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We use the trapezoid method to approximate the solution to\n\n\[ y\left( 0\right) = 2\;{y}^{\prime }\left( t\right) = t - y\;0 \leq t \leq 4 \] \n\nusing \( n = {16} \) intervals and compare the results with those obtained with Euler’s method. | Step 0. \( {t}_{k} = 0 + k * {0.25} \), for \( k = 0,1,\cdots ,{15} \) and \( {y}_{0} = 2 \) .\n\nStep 1. \( {\operatorname{slope}}_{1} = {t}_{0} - {y}_{0} = 0 - 2 = - 2 \)\n\nEuler’s projected \( {y}_{1} \) is \( {\widehat{y}}_{1} = {y}_{0} + h \times {\text{slope}}_{1} = 2 + {0.25} \times \left( {-2}\right) = {1.5} \... | Yes |
Theorem 17.6.1 Existence and Uniqueness of Solutions. If \( f\left( {t, y}\right) \) and \( {f}_{2}\left( {t, y}\right) = \frac{\partial }{\partial y}f\left( {t, y}\right) \) are continuous on a rectangle\n\n\[ a - d \leq t \leq a + d\;{y}_{a} - d \leq y \leq {y}_{a} + d\;d > 0, \]\n\nthen on an interval \( a - e \leq ... | Under the hypothesis that \( {f}_{2} \) is continuous on the rectangle, it follows that \( {f}_{2} \) is bounded on the rectangle, and the proof of Theorem 17.6.1 hinges on this fact. | Yes |
Example 17.6.1 Although proof of Theorem 17.6.1 is beyond our scope, existence and uniqueness of solutions to a large number of differential equations follows rather easily from simple anti-derivative formulas and the Parallel Graph Theorem. For example, we know by simple substitution that the growth equation \( \left(... | Then\n\n\[ \n{Q}^{\prime }\left( t\right) = {kQ}\left( t\right) \n\]\nHypothesis\n\n\[ \n{e}^{-{kt}}{Q}^{\prime }\left( t\right) - k{e}^{-{kt}}Q\left( t\right) = 0 \n\]\nBlue Sky.\n\n\[ \n{\left\lbrack {e}^{-{kt}}Q\left( t\right) \right\rbrack }^{\prime } = {\left\lbrack 1\right\rbrack }^{\prime } \n\]\nDerivative form... | Yes |
Theorem 17.7.1\n\nStep 1. Define \( \;u\left( t\right) = {\int }_{a}^{t}p\left( s\right) {ds} \) .\n\nStep 2. Define \( \;v\left( t\right) = {\int }_{a}^{t}{e}^{u\left( s\right) } \times q\left( s\right) {ds} \) .\n\nThen the solution to Equation 17.19 is | \[ y\left( t\right) = v\left( t\right) {e}^{-u\left( t\right) } + {y}_{a}{e}^{-u\left( t\right) } \] | Yes |
Consider a case in which \( p\left( t\right) = 3 \) and \( q\left( t\right) = 5 \) are constant. Solve \[ y\left( 1\right) = 5\;{y}^{\prime }\left( t\right) + {3y}\left( t\right) = 2,\;a = 1,\;{y}_{a} = 5,\;p\left( t\right) = 3,\;q\left( t\right) = 5 \] | Define \[ u\left( t\right) = {\int }_{0}^{t}{3dt} = {3t} \] \[ v\left( t\right) = {\int }_{0}^{t}{e}^{3t} \times {2dt} = \frac{2}{3}\left( {{e}^{3t} - 1}\right) . \] Then \[ y\left( t\right) = \frac{2}{3}\left( {{e}^{3t} - 1}\right) \times {e}^{-{3t}} + 5{e}^{-{3t}} \] \[ = \frac{2}{3} + \frac{13}{3}{e}^{-{3t}} \] To c... | Yes |
The formulas for \( u\left( t\right) \) and \( v\left( t\right) \) are explicit, but the integrals may not be computable in terms of familiar functions. In the equation\n\n\[ y\left( 0\right) = 5\;{y}^{\prime }\left( t\right) - {2ty}\left( t\right) = 1, \]\n\n\[ p\left( t\right) = - {2t},\;u\left( t\right) = {\int }_{0... | There is no formula for \( v\left( t\right) \) in familiar terms \( {}^{9} \) for \( v\left( t\right) \) . It can be numerically approximated as you did in Chapter 11, The Integral, of Volume I, and is an important formula in statistics, but there is no expression for \( {\int }_{0}^{t}{e}^{-{s}^{2}}{ds} \) in familiar... | No |
Problem 1. Solve \( y\left( 1\right) = 1,\;{y}^{\prime }\left( t\right) + \frac{1}{t}y\left( t\right) = {e}^{t} \) . | \[ a = 1,\;{y}_{1} = 1,\;p\left( t\right) = \frac{1}{t},\;q\left( t\right) = {e}^{t} \]\n\n\[ u\left( t\right) = {\int }_{0}^{t}p\left( s\right) {ds} = {\int }_{0}^{t}\frac{1}{s}{ds} = {\left. \ln s\right| }_{s = 1}^{t} = \ln t - 0 = \ln t \]\n\n\[ v\left( t\right) = {\int }_{0}^{t}{e}^{u\left( s\right) }q\left( s\righ... | Yes |
Problem 2. Solve \( y\left( 0\right) = 2,\;{y}^{\prime }\left( t\right) + {2y}\left( t\right) = \sin {3t} \) . | \[ a = 0,\;{y}_{0} = 2,\;p\left( t\right) = 2,\;q\left( t\right) = \sin {3t} \]\n\n\[ u\left( t\right) = {\int }_{0}^{t}p\left( s\right) {ds} = {\int }_{0}^{t}{2ds} = {\left. 2s\right| }_{s = 0}^{t} = {2t} - 0 = {2t} \]\n\n\[ v\left( t\right) = {\int }_{0}^{t}{e}^{u\left( s\right) }q\left( s\right) {ds} = {\int }_{0}^{... | Yes |
Consider\n\n\[ \n{y}^{\prime }\left( t\right) = {2t} \times y \n\] | Then\n\n\[ \n\frac{1}{y}{y}^{\prime } = {2t} \n\]\n\nSV Step 1, LHS Find \( H\left( y\right) \) such that\n\n\[ \n\frac{{dH}\left( y\right) }{dy} = \frac{1}{y}.\;\text{ Choose }\;H\left( y\right) = \ln y.\n\]\n\nSV Step 2, RHS Find \( G\left( t\right) \) such that \( \frac{d}{dt}G\left( t\right) = {2t} \).\n\nThen\n\n\... | Yes |
Of the following six equations, the variables can be separated in only two.\n\n\[ \n{y}^{\prime }\left( t\right) = t + y\;{y}^{\prime } = {e}^{t + y} \]\n\n\[ \n{y}^{\prime }\left( t\right) = \ln \left( {t + y}\right) \;{y}^{\prime } = {e}^{t \times y} \]\n\n\[ \n{y}^{\prime }\left( t\right) = \ln \left( {t \times y}\r... | The two equations in which variables are separable are shown below.\n\n\[ \n{y}^{\prime } = {e}^{t + y} = {e}^{t} \times {e}^{y}\;{y}^{\prime } = \ln {t}^{y} = y \times \ln t \]\n\n\[ \n{e}^{-y} \times {y}^{\prime } = {e}^{t}\;\frac{1}{y}{y}^{\prime } = \ln t \]\n\n\[ \n{\left\lbrack -{e}^{-y}\right\rbrack }^{\prime } ... | Yes |
Example 17.8.3 The variables can be separated in every autonomous differential equation\n\n\[ \n{y}^{\prime } = f\left( y\right) \;\frac{1}{f\left( y\right) }{y}^{\prime } = 1 \n\] | To find an implicit solution to any autonomous differential equation only the problem\n\nSV Step 1 Find \( F\left( y\right) \) such that \( \;{F}^{\prime }\left( y\right) = \frac{1}{f\left( y\right) }\; \) requires attention.\n\nSV Step 2 is easy: Find \( G\left( t\right) \) such that \( {G}^{\prime }\left( t\right) = ... | No |
To solve the autonomous equation, \( {y}^{\prime } = - {y}^{2} \) | \[ \frac{1}{{y}^{2}}{y}^{\prime } = - 1\;\frac{{dH}\left( y\right) }{dy} = \frac{1}{{y}^{2}},\;\text{ choose }\;H\left( y\right) = - \frac{1}{y} \] \[ \frac{d}{dt}\left\lbrack {-\frac{1}{y\left( t\right) }}\right\rbrack = \frac{d}{dt}\left\lbrack {-t}\right\rbrack \] \[ - \frac{1}{y\left( t\right) } = - t + C \] \[ y\l... | Yes |
\[ \frac{{3x} + 4}{\left( {x + 3}\right) \times \left( {x - 2}\right) } = \frac{A}{x + 3} + \frac{B}{x - 2} \] | To find \( A \) and \( B \), multiply by \( \left( {x + 3}\right) \times \left( {x - 2}\right) \) and get \[ {3x} + 4 = A \times \left( {x - 2}\right) + B \times \left( {x + 3}\right) \] Then substitute \[ x = - 3\;3\left( {-3}\right) + 4 = A \times \left( {-3 - 2}\right) + B \times \left( {-3 + 3}\right) \] \[ - 5 = A... | No |
