Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Evaluate \( {\left( 1 + x\right) }^{\frac{1}{x}} \) when \( x = 0 \) . | This function assumes the indeterminate form \( {1}^{\infty } \) for \( x = 0 \) . Let \( y = {\left( 1 + x\right) }^{\frac{1}{x}} \) ; then \( \log y = \frac{1}{x}\log \left( {1 + x}\right) = \infty \cdot 0 \) when \( x = 0 \) . By \( §{10.11} \) , \( y = \frac{\log \left( {1 + x}\right) }{x} = \frac{0}{0} \), when \(... | No |
Example 10.13.3. Evaluate \( \cot x\sin x \) for \( x = 0 \) . | Solution. This function assumes the indeterminate form \( {\infty }^{0} \) for \( x = 0 \) . Let \( y = {\left( \cot x\right) }^{\sin x} \) ; then \( \log y = \sin x\log \cot x = 0 \cdot \infty \) when \( x = 0 \) . By \( \$ {10.11} \) , \( \log y = \frac{\log \cot x}{\csc x} = \frac{\infty }{\infty } \) when \( x = 0 ... | No |
Consider the function \( f\left( x\right) = {e}^{x} \) . We want a polynomial approximation of this function near the point \( x = 0 \) . | Since the derivative of \( {e}^{x} \) is \( {e}^{x} \) , the value of all the derivatives at \( x = 0 \) is \( {f}^{\left( n\right) }\left( 0\right) = {e}^{0} = 1 \) . Taylor’s theorem thus\n\nstates that\n\[ \n{e}^{x} = 1 + x + \frac{{x}^{2}}{2!} + \frac{{x}^{3}}{3!} + \cdots + \frac{{x}^{n}}{n!} + \frac{{x}^{n + 1}}{... | Yes |
Consider the function \( f\left( x\right) = \cos x \) . We want a polynomial approximation of this function near the point \( x = 0 \) . | The first few derivatives of \( f \) are\n\n\[ f\left( x\right) = \cos x \]\n\n\[ {f}^{\prime }\left( x\right) = - \sin x \]\n\n\[ {f}^{\prime \prime }\left( x\right) = - \cos x \]\n\n\[ {f}^{\prime \prime \prime }\left( x\right) = \sin x \]\n\n\[ {f}^{\left( 4\right) }\left( x\right) = \cos x \]\n\nIt's easy to pick o... | Yes |
Consider the function \( f\left( x\right) = \ln x \) . We want a polynomial approximation of this function near the point \( x = 1 \) . | The first few derivatives of \( f \) are\n\n\[ f\left( x\right) = \ln x \]\n\n\[ {f}^{\prime }\left( x\right) = \frac{1}{x} \]\n\n\[ {f}^{\prime \prime }\left( x\right) = - \frac{1}{{x}^{2}} \]\n\n\[ {f}^{\prime \prime \prime }\left( x\right) = \frac{2}{{x}^{3}} \]\n\n\[ {f}^{\left( 4\right) }\left( x\right) = - \frac{... | Yes |
We wish to approximate the derivative of the function on the grid points using only the value of the function on those discrete points. | From the definition of the derivative, one is lead to the formula\n\n\[ \n{f}^{\prime }\left( x\right) \approx \frac{f\left( {x + {\Delta x}}\right) - f\left( x\right) }{\Delta x}.\n\]\n\n(10.22)\n\nTaylor's theorem states that\n\n\[ \nf\left( {x + {\Delta x}}\right) = f\left( x\right) + {\Delta x}{f}^{\prime }\left( x... | Yes |
Find the curvature of the parabola \( {y}^{2} = {4px} \) at the left-most end of the chord that passes through the focus and is perpendicular to the \( y \) -axis. | Solution. \( \frac{dy}{dx} = \frac{2p}{y};\frac{{d}^{2}y}{d{x}^{2}} = - \frac{2p}{{y}^{2}}\frac{dy}{dx} = - \frac{4{p}^{2}}{{y}^{3}} \) . Substituting in (11.3), \( K = \) \( - \frac{{40} - {p}^{2}}{{\left( {y}^{2} + 4{p}^{2}\right) }^{\frac{3}{2}}} \), giving the curvature at any point. At the left-most end of the cho... | Yes |
Find the curvature of the logarithmic spiral \( \rho = {e}^{a\theta } \) at any point. | Solution. \( \frac{d\rho }{d\theta } = a{e}^{a\theta } = {a\rho };\frac{{d}^{2}\rho }{d{\theta }^{2}} = {a}^{2}{e}^{a\theta } = {a}^{2}\rho \) . \n\nSubstituting in (11.4), \( K = \frac{1}{\rho \sqrt{1 + {a}^{2}}} \) . | Yes |
The transition curve on a railway track has the shape of an arc of the cubical parabola \( y = \frac{1}{3}{x}^{3} \) . At what rate is a car on this track changing its direction \( \left( {1\mathrm{{mi}}\text{.} = \text{unit of length}}\right) \) when it is passing through (a) the point \( \left( {3,9}\right) \) ? (b) ... | Solution. \( \frac{dy}{dx} = {x}^{2},\frac{{d}^{2}y}{d{x}^{2}} = {2x} \) . Substituting in (11.3), \( K = \frac{2x}{{\left( 1 + {x}^{4}\right) }^{\frac{3}{2}}} \) . (a) At \( \left( {3,9}\right) \) , \( K = \frac{6}{{\left( {82}\right) }^{\frac{3}{2}}} \) radians per mile \( = {28}^{\prime } \) per mile. (b) At \( \lef... | Yes |
Find the radius of curvature at any point of the catenary \( y = \) \( \frac{a}{2}\left( {{e}^{\frac{x}{a}} + {e}^{-\frac{x}{a}}}\right) \) . | Solution. \( \frac{dy}{dx} = \frac{1}{2}\left( {{e}^{\frac{x}{a}} - {e}^{-\frac{x}{a}}}\right) ;\frac{{d}^{2}y}{d{x}^{2}} = \frac{1}{2a}\left( {{e}^{\frac{x}{a}} - {e}^{-\frac{x}{a}}}\right) \) . Substituting in (11.5),\n\n\[ R = \frac{{\left\lbrack 1 + {\left( \frac{{e}^{\frac{x}{a}} - {e}^{-\frac{x}{a}}}{2}\right) }^... | Yes |
Find the radius of curvature of the cycloid \( x = a\left( {t - \sin t}\right) \) , \( y = a\left( {t - \cos t}\right) . \) | Solution. \( \frac{dx}{dt} = a\left( {1 - \cos t}\right) ,\frac{dy}{dt} = a\sin t;\frac{{d}^{2}x}{d{t}^{2}} = a\sin t,\frac{{d}^{2}y}{d{t}^{2}} = a\cos t \) . Substituting the previous example and then in (11.5), we get\n\n\[ \n\frac{dy}{dx} = \frac{\sin t}{1 - \cos t},\frac{{d}^{2}y}{d{x}^{2}} = \frac{a\left( {1 - \co... | Yes |
Find the radius of curvature at the point \( \left( {3,4}\right) \) on the equilateral hyperbola \( {xy} = {12} \), and draw the corresponding circle of curvature. | Solution. \( \frac{dy}{dx} = - \frac{y}{x},\frac{{d}^{2}y}{d{x}^{2}} = \frac{2y}{{x}^{2}} \) . For \( \left( {3,4}\right) ,\frac{dy}{dx} = - \frac{4}{3},\frac{{d}^{2}y}{d{x}^{2}} = \frac{8}{9} \), so\n\n\[ R = \frac{{\left\lbrack 1 + \frac{16}{9}\right\rbrack }^{\frac{3}{2}}}{\frac{8}{9}} = \frac{125}{24} = {25}\frac{5... | Yes |
