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The function \( f\left( x\right) = {a}^{x} \) is continuous on \( \mathbb{R} \) . | Indeed by property 3) of the exponential function (see Par. d in Subsect. 3.2.2, Example 10a), at any point \( {x}_{0} \in \mathbb{R} \) we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}{a}^{x} = {a}^{{x}_{0}} \]\n\nwhich, as we now know, is equivalent to the continuity of the function \( {a}^{x} \) at the ... | Yes |
The function \( f\left( x\right) = {\log }_{a}x \) is continuous at any point \( {x}_{0} \) in its domain of definition \( {\mathbb{R}}_{ + } = \{ x \in \mathbb{R} \mid x > 0\} \) . | In fact, by property 3) of the logarithm (see Par. d in Subsect. 3.2.2, Example 10b), at each point \( {x}_{0} \in {\mathbb{R}}_{ + } \) we have\n\n\[ \mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow {x}_{0}}}{\log }_{a}x = {\log }_{a}{x}_{0} \]\n\nwhich is equivalent to the continuity of the function \( {\... | Yes |
The function \( f\left( x\right) = \operatorname{sgn}x \) is constant and hence continuous in the neighborhood of any point \( a \in \mathbb{R} \) that is different from 0 . But in any neighborhood of 0 its oscillation equals 2. Hence 0 is a point of discontinuity for sgn \( \;x.\; \) | We remark that this function has a left-hand limit \( \mathop{\lim }\limits_{{x \rightarrow - 0}} \) sgn \( \;x = - 1 \) and a right-hand limit \( \mathop{\lim }\limits_{{x \rightarrow + 0}}\operatorname{sgn}x = 1 \) . However, in the first place, these limits are not the same; and in the second place, neither of them ... | Yes |
The function \( f\left( x\right) = \left| {\operatorname{sgn}x}\right| \) has the limit \( \mathop{\lim }\limits_{{x \rightarrow 0}}\left| {\operatorname{sgn}x}\right| = 1 \) as \( x \rightarrow 0 \), but \( f\left( 0\right) = \left| {\operatorname{sgn}0}\right| = 0 \), so that \( \mathop{\lim }\limits_{{x \rightarrow ... | We remark, however, that in this case, if we were to change the value of the function at the point 0 and set it equal to 1 there, we would obtain a function that is continuous at 0 , that is, we would remove the discontinuity. | Yes |
The function\n\n\[ f\left( x\right) = \left\{ \begin{matrix} \sin \frac{1}{x},\text{ for }x \neq 0, \\ 0,\text{ for }x = 0, \end{matrix}\right. \]\n\nis discontinuous at 0. | Moreover, it does not even have a limit as \( x \rightarrow 0 \) , since, as was shown Example 5 in Subsect. 3.2.1, \( \mathop{\lim }\limits_{{x \rightarrow 0}}\sin \frac{1}{x} \) does not exist. | Yes |
Example 11. The function\n\n\[ \n\mathcal{D}\left( x\right) = \left\{ \begin{array}{l} 1,\text{ if }x \in \mathbb{Q}, \\ 0,\text{ if }x \in \mathbb{R} \smallsetminus \mathbb{Q}, \end{array}\right.\n\]\n\nis called the Dirichlet function \( {}^{3} \) | This function is discontinuous at every point, and obviously all of its discontinuities are of second kind, since in every interval there are both rational and irrational numbers. | Yes |
Consider the Riemann function \( \mathcal{R}\left( x\right) = \left\{ \begin{array}{l} \frac{1}{n},\text{ if }x = \frac{m}{n} \in \mathbb{Q},\text{ where }\frac{m}{n}\text{ is in lowest terms. } \\ 0,\text{ if }x \in \mathbb{R} \smallsetminus \mathbb{Q}. \end{array}\right. \) | We remark that for any point \( a \in \mathbb{R} \), any bounded neighborhood \( U\left( a\right) \) of it, and any number \( N \in \mathbb{N} \), the neighborhood \( U\left( a\right) \) contains only a finite number of rational numbers \( \frac{m}{n}, m \in \mathbb{Z}, n \in \mathbb{N} \), with \( n < N \) . By shrink... | Yes |
Theorem 1. Let \( f : E \rightarrow \mathbb{R} \) be a function that is continuous at the point \( a \in E \) . Then the following statements hold.\n\n\( {1}^{0} \) The function \( f : E \rightarrow \mathbb{R} \) is bounded in some neighborhood \( {U}_{E}\left( a\right) \) of \( a \) . | Proof. To prove this theorem it suffices to recall (see Sect. 4.1) that the continuity of the function \( f \) or \( g \) at a point \( a \) of its domain of definition is equivalent to the condition that the limit of this function exists over the base \( {\mathcal{B}}_{a} \) of neighborhoods of \( a \) and is equal to... | No |
An algebraic polynomial \( P\left( x\right) = {a}_{0}{x}^{n} + {a}_{1}{x}^{n - 1} + \cdots + {a}_{n} \) is a continuous function on \( \mathbb{R} \) . | Indeed, it follows by induction from \( {3}^{0} \) of Theorem 1 that the sum and product of any finite number of functions that are continuous at a point are themselves continuous at that point. We have verified in Examples 1 and 2 of Sect. 4.1 that the constant function and the function \( f\left( x\right) = x \) are ... | Yes |
The composition of a finite number of continuous functions is continuous at each point of its domain of definition. | This follows by induction from assertion \( {4}^{0} \) of Theorem 1. | No |
Theorem 2. (The Bolzano-Cauchy intermediate-value theorem). If a function that is continuous on a closed interval assumes values with different signs at the endpoints of the interval, then there is a point in the interval where it assumes the value 0. | Proof. Let us divide the interval \( \left\lbrack {a, b}\right\rbrack \) in half. If the function does not assume the value 0 at the point of division, then it must assume opposite values at the endpoints of one of the two subintervals. In that interval we proceed as we did with the original interval, that is, we bisec... | Yes |
Theorem 3. (The Weierstrass maximum-value theorem). A function that is continuous on a closed interval is bounded on that interval. Moreover there is a point in the interval where the function assumes its maximum value and a point where it assumes its minimal value. | Proof. Let \( f : E \rightarrow \mathbb{R} \) be a continuous function on the closed interval \( E = \) \( \left\lbrack {a, b}\right\rbrack \) . By the local properties of a continuous function (see Theorem 1) for any point \( x \in E \) there exists a neighborhood \( U\left( x\right) \) such that the function is bound... | Yes |
The function \( f\left( x\right) = \sin \frac{1}{x} \) is continuous on the open interval \( \rbrack 0,1\lbrack = E \). However, in every neighborhood of 0 in the set \( E \) the function assumes both values -1 and 1. Therefore, for \( \varepsilon < 2 \), the condition \( \left| {f\left( {x}_{1}\right) - f\left( {x}_{2... | In this connection it is useful to write out explicitly the negation of the property of uniform continuity for a function:\n\n\[ \left( {f : E \rightarrow \mathbb{R}\text{ is not uniformly continuous }}\right) \mathrel{\text{:=}} \]\n\n\[ = \left( {\exists \varepsilon > 0\;\forall \delta > 0\;\exists {x}_{1} \in E\;\ex... | Yes |
