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[{\int }_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1 - {x}^{2}}} = {\int }_{-1}^{0}\frac{\mathrm{d}x}{\sqrt{1 - {x}^{2}}} + {\int }_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1 - {x}^{2}}}] | [= {\left. \arcsin x\right| }_{-1}^{0} + {\left. \arcsin x\right| }_{0}^{1} = {\left. \arcsin x\right| }_{-1}^{1} = \pi \text{.}] | Yes |
Example 14. The integral\n\n\[ \n{\int }_{-\infty }^{+\infty }{\mathrm{e}}^{-{x}^{2}}\mathrm{\;d}x \n\]\n\nis called the Euler-Poisson integral, and sometimes the Gaussian integral. It obviously converges in the sense given above. | It will be shown later that its value is \( \sqrt{\pi } \) . | No |
The integral\n\n\[ \n{\int }_{0}^{+\infty }\frac{\sin x}{{x}^{\alpha }}\mathrm{d}x \n\]\n\nconverges if each of the integrals\n\n\[ \n{\int }_{0}^{1}\frac{\sin x}{{x}^{\alpha }}\mathrm{d}x\text{ and }{\int }_{1}^{+\infty }\frac{\sin x}{{x}^{\alpha }}\mathrm{d}x \n\]\n\nconverges. | The first of these integrals converges if \( \alpha < 2 \), since\n\n\[ \n\frac{\sin x}{{x}^{\alpha }} \sim \frac{1}{{x}^{\alpha - 1}} \n\]\n\nas \( x \rightarrow + 0 \) . The second integral converges if \( \alpha > 0 \), as one can verify directly through an integration by parts similar to the one shown in Example 12... | Yes |
[ \n\\mathrm{{PV}}{\\int }_{-\\infty }^{+\\infty }x\\mathrm{\\;d}x = 0 \n] | Finally, if there are several (finitely many) singularities of one kind or another on the interval of integration, at interior points or endpoints, then the nonsingular points of the interval are divided into a finite number of such intervals, each containing only one singularity, and the integral is computed as the su... | No |
A ball \( B\left( {a;r}\right) \) is an open set in \( {\mathbb{R}}^{m} \). | Indeed, if \( x \in B\left( {a;r}\right) \), that is, \( d\left( {a, x}\right) < r \), then for \( 0 < \delta < r - d\left( {a, x}\right) \), we have \( B\left( {x;\delta }\right) \subset B\left( {a;r}\right) \), since\n\n\[ \left( {\xi \in B\left( {x;\delta }\right) }\right) \Rightarrow \left( {d\left( {x,\xi }\right)... | Yes |
A set \( G = \left\{ {x \in {\mathbb{R}}^{m} \mid d\left( {a, x}\right) > r}\right\} \), that is, the set of points whose distance from a fixed point \( a \in {\mathbb{R}}^{m} \) is larger than \( r \), is open. | This fact is easy to verify, as in Example 3, using the triangle inequality for the metric. | No |
The set \( \bar{B}\left( {a;r}\right) = \left\{ {x \in {\mathbb{R}}^{m} \mid d\left( {a, x}\right) \leq r}\right\}, r \geq 0 \), that is, the set of points whose distance from a fixed point \( a \in {\mathbb{R}}^{m} \) is at most \( r \), is closed. | as follows from Definition 3 and Example 4. | No |
Proposition 1. a) The union \( \mathop{\bigcup }\limits_{{\alpha \in A}}{G}_{\alpha } \) of the sets of any system \( \left\{ {{G}_{\alpha },\alpha \in A}\right\} \) of open sets in \( {\mathbb{R}}^{m} \) is an open set in \( {\mathbb{R}}^{m} \) . | Proof. a) If \( x \in \mathop{\bigcup }\limits_{{\alpha \in A}}{G}_{\alpha } \), then there exists \( {\alpha }_{0} \in A \) such that \( x \in {G}_{{\alpha }_{0}} \), and consequently there is a \( \delta \) -neighborhood \( B\left( {x;\delta }\right) \) of \( x \) such that \( B\left( {x;\delta }\right) \subset {G}_{... | Yes |
The set \( S\left( {a;r}\right) = \left\{ {x \in {\mathbb{R}}^{m} \mid d\left( {a, x}\right) = r}\right\}, r \geq 0 \), is called the sphere of radius \( r \) with center \( a \in {\mathbb{R}}^{m} \). | The complement of \( S\left( {a;r}\right) \) in \( {\mathbb{R}}^{m} \), by Examples 3 and 4, is the union of open sets. Hence by the proposition just proved it is open, and the sphere \( S\left( {a;r}\right) \) is closed in \( {\mathbb{R}}^{m} \). | No |
Proposition 2. \( \left( F\right. \) is closed in \( \left. {\mathbb{R}}^{m}\right) \Leftrightarrow \left( F\right. = \bar{F} \) in \( \left. {\mathbb{R}}^{m}\right) \) . | Proof. Let \( F \) be closed in \( {\mathbb{R}}^{m}, x \in {\mathbb{R}}^{m} \), and \( x \notin F \) . Then the open set \( G = {\mathbb{R}}^{m} \smallsetminus F \) is a neighborhood of \( x \) that contains no points of \( F \) . Thus we have shown that if \( x \notin F \), then \( x \) is not a limit point of \( F \)... | Yes |
We shall show that \( I \) is compact in \( {\mathbb{R}}^{m} \) . | Proof. Assume that from some open covering of \( I \) one cannot extract a finite covering. Bisecting each of the coordinate closed intervals \( {I}^{i} = \left\{ {{x}^{i} \in \mathbb{R} : {a}^{i} \leq }\right. \) \( \left. {{x}^{i} \leq {b}^{i}}\right\} ,\left( {i = 1,\ldots, m}\right) \), we break the interval \( I \... | Yes |
Proposition 3. If \( K \) is a compact set in \( {\mathbb{R}}^{m} \), then\na) \( K \) is closed in \( {\mathbb{R}}^{m} \) ;\nb) any closed subset of \( {\mathbb{R}}^{m} \) contained in \( K \) is itself compact. | Proof. a) We shall show that any point \( a \in {\mathbb{R}}^{m} \) that is a limit point of \( K \) must belong to \( K \) . Suppose \( a \notin K \) . For each point \( x \in K \) we construct a neighborhood \( G\left( x\right) \) such that \( a \) has a neighborhood disjoint from \( G\left( x\right) \) . The set \( ... | Yes |
Proposition 4. If \( K \) is a compact set in \( {\mathbb{R}}^{m} \), then \( K \) is a bounded subset of \( {\mathbb{R}}^{m} \) . | Proof. Take an arbitrary point \( a \in {\mathbb{R}}^{m} \) and consider the sequence of open balls \( \{ B\left( {a;n}\right) \} ,\left( {n = 1,2,\ldots \text{,}}\right) . \) They form an open covering of \( {\mathbb{R}}^{m} \) and consequently also of \( K \) . If \( K \) were not bounded, it would be impossible to s... | Yes |
Proposition 5. The set \( K \subset {\mathbb{R}}^{m} \) is compact if and only if \( K \) is closed and bounded in \( {\mathbb{R}}^{m} \) . | Proof. The necessity of these conditions was proved in Propositions 3 and 4.\n\nLet us verify that the conditions are sufficient. Since \( K \) is a bounded set, there exists an \( m \) -dimensional interval \( I \) containing \( K \) . As was shown in Example 13, \( I \) is compact in \( {\mathbb{R}}^{m} \) . But if \... | Yes |
Theorem 1. Let \( X \) be a set and \( \mathcal{B} \) a base in \( X \) . A function \( f : X \rightarrow {\mathbb{R}}^{n} \) has a limit over the base \( \mathcal{B} \) if and only if for every \( \varepsilon > 0 \) there exists an element \( B \in \mathcal{B} \) of the base on which the oscillation of the function is... | Thus, \[ \exists \mathop{\lim }\limits_{\mathcal{B}}f\left( x\right) \Leftrightarrow \forall \varepsilon > 0\exists B \in \mathcal{B}\left( {\omega \left( {f;B}\right) < \varepsilon }\right) . \] The proof of Theorem 1 is a verbatim repetition of the proof of the Cauchy criterion for numerical functions (Theorem 4 in S... | Yes |