We shall solve for the radius of curvature of \( y = {x}^{3} - {x}^{2} + 1 \) at \( x = 1 \) using Sage. | Sage\n\nsage: \( y = {x}^{ \land }3 - {x}^{ \land }2 + 1 \)\n\nsage: Dy = diff(y, x)\n\nsage: D2y = diff(y, x, x)\n\nsage: \( R = \left( {1 + {Dy}{}^{ \land }2}\right) {}^{ \land }\left( {3/2}\right) /{D2y} \)\n\nsage: R(1)\n\n1/sqrt(2)\n\nsage: alpha = x - Dy*(1+Dy^2)/D2y\n\nsage: beta = y + (1+Dy^2)/D2y\n\nsage: alph... | Yes |
Find the coordinates of the center of curvature of the parabola \( {y}^{2} = {4px} \) corresponding (a) to any point on the curve; (b) to the vertex. | Solution. \( \frac{dy}{dx} = \frac{2p}{y};\frac{{d}^{2}y}{d{x}^{2}} = - \frac{4{p}^{2}}{{y}^{3}} \) . \n\n(a) Substituting in (11.13) [§11.8], \n\n\[ \n\alpha = x + \frac{{y}^{2} + 4{p}^{2}}{{y}^{2}} \cdot \frac{2p}{y} \cdot \frac{{y}^{3}}{4{p}^{2}} = {3x} + {2p}. \n\] \n\n\[ \n\beta = y - \frac{{y}^{2} + 4{p}^{2}}{{y}... | Yes |
Find the equation of the evolute of the parabola \( {y}^{2} = {4px} \) . | Solution. \( \frac{dy}{dx} = \frac{2p}{y},\frac{{d}^{2}y}{d{x}^{2}} = - \frac{4{p}^{2}}{{y}^{3}} \) . \n\nFirst step. \( \alpha = {3x} + {2p},\beta = - \frac{{y}^{3}}{4{p}^{2}} \) . \n\nSecond step. \( x = \frac{\alpha - {2p}}{3}, y = - {\left( 4{p}^{2}\beta \right) }^{\frac{1}{3}} \) . \n\nThird step \( {\left( 4{p}^{... | Yes |
Find the equation of the evolute of the ellipse \( {b}^{2}{x}^{2} + {a}^{2}{y}^{2} = \) \( {a}^{2}{b}^{2} \) . | Solution. \( \frac{dy}{dx} = - \frac{{b}^{2}x}{{a}^{2}y},\frac{{d}^{2}y}{d{x}^{2}} = - \frac{{b}^{4}}{{a}^{2}{y}^{3}} \) . \n\nFirst step. \( \alpha = \frac{\left( {{a}^{2} - {b}^{2}}\right) {x}^{3}}{{a}^{4}},\beta = - \frac{\left( {{a}^{2} - {b}^{2}}\right) {y}^{3}}{{b}^{4}} \) . \n\nSecond step. \( x = {\left( \frac{... | Yes |
Find the equation of the evolute in parametric form, plot the curve and the evolute, find the radius of curvature at the point where \( t = 1 \), and draw the corresponding circle of curvature. | Solution. \( \frac{dx}{dt} = \frac{t}{2},\frac{{d}^{2}x}{d{t}^{2}} = \frac{1}{2},\frac{dy}{dt} = \frac{{t}^{2}}{2},\frac{{d}^{2}y}{d{t}^{2}} = t \) . Substituting in above formulas (11.18) and then in (11.15), gives\n\n\[ \alpha = \frac{1 - {t}^{2} - 2{t}^{4}}{4},\;\beta = \frac{4{t}^{3} + {3t}}{6}, \]\n\n(11.20)\n\nth... | Yes |
Find the parametric equations of the evolute of the cycloid, \n\n\[ \n\\left\\{ \\begin{array}{l} x = a\\left( {t - \\sin t}\\right) \\\\ y = a\\left( {1 - \\cos t}\\right) \\end{array}\\right. \n\] | Solution. As in Example 11.5.2, we get \n\n\[ \n\\frac{dy}{dx} = \\frac{\\sin t}{1 - \\cos t},\\;\\frac{{d}^{2}y}{d{x}^{2}} = - \\frac{1}{\\alpha {\\left( 1 - \\cos t\\right) }^{2}}. \n\] \n\nSubstituting these results in formulas (11.15), we get the answer: \n\n\[ \n\\left\\{ \\begin{array}{l} \\alpha = a\\left( {t + ... | Yes |
Corollary 1.13 (extension property) Let \( F : \left( {a, b}\right) \rightarrow \mathbb{R} \) be a function that is continuous on the bounded, open interval \( \left( {a, b}\right) \) . Then \( F \) can be extended to a uniformly continuous function on all of the closed, bounded interval \( \left\lbrack {a, b}\right\rb... | That extension is obtained by defining\n\n\[ F\left( a\right) = F\left( {a + }\right) = \mathop{\lim }\limits_{{x \rightarrow a + }}F\left( x\right) \text{ and }F\left( b\right) = F\left( {b - }\right) = \mathop{\lim }\limits_{{x \rightarrow b - }}F\left( x\right) \]\n\nboth of which limits exist if \( F \) is uniforml... | Yes |
Theorem 1.33 (vanishing derivatives with many exceptions) Let \( F : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be uniformly continuous on the closed, bounded interval \( \left\lbrack {a, b}\right\rbrack \) and suppose that \( {F}^{\prime }\left( x\right) = 0 \) for every \( a < x < b \) with the possib... | [The argument that was successful for Theorem 1.31 will not work for infinitely many exceptional points. A Cousin partitioning argument does work.] Answer \( ▱ \) | No |
Prove this part of Theorem 2.2: If a function \( f \) is bounded and possesses an indefinite integral \( F \) on \( \left( {a, b}\right) \) then \( F \) is Lipschitz on \( \left( {a, b}\right) \) . Deduce that \( F \) is uniformly continuous on \( \left( {a, b}\right) \) . | Answer | No |
Theorem 3.14 (integral inequalities) Suppose that the two functions \( f, g \) are both integrable on a closed, bounded interval \( \left\lbrack {a, b}\right\rbrack \) and that \( f\left( x\right) \leq g\left( x\right) \) for all \( x \in \left\lbrack {a, b}\right\rbrack \) with possibly finitely many exceptions. Then\... | The proof is an easy exercise in derivatives. We know that if \( H \) is uniformly continuous on \( \left\lbrack {a, b}\right\rbrack \) and if\n\n\[ \frac{d}{dx}H\left( x\right) \geq 0 \]\n\nfor all but finitely many points \( x \) in \( \left( {a, b}\right) \) then \( H\left( x\right) \) must be nondecreasing on \( \l... | No |
Theorem 3.15 Let \( f : \\left( {a, b}\\right) \\rightarrow \\mathbb{R} \) be integrable on \( \\left\\lbrack {a, b}\\right\\rbrack \) and suppose that \( F \) is an indefinite integral. Suppose further that \( {F}^{\\prime }\\left( x\\right) = f\\left( x\\right) \) for all \( a < x < b \) with no exceptional points. T... | \[ {\\int }_{a}^{b}f\\left( x\\right) {dx} = f\\left( \\xi \\right) \\left( {b - a}\\right) \] | No |