If the function \( f : E \rightarrow \mathbb{R} \) is unbounded in every neighborhood of a fixed point \( {x}_{0} \in E \), then it is not uniformly continuous. | Indeed, in that case for any \( \delta > 0 \) there are points \( {x}_{1} \) and \( {x}_{2} \) in every \( \frac{\delta }{2} \) -neighborhood of \( {x}_{0} \) such that \( \left| {f\left( {x}_{1}\right) - f\left( {x}_{2}\right) }\right| > 1 \) although \( \left| {{x}_{1} - {x}_{2}}\right| < \delta \) . | Yes |
The function \( f\left( x\right) = {x}^{2} \), which is continuous on \( \mathbb{R} \), is not uniformly continuous on \( \mathbb{R} \). | In fact, at the points \( {x}_{n}^{\prime } = \sqrt{n + 1} \) and \( {x}_{n}^{\prime \prime } = \sqrt{n} \), where \( n \in \mathbb{N} \), we have \( f\left( {x}_{n}^{\prime }\right) = n + 1 \) and \( f\left( {x}_{n}^{\prime \prime }\right) = n \), so that \( f\left( {x}_{n}^{\prime }\right) - f\left( {x}_{n}^{\prime \... | Yes |
The function \( f\left( x\right) = \sin \left( {x}^{2}\right) \), which is continuous and bounded on \( \mathbb{R} \), is not uniformly continuous on \( \mathbb{R} \). | Indeed, at the points \( {x}_{n}^{\prime } = \sqrt{\frac{\pi }{2}\left( {n + 1}\right) } \) and \( {x}_{n}^{\prime \prime } = \sqrt{\frac{\pi }{2}n} \), where \( n \in \mathbb{N} \), we have \( \left| {f\left( {x}_{n}^{\prime }\right) - f\left( {x}_{n}^{\prime \prime }\right) }\right| = 1 \), while \( \mathop{\lim }\li... | Yes |
Proposition 1. A continuous mapping \( f : E \rightarrow \mathbb{R} \) of a closed interval \( E = \) \( \left\lbrack {a, b}\right\rbrack \) into \( \mathbb{R} \) is injective if and only if the function \( f \) is strictly monotonic on \( \left\lbrack {a, b}\right\rbrack \) . | Proof. If \( f \) is increasing or decreasing on any set \( E \subset \mathbb{R} \) whatsoever, the mapping \( f : E \rightarrow \mathbb{R} \) is obviously injective: at different points of \( E \) the function assumes different values.\n\nThus the more substantive part of Proposition 1 consists of the assertion that e... | Yes |
Proposition 2. Each strictly monotonic function \( f : X \rightarrow \mathbb{R} \) defined on a numerical set \( X \subset \mathbb{R} \) has an inverse \( {f}^{-1} : Y \rightarrow \mathbb{R} \) defined on the set \( Y = f\left( X\right) \) of values of \( f \), and has the same kind of monotonicity on \( Y \) that \( f... | Proof. The mapping \( f : X \rightarrow Y = f\left( X\right) \) is surjective, that is, it is a mapping of \( X \) onto \( Y \) . For definiteness assume that \( f : X \rightarrow Y \) is increasing on \( X \) . In that case\n\n\[ \forall {x}_{1} \in X\forall {x}_{2} \in X\left( {{x}_{1} < {x}_{2} \Leftrightarrow f\lef... | Yes |
Proposition 3. The discontinuities of a function \( f : E \rightarrow \mathbb{R} \) that is monotonic on the set \( E \subset \mathbb{R} \) can be only discontinuities of first kind. | Proof. For definiteness let \( f \) be nondecreasing. Assume that \( a \in E \) is a point of discontinuity of \( f \) . Since \( a \) cannot be an isolated point of \( E, a \) must be a limit point of at least one of the two sets \( {E}_{a}^{ - } = \{ x \in E \mid x < a\} \) and \( {E}_{a}^{ + } = \{ x \in E \mid x > ... | Yes |
Corollary 1. If \( a \) is a point of discontinuity of a monotonic function \( f \) : \( E \rightarrow \mathbb{R} \), then at least one of the limits\n\n\[ \mathop{\lim }\limits_{{E \ni x \rightarrow a - 0}}f\left( x\right) = f\left( {a - 0}\right) ,\;\mathop{\lim }\limits_{{E \ni x \rightarrow a + 0}}f\left( x\right) ... | Proof. Indeed, if \( a \) is a point of discontinuity, it must be a limit point of the set \( E \), and by Proposition 3 is a discontinuity of first kind. Thus at least one of the bases \( E \ni x \rightarrow a - 0 \) and \( E \ni x \rightarrow a + 0 \) is defined, and the limit of the function over that base exists. (... | Yes |
Corollary 2. The set of points of discontinuity of a monotonic function is at most countable. | Proof. With each point of discontinuity of a monotonic function we associate the corresponding open interval in Corollary 1 containing no values of \( f \) . These intervals are pairwise disjoint. But on the line there cannot be more than a countable number of pairwise disjoint open intervals. In fact, one can choose a... | Yes |
Proposition 4. (A criterion for continuity of a monotonic function.) \( A \) monotonic function \( f : E \rightarrow \mathbb{R} \) defined on a closed interval \( E = \left\lbrack {a, b}\right\rbrack \) is continuous if and only if its set of values \( f\left( E\right) \) is the closed interval with endpoints \( f\left... | Proof. If \( f \) is a continuous monotonic function, the monotonicity implies that all the values that \( f \) assumes on the closed interval \( \left\lbrack {a, b}\right\rbrack \) lie between the values \( f\left( a\right) \) and \( f\left( b\right) \) that it assumes at the endpoints. By continuity, the function mus... | Yes |
Theorem 5. (The inverse function theorem). A function \( f : X \rightarrow \mathbb{R} \) that is strictly monotonic on a set \( X \subset \mathbb{R} \) has an inverse \( {f}^{-1} : Y \rightarrow \mathbb{R} \) defined on the set \( Y = f\left( X\right) \) of values of \( f \) . The function \( {f}^{-1} : Y \rightarrow \... | Proof. The assertion that the set \( Y = f\left( X\right) \) is the closed interval with endpoints \( f\left( a\right) \) and \( f\left( b\right) \) when \( X = \left\lbrack {a, b}\right\rbrack \) and \( f \) is continuous follows from Proposition 4 proved above. It remains to be verified that \( {f}^{-1} : Y \rightarr... | Yes |
The restriction of the function \( y = \tan x \) to the open interval \( X = \rbrack - \frac{\pi }{2},\frac{\pi }{2}\lbrack \) is a continuous function that increases from \( - \infty \) to \( + \infty \) . By the first part of Theorem 5 it has an inverse denoted \( x = \arctan y \), defined for all \( y \in \mathbb{R}... | we take the point \( {x}_{0} = \arctan {y}_{0} \) and a closed interval \( \left. {\left\lbrack {{x}_{0} - \varepsilon ,{x}_{0} + \varepsilon }\right\rbrack \text{containing}{x}_{0}\text{and contained in the open interval}}\right\rbrack - \frac{\pi }{2},\frac{\pi }{2}\lbrack \) . If \( {x}_{0} - \varepsilon = \arctan \... | Yes |
Proposition 1. A function \( f : E \rightarrow \mathbb{R} \) that is continuous at a point \( {x}_{0} \in E \) that is a limit point of \( E \subset \mathbb{R} \) admits a linear approximation (5.18) if and only if it is differentiable at the point. | The function\n\n\[ \varphi \left( x\right) = {c}_{0} + {c}_{1}\left( {x - {x}_{0}}\right) \]\n\n(5.20)\n\nwith \( {c}_{0} = f\left( {x}_{0}\right) \) and \( {c}_{1} = {f}^{\prime }\left( {x}_{0}\right) \) is the only function of the form (5.20) that satisfies (5.18).\n\nThus the function\n\n\[ \varphi \left( x\right) =... | Yes |
