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Chapter 8: Q12E (page 437)

Question: a. Determine the number of k-faces of the 5-dimensional simplex \({S^{\bf{5}}}\) for \(k = {\bf{0}},{\bf{1}},.....,{\bf{4}}\). Verify that your answer satisfies Euler’s formula.

b. Make a chart of the values of \({f_k}\left( {{S^n}} \right)\) for \(n = {\bf{1}},.....,{\bf{5}}\) and \(k = {\bf{0}},{\bf{1}},.....,{\bf{4}}\). Can you see a pattern? Guess a general formula for \({f_k}\left( {{S^n}} \right)\).

Short Answer

Expert verified

a. 6, 15, 20, 15, 6

b.

\({f_0}\)

\({f_1}\)

\({f_2}\)

\({f_3}\)

\({f_4}\)

\({S^1}\)

2

\({S^2}\)

3

3

\({S^3}\)

4

6

4

\({S^4}\)

5

10

10

5

\({S^5}\)

6

15

20

15

6

There exist a pattern for \({f_k}\left( {{S^n}} \right)\) and it is given by the formula \({f_k}\left( {{S^n}} \right) = \left( {\begin{array}{*{20}{c}}{n + 1}\\{k + 1}\end{array}} \right)\), 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 number of faces for \(k = 0\):

\({f_0}\left( {{S^5}} \right) = 6\)

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

\({f_1}\left( {{S^5}} \right) = 15\)

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

\({f_2}\left( {{S^5}} \right) = 20\)

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

\({f_3}\left( {{S^5}} \right) = 15\)

The number of faces for \(k = 4\):

\({f_4}\left( {{S^5}} \right) = 6\)

02

Verify the Euler’s formula

Euler’s formula can be verified as follows:

\(\begin{array}{c}{f_0}\left( {{S^5}} \right) - {f_1}\left( {{S^5}} \right) + {f_2}\left( {{S^5}} \right) - {f_3}\left( {{S^5}} \right) + {f_4}\left( {{S^5}} \right) = 6 - 15 + 20 - 15 + 6\\ = 2\end{array}\)

So, Euler’s formula is verified.

03

Find the solution for part (b)

The table below represents the values of \({f_k}\left( {{S^n}} \right)\).

\({f_0}\)

\({f_1}\)

\({f_2}\)

\({f_3}\)

\({f_4}\)

\({S^1}\)

2

\({S^2}\)

3

3

\({S^3}\)

4

6

4

\({S^4}\)

5

10

10

5

\({S^5}\)

6

15

20

15

6

There exist a pattern for the values in the chart. The pattern for the above chart is given by the formula \({f_k}\left( {{S^n}} \right) = \left( {\begin{array}{*{20}{c}}{n + 1}\\{k + 1}\end{array}} \right)\). 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

In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{aligned}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{aligned}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each of the given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)

a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{8}}\\{\bf{4}}\end{aligned}} \right)\)

b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{6}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right)\)

c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{ - {\bf{1}}}\\{ - {\bf{5}}}\end{aligned}} \right)\)

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?

Question: 1. 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 \(\left( {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{1}}\end{array}} \right)\). Find a linear functional f and a real number d such that \(L = \left( {f:d} \right)\).

Let \({\bf{x}}\left( t \right)\) be a B-spline in Exercise 2, with control points \({{\bf{p}}_o}\), \({{\bf{p}}_1}\) , \({{\bf{p}}_2}\) , and \({{\bf{p}}_3}\).

a. Compute the tangent vector \({\bf{x}}'\left( t \right)\) and determine how the derivatives \({\bf{x}}'\left( 0 \right)\) and \({\bf{x}}'\left( 1 \right)\) are related to the control points. Give geometric descriptions of the directions of these tangent vectors. Explore what happens when both \({\bf{x}}'\left( 0 \right)\)and \({\bf{x}}'\left( 1 \right)\)equal 0. Justify your assertions.

b. Compute the second derivative and determine how and are related to the control points. Draw a figure based on Figure 10, and construct a line segment that points in the direction of . [Hint: Use \({{\bf{p}}_2}\) as the origin of the coordinate system.]

Question: In Exercise 8, let Hbe the hyperplane through the listed points. (a) Find a vector n that is normal to the hyperplane. (b) Find a linear functional f and a real number d such that \(H = \left( {f:d} \right)\).

8. \(\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{2}}}\\{\bf{1}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{4}}\\{ - {\bf{2}}}\\{\bf{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\bf{7}}\\{ - {\bf{4}}}\\{\bf{4}}\end{array}} \right)\)
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