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Lemma 2.7.11. If \( A > 0 \) and \( 0 \leq \overrightarrow{\mathbf{v}} \leq \overrightarrow{\mathbf{w}} \) with \( \overrightarrow{\mathbf{v}} \neq \overrightarrow{\mathbf{w}} \), then \( A\overrightarrow{\mathbf{v}} < A\overrightarrow{\mathbf{w}} \) . | Proof. In order to have \( \overrightarrow{\mathbf{v}} \neq \overrightarrow{\mathbf{w}} \), at least one entry of \( \overrightarrow{\mathbf{w}} \) must be larger than the corresponding entry of \( \overrightarrow{\mathbf{v}} \) . That entry will be multiplied by some strictly positive entry of \( A > \mathbf{0} \), le... | Yes |
How do calculators compute the square root of a positive number \( b \) ? | They apply Newton’s method to the equation \( f\left( x\right) = {x}^{2} - b = 0 \). In this case, this means the following: choose \( {a}_{0} \) and plug it into equation 2.8.3. Our equation is in one variable, so we can replace \( \left\lbrack {\mathbf{D}\overrightarrow{\mathbf{f}}\left( {\mathbf{a}}_{0}\right) }\rig... | Yes |
Let's apply Newton's method to the equation\n\n\\[ \n{x}^{3} - x + \\frac{\\sqrt{2}}{2} = 0 \n\\] | starting at \\( x = 0 \\) (i.e., our \ | No |
Consider the mapping \( \mathbf{f} : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) \( \mathbf{f}\left( \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right) = \left( \begin{array}{l} {x}_{1} - {x}_{2}^{2} \\ {x}_{1}^{2} + {x}_{2} \end{array}\right) \) with derivative \( \left\lbrack {\mathbf{{Df}}\left( \begin{arra... | Given two points \( \mathbf{x} \) and \( \mathbf{y} \) , \[ \left\lbrack {\mathbf{{Df}}\left( \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right) }\right\rbrack - \left\lbrack {\mathbf{{Df}}\left( \begin{array}{l} {y}_{1} \\ {y}_{2} \end{array}\right) }\right\rbrack = \left\lbrack \begin{matrix} 0 & - 2\left( {{x}_{... | Yes |
Example 2.8.6 (Lipschitz ratio: a more complicated case). Consider the mapping \( \mathbf{f} : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) given by \[ \mathbf{f}\left( \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right) = \left( \begin{array}{l} {x}_{1} - {x}_{2}^{3} \\ {x}_{1}^{3} + {x}_{2} \end{array}\right) ... | Given two points \( \mathbf{x} \) and \( \mathbf{y} \), we have \[ \left\lbrack {\mathbf{{Df}}\left( \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right) }\right\rbrack - \left\lbrack {\mathbf{{Df}}\left( \begin{array}{l} {y}_{1} \\ {y}_{2} \end{array}\right) }\right\rbrack = \left\lbrack \begin{matrix} 0 & - 3\left(... | Yes |
Let \( f \) be the function\n\n\[ f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {2x} + x{y}^{3} + {2y}{z}^{2}. \text{ Then } {D}_{2}\left( {{D}_{1}f}\right) \left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {D}_{2}\underset{{D}_{1}f}{\underbrace{\left( 2 + {y}^{3}\right) }} = 3{y}^{2}. \] | \[ \text{Similarly,}{D}_{3}\left( {{D}_{2}f}\right) \left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {D}_{3}\underset{{D}_{2}f}{\underbrace{\left( 3x{y}^{2} + 2{z}^{2}\right) }} = {4z}. \bigtriangleup \] | No |
Let's see how much easier it is to find a Lipschitz ratio in Example 2.8.6 using second partial derivatives. First we compute the first and second derivatives, for \( {f}_{1} = {x}_{1} - {x}_{2}^{3} \) and \( {f}_{2} = {x}_{1}^{3} + {x}_{2} \) : | \[ {D}_{1}{f}_{1} = 1;\;{D}_{2}{f}_{1} = - 3{x}_{2}^{2};\;{D}_{1}{f}_{2} = 3{x}_{1}^{2};\;{D}_{2}{f}_{2} = 1. \] This gives \[ {D}_{1}{D}_{1}{f}_{1} = 0;\;{D}_{1}{D}_{2}{f}_{1} = {D}_{2}{D}_{1}{f}_{1} = 0;\;{D}_{2}{D}_{2}{f}_{1} = - 6{x}_{2} \] \[ {D}_{1}{D}_{1}{f}_{2} = 6{x}_{1};\;{D}_{1}{D}_{2}{f}_{2} = {D}_{2}{D}_{1... | Yes |
Example 2.8.11 (Finding a Lipschitz ratio using second derivatives: a second example). Let us find a Lipschitz ratio for the derivative of the function\n\n\\[ \n\\mathbf{F}\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) = \\left( \\begin{matrix} \\sin \\left( {x + y}\\right) \\\\ \\cos \\left( {xy}\\right) \\en... | We compute\n\n\\[ \n{D}_{1}{D}_{1}{F}_{1} = {D}_{2}{D}_{2}{F}_{1} = {D}_{2}{D}_{1}{F}_{1} = {D}_{1}{D}_{2}{F}_{1} = - \\sin \\left( {x + y}\\right) ,\n\\]\n\n\\[ \n{D}_{1}{D}_{1}{F}_{2} = - {y}^{2}\\cos \\left( {xy}\\right) ,\\;{D}_{2}{D}_{1}{F}_{2} = {D}_{1}{D}_{2}{F}_{2} = - \\left( {\\sin \\left( {xy}\\right) + {yx}... | Yes |
Proposition 2.8.14 (Unique solution in \( \overline{{U}_{0}} \) ). Define\n\n\[ \n{U}_{0}\overset{\text{ def }}{ = }\left\{ {\mathbf{x}\left| \right| \mathbf{x} - {\mathbf{a}}_{0}\left| { < 2}\right| {\overrightarrow{\mathbf{h}}}_{0} \mid }\right\} \n\]\n\nThen if the Lipschitz condition\n\n\[ \n\left| {\left\lbrack {\... | The proposition is justified in Appendix A5, when we prove the Kan-torovich theorem. | Yes |
Example 2.8.16: If we wanted to solve equation 2.8.64 without Newton's method, by considering \( X = \left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack \) as a point in \( {\mathbb{R}}^{4} \), we would need to solve four quadratic equations in four unknowns. It's not obvious how to go about it. | There is an obvious initial guess \( {A}_{0} = \left\lbrack \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}\right\rbrack . \) Since \( f\left( {A}_{0}\right) = \left\lbrack \begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}\right\rbrack , \) we have \( \left| {f\left( {A}_{0}\right) }\right| = 2 \) . Now we compute the other qu... | Yes |
Example 2.8.17 (Newton's method, using a computer). Now we will use the MATLAB program NEWTON. \( {\mathrm{M}}^{23} \) to solve the equations\n\n\[ \n{x}^{2} - y + \sin \left( {x - y}\right) = 2\text{ and }{y}^{2} - x = 3 \n\]\n\nstarting at \( \left( \begin{array}{l} 2 \\ 2 \end{array}\right) \) and at \( \left( \begi... | The equation we are solving is\n\n\[ \n\overrightarrow{\mathbf{f}}\left( \begin{array}{l} x \\ y \end{array}\right) = \left\lbrack \begin{matrix} {x}^{2} - y + \sin \left( {x - y}\right) - 2 \\ {y}^{2} - x - 3 \end{matrix}\right\rbrack = \left\lbrack \begin{array}{l} 0 \\ 0 \end{array}\right\rbrack . \n\]\n\nStarting a... | Yes |
Let us solve \( f\left( x\right) = {\left( x - 1\right) }^{2} = 0 \) by Newton’s method, starting at \( {a}_{0} = 0 \) . As Exercise 2.9.1 asks you to show, the best Lipschitz ratio for \( {f}^{\prime } \) is 2, so | \[
\left| {f\left( {a}_{0}\right) }\right| {\left| {\left( {f}^{\prime }\left( {a}_{0}\right) \right) }^{-1}\right| }^{2}M = 1 \cdot {\left( -\frac{1}{2}\right) }^{2} \cdot 2 = \frac{1}{2},
\]
and Theorem 2.8.13 guarantees that Newton's method will converge to the unique root \( a = 1 \), which is on the boundary of \... | No |
Example 2.9.3 (Superconvergence). The sequence \( {x}_{n + 1} = {x}_{n}^{2} \), starting with \( \;{x}_{0} = 1/2 \) (written \( .1 \) in base \( 2 \) ), superconverges to zero, as shown in the left side of Table 2.9.1. | \[ \n{x}_{0} = {.1}\;{x}_{0} = {.1} \]\n\n\[ \n{x}_{1} = {.01}\;{x}_{1} = {.01} \]\n\n\[ \n{x}_{2} = {.0001}\;{x}_{2} = {.001} \]\n\n\[ \n{x}_{3} = {.00000001}\;{x}_{3} = {.0001} \]\n\n\[ \n{x}_{4} = {.0000000000000001}\;{x}_{4} = {.00001} \]\n\nTABLE 2.9.1. LEFT: Superconvergence. RIGHT: The convergence guaranteed by ... | No |
