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Theorem 12.6. Suppose \( q : E \rightarrow X \) is a covering map and \( x \) is any point in \( X \) . The restriction map \( \varphi \mapsto {\left. \varphi \right| }_{{q}^{-1}\left( x\right) } \) is a group isomorphism between \( {\operatorname{Aut}}_{q}\left( E\right) \) and the group \( {\operatorname{Aut}}_{{\pi ... | Proof. Proposition 12.1(b) shows that each covering automorphism restricts to a \( {\pi }_{1}\left( {X, x}\right) \) -automorphism of \( {q}^{-1}\left( x\right) \) . Since \( {\left. \left( {\varphi }_{1} \circ {\varphi }_{2}\right) \right| }_{{q}^{-1}\left( x\right) } = {\left. {\varphi }_{1}\right| }_{{q}^{-1}\left( ... | No |
Proposition 12.1(a) shows that two covering automorphisms whose restrictions to \( {q}^{-1}\left( x\right) \) agree must be identical, so the restriction homomorphism is injective. To see that it is surjective, suppose \( \eta : {q}^{-1}\left( x\right) \rightarrow {q}^{-1}\left( x\right) \) is any \( {\pi }_{1}\left( {... | If \( {e}_{1} \) is any point in \( {q}^{-1}\left( x\right) \) and \( {e}_{2} = \eta \left( {e}_{1}\right) \), then the orbit criterion for \( G \) - automorphisms (Proposition 11.27) shows that the isotropy groups of \( {e}_{1} \) and \( {e}_{2} \) are the same. Since these isotropy groups are exactly \( {q}_{ * }{\pi... | Yes |
Theorem 12.7 (Covering Automorphism Group Structure Theorem). Suppose \( q : E \rightarrow X \) is a covering map, \( e \in E \), and \( x = q\left( e\right) \) . Let \( G = {\pi }_{1}\left( {X, x}\right) \) and \( H = \) \( {q}_{ * }{\pi }_{1}\left( {E, e}\right) \subseteq {\pi }_{1}\left( {X, x}\right) \) . For each ... | Proof. We have two isomorphisms:\n\n\[ \n{N}_{G}\left( H\right) /H\overset{ \cong }{ \rightarrow }{\operatorname{Aut}}_{G}\left( {{q}^{-1}\left( x\right) }\right) \overset{ \cong }{ \rightarrow }{\operatorname{Aut}}_{q}\left( E\right) .\n\]\n\nThe first isomorphism is induced by the map of Theorem 11.28, which sends an... | Yes |
Theorem 12.14 (Covering Space Quotient Theorem). Let \( E \) be a connected, locally path-connected space, and suppose we are given an effective action of a group \( \Gamma \) on \( E \) by homeomorphisms. Then the quotient map \( q : E \rightarrow E/\Gamma \) is a covering map if and only if the action is a covering s... | Proof. Assume first that \( q \) is a covering map. Then the action of each \( g \in \Gamma \) is an automorphism of the covering, because it is a homeomorphism satisfying \( q\left( {g \cdot e}\right) = \) \( q\left( e\right) \), so we can identify \( \Gamma \) with a subgroup of \( {\operatorname{Aut}}_{q}\left( E\ri... | No |
Proposition 12.15. Let \( \Gamma \) be a discrete subgroup of a connected and locally path-connected topological group \( G \) . Then the action of \( \Gamma \) on \( G \) by right translations is a covering space action, so the quotient map \( q : G \rightarrow G/\Gamma \) is a normal covering map. | Proof. Because \( \Gamma \) is discrete, there is a neighborhood \( V \) of 1 in \( G \) such that \( V \cap \) \( \Gamma = \{ 1\} \) . Consider the continuous map \( F : G \times G \rightarrow G \) given by \( F\left( {g, h}\right) = {g}^{-1}h \) . Since \( {F}^{-1}\left( V\right) \) is a neighborhood of \( \left( {1,... | Yes |
Corollary 12.16. Suppose \( G \) and \( H \) are connected and locally path-connected topological groups, and \( \varphi : G \rightarrow H \) is a surjective continuous homomorphism with discrete kernel. If \( \varphi \) is an open or closed map, then it is a normal covering map. | Proof. Let \( \Gamma = \operatorname{Ker}\varphi \) . By the preceding proposition, the quotient map \( q : G \rightarrow \) \( G/\Gamma \) is a normal covering map. The assumption that \( \varphi \) is either open or closed implies that it is a quotient map, and by the first isomorphism theorem the identifications mad... | Yes |
For any integers \( a, b, c, d \) such that \( {ad} - {bc} \neq 0 \), consider the map \( q : {\mathbb{T}}^{2} \rightarrow {\mathbb{T}}^{2} \) given by \( q\left( {z, w}\right) = \left( {{z}^{a}{w}^{b},{z}^{c}{w}^{d}}\right) \). This is easily seen to be a surjective continuous homomorphism, and it is a closed map by t... | Let \( A \) denote the invertible linear transformation of \( {\mathbb{R}}^{2} \) whose matrix is \( \left( \begin{array}{ll} a & b \\ c & d \end{array}\right) \). Then we have a commutative diagram\n\n\n\n(12.2)\n\n... | Yes |
Proposition 12.21 (Hausdorff Criterion for Orbit Spaces). Suppose \( E \) is a topological space and \( \Gamma \) is a group acting \( E \) by homeomorphisms. Then \( E/\Gamma \) is Haus-dorffif and only if the action satisfies the following condition:\n\n\[ \text{if}e,{e}^{\prime } \in E\text{lie in different orbits, ... | Proof. Let \( q : E \rightarrow E/\Gamma \) denote the quotient map. If \( E/\Gamma \) is Hausdorff, then given \( e,{e}^{\prime } \) in different orbits, there are disjoint neighborhoods \( U \) of \( q\left( e\right) \) and \( {U}^{\prime } \) of \( q\left( {e}^{\prime }\right) \) , and then \( V = {q}^{-1}\left( U\r... | Yes |
Proposition 12.22. Every continuous action of a compact topological group on a Hausdorff space is proper: | Proof. Suppose \( G \) is a compact group acting continuously on a Hausdorff space \( E \), and let \( \Theta : G \times E \rightarrow E \times E \) be the map defined by (12.5). Given a compact set \( L \subseteq E \times E \), let \( K = {\pi }_{2}\left( L\right) \), where \( {\pi }_{2} : E \times E \rightarrow E \) ... | Yes |
Proposition 12.23. Suppose we are given a continuous action of a topological group \( G \) on a Hausdorff space \( E \) . The action is proper if and only if for every compact subset \( K \subseteq E \), the set \( {G}_{K} = \{ g \in G : \left( {g \cdot K}\right) \cap K \neq \varnothing \} \) is compact. | Proof. Let \( \Theta : G \times E \rightarrow E \times E \) be the map defined by (12.5). Suppose first that \( \Theta \) is proper. Then for any compact set \( K \subseteq E \), we have\n\n\[{G}_{K} = \{ g \in G : \text{ there exists }e \in K\text{ such that }g \cdot e \in K\}\]\n\n\[= \{ g \in G : \text{ there exists... | Yes |
Proposition 12.24. If a topological group \( G \) acts continuously and properly on a locally compact Hausdorff space \( E \), then the orbit space \( E/G \) is Hausdorff. | Proof. Let \( \mathcal{O} \subseteq E \times E \) be the orbit relation defined in Problem 3-22. By the result of that problem, the orbit space is Hausdorff if and only if \( \mathcal{O} \) is closed in \( E \times E \). But \( \mathcal{O} \) is just the image of the map \( \Theta : G \times E \rightarrow E \times E \)... | Yes |
