Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Let \( z = 2\sqrt{3} + {2i} \) and \( w = - 1 + i\sqrt{3} \). Use Theorem 11.16 to find the following\n\n1. \( {zw} \) 2. \( {w}^{5} \) 3. \( \frac{z}{w} \n\nWrite your final answers in rectangular form. | Solution. In order to use Theorem 11.16, we need to write \( z \) and \( w \) in polar form. For \( z = 2\sqrt{3} + {2i} \), we find \( \left| z\right| = \sqrt{{\left( 2\sqrt{3}\right) }^{2} + {\left( 2\right) }^{2}} = \sqrt{16} = 4 \). If \( \theta \in \arg \left( z\right) \), we know \( \tan \left( \theta \right) = \... | Yes |
Theorem 11.17. The \( {n}^{\text{th }} \) roots of a Complex Number: Let \( z \neq 0 \) be a complex number with polar form \( z = r\operatorname{cis}\left( \theta \right) \). For each natural number \( n, z \) has \( n \) distinct \( {n}^{\text{th }} \) roots, which we denote by \( {w}_{0},{w}_{1},\ldots ,{w}_{n - 1} ... | The proof of Theorem 11.17 breaks into to two parts: first, showing that each \( {w}_{k} \) is an \( {n}^{\text{th }} \) root, and second, showing that the set \( \left\{ {{w}_{k} \mid k = 0,1,\ldots ,\left( {n - 1}\right) }\right\} \) consists of \( n \) different complex numbers. To show \( {w}_{k} \) is an \( {n}^{\... | Yes |
1. both square roots of \( z = - 2 + {2i}\sqrt{3} \) | We start by writing \( z = - 2 + {2i}\sqrt{3} = 4\operatorname{cis}\left( \frac{2\pi }{3}\right) \) . To use Theorem 11.17, we identify \( r = 4 \) , \( \theta = \frac{2\pi }{3} \) and \( n = 2 \) . We know that \( z \) has two square roots, and in keeping with the notation in Theorem 11.17, we’ll call them \( {w}_{0} ... | Yes |
A plane leaves an airport with an airspeed \( {}^{5} \) of 175 miles per hour at a bearing of \( \mathrm{N}{40}^{ \circ }\mathrm{E} \) . A \( {35}\mathrm{{mile}} \) per hour wind is blowing at a bearing of \( \mathrm{S}{60}^{ \circ }\mathrm{E} \) . Find the true speed of the plane, rounded to the nearest mile per hour,... | Solution: For both the plane and the wind, we are given their speeds and their directions. Coupling speed (as a magnitude) with direction is the concept of velocity which we've seen a few times before in this textbook. \( {}^{6} \) We let \( \overrightarrow{v} \) denote the plane’s velocity and \( \overrightarrow{w} \)... | Yes |
Let \( \overrightarrow{v} = \langle 3,4\rangle \) and suppose \( \overrightarrow{w} = \overrightarrow{PQ} \) where \( P\left( {-3,7}\right) \) and \( Q\left( {-2,5}\right) \) . Find \( \overrightarrow{v} + \overrightarrow{w} \) and interpret this sum geometrically. | Solution. Before can add the vectors using Definition 11.6, we need to write \( \overrightarrow{w} \) in component form. Using Definition 11.5, we get \( \overrightarrow{w} = \langle - 2 - \left( {-3}\right) ,5 - 7\rangle = \langle 1, - 2\rangle \) . Thus\n\n\[ \overrightarrow{v} + \overrightarrow{w} = \langle 3,4\rang... | Yes |
Theorem 11.19. Properties of Scalar Multiplication\n\n- Associative Property: For every vector \( \overrightarrow{v} \) and scalars \( k \) and \( r,\left( {kr}\right) \overrightarrow{v} = k\left( {r\overrightarrow{v}}\right) \) . | The proof of Theorem 11.19, like the proof of Theorem 11.18, ultimately boils down to the definition of scalar multiplication and properties of real numbers. For example, to prove the associative property, we let \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \) . If \( k \) and \( r \) are scalars... | Yes |
Solve \( 5\overrightarrow{v} - 2\left( {\overrightarrow{v}+\langle 1, - 2\rangle }\right) = \overrightarrow{0} \) for \( \overrightarrow{v} \) . | \[ 5\overrightarrow{v} - 2\left( {\overrightarrow{v}+\langle 1, - 2\rangle }\right) = \overrightarrow{0} \] \[ 5\overrightarrow{v} + \left( {-1}\right) \left\lbrack {2\left( {\overrightarrow{v}+\langle 1, - 2\rangle }\right) }\right\rbrack = \overrightarrow{0} \] \[ 5\overrightarrow{v} + \left\lbrack {\left( {-1}\right... | Yes |
Theorem 11.20. Properties of Magnitude and Direction: Suppose \( \overrightarrow{v} \) is a vector.\n\n- \( \parallel \overrightarrow{v}\parallel \geq 0 \) and \( \parallel \overrightarrow{v}\parallel = 0 \) if and only if \( \overrightarrow{v} = \overrightarrow{0} \) | The proof of the first property in Theorem 11.20 is a direct consequence of the definition of \( \parallel \overrightarrow{v}\parallel \) . If \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \), then \( \parallel \overrightarrow{v}\parallel = \sqrt{{v}_{1}^{2} + {v}_{2}^{2}} \) which is by definitio... | Yes |
Find the component form of the vector \( \overrightarrow{v} \) with \( \parallel \overrightarrow{v}\parallel = 5 \) so that when \( \overrightarrow{v} \) is plotted in standard position, it lies in Quadrant II and makes a \( {60}^{ \circ } \) angle \( {}^{12} \) with the negative \( x \) -axis. | We are told that \( \parallel \overrightarrow{v}\parallel = 5 \) and are given information about its direction, so we can use the formula \( \overrightarrow{v} = \parallel \overrightarrow{v}\parallel \widehat{v} \) to get the component form of \( \overrightarrow{v} \) . To determine \( \widehat{v} \), we appeal to Defi... | Yes |
Theorem 11.21. Principal Vector Decomposition Theorem: Let \( \overrightarrow{v} \) be a vector with component form \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \) . Then \( \overrightarrow{v} = {v}_{1}\widehat{\imath } + {v}_{2}\widehat{\jmath } \) . | The proof of Theorem 11.21 is straightforward. Since \( \widehat{\imath } = \langle 1,0\rangle \) and \( \widehat{\jmath } = \langle 0,1\rangle \), we have from the definition of scalar multiplication and vector addition that\n\n\[ \n{v}_{1}\widehat{\imath } + {v}_{2}\widehat{\jmath } = {v}_{1}\langle 1,0\rangle + {v}_... | Yes |
Commutative Property: For all vectors \( \overrightarrow{v} \) and \( \overrightarrow{w},\overrightarrow{v} \cdot \overrightarrow{w} = \overrightarrow{w} \cdot \overrightarrow{v} \) . | To show the commutative property for instance, let \( \overrightarrow{v} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \) and \( \overrightarrow{w} = \left\langle {{w}_{1},{w}_{2}}\right\rangle \) . Then\n\n\[ \overrightarrow{v} \cdot \overrightarrow{w} = \left\langle {{v}_{1},{v}_{2}}\right\rangle \cdot \left\langle {... | Yes |
Prove the identity: \( \parallel \overrightarrow{v} - \overrightarrow{w}{\parallel }^{2} = \parallel \overrightarrow{v}{\parallel }^{2} - 2\left( {\overrightarrow{v} \cdot \overrightarrow{w}}\right) + \parallel \overrightarrow{w}{\parallel }^{2} \) . | Solution. We begin by rewriting \( \parallel \overrightarrow{v} - \overrightarrow{w}{\parallel }^{2} \) in terms of the dot product using Theorem 11.22.\n\n\[ \parallel \overrightarrow{v} - \overrightarrow{w}{\parallel }^{2} = \left( {\overrightarrow{v} - \overrightarrow{w}}\right) \cdot \left( {\overrightarrow{v} - \o... | Yes |