Theorem 3.29 Let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a uniformly continuous function. Then the integral may be uniformly approximated by unstraddled Riemann sums: for every \( \varepsilon > 0 \) there is a \( \delta > 0 \) so that\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}\left| {{\int }... | The proof is sufficiently similar to that for Theorem 3.19 that the reader need not trouble over it. The only moral here is that one should remain alert to other formulations of technical ideas and be prepared to exploit them (as did McShane) in other contexts. | No |
Theorem 3.34 Suppose that a function \( f : \left( {a, b}\right) \rightarrow \mathbb{R} \) is absolutely integrable on a closed, bounded interval \( \left\lbrack {a, b}\right\rbrack \) . Then its indefinite integral \( F \) must be a function of bounded variation there and, moreover, | \[ V\left( {F,\left\lbrack {a, b}\right\rbrack }\right) = {\int }_{a}^{b}\left| {f\left( x\right) }\right| {dx}. \] | Yes |
Theorem 3.39 Let \( \\left\\{ {g}_{k}\\right\\} \) be a sequence of functions defined on an interval I. Then the series \( \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }{f}_{k} \) converges uniformly to some function \( f \) on the interval \( I \) if and only if for every \( \\varepsilon > 0 \) there is an integer \( N... | \[ \\left| {\\mathop{\\sum }\\limits_{{j = m}}^{n}{f}_{j}\\left( x\\right) }\\right| < \\varepsilon \] for all \( n \\geq m \\geq N \) and all \( x \\in I \) . | Yes |
Theorem 3.41 (Abel) Let \( \\left\\{ {a}_{k}\\right\\} \) and \( \\left\\{ {b}_{k}\\right\\} \) be sequences of functions on an interval I. Suppose that there is a number \( M \) so that\n\n\[ \n- M \\leq {s}_{N}\\left( x\\right) = \\mathop{\\sum }\\limits_{{k = 1}}^{N}{a}_{k}\\left( x\\right) \\leq M \n\]\n\nfor all \... | Answer \( ▱ \) | No |
Theorem 3.44 Let \( {f}_{1},{f}_{2},{f}_{3},\ldots \) be a sequence of functions defined and integrable on a closed, bounded interval \( \left\lbrack {a, b}\right\rbrack \) . Suppose that \( \left\{ {f}_{n}\right\} \) converges uniformly on \( \left\lbrack {a, b}\right\rbrack \) to a function \( f \) . Then, provided w... | Answer \( ▱ \) | No |
Theorem 3.53 (integration of power series) Let\n\n\\[ \nf\\left( x\\right) = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }{a}_{n}{x}^{n} = {a}_{0} + {a}_{1}x + {a}_{2}{x}^{2} + {a}_{3}{x}^{3} + \\ldots \n\\]\n\nbe a power series and let\n\n\\[ \nF\\left( x\\right) = \\mathop{\\sum }\\limits_{{n = 0}}^{\\infty }\\frac{{... | Note that the integration theorem uses the interval of convergence of the integrated series. It is not a concern whether the original series for \\( f \\) converges at the endpoints of the interval of convergence, but it is essential to look at these endpoints for the integrated series. The proofs of the separate state... | No |
Theorem 3.60 Suppose that \( f \) is twice continuously differentiable at all points of the interval \( \left\lbrack {a, b}\right\rbrack \) . Let\n\n\[ \n{T}_{n} = \frac{b - a}{n}\left\lbrack {\frac{f\left( a\right) + f\left( b\right) }{2} + \mathop{\sum }\limits_{{k = 1}}^{{n - 1}}f\left( {a + k\frac{b - a}{n}}\right)... | Answer \( ▱ \) | No |
Show that if \( H \) is a point set and \( K \) is a point set and \( p \) is a limit point of \( H \cap K \) | then \( p \) is a limit point of \( H \) and \( p \) is a limit point of \( K \) | Yes |
Theorem 67. Let \( {f}_{1},{f}_{2},\ldots \) be the sequence of functions defined in the previous problem and show that if \( x \in \left\lbrack {0,1}\right\rbrack \) then \( {f}_{1}\left( x\right) ,{f}_{2}\left( x\right) ,{f}_{3}\left( x\right) \ldots \) converges to some number. | Since for each \( x \in \left\lbrack {0,1}\right\rbrack \) the sequence \( {f}_{1}\left( x\right) ,{f}_{2}\left( x\right) ,{f}_{3}\left( x\right) \ldots \), converges we may define a function \( f \) on \( \left\lbrack {0,1}\right\rbrack \) as follows: for each \( x \in \left\lbrack {0,1}\right\rbrack \) let \( f\left(... | Yes |
Theorem 73. Show there is a unique solution to the initial value problem \( {y}^{\prime \prime } + y = 0, y\left( 0\right) = 0,{y}^{\prime }\left( 0\right) = 1 \) as follows: | 1. Convert the second order equation to a first order system,\n\n\[ \n{\left( \begin{array}{l} u \\ v \end{array}\right) }^{\prime } = A\left( \begin{array}{l} u \\ v \end{array}\right) ,\left( \begin{array}{l} u \\ v \end{array}\right) \left( 0\right) = \left( \begin{array}{l} 0 \\ 1 \end{array}\right) \text{where}A\t... | No |
Theorem 85. Assume that each of \( A, B \) and \( C \) are subsets of the set \( X \). | 1. \( A \cup B = B \cup A \) and \( A \cap B = B \cap A \)\n2. \( A \cup \varnothing = A \) and \( A \cap \varnothing = \varnothing \)\n3. \( A \cup X = X \) and \( A \cap X = A \)\n4. \( A \cap \left( {B \cup C}\right) = \left( {A \cap B}\right) \cup \left( {A \cap C}\right) \) and \( A \cup \left( {B \cap C}\right) =... | Yes |
Let \( y \) be defined by the following formulas, each applying to just one range of inputs.\n\n\[ y = \left\{ \begin{array}{ll} {x}^{2} & \text{ if }x \leq 0 \\ - {x}^{2} & \text{ if }0 < x \leq 3 \\ {e}^{x} & \text{ if }3 < x \end{array}\right. \] | Which formula you use depends upon which \( x \) -value you are plugging in. To plug in \( x = - 1 \) we use the first formula. So an input of \( x = - 1 \) has an output of \( {\left( -1\right) }^{2} = 1 \) . To plug in \( x = 2 \) we use the second formula, so the output is \( - {2}^{2} = - 4 \) . Similarly an input ... | Yes |