\[ \frac{3{x}^{2} - {5x} + 1}{\left( {x + 1}\right) \times {\left( x - 2\right) }^{2}} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{{\left( x - 2\right) }^{2}} \] | To find \( A, B \) and \( C \), multiply by \( \left( {x + 1}\right) \times {\left( x - 2\right) }^{2} \) and get\n\n\[ 3{x}^{2} - {5x} + 1 = A \times {\left( x - 2\right) }^{2} + B \times \left( {x - 2}\right) \times \left( {x + 1}\right) + C \times \left( {x + 1}\right) \]\n\nThen substitute\n\n\[ x = - 1\;3{\left( -... | Yes |
For Candidate 3 the method of partial fractions converts\n\n\\[ {u}^{\\prime } = u \\times \\left( {u - \\epsilon }\\right) \\times \\left( {1 - u}\\right) \\;\\frac{{u}^{\\prime }}{u \\times \\left( {u - \\epsilon }\\right) \\times \\left( {1 - u}\\right) } = 1 \\]\n\ninto\n\n\\[ \\left( {\\frac{-1/\\epsilon }{u} + \\... | which can be integrated to obtain\n\n\\[ - \\frac{1}{\\epsilon }\\ln \\left| {u\\left( t\\right) }\\right| + \\frac{1}{\\epsilon \\left( {1 - \\epsilon }\\right) }\\ln \\left| {u\\left( t\\right) - \\epsilon }\\right| - \\frac{1}{1 - \\epsilon }\\ln \\left| {1 - u\\left( t\\right) }\\right| = t + C \\] | Yes |
Suppose the one gram of carbon from deer bone recently found among American Indian artifacts is emits \( 7{\beta }^{ - } \) particles per minute. How old is the bone? | Solution. Let \( {t}_{0} \) be the time at which the deer died. Assume that \( E\left( {t}_{0}\right) = E\left( 0\right) = {15.3} \) . Then\n\n\[ \n{\left. {E}_{{t}_{0}}\left( t\right) \right| }_{t = 0} = {E}_{{t}_{0}}\left( 0\right) = E\left( {t}_{0}\right) {e}^{-\frac{\ln 2}{5730}\left( {0 - {t}_{0}}\right) } \n\]\n\... | Yes |
Example 17.9.2 Data from Reaction 1 of Exercise Table 17.9.9 are plotted in Figure 17.20 and it is clear that the data are from a second order reaction, and \( m = 2 \) . The reaction is thus\n\n\[ \n{2A} + {nB} \rightarrow {A}_{2}{B}_{n} \n\]\n\nfor some \( n \) (found below). | Figure 17.20: Graphs of Reaction 1 data. A. In Concentration vs \( t \) . B. 1/Concentration vs \( t \) . C. \( 1/{\left( \text{Concentration}\right) }^{2} \) vs \( t \) . B. is linear and the line \( \mathrm{y} = {99.96} + {4t} \) fits the data. Therefore, the reaction is second order, \( \widehat{K} = 4 \), and \( \f... | Yes |
For the Reaction 1 with \( \left\lbrack B\right\rbrack = {0.2}\mathrm{\;{mol}} \), we found that the reaction was second order, \( m = 2 \) and \( \widehat{K} = 4 \) . The data for Reaction 1 with \( \left\lbrack B\right\rbrack = {0.4}\mathrm{\;{mol}} \) is plotted in Figure 17.21 as for a second order reaction, \( 1/ ... | \[ {2A} + {3B} \rightarrow {A}_{2}{B}_{3} \] | Yes |
Theorem 18.1.1 Superposition. Suppose \( {y}_{1}\left( t\right) \) and \( {y}_{2}\left( t\right) \) are two solutions to the homogeneous equation \( {y}^{\prime \prime }\left( t\right) + p{y}^{\prime }\left( t\right) + {qy}\left( t\right) = 0 \) and \( {C}_{1} \) is a number and \( {C}_{2} \) is a number. Then\n\n\[ y\... | Proof. The proof of Theorem 18.1.1 is Exercise 18.1.1. | No |
Theorem 18.1.2 If \( {y}_{p,1}\left( t\right) \) solves \( {y}^{\prime \prime }\left( t\right) - p{y}^{\prime }\left( t\right) + {qy}\left( t\right) = {f}_{1}\left( t\right) \) and \( {y}_{p,2}\left( t\right) \) solves\n\n\( {y}^{\prime \prime }\left( t\right) - p{y}^{\prime }\left( t\right) + {qy}\left( t\right) = {f}... | Proof. Exercise 18.1.4. | No |
Consider the equations\n\n\\[ \nx\\left( 0\\right) = {x}_{0}\\;{x}^{\\prime }\\left( t\\right) = y \n\\]\n\n(18.25)\n\n\\[ \ny\\left( 0\\right) = {y}_{0}\\;{y}^{\\prime }\\left( t\\right) = {2xy} \n\\] | First observe that every point with \\( y \\) -coordinate \\( 0,\\left( {{x}_{e},0}\\right) \\), is an equilibrium point. Also observe the direction field in, Figure 18.4A; the direction of motion in the four quarters of the plane is determined by whether \\( {x}^{\\prime } \\) and \\( {y}^{\\prime } \\) are positive o... | Yes |
It appears from the direction field in Figure 18.6, that the origin, (0,0), is an asymptotically stable equilibrium of\n\n\\[ \nx\\left( 0\\right) = {x}_{0}\\;{x}^{\\prime } = - x \n\\] \n\n(18.27) \n\n\\[ \ny\\left( 0\\right) = {y}_{0}\\;{y}^{\\prime } = - y. \n\\] | The solution to Equations 18.27 is \n\n\\[ \nx\\left( t\\right) = {x}_{0}{e}^{-t},\\;y\\left( t\\right) = {y}_{0}{e}^{-t} \n\\] \n\nThe origin is stable: Suppose \\( \\epsilon > 0 \\), choose \\( \\delta = \\epsilon \\) . If for some \\( {t}_{1} \\) \n\n\\[ \n\\sqrt{{\\left( x\\left( {t}_{1}\\right) - 0\\right) }^{2} +... | Yes |
Theorem 18.3.1 Asymptotical Stability of a pair of constant coefficient homogeneous differential equations. The origin \( \left( {0,0}\right) \) is an asymptotically stable equilibrium point of\n\n\[ \n{x}^{\prime }\left( t\right) = {a}_{1,1}x\left( t\right) + {a}_{1,2}y\left( t\right) \n\]\n\n(18.37)\n\n\[ \n{y}^{\pri... | Proof. Suppose the roots are real and distinct and negative. We first show that \( \left( {0,0}\right) \) is a stable equilibrium. From Equations 18.34 we observe that\n\n\[ \n\left| {x\left( t\right) }\right| = \left| {{C}_{1}{e}^{{r}_{1}t} + {C}_{2}{e}^{{r}_{2}t}}\right| \n\]\n\n\[ \n\leq \left| {C}_{1}\right| + \lef... | Yes |
Explore 18.4.1 Show that \( x \equiv 5, y \equiv {12} \) is a solution to\n\n\[ {x}^{\prime } = \left( {{169} - {x}^{2} - {y}^{2}}\right) /{10},\;{y}^{\prime } = {17} - x - y. \] | - Direction fields near \( {e}_{1} = \left( {5,{12}}\right) \) and near \( {e}_{2} = \left( {{12},5}\right) \) for \( {x}^{\prime } = \left( {{169} - {x}^{2} - {y}^{2}}\right) /{10},{y}^{\prime } = {17} - x - y \) are shown in Figure 18.11A and B, respectively. They are quite different. Near \( {e}_{2} = \left( {{12},5... | No |
Example 18.4.1 (Continued) Equations 18.50,\n\n\[ \n{x}^{\prime } = \left( {{169} - {x}^{2} - {y}^{2}}\right) /{10} \n\]\n\n(18.55)\n\n\[ \n{y}^{\prime } = {17} - x - y \n\]\n\nhas two equilibrium points, \( {e}_{1} = \left( {5,{12}}\right) \) and \( {e}_{2} = \left( {{12},5}\right) \) . The Jacobian of Equations 18.55... | For \( {e}_{1} = \left( {5,{12}}\right) \) the Jacobian is\n\n\[ \n{\left\lbrack \begin{array}{rr} - x/5 & - y/5 \\ - 1 & - 1 \end{array}\right\rbrack }_{\left( {x, y}\right) = \left( {5,{12}}\right) }\; = \;\left\lbrack \begin{array}{rr} - 1 & - {12}/5 \\ - 1 & - 1 \end{array}\right\rbrack .\n\]\n\nThe trace and deter... | Yes |