Let \( f\left( x\right) = \sin x \) . We shall show that \( {f}^{\prime }\left( x\right) = \cos x \) . | \[ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\sin \left( {x + h}\right) - \sin x}{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{2\sin \left( \frac{h}{2}\right) \cos \left( {x + \frac{h}{2}}\right) }{h} = \] \[ = \mathop{\lim }\limits_{{h \rightarrow 0}}\cos \left( {x + \frac{h}{2}}\right) \cdot \mathop{\lim ... | Yes |
Example 2. We shall show that \( {\cos }^{\prime }x = - \sin x \) . | Proof.\n\n\[ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{\cos \left( {x + h}\right) - \cos x}{h} = \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{-2\sin \left( \frac{h}{2}\right) \sin \left( {x + \frac{h}{2}}\right) }{h} = \]\n\n\[ = - \mathop{\lim }\limits_{{h \rightarrow 0}}\sin \left( {x + \frac{h}{2}}\right) \cd... | Yes |
We shall show that if \( f\left( t\right) = r\cos {\omega t} \), then \( {f}^{\prime }\left( t\right) = - {r\omega }\sin {\omega t} \) . | \[ \mathop{\lim }\limits_{{h \rightarrow 0}}\frac{r\cos \omega \left( {t + h}\right) - r\cos {\omega t}}{h} = r\mathop{\lim }\limits_{{h \rightarrow 0}}\frac{-2\sin \left( \frac{\omega h}{2}\right) \sin \omega \left( {t + \frac{h}{2}}\right) }{h} = \] \[ = - {r\omega }\mathop{\lim }\limits_{{h \rightarrow 0}}\sin \omeg... | Yes |
Example 4. If \( f\left( t\right) = r\sin {\omega t} \), then \( {f}^{\prime }\left( t\right) = {r\omega }\cos {\omega t} \) . | Proof. The proof is analogous to that of Examples 1 and 3. | No |
The instantaneous velocity and instantaneous acceleration of a point mass. Suppose a point mass is moving in a plane and that in some given coordinate system its motion is described by differentiable functions of time | \[ \mathbf{r}\left( t\right) = \left( {x\left( t\right), y\left( t\right) }\right) \] As we have explained in Subsect. 5.1.1, the velocity of the point at time \( t \) is the vector \[ \mathbf{v}\left( t\right) = \dot{\mathbf{r}}\left( t\right) = \left( {\dot{x}\left( t\right) ,\dot{y}\left( t\right) }\right) \] where ... | Yes |
The optic property of a parabolic mirror. Let us consider the parabola \( y = \frac{1}{2p}{x}^{2}\left( {p > 0\text{, see Fig. 5.4}}\right) \), and construct the tangent to it at the point \( \left( {{x}_{0},{y}_{0}}\right) = \left( {{x}_{0},\frac{1}{2p}{x}_{0}^{2}}\right) \). | Since \( f\left( x\right) = \frac{1}{2p}{x}^{2} \), we have\n\n\[ \n{f}^{\prime }\left( {x}_{0}\right) = \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}\frac{\frac{1}{2p}{x}^{2} - \frac{1}{2p}{x}_{0}^{2}}{x - {x}_{0}} = \frac{1}{2p}\mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}\left( {x + {x}_{0}}\right) = \frac{1}{p}... | Yes |
With this example we shall show that the tangent is merely the best linear approximation to the graph of a function in a neighborhood of the point of tangency and does not necessarily have only one point in common with the curve, as was the case with a circle, or in general, with convex curves. | Let the function be given by\n\n\\[ \nf\\left( x\\right) = \\left\\{ \\begin{matrix} {x}^{2}\\sin \\frac{1}{x},\\text{ if }x \\neq 0, \\\\ 0,\\text{ if }x = 0. \\end{matrix}\\right.\n\\]\n\nThe graph of this function is shown by the thick line in Fig. 5.5.\n\n = \left| x\right| \) ,(Fig. 5.6). Then at the point \( {x}_{0} = 0 \) we have | \[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0} - 0}}\frac{f\left( x\right) - f\left( {x}_{0}\right) }{x - {x}_{0}} = \mathop{\lim }\limits_{{x \rightarrow - 0}}\frac{\left| x\right| - 0}{x - 0} = \mathop{\lim }\limits_{{x \rightarrow - 0}}\frac{-x}{x} = - 1, \] \[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0} + 0}}\f... | Yes |
We shall show that \( {\mathrm{e}}^{x + h} - {\mathrm{e}}^{x} = {\mathrm{e}}^{x}h + o\left( h\right) \) as \( h \rightarrow 0 \) . | \[ {\mathrm{e}}^{x + h} - {\mathrm{e}}^{x} = {\mathrm{e}}^{x}\left( {{\mathrm{e}}^{h} - 1}\right) = {\mathrm{e}}^{x}\left( {h + o\left( h\right) }\right) = {\mathrm{e}}^{x}h + o\left( h\right) . \] Here we have used the formula \( {\mathrm{e}}^{h} - 1 = h + o\left( h\right) \) obtained in Example 39 of Subsect. 3.2.4. | Yes |
If \( a > 0 \), then \( {a}^{x + h} - {a}^{x} = {a}^{h}\left( {\ln a}\right) h + o\left( h\right) \) as \( h \rightarrow 0 \) . Thus \( \mathrm{d}{a}^{x} = {a}^{x}\left( {\ln a}\right) \mathrm{d}x \) and \( \frac{\mathrm{d}{a}^{x}}{\mathrm{\;d}x} = {a}^{x}\ln a. \) | \[ {a}^{x + h} - {a}^{x} = {a}^{x}\left( {{a}^{h} - 1}\right) = {a}^{x}\left( {{\mathrm{e}}^{h\ln a} - 1}\right) = \] \[ = {a}^{x}\left( {h\ln a + o\left( {h\ln a}\right) }\right) = {a}^{x}\left( {\ln a}\right) h + o\left( h\right) \text{ as }h \rightarrow 0. \] | Yes |
If \( x \neq 0 \), then \( \ln \left| {x + h}\right| - \ln \left| x\right| = \frac{1}{x}h + o\left( h\right) \) as \( h \rightarrow 0 \) . Thus \( \mathrm{d}\ln \left| x\right| = \frac{1}{x}\mathrm{d}x \) and \( \frac{\mathrm{d}\ln \left| x\right| }{\mathrm{d}x} = \frac{1}{x} \). | \[ \ln \left| {x + h}\right| - \ln \left| x\right| = \ln \left| {1 + \frac{h}{x}}\right| . \] For \( \left| h\right| < \left| x\right| \) we have \( \left| {1 + \frac{h}{x}}\right| = 1 + \frac{h}{x} \), and so for sufficiently small values of \( h \) we can write \[ \ln \left| {x + h}\right| - \ln \left| x\right| = \ln... | Yes |
If \( x \neq 0 \) and \( 0 < a \neq 1 \), then \( {\log }_{a}\left| {x + h}\right| \) - \( {\log }_{a}\left| x\right| = \frac{1}{x\ln a}h \) + \( o\left( h\right) \) as \( h \rightarrow 0. \) Thus, \( {\mathrm{d}}_{▱}{\log }_{a}\left| x\right| = \frac{1}{x\ln a}\mathrm{d}x \) and \( \frac{\mathrm{d}{\log }_{a}\left| x\... | \[ {\log }_{a}\left| {x + h}\right| - {\log }_{a}\left| x\right| = {\log }_{a}\left| {1 + \frac{h}{x}}\right| = {\log }_{a}\left( {1 + \frac{h}{x}}\right) = \] \[ = \frac{1}{\ln a}\ln \left( {1 + \frac{h}{x}}\right) = \frac{1}{\ln a}\left( {\frac{h}{x} + o\left( \frac{h}{x}\right) }\right) = \frac{1}{x\ln a}h + o\left(... | Yes |
Theorem 1. If functions \( f : X \rightarrow \mathbb{R} \) and \( g : X \rightarrow \mathbb{R} \) are differentiable at a point \( x \in X \), then\na) their sum is differentiable at \( x \), and\n\[{\left( f + g\right) }^{\prime }\left( x\right) = \left( {{f}^{\prime } + {g}^{\prime }}\right) \left( x\right)\] | Proof. In the proof we shall rely on the definition of a differentiable function and the properties of the symbol \( o\left( \cdot \right) \) proved in Subsect. 3.2.4.\n\n\[ \text{a)}\;\left( {f + g}\right) \left( {x + h}\right) - \left( {f + g}\right) \left( x\right) = \left( {f\left( {x + h}\right) + g\left( {x + h}\... | Yes |