Proposition 12.25. Suppose we are given a covering space action of a group \( \Gamma \) on a topological space \( E \), and \( E/\Gamma \) is Hausdorff. Then with the discrete topology, \( \Gamma \) acts properly on \( E \) . | Proof. For convenience, write \( X = E/\Gamma \), and let \( q : E \rightarrow X \) be the quotient map, which is a normal covering map by the covering space quotient theorem. It follows from Proposition 3.57 and Problem 3-22 that the orbit relation \( \mathcal{O} \) defined by (3.6) is closed in \( E \times E \) . Als... | Yes |
Theorem 12.26. Suppose \( E \) is a connected, locally path-connected, and locally compact Hausdorff space, and a discrete group \( \Gamma \) acts continuously, freely, and properly on \( E \). Then the action is a covering space action, \( E/\Gamma \) is Hausdorff, and the quotient map \( q : E \rightarrow E/\Gamma \)... | Proof. We need only show that the action is a covering space action, for then Proposition 12.24 shows that \( E/\Gamma \) is Hausdorff, and the covering space quotient theorem shows that \( q \) is a normal covering map.\n\nSuppose \( {e}_{0} \in E \) is arbitrary. Because \( E \) is locally compact, \( {e}_{0} \) has ... | Yes |
Example 12.28 (Lens Spaces). By identifying \( {\mathbb{R}}^{4} \) with \( {\mathbb{C}}^{2} \) in the usual way, we can consider \( {\mathbb{S}}^{3} \) as the following subset of \( {\mathbb{C}}^{2} \) :\n\n\[ \n{\mathbb{S}}^{3} = \left\{ {\left( {{z}_{1},{z}_{2}}\right) \in {\mathbb{C}}^{2} : {\left| {z}_{1}\right| }^... | By the classification theorem, the coverings of \( L\left( {n, m}\right) \) are in one-to-one correspondence with subgroups of \( \mathbb{Z}/n \) . Since every subgroup of a cyclic group is cyclic (Exercise C.15), the only possibilities for subgroups \( G \subseteq {\pi }_{1}\left( {L\left( {n, m}\right) }\right) \) ar... | Yes |
Theorem 12.29. Let \( M \) be a compact surface. The universal covering space of \( M \) is homeomorphic to\n\n(a) \( {\mathbb{S}}^{2} \) if \( M \approx {\mathbb{S}}^{2} \) or \( {\mathbb{P}}^{2} \) ,\n\n(b) \( {\mathbb{R}}^{2} \) if \( M \approx {\mathbb{T}}^{2} \) or \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) ,\n\n(c... | Proof. Because \( {\mathbb{S}}^{2} \) is simply connected, it is its own universal covering space. It was shown in Example 11.42 that the universal covering space of \( {\mathbb{T}}^{2} \) is \( {\mathbb{R}}^{2} \), and that of \( {\mathbb{P}}^{2} \) is \( {\mathbb{S}}^{2} \) . If \( M \) is a connected sum of \( n \ge... | No |
Lemma 13.1. Ifc is a singular chain, then \( \partial \left( {\partial c}\right) = 0 \) . | Proof. Since each chain group \( {C}_{p}\left( X\right) \) is generated by singular simplices, it suffices to show this in the case in which \( c = \sigma \) is a singular \( p \) -simplex.\n\nFirst we note that the face maps satisfy the commutation relation\n\n\[ \n{F}_{i, p} \circ {F}_{j, p - 1} = {F}_{j, p} \circ {F... | Yes |
Proposition 13.2 (Functorial Properties of Homology). Let \( X \) , \( Y \) , and \( Z \) be topological spaces.\n\n(a) The homomorphism \( {\left( {\operatorname{Id}}_{X}\right) }_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p}\left( X\right) \) induced by the identity map of \( X \) is the identity of \( {H}_{p}\l... | Proof. It is easy to check that both properties hold already for \( {f}_{\# } \) . | No |
Proposition 13.5. Let \( X \) be a space, let \( {\left\{ {X}_{\alpha }\right\} }_{\alpha \in A} \) be the set of path components of \( X \), and let \( {\iota }_{\alpha } : {X}_{\alpha } \hookrightarrow X \) be inclusion. Then for each \( p \geq 0 \) the map\n\n\[ \n{\bigoplus }_{\alpha \in A}{H}_{p}\left( {X}_{\alpha... | Proof. Since the image of any singular simplex must lie entirely in one path component, the chain maps \( {\left( {\iota }_{\alpha }\right) }_{\# } : {C}_{p}\left( {X}_{\alpha }\right) \rightarrow {C}_{p}\left( X\right) \) already induce isomorphisms\n\n\[ \n{\bigoplus }_{\alpha \in A}{C}_{p}\left( {X}_{\alpha }\right)... | Yes |
Proposition 13.6 (Zero-Dimensional Homology). For any topological space \( X \) , \( {H}_{0}\left( X\right) \) is a free abelian group with basis consisting of an arbitrary point in each path component. | Proof. It suffices to show that \( {H}_{0}\left( X\right) \) is the infinite cyclic group generated by the class of any point when \( X \) is path-connected, for then in the general case Proposition 13.5 guarantees that \( {H}_{0}\left( X\right) \) is the direct sum of infinite cyclic groups, one for each path componen... | Yes |
Proposition 13.7 (Homology of a Discrete Space). If \( X \) is a discrete space, then \( {H}_{0}\left( X\right) \) is a free abelian group with one generator for each point of \( X \), and \( {H}_{p}\left( X\right) = 0 \) for \( p > 0 \) . | Proof. The case \( p = 0 \) follows from the preceding proposition, so we concentrate on \( p > 0 \) . By Proposition 13.5, it suffices to show that \( {H}_{p}\left( *\right) = 0 \) when \( * \) is a one-point space. In that case, there is exactly one singular simplex in each dimension, namely the constant map \( {\sig... | Yes |
Corollary 13.9 (Homotopy Invariance of Singular Homology). Suppose \( f : X \rightarrow Y \) is a homotopy equivalence. Then for each \( p \geq 0, {f}_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p}\left( Y\right) \) is an isomorphism. | Proof of Theorem 13.8. We begin by considering the special case in which \( Y = X \times I \) and \( {f}_{i} = {\iota }_{i} \), where \( {\iota }_{0},{\iota }_{1} : X \rightarrow X \times I \) are the maps \[ {\iota }_{0}\left( x\right) = \left( {x,0}\right) ,\;{\iota }_{1}\left( x\right) = \left( {x,1}\right) . \] (Se... | No |
Lemma 13.13. Suppose \( {f}_{0} \) and \( {f}_{1} \) are paths in \( X \), and \( {f}_{0} \sim {f}_{1} \) . Then, considered as a singular chain, \( {f}_{0} - {f}_{1} \) is a boundary. | Proof. We must show there is a singular 2-chain whose boundary is the 1-chain \( {f}_{0} - {f}_{1} \) . Let \( H : {f}_{0} \sim {f}_{1} \), and let \( b : I \times I \rightarrow {\Delta }_{2} \) be the map\n\n\[ b\left( {x, y}\right) = \left( {x - {xy},{xy}}\right) ,\]\n\n(13.7)\n\nwhich maps the square onto the triang... | Yes |