Lemma 2.11 If \( {\mathcal{L}}_{\text{left }} \equiv {\mathcal{L}}_{\text{right }} \) then, for any library \( {\mathcal{L}}^{ * } \), we have \( {\mathcal{L}}^{ * }\diamond {\mathcal{L}}_{\text{left }} \equiv {\mathcal{L}}^{ * }\diamond {\mathcal{L}}_{\text{right }} \) . | of Note that we are comparing \( {\mathcal{L}}^{ * }\diamond {\mathcal{L}}_{\text{left }} \) and \( {\mathcal{L}}^{ * }\diamond {\mathcal{L}}_{\text{right }} \) as compound libraries. Hence we consider a calling program \( \mathcal{A} \) that is linked to either \( {\mathcal{L}}^{ * }\diamond {\mathcal{L}}_{\text{left ... | Yes |
Theorem 2.16 There is an encryption scheme that satisfies one-time secrecy (Definition 2.6) but not one-time uniform ciphertexts (Definition 2.5). In other words, one-time secrecy does not necessarily imply one-time uniform ciphertexts. | Proof\n\nOne such encryption scheme is given below:\n\n\[ \begin{array}{lllll} \mathcal{K} = \{ 0,1{\} }^{\lambda } & \text{ keyGen: } & \frac{\operatorname{Enc}\left( {k, m \in \{ 0,1{\} }^{\lambda }}\right) : }{{c}^{\prime } \mathrel{\text{:=}} k \oplus m} & \frac{\operatorname{Dec}\left( {k, c \in \{ 0,1{\} }^{\lamb... | No |
Let \( \\left\\{ {\\left( {{x}_{1},{y}_{1}}\\right) ,\\ldots ,\\left( {{x}_{d + 1},{y}_{d + 1}}\\right) }\\right\\} \\subseteq {\\mathbb{R}}^{2} \) be a set of points whose \( {x}_{i} \) values are all distinct. Then there is a unique degree-d polynomial \( f \) with real coefficients that satisfies \( {y}_{i} = f\\lef... | Proof To start, consider the following polynomial:\n\n\[ \n{\\ell }_{1}\\left( \\mathbf{x}\\right) = \\frac{\\left( {\\mathbf{x} - {x}_{2}}\\right) \\left( {\\mathbf{x} - {x}_{3}}\\right) \\cdots \\left( {\\mathbf{x} - {x}_{d + 1}}\\right) }{\\left( {{x}_{1} - {x}_{2}}\\right) \\left( {{x}_{1} - {x}_{3}}\\right) \\cdot... | Yes |
Let \( p \) be a prime, and let \( \left\{ {\left( {{x}_{1},{y}_{1}}\right) ,\ldots ,\left( {{x}_{d + 1},{y}_{d + 1}}\right) }\right\} \subseteq {\left( {\mathbb{Z}}_{p}\right) }^{2} \) be a set of points whose \( {x}_{i} \) values are all distinct. Then there is a unique degree-d polynomial \( f \) with coefficients f... | The proof is the same as the one for Theorem 3.8, if you interpret all arithmetic modulo \( p \) . Addition, subtraction, and multiplication \( {\;\operatorname{mod}\;p} \) are straight forward; the only nontrivial question is how to interpret \ | No |
Corollary 3.10 Let \( \mathcal{P} = \left\{ {\left( {{x}_{1},{y}_{1}}\right) ,\ldots ,\left( {{x}_{k},{y}_{k}}\right) }\right\} \subseteq {\left( {\mathbb{Z}}_{p}\right) }^{2} \) be a set of points whose \( {x}_{i} \) values are distinct. Let \( d \) satisfy \( k \leq d + 1 \) and \( p > d \) . Then the number of degre... | Proof The proof is by induction on the value \( d + 1 - k \) . The base case is when \( d + 1 - k = 0 \) . Then we have \( k = d + 1 \) distinct points, and Theorem 3.9 says that there is a unique polynomial satisfying the condition. Since \( {p}^{d + 1 - k} = {p}^{0} = 1 \), the base case is true.\n\nFor the inductive... | Yes |
Lemma 3.12 2 Let \( p \) be a prime and define the following two libraries: \n\n\( {\mathcal{L}}_{\text{shamir-real }} \) chooses a random degree- \( \left( {t - 1}\right) \) polynomial that passes through the point \(... | Proof Fix a message \( m \in {\mathbb{Z}}_{p} \), fix set \( U \) of users with \( \left| U\right| < t \), and for each \( i \in U \) fix a value \( {y}_{i} \in {\mathbb{Z}}_{p} \) . We wish to consider the probability that a call to \( \operatorname{POLY}\left( {m, t, U}\right) \) outputs \( \left\{ {\left( {i,{y}_{i}... | Yes |
Theorem 3.13 13 Shamir’s secret-sharing scheme (Construction 3.11) is secure according to Definition 3.3. | Proof Let \( \mathcal{S} \) denote the Shamir secret-sharing scheme. We prove that \( {\mathcal{L}}_{\text{tsss-L }}^{\mathcal{S}} \equiv {\mathcal{L}}_{\text{tsss-R }}^{\mathcal{S}} \) via a hybrid argument. \( \operatorname{SHARE}\left( {{m}_{L},{m}_{R}, U}\right) \) : Our starting point is \( {\mathcal{L}}_{\text{ts... | Yes |
Lemma 4.9 BirthdayProb \( \left( {q, N}\right) = 1 - \mathop{\prod }\limits_{{i = 1}}^{{q - 1}}\left( {1 - \frac{i}{N}}\right) \) . | Proof Let us instead compute the probability that \( \mathcal{B} \) outputs 0, which will allow us to then solve for the probability that it outputs 1 . In order for \( \mathcal{B} \) to output 0, it must avoid the early termination conditions in each iteration of the main loop. Therefore:\n\n\[ \Pr \left\lbrack {\math... | Yes |
If \( q \leq \sqrt{2N} \), then\n\n\[ \n{0.632}\frac{q\left( {q - 1}\right) }{2N} \leq \text{ Birthday Prob }\left( {q, N}\right) \leq \frac{q\left( {q - 1}\right) }{2N}.\n\] | Proof We split the proof into two parts.\n\n- To prove the upper bound, we use the fact that when \( x \) and \( y \) are positive,\n\n\[ \n\left( {1 - x}\right) \left( {1 - y}\right) = 1 - \left( {x + y}\right) + {xy}\n\]\n\n\[ \n\geq 1 - \left( {x + y}\right)\n\]\n\nMore generally, when all terms \( {x}_{i} \) are po... | Yes |
Let \( {\mathcal{L}}_{\text{samp-L }} \) and \( {\mathcal{L}}_{\text{samp-R }} \) be defined as above. Then for all calling programs \( \mathcal{A} \) that make \( q \) queries to the SAMP subroutine, the advantage of \( \mathcal{A} \) in distinguishing the libraries is at most BirthdayProb \( \left( {q,{2}^{\lambda }}... | Proof Consider the following hybrid libraries:\n\n\n\nFirst, let us prove some simple observations about these libraries:\n\n\( {\mathcal{L}}_{\mathrm{{hyb}} - \mathrm{L}} \equiv {\mathcal{L}}_{\text{samp } - \mathrm{L... | Yes |
Lemma 4.12 The following two libraries are indistinguishable, provided that the argument \( \mathcal{R} \) to SAMP is passed as an explicit list of items. | Suppose the calling program makes \( q \) calls to SAMP, and in the \( i \) th call it uses an argument \( \mathcal{R} \) with \( {n}_{i} \) items. Then the advantage of the calling program is at most:\n\n\[ 1 - \mathop{\prod }\limits_{{i = 1}}^{q}\left( {1 - \frac{{n}_{i}}{{2}^{\lambda }}}\right) \]\n\nWe can bound th... | Yes |
Let \( {\mathcal{L}}_{\text{prf-rand }} \) and \( {\mathcal{L}}_{\text{prp-rand }} \) be defined as in Definitions 6.1 & 6.6, with parameters in \( = \) out \( = \) blen \( = \lambda \) (so that the interfaces match up). Then \( {\mathcal{L}}_{\text{prf-rand }} \approx {\mathcal{L}}_{\text{prp-rand }} \) . | Recall the replacement-sampling lemma, Lemma 4.11, which showed that the following libraries are indistinguishable:\n\n\n\n\( {\mathcal{L}}_{\text{samp-L }} \) samples values with replacement, and \( {\mathcal{L}}_{\... | Yes |
Corollary 6.8 Let \( F : \{ 0,1{\} }^{\lambda } \times \{ 0,1{\} }^{\lambda } \rightarrow \{ 0,1{\} }^{\lambda } \) be a secure PRP (with blen \( = \lambda \) ). Then \( F \) is also a secure PRF. | Proof As we have observed above, \( {\mathcal{L}}_{\text{prf-real }}^{F} \) and \( {\mathcal{L}}_{\text{prp-real }}^{F} \) are literally the same library. Since \( F \) is a secure PRP, \( {\mathcal{L}}_{\text{prp-real }}^{F} \approx {\mathcal{L}}_{\text{prp-rand }}^{F} \) . Finally, by the switching lemma, \( {\mathca... | Yes |