Corollary 1. The derivative of a linear combination of differentiable functions equals the same linear combination of the derivatives of these functions. | Proof. Since a constant function is obviously differentiable and has a derivative equal to 0 at every point, taking \( f \equiv \) const \( = c \) in statement b) of Theorem 1, we find \( {\left( cg\right) }^{\prime }\left( x\right) = c{g}^{\prime }\left( x\right) \) .\n\nNow, using statement a) of Theorem 1, we can wr... | Yes |
Corollary 2. If the functions \( {f}_{1},\ldots ,{f}_{n} \) are differentiable at \( x \), then\n\n\[ \n{\left( {f}_{1}\cdots {f}_{n}\right) }^{\prime }\left( x\right) = {f}_{1}^{\prime }\left( x\right) {f}_{2}\left( x\right) \cdots {f}_{n}\left( x\right) +\n\]\n\n\[ \n+ {f}_{1}\left( x\right) {f}_{2}^{\prime }\left( x... | Proof. For \( n = 1 \) the statement is obvious.\n\nIf it holds for some \( n \in \mathbb{N} \), then by statement b) of Theorem 1 it also holds for \( \left( {n + 1}\right) \in \mathbb{N} \) . By the principle of induction, we conclude that the formula is valid for any \( n \in \mathbb{N} \) . | No |
Invariance of the definition of velocity. We are now in a position to verify that the instantaneous velocity vector of a point mass defined in Subsect. 5.1.1 is independent of the Cartesian coordinate system used to define it. In fact we shall verify this for all affine coordinate systems. | Let \( \left( {{x}^{1},{x}^{2}}\right) \) and \( \left( {{\widetilde{x}}^{1},{\widetilde{x}}^{2}}\right) \) be the coordinates of the same point of the plane in two different coordinate systems connected by the relations\n\n\[ \n{\widetilde{x}}^{1} = {a}_{1}^{1}{x}^{1} + {a}_{2}^{1}{x}^{2} + {b}^{1}, \n\]\n\n\[ \n{\wid... | Yes |
Let \( f\left( x\right) = \tan x \) . We shall show that \( {f}^{\prime }\left( x\right) = \frac{1}{{\cos }^{2}x} \) at every point where \( \cos x \neq 0 \), that is, in the domain of definition of the function \( \tan x = \frac{\sin x}{\cos x} \) . | It was shown in Examples 1 and 2 of Sect. 5.1 that \( {\sin }^{\prime }\left( x\right) = \cos x \) and \( {\cos }^{\prime }x = - \sin x \), so that by statement c) of Theorem 1 we find, when \( \cos x \neq 0 \), \[ {\tan }^{\prime }x = {\left( \frac{\sin }{\cos }\right) }^{\prime }\left( x\right) = \frac{{\sin }^{\prim... | Yes |
Example 3. \( {\cot }^{\prime }x = - \frac{1}{{\sin }^{2}x} \) wherever \( \sin x \neq 0 \), that is, in the domain of definition of \( \cot x = \frac{\cos x}{\sin x} \) . | Indeed,\n\n\[ \n{\cot }^{\prime }x = {\left( \frac{\cos }{\sin }\right) }^{\prime }\left( x\right) = \frac{{\cos }^{\prime }x\sin x - \cos x{\sin }^{\prime }x}{{\sin }^{2}x} = \n\]\n\n\[ \n= \frac{-\sin x\sin x - \cos x\cos x}{{\sin }^{2}x} = - \frac{1}{{\sin }^{2}x}\text{. } \n\] | Yes |
Example 4. If \( P\left( x\right) = {c}_{0} + {c}_{1}x + \cdots + {c}_{n}{x}^{n} \) is a polynomial, then \( {P}^{\prime }\left( x\right) = \) \( {c}_{1} + 2{c}_{2}x + \cdots + n{c}_{n}{x}^{n - 1}. \) | Indeed, since \( \frac{\mathrm{d}x}{\mathrm{\;d}x} = 1 \), by Corollary 2 we have \( \frac{\mathrm{d}{x}^{n}}{\mathrm{\;d}x} = n{x}^{n - 1} \), and the statement now follows from Corollary 1. | Yes |
The derivative \( {\left( g \circ f\right) }^{\prime }\left( x\right) \) of the composition of differentiable real-valued functions equals the product \( {g}^{\prime }\left( {f\left( x\right) }\right) \cdot {f}^{\prime }\left( x\right) \) of the derivatives of these functions computed at the corresponding points. | There is a strong temptation to give a short proof of this last statement in Leibniz’ notation for the derivative, in which if \( z = z\left( y\right) \) and \( y = y\left( x\right) \), we have\n\n\[\n\frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx}\n\]\nwhich appears to be completely natural, if one regards the symbo... | Yes |
Corollary 5. If the composition \( \left( {{f}_{n} \circ \cdots \circ {f}_{1}}\right) \left( x\right) \) of differentiable functions \( {y}_{1} = {f}_{1}\left( x\right) ,\ldots ,{y}_{n} = {f}_{n}\left( {y}_{n - 1}\right) \) exists, then\n\n\[{\left( {f}_{n} \circ \cdots \circ {f}_{1}\right) }^{\prime }\left( x\right) =... | Proof. The statement is obvious if \( n = 1 \) .\n\nIf it holds for some \( n \in \mathbb{N} \), then by Theorem 2 it also holds for \( n + 1 \), so that by the principle of induction, it holds for any \( n \in \mathbb{N} \) . | Yes |
Let us show that for \( \alpha \in \mathbb{R} \) we have \( \frac{\mathrm{d}{x}^{\alpha }}{\mathrm{d}x} = \alpha {x}^{\alpha - 1} \) in the domain \( x > 0 \), that is, \( \mathrm{d}{x}^{\alpha } = \alpha {x}^{\alpha - 1}\mathrm{\;d}x \) and\n\n\[{\left( x + h\right) }^{\alpha } - {x}^{\alpha } = \alpha {x}^{\alpha - 1... | Proof. We write \( {x}^{\alpha } = {\mathrm{e}}^{\alpha \ln x} \) and apply the theorem, taking account of the results of Examples 9 and 11 from Sect. 5.1 and statement b) of Theorem 1. Let \( g\left( y\right) = {\mathrm{e}}^{y} \) and \( y = f\left( x\right) = \alpha \ln \left( x\right) \) . Then \( {x}^{\alpha } = \l... | Yes |
The derivative of the logarithm of the absolute value of a differentiable function is often called its logarithmic derivative. | Since \( F\left( x\right) = \ln \left| {f\left( x\right) }\right| = \left( {\ln \circ \left| \right| \circ f}\right) \left( x\right) \), by Example 11 of Sect. 5.1, we have \( {F}^{\prime }\left( x\right) = {\left( \ln \left| f\right| \right) }^{\prime }\left( x\right) = \frac{\dot{{f}^{\prime }\left( x\right) }}{f\lef... | Yes |
The absolute and relative errors in the value of a differentiable function caused by errors in the data for the argument. | If the function \( f \) is differentiable at \( x \), then\n\n\[ f\left( {x + h}\right) - f\left( x\right) = {f}^{\prime }\left( x\right) h + \alpha \left( {x;h}\right) ,\]\n\nwhere \( \alpha \left( {x;h}\right) = o\left( h\right) \) as \( h \rightarrow 0 \) .\n\nThus, if in computing the value \( f\left( x\right) \) o... | Yes |
Let us differentiate a function \( u{\left( x\right) }^{v\left( x\right) } \), where \( u\left( x\right) \) and \( v\left( x\right) \) are differentiable functions and \( u\left( x\right) > 0 \). | We write \( u{\left( x\right) }^{v\left( x\right) } = {\mathrm{e}}^{v\left( x\right) \ln u\left( x\right) } \) and use Corollary 5. Then\n\n\[ \frac{\mathrm{d}{\mathrm{e}}^{v\left( x\right) \ln u\left( x\right) }}{\mathrm{d}x} = {\mathrm{e}}^{v\left( x\right) \ln u\left( x\right) }\left( {{v}^{\prime }\left( x\right) \... | Yes |