Theorem 13.16 (Mayer-Vietoris). Let \( X \) be a topological space, and let \( U, V \) be open subsets of \( X \) whose union is \( X \) . Then for each \( p \) there is a homomorphism \( {\partial }_{ * } : {H}_{p}\left( X\right) \rightarrow {H}_{p - 1}\left( {U \cap V}\right) \) such that the following sequence is ex... | The exact sequence (13.10) is called the Mayer-Vietoris sequence of the triple \( \left( {X, U, V}\right) \), and \( {\partial }_{ * } \) is called the connecting homomorphism. The other maps are defined by \( \left( {{i}_{ * } \oplus {j}_{ * }}\right) \left\lbrack c\right\rbrack = \left( {{i}_{ * }\left\lbrack c\right... | Yes |
Proposition 13.18 (Naturality of the Connecting Homomorphism). Suppose \n\n(13.12)\n\nis a commutative diagram of chain maps in which the horizontal rows are exact. Then the following diagram commutes for each \( p \... | Proof. Let \( \left\lbrack {e}_{p}\right\rbrack \in {H}_{p}\left( {E}_{ * }\right) \) be arbitrary. Then \( {\partial }_{ * }\left\lbrack {e}_{p}\right\rbrack = \left\lbrack {c}_{p - 1}\right\rbrack \), where \( F{c}_{p - 1} = \partial {d}_{p} \) for some \( {d}_{p} \) such that \( G{d}_{p} = {e}_{p} \) . Then by commu... | Yes |
Proposition 13.19. Suppose \( \mathcal{U} \) is any open cover of \( X \) . Then the inclusion map \( {C}_{ * }^{\mathcal{U}}\left( X\right) \rightarrow {C}_{ * }\left( X\right) \) induces a homology isomorphism \( {H}_{p}^{\mathcal{U}}\left( X\right) \cong {H}_{p}\left( X\right) \) for all \( p \) . | The idea of the proof is simple, although the technical details are somewhat involved. If \( \sigma : {\Delta }_{p} \rightarrow X \) is any singular \( p \) -simplex, the plan is to show that there is a homologous \( p \) -chain obtained by \ | No |
If \( c \) is an affine chain, then\n\n\[ \partial \left( {w * c}\right) + w * \partial c = c. \] | Proof. For an affine simplex \( \alpha = A\left( {{v}_{0},\ldots ,{v}_{p}}\right) \), this is just a computation:\n\n\[ \partial \left( {w * \alpha }\right) = \partial A\left( {w,{v}_{0},\ldots ,{v}_{p}}\right) \]\n\n\[ = \mathop{\sum }\limits_{{i = 0}}^{{p + 1}}{\left( -1\right) }^{i}A\left( {w,{v}_{0},\ldots ,{v}_{p}... | Yes |
Lemma 13.21. Suppose \( \alpha : {\Delta }_{p} \rightarrow {\mathbb{R}}^{n} \) is an affine simplex that is a homeomorphism onto a p-simplex \( \sigma \subseteq {\mathbb{R}}^{n} \) . Let \( \beta : {\Delta }_{p} \rightarrow {\mathbb{R}}^{n} \) be any one of the affine singular \( p \) - simplices that appear in the cha... | Proof. Part (a) follows immediately from the definition of the subdivision operator and an easy induction on \( p \) (see Fig. 13.9).\n\nTo prove (b), write \( \sigma = \alpha \left( {\Delta }_{p}\right) = \left\lbrack {{v}_{0},\ldots ,{v}_{p}}\right\rbrack \) and \( \tau = \beta \left( {\Delta }_{p}\right) = \left\lbr... | Yes |
Lemma 13.22. The singular subdivision operators \( s : {C}_{p}\left( X\right) \rightarrow {C}_{p}\left( X\right) \) have the following properties.\n\n(a) \( s \circ {f}_{\# } = {f}_{\# } \circ s \) for any continuous map \( f \) .\n\n(b) \( \partial \circ s = s \circ \partial \) .\n\n(c) Given an open cover \( \mathcal... | Proof. The first identity follows immediately from the definition of \( s \) :\n\n\[ s\left( {{f}_{\# }\sigma }\right) = s\left( {f \circ \sigma }\right) = {\left( f \circ \sigma \right) }_{\# }\left( {s{i}_{p}}\right) = {f}_{\# }{\sigma }_{\# }\left( {s{i}_{p}}\right) = {f}_{\# }\left( {s\sigma }\right) . \]\n\nThe se... | Yes |
Theorem 13.23 (Homology Groups of Spheres). For \( n \geq 1,{\mathbb{S}}^{n} \) has the following singular homology groups: | Proof. We use the Mayer-Vietoris sequence as follows. Let \( N \) and \( S \) denote the north and south poles, and let \( U = {\mathbb{S}}^{n} \smallsetminus \{ N\}, V = {\mathbb{S}}^{n} \smallsetminus \{ S\} \) . Part of the Mayer-Vietoris sequence reads\n\n\[{H}_{p}\left( U\right) \oplus {H}_{p}\left( V\right) \righ... | Yes |
Corollary 13.24 (Homology Groups of Punctured Euclidean Spaces). For \( n \geq 2 \) , \( {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) has the following singular homology groups:\n\n\[ \n{H}_{p}\left( {{\mathbb{R}}^{n}\smallsetminus \{ 0\} }\right) \cong \left\{ \begin{array}{ll} \mathbb{Z} & \text{ if }p = 0, \\ 0 & \text... | Proof. Inclusion \( {\mathbb{S}}^{n - 1} \hookrightarrow {\mathbb{R}}^{n} \smallsetminus \{ 0\} \) is a homotopy equivalence. | No |
Proposition 13.25. Suppose \( n \geq 1 \) and \( f, g : {\mathbb{S}}^{n} \rightarrow {\mathbb{S}}^{n} \) are continuous maps.\n\n(a) \( \deg \left( {g \circ f}\right) = \left( {\deg g}\right) \left( {\deg f}\right) \).\n\n(b) If \( f \simeq g \), then \( \deg f = \deg g \) . | Proof. Part (a) follows from the fact that \( {\left( g \circ f\right) }_{ * } = {g}_{ * } \circ {f}_{ * } \), and part (b) from the fact that homotopic maps induce the same homology homomorphism. | Yes |
Lemma 13.26. The homological degree and the homotopic degree of a continuous map \( f : {\mathbb{S}}^{1} \rightarrow {\mathbb{S}}^{1} \) are equal. | Proof. By (13.8), the following diagram commutes:\n\n\[ \n{\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \xrightarrow[]{{\left( \rho \circ f\right) }_{ * }}{\pi }_{1}\left( {{\mathbb{S}}^{1},1}\right) \n\]\n\n\[ \n\gamma \n\]\n\n\[ \n{H}_{1}\left( {\mathbb{S}}^{1}\right) \xrightarrow[{\left( \rho \circ f\right) }_{ * }]{... | No |
Proposition 13.31. The antipodal map \( \alpha : {\mathbb{S}}^{n} \rightarrow {\mathbb{S}}^{n} \) is homotopic to the identity map if and only if \( n \) is odd. | Proof. If \( n = {2k} - 1 \) is odd, an explicit homotopy \( H : \mathrm{{Id}} \simeq \alpha \) is given by\n\n\[ H\left( {x, t}\right) = \left( {\left( {\cos {\pi t}}\right) {x}_{1} + \left( {\sin {\pi t}}\right) {x}_{2},\left( {\cos {\pi t}}\right) {x}_{2} - \left( {\sin {\pi t}}\right) {x}_{1},}\right.\n\n\[ \left. ... | Yes |
There exists a nowhere vanishing vector field on \( {\mathbb{S}}^{n} \) if and only if \( n \) is odd. | Suppose there exists such a vector field \( V \) . By replacing \( V \) with \( V/\left| V\right| \), we can assume \( \left| {V\left( x\right) }\right| = 1 \) everywhere. We use \( V \) to construct a homotopy between the identity map and the antipodal map as follows:\n\n\[ H\left( {x, t}\right) = \left( {\cos {\pi t}... | Yes |