Theorem 2. Let \( Y \) be a set, \( {\mathcal{B}}_{Y} \) a base in \( Y \), and \( g : Y \rightarrow {\mathbb{R}}^{n} \) a mapping having a limit over the base \( {\mathcal{B}}_{Y} \). Let \( X \) be a set, \( {\mathcal{B}}_{X} \) a base in \( X \), and \( f : X \rightarrow Y \) a mapping of \( X \) into \( Y \) such t... | The proof of Theorem 2 can be carried out either by repeating the proof of Theorem 5 of Sect. 3.2, replacing \( \mathbb{R} \) by \( {\mathbb{R}}^{n} \), or by invoking that theorem and using relation (7.4). | Yes |
Let \( x \mapsto {\pi }^{i}\left( x\right) \) be the mapping \( {\pi }^{i} : {\mathbb{R}}^{m} \rightarrow \mathbb{R} \) assigning to each \( x = \left( {{x}^{1},\ldots ,{x}^{m}}\right) \) in \( {\mathbb{R}}^{m} \) its \( i \) th coordinate \( {x}^{i} \) . Thus\n\n\[ \n{\pi }^{i}\left( x\right) = {x}^{i}.\n\] | If \( a = \left( {{a}^{1},\ldots ,{a}^{m}}\right) \), then obviously\n\n\[ \n{\pi }^{i}\left( x\right) \rightarrow {a}^{i}\text{ as }x \rightarrow a.\n\]\n\nThe function \( x \mapsto {\pi }^{i}\left( x\right) \) does not tend to any finite value nor to infinity as \( x \rightarrow \infty \) if \( m > 1 \) . | Yes |
Let the function \( f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) be defined at the point \( \left( {x, y}\right) \in {\mathbb{R}}^{2} \) as follows: \[ f\left( {x, y}\right) = \left\{ \begin{matrix} \frac{xy}{{x}^{2} + {y}^{2}} & \text{if }{x}^{2} + {y}^{2} \neq 0, \\ 0, & \text{if }{x}^{2} + {y}^{2} = 0. \end{matrix... | Then \( f\left( {0, y}\right) = f\left( {x,0}\right) = 0 \), while \( f\left( {x, x}\right) = \frac{1}{2} \) for \( x \neq 0 \) . Hence this function has no limit as \( \left( {x, y}\right) \rightarrow \left( {0,0}\right) \) . | Yes |
For the function\n\n\[ f\left( {x, y}\right) = \left\{ \begin{matrix} \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}},\text{ if }{x}^{2} + {y}^{2} \neq 0, \\ 0,\;\text{ if }{x}^{2} + {y}^{2} = 0, \end{matrix}\right. \]\n\nwe have | \n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\mathop{\lim }\limits_{{y \rightarrow 0}}f\left( {x, y}\right) }\right) = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( \frac{{x}^{2}}{{x}^{2}}\right) = 1 \]\n\n\[ \mathop{\lim }\limits_{{y \rightarrow 0}}\left( {\mathop{\lim }\limits_{{x \rightarrow 0}}f\left( {x... | Yes |
The functions \( \left( {{x}^{1},\ldots ,{x}^{m}}\right) \overset{{\pi }^{i}}{ \mapsto }{x}^{i}\left( {i = 1,\ldots, m}\right) \), mapping \( {\mathbb{R}}^{m} \) onto \( \mathbb{R} \) (projections) are obviously continuous at each point \( a \) = \( \left( {{a}^{1},\ldots ,{a}^{m}}\right) \in {\mathbb{R}}^{m} \) | since \( \mathop{\lim }\limits_{{x \rightarrow a}}{\pi }^{i}\left( x\right) = {a}^{i} = {\pi }^{i}\left( a\right) \) | Yes |
Any function \( x \mapsto f\left( x\right) \) defined on \( \mathbb{R} \), for example \( x \mapsto \sin x \), can also be regarded as a function \( \left( {x, y}\right) \overset{F}{ \mapsto }f\left( x\right) \) defined, say, on \( {\mathbb{R}}^{2} \) . In that case, if \( f \) was continuous as a function on \( \mathb... | This can be verified either directly from the definition of continuity or by remarking that the function \( F \) is the composition \( \left( {f \circ {\pi }^{1}}\right) \left( {x, y}\right) \) of continuous functions. | Yes |
The function \( f\left( {x, y}\right) \) of Example 2 is continuous at any point of the space \( {\mathbb{R}}^{2} \) except \( \left( {0,0}\right) \). | We remark that, despite the discontinuity of \( f\left( {x, y}\right) \) at this point, the function is continuous in either of its two variables for each fixed value of the other variable. | No |
If a function \( f : E \rightarrow {\mathbb{R}}^{n} \) is continuous on the set \( E \) and \( \widetilde{E} \) is a subset of \( E \), then the restriction \( {\left. f\right| }_{\widetilde{E}} \) of \( f \) to this subset is continuous on \( \widetilde{E} \) | as follows immediately from the definition of continuity of a function at a point. | No |
An open ball \( B\left( {a;r}\right), r > 0 \), in \( {\mathbb{R}}^{m} \) is a domain. | We already know that \( B\left( {a;r}\right) \) is open in \( {\mathbb{R}}^{m} \) . Let us verify that the ball is connected. Let \( {x}_{0} = \left( {{x}_{0}^{1}\ldots ,{x}_{0}^{m}}\right) \) and \( {x}_{1} = \left( {{x}_{1}^{1},\ldots ,{x}_{1}^{m}}\right) \) be two points of the ball. The path defined by the function... | Yes |
The sphere \( S\left( {0;r}\right) \) defined in \( {\mathbb{R}}^{m} \) by the equation\n\n\[ \n{\left( {x}^{1}\right) }^{2} + \cdots + {\left( {x}^{m}\right) }^{2} = {r}^{2} \n\]\n\nis a compact set. | Indeed, it follows from the continuity of the function\n\n\[ \n\left( {{x}^{1},\ldots ,{x}^{m}}\right) \mapsto {\left( {x}^{1}\right) }^{2} + \cdots + {\left( {x}^{m}\right) }^{2} \n\]\n\nthat the sphere is closed, and from the fact that \( \left| {x}^{i}\right| \leq r\left( {i = 1,\ldots, m}\right) \) on the sphere th... | Yes |
The open set \( {\mathbb{R}}^{m} \smallsetminus S\left( {0;r}\right) \) for \( r > 0 \) is not a domain, since it is not connected. | Indeed, if \( \Gamma : I \rightarrow {\mathbb{R}}^{m} \) is a path one end of which is at the point \( {x}_{0} = \) \( \left( {0,\ldots ,0}\right) \) and the other at some point \( {x}_{1} = \left( {{x}_{1}^{1},\ldots ,{x}_{1}^{m}}\right) \) such that \( {\left( {x}_{1}^{1}\right) }^{2} + \cdots + \) \( {\left( {x}_{1}... | Yes |
Proposition 1. A mapping \( f : E \rightarrow {\mathbb{R}}^{n} \) of a set \( E \subset {\mathbb{R}}^{m} \) is differentiable at a point \( x \in E \) that is a limit point of \( E \) if and only if the functions \( {f}^{i} : E \rightarrow \mathbb{R} \) \( \left( {i = 1,\ldots, n}\right) \) that define the coordinate r... | Since relations (8.21) and (8.22) are equivalent, to find the differential \( L\left( x\right) \) of a mapping \( f : E \rightarrow {\mathbb{R}}^{n} \) it suffices to learn how to find the differentials \( {L}^{i}\left( x\right) \) of its coordinate functions \( {f}^{i} : E \rightarrow \mathbb{R} \) . | No |
If \( f\left( {u, v}\right) = {u}^{3} + {v}^{2}\sin u \), then | \[ {\partial }_{1}f\left( {u, v}\right) = \frac{\partial f}{\partial u}\left( {u, v}\right) = 3{u}^{2} + {v}^{2}\cos u \] \[ {\partial }_{2}f\left( {u, v}\right) = \frac{\partial f}{\partial v}\left( {u, v}\right) = {2v}\sin u. \] | Yes |
If \( f\left( {x, y, z}\right) = \arctan \left( {x{y}^{2}}\right) + {\mathrm{e}}^{z} \), then | \[