Theorem 11.24. Let \( \overrightarrow{v} \) and \( \overrightarrow{w} \) be nonzero vectors and let \( \theta \) the angle between \( \overrightarrow{v} \) and \( \overrightarrow{w} \) . Then\n\n\[ \theta = \arccos \left( \frac{\overrightarrow{v} \cdot \overrightarrow{w}}{\parallel \overrightarrow{v}\parallel \parallel... | We obtain the formula in Theorem 11.24 by solving the equation given in Theorem 11.23 for \( \theta \) . Since \( \overrightarrow{v} \) and \( \overrightarrow{w} \) are nonzero, so are \( \parallel \overrightarrow{v}\parallel \) and \( \parallel \overrightarrow{w}\parallel \) . Hence, we may divide both sides of \( \ov... | Yes |
Find the angle between the following pairs of vectors. | Solution. We use the formula \( \theta = \arccos \left( \frac{\overrightarrow{v} \cdot \overrightarrow{w}}{\parallel \overrightarrow{v}\parallel \parallel \overrightarrow{w}\parallel }\right) \) from Theorem 11.24 in each case below.\n\n1. We have \( \overrightarrow{v} \cdot \overrightarrow{w} = \langle 3, - 3\sqrt{3}\... | Yes |
Theorem 11.25. The Dot Product Detects Orthogonality: Let \( \overrightarrow{v} \) and \( \overrightarrow{w} \) be nonzero vectors. Then \( \overrightarrow{v} \bot \overrightarrow{w} \) if and only if \( \overrightarrow{v} \cdot \overrightarrow{w} = 0 \) . | To prove Theorem 11.25, we first assume \( \overrightarrow{v} \) and \( \overrightarrow{w} \) are nonzero vectors with \( \overrightarrow{v} \bot \overrightarrow{w} \) . By definition, the angle between \( \overrightarrow{v} \) and \( \overrightarrow{w} \) is \( \frac{\pi }{2} \) . By Theorem 11.23, \( \overrightarrow{... | Yes |
Let \( {L}_{1} \) be the line \( y = {m}_{1}x + {b}_{1} \) and let \( {L}_{2} \) be the line \( y = {m}_{2}x + {b}_{2} \) . Prove that \( {L}_{1} \) is perpendicular to \( {L}_{2} \) if and only if \( {m}_{1} \cdot {m}_{2} = - 1 \) . | Solution. Our strategy is to find two vectors: \( \overrightarrow{{v}_{1}} \), which has the same direction as \( {L}_{1} \), and \( \overrightarrow{{v}_{2}} \) , which has the same direction as \( {L}_{2} \) and show \( \overrightarrow{{v}_{1}} \bot \overrightarrow{{v}_{2}} \) if and only if \( {m}_{1}{m}_{2} = - 1 \)... | Yes |
Let \( \overrightarrow{v} = \langle 1,8\rangle \) and \( \overrightarrow{w} = \langle - 1,2\rangle \) . Find \( \overrightarrow{p} = {\operatorname{proj}}_{\overrightarrow{w}}\left( \overrightarrow{v}\right) \), and plot \( \overrightarrow{v},\overrightarrow{w} \) and \( \overrightarrow{p} \) in standard position. | Solution. We find \( \overrightarrow{v} \cdot \overrightarrow{w} = \langle 1,8\rangle \cdot \langle - 1,2\rangle = \left( {-1}\right) + {16} = {15} \) and \( \overrightarrow{w} \cdot \overrightarrow{w} = \langle - 1,2\rangle \cdot \langle - 1,2\rangle = 1 + 4 = 5 \) . Hence, \( \overrightarrow{p} = \frac{\overrightarro... | Yes |
Theorem 11.27. Generalized Decomposition Theorem: Let \( \overrightarrow{v} \) and \( \overrightarrow{w} \) be nonzero vectors. There are unique vectors \( \overrightarrow{p} \) and \( \overrightarrow{q} \) such that \( \overrightarrow{v} = \overrightarrow{p} + \overrightarrow{q} \) where \( \overrightarrow{p} = k\over... | To prove Theorem 11.27, we take \( \overrightarrow{p} = {\operatorname{proj}}_{\overrightarrow{w}}\left( \overrightarrow{v}\right) \) and \( \overrightarrow{q} = \overrightarrow{v} - \overrightarrow{p} \) . Then \( \overrightarrow{p} \) is, by definition, a scalar multiple of \( \overrightarrow{w} \) . Next, we compute... | Yes |
Taylor exerts a force of 10 pounds to pull her wagon a distance of 50 feet over level ground. If the handle of the wagon makes a \( {30}^{ \circ } \) angle with the horizontal, how much work did Taylor do pulling the wagon? Assume Taylor exerts the force of 10 pounds at a \( {30}^{ \circ } \) angle for the duration of ... | Solution. There are two ways to attack this problem. One way is to find the vectors \( \overrightarrow{F} \) and \( \overrightarrow{PQ} \) mentioned in Theorem 11.28 and compute \( W = \overrightarrow{F} \cdot \overrightarrow{PQ} \). To do this, we assume the origin is at the point where the handle of the wagon meets t... | Yes |
Sketch the curve described by \( \\left\\{ \\begin{array}{l} x = {t}^{2} - 3 \\\\ y = {2t} - 1 \\end{array}\\right. \) for \( t \\geq - 2 \) . | Solution. We follow the same procedure here as we have time and time again when asked to graph anything new - choose friendly values of \( t \), plot the corresponding points and connect the results in a pleasing fashion. Since we are told \( t \\geq - 2 \), we start there and as we plot successive points, we draw an a... | Yes |
Sketch the curves described by the following parametric equations. 1. \( \\left\\{ \\begin{array}{l} x = {t}^{3} \\\\ y = 2{t}^{2} \\end{array}\\right. \) for \( - 1 \\leq t \\leq 1 \) | To get a feel for the curve described by the system \( \\left\\{ {x = {t}^{3}, y = 2{t}^{2}}\\right. \) we first sketch the graphs of \( x = {t}^{3} \) and \( y = 2{t}^{2} \) over the interval \( \\left\\lbrack {-1,1}\\right\\rbrack \) . We note that as \( t \) takes on values in the interval \( \\left\\lbrack {-1,1}\\... | Yes |
Find a parametrization for each of the following curves and check your answers.\n\n1. \( y = {x}^{2} \) from \( x = - 3 \) to \( x = 2 \) | Since \( y = {x}^{2} \) is written in the form \( y = f\left( x\right) \), we let \( x = t \) and \( y = f\left( t\right) = {t}^{2} \) . Since \( x = t \), the bounds on \( t \) match precisely the bounds on \( x \) so we get \( \left\{ {x = t, y = {t}^{2}}\right. \) for \( - 3 \leq t \leq 2 \) . The check is almost tr... | Yes |
Find a parametrization for the following curves.\n\n1. The curve which starts at \( \left( {2,4}\right) \) and follows the parabola \( y = {x}^{2} \) to end at \( \left( {-1,1}\right) \) . Shift the parameter so that the path starts at \( t = 0 \) . | Solution.\n\n1. We can parametrize \( y = {x}^{2} \) from \( x = - 1 \) to \( x = 2 \) using the formula given on Page 1053 as \( \left\{ {x = t, y = {t}^{2}}\right. \) for \( - 1 \leq t \leq 2 \) . This parametrization, however, starts at \( \left( {-1,1}\right) \) and ends at \( \left( {2,4}\right) \) . Hence, we nee... | Yes |
Find the parametric equations of a cycloid which results from a circle of radius 3 rolling down the positive \( x \) -axis as described above. | Solution. We have \( r = 3 \) which gives the equations \( \{ x = 3\left( {t - \sin \left( t\right) }\right), y = 3\left( {1 - \cos \left( t\right) }\right) \) for \( t \geq 0 \) . (Here we have returned to the convention of using \( t \) as the parameter.) Sketching the cycloid by hand is a wonderful exercise in Calcu... | Yes |
