B\)is convex.

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Most popular questions from this chapter

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 \({{\rm{v}}_{\rm{3}}}\).

Question: 15. Let \(A\) be an \({\rm{m}} \times {\rm{n}}\) matrix and, given \({\rm{b}}\) in \({\mathbb{R}^m}\), show that the set \(S\) of all solutions of \(A{\rm{x}} = {\rm{b}}\) is an affine subset of \({\mathbb{R}^n}\).

Question: 2. Let Lbe the line in \({\mathbb{R}^{\bf{2}}}\) through the points \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{\bf{4}}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{ - {\bf{1}}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the “left” side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.

Question: In Exercises 11 and 12, mark each statement True or False. Justify each answer.

11.a. The cubic Bezier curve is based on four control points.

b. Given a quadratic Bezier curve \({\mathop{\rm x}\nolimits} \left( t \right)\) with control points \({{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},\) and \({{\mathop{\rm p}\nolimits} _2}\), the directed line segment \({{\mathop{\rm p}\nolimits} _1} - {{\mathop{\rm p}\nolimits} _0}\) (from \({{\mathop{\rm p}\nolimits} _0}\) to \({{\mathop{\rm p}\nolimits} _1}\)) is the tangent vector to the curve at \({{\mathop{\rm p}\nolimits} _0}\).

c. When two quadratic Bezier curves with control points \(\left\{ {{{\mathop{\rm p}\nolimits} _0},{{\mathop{\rm p}\nolimits} _1},{{\mathop{\rm p}\nolimits} _2}} \right\}\) and \(\left\{ {{{\mathop{\rm p}\nolimits} _2},{{\mathop{\rm p}\nolimits} _3},{{\mathop{\rm p}\nolimits} _4}} \right\}\) are joined at \({{\mathop{\rm p}\nolimits} _2}\), the combined Bezier curve will have \({C^1}\) continuity at \({{\mathop{\rm p}\nolimits} _2}\)if\({{\mathop{\rm p}\nolimits} _2}\) is the midpoint of the line segment between \({{\mathop{\rm p}\nolimits} _1}\) and \({{\mathop{\rm p}\nolimits} _3}\).
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