Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
\[ \text{(a)}{\mathcal{L}}^{-1}\left\{ \frac{s}{{\left( {s}^{2} + 1\right) }^{2}}\right\} \] | Solution (a) We cannot evaluate this inverse Laplace transform in any direct fashion. However we do know the standard forms\n\n\[ {\mathcal{L}}^{-1}\left\{ \frac{s}{\left( {s}^{2} + 1\right) }\right\} = \cos t\text{ and }{\mathcal{L}}^{-1}\left\{ \frac{1}{\left( {s}^{2} + 1\right) }\right\} = \sin t. \]\n\nHence\n\[ \m... | Yes |
Example 3.4 Use a suitable double integral to evaluate the improper integral\n\n\[ \n{\int }_{0}^{\infty }{e}^{-{t}^{2}}{dt} \n\] | Solution Consider the double integral\n\n\[ \n{\iint }_{S}{e}^{-\left( {{x}^{2} + {y}^{2}}\right) }{dS} \n\]\n\nwhere \( S \) is the quarter disc \( x \geq 0, y \geq 0,{x}^{2} + {y}^{2} \leq {a}^{2} \) . Converting to polar co-ordinates \( \left( {R,\theta }\right) \) this integral becomes\n\n\[ \n{\int }_{0}^{a}{\int ... | Yes |
\[ {\mathcal{L}}^{-1}\left\{ \frac{1}{\sqrt{s}\left( {s - 1}\right) }\right\} \] | The only sensible way to proceed using our present knowledge is to use\n\n\[ {\mathcal{L}}^{-1}\left\{ \frac{1}{\sqrt{s}}\right\} = \frac{1}{\sqrt{\pi t}} \]\n\nand (by the shifting property)\n\n\[ {\mathcal{L}}^{-1}\left\{ \frac{1}{s - 1}\right\} = {e}^{t} \]\n\nWhence, using the convolution theorem\n\n\[ {\mathcal{L}... | Yes |
Example 3.7 Use the convolution theorem to find\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \frac{{e}^{-k\sqrt{s}}}{s}\right\} \n\] | Solution We note the result just derived, namely\n\n\[ \n{\mathcal{L}}^{-1}\left\{ {e}^{-k\sqrt{s}}\right\} = \frac{k}{2\sqrt{\pi {t}^{3}}}{e}^{-{k}^{2}/{4t}}. \n\]\n\ntogether with the standard result\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \frac{1}{s}\right\} = 1 \n\]\n\nto deduce that\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \fr... | Yes |
Solve the first order differential equation\n\n\\[ \n\\frac{dx}{dt} + {3x} = 0\\text{ where }x\\left( 0\\right) = 1.\n\\] | Solution Note that we have abandoned \\( f\\left( t\\right) \\) for the more usual \\( x\\left( t\\right) \\), but this should be regarded as a trivial change of dummy variable. This rather simple differential equation can in fact be solved by a variety of methods. Of course we use Laplace transforms, but it is useful ... | Yes |
Solve the first order differential equation\n\n\\[ \n\\frac{dx}{dt} + {3x} = 0\\text{ where }x\\left( 1\\right) = 1.\n\\] | Solution Proceeding as before, we now cannot insert the value of \\( x\\left( 0\\right) \\) so we arrive\n\nat the solution\n\\[\n x\\left( t\\right) = {\\mathcal{L}}^{-1}\\left\\{ \\frac{x\\left( 0\\right) }{s + 3}\\right\\} = x\\left( 0\\right) {e}^{-{3t}}.\n\\]\n\nWe now use the boundary condition we do have to give... | Yes |
Example 3.10 Solve the differential equation\n\n\[ \frac{dx}{dt} + {3x} = \cos {3t}\text{ given }x\left( 0\right) = 0. \] | Solution Taking the Laplace transform (we have already done this in Example 3.8 for the left hand side) we obtain\n\n\[ s\bar{x}\left( s\right) - x\left( 0\right) + 3\bar{x}\left( s\right) = \frac{s}{{s}^{2} + 9} \]\n\nusing standard forms. With the zero initial condition solving this for \( \bar{x}\left( s\right) \) y... | Yes |