1. Find \( {P}_{3}\left( x\right) \) about \( {x}_{0} = \pi /2 \) and use it to approximate \( f\left( {0.8}\right) \) . | Solution. 1. First note that \( f\left( {\pi /2}\right) = - \pi /2 \) . Differentiating \( f \) we get:\n\n\[ \n{f}^{\prime }\left( x\right) = \cos x - x\sin x - 1 \Rightarrow {f}^{\prime }\left( {\pi /2}\right) = - \pi /2 - 1 \n\]\n\n\[ \n{f}^{\prime \prime }\left( x\right) = - 2\sin x - x\cos x \Rightarrow {f}^{\prim... | Yes |
Example 9. Find the floating-point representation of 10.375. | Solution. You can check that \( {10} = {\left( {1010}\right) }_{2} \) and \( {0.375} = {\left( {.011}\right) }_{2} \) by computing\n\n\[ \n{10} = 0 \times {2}^{0} + 1 \times {2}^{1} + 0 \times {2}^{2} + 1 \times {2}^{3} \n\] \n\n\[ \n{0.375} = 0 \times {2}^{-1} + 1 \times {2}^{-2} + 1 \times {2}^{-3}. \n\] \n\nThen \n\... | Yes |
Find 5-digit \( \left( {k = 5}\right) \) chopping and rounding values of the numbers below:\n\n- \( \pi = {0.314159265}\ldots \times {10}^{1} \) | Chopping gives \( {fl}\left( \pi \right) = {0.31415} \) and rounding gives \( {fl}\left( \pi \right) = {0.31416} \) . | Yes |
Find absolute and relative errors of\n\n1. \( x = {0.20} \times {10}^{1},{x}^{ * } = {0.21} \times {10}^{1} \)\n\n2. \( x = {0.20} \times {10}^{-2},{x}^{ * } = {0.21} \times {10}^{-2} \)\n\n3. \( x = {0.20} \times {10}^{5},{x}^{ * } = {0.21} \times {10}^{5} \) | Notice how the only difference in the three cases is the exponent of the numbers. The absolute errors are: \( {0.01} \times {10},{0.01} \times {10}^{-2},{0.01} \times {10}^{5} \) . The absolute errors are different since the exponents are different. However, the relative error in each case is the same: 0.05 . | Yes |
Lemma 15. The relative error of approximating \( x \) by \( {fl}\left( x\right) \) in the \( k \) -digit normalized decimal floating-point representation satisfies | Proof. We will give the proof for chopping; the proof for rounding is similar but tedious. Let\n\n\[ x = 0.{d}_{1}{d}_{2}\ldots {d}_{k}{d}_{k + 1}\ldots \times {10}^{n}. \]\n\nThen\n\n\[ {fl}\left( x\right) = 0.{d}_{1}{d}_{2}\ldots {d}_{k} \times {10}^{n} \]\n\nif chopping is used. Observe that\n\n\[ \frac{\left| x - f... | Yes |
Let's revisit the calculation of\n\n\[ \frac{1 - \cos x}{\sin x}. \] | Observe that using the algebraic identity\n\n\[ \frac{1 - \cos x}{\sin x} = \frac{\sin x}{1 + \cos x} \]\n\nremoves both difficulties encountered before: there is no cancellation of significant digits and division by a small number. Using five-digit rounding, we have\n\n\[ {fl}\left( \frac{\sin {0.1}}{1 + \cos {0.1}}\r... | Yes |
Consider the quadratic formula: the solution of \( a{x}^{2} + {bx} + c = 0 \) is\n\n\[ \n{r}_{1} = \frac{-b + \sqrt{{b}^{2} - {4ac}}}{2a},{r}_{2} = \frac{-b - \sqrt{{b}^{2} - {4ac}}}{2a}.\n\]\n\nIf \( \left| b\right| \approx \sqrt{{b}^{2} - {4ac}} \), then we have a potential loss of precision in computing one of the r... | Next will use four-digit arithmetic with rounding to compute the roots:\n\n\[ \n{fl}\left( \sqrt{117}\right) = {10.82}\n\]\n\n\[ \n{fl}({r}_{1}) = {fl}\left( \frac{{fl}({fl}({11.0}) + {fl}(\sqrt{117}))}{{fl}({2.0})}\right) = {fl}\left( \frac{{fl}({11.0} + {10.82})}{2.0}\right) = {fl}\left( \frac{21.82}{2.0}\right) = {1... | Yes |
For a simple example, consider using four-digit arithmetic with rounding, and computing the average of two numbers, \( \frac{a + b}{2} \) . For \( a = {2.954} \) and \( b = {100.9} \), the true average is 51.927 . However, four-digit arithmetic with rounding yields: | \[ {fl}\left( \frac{{100.9} + {2.954}}{2}\right) = {fl}\left( \frac{{fl}\left( {103.854}\right) }{2}\right) = {fl}\left( \frac{103.9}{2}\right) = {51.95} \] which has a relative error of \( {4.43} \times {10}^{-4} \) . If we rewrite the averaging formula as \( a + \frac{b - a}{2} \), on the other hand, we obtain 51.93,... | Yes |
There are two standard formulas given in textbooks to compute the sample variance \( {s}^{2} \) of the numbers \( {x}_{1},\ldots ,{x}_{n} \) :\n\n1. \( {s}^{2} = \frac{1}{n - 1}\left\lbrack {\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}^{2} - \frac{1}{n}{\left( \mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i}\right) }^{2}}\righ... | For an example, consider four-digit rounding arithmetic, and let the data be \( {1.253},{2.411},{3.174} \) . The sample variance from formula 1 and formula 2 are,0.93 and 0.9355, respectively. The exact value, up to 6 digits, is 0.935562 . Formula 2 is a numerically more stable choice for computing the variance than th... | No |
Example 22. We have a sum to compute:\n\n\[ \n{e}^{-7} = 1 + \frac{-7}{1} + \frac{{\\left( -7\\right) }^{2}}{2!} + \frac{{\\left( -7\\right) }^{3}}{3!} + \\ldots + \frac{{\\left( -7\\right) }^{n}}{n!}.\n\] | Julia reports the \ | No |
A quantity and its \( 1/4 \) added together become 15 . What is the quantity? | Assume 4.\n\n\( \smallsetminus 1 \) 4\n\n\( \smallsetminus 1/4 \) 1\n\nTotal 5\n\nAs many times as 5 must be multiplied to give 15, so many times 4 must be multiplied to give the required number. Multiply 5 so as to get 15.\n\n\\1 5\n\n10\n\nTotal 3\n\nMultiply 3 by 4.\n\n1 3\n\n2\n\n6\n\n\\4 12\n\nThe quantity is\n\n\... | No |
Consider the sequences defined by\n\n\[ \n{p}_{n + 1} = {0.7}{p}_{n}\\text{and}{p}_{1} = 1 \n\]\n\n\[ \n{p}_{n + 1} = {0.7}{p}_{n}^{2}\\text{ and }{p}_{1} = 1 \n\] | The first sequence converges to 0 linearly, and the second quadratically. Here are a few iterations of the sequences:\n\n<table><thead><tr><th>\\( n \\)</th><th>Linear</th><th>Quadratic</th></tr></thead><tr><td>1</td><td>0.7</td><td>0.7</td></tr><tr><td>4</td><td>0.24</td><td>\\( {4.75} \\times {10}^{-3} \\)</td></tr><... | No |
Compute the first three iterations by hand for the function plotted in Figure (2.2). | Step 1 : To start, we need to pick an interval \( \left\lbrack {a, b}\right\rbrack \) that contains the root, that is, \( f\left( a\right) f\left( b\right) < 0 \) . From the plot, it is clear that \( \left\lbrack {0,4}\right\rbrack \) is a possible choice. In the next few steps, we will be working with a sequence of in... | Yes |
Theorem 28. Suppose that \( f \in {C}^{0}\left\lbrack {a, b}\right\rbrack \) and \( f\left( a\right) f\left( b\right) < 0 \) . The bisection method generates a sequence \( \left\{ {p}_{n}\right\} \) approximating a zero \( p \) of \( f\left( x\right) \) with\n\n\[ \left| {{p}_{n} - p}\right| \leq \frac{b - a}{{2}^{n}},... | Proof. Let the sequences \( \left\{ {a}_{n}\right\} \) and \( \left\{ {b}_{n}\right\} \) denote the left-end and right-end points of the subintervals generated by the bisection method. Since at each step the interval is halved, we have\n\n\[ {b}_{n} - {a}_{n} = \frac{1}{2}\left( {{b}_{n - 1} - {a}_{n - 1}}\right) .\n\n... | Yes |