Theorem 3. (The derivative of an inverse function). Let the functions \( f \) : \( X \rightarrow Y \) and \( {f}^{-1} : Y \rightarrow X \) be mutually inverse and continuous at points \( {x}_{0} \in X \) and \( f\left( {x}_{0}\right) = {y}_{0} \in Y \) respectively. If \( f \) is differentiable at \( {x}_{0} \) and \( ... | Proof. Since the functions \( f : X \rightarrow Y \) and \( {f}^{-1} : Y \rightarrow X \) are mutually inverse, the quantities \( f\left( x\right) - f\left( {x}_{0}\right) \) and \( {f}^{-1}\left( y\right) - {f}^{-1}\left( {y}_{0}\right) \), where \( y = f\left( x\right) \), are both nonzero if \( x \neq {x}_{0} \) . I... | Yes |
We shall show that \( {\arcsin }^{\prime }y = \frac{1}{\sqrt{1 - {y}^{2}}} \) for \( \left| y\right| < 1 \). | The functions \( \sin : \left\lbrack {-\pi /2,\pi /2}\right\rbrack \rightarrow \left\lbrack {-1,1}\right\rbrack \) and arcsin \( : \left\lbrack {-1,1}\right\rbrack \rightarrow \left\lbrack {-\pi /2,\pi /2}\right\rbrack \) are mutually inverse and continuous (see Example 8 of Sect. 4.2) and \( {\sin }^{\prime }\left( x\... | Yes |
\[ {\arccos }^{\prime }y = - \frac{1}{\sqrt{1 - {y}^{2}}}\text{ for }\left| y\right| < 1. \] | Indeed, \[ {\arccos }^{\prime }y = \frac{1}{{\cos }^{\prime }x} = - \frac{1}{\sin x} = - \frac{1}{\sqrt{1 - {\cos }^{2}x}} = - \frac{1}{\sqrt{1 - {y}^{2}}}. \] The sign in front of the radical is chosen taking account of the inequality \( \sin x > 0 \) if \( 0 < x < \pi \) . | Yes |
Example 11. \( {\arctan }^{\prime }y = \frac{1}{1 + {y}^{2}}, y \in \mathbb{R} \) | Indeed,\n\n\[ \n{\arctan }^{\prime }y = \frac{1}{{\tan }^{\prime }x} = \frac{1}{\left( \frac{1}{{\cos }^{2}x}\right) } = {\cos }^{2}x = \frac{1}{1 + {\tan }^{2}x} = \frac{1}{1 + {y}^{2}}. \n\] | Yes |
Example 12. \( {\operatorname{arccot}}^{\prime }y = - \frac{1}{1 + {y}^{2}}, y \in \mathbb{R} \) . | Indeed\n\n\[ \n{\operatorname{arccot}}^{\prime }y = \frac{1}{{\cot }^{\prime }x} = \frac{1}{\left( \text{ } - \frac{1}{{\sin }^{2}x}\right) } = - {\sin }^{2}x = - \frac{1}{1 + {\cot }^{2}x} = - \frac{1}{1 + {y}^{2}}. \n\] | Yes |
We already know (see Examples 10 and 12 of Sect. 5.1) that the functions \( y = f\left( x\right) = {a}^{x} \) and \( x = {f}^{-1}\left( y\right) = {\log }_{a}y \) have the derivatives \( {f}^{\prime }\left( x\right) = {a}^{x}\ln a \) and \( {\left( {f}^{-1}\right) }^{\prime }\left( y\right) = \frac{1}{y\ln a}. \) | Let us see how this is consistent with Theorem 3:\n\n\[ \n{\left( {f}^{-1}\right) }^{\prime }\left( y\right) = \frac{1}{{f}^{\prime }\left( x\right) } = \frac{1}{{a}^{x}\ln a} = \frac{1}{y\ln a}, \n\] \n\n\[ \n{f}^{\prime }\left( x\right) = \frac{1}{{\left( {f}^{-1}\right) }^{\prime }\left( y\right) } = \frac{1}{\left(... | Yes |
The hyperbolic and inverse hyperbolic functions and their derivatives. | The functions\n\n\[ \sinh x = \frac{1}{2}\left( {{\mathrm{e}}^{x} - {\mathrm{e}}^{-x}}\right) \]\n\n\[ \cosh x = \frac{1}{2}\left( {{\mathrm{e}}^{x} + {\mathrm{e}}^{-x}}\right) \]\n\nare called respectively the hyperbolic sine and hyperbolic cosine \( {}^{8} \) of \( x \) .\n\nThese functions, which for the time being ... | Yes |
The law of addition of velocities. The motion of a point along a line is completely determined if we know the coordinate \( x \) of the point in our chosen coordinate system (the real line) at each instant \( t \) in a system we have chosen for measuring time. Thus the pair of numbers \( \left( {x, t}\right) \) determi... | Using the rule for differentiating an implicit function and formula (5.31), we have \[ \frac{\mathrm{d}\widetilde{x}}{\mathrm{\;d}\widetilde{t}} = \frac{\frac{\mathrm{d}\widetilde{x}}{\mathrm{\;d}t}}{\frac{\mathrm{d}\widetilde{t}}{\mathrm{\;d}t}} = \frac{\alpha \frac{\mathrm{d}x}{\mathrm{\;d}t} + \beta }{\gamma \frac{\... | Yes |
Let \( u\left( x\right) \) and \( v\left( x\right) \) be functions having derivatives up to order \( n \) inclusive on a common set \( E \) . The following formula of Leibniz holds for the \( n \) th derivative of their product:\n\n\[ \n{\left( uv\right) }^{\left( n\right) } = \mathop{\sum }\limits_{{m = 0}}^{n}\left( ... | Proof. For \( n = 1 \) formula (5.44) agrees with the rule already established for the derivative of a product.\n\nIf the functions \( u \) and \( v \) have derivatives up to order \( n + 1 \) inclusive, then, assuming that formula (5.44) holds for order \( n \), after differentiating the left-and right-hand sides, we ... | Yes |
If \( {P}_{n}\left( x\right) = {c}_{0} + {c}_{1}x + \cdots + {c}_{n}{x}^{n} \), then | \[ {P}_{n}\left( 0\right) = {c}_{0} \]\n\[ {P}_{n}^{\prime }\left( x\right) = {c}_{1} + 2{c}_{2}x + \cdots + n{c}_{n}{x}^{n - 1}\text{ and }{P}_{n}^{\prime }\left( 0\right) = {c}_{1}, \]\n\[ {P}_{n}^{\prime \prime }\left( x\right) = 2{c}_{2} + 3 \cdot 2{c}_{3}x + \cdots + n\left( {n - 1}\right) {c}_{n}{x}^{n - 2}\text{... | Yes |
Using Leibniz' formula and the fact that all the derivatives of a polynomial of order higher than the degree of the polynomial are zero, we can find the \( n \) th derivative of \( f\left( x\right) = {x}^{2}\sin x \) : | \[ {f}^{\left( n\right) }\left( x\right) = {\sin }^{\left( n\right) }\left( x\right) \cdot {x}^{2} + \left( \begin{array}{l} n \\ 1 \end{array}\right) {\sin }^{\left( n - 1\right) }x \cdot {2x} + \left( \begin{array}{l} n \\ 2 \end{array}\right) {\sin }^{\left( n - 2\right) }x \cdot 2 = \] \[ = {x}^{2}\sin \left( {x + ... | Yes |
Let \( f\left( x\right) = \arctan x \) . Let us find the values \( {f}^{\left( n\right) }\left( 0\right) \) \( \left( {n = 1,2,\ldots }\right) \) . | Since \( {f}^{\prime }\left( x\right) = \frac{1}{1 + {x}^{2}} \), it follows that \( \left( {1 + {x}^{2}}\right) {f}^{\prime }\left( x\right) = 1. \n\nApplying Leibniz' formula to this last equality, we find the recursion relation\n\n\[ \left( {1 + {x}^{2}}\right) {f}^{\left( n + 1\right) }\left( x\right) + {2nx}{f}^{\... | Yes |
If \( x = x\left( t\right) \) denotes the time dependence of a point mass moving along the real line, then \( \frac{\mathrm{d}x\left( t\right) }{\mathrm{d}t} = \dot{x}\left( t\right) \) is the velocity of the point, and then \( \frac{\mathrm{d}\dot{x}\left( t\right) }{\mathrm{d}t} = \frac{{\mathrm{d}}^{2}x\left( t\righ... | If \( x\left( t\right) = {\alpha t} + \beta \), then \( \dot{x}\left( t\right) = \alpha \) and \( \ddot{x}\left( t\right) \equiv 0 \), that is, the acceleration in a uniform motion is zero. We shall soon verify that if the second derivative equals zero, then the function itself has the form \( {\alpha t} + \beta \) . T... | Yes |
Example 29. The second derivative of a simple implicit function. Let \( y = y\left( t\right) \) and \( x = x\left( t\right) \) be twice-differentiable functions. Assume that the function \( x = x\left( t\right) \) has a differentiable inverse function \( t = t\left( x\right) \) . Then the quantity \( y\left( t\right) \... | By the rule for differentiating such a function, studied in Subsect. 5.2.5, we have\n\n\[ \n{y}_{x}^{\prime } = \frac{{y}_{t}^{\prime }}{{x}_{t}^{\prime }} \n\] \n\nso that \n\n\[ \n{y}_{xx}^{\prime \prime } = {\left( {y}_{x}^{\prime }\right) }_{x}^{\prime } = \frac{{\left( {y}_{x}^{\prime }\right) }_{t}^{\prime }}{{x}... | Yes |