Example 2.9.7 (Norm of a matrix). Take \( A = \left\lbrack \begin{array}{ll} 2 & 0 \\ 0 & 1 \end{array}\right\rbrack \) . | Then\n\n\( A\overrightarrow{\mathbf{x}} = A\left\lbrack \begin{array}{l} x \\ y \end{array}\right\rbrack = \left\lbrack \begin{matrix} {2x} \\ y \end{matrix}\right\rbrack \), so\n\n\[ \parallel A\parallel = \mathop{\sup }\limits_{{\left| \overrightarrow{\mathbf{x}}\right| = 1}}\left| {A\overrightarrow{\mathbf{x}}}\righ... | Yes |
Theorem 2.9.8 (Kantorovich’s theorem: a stronger version). Kantorovich's theorem 2.8.13 still holds if you replace both lengths of matrices by norms of matrices: \( \left| {\left\lbrack {\mathbf{D}\overrightarrow{\mathbf{f}}\left( {\mathbf{u}}_{1}\right) }\right\rbrack - \left\lbrack {\mathbf{D}\overrightarrow{\mathbf{... | Proof. In the proof of Theorem 2.8.13 we only used the triangle inequality and Proposition 1.4.11, and these hold for the norm \( \parallel A\parallel \) of a matrix \( A \) as well as for its length \( \left| A\right| \), as Exercises 2.9.2 and 2.9.3 ask you to show. | No |
The length of the matrix \( A = \left\lbrack \begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right\rbrack \) is \( \sqrt{{1}^{2} + {1}^{2} + {1}^{2}} = \sqrt{3} \), or about 1.732 . The norm is \( \frac{1 + \sqrt{5}}{2} \), or about 1.618; arriving at that figure takes some work, as follows. A vector \( \left\lbrack \begi... | At a minimum or maximum of a function, its derivative is 0 , so we need to see where the derivative of \( {\left( \cos t + \sin t\right) }^{2} + {\sin }^{2}t \) vanishes. That derivative is \( 2\cos {2t} + \sin {2t} \), which vanishes for \( {2t} = \arctan \left( {-2}\right) \) . We have two possible angles to look for... | Yes |
Example 2.9.10 (Using the norm in Newton's method). Suppose we want to find a \( 2 \times 2 \) matrix \( A \) such that \( {A}^{2} = \left\lbrack \begin{array}{rr} 8 & 1 \\ - 1 & {10} \end{array}\right\rbrack \) . We define \( F : \mathrm{{Mat}}\left( {2,2}\right) \rightarrow \mathrm{{Mat}}\left( {2,2}\right) \) by\n\n... | First, compute the derivative:\n\n\[ \left\lbrack {\mathbf{D}F\left( A\right) }\right\rbrack B = {AB} + {BA}. \]\n\nThe following computation shows that \( A \mapsto \left\lbrack {\mathbf{D}F\left( A\right) }\right\rbrack \) is Lipschitz with respect to the norm, with Lipschitz ratio 2 on all of \( \operatorname{Mat}\l... | Yes |
Consider \( f\left( x\right) = {2x} + \sin x \), shown in Figure 2.10.1, and choose \( \left\lbrack {a, b}\right\rbrack = \left\lbrack {-{k\pi },{k\pi }}\right\rbrack \) for some positive integer \( k \) . Then | \[ f\left( a\right) = f\left( {-{k\pi }}\right) = - {2k\pi } + \overset{ = 0}{\overbrace{\sin \left( {-{k\pi }}\right) }},\;f\left( b\right) = f\left( {k\pi }\right) = {2k\pi } + \overset{ = 0}{\overbrace{\sin \left( {k\pi }\right) }}; \] i.e., \( f\left( a\right) = {2a} \) and \( f\left( b\right) = {2b} \) . Since \( ... | Yes |
Where is the function\n\n\[ \mathbf{f}\left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{matrix} \sin \left( {x + y}\right) \\ {x}^{2} - {y}^{2} \end{matrix}\right) \]\n\nlocally invertible? | The derivative is\n\n\[ \left\lbrack {\mathbf{{Df}}\left( \begin{array}{l} x \\ y \end{array}\right) }\right\rbrack = \left\lbrack \begin{matrix} \cos \left( {x + y}\right) & \cos \left( {x + y}\right) \\ {2x} & - {2y} \end{matrix}\right\rbrack \]\n\nwhich is invertible if \( - {2y}\cos \left( {x + y}\right) - {2x}\cos... | Yes |
Let \( {C}_{1} \) be the circle of radius 3 centered at the origin in \( {\mathbb{R}}^{2} \), and let \( {C}_{2} \) be the circle of radius 1 centered at \( \left( \begin{matrix} {10} \\ 0 \end{matrix}\right) \) . What is the locus of centers of line segments drawn from a point of \( {C}_{1} \) to a point of \( {C}_{2}... | The center of the segment joining\n\n\[ \left( \begin{matrix} 3\cos \theta \\ 3\sin \theta \end{matrix}\right) \in {C}_{1}\;\text{ to }\;\left( \begin{matrix} \cos \varphi + {10} \\ \sin \varphi \end{matrix}\right) \in {C}_{2} \]\n\n\nis the point\n\n\[ F\left( \begin{array}{l} \theta \\ \varphi \end{array}\right) \ove... | Yes |
Theorem 2.10.7 (The inverse function theorem). Let \( W \subset {\mathbb{R}}^{m} \) be an open neighborhood of \( {\mathbf{x}}_{0} \), and let \( \mathbf{f} : W \rightarrow {\mathbb{R}}^{m} \) be a continuously differentiable function. Set \( {\mathbf{y}}_{0} = \mathbf{f}\left( {\mathbf{x}}_{0}\right) \). If the deriva... | To quantify this statement, we will specify the radius \( R \) of a ball \( V \) centered at \( {\mathbf{y}}_{0} \), in which the inverse function is defined. First simplify notation by setting \( L = \left\lbrack {\mathbf{{Df}}\left( {\mathbf{x}}_{0}\right) }\right\rbrack \). Now find \( R > 0 \) satisfying the follow... | Yes |
Let's return to the function\n\n\\[ \n\\mathbf{f}\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) = \\left( \\begin{matrix} \\sin \\left( {x + y}\\right) \\\\ {x}^{2} - {y}^{2} \\end{matrix}\\right) \n\\]\n\nof Example 2.10.5. Let’s choose a point \\( {\\mathbf{x}}_{0} \\) where the derivative is invertible and ... | We know from Example 2.10.5 that the derivative is invertible at \\( {\\mathbf{x}}_{0} = \\left( \\begin{array}{l} 0 \\\\ \\pi \\end{array}\\right) \\) . This gives\n\n\\[ \nL = \\left\\lbrack {\\operatorname{Df}\\left( \\begin{array}{l} 0 \\\\ \\pi \\end{array}\\right) }\\right\\rbrack = \\left\\lbrack \\begin{matrix}... | Yes |
The equation \( {x}^{2} + {y}^{2} + {z}^{2} - 1 = 0 \) expresses \( z \) as an implicit function of \( \left( \begin{array}{l} x \\ y \end{array}\right) \) near \( {\overrightarrow{\mathbf{e}}}_{3} \). | This implicit function can be made explicit: \( z = \sqrt{1 - {x}^{2} - {y}^{2}} \) ; you can solve for \( z \) as a function of \( x \) and \( y \) (see Figure 2.10.9). | Yes |
Example 2.10.12 (The unit circle and the short implicit function theorem). The unit circle is the set of points \( \mathbf{c} = \left( \begin{array}{l} x \\ y \end{array}\right) \) such that \( F\left( \mathbf{c}\right) = 0 \) when \( F \) is the function \( F\left( \begin{array}{l} x \\ y \end{array}\right) = {x}^{2} ... | \[ \left\lbrack {\mathbf{D}F\left( \begin{array}{l} a \\ b \end{array}\right) }\right\rbrack = \left\lbrack {{2a},{2b}}\right\rbrack \] This derivative will be onto \( \mathbb{R} \) as long as it is not \( \left\lbrack {0,0}\right\rbrack \), which it will never be on the unit circle. If \( b \neq 0 \), we may consider ... | Yes |
Theorem 2.10.14 (The implicit function theorem). Let \( W \) be an open neighborhood of \( \mathbf{c} \in {\mathbb{R}}^{n} \), and let \( \mathbf{F} : W \rightarrow {\mathbb{R}}^{n - k} \) be a differentiable function, with \( \mathbf{F}\left( \mathbf{c}\right) = \mathbf{0} \) and \( \left\lbrack {\mathbf{{DF}}\left( \... | To specify the domain of \( \mathbf{g} \), let \( L \) be the \( n \times n \) matrix\n\n\[ L = \left\lbrack \begin{matrix} \left\lbrack {{D}_{1}\mathbf{F}\left( \mathbf{c}\right) ,\ldots ,{D}_{n - k}\mathbf{F}\left( \mathbf{c}\right) }\right\rbrack & \left\lbrack {{D}_{n - k + 1}\mathbf{F}\left( \mathbf{c}\right) ,\ld... | Yes |