Corollary 29. The bisection method has linear convergence. | Proof. The bisection method does not satisfy (2.1) for any \( C < 1 \), but it satisfies a variant of (2.2) with \( C = 1/2 \) from the previous theorem. | Yes |
Example 30. Determine the number of iterations necessary to solve \( f\\left( x\\right) = {x}^{5} + 2{x}^{3} - {5x} - 2 = 0 \) with accuracy \( {10}^{-4}, a = 0, b = 2 \). | Solution. Since \( n \\geq {\\log }_{2}\\left( \\frac{2}{{10}^{-4}}\\right) = 4{\\log }_{2}{10} + 1 = {14.3} \), the number of required iterations is 15. | Yes |
Corollary 33. Newton's method has quadratic convergence. | Proof. Recall that quadratic convergence means\n\n\[ \left| {{p}_{n + 1} - p}\right| \leq C{\left| {p}_{n} - p\right| }^{2} \]\n\nfor some constant \( C > 0 \) . Taking the absolute values of the limit established in the previous theorem, we obtain\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| \frac{p - {... | Yes |
Theorem 35. Let \( f \in {C}^{2}\left\lbrack {a, b}\right\rbrack \) and assume \( f\left( p\right) = 0,{f}^{\prime }\left( p\right) \neq 0 \), for \( p \in \left( {a, b}\right) \) . If the initial guesses \( {p}_{0},{p}_{1} \) are sufficiently close to \( p \), then the iterates of the secant method converge to \( p \) | with\n\[\n\mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\left| p - {p}_{n + 1}\right| }{{\left| p - {p}_{n}\right| }^{{r}_{0}}} = {\left| \frac{{f}^{\prime \prime }\left( p\right) }{2{f}^{\prime }\left( p\right) }\right| }^{{r}_{1}}\n\]\n\nwhere \( {r}_{0} = \frac{\sqrt{5} + 1}{2} \approx {1.62},{r}_{1} = \frac{... | Yes |
If \( g \) is a continuous function on \( \left\lbrack {a, b}\right\rbrack \) and \( g\left( x\right) \in \left\lbrack {a, b}\right\rbrack \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) , then \( g \) has at least one fixed-point in \( \left\lbrack {a, b}\right\rbrack \) . | Consider \( f\left( x\right) = g\left( x\right) - x \) . Assume \( g\left( a\right) \neq a \) and \( g\left( b\right) \neq b \) (otherwise the proof is over.) Then \( f\left( a\right) = g\left( a\right) - a > 0 \) since \( g\left( a\right) \) must be greater than \( a \) if it’s not equal to \( a \) . Similarly, \( f\l... | Yes |
Theorem 40. If \( g \) is a continuous function on \( \left\lbrack {a, b}\right\rbrack \) satisfying the conditions\n\n1. \( g\left( x\right) \in \left\lbrack {a, b}\right\rbrack \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) ,\n\n2. \( \left| {g\left( x\right) - g\left( y\right) }\right| \leq \lambda \left| {... | Proof. Since \( {p}_{0} \in \left\lbrack {a, b}\right\rbrack \) and \( g\left( x\right) \in \left\lbrack {a, b}\right\rbrack \) for all \( x \in \left\lbrack {a, b}\right\rbrack \), all iterates \( {p}_{n} \in \left\lbrack {a, b}\right\rbrack \) . Observe that\n\n\[ \n\left| {p - {p}_{n}}\right| = \left| {g\left( p\rig... | Yes |
Consider the root-finding problem \( {x}^{3} - 2{x}^{2} - 1 = 0 \) on \( \left\lbrack {1,3}\right\rbrack \). | 1. There are several ways we can write this problem as \( g\left( x\right) = x \):\n\n(a) Let \( f\left( x\right) = {x}^{3} - 2{x}^{2} - 1 \), and \( p \) be its root, that is, \( f\left( p\right) = 0 \). If we let \( g\left( x\right) = \) \( x - f\left( x\right) \), then \( g\left( p\right) = p - f\left( p\right) = p ... | Yes |
Theorem 44. Assume \( p \) is a solution of \( g\left( x\right) = x \), and suppose \( g\left( x\right) \) is continuously differentiable in some interval about \( p \) with \( \left| {{g}^{\prime }\left( p\right) }\right| < 1 \) . Then the fixed-point iteration converges to \( p \) , provided \( {p}_{0} \) is chosen s... | Proof. Since \( {g}^{\prime } \) is continuous and \( \left| {{g}^{\prime }\left( p\right) }\right| < 1 \), there exists an interval \( I = \left\lbrack {p - \epsilon, p + \epsilon }\right\rbrack \) such that \( \left| {{g}^{\prime }\left( x\right) }\right| \leq k \) for all \( x \in I \), for some \( k < 1 \) . Then, ... | Yes |
Let \( g\left( x\right) = x + c\left( {{x}^{2} - 2}\right) \), which has the fixed-point \( p = \sqrt{2} \approx {1.4142} \). Pick a value for \( c \) to ensure the convergence of fixed-point iteration. For the picked value \( c \), determine the interval of convergence \( I = \left\lbrack {a, b}\right\rbrack \), that ... | Solution. Theorem 44 requires \( \left| {{g}^{\prime }\left( p\right) }\right| < 1 \). We have \( {g}^{\prime }\left( x\right) = 1 + {2xc} \), and thus \( {g}^{\prime }\left( \sqrt{2}\right) = 1 + 2\sqrt{2}c \). Therefore\n\n\[ \left| {{g}^{\prime }\left( \sqrt{2}\right) }\right| < 1 \Rightarrow - 1 < 1 + 2\sqrt{2}c < ... | Yes |
Theorem 46. Assume \( p \) is a solution of \( g\left( x\right) = x \) where \( g \in {C}^{\alpha }\left( I\right) \) for some interval \( I \) that contains \( p \), and for some \( \alpha \geq 2 \) . Furthermore assume\n\n\[ \n{g}^{\prime }\left( p\right) = {g}^{\prime \prime }\left( p\right) = \ldots = {g}^{\left( \... | Proof. From Taylor's theorem,\n\n\[ \n{p}_{n + 1} = g\left( {p}_{n}\right) = g\left( p\right) + \left( {{p}_{n} - p}\right) {g}^{\prime }\left( p\right) + \ldots + \frac{{\left( {p}_{n} - p\right) }^{\alpha - 1}}{\left( {\alpha - 1}\right) !}{g}^{\left( \alpha - 1\right) }\left( p\right) + \frac{{\left( {p}_{n} - p\rig... | Yes |
Find the interpolating polynomial using the monomial basis and Lagrange basis functions for the data: \( \\left( {-1, - 6}\\right) ,\\left( {1,0}\\right) ,\\left( {2,6}\\right) \) . | - Monomial basis: \( {p}_{2}\\left( x\\right) = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} \)\n\nWe can use Gaussian elimination to solve this matrix equation, or get help from Julia:\n\nIn [1]: A \( = \\left\\lbrack {1 - 1}\\right\\rbrack 1;1\\dot{1}\\dot{1};\\dot{1}\\dot{2}\\dot{4} \) ]\n\nOut \( \\left\\lbrack 1\\right\\rb... | Yes |
Find the interpolating polynomial using Newton's basis for the data: \( \\left( {-1, - 6}\\right) ,\\left( {1,0}\\right) ,\\left( {2,6}\\right) \) . | Solution. We have \( {p}_{2}\\left( x\\right) = {a}_{0} + {a}_{1}{\\pi }_{1}\\left( x\\right) + {a}_{2}{\\pi }_{2}\\left( x\\right) = {a}_{0} + {a}_{1}\\left( {x + 1}\\right) + {a}_{2}\\left( {x + 1}\\right) \\left( {x - 1}\\right) \) . Find \( {a}_{0},{a}_{1},{a}_{2} \) from\n\n\[ \n{p}_{2}\\left( {-1}\\right) = - 6 \... | Yes |
Example 50. Write \( {p}_{2}\left( x\right) = - 6 + 3\left( {x + 1}\right) + \left( {x + 1}\right) \left( {x - 1}\right) \) using the nested form. | Solution. \( - 6 + 3\left( {x + 1}\right) + \left( {x + 1}\right) \left( {x - 1}\right) = - 6 + \left( {x + 1}\right) \left( {2 + x}\right) \) ; note that the left-hand side has 2 multiplications, and the right-hand side has 1 . | No |
