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Chapter 8: Q14E (page 437)

Suppose \({{\bf{v}}_{\bf{1}}}\),鈥�.., \({{\bf{v}}_{\bf{k}}}\) are linearly independent vectors in \({\mathbb{R}^n}\left( {{\bf{1}} \le k \le n} \right)\). Then the set \({X^k} = {\bf{conv}}\;\left\{ { \pm {{\bf{v}}_{\bf{1}}},......, \pm {{\bf{v}}_k}} \right\}\) is called a k-crosspolytope.

a. Sketch \({X^{\bf{1}}}\) and \({X^{\bf{2}}}\).

b. Determine the number of k-faces of the 3-dimensional crosspolytope \({X^{\bf{3}}}\) for \(k = {\bf{0}},{\bf{1}},\,{\bf{2}}\). What is the another name of \({X^{\bf{3}}}\).

c. Determine the number of k-faces of the 4-dimensional crosspolytope \({X^{\bf{4}}}\) for \(k = {\bf{0}},{\bf{1}},\,{\bf{2}},{\bf{3}}\). Verify that your answer satisfies Euler鈥檚 formula.

d. Find a formula for \({f_k}\left( {{X^n}} \right)\), the number of k-faces of \({X^n}\) for \({\bf{0}} \le k \le n - {\bf{1}}\).

Short Answer

Expert verified

b. 6, 12, 8, and another name of \({X^3}\) is an octahedron.

c. 8, 24, 32, 16

d. \({f_k}\left( {{X^n}} \right) = {2^{k + 1}}\left( {\begin{array}{*{20}{c}}n\\{k + 1}\end{array}} \right)\). Here \(\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right) = \frac{{a!}}{{b!\left( {a - b} \right)!}}\) is the binomial coefficient.

Step by step solution

01

Find the solution for part (a)

The plot of \({X^1}\) is a straight line. The figure below represents the sketch for \({X^1}\) passing through vectors 0 and \({{\bf{v}}_1}\).

The plot of \({X^2}\) is a parallelogram for \(k = 2\), and the vectors \({{\bf{v}}_1}\) and \({{\bf{v}}_2}\) that is shown below:

02

Find the solution for part (b)

The number offaces for \(k = 0\):

\({f_0}\left( {{X^3}} \right) = 6\)

The number of faces for \(k = 1\):

\({f_1}\left( {{X^3}} \right) = 12\)

The number of faces for \(k = 2\):

\({f_2}\left( {{X^3}} \right) = 8\)

Euler鈥檚 formulacan be verified as follows:

\(\begin{array}{c}{f_0}\left( {{X^3}} \right) - {f_1}\left( {{X^3}} \right) + {f_2}\left( {{X^3}} \right) = 6 - 12 + 8\\ = 2\end{array}\)

Another name for \({X^3}\) is Octahedron.

03

Find the solution for part (c)

The number facesfor \(k = 0\):

\({f_0}\left( {{X^4}} \right) = 8\)

The number of faces for \(k = 1\):

\({f_1}\left( {{X^4}} \right) = 24\)

The number of faces for \(k = 2\):

\({f_2}\left( {{X^4}} \right) = 32\)

The number of faces for \(k = 3\):

\({f_3}\left( {{X^4}} \right) = 16\)

Euler鈥檚 formula can be verified as follows:

\(\begin{array}{c}{f_0}\left( {{X^4}} \right) - {f_1}\left( {{X^4}} \right) + {f_2}\left( {{X^4}} \right) + {f_3}\left( {{X^4}} \right) = 8 - 24 + 32 - 16\\ = 0\end{array}\)

04

Find the solution for part (d)

The pattern for the above chart is given by the formula \({f_k}\left( {{X^n}} \right) = {2^{k + 1}}\left( {\begin{array}{*{20}{c}}n\\{k + 1}\end{array}} \right)\). Here \(\left( {\begin{array}{*{20}{c}}a\\b\end{array}} \right) = \frac{{a!}}{{b!\left( {a - b} \right)!}}\) is the binomial coefficient.

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Most popular questions from this chapter

Question: 24. Repeat Exercise 23 for \({v_1} = \left( \begin{array}{l}1\\2\end{array} \right)\), \({v_2} = \left( \begin{array}{l}5\\1\end{array} \right)\), \({v_3} = \left( \begin{array}{l}4\\4\end{array} \right)\) and \(p = \left( \begin{array}{l}2\\3\end{array} \right)\).

Explain why any set of five or more points in \({\mathbb{R}^3}\) must be affinely dependent.

The parametric vector form of a B-spline curve was defined in the Practice Problems as

\({\bf{x}}\left( t \right) = \frac{1}{6}\left[ \begin{array}{l}{\left( {1 - t} \right)^3}{{\bf{p}}_o} + \left( {3t{{\left( {1 - t} \right)}^2} - 3t + 4} \right){{\bf{p}}_1}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {3{t^2}\left( {1 - t} \right) + 3t + 1} \right){{\bf{p}}_2} + {t^3}{{\bf{p}}_3}\end{array} \right]\;\), for \(0 \le t \le 1\) where \({{\bf{p}}_o}\) , \({{\bf{p}}_1}\), \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\) are the control points.

a. Show that for \(0 \le t \le 1\), \({\bf{x}}\left( t \right)\) is in the convex hull of the control points.

b. Suppose that a B-spline curve \({\bf{x}}\left( t \right)\)is translated to \({\bf{x}}\left( t \right) + {\bf{b}}\) (as in Exercise 1). Show that this new curve is again a B-spline.

Question: Suppose that the solutions of an equation \(A{\bf{x}} = {\bf{b}}\) are all of the form \({\bf{x}} = {x_{\bf{3}}}{\bf{u}} + {\bf{p}}\), where \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\end{array}} \right)\) and \({\bf{p}} = \left( {\begin{array}{*{20}{c}}{ - {\bf{3}}}\\{\bf{0}}\end{array}} \right)\). Find points \({{\bf{v}}_{\bf{1}}}\) and \({{\bf{v}}_{\bf{2}}}\) such that the solution set of \(A{\bf{x}} = {\bf{b}}\) is \({\bf{aff}}\left\{ {{{\bf{v}}_{\bf{1}}},\,{{\bf{v}}_{\bf{2}}}} \right\}\).

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 鈥渓eft鈥� 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.
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