Example 1. Let\n\n\[ f\\left( x\\right) = \\left\\{ \\begin{array}{l} {x}^{2},\\text{ if } - 1 \\leq x < 2 \\\\ 4,\\text{ if }2 \\leq x \\end{array}\\right. \]\n\n(see Fig. 5.8). For this function | \n\( x = - 1 \) is a strict local maximum;\n\n\( x = 0 \) is a strict local minimum;\n\n\( x = 2 \) is a local maximum;\n\nthe points \( x > 2 \) are all local extrema, being simultaneously maxima and minima, since the function is locally constant at these points. | Yes |
Example 2. Let \( f\left( x\right) = \sin \frac{1}{x} \) on the set \( E = \mathbb{R} \smallsetminus 0 \). | The points \( x = {\left( \frac{\pi }{2} + 2k\pi \right) }^{-1}, k \in \mathbb{Z} \), are strict local maxima, and the points \( x = {\left( -\frac{\pi }{2} + 2k\pi \right) }^{-1}, k \in \mathbb{Z} \), are strict local minima for \( f\left( x\right) \) (see Fig. 4.1). | Yes |
Lemma 1. (Fermat). If a function \( f : E \rightarrow \mathbb{R} \) is differentiable at an interior extremum, \( {x}_{0} \in E \), then its derivative at \( {x}_{0} \) is \( 0 : {f}^{\prime }\left( {x}_{0}\right) = 0 \) . | Proof. By definition of differentiability at \( {x}_{0} \) we have\n\n\[ f\left( {{x}_{0} + h}\right) - f\left( {x}_{0}\right) = {f}^{\prime }\left( {x}_{0}\right) h + \alpha \left( {{x}_{0};h}\right) h, \]\n\nwhere \( \alpha \left( {{x}_{0};h}\right) \rightarrow 0 \) as \( h \rightarrow x,{x}_{0} + h \in E \) .\n\nLet... | Yes |
Proposition 1. (Rolle’s \( {}^{10} \) theorem). If a function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is continuous on a closed interval \( \left\lbrack {a, b}\right\rbrack \) and differentiable on the open interval \( \rbrack a, b\lbrack \) and \( f\left( a\right) = f\left( b\right) \), then ... | Proof. Since the function \( f \) is continuous on \( \left\lbrack {a, b}\right\rbrack \), there exist points \( {x}_{m},{x}_{M} \in \) \( \left\lbrack {a, b}\right\rbrack \) at which it assumes its minimal and maximal values respectively. If \( f\left( {x}_{m}\right) = f\left( {x}_{M}\right) \), then the function is c... | Yes |
Theorem 1. (Lagrange’s finite-increment theorem). If a function \( f : \left\lbrack {a, b}\right\rbrack \) \( \rightarrow \mathbb{R} \) is continuous on a closed interval \( \left\lbrack {a, b}\right\rbrack \) and differentiable on the open interval \( \rbrack a, b\left\lbrack \right. \), there exists a point \( \xi \i... | Proof. Consider the auxiliary function\n\n\[ F\left( x\right) = f\left( x\right) - \frac{f\left( b\right) - f\left( a\right) }{b - a}\left( {x - a}\right) ,\]\n\nwhich is obviously continuous on the closed interval \( \left\lbrack {a, b}\right\rbrack \) and differentiable on the open interval \( \rbrack a, b\lbrack \) ... | Yes |
Corollary 1. (Criterion for monotonicity of a function). If the derivative of a function is nonnegative (resp. positive) at every point of an open interval, then the function is nondecreasing (resp. increasing) on that interval. | Proof. Indeed, if \( {x}_{1} \) and \( {x}_{2} \) are two points of the interval and \( {x}_{1} < {x}_{2} \), that is, \( {x}_{2} - {x}_{1} > 0 \), then by formula (5.46)\n\n\[ f\left( {x}_{2}\right) - f\left( {x}_{1}\right) = {f}^{\prime }\left( \xi \right) \left( {{x}_{2} - {x}_{1}}\right) ,\text{ where }{x}_{1} < \x... | Yes |
Corollary 2. (Criterion for a function to be constant). A function that is continuous on a closed interval \( \left\lbrack {a, b}\right\rbrack \) is constant on it if and only if its derivative equals zero at every point of the interval \( \left\lbrack {a, b}\right\rbrack \) (or only the open interval \( \rbrack a, b\l... | Proof. Only the fact that \( {f}^{\prime }\left( x\right) \equiv 0 \) on \( \rbrack a, b\lbrack \) implies that \( f\left( {x}_{1}\right) = f\left( {x}_{2}\right) \) for all \( {x}_{1},{x}_{2}, \in \left\lbrack {a, b}\right\rbrack \) is of interest. But this follows from Lagrange’s formula, according to which\n\n\[ f\l... | Yes |
Proposition 2. (Cauchy’s finite-increment theorem). Let \( x = x\left( t\right) \) and \( y = y\left( t\right) \) be functions that are continuous on a closed interval \( \left\lbrack {\alpha ,\beta }\right\rbrack \) and differentiable on the open interval \( \rbrack \alpha ,\beta \lbrack \) . Then there exists a point... | Proof. The function \( F\left( t\right) = x\left( t\right) \left( {y\left( \beta \right) - y\left( \alpha \right) }\right) - y\left( t\right) \left( {x\left( \beta \right) - x\left( \alpha \right) }\right) \) satisfies the hypotheses of Rolle’s theorem on the closed interval \( \left\lbrack {\alpha ,\beta }\right\rbrac... | Yes |
Theorem 2. If the function \( f \) is continuous on the closed interval with endpoints \( {x}_{0} \) and \( x \) along with its first \( n \) derivatives, and it has a derivative of order \( n + 1 \) at the interior points of this interval, then for any function \( \varphi \) that is continuous on this closed interval ... | Proof. On the closed interval \( I \) with endpoints \( {x}_{0} \) and \( x \) we consider the auxiliary function\n\n\[ \nF\left( t\right) = f\left( x\right) - {P}_{n}\left( {t;x}\right) \n\]\n\n\( \left( {5.53}\right) \)\n\nof the argument \( t \) . We now write out the definition of the function \( F\left( t\right) \... | Yes |
For the function \( f\left( x\right) = {\mathrm{e}}^{x} \) with \( {x}_{0} = 0 \) Taylor’s formula has the form \[ {\mathrm{e}}^{x} = 1 + \frac{1}{1!}x + \frac{1}{2!}{x}^{2} + \cdots + \frac{1}{n!}{x}^{n} + {r}_{n}\left( {0;x}\right) ,\] and by (5.56) we can assume that \[ {r}_{n}\left( {0;x}\right) = \frac{1}{\left( {... | Thus \[ \left| {{r}_{n}\left( {0;x}\right) }\right| = \frac{1}{\left( {n + 1}\right) !}{\mathrm{e}}^{\xi } \cdot {\left| x\right| }^{n + 1} < \frac{{\left| x\right| }^{n + 1}}{\left( {n + 1}\right) !}{\mathrm{e}}^{\left| x\right| }. \] But for each fixed \( x \in \mathbb{R} \), if \( n \rightarrow \infty \), the quanti... | Yes |
We obtain the expansion of the function \( {a}^{x} \) for any \( a,0 < a \) , \( a \neq 1 \) | \[ {a}^{x} = 1 + \frac{\ln a}{1!}x + \frac{{\ln }^{2}a}{2!}{x}^{2} + \cdots + \frac{{\ln }^{n}a}{n!}{x}^{n} + \cdots . \] | Yes |
Let \( f\left( x\right) = \sin x \) . We know (see Example 18 of Subsect. 5.2.6) that \( {f}^{\left( n\right) }\left( x\right) = \sin \left( {x + \frac{\pi }{2}n}\right) ,\;n \in \mathbb{N} \), and so by Lagrange’s formula (5.56) with \( {x}_{0} = 0 \) and any \( x \in \mathbb{R} \) we find | \[ {r}_{n}\left( {0;x}\right) = \frac{1}{\left( {n + 1}\right) !}\sin \left( {\xi + \frac{\pi }{2}\left( {n + 1}\right) }\right) {x}^{n + 1}, \] from which it follows that \( {r}_{n}\left( {0;x}\right) \) tends to zero for any \( x \in \mathbb{R} \) as \( n \rightarrow \infty \) . Thus we have the expansion \[ \sin x =... | Yes |