Proposition 13.33 (Homology Effect of Attaching a Cell). Let \( X \) be any topological space, and let \( Y \) be obtained from \( X \) by attaching a closed cell \( D \) of dimension \( n \geq 2 \) along the attaching map \( \varphi : \partial D \rightarrow X \) . Let \( K \) and \( L \) denote the kernel and image, r... | Proof. First, assume that \( p \geq 2 \) . Let \( q : X \coprod D \rightarrow Y \) be a quotient map realizing \( Y \) as an adjunction space. Choose a point \( z \in \operatorname{Int}D \), and define open subsets \( U, V \subseteq Y \) by \( U = q\left( {\operatorname{Int}D}\right) \) and \( V = q\left( {X \coprod \l... | Yes |
Theorem 13.34 (Homology Properties of CW Complexes). Let \( X \) be a finite \( n \) - dimensional CW complex.\n\n(a) Inclusion \( {X}_{k} \hookrightarrow X \) induces isomorphisms \( {H}_{p}\left( {X}_{k}\right) \cong {H}_{p}\left( X\right) \) for \( p \leq k - 1 \) . | Proof. Part (a) follows immediately from Theorem 13.33, because attaching an \( m \) - cell cannot change \( {H}_{p}\left( X\right) \) if \( p < m - 1 \) . | Yes |
Theorem 13.36. If \( X \) is a finite \( {CW} \) complex,\n\n\[ \chi \left( X\right) = \mathop{\sum }\limits_{p}{\left( -1\right) }^{p}\operatorname{rank}{H}_{p}\left( X\right) . \]\n\n(13.21)\n\nTherefore, the Euler characteristic is a homotopy invariant. | Proof. First let us assume that \( X \) is connected. We prove (13.21) by induction on the number of cells of dimension 2 or more. If \( X \) has no such cells, then it is a connected graph. Problem 10-20 shows that \( {\pi }_{1}\left( X\right) \) is a free group on \( 1 - \chi \left( X\right) \) generators, and then T... | Yes |
Proposition 13.37 (Functorial Properties of Cohomology). The induced cohomology homomorphism satisfies the following properties.\n\n(a) If \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) are continuous, then \( {\left( g \circ f\right) }^{ * } = {f}^{ * } \circ {g}^{ * } \).\n\n(b) The homomorphism induced by t... | Therefore, the assignments \( X \mapsto {H}^{p}\left( {X;G}\right), f \mapsto {f}^{ * } \) define a contravariant functor from the category of topological spaces to the category of abelian groups. | Yes |
Lemma 13.40. Let \( \mathbb{F} \) be a field of characteristic zero.\n\n(a) For any abelian group \( G \), the set \( \operatorname{Hom}\left( {G,\mathbb{F}}\right) \) of group homomorphisms from G to \( \mathbb{F} \) is a vector space over \( \mathbb{F} \) with scalar multiplication defined pointwise: \( \left( {a\var... | Proof. The proofs of (a) and (b) are straightforward (and hold for any field, not just one of characteristic zero), and are left as an exercise. | No |
Corollary 13.44. If \( X \) is a topological space such that \( {H}_{p}\left( X\right) \) is finitely generated for all \( p \) and zero for \( p \) sufficiently large, then for any field \( \mathbb{F} \) of characteristic zero, | \[ \chi \left( X\right) = \mathop{\sum }\limits_{p}{\left( -1\right) }^{p}\dim {H}^{p}\left( {X;\mathbb{F}}\right) . \] | No |
Build a differential equation relating a dead body's cooling temperature to time in order to find the time of death. | Newton's Law of Cooling states that the surface temperature of an object changes at a rate proportional to its relative temperature. That is, the difference between its temperature and the surrounding environment temperature \( A \) . Let \( T\left( t\right) = \) the temperature of the object at time \( t \) . Accordin... | No |
Relate the concentration of barley in the vat to time \( t \) . | The rate of change of concentration is measured as the difference between the rate pumped in and the mixture being pumped out of the vat. Let \( C\left( t\right) \) be the concentration or amount of barley in the vat after time \( t \), then\n\n\[ \n{C}^{\prime }\left( t\right) = {r}_{\text{in }} - {r}_{\text{out }}\n\... | No |
Consider an experiment measuring the growth of a yeast culture. An analysis yields the growth rate as approximately 0.6 . We might try as our ODE \( {Y}^{\prime }\left( t\right) = \) \( {Y}_{0} \cdot {e}^{0.6t} \) . | Note that this model will predict a population that increases forever. We might think that is not possible. | No |
Suppose there are 400 students in a college dormitory and that one or more of the students has a severe case of the flu. Let \( {i}_{n} \) represent the number of infected students after \( n \) time periods. Assume some interaction between those infected and those not infected is required to pass the disease. If all a... | \[ {i}_{n + 1} - {i}_{n} = k \cdot {i}_{n} \cdot \left( {{400} - {i}_{n}}\right) \] | Yes |
[
\\frac{dy}{dt} = t \\cdot y
] | This ODE is separable because the function \\( f\\left( {t, y}\\right) = t \\cdot y \\) can be written as the product of \\( h\\left( t\\right) = t \\) and \\( g\\left( y\\right) = y \\) . | No |
\[ \frac{dy}{dt} = t + {9y} \] | This ODE is not separable since \( f\left( {t, y}\right) = t + {9y} \) cannot be written as the product of two functions \( h\left( t\right) \) of \( t \) only and \( g\left( y\right) \) of \( y \) only. | Yes |
[
\frac{dy}{dt} = {8ty} - {12y} - {2t} + 3
\]
This ODE can be written as \( \frac{dy}{dt} = \left( {{2t} - 3}\right) \cdot \left( {{4y} - 1}\right) \) and is seen to be separable
when in this form. | In order to solve separable ODEs, we perform the following steps:
1. Rewrite the differential equation into a clearly separable form of
\[
\frac{dy}{dt} = h\left( t\right) \cdot g\left( y\right)
\]
2. Separate the differential equation as
\[
\frac{dy}{g\left( y\right) } = h\left( t\right) {dt}
\]
3. Integrate both... | Yes |
Solve \( \frac{dy}{dt} = t \cdot y\left( t\right) \). | 1. This ODE is separable with \( h\left( t\right) = t \) and \( g\left( y\right) = y \). \n\n2. Separate: \n\n\[ \frac{dy}{y} = {tdt} \] \n\n3. Integrate: \n\n\[ \int \frac{1}{y}{dy} = \int {tdt} \] \n\n\[ \ln \left( y\right) + C = \frac{{t}^{2}}{2} + C \] \n\n\[ {e}^{\ln \left( y\right) } = {e}^{{t}^{2}/2 + C} \] \n\n... | Yes |
Model the population of the US provided in Table 2.3. (Data source: U.S. Census Bureau.) | The model that we will initially use is our simple growth model \( {dP}/{dt} = {kP} \) , \( k > 0 \), where \( t \) is the year past 1990 . Solve this separable ODE with our standard procedure:\n\n1. Separate:\n\n\[ \frac{dP}{P} = {kdt} \]\n\n2. Integrate:\n\n\[ \int \frac{1}{P}{dP} = \int {kdt} \]\n\n\[ \ln \left( P\r... | Yes |