Lemma 52. Consider the partition of \( \left\lbrack {a, b}\right\rbrack \) as \( {x}_{0} = a,{x}_{1} = a + h,\ldots ,{x}_{n} = a + {nh} = b \) . More succinctly, \( {x}_{i} = a + {ih} \) for \( i = 0,1,\ldots, n \) and \( h = \frac{b - a}{n} \) . Then for any \( x \in \left\lbrack {a, b}\right\rbrack \)\n\n\[ \mathop{\... | Proof. Since \( x \in \left\lbrack {a, b}\right\rbrack \), it falls into one of the subintervals: let \( x \in \left\lbrack {{x}_{j},{x}_{j + 1}}\right\rbrack \) . Consider the product \( \left| {x - {x}_{j}}\right| \left| {x - {x}_{j + 1}}\right| \) . Put \( s = \left| {x - {x}_{j}}\right| \) and \( t = \left| {x - {x... | Yes |
Find an upper bound for the absolute error when \( f\left( x\right) = \cos x \) is approximated by its interpolating polynomial \( {p}_{n}\left( x\right) \) on \( \left\lbrack {0,\pi /2}\right\rbrack \) . For the interpolating polynomial, use 5 equally spaced nodes \( \left( {n = 4}\right) \) in \( \left\lbrack {0,\pi ... | Solution. From Theorem 51,\n\n\[ \left| {f\left( x\right) - {p}_{4}\left( x\right) }\right| = \frac{\left| {f}^{\left( 5\right) }\left( \xi \right) \right| }{5!}\left| {\left( {x - {x}_{0}}\right) \cdots \left( {x - {x}_{4}}\right) }\right| .\n\]\n\nWe have \( \left| {{f}^{\left( 5\right) }\left( \xi \right) }\right| \... | Yes |
Theorem 55. The ordering of the data in constructing divided differences is not important, that is, the divided difference \( f\left\lbrack {{x}_{0},\ldots ,{x}_{k}}\right\rbrack \) is invariant under all permutations of the arguments \( {x}_{0},\ldots ,{x}_{k} \) . | Proof. Consider the data \( \left( {{x}_{0},{y}_{0}}\right) ,\left( {{x}_{1},{y}_{1}}\right) ,\ldots ,\left( {{x}_{k},{y}_{k}}\right) \) and let \( {p}_{k}\left( x\right) \) be its interpolating polynomial:\n\n\[ \n{p}_{k}\left( x\right) = f\left\lbrack {x}_{0}\right\rbrack + f\left\lbrack {{x}_{0},{x}_{1}}\right\rbrac... | Yes |
Find the interpolating polynomial for the data \( \\left( {-1, - 6}\\right) ,\\left( {1,0}\\right) ,\\left( {2,6}\\right) \) using Newton's form and divided differences. | Solution. We want to compute\n\n\[ \n{p}_{2}\\left( x\\right) = f\\left\\lbrack {x}_{0}\\right\\rbrack + f\\left\\lbrack {{x}_{0},{x}_{1}}\\right\\rbrack \\left( {x - {x}_{0}}\\right) + f\\left\\lbrack {{x}_{0},{x}_{1},{x}_{2}}\\right\\rbrack \\left( {x - {x}_{0}}\\right) \\left( {x - {x}_{1}}\\right) .\n\]\n\nHere are... | Yes |
Use polynomial interpolation to estimate \( \Gamma \left( {1.761}\right) \) . | Solution. The finite differences, with five-digit rounding, are:\n\n<table><thead><tr><th>\( i \)</th><th>\( {x}_{i} \)</th><th>\( f\left\lbrack {x}_{i}\right\rbrack \)</th><th>\( f\left\lbrack {{x}_{i},{x}_{i + 1}}\right\rbrack \)</th><th>\( f\left\lbrack {{x}_{i - 1},{x}_{i},{x}_{i + 1}}\right\rbrack \)</th><th>\( f\... | No |
Theorem 58. Suppose \( f \in {C}^{n}\left\lbrack {a, b}\right\rbrack \) and \( {x}_{0},{x}_{1},\ldots ,{x}_{n} \) are distinct numbers in \( \left\lbrack {a, b}\right\rbrack \) . Then there exists \( \xi \in \left( {a, b}\right) \) such that\n\n\[ f\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack = \frac{{f}^{\left(... | To prove this theorem, we need the generalized Rolle's theorem. | Yes |
Theorem 61. If \( f \in {C}^{1}\left\lbrack {a, b}\right\rbrack \) and \( {x}_{0},\ldots ,{x}_{n} \in \left\lbrack {a, b}\right\rbrack \) are distinct, then there is a unique polynomial \( {H}_{{2n} + 1}\left( x\right) \), of degree at most \( {2n} + 1 \), agreeing with \( f \) and \( {f}^{\prime } \) at \( {x}_{0},\ld... | The polynomial can be written as:\n\n\[ \n{H}_{{2n} + 1}\left( x\right) = \mathop{\sum }\limits_{{i = 0}}^{n}{y}_{i}{h}_{i}\left( x\right) + \mathop{\sum }\limits_{{i = 0}}^{n}{y}_{i}^{\prime }{\widetilde{h}}_{i}\left( x\right) \]\n\nwhere\n\n\[ \n{h}_{i}\left( x\right) = \left( {1 - 2\left( {x - {x}_{i}}\right) {l}_{i... | Yes |
We want to interpolate the following data:\n\n\[ \nx\\text{-coordinates :} - {1.5},{1.6},{4.7} \]\n\n\[ \ny\\text{-coordinates :}{0.071}, - {0.029}, - {0.012}\\text{.} \]\n\nThe underlying function the data comes from is \( \\cos x \), but we pretend we do not know this. Figure (3.2) plots the underlying function, the ... | Now let's assume we know the derivative of the underlying function at these nodes:\n\n\[ \nx\\text{-coordinates :} - {1.5},{1.6},{4.7} \]\n\n\[ \ny\\text{-coordinates :}{0.071}, - {0.029}, - {0.012} \]\n\n\[ \n{y}^{\\prime }\\text{-values : }1, - 1,1\\text{. } \]\n\nWe then construct the Hermite interpolating polynomia... | Yes |
Let's compute the Hermite polynomial of Example 62. The data is:\n\n\\[\\begin{array}{llll} i & {x}_{i} & {y}_{i} & {y}_{i}^{\prime } \\end{array}\\]\n\n\\[\\begin{array}{llll} 0 & \\text{-}{1.5} & {0.071} & 1 \\end{array}\\]\n\n\\[\\begin{array}{llll} 1 & {1.6} & - {0.029} & - 1 \\end{array}\\]\n\n\\[\\begin{array}{ll... | The divided differences are:\n\n<table><thead><tr><th>\\( z \\)</th><th>\\( f\\left( z\\right) \\)</th><th>1st diff</th><th>2nd diff</th><th>3rd diff</th><th>4th diff</th><th>5th diff</th></tr></thead><tr><td>\\( {z}_{0} = - {1.5} \\)</td><td>0.071</td><td>\\( {f}^{\\prime }\\left( {z}_{0}\\right) = 1 \\)</td><td></td>... | No |
Estimate \( {\int }_{0.5}^{1}{x}^{x}{dx} \) using the midpoint, trapezoidal, and Simpson’s rules. | Solution. Let \( f\left( x\right) = {x}^{x} \) . The midpoint estimate for the integral is \( {2hf}\left( {x}_{0}\right) \) where \( h = \) \( \left( {b - a}\right) /2 = 1/4 \) and \( {x}_{0} = {0.75} \) . Then the midpoint estimate, using 6-digits, is \( f\left( {0.75}\right) /2 = \) \( {0.805927}/2 = {0.402964} \) . ... | Yes |
Find the constants \( {c}_{0},{c}_{1},{x}_{1} \) so that the quadrature formula\n\n\[ \n{\int }_{0}^{1}f\left( x\right) {dx} = {c}_{0}f\left( 0\right) + {c}_{1}f\left( {x}_{1}\right) \n\]\n\nhas the highest possible degree of accuracy. | Solution. We will find how many of the polynomials \( 1, x,{x}^{2},\ldots \) the rule can integrate exactly.\n\nIf \( p\left( x\right) = 1 \), then\n\n\[ \n{\int }_{0}^{1}p\left( x\right) {dx} = {c}_{0}p\left( 0\right) + {c}_{1}p\left( {x}_{1}\right) \Rightarrow 1 = {c}_{0} + {c}_{1}.\n\]\n\nIf \( p\left( x\right) = x ... | Yes |
Let’s compute \( {\int }_{0}^{2}{e}^{x}\sin {xdx} \) . The antiderivative can be computed using integration by parts, and the true value of the integral to 6 digits is 5.39689 . | If we apply the Simpson's rule we get:\n\n\[ \n{\int }_{0}^{2}{e}^{x}\sin {xdx} \approx \frac{1}{3}\left( {{e}^{0}\sin 0 + {4e}\sin 1 + {e}^{2}\sin 2}\right) = {5.28942}. \n\]\n\nIf we partition the integration domain \( \left( {0,2}\right) \) into \( \left( {0,1}\right) \) and \( \left( {1,2}\right) \), and apply Simp... | No |