Claim 11.3 Suppose \( h \) is a compression function and \( M{D}_{h} \) is the Merkle-Damgård construction applied to \( h \) . Given a collision \( x,{x}^{\prime } \) in \( M{D}_{h} \), it is easy to find a collision in \( h \) . In other words, if it is hard to find a collision in \( h \), then it must also be hard t... | Proof Suppose that \( x,{x}^{\prime } \) are a collision under \( {\mathrm{{MD}}}_{h} \) . Define the values \( {x}_{1},\ldots ,{x}_{k + 1} \) and \( {y}_{1},\ldots ,{y}_{k + 1} \) as in the computation of \( {\mathrm{{MD}}}_{h}\left( x\right) \) . Similarly, define \( {x}_{1}^{\prime },\ldots ,{x}_{{k}^{\prime } + 1}^... | Yes |
If \( x \in {\mathbb{Z}}_{n}^{ * } \) then \( {x}^{\phi \left( n\right) }{ \equiv }_{n}1 \) . | Using the formula for \( \phi \left( n\right) \), we can see that \( \phi \left( {15}\right) = \phi \left( {3 \cdot 5}\right) = \left( {3 - 1}\right) \left( {5 - 1}\right) = 8 \) . Euler’s theorem says that raising any element of \( {\mathbb{Z}}_{15}^{ * } \) to the 8 power results in 1: We can use Sage to verify this:... | No |
Suppose \( \gcd \left( {r, s}\right) = 1 \) . Then for all integers \( u, v \), there is a solution for \( x \) in the following system of equations:\n\n\[ x{ \equiv }_{r}u \]\n\n\[ x{ \equiv }_{s}v \]\n\nFurthermore, this solution is unique modulo \( {rs} \) . | Proof \( f\; \) Since \( \gcd \left( {r, s}\right) = 1 \), we have by Bezout’s theorem that \( 1 = {ar} + {bs} \) for some integers \( a \) and \( b \) . Furthermore, \( b \) and \( s \) are multiplicative inverses modulo \( r \) . Now choose \( x = {var} + {ubs} \) . Then,\n\n\[ x = {var} + {ubs}{ \equiv }_{r}\left( {... | Yes |
Theorem 2.1.1 (Superposition). Suppose \( {y}_{1} \) and \( {y}_{2} \) are two solutions of the homogeneous equation (2.2). Then\n\n\[ y\left( x\right) = {C}_{1}{y}_{1}\left( x\right) + {C}_{2}{y}_{2}\left( x\right) \]\n\nalso solves (2.2) for arbitrary constants \( {C}_{1} \) and \( {C}_{2} \) . | Proof: Let \( y = {C}_{1}{y}_{1} + {C}_{2}{y}_{2} \) . Then\n\n\[ {y}^{\prime \prime } + p{y}^{\prime } + {qy} = {\left( {C}_{1}{y}_{1} + {C}_{2}{y}_{2}\right) }^{\prime \prime } + p{\left( {C}_{1}{y}_{1} + {C}_{2}{y}_{2}\right) }^{\prime } + q\left( {{C}_{1}{y}_{1} + {C}_{2}{y}_{2}}\right) \]\n\n\[ = {C}_{1}{y}_{1}^{\... | Yes |
Theorem 2.1.3. Let \( p, q \) be continuous functions. Let \( {y}_{1} \) and \( {y}_{2} \) be two linearly independent solutions to the homogeneous equation (2.2). Then every other solution is of the form\n\n\[ y = {C}_{1}{y}_{1} + {C}_{2}{y}_{2} \]\n\nThat is, \( y = {C}_{1}{y}_{1} + {C}_{2}{y}_{2} \) is the general s... | For example, we found the solutions \( {y}_{1} = \sin x \) and \( {y}_{2} = \cos x \) for the equation \( {y}^{\prime \prime } + y = 0 \) . It is not hard to see that sine and cosine are not constant multiples of each other. If \( \sin x = A\cos x \) for some constant \( A \), we let \( x = 0 \) and this would imply \(... | Yes |
Theorem 2.2.1. Suppose that \( {r}_{1} \) and \( {r}_{2} \) are the roots of the characteristic equation.\n\n(i) If \( {r}_{1} \) and \( {r}_{2} \) are distinct and real (when \( {b}^{2} - {4ac} > 0 \) ), then (2.3) has the general solution\n\n\[ y = {C}_{1}{e}^{{r}_{1}x} + {C}_{2}{e}^{{r}_{2}x}. \]\n\n(ii) If \( {r}_{... | Example 2.2.1: Solve\n\n\[ {y}^{\prime \prime } - {k}^{2}y = 0. \]\n\nThe characteristic equation is \( {r}^{2} - {k}^{2} = 0 \) or \( \left( {r - k}\right) \left( {r + k}\right) = 0 \) . Consequently, \( {e}^{-{kx}} \) and \( {e}^{kx} \) are the two linearly independent solutions, and the general solution is\n\n\[ y =... | Yes |
Theorem 2.2.3. Take the equation\n\n\\[ \na{y}^{\\prime \\prime } + b{y}^{\\prime } + {cy} = 0.\n\\]\n\nIf the characteristic equation has the roots \\( \\alpha \\pm {i\\beta } \\) (when \\( {b}^{2} - {4ac} < 0 \\) ), then the general solution is\n\n\\[ \ny = {C}_{1}{e}^{\\alpha x}\\cos \\left( {\\beta x}\\right) + {C}... | Example 2.2.3: Find the general solution of \\( {y}^{\\prime \\prime } + {k}^{2}y = 0 \\), for a constant \\( k > 0 \\) .\n\nThe characteristic equation is \\( {r}^{2} + {k}^{2} = 0 \\) . Therefore, the roots are \\( r = \\pm {ik} \\), and by the theorem, we have the general solution\n\n\\[ \ny = {C}_{1}\\cos \\left( {... | Yes |
Theorem 2.5.1. Let \( {Ly} = f\left( x\right) \) be a linear ODE (not necessarily constant coefficient). Let \( {y}_{c} \) be the complementary solution (the general solution to the associated homogeneous equation \( {Ly} = 0 \) ) and let \( {y}_{p} \) be any particular solution to \( {Ly} = f\left( x\right) \) . Then ... | The moral of the story is that we can find the particular solution in any old way. If we find a different particular solution (by a different method, or simply by guessing), then we still get the same general solution. The formula may look different, and the constants we have to choose to satisfy the initial conditions... | No |
Theorem 3.2.1. An \( n \times n \) matrix \( A \) is invertible if and only if \( \det \left( A\right) \neq 0 \) . | In fact, \( \det \left( {A}^{-1}\right) \det \left( A\right) = 1 \) says that \( \det \left( {A}^{-1}\right) = \frac{1}{\det \left( A\right) } \) . So we even know what the determinant of \( {A}^{-1} \) is before we know how to compute \( {A}^{-1} \) . | No |
Theorem 3.3.2. Let \( {\overrightarrow{x}}^{\prime } = P\overrightarrow{x} + \overrightarrow{f} \) be a linear system of ODEs. Suppose \( {\overrightarrow{x}}_{p} \) is one particular solution. Then every solution can be written as\n\n\[ \overrightarrow{x} = {\overrightarrow{x}}_{c} + {\overrightarrow{x}}_{p} \]\n\nwhe... | The procedure for systems is the same as for single equations. We find a particular solution to the nonhomogeneous equation, then we find the general solution to the associated homogeneous equation, and finally we add the two together. | No |
Theorem 3.4.1. Take \( {\overrightarrow{x}}^{\prime } = P\overrightarrow{x} \) . If \( P \) is an \( n \times n \) constant matrix that has \( n \) distinct real eigenvalues \( {\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{n} \), then there exist \( n \) linearly independent corresponding eigenvectors \( {\overrigh... | The corresponding fundamental matrix solution is\n\n\[ X\left( t\right) = \left\lbrack \begin{array}{llll} {\overrightarrow{v}}_{1}{e}^{{\lambda }_{1}t} & {\overrightarrow{v}}_{2}{e}^{{\lambda }_{2}t} & \cdots & {\overrightarrow{v}}_{n}{e}^{{\lambda }_{n}t} \end{array}\right\rbrack .\n\nThat is, \( X\left( t\right) \) ... | Yes |