Suppose a cold can of soda is taken from a refrigerator and placed in a warm classroom. The temperature is measured periodically. The temperature of the soda is initially \( {40}^{ \circ } \) Fahrenheit; room temperature is a cozy \( {72}^{ \circ }\mathrm{F} \) . Model the temperature of the soda. | Using Newton's Law of Cooling gives\n\n\[ \frac{dT}{dt} = - k\left( {{T}_{a} - T}\right) ,\;T\left( 0\right) = {40}. \]\n\nAssume that we have measured \( k = {0.05} \) . Substituting our values for \( k \) and room temperature, \( {T}_{a} \), gives\n\n\[ \frac{dT}{dt} = - {0.05}\left( {{72} - T}\right) ,\;T\left( 0\ri... | Yes |
Determine the amount of salt present at any time \( t \) if the concentration of the entering solution is two pounds per gallon. | Write a model for the problem as\n\n\[ \frac{dA}{dt} = {R}_{\text{in }} - {R}_{\text{out }} \]\n\nwhere \( A \) represents the amount of salt (pounds) in the tank at time \( t \) . Now, determine the \( R\mathrm{\;s} \), the rate salt enters/leaves \( \left( {\mathrm{{lb}}/\mathrm{{min}}}\right) \), respectively, from\... | Yes |
A logistic growth model defined for blue crabs in Venezuela is\n\n\\[ \n\\frac{dB}{dt} = {0.25B}\\left( {{10} - B}\\right) ,\\;B\\left( 0\\right) = 4 \n\\]\n\nwhere \\( B \\) is in millions of bushels. Determine the maximum sustainable amount (millions of bushels). | The logistic DE is separable, so we can apply our standard technique.\n\n1. Separate:\n\n\\[ \n\\frac{dB}{B\\left( {{10} - B}\\right) } = {0.25dt} \n\\]\n\n2. Integrate:\n\n\\[ \n\\int \\frac{dB}{B\\left( {{10} - B}\\right) } = \\int {0.25dt} = {0.25t} + C \n\\]\n\nThe integral on the left will require partial fraction... | Yes |
Solving a Linear Differential Equation. Solve the linear DE from the slope field shown in Figure 2.2,\n\n\[ \frac{dy}{dt} = \frac{t - y}{2} \] | Step 1. We put the linear differential equation in the proper form.\n\n\[ \frac{dy}{dt} + \frac{1}{2}y = \frac{1}{2}t \]\n\nidentifying \( P\left( t\right) = 1/2 \) and \( f\left( t\right) = t/2 \) .\n\nStep 2. Calculate the integrating factor \( u \) .\n\n\[ u\left( t\right) = {e}^{\int 1/{2dt}} = {e}^{t/2} \]\n\nStep... | Yes |
We know that Newton’s Law of Cooling gives the temperature \( T \) of a mass being cooled by the ODE\n\n\[ \frac{dT}{dt} = k\left( {{T}_{a} - T}\right) ,\;k > 0 \]\n\nwhere \( {T}_{a} \) is the ambient temperature or the constant temperature of the surroundings, and \( k > 0 \) is called the heat transfer coefficient.\... | Observe that the ODE from Newton's Law of Cooling is both linear and separable. We’ll find \( T \) using the integrating factor technique on this linear ODE.\n\nStep 1. We put the linear differential equation in the proper form\n\n\[ \frac{dT}{dt} + {0.19T} = {13.3} \]\n\nidentifying \( P\left( t\right) = {0.19} \) and... | Yes |
Solve the ODE \( \frac{dy}{dt} = y \cdot t \) . | \[ \begin{array}{l} y\left( t\right) = \operatorname{Coff}\left( {y\left( t\right), t}\right) = y\left( t\right) \cdot t \\ \text{ deq }n \mathrel{\text{:=}} \frac{d}{dt}y\left( t\right) = y\left( t\right) \cdot t \\ y\left( t\right) = \operatorname{C1}{e}^{\frac{{t}^{2}}{2}} \end{array} \] | Yes |
Solving the Refined Population Model with Maple. | \n\[
\begin{array}{l} \left\lbrack \begin{matrix} > \text{ deqn } \mathrel{\text{:=}} \operatorname{diff}\left( {B\left( t\right), t}\right) = {0.25} \cdot B\left( t\right) \cdot \left( {{10} - B\left( t\right) }\right) ; \\ \text{ deqn } \mathrel{\text{:=}} \operatorname{diff}\left( {B\left( t\right), t}\right) = {0.2... | Yes |
Estimate the solution from \( y\left( 0\right) \) to \( y\left( 3\right) \) to the IVP\n\n\[ \frac{dy}{dt} = {0.25} \cdot y \cdot t,\;y\left( 0\right) = 2 \]\n\nusing Euler's Method. | The interval for the problem is \( \left\lbrack {0,3}\right\rbrack \) and \( g\left( {t, y}\right) = {0.25} \cdot y \cdot t \) . Let’s calculate a few steps by hand.\n\nStep 1. Take a step size of \( h = 1 \) . Then \( \left( {b - {t}_{0}}\right) /h = \left( {3 - 0}\right) /1 = 3 \) steps. Let \( k = 1 \) .\n\nStep 2. ... | Yes |
Problem Identification: Build a model for price and supply for a specific product. | Assumptions and variables: Assume the price is proportional to the quantity supplied. Also assume the change in the quantity supplied is proportional to the price. We define the following variables.\n\n\[ P\left( t\right) = \text{the price of the product at time}t\text{,}\]\n\n\[ Q\left( t\right) = \text{the quantity s... | No |
Example 3.2. An Electrical Network.\n\nElectrical networks with more than one loop give rise to systems of differential equations. Consider the electrical network displayed in Figure 3.1 where there are two resisters and two inductors. We apply Kirchhoff's Voltage Law (the sum of the voltage drops in a closed circuit i... | We substitute this expression for \( {i}_{1} \) into the loop equations to obtain the model:\n\n\[ \frac{d{i}_{2}}{dt} = - \frac{{R}_{1}}{{L}_{1}}{i}_{2} - \frac{{R}_{1}}{{L}_{1}}{i}_{3} + \frac{E\left( t\right) }{{L}_{1}} \]\n\n\[ \frac{d{i}_{3}}{dt} = - \frac{{R}_{1}}{{L}_{2}}{i}_{2} - \frac{{R}_{1} + {R}_{2}}{{L}_{2... | Yes |
Imagine a small fish pond supporting both brown trout and rainbow trout. Let \( B\left( t\right) \) denote the population of brown trout at time \( t \) and \( R\left( t\right) \) denote the population of rainbow trout at time \( t \) . We want to know if both can coexist in the pond. Although population growth depends... | We assume that the species grow in isolation. The level of the population of the rainbow trout or the brown trout, \( B\left( t\right) \) and \( R\left( t\right) \), depend on many variables such as their initial numbers, the amount of competition, the existence of predators, their individual species birth and death ra... | Yes |
First, find the rest points by solving \( {dB}/{dt} = 0 \) and \( {dR}/{dt} = 0 \) simultaneously. | \[ \left\lbrack \begin{matrix} > \operatorname{solve}\left( {\{ {0.8B} - {0.02B} \cdot R = 0,{0.6R} - {0.01B} \cdot R\} ,\{ B, R\} }\right) ; \\ \{ B = 0., R = 0.\} ,\{ B = {60}., R = {40}.\} \end{matrix}\right. \] | Yes |
A Homogenous System with Complex Eigenvalues \( \lambda = a \pm {bi} \) . | Complex eigenvalues for a real \( 2 \times 2 \) matrix must always come in conjugate pairs, so we can use a moderate amount of algebra with a little analysis to eliminate the \( i \) terms leaving only real functions in our solution. The key to finding two real linearly independent solutions from complex eigenvalues is... | Yes |