In Example 2.10.12 we showed that if \( a \neq 0 \), then in some neighborhood of \( \left( \begin{array}{l} a \\ b \end{array}\right) \), the equation \( {x}^{2} + {y}^{2} - 1 = 0 \) implicitly expresses \( x \) as a function of \( y \) . Let’s see what the strong version of the implicit function theorem says about th... | The matrix \( L \) of equation 2.10.25 is\n\n\[ L = \left\lbrack \begin{matrix} {2a} & {2b} \\ 0 & 1 \end{matrix}\right\rbrack ,\;\text{ and }\;{L}^{-1} = \frac{1}{2a}\left\lbrack \begin{array}{rr} 1 & - {2b} \\ 0 & {2a} \end{array}\right\rbrack . \]\n\nSo we have\n\n\[ \left| {L}^{-1}\right| = \frac{1}{2\left| a\right... | Yes |
In what neighborhood of \( \mathbf{c}\overset{\text{ def }}{ = }\left( \begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right) \) do the equations \( \left\{ \begin{array}{l} {x}^{2} - y = a \\ {y}^{2} - z = b \\ {z}^{2} - x = 0 \end{array}\right. \) determine \( \left( \begin{array}{l} x \\ y \\ z \end{array}\right) ... | Here, \( n = 5, k = 2 \), and \( \mathbf{F}\left( \begin{array}{l} x \\ y \\ z \\ a \\ b \end{array}\right) = \left( \begin{matrix} {x}^{2} - y - a \\ {y}^{2} - z - b \\ {z}^{2} - x \end{matrix}\right) , \) with derivative \( \left\lbrack \begin{array}{rrrrr} {2x} & - 1 & 0 & - 1 & 0 \\ 0 & {2y} & - 1 & 0 & - 1 \\ - 1 ... | Yes |
In what neighborhood of \( \\mathbf{0} = \\left( \\begin{array}{l} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\end{array}\\right) \) do the equations \( \\left\\{ \\begin{array}{l} {x}^{3} - y = a \\ {y}^{3} - z = b \\ {z}^{3} - x = 0 \\end{array}\\right. \) determine \( \\left( \\begin{array}{l} x \\ y \\ z \\end{array}\\right) \) implic... | When we evaluate the first three columns of the derivative at \( x = y = z = 0 \) , we clearly get an invertible matrix, so already we know that in some neighborhood, \( x, y, z \) can be expressed implicitly as a function of \( a \) and \( b \) . | Yes |
The graph of any smooth function is a smooth manifold. | The curve of equation \( y = {x}^{2} \) is a one-dimensional manifold: the graph of \( y \) as the function \( f\left( x\right) = {x}^{2} \) . The curve of equation \( x = {y}^{2} \) is also a one-dimensional manifold: the graph of a function representing \( x \) as a function of \( y \) . Each surface at the top of Fi... | No |
The unit circle of equation \( {x}^{2} + {y}^{2} = 1 \) is a smooth curve. | Here we need the graphs of four functions to cover the entire circle: the unit circle is only locally the graph of a function. As discussed in Example 2.10.12, the upper half of the circle, made up of points \( \left( \begin{array}{l} x \\ y \end{array}\right) \) with \( y > 0 \), is the graph of the function \( \sqrt{... | No |
Example 3.1.6 (Surface in \( {\mathbb{R}}^{3} \) ). By Definition 3.1.2, a subset \( S \subset {\mathbb{R}}^{3} \) is a smooth surface (two-dimensional manifold) in \( {\mathbb{R}}^{3} \) if locally it is the graph of a \( {C}^{1} \) mapping expressing one variable as a function of the other two variables. That is, \( ... | For instance, the unit sphere\n\n\[ \n{S}^{2}\\overset{\\text{ def }}{ = }\\left\\{ {\\left( \\begin{array}{l} x \\ y \\ z \\end{array}\\right) \\text{ such that }{x}^{2} + {y}^{2} + {z}^{2} = 1}\\right\\} \n\] \n\nis a smooth surface. Let\n\n\[ \n{D}_{x, y} = \\left\\{ {\\left. \\left( \\begin{array}{l} x \\ y \\end{a... | No |
Consider the configuration space (set of positions) of a bicycle as a \( k \) -dimensional manifold. What number is \( k \) ? | To specify a position, you must say where the center of gravity is (three numbers) and specify the orientation of the bicycle, say the direction in which the horizontal bar is pointing (two numbers) and the angle that the bar supporting the saddle makes with the vertical (one number). Then you need to give the angle th... | No |
Example 3.1.11 (Determining that a locus is a smooth curve). For a one-dimensional manifold in \( {\mathbb{R}}^{2} \) (i.e., a plane curve), the function of Theorem 3.1.10 goes from an open subset of \( {\mathbb{R}}^{2} \) to \( \mathbb{R} \), and the requirement that \( \left\lbrack {\mathbf{{DF}}\left( \mathbf{z}\rig... | For instance, we have no idea what the locus \( {X}_{c} \) defined by\n\n\[ {x}^{8} + 2{x}^{3} + y + {y}^{5} = c \]\n\nlooks like, but we know it is a smooth curve for all \( c \), since the derivative of the function \( F\left( \begin{array}{l} x \\ y \end{array}\right) = {x}^{8} + 2{x}^{3} + y + {y}^{5} - c \) is\n\n... | Yes |
Example 3.1.13 (Smooth surface in \( {\mathbb{R}}^{3} \) ). When we use Theorem 3.1.10 to determine whether a locus is a smooth surface in \( {\mathbb{R}}^{3} \), the function \( \mathbf{F} \) goes from an open subset of \( {\mathbb{R}}^{3} \) to \( \mathbb{R} \) . In this case the requirement that \( \left\lbrack {\ma... | For instance, consider the set \( X \) defined by the equation\n\n\[ F\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \sin \left( {x + {yz}}\right) = 0 \]\n\nIs it a smooth surface? The derivative is\n\n\[ \left\lbrack {\mathbf{D}F\left( \begin{array}{l} a \\ b \\ c \end{array}\right) }\right\rbrack = \left\lb... | Yes |
Theorem 3.1.10 suggests that a natural way to think of a curve \( C \) in \( {\mathbb{R}}^{3} \) is as the intersection of two surfaces. Here \( n = 3, k = 1 \) . If surfaces \( {S}_{1} \) and \( {S}_{2} \) are given by equations \( {F}_{1}\left( \mathbf{z}\right) = 0 \) and \( {F}_{2}\left( \mathbf{z}\right) = 0 \), e... | Thus \( \left\lbrack {\mathbf{{DF}}\left( \mathbf{z}\right) }\right\rbrack \) is a \( 2 \times 3 \) matrix:\n\n\[ \left\lbrack {\mathbf{{DF}}\left( \mathbf{z}\right) }\right\rbrack = \left\lbrack \begin{array}{lll} {D}_{1}{F}_{1}\left( \mathbf{z}\right) & {D}_{2}{F}_{1}\left( \mathbf{z}\right) & {D}_{3}{F}_{1}\left( \m... | Yes |
Example 3.1.15 (Checking that the linkage space is a manifold). In Example 3.1.8, \( {X}_{2} \) is the set of positions of four rigid rods restricted to | \[ \mathbf{f}\left( \mathbf{z}\right) = \mathbf{f}\left( \begin{array}{l} {x}_{1} \\ {y}_{1} \\ {x}_{2} \\ {y}_{2} \\ {x}_{3} \\ {y}_{3} \\ {x}_{4} \\ {y}_{4} \end{array}\right) = \left\lbrack \begin{array}{l} {\left( {x}_{2} - {x}_{1}\right) }^{2} + {\left( {y}_{2} - {y}_{1}\right) }^{2} - {l}_{1}^{2} \\ {\left( {x}_{... | "No" |
Let \( M \subset {\mathbb{R}}^{m} \) be a \( k \) -dimensional manifold, \( U \) an open subset of \( {\mathbb{R}}^{n} \), and \( f : U \rightarrow {\mathbb{R}}^{m}\;a\;{C}^{1}\; \) mapping whose derivative \( \;\left\lbrack {\mathbf{D}\mathbf{f}\left( \mathbf{x}\right) }\right\rbrack \; \) is surjective at every \( \m... | Proof. Let \( \mathbf{x} \) be a point of \( {\mathbf{f}}^{-1}\left( M\right) \) . By part 2 of Theorem 3.1.10, there exists a neighborhood \( V \) of \( \mathbf{f}\left( \mathbf{x}\right) \) such that \( M \cap V \) is defined by the equation \( \mathbf{F}\left( \mathbf{y}\right) = \mathbf{0} \), where \( \mathbf{F} :... | Yes |