Theorem 3.4.2. Let \( P \) be a real-valued constant matrix. If \( P \) has a complex eigenvalue \( a + {ib} \) and a corresponding eigenvector \( \overrightarrow{v} \), then \( P \) also has a complex eigenvalue \( a - {ib} \) with a corresponding eigenvector \( \overline{\overrightarrow{v}} \) . Furthermore, \( {\ove... | \[ {\overrightarrow{x}}_{1} = \operatorname{Re}\overrightarrow{v}{e}^{\left( {a + {ib}}\right) t},\;\text{ and }\;{\overrightarrow{x}}_{2} = \operatorname{Im}\overrightarrow{v}{e}^{\left( {a + {ib}}\right) t}. \] | Yes |
Let \( P \) be an \( n \times n \) matrix. Then the general solution to \( {\overrightarrow{x}}^{\prime } = P\overrightarrow{x} \) is\n\n\[ \overrightarrow{x} = {e}^{tP}\overrightarrow{c} \]\n\nwhere \( \overrightarrow{c} \) is an arbitrary constant vector. In fact, \( \overrightarrow{x}\left( 0\right) = \overrightarro... | Let us check:\n\n\[ \frac{d}{dt}\overrightarrow{x} = \frac{d}{dt}\left( {{e}^{tP}\overrightarrow{c}}\right) = P{e}^{tP}\overrightarrow{c} = P\overrightarrow{x}. \]\n\nHence \( {e}^{tP} \) is a fundamental matrix solution of the homogeneous system. So if we can compute the matrix exponential, we have another method of s... | Yes |
Theorem 4.1.1. Suppose that \( {x}_{1}\left( t\right) \) and \( {x}_{2}\left( t\right) \) are two eigenfunctions of the problem (4.1),(4.2) or (4.3) for two different eigenvalues \( {\lambda }_{1} \) and \( {\lambda }_{2} \) . Then they are orthogonal in the sense that\n\n\[ \n{\int }_{a}^{b}{x}_{1}\left( t\right) {x}_... | The terminology comes from the fact that the integral is a type of inner product. We will expand on this in the next section. The theorem has a very short, elegant, and illuminating proof so let us give it here. First, we have the following two equations.\n\n\[ \n{x}_{1}^{\prime \prime } + {\lambda }_{1}{x}_{1} = 0\;\t... | Yes |
Theorem 4.1.2 (Fredholm alternative*). Exactly one of the following statements holds. Either\n\n\\[ \n{x}^{\\prime \\prime } + {\\lambda x} = 0,\\;x\\left( a\\right) = 0,\\;x\\left( b\\right) = 0 \n\\]\n\n(4.4)\n\nhas a nonzero solution, or\n\n\\[ \n{x}^{\\prime \\prime } + {\\lambda x} = f\\left( t\\right) ,\\;x\\left... | The theorem is also true for the other types of boundary conditions we considered. The theorem means that if \\( \\lambda \\) is not an eigenvalue, the nonhomogeneous equation (4.5) has a unique solution for every right-hand side. On the other hand if \\( \\lambda \\) is an eigenvalue, then (4.5) need not have a soluti... | Yes |
Theorem 4.3.1. Suppose \( f\left( t\right) \) is a 2L-periodic piecewise smooth function. Let\n\n\[ \n\frac{{a}_{0}}{2} + \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}\cos \left( {\frac{n\pi }{L}t}\right) + {b}_{n}\sin \left( {\frac{n\pi }{L}t}\right)\n\]\n\nbe the Fourier series for \( f\left( t\right) \) . Then th... | If we happen to have that \( f\left( t\right) = \frac{f\left( {t - }\right) + f\left( {t + }\right) }{2} \) at all the discontinuities, the Fourier series converges to \( f\left( t\right) \) everywhere. We can always just redefine \( f\left( t\right) \) by changing the value at each discontinuity appropriately. Then we... | Yes |
Theorem 4.3.2. Suppose\n\n\[ f\left( t\right) = \frac{{a}_{0}}{2} + \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n}\cos \left( {\frac{n\pi }{L}t}\right) + {b}_{n}\sin \left( {\frac{n\pi }{L}t}\right) \]\n\nis a piecewise smooth continuous function and the derivative \( {f}^{\prime }\left( t\right) \) is piecewise smoo... | \[ {f}^{\prime }\left( t\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{-{a}_{n}{n\pi }}{L}\sin \left( {\frac{n\pi }{L}t}\right) + \frac{{b}_{n}{n\pi }}{L}\cos \left( {\frac{n\pi }{L}t}\right) . \] | Yes |
Theorem 4.3.3. Suppose\n\n\\[ \nf\\left( t\\right) = \\frac{{a}_{0}}{2} + \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }{a}_{n}\\cos \\left( {\\frac{n\\pi }{L}t}\\right) + {b}_{n}\\sin \\left( {\\frac{n\\pi }{L}t}\\right) \n\\]\n\nis a piecewise smooth function. Then the antiderivative is obtained by antidifferentiating... | Note that the series for \\( F\\left( t\\right) \\) is no longer a Fourier series as it contains the \\( \\frac{{a}_{0}t}{2} \\) term. The antiderivative of a periodic function need no longer be periodic and so we should not expect a Fourier series. | Yes |
Theorem 4.7.1. Take the equation\n\n\[ \n{y}_{tt} = {a}^{2}{y}_{xx} \]\n\n\[ \ny\left( {0, t}\right) = y\left( {L, t}\right) = 0, \]\n\n(4.14)\n\n\[ \ny\left( {x,0}\right) = f\left( x\right) \;\text{for}\;0 < x < L, \]\n\n\[ \n{y}_{t}\left( {x,0}\right) = g\left( x\right) \;\text{ for }0 < x < L, \]\n\nwhere\n\n\[ \nf\... | \[ \ny\left( {x, t}\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}\frac{L}{n\pi a}\sin \left( {\frac{n\pi }{L}x}\right) \sin \left( {\frac{n\pi a}{L}t}\right) + {c}_{n}\sin \left( {\frac{n\pi }{L}x}\right) \cos \left( {\frac{n\pi a}{L}t}\right) \]\n\n\[ \n= \mathop{\sum }\limits_{{n = 1}}^{\infty }\sin \left... | Yes |
Put the following equation into the form (5.1):\n\n\[ \n{x}^{2}{y}^{\prime \prime } + x{y}^{\prime } + \left( {\lambda {x}^{2} - {n}^{2}}\right) y = 0.\n\] | Multiply both sides by \( \frac{1}{x} \) to obtain\n\n\[ \n\frac{1}{x}\left( {{x}^{2}{y}^{\prime \prime } + x{y}^{\prime } + \left( {\lambda {x}^{2} - {n}^{2}}\right) y}\right) = x{y}^{\prime \prime } + {y}^{\prime } + \left( {{\lambda x} - \frac{{n}^{2}}{x}}\right) y\n\]\n\n\[ \n= \frac{d}{dx}\left( {x\frac{dy}{dx}}\r... | Yes |
Theorem 5.1.1. Suppose \( p\left( x\right) ,{p}^{\prime }\left( x\right), q\left( x\right) \) and \( r\left( x\right) \) are continuous on \( \left\lbrack {a, b}\right\rbrack \) and suppose \( p\left( x\right) > 0 \) and \( r\left( x\right) > 0 \) for all \( x \) in \( \left\lbrack {a, b}\right\rbrack \) . Then the Stu... | \[ {\lambda }_{1} < {\lambda }_{2} < {\lambda }_{3} < \cdots \] such that \[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\lambda }_{n} = + \infty \] and such that to each \( {\lambda }_{n} \) there is (up to a constant multiple) a single eigenfunction \( {y}_{n}\left( x\right) \) . Moreover, if \( q\left( x\right) ... | Yes |
Theorem 5.1.4. Suppose \( f \) is a piecewise smooth continuous function on \( \left\lbrack {a, b}\right\rbrack \) . If \( {y}_{1},{y}_{2},\ldots \) are eigenfunctions of a regular Sturm-Liouville problem, one for each eigenvalue, then there exist real constants \( {c}_{1},{c}_{2},\ldots \) given by (5.4) such that (5.... | Example 5.1.4: Consider\n\n\[ \n{y}^{\prime \prime } + {\lambda y} = 0,\;0 < x < \pi /2, \]\n\n\[ \ny\left( 0\right) = 0,\;{y}^{\prime }\left( {\pi /2}\right) = 0. \]\n\nThe above is a regular Sturm-Liouville problem, and Theorem 5.1.1 on page 275 says that if \( \lambda \) is an eigenvalue then \( \lambda \geq 0 \) .\... | Yes |