The distance between the points \( \left( {2,5}\right) \) and \( \left( {5,1}\right) \) (see figure 1.13) is | \[ \sqrt{{\left( 2 - 5\right) }^{2} + {\left( 5 - 1\right) }^{2}} = \sqrt{9 + {16}} = {25} \] | No |
What is the limit of \( f\left( x\right) = x + 7 \) as \( x \) approaches 4 ? | There are two steps to answering such a question first we must determine the answer - this is where intuition and guessing is useful, as well as the informal definition of a limit. Then, we must prove that the answer is right. For this problem, the answer happens to be 11 . Now, we must prove it using the definition of... | Yes |
What is the limit of \( x\sin \left( {1/x}\right) \) as \( x \) approaches 0 ? | We will prove that the limit is 0 . For every \( \epsilon > 0 \), choose \( \delta = \epsilon \) so that for all \( x \), if \( 0 < \left| x\right| < \delta \), then \( \left| {x\sin x - 0}\right| < = \left| x\right| < \epsilon \) as required. | No |
Every non-empty set of real numbers which is bounded below has an greatest lower bound. | Let \( E \) be a non-empty set of of real numbers which is bounded below. Then \( - E \) is bounded above (check this assertion). Let \( M \) be a least upper bound for \( - E \) . Then \( - M \) is a greatest lower bound for \( E \) (check this assertion). | No |
\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{2x} + 3}{5 - x} \] | \[ \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{{2x} + 3}{5 - x} = \mathop{\lim }\limits_{{x \rightarrow \infty }}\frac{2 + \frac{3}{x}}{\frac{5}{x} - 1} = \mathop{\lim }\limits_{{y \rightarrow 0}}\frac{2 + {3y}}{{5y} - 1} = \frac{2 + 3 \cdot 0}{5 \cdot 0 - 1} = - 2 \] | Yes |
Example what is \( \frac{d}{dx}\left( {3{x}^{2} + {5x}}\right) \) | \[ \frac{d}{dx}\left( {3{x}^{2} + {5x}}\right) = \frac{d}{dx}3{x}^{2} + \frac{d}{dx}{5x} \]\n\[ = {6x} + \frac{d}{dx}{5x} \]\n\[ = {6x} + 5 \] | Yes |
If \( {f}^{\prime }\left( x\right) = 0 \) for all \( x \), then \( f \) is a constant function. | Let \( f \) be ULD on \( \left\lbrack {A, B}\right\rbrack \) and \( {f}^{\prime } = 0 \) . IFT tells us that \( f\left( A\right) \geq f\left( B\right) \) . But \( {\left( -f\right) }^{\prime } = 0 \) too, so \( - f\left( A\right) \geq - f\left( B\right) \), and \( f\left( A\right) \leq f\left( B\right) \), therefore \(... | No |
The derivative of a ULD function is ULC. | For \( x \neq a \), by dividing both sides of ?? by \( \left| {x - a}\right| \), we get\n\n\[ \left| {\frac{f\left( x\right) - f\left( a\right) }{x - a} - {f}^{\prime }\left( a\right) }\right| \leq K\left| {x - a}\right| . \]\n\n(3.3)\n\nThis estimate may be handy to check your differentiation. If your formula for \( {... | No |
[author=garrett, file =text_files/quotient_rule]\n\n\[ \n\frac{d}{dx}\left( \frac{1}{x - 2}\right) \n\] | \[ \n\frac{d}{dx}\left( \frac{1}{x - 2}\right) = \frac{\frac{d}{dx}1 \cdot \left( {x - 2}\right) - 1 \cdot \frac{d}{dx}\left( {x - 2}\right) }{{\left( x - 2\right) }^{2}} = \frac{0 \cdot \left( {x - 2}\right) - 1 \cdot 1}{{\left( x - 2\right) }^{2}} = \frac{-1}{{\left( x - 2\right) }^{2}} \n\] | Yes |
[author=garrett, file =text_files/quotient_rule]\n\n\[ \n\frac{d}{dx}\left( \frac{x - 1}{x - 2}\right) \n\] | \[ \n\frac{d}{dx}\left( \frac{x - 1}{x - 2}\right) = \frac{{\left( x - 1\right) }^{\prime }\left( {x - 2}\right) - \left( {x - 1}\right) {\left( x - 2\right) }^{\prime }}{{\left( x - 2\right) }^{2}} = \frac{1 \cdot \left( {x - 2}\right) - \left( {x - 1}\right) \cdot 1}{{\left( x - 2\right) }^{2}} \n\]\n\n\[ \n= \frac{\... | Yes |
\[ \frac{d}{dx}\left( \frac{5{x}^{3} + x}{2 - {x}^{7}}\right) = \frac{{\left( 5{x}^{3} + x\right) }^{\prime } \cdot \left( {2 - {x}^{7}}\right) - \left( {5{x}^{3} + x}\right) \cdot \left( {2 - {x}^{7}}\right) }{{\left( 2 - {x}^{7}\right) }^{2}} \] | \[ = \frac{\left( {{15}{x}^{2} + 1}\right) \cdot \left( {2 - {x}^{7}}\right) - \left( {5{x}^{3} + x}\right) \cdot \left( {-7{x}^{6}}\right) }{{\left( 2 - {x}^{7}\right) }^{2}} \] | Yes |
It is very important to recognize situations like\n\n\\[ \n\\frac{d}{dx}{\\left( ax + b\\right) }^{n} = n{\\left( ax + b\\right) }^{n - 1} \\cdot a \n\\]\n\nfor any constants \\( a, b, n \\) . | And, of course, this includes\n\n\\[ \n\\frac{d}{dx}\\sqrt{{ax} + b} = \\frac{1}{2}{\\left( ax + b\\right) }^{-1/2} \\cdot a \n\\]\n\n\\[ \n\\frac{d}{dx}\\frac{1}{{ax} + b} = - {\\left( ax + b\\right) }^{-2} \\cdot a = \\frac{-a}{{\\left( ax + b\\right) }^{2}} \n\\] | Yes |
If \( x = c \) is a local min/max then \( {f}^{\prime }\left( c\right) = 0 \) or \( {f}^{\prime }\left( c\right) \) is undefined. | Look at some pictures of local min/max. If \( {f}^{\prime }\left( c\right) = 0 \) or \( {f}^{\prime }\left( c\right) \) is undefined we call \( c \) a critical point. This fact justifies our approach to finding local min/max's which always starts with finding the critical points. | No |
Let \( f\left( x\right) = \sin \left( x\right) \). Then the equation of the tangent line at \( x = 0 \) is \( L\left( x\right) = x \). Then \( \sin \left( x\right) \cong x \) for \( x \) near 0. | If you like, make a table of some values of \( {y}_{1} = \sin \left( x\right) \) and \( {y}_{2} = x \) for \( x \) near 0. (By the way, this explains why \( \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin \left( x\right) }{x} = 1 \).) | No |