In the case of a surface in \( {\mathbb{R}}^{3} \), the \( U \) of Definition 3.1.18 is an open subset of \( {\mathbb{R}}^{2} \), and \( \left\lbrack {\mathbf{D}\gamma \left( \mathbf{u}\right) }\right\rbrack \) is a \( 3 \times 2 \) matrix. Saying that it is one to one is the same as saying that \( \overrightarrow{{D}_... | To check that \( \gamma \) is one to one and onto \( M \), first note that the image of \( \gamma \) is part of the unit sphere, since\n\n\[ {\cos }^{2}\theta {\cos }^{2}\varphi + {\sin }^{2}\theta {\cos }^{2}\varphi + {\sin }^{2}\varphi = 1 \]\n\nSince \( z \in \left( {0,1}\right) \), there exists a unique \( \varphi ... | Yes |
A subset \( S \subset {\mathbb{R}}^{3} \) is a smooth surface if at every point \( \mathbf{a} \in S \), the subset \( S \) is the graph of a function expressing one variable in terms of two others. The tangent space to a smooth surface \( S \) at \( \mathbf{a} \) is the plane composed of the vectors anchored at \( \mat... | If at \( \mathbf{a} = \left( \begin{array}{l} a \\ b \\ c \end{array}\right) \) the surface is the graph of a function \( f \) expressing \( z \) in terms of \( x \) and \( y \), then \( z \) is the passive variable: \( z = f\left( \begin{array}{l} x \\ y \end{array}\right) \) . The equation for the tangent plane then ... | No |
Theorem 3.2.4 (Tangent space to a manifold given by equations). If \( \mathbf{F}\left( \mathbf{z}\right) = \mathbf{0} \) describes a manifold \( M \), and \( \left\lbrack {\mathbf{{DF}}\left( {\mathbf{z}}_{0}\right) }\right\rbrack \) is onto for some \( {\mathbf{z}}_{0} \in M \), then the tangent space \( {T}_{{\mathbf... | \[ {T}_{{\mathbf{z}}_{0}}M = \ker \left\lbrack {\mathbf{{DF}}\left( {\mathbf{z}}_{0}\right) }\right\rbrack \] | Yes |
Example 3.2.5: Saying that \( \ker \left\lbrack {\mathbf{D}F\left( \mathbf{a}\right) }\right\rbrack \) is the tangent space to \( {X}_{c} \) at a says that every vector \( \overrightarrow{\mathbf{v}} \) tangent to \( {X}_{c} \) at a satisfies the equation\n\n\[ \left\lbrack {\mathbf{D}F\left( \mathbf{a}\right) }\right\... | This puzzled one student, who argued that for this equation to be true, either [DF(a)] must be [0] or \( \overrightarrow{\mathbf{v}} \) must be \( \overrightarrow{\mathbf{0}} \), yet the requirement that the derivative be onto says that \( \left\lbrack {\mathbf{D}F\left( \mathbf{a}\right) }\right\rbrack \neq \left\lbra... | Yes |
Example 3.2.6 (Tangent space to smooth surface). We saw in Example 3.1.13 that the set defined by \( F\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \sin \left( {x + {yz}}\right) = 0 \) is a smooth surface, since the derivative of \( F \) is | \[ \left\lbrack {\mathbf{D}F\left( \begin{array}{l} a \\ b \\ c \end{array}\right) }\right\rbrack = \left\lbrack {\underset{{D}_{1}F}{\underbrace{\cos \left( {a + {bc}}\right) }},\underset{{D}_{2}F}{\underbrace{c\cos \left( {a + {bc}}\right) }},\underset{{D}_{3}F}{\underbrace{b\cos \left( {a + {bc}}\right) }}}\right\rb... | Yes |
Proposition 3.2.7 (Tangent space of manifold given by parametrization). Let \( U \subset {\mathbb{R}}^{k} \) be open, and let \( \gamma : U \rightarrow {\mathbb{R}}^{n} \) be a parametrization of a manifold \( M \) . Then\n\n\[ \n{T}_{\gamma \left( \mathbf{u}\right) }M = \operatorname{img}\left\lbrack {\mathbf{D}\gamma... | Proof. Suppose that in some neighborhood \( V \) of \( \gamma \left( \mathbf{u}\right) \in {\mathbb{R}}^{n} \) the manifold \( M \) is given by \( \mathbf{F}\left( \mathbf{x}\right) = \mathbf{0} \), where \( \mathbf{F} : V \rightarrow {\mathbb{R}}^{n - k} \) is a \( {C}^{1} \) mapping whose derivative is onto, so its k... | Yes |
The map \( \gamma : \left( \begin{array}{l} u \\ v \end{array}\right) \mapsto \left( \begin{array}{l} {u}^{2} \\ {uv} \\ {v}^{2} \end{array}\right) \), for \( 0 < u < \infty \) and \( 0 < v < \infty \), parametrizes a surface \( M \) of equation \( F\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {xz} - {y}^{2... | Since \[ \left\lbrack {\mathbf{D}\gamma \left( \begin{array}{l} u \\ v \end{array}\right) }\right\rbrack = \left\lbrack \begin{matrix} {2u} & 0 \\ v & u \\ 0 & {2v} \end{matrix}\right\rbrack \] the tangent space to \( M \) at \( \gamma \left( \begin{array}{l} 1 \\ 1 \end{array}\right) = \left( \begin{array}{l} 1 \\ 1 \... | Yes |
Proposition 3.2.10. Let \( M \subset {\mathbb{R}}^{n} \) be an \( m \) -dimensional manifold, and \( \mathbf{f} : M \rightarrow {\mathbb{R}}^{k} \) a \( {C}^{1} \) map; let \( P \subset M \) be the set where \( \mathbf{f} = \mathbf{0} \) . If \( \left\lbrack {\mathbf{D}\mathbf{f}\left( \mathbf{x}\right) }\right\rbrack ... | Proof. In a neighborhood \( U \) of \( \mathbf{x} \in P \) define \( M \cap U \) by \( \mathbf{g} = \mathbf{0} \), where \( \mathbf{g} : U \rightarrow {\mathbb{R}}^{n - m} \) is of class \( {C}^{1} \) with \( \left\lbrack {\mathbf{{Dg}}\left( \mathbf{y}\right) }\right\rbrack \) onto for every \( \mathbf{y} \in M \cap U... | Yes |
Proposition 3.2.11. Let \( M \subset {\mathbb{R}}^{n} \) be a manifold, and \( \mathbf{f} : M \rightarrow {\mathbb{R}}^{k} \) a \( {C}^{1} \) map. Let \( U \subset {\mathbb{R}}^{\ell } \) be open and \( \mathbf{g} : U \rightarrow M \) a \( {C}^{1} \) map. Then\n\n\[ \left\lbrack {\mathbf{D}\left( {\mathbf{f} \circ \mat... | Proof. Choose \( \mathbf{x} \in \mathbf{g}\left( U\right) \), and let \( \widetilde{\mathbf{f}} \) be an extension of \( \mathbf{f} \) to a neighborhood | No |
Proposition 3.2.12. If \( V \subset {\mathbb{R}}^{m} \) is open and \( \gamma : V \rightarrow {\mathbb{R}}^{n} \) is a \( {C}^{1} \) parametrization of a manifold \( M \subset {\mathbb{R}}^{n} \), then \( {\gamma }^{-1} : M \rightarrow V \) is of class \( {C}^{1} \) . | Proof. To apply Proposition and Definition 3.2.9, we need, for each \( \mathbf{x} \in M \) , to extend \( {\gamma }^{-1} \) to a neighborhood \( U \subset {\mathbb{R}}^{n} \) of \( \mathbf{x} \) ; this requires the inverse function theorem. Let \( \mathbf{y} = {\gamma }^{-1}\left( \mathbf{x}\right) \) . Define \( \Psi ... | Yes |
Theorem 3.3.1 (Taylor’s theorem without remainder in one variable). If \( U \subset \mathbb{R} \) is an open subset and \( f : U \rightarrow \mathbb{R} \) is \( k \) times continuously differentiable on \( U \), then the polynomial\n\n\[ \n\\underset{\\begin{matrix} \\text{Taylor polynomial} \\\\ \\text{of degree }k \\... | One proof, sketched in Exercise 3.3.11, consists of using l'Hôpital's rule \( k \) times. | No |
What are the elements of the set \( {\mathcal{I}}_{2}^{3} \) ? Of \( {\mathcal{I}}_{3}^{3} \) ? | \[ \overline{{}^{6}{\mathcal{I}}_{2}^{3} = \{ \left( {1,2}\right) ,\left( {2,1}\right) ,\left( {0,3}\right) ,\left( {3,0}\right) \} ;}\;{\mathcal{I}}_{3}^{3} = \{ \left( {1,1,1}\right) ,\left( {2,1,0}\right) ,\left( {2,0,1}\right) ,\left( {1,2,0}\right) ,\left( {1,0,2}\right) ,\left( {0,2,1}\right) ,\left( {0,1,2}\righ... | Yes |