Theorem 6.1.1 (Linearity of the Laplace transform). Suppose that \( A, B \), and \( C \) are constants, then\n\n\[ \mathcal{L}\{ {Af}\left( t\right) + {Bg}\left( t\right) \} = A\mathcal{L}\{ f\left( t\right) \} + B\mathcal{L}\{ g\left( t\right) \} \]\n\nand in particular\n\n\[ \mathcal{L}\{ {Cf}\left( t\right) \} = C\m... | Exercise 6.1.2: Verify the theorem. That is, show that \( \mathcal{L}\{ {Af}\left( t\right) + {Bg}\left( t\right) \} = A\mathcal{L}\{ f\left( t\right) \} + \) \( B\mathcal{L}\{ g\left( t\right) \} \) . | No |
Theorem 6.1.2 (Existence). Let \( f\left( t\right) \) be continuous and of exponential order for a certain constant c. Then \( F\left( s\right) = \mathcal{L}\{ f\left( t\right) \} \) is defined for all \( s > c \) . | The existence is not difficult to see. Let \( f\left( t\right) \) be of exponential order, that is \( \left| {f\left( t\right) }\right| \leq M{e}^{ct} \) for all \( t > 0 \) (for simplicity \( {t}_{0} = 0 \) ). Let \( s > c \), or in other words \( \left( {c - s}\right) < 0 \) . By the comparison theorem from calculus,... | Yes |
Theorem 7.1.1. For a power series (7.1), there exists a number \( \rho \) (we allow \( \rho = \infty \) ) called the radius of convergence such that the series converges absolutely on the interval \( \left( {{x}_{0} - \rho ,{x}_{0} + \rho }\right) \) and diverges for \( x < {x}_{0} - \rho \) and \( x > {x}_{0} + \rho \... | A useful test for convergence of a series is the ratio test. Suppose that\n\n\[ \mathop{\sum }\limits_{{k = 0}}^{\infty }{c}_{k} \]\n\nis a series and the limit\n\n\[ L = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| \frac{{c}_{k + 1}}{{c}_{k}}\right| \]\n\nexists. Then the series converges absolutely if \( L <... | Yes |
Theorem 7.1.2 (Ratio and root tests for power series). Consider a power series\n\n\\[ \n\\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{a}_{k}{\\left( x - {x}_{0}\\right) }^{k}\n\\]\n\n such that\n\\[ \nA = \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left| \\frac{{a}_{k + 1}}{{a}_{k}}\\right| \\;\\text{ or }\\;... | Example 7.1.3: Suppose we have the series\n\n\\[ \n\\mathop{\\sum }\\limits_{{k = 0}}^{\\infty }{2}^{-k}{\\left( x - 1\\right) }^{k}\n\\]\n\nFirst we compute the limit in the ratio test,\n\n\\[ \nA = \\mathop{\\lim }\\limits_{{k \\rightarrow \\infty }}\\left| \\frac{{a}_{k + 1}}{{a}_{k}}\\right| = \\mathop{\\lim }\\lim... | Yes |
Theorem 7.3.1 (Method of Frobenius). Suppose that\n\n\[ p\left( x\right) {y}^{\prime \prime } + q\left( x\right) {y}^{\prime } + r\left( x\right) y = 0 \]\n\nhas a regular singular point at \( x = 0 \), then there exists at least one solution of the form\n\n\[ y = {x}^{r}\mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}... | A solution of this form is called a Frobenius-type solution.\n\n---\n\n*Named after the German mathematician Ferdinand Georg Frobenius (1849-1917).\n\n---\n\nThe method usually breaks down like this:\n\n(i) We seek a Frobenius-type solution of the form\n\n\[ y = \mathop{\sum }\limits_{{k = 0}}^{\infty }{a}_{k}{x}^{k + ... | Yes |
Theorem 8.4.1 (Poincarè-Bendixson*). Suppose \( R \) is a closed bounded region (a region in the plane that includes its boundary and does not have points arbitrarily far from the origin). Suppose \( \left( {x\left( t\right), y\left( t\right) }\right) \) is a solution of (8.2) in \( R \) that exists for all \( t \geq {... | The main point of the theorem is that if you find one solution that exists for all \( t \) large enough (that is, as \( t \) goes to infinity) and stays within a bounded region, then you have found either a periodic orbit, or a solution that spirals towards a limit cycle or tends to a critical point. That is, in the lo... | Yes |
Theorem 8.4.2 (Bendixson-Dulac*). Suppose \( R \) is a simply connected region, and the expression \( {}^{ \dagger } \)\n\n\[ \frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} \]\n\nis either always positive or always negative on \( R \) (except perhaps a small set such as on isolated points or curves) then... | Example 8.4.3: Let us look at \( {x}^{\prime } = y + {y}^{2}{e}^{x},{y}^{\prime } = x \) in the entire plane (see Example 8.2.2 on page 359). The entire plane is simply connected and so we can apply the theorem. We compute \( \frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = {y}^{2}{e}^{x} + 0 \) . The fu... | Yes |
Proposition 2.19. Suppose that \( \sim \) is an equivalence relation on \( X \) . Let \( x, y \in X \) . If \( x \sim y \), then\n\n\[ \left\lbrack x\right\rbrack = \left\lbrack y\right\rbrack \text{.} \]\n\n\( \left( {2.20}\right) \)\n\nIf \( x \) is not equivalent to \( y\left( {x \sim y}\right) \), then\n\n\[ \left\... | Proof. (i) Assume \( x \sim y \) . Let us show that \( \left\lbrack x\right\rbrack \subseteq \left\lbrack y\right\rbrack \) . Let \( z \in \left\lbrack x\right\rbrack \) . This means that \( x \sim z \) . Since \( \sim \) is symmetric, and \( x \sim y \), we have \( y \sim x \) . As \( y \sim x \) and \( x \sim z \), b... | Yes |
Proposition 2.26. If \( a \equiv r{\;\operatorname{mod}\;n} \) and \( b \equiv s{\;\operatorname{mod}\;n} \), then\n\n\[ \n\text{(i)}\;a + b \equiv r + s\;{\;\operatorname{mod}\;n} \n\]\nand\n\n\[ \n\text{(ii)}{ab} \equiv {rs}{\;\operatorname{mod}\;n}\text{.} \n\] | Proof. (i) Assume that \( a \equiv r{\;\operatorname{mod}\;n} \) and \( b \equiv s{\;\operatorname{mod}\;n} \) . Then \( n\left| {\left( {a - r}\right) \text{ and }n}\right| \left( {b - s}\right) \) . So\n\n\[ \nn \mid \left( {a + b - \left( {r + s}\right) }\right) .\n\]\n\nTherefore\n\n\[ \na + b \equiv r + s{\;\opera... | Yes |
[P \Rightarrow Q] \equiv [(\neg Q) \Rightarrow (\neg P)]. | This is a very important example of a propositional equivalence. We will show this by considering all possible assignments of truth values to \( P \) and \( Q \) . Let’s set this up in what is popularly called a truth table. We consider all possible assignments of truth values to \( P \) and \( Q \), and compare the tr... | Yes |
Example 3.17. Prove (3.16) directly. | Let \( x \in \mathbb{N} \) (we treat \( x \) as a fixed but arbitrary element of the natural numbers). If \( x = {4n} \), then\n\n\[ \n x = 2 \cdot \left( {2n}\right) \n\]\n\nand is therefore even. | No |