{\partial }_{1}f\left( {x, y, z}\right) = \frac{\partial f}{\partial x}\left( {x, y, z}\right) = \frac{{y}^{2}}{1 + {x}^{2}{y}^{4}},
\]
\[
{\partial }_{2}f\left( {x, y, z}\right) = \frac{\partial f}{\partial y}\left( {x, y, z}\right) = \frac{2xy}{1 + {x}^{2}{y}^{4}},
\]
\[
{\partial }_{3}f\left( {x, y, z}\right) =... | Yes |
Proposition 2. If a function \( f : E \rightarrow {\mathbb{R}}^{n} \) defined on a set \( E \subset {\mathbb{R}}^{m} \) is differentiable at an interior point \( x \in E \) of that set, then the function has a partial derivative at that point with respect to each variable, and the differential of the function is unique... | Using the convention of summation on an index that appears as both a subscript and a superscript, we can write formula (8.27) succinctly:\n\n\[ \mathrm{d}f\left( x\right) h = {\partial }_{i}f\left( x\right) {h}^{i}. \]\n\n(8.28) | Yes |
If we had known (as we soon will know) that the function \( f\left( {x, y, z}\right) \) considered in Example 2 is differentiable at the point \( \left( {0,1,0}\right) \), we could have written immediately | \[ \frac{{df}\left( {0,1,0}\right) }{h} = \frac{1 \cdot {h}^{1} + 0 \cdot {h}^{2} + 1 \cdot {h}^{3} = {h}^{1} + {h}^{3}}{} \] and accordingly \[ f\left( {{h}^{1},1 + {h}^{2},{h}^{3}}\right) - f\left( {0,1,0}\right) = \mathrm{d}f\left( {0,1,0}\right) h + o\left( h\right) \] or \[ \arctan \left( {{h}^{1}{\left( 1 + {h}^{... | No |
For the function \( x = \left( {{x}^{1},\ldots ,{x}^{m}}\right) \overset{{\pi }^{i}}{ \mapsto }{x}^{i} \), which assigns to the point \( x \in {\mathbb{R}}^{m} \) its \( i \) th coordinate, we have | \[ \Delta {\pi }^{i}\left( {x;h}\right) = \left( {{x}^{i} + {h}^{i}}\right) - {x}^{i} = {h}^{i}, \] that is, the increment of this function is itself a linear function in \( h:h\overset{{\pi }^{i}}{ \mapsto }{h}^{i}. \) Thus, \( \Delta {\pi }^{i}\left( {x;h}\right) = \mathrm{d}{\pi }^{i}\left( x\right) h \), and the ma... | Yes |
The function\n\n\[ f\left( {{x}^{1},{x}^{2}}\right) = \left\{ \begin{array}{l} 0,\text{ if }{x}^{1}{x}^{2} = 0, \\ 1,\text{ if }{x}^{1}{x}^{2} \neq 0, \end{array}\right. \]\n\nequals 0 on the coordinate axes and therefore has both partial derivatives at the point \( \left( {0,0}\right) \). | \[ {\partial }_{1}f\left( {0,0}\right) = \mathop{\lim }\limits_{{{h}^{1} \rightarrow 0}}\frac{f\left( {{h}^{1},0}\right) - f\left( {0,0}\right) }{{h}^{1}} = \mathop{\lim }\limits_{{{h}^{1} \rightarrow 0}}\frac{0 - 0}{{h}^{1}} = 0 \]\n\n\[ {\partial }_{2}f\left( {0,0}\right) = \mathop{\lim }\limits_{{{h}^{2} \rightarrow... | Yes |
Theorem 1. If the mappings \( {f}_{1} : E \rightarrow {\mathbb{R}}^{n} \) and \( {f}_{2} : E \rightarrow {\mathbb{R}}^{n} \), defined on a set \( E \subset {\mathbb{R}}^{m} \), are differentiable at a point \( x \in E \), then a linear combination of them \( \left( {{\lambda }_{1}{f}_{1} + {\lambda }_{2}{f}_{2}}\right)... | Proof.\n\n\[ \n\left( {{\lambda }_{1}{f}_{1} + {\lambda }_{2}{f}_{2}}\right) \left( {x + h}\right) - \left( {{\lambda }_{1}{f}_{2} + {\lambda }_{2}{f}_{2}}\right) \left( x\right) = \n\]\n\n\[ \n= \left( {{\lambda }_{1}{f}_{1}\left( {x + h}\right) + {\lambda }_{2}{f}_{2}\left( {x + h}\right) }\right) - \left( {{\lambda ... | Yes |
Theorem 2. If the functions \( f : E \rightarrow \mathbb{R} \) and \( g : E \rightarrow \mathbb{R} \), defined on a set \( E \subset {\mathbb{R}}^{m} \), are differentiable at the point \( x \in E \), then a) their product is differentiable at \( x \) and\n\n\[ \n{\left( f \cdot g\right) }^{\prime }\left( x\right) = g\... | The proof of this theorem is the same as the proof of the corresponding parts of Theorem 1 in Sect. 5.2, so that we shall omit the details. | No |
Theorem 3. If the mapping \( f : X \rightarrow Y \) of a set \( X \subset {\mathbb{R}}^{m} \) into a set \( Y \subset \) \( {\mathbb{R}}^{n} \) is differentiable at a point \( x \in X \), and the mapping \( f : Y \rightarrow {\mathbb{R}}^{k} \) is differentiable at the point \( y = f\left( x\right) \in Y \), then their... | Proof. Using the differentiability of the mappings \( f \) and \( g \) at the points \( x \) and \( y = f\left( x\right) \), and also the linearity of the differential \( {g}^{\prime }\left( x\right) \), we can write\n\n\[ \left( {g \circ f}\right) \left( {x + h}\right) - \left( {g \circ f}\right) \left( x\right) = g\l... | Yes |
Theorem 4. Let \( f : U\left( x\right) \rightarrow V\left( y\right) \) be a mapping of a neighborhood \( U\left( x\right) \subset \) \( {\mathbb{R}}^{m} \) of the point \( x \) onto a neighborhood \( V\left( y\right) \subset {\mathbb{R}}^{m} \) of the point \( y = f\left( x\right) \) . Assume that \( f \) is continuous... | Proof. We use the following notation:\n\n\[f\left( x\right) = y,\;f\left( {x + h}\right) = y + t,\;t = f\left( {x + h}\right) - f\left( x\right) ,\]\n\nso that\n\n\[{f}^{-1}\left( y\right) = x,\;{f}^{-1}\left( {y + t}\right) = x + h,\;h = {f}^{-1}\left( {y + t}\right) - {f}^{-1}\left( y\right) .\]\n\nWe shall assume th... | Yes |
Theorem 1. Let \( f : G \rightarrow \mathbb{R} \) be a real-valued function defined in a region \( G \subset {\mathbb{R}}^{m} \), and let the closed line segment \( \left\lbrack {x, x + h}\right\rbrack \) with endpoints \( x \) and \( x + h \) be contained in \( G \) . If the function \( f \) is continuous at the point... | Proof. Consider the auxiliary function\n\n\[ F\left( t\right) = f\left( {x + {th}}\right) \]\n\ndefined on the closed interval \( 0 \leq t \leq 1 \) . This function satisfies all the hypotheses of Lagrange’s theorem: it is continuous on \( \left\lbrack {0,1}\right\rbrack \), being the composition of continuous mappings... | Yes |
Proposition 1. If \( f \in {C}^{\left( k\right) }\left( {G;\mathbb{R}}\right) \), the value \( {\partial }_{{i}_{1}\ldots {i}_{k}}f\left( x\right) \) of the partial derivative is independent of the order \( {i}_{1},\ldots ,{i}_{k} \) of differentiation, that is, remains the same for any permutation of the indices \( {i... | Proof. In the case \( k = 2 \) this proposition is contained in Theorem 3.\n\nLet us assume that the proposition holds up to order \( n \) inclusive. We shall show that then it also holds for order \( n + 1 \) .\n\nBut \( {\partial }_{{i}_{1}{i}_{2}\cdots {i}_{n + 1}}f\left( x\right) = {\partial }_{{i}_{1}}\left( {{\pa... | Yes |