Example 3.3.7 (Multi-exponent notation). Using this notation, we can write the polynomial\n\n\\[ \n2 + {x}_{1} - {x}_{2}{x}_{3} + 4{x}_{1}{x}_{2}{x}_{3} + 2{x}_{1}^{2}{x}_{2}^{2}\\;\\text{ as }\\;\\mathop{\\sum }\\limits_{{k = 0}}^{4}\\mathop{\\sum }\\limits_{{I \\in {\\mathcal{I}}_{3}^{k}}}{a}_{I}{\\mathbf{x}}^{I},\\;... | \\[ \n{a}_{\\left( 0,0,0\\right) } = 2,\\;{a}_{\\left( 1,0,0\\right) } = 1,\\;{a}_{\\left( 0,1,1\\right) } = - 1,\n\\]\n\n\\[ \n{a}_{\\left( 1,1,1\\right) } = 4,\\;{a}_{\\left( 2,2,0\\right) } = 2,\n\\]\n\nand all other \\( {a}_{I} = 0 \\), for \\( I \\in {\\mathcal{I}}_{3}^{k}, k \\leq 4 \\) . (There are 35 terms in a... | Yes |
Theorem 3.3.8 Equality of crossed partials). Let \( U \) be an open subset of \( {\mathbb{R}}^{n} \) and \( f : U \rightarrow \mathbb{R} \) a function such that all first partial derivatives \( {D}_{i}f \) are themselves differentiable at \( \mathbf{a} \in U \) . Then for every pair of variables \( {x}_{i},{x}_{j} \), ... | \[ {D}_{j}\left( {{D}_{i}f}\right) \left( \mathbf{a}\right) = {D}_{i}\left( {{D}_{j}f}\right) \left( \mathbf{a}\right) . \] | Yes |
Example 3.3.9 (A case where crossed partials aren't equal). Consider the function\n\n\\[ \nf\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) = \\left\\{ {\\begin{array}{ll} {xy}\\frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} & \\text{ if }\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) \\neq \\left( \\begin... | Then when \\( \\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) \\neq \\left( \\begin{array}{l} 0 \\\\ 0 \\end{array}\\right) \\), we have\n\n\\[ \n{D}_{1}f\\left( \\begin{array}{l} x \\\\ y \\end{array}\\right) = \\frac{4{x}^{2}{y}^{3} + {x}^{4}y - {y}^{5}}{{\\left( {x}^{2} + {y}^{2}\\right) }^{2}}\\text{ and }{... | Yes |
Proposition 3.3.11 (Coefficients expressed in terms of partial derivatives at 0). Let \( p \) be the polynomial\n\n\[ p\left( \mathbf{x}\right) \overset{\text{ def }}{ = }\mathop{\sum }\limits_{{m = 0}}^{k}\mathop{\sum }\limits_{{J \in {\mathcal{I}}_{n}^{m}}}{a}_{J}{\mathbf{x}}^{J} \]\n\nThen for any \( I \in {\mathcal... | Proof. First, let us see that it is sufficient to show that\n\n\[ {D}_{I}{\mathbf{x}}^{I}\left( \mathbf{0}\right) = I!\text{ and }{D}_{I}{\mathbf{x}}^{J}\left( \mathbf{0}\right) = 0\text{ for all }J \neq I. \]\n\nWe can see this by writing\n\n\[ {D}_{I}p\left( \mathbf{0}\right) = {D}_{I}\overset{\begin{matrix} \text{ m... | Yes |
Let \( p = 3{x}_{1}^{2}{x}_{2}^{3} \). What is \( {D}_{1}^{2}{D}_{2}^{3}p \) ? | We have\n\n\[ \n{D}_{2}p = 9{x}_{2}^{2}{x}_{1}^{2},\;{D}_{2}^{2}p = {18}{x}_{2}{x}_{1}^{2}, \n\]\n\nand so on, ending with \( {D}_{1}^{2}{D}_{2}^{3}p = {36} \) . | Yes |
Example 3.3.14 (Multi-exponent notation for a Taylor polynomial of a function in two variables). Let \( f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) be a function. The formula for the Taylor polynomial of degree 2 of \( f \) at \( \mathbf{a} \) is then\n\n\[ \n{P}_{f,\mathbf{a}}^{2}\left( {\mathbf{a} + \overrightarro... | \[ \n= \underset{m = 0;\text{ i.e.,}f\left( \mathbf{a}\right) }{\underbrace{\frac{1}{0!0!}{D}_{\left( 0,0\right) }f\left( \mathbf{a}\right) {h}_{1}^{0}{h}_{2}^{0}}} + \underset{\text{terms of degree }1 : \text{ first derivatives }}{\underbrace{\frac{1}{1!0!}{D}_{\left( 1,0\right) }f\left( \mathbf{a}\right) {h}_{1}^{1}{... | Yes |
What is the Taylor polynomial of degree 2 of the function \( f\left( \begin{array}{l} x \\ y \end{array}\right) = \sin \left( {x + {y}^{2}}\right) \), at \( \mathbf{0} = \left( \begin{array}{l} 0 \\ 0 \end{array}\right) \) ? | The first term, of degree 0, is \( f\left( \mathbf{0}\right) = \sin 0 = 0 \) . The terms of degree 1 are \( {D}_{\left( 1,0\right) }f\left( \begin{array}{l} x \\ y \end{array}\right) = \cos \left( {x + {y}^{2}}\right) \) and \( {D}_{\left( 0,1\right) }f\left( \begin{array}{l} x \\ y \end{array}\right) = {2y}\cos \left(... | Yes |
Theorem 3.3.16 (Taylor’s theorem without remainder in higher dimensions). Let \( U \subset {\mathbb{R}}^{n} \) be open, \( \mathbf{a} \in U \) a point, and \( f : U \rightarrow \mathbb{R} \) a \( {C}^{k} \) function. | 1. The polynomial \( {P}_{f,\mathbf{a}}^{k}\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}\right) \) is the unique polynomial of degree \( k \) with the same partial derivatives up to order \( k \) at \( \mathbf{a} \) as \( f \) . 2. It best approximates \( f \) near a: it is the unique polynomial of degree at most \... | No |
Proposition 3.3.17 (Size of a function with many vanishing partial derivatives). Let \( U \) be an open subset of \( {\mathbb{R}}^{n} \) and let \( g : U \rightarrow \mathbb{R} \) be a \( {C}^{k} \) function. If at \( \mathbf{a} \in U \) all partial derivatives of \( g \) up to order \( k \) vanish (including \( g\left... | \[ \mathop{\lim }\limits_{{\overrightarrow{\mathbf{h}} \rightarrow \overrightarrow{\mathbf{0}}}}\frac{g\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}\right) }{{\left| \overrightarrow{\mathbf{h}}\right| }^{k}} = 0 \] | Yes |
Proposition 3.4.2 (Taylor polynomials of some standard functions). The following formulas give the Taylor polynomials at 0 of the corresponding functions: | \[ {e}^{x} = 1 + x + \frac{{x}^{2}}{2!} + \cdots + \frac{{x}^{n}}{n!} + o\left( {\left| x\right| }^{n}\right) \] | Yes |
Proposition 3.4.4 (Chain rule for Taylor polynomials). Let \( U \subset {\mathbb{R}}^{n} \) and \( V \subset \mathbb{R} \) be open, and \( g : U \rightarrow V \) and \( f : V \rightarrow \mathbb{R} \) be of class \( {C}^{k} \). Then \( f \circ g : U \rightarrow \mathbb{R} \) is of class \( {C}^{k} \), and if \( g\left(... | \[ \overrightarrow{\mathbf{h}} \mapsto {P}_{f, b}^{k}\left( {{P}_{g,\mathbf{a}}^{k}\left( {\mathbf{a} + \overrightarrow{\mathbf{h}}}\right) }\right) \] and discarding the terms of degree \( > k \) . | Yes |
Let's use these rules to compute the Taylor polynomial of degree 3 of the function \( f\left( \begin{array}{l} x \\ y \end{array}\right) = \sin \left( {x + {y}^{2}}\right) \) at 0, which we already saw in Example 3.3.15. | Using Proposition 3.4.4, we simply substitute \( x + {y}^{2} \) for \( u \) in \( \sin u = u - {u}^{3}/6 + o\left( {u}^{3}\right) \), omitting all the terms of degree \( > 3 \) :\n\n\[ \sin \left( {x + {y}^{2}}\right) = \left( {x + {y}^{2}}\right) - \frac{{\left( x + {y}^{2}\right) }^{3}}{6} + \;\text{terms of degree} ... | Yes |
Example 3.4.6 (A harder example). Let \( U \subset \mathbb{R} \) be open, and let \( f : U \rightarrow \mathbb{R} \) be of class \( {C}^{2} \) . Let \( V \subset U \times U \) be the subset of \( {\mathbb{R}}^{2} \) where \( f\left( x\right) + f\left( y\right) \neq 0 \) . Compute the Taylor polynomial of degree 2, at a... | Set \( \left( \begin{array}{l} x \\ y \end{array}\right) = \left( \begin{array}{l} a + u \\ b + v \end{array}\right) \) . First we write \( f\left( {a + u}\right) \) and \( f\left( {b + v}\right) \) in terms of the Taylor polynomial of \( f \) evaluated at \( a + u \) and \( b + v \) :\n\n\[ \begin{matrix} F\left( \beg... | Yes |
Theorem 3.4.7 (Taylor polynomial of implicit function). Let \( \mathbf{F} \) be a function of class \( {C}^{k} \) for some \( k \geq 1 \), such that \( \mathbf{F}\left( \begin{array}{l} \mathbf{a} \\ \mathbf{b} \end{array}\right) = \mathbf{0} \) . Then the implicit function \( \mathbf{g} \) is also of class \( {C}^{k} ... | Proof. By Proposition 3.4.4, it is enough to show that \( \mathbf{g} \) is of class \( {C}^{k} \) . By equation 2.10.29, if \( \mathbf{F} \) is of class \( {C}^{k} \), then \( \left\lbrack \mathbf{{Dg}}\right\rbrack \) is of class \( {C}^{k - 1} \), hence \( \mathbf{g} \) is of class \( {C}^{k} \) . | No |