A Homogeneous System with Repeated Eigenvalues. When eigenvalues are repeated we must find a method to obtain linearly independent solutions. (A system must be at least \( 4 \times 4 \) to have repeated complex eigenvalues since we know that systems with real coefficients have complex eigenvalues that must occur as con... | The following is a summary for repeated real eigenvalues for the system \( {X}^{\prime } = {AX} \) with \( A = \left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack \) . Step 1. Find the repeated eigenvalues \( {\lambda }_{1} = {\lambda }_{2} = \lambda \) . Step 2. One solution is \( {X}_{1} = K{e}^{\l... | No |
Example 3.11. Chemical Diffusion through a Single Membrane. Consider the diffusion initial value problem:\n\n\[ \n\\frac{d{x}_{1}}{dt} = \\frac{P}{{V}_{1}}\\left( {{x}_{2} - {x}_{1}}\\right) = - {x}_{1} + {x}_{2} \n\]\n\n\[ \n\\frac{d{x}_{2}}{dt} = \\frac{P}{{V}_{2}}\\left( {{x}_{1} - {x}_{2}}\\right) = {x}_{1} - {x}_{... | First, we set up the differential equation.\n\n\[ \n\\begin{array}{l} \\left\\lbrack \\begin{array}{l} > \\text{ Chem }1 \\mathrel{\\text{:=}} \\operatorname{Matrix}(\\left\\lbrack {\\left\\lbrack {1,1}\\right\\rbrack ,\\left\\lbrack {-1,1}\\right\\rbrack }\\right\\rbrack ); \\\\ \\text{ Chem }1 \\mathrel{\\text{:=}} \... | Yes |
Analyze the behavior of the electric circuit described by the system\n\n\[ \frac{dQ}{dt} = - Q - {i}_{2} \]\n\n\[ \frac{d{i}_{2}}{dt} = Q - {i}_{2} \]\n\nand\n\n\[ Q\left( 0\right) = 3,{i}_{2}\left( 0\right) = 1 \] | \n\n\[ \begin{array}{l} X \mathrel{\text{:=}} \operatorname{Matrix}\left( \left\lbrack {\left\lbrack {-1, - 1}\right\rbrack ,\left\lbrack {1, - 1}\right\rbrack }\right\rbrack \right) ; \\ A \mathrel{\text{:=}} \left\lbrack \begin{matrix} - 1 & - 1 \\ 1 & - 1 \end{matrix}\right\rbrack \\ X \mathrel{\text{:=}} \left\{ {Q... | Yes |
Analyze the behavior of the electric circuit described by the system\n\n\[ \n\frac{dQ}{dt} = - Q - {i}_{2} + {10}{e}^{-t} \]\n\n\[ \n\frac{d{i}_{2}}{dt} = Q - {i}_{2} \]\n\n\nand\n\n\[ \nQ\left( 0\right) = 3,{i}_{2}\left( 0\right) = 1 \]\n | The graph tells us that both \( Q \) and \( {i}_{2} \) go to zero, but less quickly than the unforced circuit. Adjusting the plot range will show that \( {i}_{2}\left( t\right) \) no longer goes negative. | No |
Shade the feasible region defined by the set of constraints\n\n\[ {x}_{1} + 4{x}_{2} \leq {40} \]\n\n\[ 4{x}_{1} + {x}_{2} \leq {40} \]\n\n\[ 2{x}_{1} + 3{x}_{2} \leq {35} \]\n\n\[ {x}_{1} \geq 0,{x}_{2} \geq 0 \] | Solution: The feasible region is the set of ordered pairs \( \left( {{x}_{1},{x}_{2}}\right) \) that satisfy all four constraints simultaneously. They are points that lie below \( {x}_{1} + 4{x}_{2} = \) 40, below \( 4{x}_{1} + {x}_{2} = {40} \), below \( 2{x}_{1} + 3{x}_{2} = {35} \), to the right of \( {x}_{1} = 0 \)... | No |
Minimize \( Z = {5x} + {7y} \) subject to: \[ {2x} + {3y} \geq 6 \] \[ {3x} - y \leq {15} \] \[ - x + y \leq 4 \] \[ {2x} + {5y} \leq {27} \] \[ x \geq 0\text{and}y \geq 0 \] | The corner points in Figure 4.3 are \( \left( {0,2}\right) ,\left( {0,4}\right) ,\left( {1,5}\right) ,\left( {6,3}\right) ,\left( {5,0}\right) \), and \( \left( {3,0}\right) \). Examining the values of \( Z \) in Table 4.9 shows us that the minimum value occurs at \( \left( {0,2}\right) \) with a \( Z \) value of 14. N... | Yes |
Maximize \( Z = 3{x}_{1} + {x}_{2} \) subject to: \[ 2{x}_{1} + {x}_{2} \leq 6 \] \[ {x}_{1} + 3{x}_{2} \leq 9 \] \[ {x}_{1},{x}_{2} \geq 0 \] | The Tableau Format with slack variables \( {s}_{1} \) and \( {s}_{2} \) is <table><thead><tr><th>Basic Variable</th><th>\( Z \)</th><th>\( {x}_{1} \)</th><th>\( {x}_{2} \)</th><th>\( {s}_{1} \)</th><th>\( {s}_{2} \)</th><th>RHS</th></tr></thead><tr><td>\( Z \)</td><td>1</td><td>\( - 3 \)</td><td>\( - 1 \)</td><td>0</td... | No |
Maximize \( Z = x + y \) subject to: \[ x + {3y} \leq 9 \] \[ {2x} + y \leq 8 \] \[ x, y \geq 0 \] | \[ \begin{array}{l} \lbrack > \text{ with (simplex): } \\ \left\lbrack { > \text{ Objective } \mathrel{\text{:=}} x + y : }\right. \\ \left\lbrack { > \text{ Constraints } \mathrel{\text{:=}} \{ x + {3y} \leq 9,{2x} + y \leq 8\} : }\right. \\ \left\lbrack { > \text{ Soln } \mathrel{\text{:=}} \text{ maximize (Objective... | Yes |
Solve the LP from Example 4.11 using LPSolve. | Method 2. We use the objective function and constraints defined in Method 1. First load the Optimization package. with(Optimization) : \( \lceil > \) Soln2 \( \mathrel{\text{:=}} \) LPSolve \( ( \) Objective, Constraints, assume \( = \) nonnegative, 'maximize'); \[ \text{Soln2 :=}\left\lbrack {5.,\left\lbrack {x = 3., ... | Yes |
Solve the LP from Example 4.11 using matrices in LPSolve. | \[ \left\lbrack \begin{matrix} > \text{ Soln } \mathrel{\text{:=}} \text{ LPSolve }(c,\left\lbrack {A, b}\right\rbrack ,\text{ assume } = \text{ nonnegative, maximize } \\ \text{ Soln } \mathrel{\text{:=}} \left\lbrack {5.,\left\lbrack \begin{matrix} 3. \\ {2.00000000000000} \end{matrix}\right\rbrack }\right\rbrack \en... | Yes |
Maximize profit \( Z = {150}{x}_{1} + {130}{x}_{2} \) subject to: \[ 2{x}_{1} + 3{x}_{2} \leq {1500}\;\text{ (assembly time) } \] \[ 4{x}_{1} + 3{x}_{2} \leq {1600}\;\text{ (installation time) } \] \[ {x}_{1} \geq 0,{x}_{2} \geq 0 \] | \[ \left\lbrack {{68166.6666666667},\left\lbrack {{x}_{1} = {49.9999999999998},{x}_{2} = {466.666666666667}}\right\rbrack }\right\rbrack \] The output from LPSolve shows that our LP’s solution is \( Z = {68166.66} \) when \( {x}_{1} = {49.9999} \) and \( {x}_{2} = {466.6666} \). Our solution is not an integer solution,... | Yes |
The Emergency Service Coordinator (ESC) for a county is interested in locating the county's three ambulances to maximize the number of residents that can be reached within 8 minutes in emergency situations. The county is divided into 6 zones; the average times required to travel from one region to the next under semi-p... | Solution: We assume that due to the nature of the problem, a facility location problem, we should decide to employ integer programming to solve the problem.\n\nDecision Variables:\n\n\[ \n{x}_{i} = \left\{ \begin{array}{ll} 1 & \text{ if ambulance is located on Zone }i \\ 0 & \text{ if not } \end{array}\right. \]\n\n\[... | Yes |
Consider a model whose requirement is to find the route from Node 1 to Node 10 that minimizes the probability of a vehicle accident. A primary concern is the I-95 and I-20 corridor where both interstate highways meet and converge in Florence, SC. | To simplify the problem using technology, we transform the model to maximize the probability of not having an accident after using LPSolve. If \( {p}_{i, j} \) is the probability of an accident on the route from Node \( i \) to Node \( j \), then \( \left( {1 - {p}_{i, j}}\right) \) is the probability of not having an ... | Yes |