Determine \( n \) that ensures the composite Simpson’s rule approximates \( {\int }_{1}^{2}x\log {xdx} \) with an absolute error of at most \( {10}^{-6} \) . | The error term for the composite Simpson’s rule is \( \frac{b - a}{180}{h}^{4}{f}^{\left( 4\right) }\left( \xi \right) \) where \( \xi \) is some number between \( a = 1 \) and \( b = 2 \), and \( h = \left( {b - a}\right) /n \) . Differentiate to get \( {f}^{\left( 4\right) }\left( x\right) = \frac{2}{{x}^{3}} \) . Th... | Yes |
Approximate \( {\int }_{-1}^{1}\cos {xdx} \) using Gauss-Legendre quadrature with \( n = 3 \) nodes. | Solution. From Table 4.1, and using two-digit rounding, we have\n\n\[ \n{\int }_{-1}^{1}\cos {xdx} \approx {0.56}\cos \left( {-{0.77}}\right) + {0.89}\cos 0 + {0.56}\cos \left( {0.77}\right) = {1.69} \n\] \n\nand the true solution is \( \sin \left( 1\right) - \sin \left( {-1}\right) = {1.68} \) . | Yes |
Approximate \( {\int }_{0.5}^{1}{x}^{x}{dx} \) using Gauss-Legendre quadrature with \( n = 2 \) nodes. | Transform the integral using \( x = \frac{1}{2}\left( {{0.5t} + {1.5}}\right) = \frac{1}{2}\left( {\frac{t}{2} + \frac{3}{2}}\right) = \frac{t + 3}{4},{dx} = \frac{dt}{4} \) to get:\n\n\[ \n{\int }_{0.5}^{1}{x}^{x}{dx} = \frac{1}{4}{\int }_{-1}^{1}{\left( \frac{t + 3}{4}\right) }^{\frac{t + 3}{4}}{dt} \n\]\n\nFor \( n ... | Yes |
Theorem 80. Let \( f \in {C}^{2n}\left\lbrack {-1,1}\right\rbrack \) . The error of Gauss-Legendre rule satisfies\n\n\[ \n{\int }_{a}^{b}f\left( x\right) {dx} - \mathop{\sum }\limits_{{i = 1}}^{n}{w}_{i}f\left( {x}_{i}\right) = \frac{{2}^{{2n} + 1}{\left( n!\right) }^{4}}{\left( {{2n} + 1}\right) {\left\lbrack \left( 2... | Using Stirling’s formula \( n! \sim {e}^{-n}{n}^{n}{\left( 2\pi n\right) }^{1/2} \), where the symbol \( \sim \) means the ratio of the two sides converges to 1 as \( n \rightarrow \infty \), it can be shown that\n\n\[ \n\frac{{2}^{{2n} + 1}{\left( n!\right) }^{4}}{\left( {{2n} + 1}\right) {\left\lbrack \left( 2n\right... | Yes |
One of the integrals with a known solution is\n\n\[ \n{\int }_{{\mathbb{R}}^{s}}\cos \left( {\parallel t\parallel }\right) ){e}^{-\parallel t{\parallel }^{2}}d{t}_{1}d{t}_{2}\cdots d{t}_{s} \n\] \n\nwhere \( \parallel t\parallel = {\left( {t}_{1}^{2} + \ldots + {t}_{s}^{2}\right) }^{1/2} \). | This integral can be transformed to an integral over the \( s \) -dimensional unit cube as\n\n\[ \n{\pi }^{s/2}{\int }_{{\left( 0,1\right) }^{s}}\cos \left\lbrack {\left( \frac{{\left( {F}^{-1}\left( {x}_{1}\right) \right) }^{2} + \ldots + {\left( {F}^{-1}\left( {x}_{s}\right) \right) }^{2}}{2}\right) }^{1/2}\right\rbr... | Yes |
Consider the previous integral \( {\int }_{-1}^{1}\frac{f\left( x\right) }{\sqrt{1 - {x}^{2}}}{dx} \) . Try the transformation \( \theta = \) \( {\cos }^{-1}x \) . | Then \( {d\theta } = - {dx}/\sqrt{1 - {x}^{2}} \) and\n\n\[ \n{\int }_{-1}^{1}\frac{f\left( x\right) }{\sqrt{1 - {x}^{2}}}{dx} = - {\int }_{\pi }^{0}f\left( {\cos \theta }\right) {d\theta } = {\int }_{0}^{\pi }f\left( {\cos \theta }\right) {d\theta }. \n\] \n\nThe latter integral can be evaluated using, for example, Si... | Yes |
Consider the improper integral \( {\int }_{0}^{\infty }{e}^{-{x}^{2}}{dx} \) . Write the integral as | \[ {\int }_{0}^{\infty }{e}^{-{x}^{2}}{dx} = {\int }_{0}^{t}{e}^{-{x}^{2}}{dx} + {\int }_{t}^{\infty }{e}^{-{x}^{2}}{dx} \] where \( t \) is the \ | No |
Example 84. The following table gives the values of \( f\left( x\right) = \sin x \) . Estimate \( {f}^{\prime }\left( {0.1}\right) ,{f}^{\prime }\left( {0.3}\right) \) using an appropriate three-point formula. | Solution. To estimate \( {f}^{\prime }\left( {0.1}\right) \), we set \( {x}_{0} = {0.1} \), and \( h = {0.1} \) . Note that we can only use the three-point endpoint formula.\n\n\[ \n{f}^{\prime }\left( {0.1}\right) \approx \frac{1}{0.2}\left( {-3\left( {0.09983}\right) + 4\left( {0.19867}\right) - {0.29552}}\right) = {... | Yes |
Find the least squares polynomial approximation of degree 2 to \( f\left( x\right) = {e}^{x} \) on \( \left( {0,2}\right) \) . | The normal equations are:\n\n\[ \mathop{\sum }\limits_{{j = 0}}^{2}{a}_{j}{\int }_{0}^{2}{x}^{j + k}{dx} = {\int }_{0}^{2}{e}^{x}{x}^{k}{dx} \]\n\( k = 0,1,2 \) . Here are the three equations:\n\n\[ {a}_{0}{\int }_{0}^{2}{dx} + {a}_{1}{\int }_{0}^{2}{xdx} + {a}_{2}{\int }_{0}^{2}{x}^{2}{dx} = {\int }_{0}^{2}{e}^{x}{dx}... | Yes |
Example 90 (Legendre Polynomials). If \( w\left( x\right) \equiv 1 \) and \( \left\lbrack {a, b}\right\rbrack = \left\lbrack {-1,1}\right\rbrack \), the first four polynomials obtained from the Gram-Schmidt process, when the process is applied to the monomials \( 1, x,{x}^{2},{x}^{3},\ldots \), are: | \[ {\phi }_{0}\left( x\right) = \sqrt{\frac{1}{2}},{\phi }_{1}\left( x\right) = \sqrt{\frac{3}{2}}x,{\phi }_{2}\left( x\right) = \frac{1}{2}\sqrt{\frac{5}{2}}\left( {3{x}^{2} - 1}\right) ,{\phi }_{3}\left( x\right) = \frac{1}{2}\sqrt{\frac{7}{2}}\left( {5{x}^{3} - {3x}}\right) . \] Often these polynomials are written i... | Yes |
Example 91 (Chebyshev polynomials). If we take \( w\left( x\right) = {\left( 1 - {x}^{2}\right) }^{-1/2} \) and \( \left\lbrack {a, b}\right\rbrack = \left\lbrack {-1,1}\right\rbrack \) , and again drop the orthonormal requirement in Gram-Schmidt, we obtain the following orthogonal polynomials:\n\n\[ \n{T}_{0}\left( x\... | Chebyshev polynomials also satisfy the following recursion:\n\n\[ \n{T}_{n + 1}\left( x\right) = {2x}{T}_{n}\left( x\right) - {T}_{n - 1}\left( x\right) \n\] \n\nfor \( n = 1,2,\ldots \), and\n\n\[ \n\left\langle {{T}_{j},{T}_{k}}\right\rangle = \left\{ \begin{array}{ll} 0 & \text{ if }j \neq k \\ \pi & \text{ if }j = ... | Yes |
Find the least squares polynomial approximation of degree three to \( f\left( x\right) = {e}^{x} \) on \( \left( {-1,1}\right) \) using Legendre polynomials. | Solution. Put \( n = 3 \) in Equation (5.13) and let \( {\phi }_{j} \) be \( {L}_{j} \) to get\n\n\[ \n{P}_{3}\left( x\right) = \frac{\left\langle f,{L}_{0}\right\rangle }{{\alpha }_{0}}{L}_{0}\left( x\right) + \frac{\left\langle f,{L}_{1}\right\rangle }{{\alpha }_{1}}{L}_{1}\left( x\right) + \frac{\left\langle f,{L}_{... | Yes |
Find the least squares polynomial approximation of degree three to \( f\left( x\right) = {e}^{x} \) on \( \left( {-1,1}\right) \) using Chebyshev polynomials. | Solution. As in the previous example solution, we take \( n = 3 \) in Equation (5.13)\n\n\[ \n{P}_{3}\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{3}\frac{\left\langle f,{\phi }_{j}\right\rangle }{{\alpha }_{j}}{\phi }_{j}\left( x\right) \]\n\nbut now \( {\phi }_{j} \) and \( {\alpha }_{j} \) will be replaced by \... | Yes |