Find the general solution to the differential equation\n\n\[ \frac{dx}{dt} + {3x} = f\left( t\right) \text{ where }x\left( 0\right) = 0, \] \n\nand \( f\left( t\right) \) is of exponential order (which is sufficient for the method of solution used to be valid). | Solution It is compulsory to use convolution here as the right hand side is an arbitrary function. The Laplace transform of the equation leads directly to\n\n\[ x\left( t\right) = {\mathcal{L}}^{-1}\left\{ \frac{\bar{f}\left( s\right) }{s + 3}\right\} \]\n\n\[ = {\int }_{0}^{t}{e}^{-3\left( {t - \tau }\right) }f\left( ... | Yes |
Example 3.12 Use Laplace transforms to solve the equation\n\n\\[ \n\\frac{{d}^{2}x}{d{t}^{2}} + x = 0\\text{ with }x\\left( 0\\right) = 1,{x}^{\\prime }\\left( 0\\right) = 0.\n\\] | Solution Taking the Laplace transform of this equation using the usual notation gives\n\n\\[ \n{s}^{2}\\bar{x}\\left( s\\right) - {sx}\\left( 0\\right) - {x}^{\\prime }\\left( 0\\right) + \\bar{x}\\left( s\\right) = 0.\n\\]\n\nWith \\( x\\left( 0\\right) = 1 \\) and \\( {x}^{\\prime }\\left( 0\\right) = 0 \\) we obtain... | Yes |
Find the solution to the differential equation\n\n\\[ \n\\frac{{d}^{2}x}{d{t}^{2}} + x = t\\text{ with }x\\left( 0\\right) = 1,{x}^{\\prime }\\left( 0\\right) = 0.\n\\] | Solution Apart from the trivial change of variable, we follow the last example and take Laplace transforms to obtain\n\n\\[ \n{s}^{2}\\bar{x}\\left( s\\right) - {sx}\\left( 0\\right) - {x}^{\\prime }\\left( 0\\right) + \\bar{x} = \\mathcal{L}\\{ t\\} = \\frac{1}{{s}^{2}}.\n\\]\n\nWith start conditions \\( x\\left( 0\\r... | Yes |
Use Laplace transform techniques to find the solution to the second order differential equation\n\n\[ \frac{{d}^{2}x}{d{t}^{2}} + 5\frac{dx}{dt} + {6x} = 2{e}^{-t}t \geq 0 \] \n\nsubject to the conditions \( x = 1 \) and \( {x}^{\prime } = 0 \) at \( t = 0 \) . | Solution Taking the Laplace transform of this equation we obtain using the usual overbar notation,\n\n\[ {s}^{2}\bar{x}\left( s\right) - {sx}\left( 0\right) - {x}^{\prime }\left( 0\right) + 5\left( {s\bar{x}\left( s\right) - x\left( 0\right) }\right) + 6\bar{x}\left( s\right) = \mathcal{L}\left\{ {e}^{-t}\right\} = \fr... | No |
Use Laplace transforms to solve the following ordinary differential equation\n\n\\[ \n\\frac{{d}^{2}x}{d{t}^{2}} + 6\\frac{dx}{dt} + {9x} = \\sin t\\;\\left( {t \\geq 0}\\right) ,\n\\]\n\nsubject to \\( x\\left( 0\\right) = 0 \\) and \\( {x}^{\\prime }\\left( 0\\right) = 0 \\) . | Solution With no right hand side, and with zero initial conditions, \\( x\\left( t\\right) = 0 \\) would result. However with the sinusoidal forcing, the solution turns out to be quite interesting. The formal way of tackling the problem is the same as for any second order differential equation with constant coefficient... | Yes |