Memory Chip Problem with Mixed Integer Solution. Let's assume that we must modify Constraint 1 in this problem for the assembly time of Memory Chip B from 4 hours to 3 hours. Further, we assume we only want to have \( {x}_{1} \) as an integer, but will not require \( {x}_{2} \) to be an integer. Recall the LP:\n\nMaxim... | \n\( \lceil > \) with(Optimization) :\n\n\( \lceil > \) Objective \( \mathrel{\text{:=}} {150} \cdot {x}_{1} + {130} \cdot {x}_{2}; \)\n\n\( \lbrack > \) Constraints \( \mathrel{\text{:=}} \left\{ {2 \cdot {x}_{1} + 3 \cdot {x}_{2} \leq {1500},4 \cdot {x}_{1} + 3 \cdot {x}_{2} \leq {1600}}\right\} \) ;\n\n\( \lceil > \... | Yes |
Consider snook fishing where we want to construct a simple model to predict the weight of a fish that we catch as function of its length. Assume we have developed a proportionality model, \( W = k{L}^{3} \). We have a (modified) data set from the Florida Fish and Wildlife Conservation Commission giving snook length and... | The least-squares estimate of the proportionality constant in this model is \( k = {0.0003985} \). Thus, the model is \( W = {0.0003985}{L}^{3} \). The graph of the least-squares fit with the original data above shows that the model does capture the trend of the data. | Yes |
Given certain existing conditions, we decide to fit a cubic model to the data of Table 5.5 where \( T \) represents years and \( P \) represents the size of the population. | \[ \left\lbrack \begin{aligned} > {xdata} & \mathrel{\text{:=}} \left\lbrack {7,{14},{21},{28},{35},{42}}\right\rbrack : \\ {ydata} & \mathrel{\text{:=}} \left\lbrack {{125},{275},{800},{1200},{1700},{1650}}\right\rbrack : \end{aligned}\right. \] \[ \lceil > \text{Fit}\left( {a{x}^{3} + b{x}^{2} + {cx} + d, x\text{ dat... | Yes |
Consider the following 10 data elements: \( {50},{54},{59},{63},{65},{68},{69},{72},{90},{90} \) . The mean is | \[ \bar{x} = \frac{1}{10} \cdot \mathop{\sum }\limits_{{i = 1}}^{{10}}{x}_{i} = {68} \] The variance is found by subtracting the mean, 68 , from each point, squaring the result, adding, and then dividing the total by \( 9 = n - 1 \) . Sample deviation is the square root of the sample variance. \[ {S}^{2} = \frac{1}{9} ... | Yes |
The mean is | \[ \bar{x} = \frac{1}{7} \cdot \left( {{1792} + {1666} + {1362} + {1614} + {1460} + {1867} + {1439}}\right) = \frac{11200}{7} = {1600} \]\n\nTo see clearly the nature of the variance, start with a table of the deviations of the observations from the mean.\n\nTABLE 6.4: Metabolic Rates Deviation Table\n\n<table><thead><... | Yes |
Example 6.3. Descriptive Statistics from Maple. | \[ \lbrack > \text{with(Statistics):} \] \[ \lceil > \text{weights} \mathrel{\text{:=}} \lbrack {100},{190},{168},{112},{125},{130},{175},{170},{290},{130},{176},{125},{106},{170}\text{,} \] \[ {143.3},{145},{155},{150},{156},{189}\rbrack : \] | "No" |
If a customer buys 20 light bulbs, what is the probability that all work? | \( \begin{array}{l} \lbrack > \text{ Binomial } \mathrel{\text{:=}} \left( {x, n, p}\right) \rightarrow \text{ binomial }\left( {n, x}\right) \cdot {p}^{x} \cdot {\left( 1 - p\right) }^{n - x}; \\ \lbrack > \text{ Binomial }\left( {{20},{20},1 - {0.02}}\right) ; \\ {0.6676079718} \end{array} \) | Yes |
Example 6.6. Series System.\n\nA series system is one that performs well as long as every item or component is performing well. Consider a video game with four components: the game, the controller, the CPU (such as a PlayStation), and the television as displayed in Figure 6.5.\n\n![5a517299-4058-43b3-a902-0e3fda4c2abb_... | \[ {R}_{s}\left( t\right) = {R}_{1}\left( t\right) \cdot {R}_{2}\left( t\right) \cdot {R}_{3}\left( t\right) \cdot {R}_{4}\left( t\right) = \left( {0.90}\right) \left( {0.95}\right) \left( {0.96}\right) \left( {0.99}\right) = {0.812592} \] | Yes |
A parallel system is one that performs as long as a single one of its components remains operational. Your home communications system of cordless telephone and conventional telephone is a system like Figure 6.6. Note that there are two separate and independent routes to transverse the network (input to output); either ... | We apply the definition of the system reliability for parallel components\n\n\[ \n{R}_{s}\left( t\right) = {R}_{1}\left( t\right) + {R}_{2}\left( t\right) - {R}_{1}\left( t\right) \cdot {R}_{2}\left( t\right) = {0.95} + {0.96} - \left( {0.95}\right) \left( {0.96}\right) = {0.998} \n\]\n\nNote that in parallel relations... | Yes |
We want to know the reliability of the whole system for 6 months. | Subsystem A is a standby redundant system, so we will use the Poisson model (pg. 256). We let \( X = \) the number of components which fail in one year. Since 6 months is 0.5 years, we seek \( {R}_{A}\left( {0.5}\right) = P\left( {x < 3}\right) \), where \( X \) follows a Poisson distribution with parameter \( \lambda ... | Yes |
Compute the area under the nonnegative curve \( y = {x}^{3} \) for \( 0 \leq x \leq 2 \) . | Now we are ready to approximate the area by using Monte Carlo simulation. Remember, the simulation only approximates the solution. Increase the number of trials to get closer to the value. The results are in Table 7.3. Randomness is introduced into the procedure with the Monte Carlo simulation area algorithm. In our Ma... | No |
Compute the area under the nonnegative curve \( f\left( x\right) = {e}^{{x}^{2}} \cdot \cos \left( x\right) \cdot \sqrt{x} \) for \( 0 \leq x \leq {1.4} \). | Since this integral has no closed form, we use Maple's integrator to find\n\n\[ \left\lbrack \begin{aligned} > \operatorname{int}\left( {{e}^{{x}^{2}} \cdot \cos \left( x\right) \cdot \sqrt{x}, x = 0.{.1.4}}\right) ; & \\ {1.449827600} & \end{aligned}\right. \]\n\nThe percent error for this run with \( N = {2000} \) is... | Yes |
Compute the volume in the first octant given by \( {x}^{2} + {y}^{2} + {z}^{2} \leq 1 \) . | The function defining our volume is \( f\left( {x, y}\right) = \sqrt{1 - {x}^{2} - {y}^{2}} \) . The interval \( \left\lbrack {0,1}\right\rbrack \) works for all three variables \( x, y \), and \( z \) . As the volume is \( 1/8 \) of a sphere of radius 1, the exact volume is \( \pi /6 \) . | Yes |