Since \( {\sinh }^{\prime }x = \cosh x \) and \( {\cosh }^{\prime }x = \sinh x \), formula (5.56) yields the following expression for the remainder in the Taylor series of \( f\left( x\right) = \sinh x \) : \[ {r}_{n}\left( {0;x}\right) = \frac{1}{\left( {n + 1}\right) !}{f}^{\left( n + 1\right) }\left( \xi \right) {x}... | Hence for any given value \( x \in \mathbb{R} \) we have \( {r}_{n}\left( {0;x}\right) \rightarrow 0 \) as \( n \rightarrow \infty \), and we obtain the expansion \[ \sinh x = x + \frac{1}{3!}{x}^{3} + \frac{1}{5!}{x}^{5} + \cdots + \frac{1}{\left( {{2n} + 1}\right) !}{x}^{{2n} + 1} + \cdots , \] valid for all \( x \in... | Yes |
For the function \( f\left( x\right) = \ln \left( {1 + x}\right) \) we have \( {f}^{\left( n\right) }\left( x\right) = \frac{{\left( -1\right) }^{n - 1}\left( {n - 1}\right) !}{{\left( 1 + x\right) }^{n}}, \) so that the Taylor series of this function at \( {x}_{0} = 0 \) is\n\n\[ \ln \left( {1 + x}\right) = x - \frac{... | This time we represent \( {r}_{n}\left( {0;x}\right) \) using Cauchy’s formula (5.55):\n\n\[ {r}_{n}\left( {0;x}\right) = \frac{1}{n!}\frac{{\left( -1\right) }^{n}n!}{{\left( 1 + \xi \right) }^{n}}{\left( x - \xi \right) }^{n}x \]\n\nor\n\n\[ {r}_{n}\left( {0;x}\right) = {\left( -1\right) }^{n}x{\left( \frac{x - \xi }{... | Yes |
For the function \( {\left( 1 + x\right) }^{\alpha } \), where \( \alpha \in \mathbb{R} \), we have \( {f}^{\left( n\right) }\left( x\right) = \) \( \alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - n + 1}\right) {\left( 1 + x\right) }^{\alpha - n} \), so that Taylor’s formula at \( {x}_{0} = 0 \) for this func... | Using Cauchy's formula (5.55), we find\n\n\[ \n{r}_{n}\left( {0;x}\right) = \frac{\alpha \left( {\alpha - 1}\right) \cdots \left( {\alpha - n}\right) }{n!}{\left( 1 + \xi \right) }^{\alpha - n - 1}{\left( x - \xi \right) }^{n}x,\n\] \n\nwhere \( \xi \) lies between 0 and \( x \) .\n\nIf \( \left| x\right| < 1 \), then,... | Yes |
Proposition 3. If there exists a polynomial \( {P}_{n}\left( {{x}_{0};x}\right) = {c}_{0} + {c}_{1}\left( {x - {x}_{0}}\right) + \) \( \cdots + {c}_{n}{\left( x - {x}_{0}\right) }^{n} \) satisfying condition (5.76), that polynomial is unique. | Proof. Indeed, from relation (5.76) we obtain the coefficients of the polynomial successively and completely unambiguously\n\n\[ \n{c}_{0} = \mathop{\lim }\limits_{{E \ni x \rightarrow {x}_{0}}}f\left( x\right) \n\] \n\n\[ \n\begin{matrix} {c}_{1} & = & \mathop{\lim }\limits_{{E \ni x \rightarrow {x}_{0}}}\frac{f\left(... | Yes |
We shall write a polynomial that makes it possible to compute the values of \( \sin x \) on the interval \( - 1 \leq x \leq 1 \) with absolute error at most \( {10}^{-3} \) . | One can take this polynomial to be a Taylor polynomial of suitable degree obtained from the expansion of \( \sin x \) in a neighborhood of \( {x}_{0} = 0 \) . Since\n\n\[ \sin x = x - \frac{1}{3!}{x}^{3} + \frac{1}{5!}{x}^{5} - \cdots + \frac{{\left( -1\right) }^{n}}{\left( {{2n} + 1}\right) !}{x}^{{2n} + 1} + 0 \cdot ... | Yes |
We shall show that \( \tan x = x + \frac{1}{3}{x}^{3} + o\left( {x}^{3}\right) \) as \( x \rightarrow 0 \) . | We have\n\n\[ \n{\tan }^{\prime }x = {\cos }^{-2}x \n\] \n\n\[ \n{\tan }^{\prime \prime }x = 2{\cos }^{-3}x\sin x \n\] \n\n\[ \n{\tan }^{\prime \prime \prime }x = 6{\cos }^{-4}x{\sin }^{2}x + 2{\cos }^{-2}x. \n\] \n\nThus, \( \tan 0 = 0,{\tan }^{\prime }0 = 1,{\tan }^{\prime \prime }0 = 0,{\tan }^{\prime \prime \prime ... | Yes |
Let \( \alpha > 0 \) . Let us study the convergence of the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\ln \cos \frac{1}{{n}^{\alpha }} \) . | For \( \alpha > 0 \) we have \( \frac{1}{{n}^{\alpha }} \rightarrow 0 \) as \( n \rightarrow \infty \) . Let us estimate the order of a term of the series:\n\n\[ \ln \cos \frac{1}{{n}^{\alpha }} = \ln \left( {1 - \frac{1}{2!} \cdot \frac{1}{{n}^{2\alpha }} + o\left( \frac{1}{{n}^{2\alpha }}\right) }\right) = - \frac{1}... | No |
Let us show that \( \ln \cos x = - \frac{1}{2}{x}^{2} - \frac{1}{12}{x}^{4} - \frac{1}{45}{x}^{6} + O\left( {x}^{8}\right) \) as \( x \rightarrow 0 \) . | This time, instead of computing six successive derivatives, we shall use the already-known expansions of \( \cos x \) as \( x \rightarrow 0 \) and \( \ln \left( {1 + u}\right) \) as \( u \rightarrow 0 \) :\n\n\[ \ln \cos x = \ln \left( {1 - \frac{1}{2!}{x}^{2} + \frac{1}{4!}{x}^{4} - \frac{1}{6!}{x}^{6} + O\left( {x}^{... | Yes |
Let us find the values of the first six derivatives of the function \( \ln \cos x \) at \( x = 0 \) . | We have \( {\left( \ln \cos \right) }^{\prime }x = \frac{-\sin x}{\cos x} \), and it is therefore clear that the function has derivatives of all orders at 0, since \( \cos 0 \neq 0 \) . We shall not try to find functional expressions for these derivatives, but rather we shall make use of the uniqueness of the Taylor po... | Yes |
Let \( f\left( x\right) \) be an infinitely differentiable function at the point \( {x}_{0} \) , and suppose we know the expansion\n\n\[ \n{f}^{\prime }\left( x\right) = {c}_{0}^{\prime } + {c}_{1}^{\prime }x + \cdots + {c}_{n}^{\prime }{x}^{n} + O\left( {x}^{n + 1}\right) \n\]\n\nof its derivative in a neighborhood of... | Thus for the function \( f\left( x\right) \) itself we have the expansion\n\n\[ \nf\left( x\right) = f\left( 0\right) + \frac{{c}_{0}^{\prime }}{1!}x + \frac{1!{c}_{1}^{\prime }}{2!}{x}^{2} + \cdots + \frac{n!{c}_{n}^{\prime }}{\left( {n + 1}\right) !}{x}^{n + 1} + O\left( {x}^{n + 2}\right) , \n\]\n\nor, after simplif... | Yes |
Let us find the Taylor expansion of the function \( f\\left( x\\right) = \\arctan x \) at 0 . | \n\\( {Since}\\begin{aligned} {f}^{\\prime }\\left( x\\right) & = \\frac{1}{1 + {x}^{2}} = {\\left( 1 + {x}^{2}\\right) }^{-1} \\\\\n& = 1 - {x}^{2} + {x}^{4} - \\cdots + {\\left( -1\\right) }^{n}{x}^{2n} + O\\left( {x}^{{2n} + 2}\\right) , \\end{aligned} \\) by the considerations explained in the preceding example,\n\... | Yes |
Similarly, by expanding the function \( {\arcsin }^{\prime }x = {\left( 1 - {x}^{2}\right) }^{-1/2} \) by Taylor's formula in a neighborhood of zero, we find successively, | \[ {\left( 1 + u\right) }^{-1/2} = 1 + \frac{-\frac{1}{2}}{1!}u + \frac{-\frac{1}{2}\left( {-\frac{1}{2} - 1}\right) }{2!}{u}^{2} + \cdots + \] \[ \; + \frac{-\frac{1}{2}\left( {\; - \;\frac{1}{2} - 1}\right) \cdots \left( {\; - \;\frac{1}{2} - n + 1}\right) }{n!}{u}^{n} + O\left( {u}^{n + 1}\right) \;, \] \[ {\left( 1... | Yes |