Example 3.5.2 (An integral as a quadratic form). Let \( p \) be the polynomial \( p\left( t\right) = {a}_{0} + {a}_{1}t + {a}_{2}{t}^{2} \), and let \( Q : {\mathbb{R}}^{3} \rightarrow \mathbb{R} \) be the function\n\n\[ Q\left( \mathbf{a}\right) = {\int }_{0}^{1}{\left( p\left( t\right) \right) }^{2}{dt} \] | Then \( Q \) is a quadratic form, as we can confirm by computing the integral:\n\n\[ Q\left( \mathbf{a}\right) = {\int }_{0}^{1}{\left( {a}_{0} + {a}_{1}t + {a}_{2}{t}^{2}\right) }^{2}{dt} \]\n\n\[ = {\int }_{0}^{1}\left( {{a}_{0}^{2} + {a}_{1}^{2}{t}^{2} + {a}_{2}^{2}{t}^{4} + 2{a}_{0}{a}_{1}t + 2{a}_{0}{a}_{2}{t}^{2}... | Yes |
the quadratic form \( Q\left( \mathbf{x}\right) = {x}^{2} + {xy} \) can be written as \( {\left( {\alpha }_{1}\left( \mathbf{x}\right) \right) }^{2} - {\left( {\alpha }_{2}\left( \mathbf{x}\right) \right) }^{2} \) | Since\n\n\[ \n{x}^{2} + {xy} = {x}^{2} + {xy} + \frac{1}{4}{y}^{2} - \frac{1}{4}{y}^{2} = {\left( x + \frac{y}{2}\right) }^{2} - {\left( \frac{y}{2}\right) }^{2}, \n\]\n\nwhere \( {\alpha }_{1} \) and \( {\alpha }_{2} \) are the linear functions\n\n\[ \n{\alpha }_{1}\left( \begin{array}{l} x \\ y \end{array}\right) = x... | Yes |
Consider the quadratic form\n\n\[ Q\left( \mathbf{x}\right) = {x}^{2} + {2xy} - {4xz} + {2yz} - 4{z}^{2}. \] | We take all the terms in which \( x \) appears, which gives us \( {x}^{2} + \left( {{2y} - {4z}}\right) x \) . Since \( {x}^{2} + \left( {{2y} - {4z}}\right) x + {\left( y - 2z\right) }^{2} = {\left( x + y - 2z\right) }^{2} \), adding and subtracting \( {\left( y - 2z\right) }^{2} \) yields\n\n\[ Q\left( \mathbf{x}\rig... | Yes |
Consider the quadratic form\n\n\\[ \nQ\\left( \\mathbf{x}\\right) = {xy} - {xz} + {yz}. \n\\] | One possibility is to introduce the new variable \\( u = x - y \\), so that we can trade \\( x \\) for \\( u + y \\), getting\n\n\\[ \n\\left( {u + y}\\right) y - \\left( {u + y}\\right) z + {yz} = {y}^{2} + {uy} - {uz} = {\\left( y + \\frac{u}{2}\\right) }^{2} - \\frac{{u}^{2}}{4} - {uz} - {z}^{2} + {z}^{2} \n\\]\n\n\... | Yes |
Let \( {P}_{k} \) be the space of polynomials of degree \( \leq k \), and let \( {V}_{a, b} \subset {P}_{k} \) be the space of polynomials \( p \) that vanish at \( a \) and \( b \) for some \( a < b \) . Consider the quadratic form \( Q : {V}_{a, b} \rightarrow \mathbb{R} \) given by\n\n\[ Q\left( p\right) = {\int }_{... | Using integration by parts,\n\n\[ Q\left( p\right) = {\int }_{a}^{b}p\left( t\right) {p}^{\prime \prime }\left( t\right) {dt} = \overset{ = 0\text{ by def. }}{\overbrace{p\left( b\right) {p}^{\prime }\left( b\right) - p\left( a\right) {p}^{\prime }\left( a\right) }} - {\int }_{a}^{b}{\left( {p}^{\prime }\left( t\right)... | Yes |
Proposition 3.5.15. If \( Q : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) is a positive definite quadratic form, then there exists a constant \( C > 0 \) such that\n\n\[ Q\left( \overrightarrow{\mathbf{x}}\right) \geq C{\left| \overrightarrow{\mathbf{x}}\right| }^{2}\;\text{ for all }\overrightarrow{\mathbf{x}} \in {\ma... | Proof. Since \( Q \) has rank \( n \), we can write \( Q\left( \overrightarrow{\mathbf{x}}\right) \) as a sum of squares of \( n \) linearly independent functions:\n\n\[ Q\left( \overrightarrow{\mathbf{x}}\right) = {\left( {\alpha }_{1}\left( \overrightarrow{\mathbf{x}}\right) \right) }^{2} + \cdots + {\left( {\alpha }... | Yes |
Proposition 3.5.17 (Quadratic forms and symmetric matrices). The mapping \( A \mapsto {Q}_{A} \) is a bijective map from the space of symmetric \( n \times n \) matrices to the space of quadratic forms on \( {\mathbb{R}}^{n} \) . | The symmetric matrix associated to a quadratic form is constructed as follows: each diagonal entry \( {a}_{i, i} \) is the coefficient of the corresponding variable squared in the quadratic form (i.e., the coefficient of \( {x}_{i}^{2} \) ) while each entry \( {a}_{i, j} \) is one-half the coefficient of the term \( {x... | No |
Theorem 3.6.3 (Derivative zero at extremum). Let \( U \subset {\mathbb{R}}^{n} \) be an open subset and let \( f : U \rightarrow \mathbb{R} \) be a differentiable function. If \( {\mathbf{x}}_{0} \in U \) is a local minimum or maximum of \( f \), then \( \left\lbrack {\mathbf{D}f\left( {\mathbf{x}}_{0}\right) }\right\r... | Proof. Since the derivative is given by the Jacobian matrix, it is enough to show that if \( {\mathbf{x}}_{0} \) is a local extremum of \( f \), then \( {D}_{i}f\left( {\mathbf{x}}_{0}\right) = 0 \) for all \( i = 1,\ldots, n \) . But \( {D}_{i}f\left( {\mathbf{x}}_{0}\right) = {g}^{\prime }\left( 0\right) \), where \(... | Yes |
What are the critical points of the function \( f : {\mathbb{R}}^{2} \rightarrow \mathbb{R} \) given by\n\n\[ f\left( \begin{array}{l} x \\ y \end{array}\right) = x + {x}^{2} + {xy} + {y}^{3}? \] | The partial derivatives are\n\n\[ {D}_{1}f\left( \begin{array}{l} x \\ y \end{array}\right) = 1 + {2x} + y,\;{D}_{2}f\left( \begin{array}{l} x \\ y \end{array}\right) = x + 3{y}^{2}. \]\n\nIn this case we don't need Newton's method, since the system can be solved explicitly. Substitute \( x = - 3{y}^{2} \) from the sec... | Yes |
Proposition 3.6.7. Let \( U, V \subset {\mathbb{R}}^{n} \) be open, let \( \varphi : V \rightarrow U \) be a \( {C}^{2} \) map and let \( f : U \rightarrow \mathbb{R} \) be a \( {C}^{2} \) function. Let \( {\mathbf{x}}_{0} \in U \) be a critical point of \( f \), and \( {\mathbf{y}}_{0} \in V \) a point such that \( \v... | Proof. If follows from Proposition 3.4.4 that\n\n\[ \n{P}_{f,{\mathbf{x}}_{0}}^{2} \circ {P}_{\varphi ,{\mathbf{y}}_{0}}^{1} = {P}_{f \circ \varphi ,{\mathbf{y}}_{0}}^{2}.\n\]\n\n3.6.9\n\nWhy do we write \( {P}_{\varphi ,{\mathbf{y}}_{0}}^{1} \) here rather than \( {P}_{\varphi ,{\mathbf{y}}_{0}}^{2} \) ? Since \( {\ma... | Yes |
Theorem 3.6.8 (Quadratic forms and extrema). Let \( U \subset {\mathbb{R}}^{n} \) be an open set, \( f : U \rightarrow \mathbb{R} \) of class \( {C}^{2} \), and \( \mathbf{a} \in U \) a critical point of \( f \) . | 1. If the signature of \( \mathbf{a} \) is \( \left( {n,0}\right) \), i.e., \( {Q}_{f,\mathbf{a}} \) is positive definite, then \( \mathbf{a} \) is a strict local minimum of \( f \) . If the signature of \( \mathbf{a} \) is \( \left( {k, l}\right) \) with \( l > 0 \), then \( \mathbf{a} \) is not a local minimum.\n\n2.... | Yes |
Theorem 3.7.2. Let \( X \subset {\mathbb{R}}^{n} \) be a manifold, \( f : X \rightarrow \mathbb{R} \) a \( {C}^{1} \) function, and \( \mathbf{c} \in X \) a local extremum of \( f \) . Then \( \mathbf{c} \) is a critical point of \( f \) . | Proof. Let \( \gamma : V \rightarrow X \) be a parametrization of a neighborhood of \( \mathbf{c} \in X \) , with \( \gamma \left( {\mathbf{x}}_{0}\right) = \mathbf{c} \) . Then \( \mathbf{c} \) is an extremum of \( f \) precisely if \( {\mathbf{x}}_{0} \) is an extremum of \( f \circ \gamma \) . By Theorem 3.6.3, \( \... | Yes |