Find \( \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}x\ln x \) : The \( {0}^{ + } \) means that we approach 0 from the positive side, since otherwise we won't have a real-valued logarithm. This problem illustrates the possibility as well as necessity of rearranging a limit to make it be a ratio of things, in order ... | \[ \mathop{\lim }\limits_{{x \rightarrow {0}^{ + }}}x\ln x = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\ln x}{1/x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{1/x}{-1/{x}^{2}} = \mathop{\lim }\limits_{{x \rightarrow 0}} - x = 0 \] | Yes |
The Fundamental Theorem of Calculus states that If \( {\int }_{0}^{x}f\left( t\right) {dt} = F\left( x\right) \), then \( \int f\left( x\right) {dx} = \) \( F\left( x\right) + C \), and \( {\int }_{a}^{b}f\left( x\right) {dx} = F\left( b\right) - F\left( a\right) \) for any continuous function \( f \) . | We have to establish the inequality\n\n\[ \left| {F\left( c\right) - F\left( b\right) - f\left( b\right) \left( {c - b}\right) }\right| \leq K{\left( c - b\right) }^{2} \]\n\nbut by our integration rules the LHS can be rewritten as\n\n\[ \left| {{\int }_{b}^{c}\left( {f\left( x\right) - f\left( b\right) }\right) {dx}}\... | No |
Find the distance travelled by a ball which has path given by \( y = - {x}^{2} + 4 \) . | I'll pretend I don't know how to solve this exactly and do an approximaton in 3 steps. Thus \( {\Delta x} = 4/3 \) and so I will have points at \( x \) equal to \( - 2, - 2/3,2/3,2 \) . The \( y \) -values corresponding to these \( x \) -values are \( 0,{32}/9,{32}/9,0 \) . Between these points I will use straight line... | Yes |
For example the average value of the function \( y = {x}^{2} \) over the interval \( \left\lbrack {2,3}\right\rbrack \) is | average value of \( f \) on the interval \( \left\lbrack {a, b}\right\rbrack = \frac{{\int }_{2}^{3}{x}^{2}{dx}}{3 - 2} = \frac{{\left\lbrack {x}^{3}/3\right\rbrack }_{2}^{3}}{3 - 2} = \frac{{3}^{3} - {2}^{3}}{3 \cdot \left( {3 - 2}\right) } = {19}/3 | Yes |
Let’s take the same function as in Example 6.7.1, and rotate it around the \( y \) -axis instead of the \( x \) -axis. Then we have | \[ \text{ volume } = {\int }_{\text{left }}^{\text{right }}{2\pi x}\left( {\text{ upper } - \text{ lower }}\right) {dx} \] \[ = {\int }_{0}^{1}{2\pi x}\left( {x - {x}^{2}}\right) {dx} = \pi {\int }_{0}^{1}\frac{2{x}^{3}}{3} - \frac{2{x}^{4}}{4}{dx} = {\left\lbrack \frac{2{x}^{3}}{3} - \frac{2{x}^{4}}{4}\right\rbrack }_... | Yes |
if a force of \( F\left( x\right) \) of strength \( \sin \left( x\right) \) acts on a object at position \( x \) (for \( x \) in \( \left\lbrack {0,\pi /2}\right\rbrack \) ) and the direction of \( F\left( x\right) \) is towards \( x = 0 \), find the work required to move it from \( x = 0 \) to \( x = 1 \) . | (a) Here the force is changing. Let \( {\Delta x} \) be a little distance that the object will move, at position \( x \) (for example, \( {\Delta x} = {.1} \) and \( x = 0 \) would represent the work to move from \( x = 0 \) to \( x = {.1} \) ). On this segment, the work will be \( \sin \left( x\right) \) (from \( x = ... | Yes |
Find \( \frac{123}{9} \) | We rewrite this as \( 9 \mid \overline{123} \) . We will put first a 1 on top because 9 goes into 12 once:\n\n\[ \n\begin{array}{l} 9\overline{123} \rightarrow 9\overline{123} \rightarrow 9\overline{123} \\ \end{array} \n\]\n\nSo we have a remainder of 6 . We write this as\n\n\[ \n\frac{123}{9} = {13} + \frac{6}{9} \n\... | No |
\[ \int {\sin }^{3}{xdx} \] | \[ \int {\sin }^{3}{xdx} = \int \left( {1 - {\cos }^{2}x}\right) \sin {xdx} = - \int \left( {1 - {\cos }^{2}x}\right) \left( {-\sin x}\right) {dx} \] In the latter expression, we can view the \( - \sin x \) as the derivative of \( \cos x \), so with the substitution \( u = \cos x \) this integral is \[ - \int \left( {1... | Yes |
A bigger version of this application of the half-angle formula is\n\n\\[ \n\\int {\\sin }^{6}{xdx} = \\int {\\left( \\frac{1 - \\cos {2x}}{2}\\right) }^{3}{dx} = \\int \\frac{1}{8} - {38}\\cos {2x} + \\frac{3}{8}{\\cos }^{2}{2x} - \\frac{1}{8}{\\cos }^{3}{2xdx} \n\\] | Of the four terms in the integrand in the last expression, we can do the first two directly:\n\n\\[ \n\\int \\frac{1}{8}{dx} = \\frac{x}{8} + C\\;\\int - {38}\\cos {2xdx} = \\frac{-3}{16}\\sin {2x} + C \n\\]\n\nBut the last two terms require further work: using a half-angle formula again, we\n\nhave\n\\[ \n\\int \\frac... | Yes |
Find the integral \( \int \sin \left( x\right) \sin \left( {2x}\right) {dx} \) . | \[ \text{Use}\sin x\sin {2x} = \frac{1}{2}\left( {\cos \left( {-x}\right) - \cos \left( {3x}\right) }\right) = \frac{1}{2}\left( {\cos x - \cos {3x}}\right) \text{. Then}\int \sin \left( x\right) \sin {2xdx} = \frac{1}{2}\frac{1}{2}\text{(} \] | No |
(a) Find the area under a circle with radius 1, from \( x = 0 \) to \( x = 1/2 \) . This is \( {\int }_{0}^{1/2}\sqrt{1 - {x}^{2}}{dx} \) . | The hard part is coming up with the definite integral. Let \( x = \sin \left( \theta \right) \), then \( {dx} = \cos \left( \theta \right) {d\theta } \) . Note that \( \sqrt{1 - {x}^{2}} = \sqrt{1 - {\sin }^{2}\left( \theta \right) } = \) \( \cos \left( \theta \right) \) . We also translate the endpoints of the integra... | Yes |
[author=garrett, file =text_files/trigonometric_subst]\n\nIn another example, we might have\n\n\[ \n\int \sqrt{{8x} - 4{x}^{2}}{dx} \n\] | Completing the square again, we have\n\n\[ \n{8x} - 4{x}^{2} = - 4\left( {-2 + {x}^{2}}\right) = - 4\left( {-1 + 1 - {2x} + {x}^{2}}\right) = - 4\left( {-1 + {\left( x - 1\right) }^{2}}\right) \n\]\n\nRather than put the whole ’ -4 ’ back, we only keep track of the \( \pm \), and take a ’ +4 ’ outside the square root e... | Yes |