Find the differential equation obeyed by the charge for the simple circuit shown in Fig. 3.4, and solve it by the use of Laplace transforms given \( j = \) \( 0, q = 0 \) at \( t = 0 \) . | Solution The current is \( j \) and the charge is \( q \), so the voltage drop across the three devices are\n\n\[ 2\frac{dj}{dt} = 2\frac{{d}^{2}q}{d{t}^{2}},\;{16j} = {16}\frac{dq}{dt},\text{ and }\frac{q}{0.002} = {50q}. \]\n\nThis must be equal to 300 (the voltage output of the battery), hence\n\n\[ 2\frac{{d}^{2}q}... | Yes |
Solve the same problem as in the previous example except that the battery is replaced by the oscillatory voltage source \( {100}\sin \left( {3t}\right) \) . | The differential equation is derived as before, except for the different right hand side. The equation is\n\n\[ 2\frac{{d}^{2}q}{d{t}^{2}} + {16}\frac{dq}{dt} + {50q} = {100}\sin \left( {3t}\right) .\n\]\n\nTaking the Laplace transform of this using the zero initial conditions \( q\left( 0\right) = \) \( 0,{q}^{\prime ... | Yes |
Example 3.18 Solve the simultaneous differential equations\n\n\\[ \n\\frac{dx}{dt} = {2x} - {3y},\\frac{dy}{dt} = y - {2x} \n\\]\n\nwhere \\( x\\left( 0\\right) = 8 \\) and \\( y\\left( 0\\right) = 3 \\) . | Solution Taking Laplace transforms and inserting the boundary conditions straight away gives:-\n\n\\[ \ns\\bar{x}\\left( s\\right) - 8 = 2\\bar{x}\\left( s\\right) - 3\\bar{y}\\left( s\\right) \n\\]\n\n\\[ \ns\\bar{y}\\left( s\\right) - 3 = \\bar{y}\\left( s\\right) - 2\\bar{x}\\left( s\\right) . \n\\]\n\nWhence, rearr... | Yes |
Example 3.19 Solve the simultaneous differential equations\n\n\[ \n\\frac{{d}^{2}x}{d{t}^{2}} + \\frac{dy}{dt} + {3x} = {15}{e}^{-t} \n\]\n\n\[ \n\\frac{{d}^{2}y}{d{t}^{2}} - 4\\frac{dx}{dt} + {3y} = {15}\\sin \\left( {2t}\\right) \n\]\n\nwhere \( x = {35},{x}^{\\prime } = - {48}, y = {27} \) and \( {y}^{\\prime } = - ... | Solution Taking Laplace transforms of both equations as before, retaining the standard notation for the transformed variable gives\n\n\[ \n{s}^{2}\\bar{x} - {35s} + {48} + s\\bar{y} - {27} + 3\\bar{x} = \\frac{15}{s + 1} \n\]\n\n\[ \n\\text{and}{s}^{2}\\bar{y} - {27s} + {55} - 4\\left( {s\\bar{x} - {35}}\\right) + 3\\b... | Yes |
Example 3.20 Figure 3.10 displays a mechanical system. Find the equations of motion, and solve them given that the system is initially at rest with \( x = 1 \) and \( y = 2 \) . | Solution Applying Newton's Second Law of Motion successively to each mass using Hooke's Law (there are no dampers) gives:-\n\n\[ \n{m}_{1}\ddot{x} = {k}_{2}\left( {y - x}\right) - {k}_{1}x \n\]\n\n\[ \n{m}_{2}\ddot{y} = - {k}_{3}x - {k}_{2}\left( {y - x}\right) . \n\]\n\nWith the values for the constants \( {m}_{1},{m}... | Yes |
Example 3.22 Solve the integral equation\n\n\[ \phi \left( x\right) - \lambda {\int }_{0}^{x}{e}^{x - y}\phi \left( y\right) {dy} = f\left( x\right) \]\n\nwhere \( f\left( x\right) \) is a general function of \( x \) . | Solution The integral is in the form of a convolution; it is in fact \( \lambda {e}^{x} * \phi \left( x\right) \) where * denotes the convolution operation. The integral can thus be written\n\n\[ \phi \left( x\right) - \lambda {e}^{x} * \phi \left( x\right) = f\left( x\right) . \]\n\nTaking the Laplace transform of thi... | Yes |