We use the results of Examples 5, 12, 17, and 18 and find\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\arctan x - \sin x}{\tan x - \arcsin x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\left\lbrack {x - \frac{1}{3}{x}^{3} + O\left( {x}^{5}\right) }\right\rbrack - \left\lbrack {x - \frac{1}{3!}{x}^{3} + O... | \[ = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{-\frac{1}{6}{x}^{3} + O\left( {x}^{5}\right) }{\frac{1}{6}{x}^{3} + O\left( {x}^{5}\right) } = - 1. \] | Yes |
Proposition 1. The following relations hold between the monotonicity properties of a function \( f : E \rightarrow \mathbb{R} \) that is differentiable on an open interval \( \rbrack a, b\lbrack = E \) and the sign (positivity) of its derivative \( {f}^{\prime } \) on that interval:\n\n\[ \n{f}^{\prime }\left( x\right)... | Proof. The left-hand column of implications is already known to us from Lagrange’s theorem, by virtue of which \( f\left( {x}_{2}\right) - f\left( {x}_{1}\right) = {f}^{\prime }\left( \xi \right) \left( {{x}_{2} - {x}_{1}}\right) \), where \( \left. {{x}_{1},{x}_{2} \in }\right\rbrack a, b\left\lbrack \right. \) and \(... | Yes |
Let us find the maximum of \( f\left( x\right) = {x}^{2} \) on the closed interval \( \left\lbrack {-2,1}\right\rbrack \). | It is obvious in this case that the maximum will be attained at the endpoint -2, but here is a systematic procedure for finding the maximum. We find \( {f}^{\prime }\left( x\right) = {2x} \), then we find all points of the open interval \( \rbrack - 2,1\lbrack \) at which \( {f}^{\prime }\left( x\right) = 0 \) . In thi... | Yes |
Proposition 4. (Sufficient conditions for an extremum in terms of higher-order derivatives). Suppose a function \( f : U\left( {x}_{0}\right) \rightarrow \mathbb{R} \) defined on a neighborhood \( U\left( {x}_{0}\right) \) of \( {x}_{0} \) has derivatives of order up to \( n \) inclusive at \( {x}_{0}\left( {n \geq 1}\... | Proof. Using the local Taylor formula \[ f\left( x\right) - f\left( {x}_{0}\right) = {f}^{\left( n\right) }\left( {x}_{0}\right) {\left( x - {x}_{0}\right) }^{n} + \alpha \left( x\right) {\left( x - {x}_{0}\right) }^{n}, \] (5.82) where \( \alpha \left( x\right) \rightarrow 0 \) as \( x \rightarrow {x}_{0} \), we shall... | Yes |
The law of refraction in geometric optics (Snell’s law). According to Fermat's principle, the actual trajectory of a light ray between two points is such that the ray requires minimum time to pass from one point to the other compared with all paths joining the two points. | If \( {c}_{1} \) and \( {c}_{2} \) are the velocities of light in these media, the time required to traverse the path is\n\n\[ t\left( x\right) = \frac{1}{{c}_{1}}\sqrt{{h}_{1}^{2} + {x}^{2}} + \frac{1}{{c}_{2}}\sqrt{{h}_{2}^{2} + {\left( a - x\right) }^{2}}. \]\n\nWe now find the extremum of the function \( t\left( x\... | Yes |
We shall show that for \( x > 0 \)\n\n\[ \n{x}^{\alpha } - {\alpha x} + \alpha - 1 \leq 0,\text{ when }0 < \alpha < 1,\n\]\n\n(5.84)\n\n\[ \n{x}^{\alpha } - {\alpha x} + \alpha - 1 \geq 0\text{, when }\alpha < 0\text{ or }1 < \alpha \text{. }\n\]\n\n(5.85) | Proof. Differentiating the function \( f\left( x\right) = {x}^{\alpha } - {\alpha x} + \alpha - 1 \), we find \( {f}^{\prime }\left( x\right) = \) \( \alpha \left( {{x}^{\alpha - 1} - 1}\right) \) and \( {f}^{\prime }\left( x\right) = 0 \) when \( x = 1 \) . In passing through the point 1 the derivative passes from pos... | Yes |
Let \( f\left( x\right) = \sin x \). Since \( {f}^{\prime }\left( x\right) = \cos x \) and \( {f}^{\prime \prime }\left( x\right) = - \sin x \), all the points where \( {f}^{\prime }\left( x\right) = \cos x = 0 \) are local extrema of \( \sin x \), since \( {f}^{\prime \prime }\left( x\right) = - \sin x \neq 0 \) at th... | Here \( {f}^{\prime \prime }\left( x\right) < 0 \) if \( \sin x > 0 \) and \( {f}^{\prime \prime }\left( x\right) > 0 \) if \( \sin x < 0 \). Thus the points where \( \cos x = 0 \) and \( \sin x > 0 \) are local maxima and those where \( \cos x = 0 \) and \( \sin x < 0 \) are local minima for \( \sin x \) (which, of co... | Yes |
Let us study the convexity of \( f\left( x\right) = {x}^{\alpha } \) on the set \( x > 0 \). | Since \( {f}^{\prime \prime }\left( x\right) = \alpha \left( {\alpha - 1}\right) {x}^{\alpha - 2} \), we have \( {f}^{\prime \prime }\left( x\right) > 0 \) for \( \alpha < 0 \) or \( \alpha > 1 \), that is, for these values of the exponent \( \alpha \) the power function \( {x}^{\alpha } \) is strictly convex (downward... | Yes |
Let us study the convexity of \( f\left( x\right) = \sin x \) (see Fig. 5.14). | Since \( {f}^{\prime \prime }\left( x\right) = - \sin x \), we have \( {f}^{\prime \prime }\left( x\right) < 0 \) on the intervals \( \pi \cdot {2k} < x < \pi \left( {{2k} + 1}\right) \) and \( {f}^{\prime \prime }\left( x\right) > 0 \) on \( \pi \left( {{2k} - 1}\right) < x < \pi \cdot {2k} \), where \( k \in \mathbb{... | Yes |
A function \( f : \rbrack a, b\lbrack \rightarrow \mathbb{R} \) that is differentiable on the open interval \( \rbrack a, b\lbrack \) is convex (downward) on \( \rbrack a, b\lbrack \) if and only if its graph contains no points below any tangent drawn to it. In that case, a necessary and sufficient condition for strict... | Proof. Necessity. Let \( \left. {{x}_{0} \in }\right\rbrack a, b\lbrack \) . The equation of the tangent line to the graph at \( \left( {{x}_{0}, f\left( {x}_{0}\right) }\right) \) has the form\n\n\[ y = f\left( {x}_{0}\right) + {f}^{\prime }\left( {x}_{0}\right) \left( {x - {x}_{0}}\right) ,\]\n\nso that\n\n\[ f\left(... | Yes |
The function \( f\left( x\right) = {\mathrm{e}}^{x} \) is strictly convex. | The straight line \( y = x + 1 \) is tangent to the graph of this function at \( \left( {0,1}\right) \), since \( f\left( 0\right) = {\mathrm{e}}^{0} = 1 \) and \( {f}^{\prime }\left( 0\right) = {\left. {\mathrm{e}}^{x}\right| }_{x = 0} = 1 \) . By Proposition 6 we conclude that for any \( x \in \mathbb{R} \)\n\n\[{\ma... | No |
When considering the function \( f\left( x\right) = \sin x \) in Example 12 we found the regions of convexity and concavity for its graph. We shall now show that the points of the graph with abscissas \( x = {\pi k}, k \in \mathbb{Z} \), are points of inflection. | Indeed, \( {f}^{\prime \prime }\left( x\right) = - \sin x \), so that \( {f}^{\prime \prime }\left( x\right) = 0 \) at \( x = {\pi k}, k \in \mathbb{Z} \) . Moreover, \( {f}^{\prime \prime }\left( x\right) \) changes sign as we pass through these points, which is a sufficient condition for a point of inflection (see Fi... | Yes |
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