Property 13.1.1 A property of tangents to functions of one variable. Suppose \( f \) is a function of one variable and at a number \( a \) in its domain, \( {f}^{\prime }\left( a\right) \) exists. The graph of \( L\left( x\right) = f\left( a\right) + {f}^{\prime }\left( a\right) \left( {x - a}\right) \) is the tangent ... | \[ \mathop{\lim }\limits_{{x \rightarrow a}}\frac{\left| f\left( x\right) - L\left( x\right) \right| }{\left| x - a\right| } = \mathop{\lim }\limits_{{x \rightarrow a}}\left| \frac{f\left( x\right) - f\left( a\right) - {f}^{\prime }\left( a\right) \left( {x - a}\right) }{x - a}\right| \] \[ = \left| {\mathop{\lim }\lim... | Yes |
Property 13.1.2 A property of local linear approximations to functions of two variables. Suppose \( F \) is a function of two variables, \( \left( {a, b}\right) \) is a number pair in the domain of \( F \), and \( {F}_{1} \) and \( {F}_{2} \) exist and are continuous on the interior of a circle with center \( \left( {a... | \[ \mathop{\lim }\limits_{{\left( {x, y}\right) \rightarrow \left( {a, b}\right) }}\frac{\left| F\left( x, y\right) - L\left( x, y\right) \right| }{\sqrt{{\left( x - a\right) }^{2} + {\left( y - b\right) }^{2}}} = 0 \] | Yes |
The graphs of \( F\left( {x, y}\right) = {x}^{2} + {y}^{2}, G\left( {x, y}\right) = {x}^{2} - {y}^{2} \) and \( H\left( {x, y}\right) = - {x}^{2} - {y}^{2} \) shown in Figure 13.10 illustrate three important options. The origin, \( \left( {0,0}\right) \), is a critical point of each of the graphs and the \( z = 0 \) pl... | For \( F \), for example,\n\n\[ F\left( {x, y}\right) = {x}^{2} + {y}^{2}\;{F}_{1}\left( {x, y}\right) = {2x}\;{F}_{2}\left( {x, y}\right) = {2y} \]\n\n\[ F\left( {0,0}\right) = 0\;{F}_{1}\left( {0,0}\right) = 0\;{F}_{2}\left( {0,0}\right) = 0 \]\n\nThe origin, \( \left( {0,0}\right) \), is a critical point of \( F \),... | Yes |
Fit a line to the data in Example Figure 13.2.3.3. | \[ {S}_{x} = 1 + 2 + 4 + 6 + 8 = {21}, \] \[ {S}_{y} = {0.5} + {0.8} + {1.0} + {1.7} + {1.8} = {5.8} \] \[ {S}_{xx} = {1}^{2} + {2}^{2} + {4}^{2} + {6}^{2} + {8}^{2} = {121}, \] \[ {S}_{xy} = 1 \times {0.5} + 2 \times {0.8} + 4 \times {1.0} + 6 \times {1.7} + 8 \times {1.8} = {30.7} \] \[ \Delta = n{S}_{xx} - {\left( {... | Yes |
Find the dimensions of the largest box (rectangular solid) that will fit in a hemisphere of radius \( R \) . | Solution. Assume the hemisphere is the graph of \( z = \sqrt{{R}^{2} - {x}^{2} - {y}^{2}} \) and that the optimum box has one face in the \( x, y \) -plane and the other four corners on the hemisphere (see Figure 13.2.4.4).\n\nThe volume, \( V \) of the box is\n\n\[ V\left( {x, y}\right) = {2x} \times {2y} \times z = {... | Yes |
Show that\n\n\[ {u}_{t}\left( {x, t}\right) = \frac{1}{{\pi }^{2}}{u}_{xx}\left( {x, t}\right) ,\;u\left( {x,0}\right) = {30} * \left( {1 - \cos {\pi x}}\right) ,\;\text{ and }\;\begin{array}{l} {u}_{x}\left( {0, t}\right) = 0 \\ {u}_{x}\left( {2, t}\right) = 0. \end{array} \]\n\n(13.24) | Solution. First compute some partial derivatives.\n\n\[ u\left( {x, t}\right) = {30}\left( {1 - {e}^{-t}\cos {\pi x}}\right) \]\n\n(13.25)\n\n\n\n\[ {u}_{t}\left( {x, t}\right) = {30}\left( {0 - \left( {e}^{-t}\right) \left( {-1}\right) \cos {\pi x}}\right) = {30}{e}^{-t}\cos \left( {\pi x}\right) \]\n\n(13.26)\n\n\[ {... | Yes |
Explore 13.3.5 What do you expect the 'eventual' salt concentration along the tube to be? | The diffusion equation is \( {u}_{t}\left( {x, t}\right) = k{u}_{xx}\left( {x, t}\right) \). Because initially there is no salt in the tube, \( g\left( x\right) = 0 \) for \( 0 < x < 1 \), and the reservoirs at the ends of the tube imply the fixed boundary conditions 13.22, \( u\left( {0, t}\right) = 1 \) and \( u\left... | No |
Theorem 14.4.1 If \( F \) and \( {F}^{\prime } \) are continuous on \( \left\lbrack {0,1}\right\rbrack \) and for \( x \) in \( \left\lbrack {0,1}\right\rbrack, F\left( x\right) \) is in \( \left\lbrack {0,1}\right\rbrack \), and \( E \) is a number in \( \left\lbrack {0,1}\right\rbrack \) for which \( F\left( E\right)... | Proof. The proof of Theorem 14.4.1 makes good use of the Mean Value Theorem, Theorem 9.1.1 in Volume I.\n\nSuppose \( F \) satisfies the hypothesis of Theorem 14.4.1. Let \( R = \left( {{F}^{\prime }\left( E\right) + 1}\right) /2 < 1 \) . Because \( {F}^{\prime } \) is continuous there is an subinterval \( \left( {a, b... | Yes |
We show that the number \( 1/a \) is the only locally stable equilibrium point of the iteration function\n\n\[ F\left( x\right) = x \times \left( {2 - a \times x}\right) \]\n\nfor the iteration \( {x}_{n + 1} = {x}_{n} \times \left( {2 - a * {x}_{n}}\right) \). | The equilibrium points, \( E \), are\n\n\[ E = F\left( E\right) \;E = E \times \left( {2 - a \times E}\right) ,\;E = 0\;\text{ or }\;E = \frac{1}{a} \]\n\nTo examine stability using Theorem 14.4.1 we compute\n\n\[ {F}^{\prime }\left( x\right) = {\left\lbrack x \times \left( 2 - a \times x\right) \right\rbrack }^{\prime... | Yes |
Example 14.4.2 Cray computation of \( 1/\pi \) In 24 digit binary notation, | \[ \pi \doteq {110010010000111111011010} \times {2}^{2}. \] In decimal notation, \[ \pi \doteq {0.785398} \times {2}^{2},\;\text{ and }\;\frac{1}{\pi } \doteq \frac{1}{0.785398} \times {2}^{-2}. \] An iteration sequence to compute 1/0.785398 is illustrated in Figure 14.13. The sequence is \( {x}_{0} = {1.75},{x}_{n + 1... | Yes |
If \( B > 1 \) and \( n \) is a positive integer, then \( \;\mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{{B}^{t}}{{t}^{n}} = \infty \) | We first show that\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}\frac{{e}^{t}}{t} = \infty \]\n\n\( \left( {14.20}\right) \)\n\nWe assume it to be true that\n\n\[ \mathop{\lim }\limits_{{t \rightarrow \infty }}{e}^{t} = \infty \]\n\n(14.21)\n\nbut have included a proof for you to complete in Problem 14.5.8.\n\n... | No |
Theorem 14.5.3 L’Hospital’s Theorem on a bounded interval. Suppose \( \left\lbrack {a, b}\right\rbrack \) is a number interval and \( F \) and \( G \) are continuous functions defined on the half open interval \( (a, b\rbrack \) and \( {F}^{\prime } \) and \( {G}^{\prime } \) are continuous on \( \left( {a, b}\right) \... | Proof. Because \( \mathop{\lim }\limits_{{t \rightarrow {a}^{ + }}}F\left( t\right) = \mathop{\lim }\limits_{{t \rightarrow {a}^{ + }}}G\left( t\right) = 0 \), the domain of \( F \) and \( G \) can be extended to include \( a \) by defining \( F\left( a\right) = G\left( a\right) = 0 \) and the extended \( F \) and \( G... | Yes |
We begin with the equation\n\n\[ \n{P}_{0} = {1000} \n\]\n\n\[ \n{P}_{t + 1} - {P}_{t} = {0.2} \times {P}_{t} \times \left( {1 - \frac{{P}_{t}}{1000}}\right) - h \times {P}_{t} \n\]\n\nof a population with an initial population of 1000 individuals, low density growth rate of 0.2 per time interval, carrying capacity 100... | The equilibrium population is important and we solve for \( {P}_{e} \) in\n\n\[ \n{P}_{e} - {P}_{e} = {0.2} \times {P}_{e} \times \left( {1 - \frac{{P}_{e}}{1000}}\right) - h \times {P}_{e} \n\]\n\n\[ \n0 = {0.2} \times {P}_{e} \times \left( {1 - \frac{{P}_{e}}{1000}}\right) - h \times {P}_{e} \n\]\n\n\[ \n0 = {P}_{e} ... | Yes |