Theorem 4.2 (Riemann-Lebesgue) Let \( \\left\\{ {{\\mathbf{e}}_{1},{\\mathbf{e}}_{2},\\ldots }\\right\\} \) be an orthonormal basis of infinite dimension for the inner product space \( V \) . Then, for any \( \\mathbf{a} \\in V \n\n\[ \n\\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}\\left\\langle {\\mathbf{a},{\\... | This theorem in fact follows directly from Bessel’s inequality as the \( n \) th term of the series on the right of Bessel’s inequality must tend to zero as \( n \) tends to \( \\infty \) . | Yes |
Theorem 4.3 If \( f \) is a member of the space of piecewise continuous functions which are \( {2\pi } \) periodic on the closed interval \( \left\lbrack {-\pi ,\pi }\right\rbrack \) and which has both left and right derivatives at each \( x \in \left\lbrack {-\pi ,\pi }\right\rbrack \), then for each \( x \in \left\lb... | The proof of this is beyond the scope of this book, but some comments are usefully made. If \( x \) is a point at which the function \( f \) is continuous, then\n\n\[ \frac{f\left( {x}_{ - }\right) + f\left( {x}_{ + }\right) }{2} = f\left( x\right) \]\n\nand the theorem is certainly eminently plausible as any right han... | Yes |
Determine the Fourier series for the function\n\n\[ f\left( x\right) = {2x} + 1\; - \pi < x < \pi \]\n\n\[ f\left( x\right) = f\left( {x + {2\pi }}\right) \;x \in \mathbf{R} \]\n\nwhere Theorem 4.3 applies at the end points. | Solution As \( f\left( x\right) \) is obviously piecewise continuous in \( \left\lbrack {-\pi ,\pi }\right\rbrack \), in fact the only discontinuities occurring at the end points, we simply use the formulae\n\n\[ {a}_{n} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left( x\right) \cos \left( {nx}\right) {dx} \]\n\nand\n\n\[... | Yes |
Find the Fourier series for the function\n\n\\[ \nf\\left( x\\right) = {e}^{x},\\; - \\pi < x < \\pi \n\\] \n\n\\[ \nf\\left( {x + {2\\pi }}\\right) = f\\left( x\\right) ,\\;x \\in \\mathbf{R} \n\\] \n\nwhere Theorem 4.3 applies at the end points. | Solution This problem is best tackled by using the power of complex numbers. We start with the two standard formulae:\n\n\\[ \n{a}_{n} = \\frac{1}{\\pi }{\\int }_{-\\pi }^{\\pi }{e}^{x}\\cos \\left( {nx}\\right) {dx} \n\\] \n\nand \n\n\\[ \n{b}_{n} = \\frac{1}{\\pi }{\\int }_{-\\pi }^{\\pi }{e}^{x}\\sin \\left( {nx}\\r... | Yes |
Example 4.3 Determine the Fourier Series of the function\n\n\\[ f\\left( x\\right) = \\left| x\\right| ,\\; - 3 \\leq x \\leq 3, \\]\n\n\\[ f\\left( x\\right) = f\\left( {x + 6}\\right) . \\] | Solution The function \\( f\\left( x\\right) = \\left| x\\right| \\) is continuous, therefore we can use Eqs. 4.2 and 4.3 to generate the Fourier coefficients. First of all\n\n\\[ {a}_{0} = \\frac{1}{3}{\\int }_{-3}^{3}\\left| x\\right| {dx} = 3 \\]\n\nSecondly,\n\n\\[ {a}_{n} = \\frac{1}{3}{\\int }_{-3}^{3}\\left| x\\... | Yes |