Algorithm: Simulating Flipping a Fair Coin\n\nINPUTS:\n\n\( N \) total number of random flips to generate\n\nOutputs:\n\n\( A \) estimated probabilities for heads and tails\n\nSTEPS:\n\nStep 1. Set the counters \( H \) and \( T \) to 0\n\nStep 2. For \( i \) from 1 to \( N \) do Steps 3-4\n\nStep 3. Generate a random n... | Implement the coin flipping algorithm in Maple.\n\n> \( F{lip} \mathrel{\text{:=}} \mathbf{{proc}}\left( {n : : {posint}}\right) \)\n\nlocal \( r, H, T, i, x \) ;\n\n\( r \mathrel{\text{:=}} \operatorname{rand}\left( {0.{.1.0}}\right) \)\n\n\( H, T \mathrel{\text{:=}} 0,0 \)\n\nfor \( i \) to \( n \) do\n\n\( x \mathre... | Yes |
Rolling a fair die adds the additional process of multiple assignments ( 6 for a six-sided die). We can modify the coin flipping algorithm accordingly quite easily. The probability will be (the number of occurrences of each number) / (the total number of trials). | Algorithm: Simulating Rolling a Fair Die\n\nINPUTS:\n\n\\( N\\; \\) total number of random rolls to generate\n\nOutputs:\n\nDie list of estimated probabilities for each possibility\n\nSTEPS:\n\nStep 1. Set the counters in Die to 0\n\nStep 2. For \\( i \\) from 1 to \\( N \\) do Steps 3-4\n\nStep 3. Generate a random nu... | Yes |
Lemma 2. If \( \left( {\alpha }^{\prime }\right) \prec \left( \alpha \right) \), but \( \left( {\alpha }^{\prime }\right) \) is not identical with \( \left( \alpha \right) \), then \( \left( {\alpha }^{\prime }\right) \) can be derived from \( \left( \alpha \right) \) by the successive application of a finite number of... | Proof of Lemma 2. We suppose that the condition (2.18.2) is satisfied, and call the number of the differences \( {\alpha }_{\nu } - {\alpha }_{\nu }{}^{\prime } \) which are not zero the discrepancy of \( \left( \alpha \right) \) and \( \left( {\alpha }^{\prime }\right) \) ; if the discrepancy is zero the sets are iden... | Yes |
Theorem 1 (General solution theorem) Consider the linear ODE (2.27) with \( p\left( t\right) \) and \( q\left( t\right) \) continuous on some interval containing \( {t}_{0} \) . Let\n\n\[ P\left( t\right) = {\int }_{{t}_{0}}^{t}p\left( r\right) {dr},\;R\left( t\right) = {\int }_{{t}_{0}}^{t}{e}^{P\left( s\right) }q\lef... | Note that one important consequence of this theorem is that every solution of the ODE exists at least in the interval in which \( p\left( t\right) \) and \( q\left( t\right) \) are continuous. Another important consequence is that no two solutions corresponding to different values of \( K \) can ever intersect.\n\nWhy?... | Yes |
Theorem 2 (Solution of initial value problem (IVP)) Let \( p\left( t\right) \) and \( q\left( t\right) \) be continuous in some interval containing \( {t}_{0} \), and let \( {y}_{0} \) be any constant. Then the IVP\n\n\[ \n{y}^{\prime } + p\left( t\right) y = q\left( t\right) ,\;y\left( {t}_{0}\right) = {y}_{0} \n\]\n\... | Writing \( {P}_{0}\left( t\right) = {\int }_{{t}_{0}}^{t}p\left( r\right) {dr} \), this solution is given by\n\n\[ \n\underset{\begin{matrix} \text{ total } \\ \text{ response } \end{matrix}}{\underbrace{y\left( t\right) }} = \underset{\begin{matrix} \text{ response to } \\ \text{ initial data } \end{matrix}}{\underbra... | Yes |
Theorem 4 (Linear independence of eigenvectors.) Suppose \( {v}_{1},\ldots ,{v}_{k} \) are eigenvectors that correspond to \( k \) distinct eigenvalues \( {\lambda }_{1},\ldots ,{\lambda }_{k} \) of some matrix \( A \) . Then the set of eigenvectors \( \left\{ {{\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{k}}\right\} \) is l... | Proof Suppose \( \left\{ {{\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{k}}\right\} \) is not linearly independent (i.e., linearly dependent), then there exists some \( p \) such that \( {\mathbf{v}}_{p} \) is a linear combination of the eigenvectors \( \left\{ {{\mathbf{v}}_{1},\ldots ,{\mathbf{v}}_{p - 1}}\right\} \), with ... | Yes |
Theorem 5 (Matrix invertibility.) Let \( \mathbf{A} \) be an \( n \times n \) matrix (a square matrix). Then the following are equivalent; that is, if any one of them is true, then so are the rest.\n\n(a) \( \mathbf{A} \) is invertible\n\n(b) The columns of \( \mathbf{A} \) are linearly independent\n\n(c) The columns f... | Proof. The details are omitted, but for those who are interested, see any text on linear algebra, for example Lay (1994). | No |
Evaluate the integral \( {\int }_{V}{N}_{i}\frac{\partial {N}_{j}}{\partial x}{dxdydz} \) for the linear hexahedron. | \[ \left\lbrack {{\int }_{V}{N}_{i}\frac{\partial {N}_{j}}{\partial x}{dxdydz}}\right\rbrack = {\iiint }_{-1 - 1 - 1}^{1}\left\lbrack {{N}_{i}\left( {\xi ,\eta ,\zeta }\right) \frac{\partial {N}_{j}}{\partial \xi }}\right\rbrack \left| \mathbf{J}\right| {d\xi d\eta d\zeta } \]\n\[ = \frac{V}{72a}\left\lbrack \begin{mat... | Yes |
Proposition 5.1 (de Bruijn formula) Denote as above \( f\left( {x, n}\right) = \) \( \sin \left( {\sin \left( {\ldots \left( {\sin x}\right) }\right) }\right) \), n times, \( n \in {\mathbb{N}}^{ * }, x \in \rbrack - \pi ,\pi \lbrack \) . Then,\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt{n/3}f\left( {x, ... | Proof Obviously \( f\left( {x, n + 1}\right) = \sin f\left( {x, n}\right), n \in \mathbb{N} \) and \( x \in \mathbb{R} \) . Choose \( x \in \rbrack 0,\pi \lbrack \) . Then \( f\left( {x,1}\right) \in \rbrack 0,1\rbrack \subset \rbrack 0,\pi /2\rbrack ,0 \leq f\left( {n + 1, x}\right) < f\left( {n, x}\right) \), and \( ... | Yes |
Lemma 9.1 The tangential velocity function \( u \) is of class \( {C}^{1} \), positive and increasing on the interval \( \left\lbrack {0,{t}_{f}}\right\rbrack \) . | Proof It is enough to note that \( {u}^{\prime } \) in (9.25) is continuous and positive. | Yes |
Lemma 9.3 Assume that \( t \geq 0 \) . The following two integral equalities hold\n\n\[ \n{\int }_{0}^{t}A\left( s\right) \mathrm{d}s = {tA}\left( t\right) - B\left( t\right) \text{ and }{\int }_{0}^{t}B\left( s\right) \mathrm{d}s = {tB}\left( t\right) - C\left( t\right) .\n\] | Proof The two integral equalities are obtained by integration by parts; see [53, Chap. 6]. | No |
Theorem 9.2 From (9.25) with (9.26) and (9.27) by integration, we get the equations of velocities and states. | \[ u\left( t\right) = u\left( 0\right) - {V}_{e}{c}_{1}B + {V}_{e}{c}_{4}A, \] \[ v\left( t\right) = v\left( 0\right) - {V}_{e}{c}_{2}B + {V}_{e}{c}_{5}A - {g}_{m}t \] \[ w\left( t\right) = w\left( 0\right) - {V}_{e}{c}_{3}B + {V}_{e}{c}_{6}A, \] (9.28) \[ x\left( t\right) = x\left( 0\right) + {tu}\left( t\right) + {V}... | Yes |
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