Find the integral of \( \sqrt{1 - {x}^{2}} \) | \[ {\int }_{0}^{1}\sqrt{1 - {x}^{2}}{dx} = {\int }_{0}^{\pi /2}\sqrt{1 - {\sin }^{2}\theta }\cos {\theta d\theta } \] \[ = \;{\int }_{0}^{\pi /2}{\cos }^{2}{\theta d\theta } \] \[ = \;\frac{1}{2}{\int }_{0}^{\pi /2}1 + \cos {2\theta d\theta } \] | Yes |
Find the integral of \( \sqrt{}\left( {1 + x}\right) /\sqrt{}\left( {1 - x}\right) \) . | We first rewrite this as\n\n\[ \sqrt{\frac{1 + x}{1 - x}} = \sqrt{\frac{1 + x}{1 + x}}\frac{1 + x}{1 - x} = \frac{1 + x}{\sqrt{1 - {x}^{2}}} \]\n\nThen we can make the substitution\n\n\[ {\int }_{0}^{a}\frac{1 + x}{\sqrt{1 - {x}^{2}}}{dx}\; = \;{\int }_{0}^{\alpha }\frac{1 + \sin \theta }{\cos \theta }\cos \theta \;{d\... | Yes |
Find the integral of \( {\left( {x}^{2} + {a}^{2}\right) }^{ - }3/2 \) . | We make the substitution:\n\n\[ \n{\int }_{0}^{z}{\left( {x}^{2} + {a}^{2}\right) }^{-\frac{3}{2}}{dx} = {a}^{-2}{\int }_{0}^{\alpha }\cos {\theta d\theta }\;z > 0 \n\] \n\n\[ \n= \;{a}^{-2}{\left\lbrack \sin \theta \right\rbrack }_{0}^{\alpha }\;\alpha = {\tan }^{-1}\left( {z/a}\right) \n\] \n\n\[ \n= \;{a}^{-2}\sin \... | Yes |
Consider the problem\n\n\\[ \n\\int \\frac{1}{{x}^{2} + {a}^{2}}{dx} \n\\] | with the substitution \\( x = a\\tan \\left( \\theta \\right) \\), we have \\( {dx} = {ase}{c}^{2}{\\theta d\\theta } \\), so that\n\n\\[ \n\\int \\frac{1}{{x}^{2} + {a}^{2}}{dx} = \\frac{\\arctan \\left( {x/a}\\right) }{a} \n\\] | Yes |
[author=duckworth, file =text_files/new_taylor_series_from_old] Find the Taylor polynomial at \( x = 1 \) for \( \ln \left( x\right) \) . | This problem we could do by taking lots of derivatives, but it's easier to do it by starting with an example we already know. Let’s start with \( 1/x \) and take the anti-derivative.\n\n\[ \ln \left( x\right) = \int \frac{1}{x}{dx} + C \]\n\n\[ = \int 1 - \left( {x - 1}\right) + {\left( x - 1\right) }^{2} - {\left( x -... | Yes |
Being a little more careful, let's keep track of the error term in the example we've been doing: we have\n\n\[ \frac{1}{1 - x} = 1 + x + {x}^{2} + \ldots + {x}^{n} + \frac{1}{\left( n + 1\right) }\frac{1}{{\left( 1 - c\right) }^{n + 1}}{x}^{n + 1} \] | for some \( c \) between 0 and \( x \), and also depending upon \( x \) and \( n \) . One way to avoid having the \( \frac{1}{{\left( 1 - c\right) }^{n + 1}} \) ’blow up’ on us, is to keep \( x \) itself in the range \( \lbrack 0,1) \) so that \( c \) is in the range \( \lbrack 0, x) \) which is inside \( \lbrack 0,1) ... | Yes |
For any real number \( n \), and for \( \left| x\right| < 1 \), we have\n\n\[{\left( 1 + x\right) }^{n} = 1 + {nx} + \left( \begin{array}{l} n \\ 2 \end{array}\right) {x}^{2} + \left( \begin{array}{l} n \\ 3 \end{array}\right) {x}^{3} + \ldots\] | To prove that the binomial series is correct one just applies the Maclaurin series to \( {\left( 1 + x\right) }^{n} \) . To use the binomial series for something like \( {\left( a + b\right) }^{n} \) you factor out the larger number. So suppose \( a \geq b \), then we write \( {\left( a + b\right) }^{n} = {a}^{n}{\left... | No |
Let \( r \) be a real number. Then \( \mathop{\sum }\limits_{{i = 0}}^{\infty }a{r}^{i} \) equals \( \frac{a}{1 - r} \) if \( \left| r\right| < 1 \), and does not exist otherwise. | Note: this is proven in an ad hoc manner, meaning, the proof is made up just for this series and does not follow a general strategy (essentially you multiply the partial sum \( {r}^{0} + {r}^{1} + \cdots + {r}^{n} \) by \( r - 1 \) and simplify). | Yes |
Let \( {a}_{n} = {n}^{-\frac{n + 1}{n}} \)\n\nFor large \( n \), the terms of this series are similar to, but smaller than, those of the harmonic series. We compare the limits. | \[ \lim \frac{\left| {a}_{n}\right| }{{c}_{n}} = \lim \frac{n}{{n}^{\frac{n + 1}{n}}} = \lim \frac{1}{{n}^{\frac{1}{n}}} = 1 > 0 \]\n\nso this series diverges. | Yes |
The series sum converges provided that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) . | The error in a partial sum of an alternating series is smaller than the first omitted term. \( \left| {\mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} - \mathop{\sum }\limits_{{n = 1}}^{m}{a}_{n}}\right| < \left| {a}_{m + 1}\right| \) | No |
Radius of convergence We can only use the equation \( f\left( x\right) = \mathop{\sum }\limits_{{j = 0}}^{n}{a}_{j}{x}^{j} \) to study \( f\left( x\right) \) when the power series converges. This may happen for a finite range, or for all real numbers. | If we use the ratio test on an arbitary power series, we find it converges when \( \lim \frac{\left| {a}_{n + 1}x\right| }{\left| {a}_{n}\right| } < 1 \) and diverges when \( \lim \frac{\left| {a}_{n + 1}x\right| }{\left| {a}_{n}\right| } > 1 \) The radius of convergence is therefore \( r = \lim \frac{\left| {a}_{n}\ri... | Yes |
(The sandwich theorem) Given three sequences, \( R, S, T \), If \( R \) and \( T \) both converge, \( \lim R = \lim T \) and \( \exists N\forall n > N\;{r}_{n} \leq {s}_{n} \leq {t}_{n} \) Sequence \( S \) converges to the same limit | Let \( s = \lim R = \lim T \) . For any \( \epsilon > 0 \), by definition of convergence, there exist \( \mathrm{M} \) , \( \mathrm{N} \) such that \( \forall n > M\;\left| {{r}_{n} - s}\right| < \epsilon \forall n > N\;\left| {{t}_{n} - s}\right| < \epsilon \) Combing these two inequalities with the conditions on \( \... | Yes |
A sequence \( S \) is convergent if and only if it is cauchy. | Convergence implies cauchy. Assume \( \mathrm{S} \) is convergent, with limit \( \mathrm{s} \) For a \( \epsilon > \) 0, choose \( \mathrm{n} \) such that \( \forall k > n\;\left| {{s}_{k} - s}\right| < \epsilon /2 \) (always possible by defintion of convegence) Via triangle inequality, \( \left| {{s}_{k} - {s}_{j}}\ri... | Yes |