Suppose we wish to maximize the function \( f\left( \begin{array}{l} x \\ y \end{array}\right) = {xy} \) on the first quadrant of the circle \( {x}^{2} + {y}^{2} = 1 \), which we will denote by \( X \) . As shown in Figure 3.7.2, some level sets of that function do not intersect the circle, and some intersect it in two... | Since \( F\left( \begin{array}{l} x \\ y \end{array}\right) = {x}^{2} + {y}^{2} - 1 \) is the function defining the circle, we have\n\n\[ \n{T}_{\mathbf{c}}X = \ker \left\lbrack {\mathbf{D}F\left( \mathbf{c}\right) }\right\rbrack = \ker \left\lbrack {2{c}_{1},2{c}_{2}}\right\rbrack = \ker \left\lbrack {\frac{2}{\sqrt{2... | Yes |
Let us find the minimum of the function \( f\left( \mathbf{x}\right) = {x}_{1}^{2} + {x}_{2}^{2} + {x}_{3}^{2} \), when it is constrained to the ellipse \( X \) that is the intersection of the cylinder \( {x}_{1}^{2} + {x}_{2}^{2} = 1 \) and the plane of equation \( {x}_{1} = {x}_{3} \). | Since \( f \) measures the square of the distance from the origin, we are looking for the points on the ellipse that are closest to the origin. Clearly they are \( \mathbf{a} = \left( \begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right) \) and \( - \mathbf{a} = \left( \begin{array}{r} 0 \\ - 1 \\ 0 \end{array}\right) \), wh... | Yes |
Suppose we want to maximize \( f\left( \begin{array}{l} x \\ y \end{array}\right) = x + y \) on the ellipse \( {x}^{2} + 2{y}^{2} = 1 \) . | We have\n\n\[\n\underset{\text{constraint function }}{\underbrace{F\left( \begin{array}{l} x \\ y \end{array}\right) = {x}^{2} + 2{y}^{2} - 1}}\text{ and }\left\lbrack {\mathbf{D}F\left( \begin{array}{l} x \\ y \end{array}\right) }\right\rbrack = \left\lbrack {{2x},{4y}}\right\rbrack ,\n\]\n\nwhile \( \left\lbrack {\ma... | Yes |
Let us follow this procedure for the function \( f\left( \mathbf{x}\right) = {x}^{2} + {y}^{2} + {z}^{2} \) of Example 3.7.4, constrained as before to the ellipse given by \n\nWe have\n\[ \n\mathbf{F}\left( \mathbf{x}\right) = \mathbf{F}\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = \left( \begin{matrix} {x}^... | \[ \n\left\lbrack {\mathbf{D}f\left( \mathbf{x}\right) }\right\rbrack = \left\lbrack {{2x},{2y},{2z}}\right\rbrack ,\left\lbrack {\mathbf{D}{F}_{1}\left( \mathbf{x}\right) }\right\rbrack = \left\lbrack {{2x},{2y},0}\right\rbrack ,\left\lbrack {\mathbf{D}{F}_{2}\left( \mathbf{x}\right) }\right\rbrack = \left\lbrack {1,0... | Yes |
What is the smallest number \( A \) such that any two squares \( {S}_{1},{S}_{2} \) of total area 1 can be put disjointly into a rectangle of area \( A \) ? | Let us call \( a \) and \( b \) the lengths of the sides of \( {S}_{1} \) and \( {S}_{2} \), and we may assume that \( a \geq b \geq 0 \) . Then the smallest rectangle that will contain the two squares disjointly has sides \( a \) and \( a + b \), and area \( a\left( {a + b}\right) \), as shown in Figure 3.7.5. The lar... | Yes |
Find the critical points of \( F\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {xyz} \) on the plane given by the equation \( f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = x + {2y} + {3z} - 1 = 0 \) . | Theorem 3.7.5 asserts that a critical point is a solution to\n\n\[ \text{1.}\underset{\text{deriv. of }F}{\underbrace{\left\lbrack yz, xz, xy\right\rbrack }} = \lambda \underset{\begin{matrix} \text{deriv. of }f \\ \text{(constraint) } \\ \text{ or } \end{matrix}}{\underbrace{\left\lbrack 1,2,3\right\rbrack }}\;{yz} = ... | Yes |
Example 3.7.10 (Checking boundary values). Let \( {X}_{a} \) be the part of the sphere of equation \( {x}^{2} + {y}^{2} + {z}^{2} = 1 \) where \( x + y + z \geq a \), as shown in Figure 3.7.6 (the analogous case for the circle, which may be easier to visualize, is shown in Figure 3.7.7). In terms of the parameter \( a ... | Since \( {X}_{a} \) is compact and \( f \) is continuous, there is a maximum and a minimum for every \( a \) satisfying \( \left| a\right| \leq \sqrt{3} \) . Each maximum and each minimum occurs either in the interior of \( {X}_{a} \) or on the boundary.\n\nA first problem is to find the critical points of \( f \) on t... | No |
Lemma 3.7.11. Let \( A = \left\lbrack \begin{matrix} {\alpha }_{1} \\ \vdots \\ {\alpha }_{m} \end{matrix}\right\rbrack : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) and \( \beta : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) be linear transformations. Then\n\n\( \ker A \subset \ker \beta \)\n\nif and only if there ... | Proof of Lemma 3.7.11. In one direction, if\n\n\[ \beta = {\lambda }_{1}{\alpha }_{1} + \cdots + {\lambda }_{m}{\alpha }_{m} \]\n\nand \( \overrightarrow{\mathbf{v}} \in \ker A \), then \( \overrightarrow{\mathbf{v}} \in \ker {\alpha }_{i} \) for \( i = 1,\ldots, m \), so \( \overrightarrow{\mathbf{v}} \in \ker \beta \... | Yes |
Theorem 3.7.13 (Computing the signature of a constrained critical point). Let \( U \subset {\mathbb{R}}^{n + m} \) be open, \( f : U \rightarrow \mathbb{R} \) a \( {C}^{2} \) function, and \( \mathbf{F} : U \rightarrow {\mathbb{R}}^{m} \) a \( {C}^{2} \) map defining a manifold \( Z\overset{\text{ def }}{ = }{\mathbf{F... | Theorem 3.7.13 is proved in Appendix A14. | Yes |
We will find and classify the critical points of \( f\left( \begin{array}{l} x \\ y \\ z \end{array}\right) = {ax} + {by} + {cz} \) on the surface of equation \( 1/x + 1/y + 1/z = 1 \), all in terms of the parameters \( a, b, c \) . | In this case the Lagrange multiplier equation becomes\n\n\[ \left\lbrack {a, b, c}\right\rbrack = \left\lbrack {-\frac{\lambda }{{x}^{2}}, - \frac{\lambda }{{y}^{2}}, - \frac{\lambda }{{z}^{2}}}\right\rbrack ,\]\n\nleading to the four equations\n\n\[ {x}^{2} = - \frac{\lambda }{a},\;{y}^{2} = - \frac{\lambda }{b},\;{z}... | Yes |
Theorem 3.7.15 (Spectral theorem). Let \( A \) be a symmetric \( n \times n \) matrix with real entries. Then there exists an orthonormal basis \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{n} \) of \( {\mathbb{R}}^{n} \) and numbers \( {\lambda }_{1},\ldots ,{\lambda }_{n} \in \mathbb{R} ... | Proof. Our strategy will be to consider the function \( {Q}_{A} : {\mathbb{R}}^{n} \rightarrow \mathbb{R} \) given by \( {Q}_{A}\left( \overrightarrow{\mathbf{x}}\right) = \overrightarrow{\mathbf{x}} \cdot A\overrightarrow{\mathbf{x}} \), subject to various constraints. The first constraint ensures that the first basis... | No |
Theorem 3.7.16. Let \( A \) be a real \( n \times n \) symmetric matrix. The quadratic form \( {Q}_{A} \) has signature \( \left( {k, l}\right) \) if and only if there exists an orthonormal basis \( {\overrightarrow{\mathbf{v}}}_{i},\ldots ,{\overrightarrow{\mathbf{v}}}_{n} \) of \( {\mathbb{R}}^{n} \) with \( A{\overr... | Proof. By the spectral theorem, there exists an orthonormal eigenbasis \( {\overrightarrow{\mathbf{v}}}_{1},\ldots ,{\overrightarrow{\mathbf{v}}}_{n} \) of \( {\mathbb{R}}^{n} \) for \( A \) ; denote by \( {\lambda }_{1},\ldots ,{\lambda }_{n} \) the corresponding eigenvalues. Then (Proposition 2.4.18) any vector \( \o... | Yes |