Example 3.26. Suppose there are 30 students in a class. Show that at least two of them share the same last initial. | Proof. For each letter A, B,... group all the students with that letter as their last initial. As there are only 26 groups and \( {30} > {26} \) students, at least one group must have more than on student in it. | Yes |
Proposition 4.5. Let \( N \in \mathbb{N} \) . Then\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{N}n = \frac{N\left( {N + 1}\right) }{2} \] | Proof. Base case: \( N = 0 \) .\n\nDiscussion. Note that the base case is the statement \( P\left( 0\right) \) .\n\nSince\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{0}n = 0 = \frac{\left( 0\right) \left( 1\right) }{2} \]\n\n\( P\left( 0\right) \) holds.\n\nInduction step:\n\nDiscussion. We prove the universal statement\n\n... | Yes |
Proposition 4.6. Let \( N \in \mathbb{N} \) . Then\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{N}{n}^{2} = \frac{N\left( {N + 1}\right) \left( {{2N} + 1}\right) }{6}. \] | Proof. The assertion \( P\left( N\right) \) is that the equation (4.7) holds. The base case, \( N = 0 \), is obvious:\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{0}{n}^{2} = \frac{0\left( {0 + 1}\right) \left( {2 \cdot 0 + 1}\right) }{6}. \]\n\nInduction step:\n\nAssume that \( N \in \mathbb{N} \) and\n\n\[ \mathop{\sum }\l... | Yes |
Lemma 4.11. Let \( N \in {\mathbb{N}}^{ + } \) and, for \( 0 \leq n \leq N,{a}_{n} \in \mathbb{R} \) . If \( c \in \mathbb{R} \) , then\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{N}c{a}_{n} = c\left( {\mathop{\sum }\limits_{{n = 0}}^{N}{a}_{n}}\right) . \] | Proof. We argue by induction on \( N \) .\n\nBase case: \( N = 1 \)\n\nLet \( c,{a}_{0},{a}_{1} \in \mathbb{R} \) . By the distributive property,\n\n\[ \mathop{\sum }\limits_{{n = 0}}^{1}c{a}_{n} = c{a}_{0} + c{a}_{1} \]\n\n\[ = c\left( {{a}_{0} + {a}_{1}}\right) \]\n\n\[ = c\left( {\mathop{\sum }\limits_{{n = 0}}^{1}{... | Yes |
Proposition 5.22. Suppose \( f : X \rightarrow \mathbb{R} \) and \( g : X \rightarrow \mathbb{R} \) are real functions that are continuous at \( a \in X \) . Let \( c \) and \( d \) be scalars \( {}^{2} \) . Then \( {cf} + {dg} \) and \( {fg} \) are both continuous at \( a \), and so is \( f/g \) if \( g\left( a\right)... | Proof. Exercise. | No |
Proposition 5.23. Every polynomial is continuous on \( \mathbb{R} \) . Every rational function is continuous wherever the denominator is non-zero. | What about the exponential function\n\n\[ \n{e}^{x} \mathrel{\text{:=}} \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{x}^{n}}{n!}?\n\]\n\nEach partial sum is a polynomial, and hence continuous; so if we knew that the limit of a sequence of continuous functions were continuous, we would be done. This turns out, howeve... | No |
Proposition 5.31. The exponential function is continuous on \( \mathbb{R} \) . | Proof. Let\n\n\[ \n{p}_{n}\left( x\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{k = 0}}^{n}\frac{{x}^{k}}{k!} \n\] \n\nbe the \( {n}^{\text{th }} \) -order Taylor polynomial. We know each \( {p}_{n} \) is continuous, by Proposition 5.23. If we knew that \( {p}_{n}\left( x\right) \) converged uniformly to \( {e}^{... | No |
Proposition 6.1. Let \( m, n \in \mathbb{N} \). Then \[ \left( {\left| {\ulcorner m\urcorner }\right| = \left| {\ulcorner n\urcorner }\right| : }\right) \Leftrightarrow \left( {m = n}\right) . \] | DISCUSSION. We prove the non-trivial direction of this biconditional by induction on one of the integers in the statement.\n\nProof. \( \Leftarrow \)\n\nLet \( m = n \). Then it is obvious that \[ \left| {\ulcorner m\urcorner }\right| = \left| {\ulcorner n\urcorner }\right| \text{.} \]\n\n\( \Rightarrow \)\n\nWe argue ... | Yes |
Proposition 6.8. Let \( X \) be a set, and define \( F : \ulcorner {2}^{\neg X} \rightarrow P\left( X\right) \) by: for \( \chi \in \ulcorner {2}^{\neg X} \) , \[ F\left( \chi \right) = {\chi }^{-1}\left( 1\right) \] That is \( F\left( \chi \right) = \{ x \in X \mid \chi \left( x\right) = 1\} \) . Then \( F : \ulcorner... | Proof. The proof is left as an exercise. | No |
Proposition 6.15. Let \( X \) and \( Y \) be sets. Then there is a surjection \( f : X \rightarrow Y \) iff \( \left| Y\right| \leq \left| X\right| \) . | Proof. \( \left( \Rightarrow \right) \n\nLet \( X, Y \) and \( f \) be as in the statement of the proposition. Let\n\n\[ \n\widehat{f} : X/f \rightarrow Y \n\]\n\nbe the canonical bijection associated with \( f \) that was defined in Section 2.3. We ask whether there is an injection \( g : X/f \rightarrow X \) where \(... | No |
Corollary 6.23. \( \mathbb{K} \neq \mathbb{R} \) | Since \( \mathbb{K} \) is countable and \( \mathbb{R} \) is uncountable, \( \mathbb{K} \) is a proper subset of \( \mathbb{R} \) . | Yes |
Proposition 7.1. Let \( a, b \in \mathbb{Z} \) . If \( a \) and \( b \) are relatively prime, then \( a - b \) and \( b \) are relatively prime. | Proof. Let \( c > 1 \) be a common factor of \( b \) and \( a - b \) . So\n\n\[ \left( {\exists m \in \mathbb{Z}}\right) b = {cm} \]\n\nand\n\n\[ \left( {\exists n \in \mathbb{Z}}\right) a - b = {cn}. \]\n\nThen\n\n\[ c\left( {m + n}\right) = a \]\n\nand so \( c \mid a \) . Therefore if \( a \) and \( b \) are relative... | Yes |
Proposition 7.2. Let \( a \) and \( b \) be integers. If \( a \) and \( b \) are relatively prime, then\n\n\[ \left( {\exists m, n \in \mathbb{Z}}\right) {ma} + {nb} = 1. \] | Proof. We may assume that \( a > b > 0 \) . We argue by induction on \( a + b \) .\n\nBase case: \( a + b = 3 \) .\n\nThen \( a = 2 \) and \( b = 1 \) . So\n\n\[ a - b = 1.\text{.} \]\n\nInduction step:\n\nAssume that the result holds for all pairs of relatively prime natural numbers with sum less than \( a + b \) .\n\... | No |
Proposition 7.3. Let \( a, b, c \in \mathbb{Z} \), and assume that \( \gcd \left( {a, b}\right) = 1 \) . If \( a \mid {cb} \), then \( a \mid c \) . | Proof. By Proposition 7.2 there are \( m, n \in \mathbb{Z} \) such that\n\n\[ \n{ma} + {nb} = 1\text{.}\n\]\n\nTherefore\n\n\[ \n{cma} + {cnb} = c.\n\]\n\nClearly \( a \mid {cnb} \) (since \( a \mid {cb} \) ) and \( a \mid {cma} \) . So\n\n\[ \na \mid \left( {{cma} + {cnb}}\right) ,\n\]\n\nand therefore \( a \mid c \) ... | Yes |
Proposition 7.4. Let \( a, b, c \in \mathbb{Z} \) . If \( \gcd \left( {a, b}\right) = 1, a\left| {c\text{and}b}\right| c \) , then \[ {ab} \mid c\text{.} \] | Proof. Let \( m, n \in \mathbb{Z} \) be such that \( {am} = c \) and \( {bn} = c \) . Then \[ a \mid {bn}\text{.} \] By Proposition 7.3, \( a \mid n \) . Hence there is \( k \in \mathbb{Z} \) such that \[ {ak} = n\text{.} \] Therefore \[ {akb} = c \] and \[ {ab} \mid c\text{.} \] | Yes |