A hundred liter tank has salty water flowing in. Salty water with \( {.5}\mathrm{\;{kg}}/\mathrm{L} \) flows into the tank at rate of \( 7\mathrm{\;L}/\mathrm{{min}} \) . Thoroughly mixed water flows out of the tank at the same rate as water flows in. Find an equation for the amount of salt in the tank. | Translating these words into equations we have that the rate of salt in is \( {.5} \times 7 \) and the rate of salt out is the concentration times 7, which becomes \( \frac{5}{100} \times 7 \) . Thus, the differential equation would is\n\n\[ \frac{dy}{dx} = {.5} \times 7 - \frac{y}{100} \times 7 \]\n\nWe separate this ... | Yes |
Solve \( 1 + 2{y}^{2}{\mathrm{D}}^{2}y = 0 \) if at \( \mathrm{x} = 0,\mathrm{y} = \mathrm{{Dy}} = 1 \) | First, we make the substitution, getting \( 1 + 2{y}^{2}u\frac{du}{dy} = 0 \) This is a first order ODE. By rearranging terms we can separate the variables \( {udu} = - \frac{dy}{2{y}^{2}} \) Integrating this gives \( {u}^{2}/2 = c + 1/{2y} \) We know the values of \( \mathrm{y} \) and \( \mathrm{u} \) when \( \mathrm{... | Yes |
Consider \( \frac{{d}^{2}y}{d{x}^{2}} + \frac{2}{x}\frac{dy}{dx} - \frac{6}{{x}^{2}}y = 0 \) | One solution of this is \( y = {x}^{2} \), so substitute \( y = z{x}^{2} \) into this equation.\n\n\[ \left( {{x}^{2}\frac{{d}^{2}z}{d{x}^{2}} + {2x}\frac{dz}{dx} + {2z}}\right) + \frac{2}{x}\left( {{x}^{2}\frac{dz}{dx} + {2xz}}\right) - \frac{6}{{x}^{2}}{x}^{2}z = 0 \]\n\nRearrange and simplify. \( {x}^{2}{D}^{2}z + {... | Yes |
Similarly, we get\n\n\[ \n{\int }_{s}\nabla \times {uds} = {\int }_{C}n \times {uds}\;\left( 1\right) , \n\]\n\nwhere \( C \) is the boundary of \( S \) . | To see this, suppose we have different surfaces, \( {S}_{1} \) and \( {S}_{2} \), spanning the same curve \( \mathrm{C} \), then by switching the direction of the normal on one of the surfaces we\n\ncan write\n\[ \n{\int }_{{S}_{1} + {S}_{2}}\nabla \times {udS} = {\int }_{S}\nabla \times {udS} - {\int }_{S}\nabla \time... | Yes |
Proposition 1.1.3. For every nonnegative integer, \( n \), the value of \( {n}^{2} + n + {41} \) is prime. | But \( p\left( {40}\right) = {40}^{2} + {40} + {41} = {41} \cdot {41} \), which is not prime. So it’s not true that the expression is prime for all nonnegative integers. | Yes |
Proposition 1.1.4. [Euler's Conjecture] The equation\n\n\[ \n{a}^{4} + {b}^{4} + {c}^{4} = {d}^{4} \n\]\n\nhas no solution when \( a, b, c, d \) are positive integers. | Euler (pronounced \ | No |
Proposition 1.1.6 (Four Color Theorem). Every map can be colored with 4 colors so that adjacent \( {}^{2} \) regions have different colors. | Several incorrect proofs of this theorem have been published, including one that stood for 10 years in the late 19th century before its mistake was found. A laborious proof was finally found in 1976 by mathematicians Appel and Haken, who used a complex computer program to categorize the four-colorable maps. The program... | Yes |
Proposition 1.1.7 (Fermat’s Last Theorem). There are no positive integers \( x, y \) , and \( z \) such that\n\n\[ {x}^{n} + {y}^{n} = {z}^{n} \]\n\nfor some integer \( n > 2 \) . | In a book he was reading around 1630, Fermat claimed to have a proof for this proposition, but not enough space in the margin to write it down. Over the years, the Theorem was proved to hold for all \( n \) up to 4,000,000, but we’ve seen that this shouldn’t necessarily inspire confidence that it holds for all \( n \) ... | No |
Theorem 1.5.1. If \( 0 \leq x \leq 2 \), then \( - {x}^{3} + {4x} + 1 > 0 \) . | Proof. Assume \( 0 \leq x \leq 2 \) . Then \( x,2 - x \), and \( 2 + x \) are all nonnegative. Therefore, the product of these terms is also nonnegative. Adding 1 to this product gives a positive number, so:\n\n\[ x\left( {2 - x}\right) \left( {2 + x}\right) + 1 > 0 \]\n\nMultiplying out on the left side proves that\n\... | Yes |
Theorem 1.5.2. If \( r \) is irrational, then \( \sqrt{r} \) is also irrational. | Proof. We prove the contrapositive: if \( \sqrt{r} \) is rational, then \( r \) is rational.\n\nAssume that \( \sqrt{r} \) is rational. Then there exist integers \( m \) and \( n \) such that:\n\n\[ \sqrt{r} = \frac{m}{n} \]\n\nSquaring both sides gives:\n\n\[ r = \frac{{m}^{2}}{{n}^{2}} \]\n\nSince \( {m}^{2} \) and \... | Yes |
Theorem 1.6.1. The standard deviation of a sequence of values \( {x}_{1},\ldots ,{x}_{n} \) is zero iff all the values are equal to the mean. | Proof. We construct a chain of \ | No |
Theorem 1.8.1. \( \sqrt{2} \) is irrational. | Proof. We use proof by contradiction. Suppose the claim is false, and \( \sqrt{2} \) is rational. Then we can write \( \sqrt{2} \) as a fraction \( n/d \) in lowest terms.\n\nSquaring both sides gives \( 2 = {n}^{2}/{d}^{2} \) and so \( 2{d}^{2} = {n}^{2} \) . This implies that \( n \) is a multiple of 2 (see Problems ... | Yes |
It's a fact that the Arithmetic Mean is at least as large as the Geometric Mean, namely, \n\n\\[ \n\\frac{a + b}{2} \\geq \\sqrt{ab} \n\\] \n\nfor all nonnegative real numbers \\( a \\) and \\( b \\) . But there’s something objectionable about the following proof of this fact. What's the objection, and how would you fi... | Bogus proof. \n\n\\[ \n\\frac{a + b}{2}\\overset{?}{ \\geq }\\sqrt{ab} \n\\] \n\nso \n\n\\[ \na + b\\overset{?}{ \\geq }2\\sqrt{ab} \n\\] \n\nso \n\n\\[ \n{a}^{2} + {2ab} + {b}^{2}\\overset{?}{ \\geq }{4ab} \n\\] \n\nso \n\n\\[ \n{a}^{2} - {2ab} + {b}^{2}\\overset{?}{ \\geq }0 \n\\] \n\n\\[ \n{\\left( a - b\\right) }^{... | No |