Let \( f\left( x\right) = f\left( {{x}^{1},{x}^{2}}\right) \) be a function of class \( {C}^{\left( k\right) }\left( {G;\mathbb{R}}\right) \). Let \( h = \left( {{h}^{1},{h}^{2}}\right) \) be such that the closed interval \( \left\lbrack {x, x + h}\right\rbrack \) is contained in the domain \( G \). We shall show that ... | We have \[ {\varphi }^{\prime }\left( t\right) = {\partial }_{1}f\left( {{x}^{1} + t{h}^{1},{x}^{2} + t{h}^{2}}\right) {h}^{1} + {\partial }_{2}f\left( {{x}^{1} + t{h}^{1},{x}^{2} + t{h}^{2}}\right) {h}^{2}, \] \[ {\varphi }^{\prime \prime }\left( t\right) = {\partial }_{11}f\left( {x + {th}}\right) {h}^{1}{h}^{1} + {\... | Yes |
If \( f\left( x\right) = f\left( {{x}^{1},\ldots ,{x}^{m}}\right) \) and \( f \in {C}^{\left( k\right) }\left( {G;\mathbb{R}}\right) \), then, under the assumption that \( \left\lbrack {x, x + h}\right\rbrack \subset G \), for the function \( \varphi \left( t\right) = f\left( {x + {th}}\right) \) defined on the closed ... | \[ {\varphi }^{\left( k\right) }\left( t\right) = {h}^{{i}_{1}}\cdots {h}^{{i}_{k}}{\partial }_{{i}_{1}\cdots {i}_{k}}f\left( {x + {th}}\right) ,\] (8.58) where summation over all sets of indices \( {i}_{1},\ldots ,{i}_{k} \), each assuming all values from 1 to \( m \) inclusive, is meant on the right. We can also writ... | Yes |
Theorem 4. If the function \( f : U\\left( x\\right) \\rightarrow \\mathbb{R} \) is defined and belongs to class \( {C}^{\\left( n\\right) }\\left( {U\\left( x\\right) ;\\mathbb{R}}\\right) \) in a neighborhood \( U\\left( x\\right) \\subset {\\mathbb{R}}^{m} \) of the point \( x \\in {\\mathbb{R}}^{m} \), and the clos... | Proof. Taylor's formula follows immediately from the corresponding Taylor formula for a function of one variable. In fact, consider the auxiliary function\n\n\\[ \n\\varphi \\left( t\\right) = f\\left( {x + {th}}\\right)\n\\]\nwhich, by the hypotheses of Theorem 4, is defined on the closed interval \( 0 \\leq t \\leq 1... | Yes |
Theorem 5. Suppose a function \( f : U\left( {x}_{0}\right) \rightarrow \mathbb{R} \) defined in a neighborhood \( U\left( {x}_{0}\right) \subset {\mathbb{R}}^{m} \) of the point \( {x}_{0} = \left( {{x}_{0}^{1},\ldots ,{x}_{0}^{m}}\right) \) has partial derivatives with respect to each of the variables \( {x}^{1},\ldo... | Proof. Consider the function \( \varphi \left( {x}^{1}\right) = f\left( {{x}^{1},{x}_{0}^{2},\ldots ,{x}_{0}^{m}}\right) \) of one variable defined, according to the hypotheses of the theorem, in some neighborhood of the point \( {x}_{0}^{1} \) on the real line. At \( {x}_{0}^{1} \) the function \( \varphi \left( {x}^{... | Yes |
Theorem 6. Let \( f : U\left( {x}_{0}\right) \rightarrow \mathbb{R} \) be a function of class \( {C}^{\left( 2\right) }\left( {U\left( {x}_{0}\right) ;\mathbb{R}}\right) \) defined in a neighborhood \( U\left( {x}_{0}\right) \subset {\mathbb{R}}^{m} \) of the point \( {x}_{0} = \left( {{x}_{0}^{1},\ldots ,{x}_{0}^{m}}\... | Proof. Let \( h \neq 0 \) and \( {x}_{0} + h \in U\left( {x}_{0}\right) \) . Let us represent (8.67) in the form\n\n\[ f\left( {{x}_{0} + h}\right) - f\left( {x}_{0}\right) = \frac{1}{2!}\parallel h{\parallel }^{2}\left\lbrack {\mathop{\sum }\limits_{{i, j = 1}}^{m}\frac{{\partial }^{2}f}{\partial {x}^{i}\partial {x}^{... | Yes |
Let us find the extrema of the function \( f\left( {x, y}\right) = {x}^{4} + {y}^{4} - 2{x}^{2} \) , which is defined in \( {\mathbb{R}}^{2} \) . | In accordance with the necessary conditions (8.65) we write the system of equations\n\n\[ \left\{ \begin{array}{l} \frac{\partial f}{\partial x}\left( {x, y}\right) = 4{x}^{3} - {4x} = 0 \\ \frac{\partial f}{\partial y}\left( {x, y}\right) = 4{y}^{3} = 0 \end{array}\right. \]\n\nfrom which we find three critical points... | Yes |
On the basis of the laws of conservation of energy and momentum of a closed mechanical system one can show by a simple computation that when two perfectly elastic balls having mass \( {m}_{1} \) and \( {m}_{2} \) and initial velocities \( {v}_{1} \) and \( {v}_{2} \) collide, their velocities after a central collision ... | \[ {\widetilde{v}}_{1} = \frac{\left( {{m}_{1} - {m}_{2}}\right) {v}_{1} + 2{m}_{2}{v}_{2}}{{m}_{1} + {m}_{2}} \] \[ {\widetilde{v}}_{2} = \frac{\left( {{m}_{2} - {m}_{1}}\right) {v}_{2} + 2{m}_{1}{v}_{1}}{{m}_{1} + {m}_{2}}. \] | Yes |
Proposition 1. If the function \( F : U\left( {{x}_{0},{y}_{0}}\right) \rightarrow \mathbb{R} \) defined in a neighborhood \( U\left( {{x}_{0},{y}_{0}}\right) \) of the point \( \left( {{x}_{0},{y}_{0}}\right) \in {\mathbb{R}}^{2} \) is such that\n\n\( {1}^{0}F \in {C}^{\left( p\right) }\left( {U;\mathbb{R}}\right) \),... | Proof. Suppose for definiteness that \( {F}_{y}^{\prime }\left( {{x}_{0},{y}_{0}}\right) > 0 \) . Since \( F \in {C}^{\left( 1\right) }\left( {U;\mathbb{R}}\right) \) , it follows that \( {F}_{y}^{\prime }\left( {x, y}\right) > 0 \) also in some neighborhood of \( \left( {{x}_{0},{y}_{0}}\right) \) . In order to avoid ... | Yes |
Let us return to relation (8.80) studied above, which defines a circle in \( {\mathbb{R}}^{2} \), and verify Proposition 1 on this example. | In this case\n\n\[ F\left( {x, y}\right) = {x}^{2} + {y}^{2} - 1 \]\n\nand it is obvious that \( F \in {C}^{\left( \infty \right) }\left( {{\mathbb{R}}^{2};\mathbb{R}}\right) \). Next,\n\n\[ {F}_{x}^{\prime }\left( {x, y}\right) = {2x},\;{F}_{y}^{\prime }\left( {x, y}\right) = {2y}, \]\n\nso that \( {F}_{y}^{\prime }\l... | Yes |
Proposition 2. If a function \( F : U \rightarrow \mathbb{R} \) defined in a neighborhod \( U \subset {\mathbb{R}}^{m + 1} \) of the point \( \left( {{x}_{0},{y}_{0}}\right) = \left( {{x}_{0}^{1},\ldots ,{x}_{0}^{m},{y}_{0}}\right) \in {\mathbb{R}}^{m + 1} \) is such that\n\n\( {1}^{0}F \in {C}^{\left( p\right) }\left(... | Proof. The proof of the existence of the interval \( {I}^{m + 1} = {I}_{x}^{m} \times {I}_{y}^{1} \) and the existence of the function \( y = f\left( x\right) = f\left( {{x}^{1},\ldots ,{x}^{m}}\right) \) and its continuity in \( {I}_{x}^{m} \) is a verbatim repetition of the corresponding part of the proof of Proposit... | Yes |
Assume that the function \( F : G \rightarrow \mathbb{R} \) is defined in a domain \( G \subset {\mathbb{R}}^{m} \) and belongs to the class \( {C}^{\left( 1\right) }\left( {G;\mathbb{R}}\right) ;{x}_{0} = \left( {{x}_{0}^{1},\ldots ,{x}_{0}^{m}}\right) \in G \) and \( F\left( {x}_{0}\right) = F\left( {{x}_{0}^{1},\ldo... | Then, by Proposition 2, in some neighborhood of \( {x}_{0} \) the subset of \( {\mathbb{R}}^{m} \) defined by the equation \( F\left( {{x}^{1},\ldots ,{x}^{m}}\right) = 0 \) can be defined as the graph of a function \( {x}^{m} = f\left( {{x}^{1},\ldots ,{x}^{m - 1}}\right) \), defined in a neighborhood of the point \( ... | Yes |