If \( {u}_{t} \) and \( {v}_{t} \) are solutions to the linear homogeneous difference equation\n\n\[ \n{w}_{t + n} + {p}_{n - 1}{w}_{t + n - 1} + \cdots + {p}_{1}{w}_{t + 1} + {p}_{0}{w}_{t} = 0 \n\]\n\n(15.21)\n\nwhere \( {p}_{n - 1},\cdots {p}_{1} \) and \( {p}_{0} \) are constants, then for any numbers \( {C}_{1} \)... | Proof.\n\n\[ \n{C}_{1}\left( {{u}_{t + n} + {C}_{2}{v}_{t + n}}\right) + \cdots + {p}_{1}\left( {{C}_{1}{u}_{t + 1} + {C}_{2}{v}_{t + 1}}\right) + {p}_{0}\left( {{C}_{1}{u}_{t} + {C}_{2}{v}_{t}}\right) = \n\]\n\n\[ \n{C}_{1}\left( {{u}_{t + n} + {p}_{n - 1}{u}_{t + n - 1} + \cdots + {p}_{1}{u}_{t + 1} + {p}_{0}{u}_{t}}... | Yes |
Find formulas for \( {A}_{t} \) and \( {B}_{t} \) if\n\n\[ \n{A}_{0} = 1{A}_{t + 1} = {0.52}{A}_{t} + {0.04}{B}_{t} \]\n\n\[ \n\begin{matrix} {B}_{0} & = & 2 & {B}_{t + 1} & = & {0.24}{A}_{t} + {0.4}{B}_{t} \end{matrix} \]\n | Solution\n\n\[ \np = {0.52} + {0.4} = {0.92},\;q = {0.52} \cdot {0.4} - {0.24} \cdot {0.04} = {0.1984},\;{r}^{2} - {0.92r} + {0.1984} = 0 \]\n\n\[ \n{r}_{1} = \frac{{0.92} + \sqrt{{0.92}^{2} - 4 \cdot {0.1984}}}{2} = {0.46} + \sqrt{0.0132},\;{r}_{2} = {0.46} - \sqrt{0.0132} \]\n\n\[ \n{A}_{1} = {0.52} \cdot 1 + {0.04} ... | No |
Theorem 15.5.1 If \( {r}_{1} \) and \( {r}_{2} \) are the roots to \( {x}^{2} - {px} + q = 0 \) then\n\n\[ \n{r}_{1} + {r}_{2} = p\;\text{ and }\;{r}_{1} \times {r}_{2} = q.\n\] | Proof.\n\n\[ \n{r}_{1} + {r}_{2} = \frac{p + \sqrt{{p}^{2} + {4q}}}{2} + \frac{p - \sqrt{{p}^{2} - {4q}}}{2} = \frac{p + \sqrt{{p}^{2} - {4q}} + p - \sqrt{{p}^{2} - {4q}}}{2} = p \n\]\n\n\[ \n{r}_{1} \times {r}_{2} = \frac{p + \sqrt{{p}^{2} - {4q}}}{2} \times \frac{p - \sqrt{{p}^{2} - {4q}}}{2} = \frac{{p}^{2} - \left(... | Yes |
Theorem 16.2.1 For the three forms of solutions in Equations 16.8, 16.9 and 16.10, \( {x}_{t} \rightarrow 0 \) and \( {y}_{t} \rightarrow 0 \) for all choices of \( {x}_{0} \) and \( {y}_{0} \) if and only if \[ \left| {r}_{1}\right| < 1\text{ and }\left| {r}_{2}\right| < 1;\;\text{ or }\;\left| {r}_{1}\right| < 1;\;\t... | Proof. The results are valid because \( \mathop{\lim }\limits_{{t \rightarrow \infty }}{r}^{t} = 0 \) if \( \left| r\right| < 1 \) and \( \mathop{\lim }\limits_{{t \rightarrow \infty }}t \times {r}^{t} = 0 \) if \( \left| r\right| < 1 \) . (For \( \left. {\mathop{\lim }\limits_{{t \rightarrow \infty }}t \times {r}^{t} ... | No |
Theorem 16.2.2 The fate of \( {\mathbf{x}}_{\mathbf{t}} \) . The dynamical systems of Equations 16.3 and Equations 16.5, with characteristic equation\n\n\[ \n{z}^{2} - \left( {{m}_{1,1} + {m2},2}\right) \times z + {m}_{1,1}{m}_{2,2} - {m}_{2,1}{m}_{1,2} = {z}^{2} - {pz} + q = 0, \n\]\n\nare stable if and only if\n\n\[ ... | Proof. Danger: Obnubilation Zone. This argument is tedious and reading it can be delayed - a very long time.\n\nWe show that if \( 0 \leq \left| p\right| < 1 + q < 2 \) then \( \mathop{\lim }\limits_{{t \rightarrow \infty }}{x}_{t} = 0 \) . In the case of complex roots, \( {\rho }^{2} = q \)\n\n(Equation 15.26) and bec... | Yes |
Example 16.3.1 In the next section we consider two populations that have a symbiotic relationship, a special case of which is\n\n\[ \n{x}_{t + 1} = {x}_{t} + \frac{5}{98}{x}_{t}\left( {1 + \frac{4}{10}{y}_{t} - {x}_{t}}\right) = F\left( {{x}_{t},{y}_{t}}\right) \n\]\n\n(16.23)\n\n\[ \n{y}_{t + 1} = {y}_{t} + \frac{7}{1... | The basis for the previous paragraph is in Theorem 16.3.1. The idea of the theorem and of local linear approximation can be seen by an algebraic rearrangement of the nonlinear system 16.23\n\n\[ \n{x}_{t + 1} - {1.96} = {0.9}\left( {{x}_{t} - {1.96}}\right) + {0.04}\left( {{y}_{t} - {2.4}}\right) \n\]\n\n\[ \n+ \frac{2... | Yes |
Example 17.4.1 An extreme example. Shown in Figure 17.9 are the direction field and the phase plane graph of \[ {y}^{\prime } = \left( {y - 1}\right) \times \left( {y - 2}\right) \times \left( {y - 3}\right) \times \left( {y - 4}\right) \] | It is easy to solve \( f\left( y\right) = \left( {y - 1}\right) \left( {y - 2}\right) \left( {y - 3}\right) \left( {y - 4}\right) = 0 \) and see that \( 1,2,3 \) and 4 are equilibrium points, and equivalently that \( y = 1, y = 2, y = 3 \) and \( y = 4 \) are equilibrium solutions to the differential equation. Solution... | Yes |
Theorem 17.4.1 If \( f\left( y\right) \) and \( {f}^{\prime }\left( y\right) \) are continuous, an equilibrium point \( {y}_{e} \) of \( {y}^{\prime } = f\left( y\right) \) is asymptotically stable if \( {f}^{\prime }\left( {y}_{e}\right) \) is negative. | Proof: This proof is technical and your reading of it can be delayed — a long time. Suppose the hypothesis of the theorem and let \( - m = {f}^{\prime }\left( {y}_{e}\right) < 0 \) . By hypothesis, \( {y}_{e} \) is an equilibrium point so that \( f\left( {y}_{e}\right) = 0 \) ; therefore the function \( \bar{y}\left( t... | No |
Example 17.4.2 Parameter Reduction. It is customary to divide the logistic differential equation\n\n\[ \n{p}^{\prime }\left( t\right) = r \times p\left( t\right) \times \left( {1 - \frac{p\left( t\right) }{M}}\right) \n\]\n\nby \( M \) to obtain\n\n\[ \n\frac{{p}^{\prime }\left( t\right) }{M} = r \times \frac{p\left( t... | The function, \( u \), is the fraction of the carrying capacity, \( M \), used by the population. Because \( u \) is the ratio of \( p \) to \( M \), both of which have units of population numbers, \( u \) is dimensionless. | No |
We compute points to approximate the solution to\n\n\[ v\\left( 0\\right) = {0.2}\\,{v}^{\\prime }\\left( t\\right) = v\\left( t\\right) \\times {e}^{-v\\left( t\\right) } - {0.1} * v\\left( t\\right) \\;0 \\leq t \\leq {10} \] | This is Ricker's model of fish populations with parameter reduction, Equation 17.15. First we divide the time axis \( \\left\\lbrack {0,{10}}\\right\\rbrack \) into intervals of length 2 and let\n\n\[ {t}_{0} = 0\\;{t}_{1} = 2\\;{t}_{2} = 4\\;{t}_{3} = 6\\;{t}_{4} = 8\\;\\text{ and }\\;{t}_{5} = {10} \]\n\nOur objectiv... | Yes |
We find an approximate solution to\n\n\\[ \ny\\left( 0\\right) = 2\\;{y}^{\\prime }\\left( t\\right) = t - y\n\\]\n\nfor \\( 0 \\leq t \\leq 4 \\) ; the graph of the solution is shown in Figure 17.13A. Both \\( t \\) and \\( y \\) appear in the RHS in this problem. | The basic pattern is the same.\n\n\\[ \n{y}_{0} = y\\left( 0\\right) = 2\\;{y}_{k + 1} = {y}_{k} + h \\times {\\text{slope}}_{k}\n\\]\n\nThe computations are organized in Table 17.1 for time-interval size \\( h = 1 \\) .\n\nOur approximation is shown in Figure 17.13A and is not close enough to the solution to satisfy u... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.