If we raise an irrational number to an irrational power, can the result be rational? | Show that it can by considering \( {\sqrt{2}}^{\sqrt{2}} \) and arguing by cases. | No |
Prove that for any \( n > 0 \), if \( {a}^{n} \) is even, then \( a \) is even. | Hint: Contradiction. | No |
For \( n = {40} \), the value of polynomial \( p\left( n\right) : \mathrel{\text{:=}} {n}^{2} + n + {41} \) is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers. | In particular, let \( q\left( n\right) \) be a polynomial with integer coefficients, and let \( c : \mathrel{\text{:=}} q\left( 0\right) \) be the constant term of \( q \). (a) Verify that \( q\left( {cm}\right) \) is a multiple of \( c \) for all \( m \in \mathbb{Z} \). (b) Show that if \( q \) is nonconstant and \( c... | No |
Theorem 2.2.1.\n\n\[ 1 + 2 + 3 + \cdots + n = n\left( {n + 1}\right) /2 \]\n\n(2.1)\n\nfor all nonnegative integers, \( n \) . | Proof. By contradiction. Assume that Theorem 2.2.1 is false. Then, some nonnegative integers serve as counterexamples to it. Let's collect them in a set:\n\n\[ C : \mathrel{\text{:=}} \left\{ {n \in \mathbb{N} \mid 1 + 2 + 3 + \cdots + n \neq \frac{n\left( {n + 1}\right) }{2}}\right\} .\n\nAssuming there are counterexa... | Yes |
Theorem 2.3.1. Every positive integer greater than one can be factored as a product of primes. | Proof. The proof is by well ordering.\n\nLet \( C \) be the set of all integers greater than one that cannot be factored as a product of primes. We assume \( C \) is not empty and derive a contradiction.\n\nIf \( C \) is not empty, there is a least element, \( n \in C \), by well ordering. The \( n \) can’t be prime, b... | Yes |
Theorem 2.4.1. For any nonnegative integer, \( n \), the set of integers greater than or equal to \( - n \) is well ordered. | Proof. Let \( S \) be any nonempty set of integers \( \geq - n \) . Now add \( n \) to each of the elements in \( S \) ; let’s call this new set \( S + n \) . Now \( S + n \) is a nonempty set of nonnegative integers, and so by the Well Ordering Principle, it has a minimum element, \( m \) . But then it’s easy to see t... | Yes |
Corollary 2.4.3. Any set of integers with a lower bound is well ordered. | Proof. A set of integers with a lower bound \( b \in \mathbb{R} \) will also have the integer \( n = \) \( \lfloor b\rfloor \) as a lower bound, where \( \lfloor b\rfloor \), called the floor of \( b \), is gotten by rounding down \( b \) to the nearest integer. So Theorem 2.4.1 implies the set is well ordered. | Yes |
Corollary 2.4.4. Any nonempty set of integers with an upper bound has a maximum element. | Proof. Suppose a set, \( S \), of integers has an upper bound \( b \in \mathbb{R} \) . Now multiply each element of \( S \) by -1; let’s call this new set of elements \( - S \) . Now, of course, \( - b \) is a lower bound of \( - S \) . So \( - S \) has a minimum element \( - m \) by Corollary 2.4.3. But then it’s easy... | Yes |
Lemma 2.4.5. \( \mathbb{N} + \mathbb{F} \) is well ordered. | Proof. Given any nonempty subset, \( S \), of \( \mathbb{N} + \mathbb{F} \), look at all the nonnegative integers, \( n \), such that \( n + f \) is in \( S \) for some \( f \in \mathbb{F} \). This is a nonempty set nonnegative integers, so by the WOP, there is a minimum one; call it \( {n}_{s} \). By definition of \( ... | No |
For practice using the Well Ordering Principle, fill in the template of an easy to prove fact: every amount of postage that can be assembled using only 10 cent and 15 cent stamps is divisible by 5. | Let \( C \) be the set of counterexamples to (2.2), namely\n\n\[ C : \mathrel{\text{:=}} \{ n \mid \ldots \]\n\nAssume for the purpose of obtaining a contradiction that \( C \) is nonempty. Then by the WOP, there is a smallest number, \( m \in C \). This \( m \) must be positive because ....\n\nBut if \( S\left( m\righ... | Yes |
You are given a series of envelopes, respectively containing \( 1,2,4,\ldots ,{2}^{m} \) dollars. Define\n\nProperty \( m \) : For any nonnegative integer less than \( {2}^{m + 1} \), there is a selection of envelopes whose contents add up to exactly that number of dollars.\n\nUse the Well Ordering Principle (WOP) to p... | Hint: Consider two cases: first, when the target number of dollars is less than \( {2}^{m} \) and second, when the target is at least \( {2}^{m} \) . | No |
Theorem 3.4.3. Every propositional formula is equivalent to both a disjunctive normal form and a conjunctive normal form. | ## 3.4.2 Proving Equivalences\n\nA check of equivalence or validity by truth table runs out of steam pretty quickly: a proposition with \( n \) variables has a truth table with \( {2}^{n} \) lines, so the effort required to check a proposition grows exponentially with the number of variables. For a proposition with jus... | No |
Theorem 3.4.4. Any propositional formula can be transformed into disjunctive normal form or a conjunctive normal form using the equivalences listed above. | What has this got to do with equivalence? That's easy: to prove that two formulas are equivalent, convert them both to disjunctive normal form over the set of variables that appear in the terms. Then use commutativity to sort the variables and AND-terms so they all appear in some standard order. We claim the formulas a... | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.