Polar coordinates. The mapping \( f : {\mathbb{R}}_{ + }^{2} \rightarrow {\mathbb{R}}^{2} \) of the half-plane \( {\mathbb{R}}_{ + }^{2} = \left\{ {\left( {\rho ,\varphi }\right) \in {\mathbb{R}}^{2} \mid \rho \geq 0}\right\} \) onto the plane \( {\mathbb{R}}^{2} \) defined by the formula\n\n\[ x = \rho \cos \varphi \]... | The Jacobian of this mapping, as can be easily computed, is \( \rho \), that is, it is nonzero in a neighborhood of any point \( \left( {\rho ,\varphi }\right) \), where \( \rho > 0 \) . Therefore formulas (8.115) are locally invertible and hence locally the numbers \( \rho \) and \( \varphi \) can be taken as new coor... | Yes |
Suppose for example, a curve in the plane \( {\mathbb{R}}^{2} \) is defined by the equation\n\n\[ F\left( {x, y}\right) = 0. \]\n\nAssume that \( F \) is a smooth function, that the point \( \left( {{x}_{0},{y}_{0}}\right) \) lies on the curve, that is, \( F\left( {{x}_{0},{y}_{0}}\right) = 0 \), and that this point is... | Let us try to choose coordinates \( \xi ,\eta \) so that in these coordinates a closed interval of a coordinate line, for example, the line \( \eta = 0 \), corresponds to an arc of this curve.\n\nWe set\n\n\[ \xi = x - {x}_{0},\;\eta = F\left( {x, y}\right) . \]\n\nThe Jacobi matrix\n\n\[ \left( \begin{matrix} 1 & 0 \\... | Yes |
Theorem 2. (The rank theorem). Let \( f : U \rightarrow {\mathbb{R}}^{n} \) be a mapping defined in a neighborhood \( U \subset {\mathbb{R}}^{m} \) of a point \( {x}_{0} \in {\mathbb{R}}^{m} \). If \( f \in {C}^{\left( p\right) }\left( {U;{\mathbb{R}}^{n}}\right), p \geq 1 \), and the mapping \( f \) has the same rank ... | Proof. We write the coordinate representation\n\n\[ {y}^{1} = {f}^{1}\left( {{x}^{1},\ldots ,{x}^{m}}\right) \]\n\n..............................\n\n\[ {y}^{k} = {f}^{k}\left( {{x}^{1},\ldots ,{x}^{m}}\right) \]\n\n(8.120)\n\n\[ {y}^{k + 1} = {f}^{k + 1}\left( {{x}^{1},\ldots ,{x}^{m}}\right) ,\]\n\n...................... | Yes |
Proposition 1. If a system \( {f}^{i}\left( {{x}^{1},\ldots ,{x}^{m}}\right) \left( {i = 1,\ldots, n}\right) \) of smooth functions defined on a neighborhood \( U\left( {x}_{0}\right) \) of the point \( {x}_{0} \in {\mathbb{R}}^{m} \) is such that the rank of the matrix\n\n\[ \left( \begin{array}{l} \frac{\partial {f}^... | Proof. In fact, if \( k = n \), then by Remark 1 after the rank theorem, the image of a neighborhood of the point \( {x}_{0} \) under the mapping\n\n\[ {y}^{1} = {f}^{1}\left( {{x}^{1},\ldots ,{x}^{m}}\right) \]\n\n........................\n\n(8.127)\n\n\[ {y}^{n} = {f}^{n}\left( {{x}^{1},\ldots ,{x}^{m}}\right) \]\n\n... | Yes |
The space \( {\mathbb{R}}^{n} \) itself is an \( n \) -dimensional surface of class \( {C}^{\left( \infty \right) } \) . As the mapping \( \varphi : {\mathbb{R}}^{n} \rightarrow {I}^{n} \) here, one can take, for example, the mapping | \[ {\xi }^{i} = \frac{2}{\pi }\arctan {x}^{i}\;\left( {i = 1,\ldots, n}\right) \] | Yes |
The set in \( {\mathbb{R}}^{n} \) defined by the system of relations\n\n\[ \n\\left\\{ \\begin{matrix} {a}_{1}^{1}{x}^{1} + \\cdots + {a}_{k}^{1}{x}^{k} + {a}_{k + 1}^{1}{x}^{k + 1} + \\cdots + {a}_{n}^{1}{x}^{n} = 0, \\\\ \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \... | Indeed, suppose for example that the determinant\n\n\[ \n\\left| \\begin{array}{lll} {a}_{k + 1}^{1} & \\cdots & {a}_{n}^{1} \\\\ \\cdots & \\cdots & \\cdots \\cdots \\\\ {a}_{k + 1}^{n - k} & \\cdots & {a}_{n}^{n - k} \\end{array}\\right|\n\]\n\nis nonzero. Then the linear transformation\n\n\[ \n{t}^{1} = {x}^{1} \n\]... | Yes |
The graph of a smooth function \( {x}^{n} = f\left( {{x}^{1},\ldots ,{x}^{n - 1}}\right) \) defined in a domain \( G \subset {\mathbb{R}}^{n - 1} \) is a smooth \( \left( {n - 1}\right) \) -dimensional surface in \( {\mathbb{R}}^{n}. \) | Indeed, setting\n\n\[ \left\{ \begin{array}{l} {t}^{i} = {x}^{i}\;\left( {i = 1,\ldots, n - 1}\right) \\ {t}^{n} = {x}^{n} - f\left( {{x}^{1},\ldots ,{x}^{n - 1}}\right) \end{array}\right. \]\n\nwe obtain a coordinate system in which the graph of the function has the equation \( {t}^{n} = 0 \) . | Yes |
The circle \( {x}^{2} + {y}^{2} = 1 \) in \( {\mathbb{R}}^{2} \) is a one-dimensional submanifold of \( {\mathbb{R}}^{2} \) | as is established by the locally invertible conversion to polar coordinates \( \left( {\rho ,\varphi }\right) \) studied in the preceding section. In these coordinates the circle has equation \( \rho = 1 \) | Yes |
Let \( {F}^{i}\left( {{x}^{1},\ldots ,{x}^{n}}\right) \left( {i = 1,\ldots, n - k}\right) \) be a system of smooth functions of rank \( n - k \) . We shall show that the relations\n\n\[ \left\\{ \begin{array}{l} {F}^{1}\left( {{x}^{1},\ldots ,{x}^{k},{x}^{k + 1},\ldots ,{x}^{n}}\right) = 0, \\\\ \cdots \cdots \cdots \c... | Suppose the condition\n\n\[ \\left| \\begin{matrix} \\frac{\\partial {F}^{1}}{\\partial {x}^{k + 1}} & \\cdots & \\frac{\\partial {F}^{1}}{\\partial {x}^{n}} \\\\ \\cdots \\cdots & \\cdots \\cdots & \\cdots \\cdots \\\\ \\frac{\\partial {F}^{n - k}}{\\partial {x}^{k + 1}} & \\cdots & \\frac{\\partial {F}^{n - k}}{\\par... | Yes |
If a smooth mapping \( f : G \rightarrow {\mathbb{R}}^{n} \) of the domain \( G \subset {\mathbb{R}}^{n} \) defined in coordinate form by (8.136) has rank \( k \) at the point \( {t}_{0} \in G \), then there exists a neighborhood \( U\left( {t}_{0}\right) \subset G \) of this point whose image \( f\left( {U\left( {t}_{... | Indeed, as already noted above, in this case relations (8.136) can be replaced by the equivalent system\n\n\[ \left\{ \begin{matrix} {x}^{k + 1} = {\varphi }^{k + 1}\left( {{x}^{1},\ldots ,{x}^{k}}\right) \\ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ {x}^{n} = {\varphi }^{n}\left( {{x}^{1},\ldots ,{x}^{... | Yes |
Theorem 1. Let \( f : D \rightarrow \mathbb{R} \) be a function defined on an open set \( D \subset {\mathbb{R}}^{n} \) and belonging to \( {C}^{\left( 1\right) }\left( {D;\mathbb{R}}\right) \) . Let \( S \) be a smooth surface in \( D \) . A necessary condition for a point \( {x}_{0} \in S \) that is noncritical for \... | Proof. We choose an arbitrary vector \( \xi \in T{S}_{{x}_{0}} \) and a smooth path \( x = x\left( t\right) \) on \( S \) that passes through this point at \( t = 0 \) and for which the vector \( \xi \) is the velocity at \( t = 0 \), that is, \[ \frac{dx}{dt}\left( 0\right) = \xi \] If \( {x}_{0} \) is an extremum of ... | Yes |
Let us find the extrema of a symmetric quadratic form\n\n\\[ \nf\\left( x\\right) = \\mathop{\\sum }\\limits_{{i, j = 1}}^{n}{a}_{ij}{x}^{i}{x}^{j}\\;\\left( {{a}_{ij} = {a}_{ji}}\\right) \n\\]\n\non the sphere\n\n\\[ \nF\\left( x\\right) = \\mathop{\\sum }\\limits_{{i = 1}}^{n}{\\left( {x}^{i}\\right) }^{2} - 1 = 0. \... | Let us write the Lagrange function for this problem\n\n\\[ \nL\\left( {x,\\lambda }\\right) = \\mathop{\\sum }\\limits_{{i, j = 1}}^{n}{a}_{ij}{x}^{i}{x}^{j} - \\lambda \\left( {\\mathop{\\sum }\\limits_{{i = 1}}^{n}{\\left( {x}^{i}\\right) }^{2} - 1}\\right) , \n\\]\n\nand the necessary conditions for an extremum of \... | Yes |