Lemma 3.8.2. The symmetric matrices \( {A}^{\top }A \) and \( A{A}^{\top } \) have the same nonzero eigenvalues with the same multiplicities: for all \( \lambda \neq 0 \), the map\n\n\[ \overrightarrow{\mathbf{v}} \mapsto \frac{1}{\sqrt{\lambda }}A\overrightarrow{\mathbf{v}} \]\n\nis an isomorphism \( \ker \left( {{A}^... | Proof of Lemma 3.8.2. If \( \overrightarrow{\mathbf{v}} \) is in \( \ker \left( {{A}^{\top }A - {\lambda I}}\right) \), i.e., \( {A}^{\top }A\overrightarrow{\mathbf{v}} = \lambda \overrightarrow{\mathbf{v}} \), then \( A\overrightarrow{\mathbf{v}}/\sqrt{\lambda } \) is in \( \ker \left( {A{A}^{\top } - {\lambda I}}\rig... | Yes |
Proposition 3.8.3. For any real matrix \( A \), the norm of \( A \) is\n\n\[ \parallel A\parallel = \mathop{\max }\limits_{{\lambda \text{ eigenvalue of }{A}^{\top }A}}\sqrt{\lambda }.\] | Proof. By Proposition and Definition 2.4.19, an orthogonal matrix preserves lengths. Thus if \( P \) and \( Q \) are orthogonal, and \( B \) any matrix (not necessarily square) \( \parallel {QB}\parallel = \mathop{\sup }\limits_{{\left| \overrightarrow{\mathbf{v}}\right| = 1}}\left| {{QB}\overrightarrow{\mathbf{v}}}\ri... | Yes |
Proposition 3.8.7. Let \( f, g : \left( {S,\mathbf{P}}\right) \rightarrow \mathbb{R} \) be two random variables. Then\n\n1. \( \left| {\operatorname{corr}\left( {f, g}\right) }\right| \leq 1 \)\n\n2. \( \left| {\operatorname{corr}\left( {f, g}\right) }\right| = 1 \) if and only if there exist \( a, b \in \mathbb{R} \) ... | Proof. 1. By Schwarz's inequality,\n\n\[ \left| {\operatorname{cov}\left( {f, g}\right) }\right| = \left| {\langle \widetilde{f},\widetilde{g}{\rangle }_{\left( S,\mathbf{P}\right) }}\right| \underset{\left( 1\right) }{ = }\left| {\Phi \left( \widetilde{f}\right) \cdot \Phi \left( \widetilde{g}\right) }\right| \]\n\n\[... | Yes |
Example 3.8.8. Let our sample space be the subset of the plane consisting of the set of points \( \left( \begin{matrix} \pm {x}_{i} \\ {x}_{i}^{2} \end{matrix}\right) \) for some collection of \( m \) numbers \( {x}_{i} \neq 0 \), all given the same probability \( 1/\left( {2m}\right) \) . Let \( f \) and \( g \) be th... | \[ f\left( \begin{matrix} \pm {x}_{i} \\ {x}_{i}^{2} \end{matrix}\right) = \pm {x}_{i},\;g\left( \begin{matrix} \pm {x}_{i} \\ {x}_{i}^{2} \end{matrix}\right) = {x}_{i}^{2},\;\text{ so }g = {f}^{2}; \] | Yes |
Proposition 3.9.2 (Computing the curvature of a plane curve known as a graph). The curvature \( \kappa \) of the curve \( y = f\\left( x\\right) \) at \( \\left( \\begin{matrix} a \\\\ f\\left( a\\right) \\end{matrix}\\right) \) is | \[ \kappa \\left( \\begin{matrix} a \\\\ f\\left( a\\right) \\end{matrix}\\right) = \\frac{\\left| {f}^{\\prime \\prime }\\left( a\\right) \\right| }{{\\left( 1 + {\\left( {f}^{\\prime }\\left( a\\right) \\right) }^{2}\\right) }^{3/2}}. \] | Yes |
The curvature of the curve \( y = {x}^{2} \) at the point \( \left( \begin{matrix} a \\ {a}^{2} \end{matrix}\right) \) is | \[ \kappa = \frac{2}{{\left( 1 + 4{a}^{2}\right) }^{3/2}}. \] | Yes |
We saw in Example 3.4.8 that\n\n\[ \n{x}^{3} + {xy} + {y}^{3} = 3 \n\]\n\nimplicitly expresses \( x \) in terms of \( y \) near \( \left( \begin{array}{l} 1 \\ 1 \end{array}\right) \) and that the first and second derivatives at 1 of the implicit function \( g \) are \( {g}^{\prime }\left( 1\right) = - 1 \), and \( {g}... | \[ \n\kappa = \frac{\left| {f}^{\prime \prime }\left( a\right) \right| }{{\left( 1 + {\left( {f}^{\prime }\left( a\right) \right) }^{2}\right) }^{3/2}} = \frac{\left| -5/2\right| }{{2}^{3/2}} = \frac{5}{4\sqrt{2}}. \n\] | Yes |
Proposition 3.9.11 (Computing curvature of surfaces). Let \( S \) and \( \overrightarrow{\mathbf{n}} \) be as in Proposition 3.9.10.\n\n1. The Gaussian curvature of \( S \) at the origin is\n\n\[ K\left( \mathbf{0}\right) = \frac{{a}_{2,0}{a}_{0,2} - {a}_{1,1}^{2}}{{\left( 1 + {c}^{2}\right) }^{2}}. \] | Proof of Proposition 3.9.11. For the Gaussian curvature, plug into Definition 3.9.8 the values for \( {A}_{2,0},{A}_{1,1},{A}_{0,2} \) given in Proposition 3.9.10:\n\n\[ K\left( \mathbf{0}\right) = {A}_{2,0}{A}_{0,2} - {A}_{1,1}^{2} \]\n\n\[ = \frac{1}{{c}^{4}{\left( 1 + {c}^{2}\right) }^{2}}\left( {\left( {{a}_{2,0}{a... | Yes |
Suppose we want to measure the Gaussian curvature at a point \( \mathbf{a}\overset{\text{ def }}{ = }\left( \begin{matrix} a \\ b \\ {a}^{2} - {b}^{2} \end{matrix}\right) \) of the surface given by the equation \( z = {x}^{2} - {y}^{2} \) and shown in Figure 3.9.8. | To determine the Gaussian curvature at the point \( \mathbf{a} \), we make \( \mathbf{a} \) our new origin; i.e., we use new translated coordinates, \( u, v, w \), where\n\n\[ x = a + u \]\n\n\[ y = b + v \]\n\n\[ z = {a}^{2} - {b}^{2} + w \]\n\nNow we rewrite \( z = {x}^{2} - {y}^{2} \) as\n\n\[ \underset{z}{\underbra... | Yes |
Proposition 3.9.15 (Frenet frame). Let \( \delta : \mathbb{R} \rightarrow {\mathbb{R}}^{3} \) parametrize a curve \( C \) by arc length, with \( {\overrightarrow{\delta }}^{\prime \prime }\left( s\right) \neq 0 \) . Then the coordinates with respect to the Frenet frame at 0 , \[ \overrightarrow{\mathbf{t}}\left( 0\righ... | Thus the point with coordinates \( X, Y, Z \) in the new, adapted coordinates is the point \[ \delta \left( 0\right) + X\overrightarrow{\mathbf{t}}\left( 0\right) + Y\overrightarrow{\mathbf{n}}\left( 0\right) + Z\overrightarrow{\mathbf{b}}\left( 0\right) \] in the old \( \left( {x, y, z}\right) \) -coordinates. | Yes |
Proposition 3.9.16 (Frenet frame related to curvature and torsion). Let \( \delta : \mathbb{R} \rightarrow {\mathbb{R}}^{3} \) parametrize a curve \( C \) by arc length. Then the Frenet frame satisfies the following equations, where \( \kappa \) is the curvature of the curve at \( \delta \left( 0\right) \) and \( \tau ... | \[ {\overrightarrow{\mathbf{t}}}^{\prime }\left( 0\right) = \;\kappa \overrightarrow{\mathbf{n}}\left( 0\right) \] \[ {\overrightarrow{\mathbf{n}}}^{\prime }\left( 0\right) = - \kappa \overrightarrow{\mathbf{t}}\left( 0\right) \; + \tau \overrightarrow{\mathbf{b}}\left( 0\right) \] \[ {\overrightarrow{\mathbf{b}}}^{\pr... | Yes |
Proposition 3.9.18 (Torsion of a parametrized curve). The torsion \( \tau \) of a curve parametrized by \( \gamma : \mathbb{R} \rightarrow {\mathbb{R}}^{3} \) is | \[ \tau \left( {\gamma \left( t\right) }\right) = \frac{\left( {{\overrightarrow{\gamma }}^{\prime }\left( t\right) \times {\overrightarrow{\gamma }}^{\prime \prime }\left( t\right) }\right) \cdot {\overrightarrow{\gamma }}^{\prime \prime \prime }\left( t\right) }{{\left| {\overrightarrow{\gamma }}^{\prime }\left( t\ri... | Yes |
Let \( \gamma \left( t\right) = \left( \begin{matrix} t \\ {t}^{2} \\ {t}^{3} \end{matrix}\right) \). Then | \[ {\overrightarrow{\gamma }}^{\prime }\left( t\right) = \left( \begin{matrix} 1 \\ {2t} \\ 3{t}^{2} \end{matrix}\right) ,\;{\overrightarrow{\gamma }}^{\prime \prime }\left( t\right) = \left( \begin{matrix} 0 \\ 2 \\ {6t} \end{matrix}\right) ,\;{\overrightarrow{\gamma }}^{\prime \prime \prime }\left( t\right) = \left( ... | Yes |
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