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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鈥檚 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鈥檚 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鈥檚 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

Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in the statement of Exercise 25.

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: In Exercise 5, determine whether or not each set is compact and whether or not it is convex.

5. Use the sets from Exercise 3.

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

In Exercises 21鈥�24, a, b, and c are noncollinear points in\({\mathbb{R}^{\bf{2}}}\)and p is any other point in\({\mathbb{R}^{\bf{2}}}\). Let\(\Delta {\bf{abc}}\)denote the closed triangular region determined by a, b, and c, and let\(\Delta {\bf{pbc}}\)be the region determined by p, b, and c. For convenience, assume that a, b, and c are arranged so that\(\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]\)is positive, where\(\overrightarrow {\bf{a}} \),\(\overrightarrow {\bf{b}} \) and\(\overrightarrow {\bf{c}} \)are the standard homogeneous forms for the points.

21. Show that the area of\(\Delta {\bf{abc}}\)is\(det\left[ {\begin{array}{*{20}{c}}{\overrightarrow {\bf{a}} }&{\overrightarrow {\bf{b}} }&{\overrightarrow {\bf{c}} }\end{array}} \right]/2\).

[Hint:Consult Sections 3.2 and 3.3, including the Exercises.]
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