A sufficient condition for the point \( {x}_{0} \) to be an extremum of the function \( {\left. f\right| }_{S} \) is that the quadratic form \[ \frac{{\partial }^{2}L}{\partial {x}^{i}\partial {x}^{j}}\left( {x}_{0}\right) {\xi }^{i}{\xi }^{j} \] (8.176) be either positive-definite or negative-definite for vectors \( \... | Proof. We first note that \( L\left( x\right) \equiv f\left( x\right) \) for \( x \in S \), so that if we show that \( {x}_{0} \in \) \( S \) is an extremum of the function \( {\left. L\right| }_{S} \), we shall have shown simultaneously that it is an extremum of \( {\left. f\right| }_{S} \) . By hypothesis, the necess... | Yes |
Suppose we are given the function \[ f\left( {x, y, z}\right) = {x}^{2} - {y}^{2} + {z}^{2} \] in the space \( {\mathbb{R}}^{3} \) with coordinates \( x, y, z \). We seek an extremum of this function on the plane \( S \) defined by the equation \[ F\left( {x, y, z}\right) = {2x} - y - 3 = 0. \] | Writing the Lagrange function \[ L\left( {x, y, z}\right) = \left( {{x}^{2} - {y}^{2} + {z}^{2}}\right) - \lambda \left( {{2x} - y - 3}\right) \] and the necessary conditions for an extremum \[ \left\{ \begin{array}{l} \frac{\partial L}{\partial x} = {2x} - {2\lambda } = 0, \\ \frac{\partial L}{\partial y} = - {2y} + \... | Yes |
Under the hypotheses of Example 10 we replace \( {\mathbb{R}}^{3} \) by \( {\mathbb{R}}^{2} \) and the function \( f \) by\n\n\[ f\left( {x, y}\right) = {x}^{2} - {y}^{2} \]\n\nretaining the condition\n\n\[ {2x} - y - 3 = 0 \] | We find \( p = \left( {2,1}\right) \) as a possible extremum.\n\nInstead of the form (8.181) we obtain the form\n\n\[ {\left( {\xi }^{1}\right) }^{2} - {\left( {\xi }^{2}\right) }^{2} \]\n\n(8.183)\n\nwith the previous relation (8.182) between \( {\xi }^{1} \) and \( {\xi }^{2} \) .\n\nThus the form (8.183) now has the... | Yes |
On the plane \( {\mathbb{R}}^{2} \) with Cartesian coordinates \( \left( {x, y}\right) \) we are given the function\n\n\[ f\left( {x, y}\right) = {x}^{2} + {y}^{2}. \]\n\nLet us find the extremum of this function on the ellipse given by the canonical relation\n\n\[ F\left( {x, y}\right) = \frac{{x}^{2}}{{a}^{2}} + \fra... | It is obvious from geometric considerations that \( {\left. \min f\right| }_{S} = {a}^{2} \) and \( {\left. \max f\right| }_{S} = {b}^{2} \) . Let us obtain this result on the basis of the procedures recommended by Theorems 1 and 2.\n\nBy writing the Lagrange function\n\n\[ L\left( {x, y,\lambda }\right) = \left( {{x}^... | Yes |
Example 13. Let us find the extrema of the function\n\n\\[ f\\left( {x, y, z}\\right) = {x}^{2} + {y}^{2} + {z}^{2} \\]\n\non the ellipsoid \\( S \\) defined by the relation\n\n\\[ F\\left( {x, y, z}\\right) = \\frac{{x}^{2}}{{a}^{2}} + \\frac{{y}^{2}}{{b}^{2}} + \\frac{{z}^{2}}{{c}^{2}} - 1 = 0, \\]\n\nwhere \\( 0 < a... | By writing the Lagrange function\n\n\\[ L\\left( {x, y, z,\\lambda }\\right) = \\left( {{x}^{2} + {y}^{2} + {z}^{2}}\\right) - \\lambda \\left( {\\frac{{x}^{2}}{{a}^{2}} + \\frac{{y}^{2}}{{b}^{2}} + \\frac{{z}^{2}}{{c}^{2}} - 1}\\right) ,\\]\n\nin accordance with the necessary criterion for an extremum, we find the sol... | Yes |
Problem 6. Show that\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( 1 + \frac{z}{n}\right) }^{n} = {\mathrm{e}}^{x}\left( {\cos y + \mathrm{i}\sin y}\right) \;\left( {z = x + \mathrm{i}y}\right) ,\] | so that it is natural to suppose that \( {e}^{\mathrm{i}y} = \cos y + \mathrm{i}\sin y \) (Euler’s formula) and\n\n\[ {\mathrm{e}}^{z} = {\mathrm{e}}^{x}{\mathrm{e}}^{\mathrm{i}y} = {\mathrm{e}}^{x}\left( {\cos y + \mathrm{i}\sin y}\right) .\] | No |
Problem 3. The function \( \operatorname{erf}\left( x\right) = \frac{1}{\sqrt{\pi }}\mathop{\int }\limits_{{-x}}^{x}{\mathrm{e}}^{-{t}^{2}}\mathrm{\;d}t \), called the probability error integral, has limit 1 as \( x \rightarrow + \infty \) . Draw the graph of this function and find its derivative. Show that as \( x \ri... | \[ \operatorname{erf}\left( x\right) = 1 - \frac{2}{\sqrt{\pi }}{\mathrm{e}}^{-{x}^{2}}\left( {\frac{1}{2x} - \frac{1}{{2}^{2}{x}^{3}} + \frac{1 \cdot 3}{{2}^{3}{x}^{5}} - \frac{1 \cdot 3 \cdot 5}{{2}^{4}{x}^{7}} + o\left( \frac{1}{{x}^{7}}\right) }\right) . \] | No |
Let \( x \mapsto f\left( x\right) \) be a nonnegative function defined for \( x \geq 0 \) and vanishing for \( x = 0 \). If this function is strictly convex upward, then, setting \[ d\left( {{x}_{1},{x}_{2}}\right) = f\left( \left| {{x}_{1} - {x}_{2}}\right| \right) \] for points \( {x}_{1},{x}_{2} \in \mathbb{R} \), w... | Axioms a) and b) obviously hold here, and the triangle inequality follows from the easily verified fact that \( f \) is strictly monotonic and satisfies the following inequalities for \( 0 < a < b \): \[ f\left( {a + b}\right) - f\left( b\right) < f\left( a\right) - f\left( 0\right) = f\left( a\right) . \] | Yes |
Besides the traditional distance\n\n\[ d\left( {{x}_{1},{x}_{2}}\right) = \sqrt{\mathop{\sum }\limits_{{i = 1}}^{n}{\left| {x}_{1}^{i} - {x}_{2}^{i}\right| }^{2}} \]\n\n(9.3)\n\nbetween points \( {x}_{1} = \left( {{x}_{1}^{1},\ldots ,{x}_{1}^{n}}\right) \) and \( {x}_{2} = \left( {{x}_{2}^{1},\ldots ,{x}_{2}^{n}}\right... | The validity of the triangle inequality for the function (9.4) follows from Minkowski’s inequality (see Subsect. 5.4.2). | Yes |
When we encounter a word with incorrect letters while reading a text, we can reconstruct the word without too much trouble by correcting the errors, provided the number of errors is not too large. However, correcting the error and obtaining the word is an operation that is sometimes ambiguous. For that reason, other co... | Geometrically the set of such sequences can be interpreted as the set of vertices of the unit cube \( I = \left\{ {x \in {\mathbb{R}}^{n} \mid 0 \leq {x}^{i} \leq 1, i = 1,\ldots, n}\right\} \) in \( {\mathbb{R}}^{n} \) . The distance between two vertices is the number of interchanges of zeros and ones needed to obtain... | Yes |
As one can easily see, if we pass to the limit in (9.4) as \( p \rightarrow + \infty \) , we obtain the following metric in \( {\mathbb{R}}^{n} \) : | \[ d\left( {{x}_{1},{x}_{2}}\right) = \mathop{\max }\limits_{{1 \leq i \leq n}}\left| {{x}_{1}^{i} - {x}_{2}^{i}}\right| . \] | Yes |