Proposition 7.9. Let \( a, b, k \in \mathbb{Z} \) . Then\n\n\[ \gcd \left( {a, b}\right) = \gcd \left( {a - {kb}, b}\right) . \] | Proof. If \( c \in \mathbb{Z}, c\left| {a\text{and}c}\right| b \), then \( c \mid a - {kb} \) . Therefore\n\n\[ \gcd \left( {a, b}\right) \leq \gcd \left( {a - {kb}, b}\right) \]\n\nLikewise, if \( c\left| {a - {kb}\text{and}c}\right| b \), then \( c \mid a \), so we get the reverse inequality of (7.10), so the two sid... | Yes |
Proposition 7.15. Let \( a \in \mathbb{Z} \), and \( p \) be a prime number such that \( p \nmid a \) . Then \( {o}_{p}\left( a\right) < p \) . | Proof. Let \( p \) be a prime number and \( a \in \mathbb{Z} \) be such that \( a \) is not a multiple of \( p \) . By Lemma 7.5, as \( p \nmid a \), then \( p \nmid {a}^{n} \), and therefore \( \left\lbrack {a}^{n}\right\rbrack \in {\mathbb{Z}}_{p}^{ * } \) for any \( n \in \mathbb{N} \) . Since \( \left| {\mathbb{Z}}... | Yes |
Proposition 7.16. Let \( a \in \mathbb{Z} \) and \( p \) be a prime number such that \( a \) is not a multiple of \( p \) . Then the remainder classes \( \left\lbrack 1\right\rbrack ,\left\lbrack a\right\rbrack ,\left\lbrack {a}^{2}\right\rbrack ,\ldots ,\left\lbrack {a}^{{o}_{p}\left( a\right) - 1}\right\rbrack \) in ... | Proof. Exercise. | No |
Proposition 8.16. \( \\left| \\left\\lbrack {0,1}\\right\\rbrack \\right| = \\left| \\mathbb{R}\\right| \) . | Proof. Define \( f : \\lbrack 0,\\infty ) \\rightarrow (1/2,1\\rbrack \) by\n\n\[ f\\left( x\\right) = \\frac{1}{x + 2} + 1/2 \]\n\nThen \( f \) is an injection. Let \( {\\mathbb{R}}^{ - } \) be the negative real numbers, and define \( g : {\\mathbb{R}}^{ - } \\rightarrow \\lbrack 0,1/2) \) by\n\n\[ g\\left( x\\right) ... | Yes |
Corollary 8.18. \( \\left| \\left\\lbrack {0,1}\\right\\rbrack \\right| = {2}^{{\\aleph }_{0}} \) . | Proof. By Lemma 8.17, Proposition 6.15 and Theorem 6.11,\n\n\[ \n\\left| \\left\\lbrack {0,1}\\right\\rbrack \\right| \\leq \\left| {D}_{0}\\right| = \\left| {\\ulcorner {10}^{\\neg \\mathbb{N}}\\urcorner }\\right| = {2}^{{\\aleph }_{0}}.\n\]\n\nLet \( g : \\ulcorner {2}^{\\neg {\\mathbb{N}}^{ + }} \\rightarrow {D}_{0}... | Yes |
Example 9.9. Let\n\n\[ \np\left( x\right) = {x}^{3} - {3x} + 1. \]\n\n(9.10)\n\nThen \( c = 1 \), and\n\n\[ {w}^{3} = \frac{-1 \pm \sqrt{-3}}{2}. \]\n\n(9.11)\n\nNow we have a worse problem: \( {w}^{3} \) involves the square root of a negative number, and even if we make sense of that, we then have to extract a cube ro... | But this cannot be the case for \( p \) . Indeed,\n\n\[ p\left( {-2}\right) = - 1 < 0 \]\n\n\[ p\left( 0\right) = 1 > 0 \]\n\n\[ p\left( 1\right) = - 1 < 0 \]\n\n\[ p\left( 2\right) = 3 > 0. \]\n\nTherefore, by the Intermediate Value Theorem 8.10, \( p \) must have a root in each of the intervals \( \left( {-2,0}\right... | Yes |
Proposition 9.19. Let \( {z}_{1} = {r}_{1}\operatorname{Cis}\left( {\theta }_{1}\right) \) and \( {z}_{2} = {r}_{2}\operatorname{Cis}\left( {\theta }_{2}\right) \) . Then\n\n\[ \n{z}_{1}{z}_{2} = {r}_{1}{r}_{2}\operatorname{Cis}\left( {{\theta }_{1} + {\theta }_{2}}\right) \n\] | Proof. Multiplying out, we get\n\n\[ \n{z}_{1}{z}_{2} = {r}_{1}{r}_{2}\left\lbrack {\cos {\theta }_{1}\cos {\theta }_{2} - \sin {\theta }_{1}\sin {\theta }_{2}}\right.\n\]\n\n\[ \n+ i\left( {\cos {\theta }_{1}\sin {\theta }_{2} + \cos {\theta }_{2}\sin {\theta }_{1}}\right) \rbrack \text{.} \n\]\n\nThe result follows b... | Yes |
Proposition 9.34. Polynomials are continuous functions on \( \mathbb{C} \) . | Proof. Repeat the proof of Proposition 5.23 with complex numbers instead of real numbers. | No |
If we have, as in the previous example,\n\n\[ \nx\left( t\right) = {100} - {4.9}{t}^{2}\text{ meters,}\n\]\n\nthen from time \( t = 1 \) to time \( t = 1 + {\Delta t} \) we would have\n\n\[ \n{\Delta x} = x\left( {1 + {\Delta t}}\right) - x\left( 1\right)\n\] | \[ \n= \left( {{100} - {4.9}{\left( 1 + \Delta t\right) }^{2}}\right) - {95.1}\n\]\n\n\[ \n= {4.9} - {4.9}\left( {1 + {2\Delta t} + {\left( \Delta t\right) }^{2}}\right)\n\]\n\n\[ \n= - {9.8\Delta t} - {4.9}{\left( \Delta t\right) }^{2}\text{meters.}\n\]\n\nHence the average velocity over the interval \( \left\lbrack {... | Yes |
To find the velocity of the object of the previous examples at time \( t = 3 \) | \[ \frac{dx}{dt} = - {29.4} - {4.9dt}\text{ meters }/\text{ second. } \] As above, we disregard the immeasurable -4.9dt to obtain the velocity of the object at time \( t = 3 \) : \[ v\left( 3\right) = - {29.4}\text{meters/second.} \] | Yes |
For our previous example, we find\n\n\[ \n{dx} = x\\left( {t + {dt}}\\right) - x\\left( t\\right) \n\] | \n\[ \n= \\left( {{100} - {4.9}{\\left( t + dt\\right) }^{2}}\\right) - \\left( {{100} - {4.9}{t}^{2}}\\right) \n\]\n\n\[ \n= - {4.9}\\left( {t + {2tdt} + {\\left( dt\\right) }^{2}}\\right) - {4.9}{t}^{2} \n\]\n\n\[ \n= - {9.8tdt} - {4.9}{\\left( dt\\right) }^{2}\\text{meters} \n\]\n\n\[ \n= \\left( {-{9.8t} - {4.9dt}}... | Yes |
Example 1.2.5. Suppose a spherical shaped balloon is being filled with water. If \( r \) is the radius of the balloon in centimeters and \( V \) is the volume of the balloon,\nthen\n\[ V = \frac{4}{3}\pi {r}^{3}{\text{ centimeters }}^{3}. \]\n\nSince a cubic centimeter of water has a mass of 1 gram, the mass of the wat... | To find the rate of change of the mass of the balloon with respect to the radius of the balloon, we first compute\n\[ {dM} = \frac{4}{3}\pi {\left( r + dr\right) }^{3} - \frac{4}{3}\pi {r}^{3} \]\n\[ = \frac{4}{3}\pi \left( {\left( {{r}^{3} + 3{r}^{2}{dr} + {3r}{\left( dr\right) }^{2} + {\left( dr\right) }^{3}}\right) ... | Yes |
If \( f\left( x\right) = {x}^{2} \), then, for example, for any infinitesimal \( \epsilon \) , | \[ f\left( {3 + \epsilon }\right) = {\left( 3 + \epsilon \right) }^{2} = 9 + {6\epsilon } + {\epsilon }^{2} \simeq 9 = f\left( 3\right) . \] Hence \( f \) is continuous at \( x = 3 \) . More generally, for any real number \( x \) , \[ f\left( {x + \epsilon }\right) = {\left( x + \epsilon \right) }^{2} = {x}^{2} + {2x\e... | Yes |
We call the function\n\n\[ \nH\left( t\right) = \left\{ \begin{array}{ll} 0, & \text{ if }t < 0 \\ 1, & \text{ if }t \geq 0 \end{array}\right.\n\]\n\nthe Heaviside function (see Figure 1.4.1). If \( \epsilon \) is a positive infinitesimal, then\n\n\[ \nH\left( {0 + \epsilon }\right) = H\left( \epsilon \right) = 1 = H\l... | Since 0 is not infinitesimally close to 1, it follows that \( H \) is not continuous at 0 . However, for any positive real number \( a \) and any infinitesimal \( \epsilon \) (positive or negative),\n\n\[ \nH\left( {a + \epsilon }\right) = 1 = H\left( a\right)\n\]\n\nsince \( a + \epsilon > 0 \), and for any negative r... | Yes |
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