The set \( C\left\lbrack {a, b}\right\rbrack \) of functions that are continuous on a closed interval becomes a metric space if we define the distance between two functions \( f \) and \( g \) to be\n\n\[ d\left( {f, g}\right) = \mathop{\max }\limits_{{a \leq x \leq b}}\left| {f\left( x\right) - g\left( x\right) }\righ... | Axioms a) and b) for a metric obviously hold, and the triangle inequality follows from the relations\n\n\[ \left| {f\left( x\right) - h\left( x\right) \leq }\right| f\left( x\right) - g\left( x\right) \left| +\right| g\left( x\right) - h\left( x\right) \mid \leq d\left( {f, g}\right) + d\left( {g, h}\right) ,\]\n\nthat... | Yes |
Like the metric (9.4), for \( p \geq 1 \) we can introduce in \( C\left\lbrack {a, b}\right\rbrack \) the metric\n\n\[ \n{d}_{p}\left( {f, g}\right) = {\left( {\int }_{a}^{b}{\left| f - g\right| }^{p}\left( x\right) dx\right) }^{1/p}.\n\] | It follows from Minkowski’s inequality for integrals, which can be obtained from Minkowski's inequality for the Riemann sums by passing to the limit, that this is indeed a metric for \( p \geq 1 \) . | Yes |
The metric (9.7) could also have been used on the set \( \mathcal{R}\left\lbrack {a, b}\right\rbrack \) of Riemann-integrable functions on the closed interval \( \left\lbrack {a, b}\right\rbrack \) . However, since the integral of the absolute value of the difference of two functions may vanish even when the two functi... | Therefore, if we partition \( \mathcal{R}\left\lbrack {a, b}\right\rbrack \) into equivalence classes of functions, regarding two functions in \( \mathcal{R}\left\lbrack {a, b}\right\rbrack \) as equivalent if they differ on at most a set of measure zero, then the relation (9.7) really does define a metric on the set \... | Yes |
In the set \( {C}^{\left( k\right) }\left\lbrack {a, b}\right\rbrack \) of functions defined on \( \left\lbrack {a, b}\right\rbrack \) and having continuous derivatives up to order \( k \) inclusive one can define the following metric:\n\n\[ d\left( {f, g}\right) = \max \left\{ {{M}_{0},\ldots ,{M}_{k}}\right\} \]\n\nw... | Using the fact that (9.6) is a metric, one can easily verify that (9.8) is also a metric. | No |
Proposition 1. a) The union \( \mathop{\bigcup }\limits_{{\alpha \in A}}{G}_{\alpha } \) of the sets in any system \( \left\{ {{G}_{\alpha },\alpha \in A}\right\} \) of sets \( {G}_{\alpha } \) that are open in \( X \) is an open set in \( X \) . | The proof of Proposition 1 is a verbatim repetition of the proof of the corresponding proposition for open and closed sets in \( {\mathbb{R}}^{n} \), and we omit it. (See Proposition 1 in Sect. 7.1.) | No |
The open interval \( \left| x\right| < 1, y = 0 \) of the \( x \) -axis in the plane \( {\mathbb{R}}^{2} \) with the standard metric in \( {\mathbb{R}}^{2} \) is a metric space \( \left( {{X}_{1},{d}_{1}}\right) \), which, like any metric space, is closed as a subset of itself, since it contains all its limit points in... | This same example shows that openness is also a relative concept. | No |
Let \( X = {\mathbb{R}}^{n}\left( {n > 1}\right) \) . Consider the metric \( {d}_{1}\left( {{x}_{1},{x}_{2}}\right) \) defined by relation (9.5) in Sect. 9.1, and the metric \( {d}_{2}\left( {{x}_{1},{x}_{2}}\right) \) defined by formula (9.3) in Sect. 9.1.\n\nThe inequalities\n\n\[ \n{d}_{1}\left( {{x}_{1},{x}_{2}}\ri... | obviously imply that every ball \( B\left( {a, r}\right) \) with center at an arbitrary point \( a \in X \), interpreted in the sense of one of these two metrics, contains a ball with the same center, interpreted in the sense of the other metric. Hence by definition of an open subset of a metric space, it follows that ... | Yes |
If \( \left( {X, d}\right) \) is a metric space and \( \left( {x,\tau }\right) \) the topological space corresponding to it, the set \( \mathfrak{B} = \{ B\left( {a, r}\right) \} \) of all balls, where \( a \in X \) and \( r > 0 \), is obviously a base of the topology \( \tau \). | Moreover, if we take the system \( \mathfrak{B} \) of all balls with positive rational radii \( r \), this system is also a base for the topology. | Yes |
If we take the system \( \mathfrak{B} \) of balls in \( {\mathbb{R}}^{k} \) of all possible rational radii \( r = \frac{m}{n} > 0 \) with centers at all possible rational points \( \left( {\frac{{m}_{1}}{{n}_{1}},\ldots ,\frac{{m}_{k}}{{n}_{k}}}\right) \in {\mathbb{R}}^{k} \) , we obviously obtain a countable base for ... | It is not difficult to verify that it is impossible to define the standard topology in \( {\mathbb{R}}^{k} \) by exhibiting a finite system of open sets. Thus the standard topological space \( {\mathbb{R}}^{k} \) has countable weight. | No |
Consider the set \( C\left( {\mathbb{R},\mathbb{R}}\right) \) of real-valued continuous functions defined on the entire real line. Using this set as foundation, we shall construct a new set - the set of germs of continuous functions. We shall regard two functions \( f, g \in C\left( {\mathbb{R},\mathbb{R}}\right) \) as... | It is worthwhile to note that in the resulting topological space two different points (germs) \( {f}_{a} \) and \( {g}_{a} \) may not have disjoint neighborhoods (see Fig. 9.1) | Yes |
The metric space \( \left( {C\left( {\left\lbrack {0,1}\right\rbrack ,\mathbb{R}}\right), d}\right) \) with the metric defined by (9.6) is also separable. | For, as follows from the uniform continuity of the functions \( f \in C\left( {\left\lbrack {0,1}\right\rbrack ,\mathbb{R}}\right) \), the graph of any such function can be approximated as closely as desired by a broken line consisting of a finite number of segments whose nodes have rational coordinates. The set of suc... | Yes |
An interval \( \left\lbrack {a, b}\right\rbrack \) of the set \( \mathbb{R} \) of real numbers in the standard topology is a compact set. | as follows immediately from the lemma of Subsect. 2.1.3 asserting that one can select a finite covering from any covering of a closed interval by open intervals. | No |
Proposition 1. A subset \( K \) of a topological space \( \left( {X,\tau }\right) \) is a compact subset of \( X \) if and only if \( K \) is compact as a subset of itself with the topology induced from \( \left( {X,\tau }\right) \) . | Proof. This proposition follows from the definition of compactness and the fact that every set \( {G}_{K} \) that is open in \( K \) can be obtained as the intersection of \( K \) with some set \( {G}_{X} \) that is open in \( X \) . | No |
Lemma 1. (Compact sets are closed). If \( K \) is a compact set in a Hausdorff space \( \left( {X,\tau }\right) \), then \( K \) is a closed subset of \( X \) . | Proof. By the criterion for a set to be closed, it suffices to verify that every limit point of \( K,{x}_{0} \in X \), belongs to \( K \) .
Suppose \( {x}_{0} \notin K \) . For each point \( x \in K \) we construct an open neighborhood \( G\left( x\right) \) such that \( {x}_{0} \) has a neighborhood disjoint from \( ... | Yes |
Lemma 2. (Nested compact sets.) If \( {K}_{1} \supset {K}_{2} \supset \cdots \supset {K}_{n} \supset \cdots \) is a nested sequence of nonempty compact sets, then the intersection \( \mathop{\bigcap }\limits_{{i = 1}}^{\infty }{K}_{i} \) is nonempty. | Proof. By Lemma 1 the sets \( {G}_{i} = {K}_{1} \smallsetminus {K}_{i}, i = 1,\ldots, n,\ldots \) are open in \( {K}_{1} \) . If the intersection \( \mathop{\bigcap }\limits_{{i = 1}}^{\infty }{K}_{i} \) is empty, then the sequence \( {G}_{1} \subset {G}_{2} \subset \cdots \subset {G}_{n} \subset \cdots \) forms a cove... | Yes |
Lemma 3. (Closed subsets of compact sets.) A closed subset \( F \) of a compact set \( K \) is itself compact. | Proof. Let \( \left\{ {{G}_{\alpha },\alpha \in A}\right\} \) be an open covering of \( F \) . Adjoining to this collection the open set \( G = K \smallsetminus F \), we obtain an open covering of the entire compact set \( K \) . From this covering we can extract a finite covering of \( K \) . Since \( G \cap F = \varn... | Yes |
Lemma 4. (Finite \( \varepsilon \) -grids.) If a metric space \( \left( {K, d}\right) \) is compact, then for every \( \varepsilon > 0 \) there exists a finite \( \varepsilon \) -grid in \( X \) . | Proof. For each point \( x \in K \) we choose an open ball \( B\left( {x,\varepsilon }\right) \) . From the open covering of \( K \) by these balls we select a finite covering \( B\left( {{x}_{1},\varepsilon }\right) ,\ldots, B\left( {{x}_{n},\varepsilon }\right) \) . The points \( {x}_{1},\ldots ,{x}_{n} \) obviously ... | Yes |
Lemma 5. If a metric space \( \left( {K, d}\right) \) is such that from each sequence of its points one can select a subsequence that converges in \( K \), then for every \( \varepsilon > 0 \) there exists a finite \( \varepsilon \) -grid. | Proof. If there were no finite \( {\varepsilon }_{0} \) -grid for some \( {\varepsilon }_{0} > 0 \), one could construct a sequence \( \left\{ {x}_{n}\right\} \) of points in \( K \) such that \( d\left( {{x}_{n},{x}_{i}}\right) > {\varepsilon }_{0} \) for all \( n \in \mathbb{N} \) and all \( i \in \{ 1,\ldots, n - 1\... | Yes |
Lemma 6. If the metric space \( \left( {K, d}\right) \) is such that from each sequence of its points one can select a subsequence that converges in \( K \), then every nested sequence of nonempty closed subsets of the space has a nonempty intersection. | Proof. If \( {F}_{1} \supset \cdots \supset {F}_{n} \supset \cdots \) is the sequence of closed sets, then choosing one point of each, we obtain a sequence \( {x}_{1},\ldots ,{x}_{n},\ldots \), from which we extract a convergent subsequence \( \left\{ {x}_{{n}_{i}}\right\} \) . The limit \( a \in K \) of this sequence,... | Yes |
Proposition. (Connected subsets of \\mathbb{R}.)\\;A\\; nonempty set \\( E \\subset \\mathbb{R}\\; \\) is connected if and only if for any \\( x \\) and \\( z \\) belonging to \\( E \\), the inequalities \\( x < y < z \\) imply that \\( y \\in E \\) . | Proof. Necessity. Let \\( E \\) be a connected subset of \\mathbb{R} \\), and let the triple of points \\( a, b, c \\) be such that \\( a \\in E, b \\in E \\), but \\( c \\notin E \\), even though \\( a < c < b \\) . Setting \\( A = \\{ x \\in E \\mid x < c\\}, B = \\{ x \\in E \\mid x > c\\} \\), we see that \\( a \\i... | Yes |
If the number 0, for example, is removed from the set \( \mathbb{R} \), the remaining set \( \mathbb{R} \smallsetminus 0 \) will not be a complete space in the standard metric. | Indeed, the sequence \( {x}_{n} = 1/n, n \in \mathbb{N} \), is a Cauchy sequence of points of this set, but has no limit in \( \mathbb{R} \smallsetminus 0 \) . | Yes |
If instead of the metric (9.9) we consider the integral metric\n\n\[ d\left( {f, g}\right) = {\int }_{a}^{b}\left| {f - g}\right| \left( x\right) {dx} \]\n\n(9.14)\n\non the same set \( C\left\lbrack {a, b}\right\rbrack \), the resulting metric space is no longer complete. | Proof. For the sake of notational simplicity, we shall assume \( \left\lbrack {a, b}\right\rbrack = \left\lbrack {-1,1}\right\rbrack \) and consider, for example, the sequence \( \left\{ {{f}_{n} \in C\left\lbrack {-1,1}\right\rbrack ;n \in \mathbb{N}}\right\} \) of functions defined as follows:\n\n\[ {f}_{n}\left( x\r... | Yes |
It is slightly more difficult to show that even the set \( \mathcal{R}\left\lbrack {a, b}\right\rbrack \) of real-valued Riemann-integrable functions defined on the closed interval \( \left\lbrack {a, b}\right\rbrack \) is not complete in the sense of the metric 9.14. | Proof. We take \( \left\lbrack {a, b}\right\rbrack \) to be the closed interval \( \left\lbrack {0,1}\right\rbrack \), and we shall construct a Cantor set on it that is not a set of measure zero. Let \( \Delta \in \rbrack 0,1/3\lbrack \) . We remove from the interval \( \left\lbrack {0,1}\right\rbrack \) the middle pie... | Yes |
Lemma. The following inequality holds for any quadruple of points \( a, b, u \) , \( v \) of the metric space \( \left( {X, d}\right) \) :\n\n\[ \left| {d\left( {a, b}\right) - d\left( {u, v}\right) }\right| \leq d\left( {a, u}\right) + d\left( {b, v}\right) . \] | Proof. By the triangle inequality\n\n\[ d\left( {a, b}\right) \leq d\left( {a, u}\right) + d\left( {u, v}\right) + d\left( {b, v}\right) . \]\n\nBy the symmetry of the points, this relation implies (9.15). | No |
Proposition 1. (Limit of a composition of mappings.) Let \( Y \) be a set with base \( {\mathcal{B}}_{Y} \) and \( g : Y \rightarrow Z \) a mapping of \( Y \) into a topological space \( Z \) having a limit over the base \( {\mathcal{B}}_{Y} \).\n\nLet \( X \) be a set with base \( {\mathcal{B}}_{X} \) and \( f : X \ri... | For the proof see Theorem 5 of Sect. 3.2. | No |
Proposition 2. (Cauchy criterion for existence of the limit of a mapping.) Let \( X \) be a set with a base \( \mathcal{B} \), and let \( f : X \rightarrow Y \) be a mapping of \( X \) into a complete metric space \( \left( {Y, d}\right) \). A necessary and sufficient condition for the mapping \( f \) to have a limit o... | For the proof see Theorem 4 of Sect. 3.2. | No |
Theorem 1. (Criterion for continuity.) A mapping \( f : X \rightarrow Y \) of a topological space \( \left( {X,{\tau }_{X}}\right) \) into a topological space \( \left( {Y,{\tau }_{Y}}\right) \) is continuous if and only if the pre-image of every open (resp. closed) subset of \( Y \) is open (resp. closed) in \( X \) . | Proof. Since the pre-image of a complement is the complement of the preimage, it suffices to prove the assertions for open sets.\n\nWe first show that if \( f \in C\left( {X, Y}\right) \) and \( {G}_{Y} \in {\tau }_{Y} \), then \( {G}_{X} = {f}^{-1}\left( {G}_{Y}\right) \) belongs to \( {\tau }_{X} \) . If \( {G}_{X} =... | Yes |
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