Determine the Fourier series for the function\n\n\[ f\left( x\right) = {x}^{2} \]\n\n\[ f\left( x\right) = f\left( {x + {2\pi }}\right) ,\; - \pi \leq x \leq \pi . \] | Solution Since \( {x}^{2} = {\left( -x\right) }^{2}, f\left( x\right) \) is an even function. Thus the Fourier series consists solely of even functions which means \( {b}_{n} = 0 \) for all \( n \) . We therefore compute the \( {a}_{n} \) ’s as follows\n\n\[ {a}_{0} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{2}{dx} \]\... | Yes |
Find the Fourier series for the function\n\n\\[ f\\left( t\\right) = {t}^{2} + t,\\; - \\pi \\leq t \\leq \\pi ,\\]\n\n\\[ f\\left( t\\right) = f\\left( {t + {2\\pi }}\\right) .\\] | Solution We could go ahead and find the Fourier series in the usual way. However it is far easier to use the complex form but in a tailor-made way as follows. Given\n\n\\[ {a}_{n} = \\frac{1}{\\pi }{\\int }_{-\\pi }^{\\pi }f\\left( t\\right) \\cos \\left( {nt}\\right) {dt} \\]\n\nand\n\n\\[ {b}_{n} = \\frac{1}{\\pi }{\... | Yes |
Theorem 4.5 If \( f \) is continuous on \( \left\lbrack {-\pi ,\pi }\right\rbrack \) and piecewise differentiable in \( \left( {-\pi ,\pi }\right) \) which means that the derivative \( {f}^{\prime } \) is piecewise continuous on \( \left\lbrack {-\pi ,\pi }\right\rbrack \), and if \( f\left( x\right) \) has the Fourier... | The proof of this theorem follows standard analysis and is not given here. | No |
Use the Fourier series\n\n\\[ \n{x}^{2} = \\frac{{\\pi }^{2}}{3} + 4\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{{\\left( -1\\right) }^{n}}{{n}^{2}}\\cos \\left( {nx}\\right) \n\\]\n\nto deduce the value of the series\n\n\\[ \n\\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }\\frac{{\\left( -1\\right) }^{k - 1}}{{\\... | Solution Utilising the result just derived on the integration of Fourier series, we put \\( \\xi = 0 \\) and \\( t = \\pi /2 \\) so we can write\n\n\\[ \n{\\int }_{0}^{\\frac{\\pi }{2}}{x}^{2}{dx} = \\frac{{\\pi }^{2}}{3}\\left( {\\frac{\\pi }{2} - 0}\\right) + 4\\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{1}{n}... | Yes |
Theorem 4.7 If \( f\left( t\right) \) and \( g\left( t\right) \) are continuous in \( \left( {-\pi ,\pi }\right) \) and provided\n\n\[ \n{\int }_{-\pi }^{\pi }{\left| f\left( t\right) \right| }^{2}{dt} < \infty \text{ and }{\int }_{-\pi }^{\pi }{\left| g\left( t\right) \right| }^{2}{dt} < \infty ,\n\]\n\nif \( {a}_{n},... | Proof Since\n\n\[ \nf\left( t\right) \sim \frac{{a}_{0}}{2} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {{a}_{n}\cos \left( {nt}\right) + {b}_{n}\sin \left( {nt}\right) }\right)\n\]\n\nand\n\n\[ \ng\left( t\right) \sim \frac{{\alpha }_{0}}{2} + \mathop{\sum }\limits_{{n = 1}}^{\infty }\left( {{\alpha }_{n}\cos \l... | Yes |
Theorem 4.8 (Parseval) If \( f\left( t\right) \) is continuous in the range \( \left( {-\pi ,\pi }\right) \), is square integrable (i.e. \( {\int }_{-\pi }^{\pi }{\left\lbrack f\left( t\right) \right\rbrack }^{2}{dt} < \infty \) ) and has Fourier coefficients \( {a}_{n},{b}_{n} \) then | Proof This is immediate from Theorem 4.7 by putting\n\n\[ f\left( t\right) = g\left( t\right) \] | No |
Given the Fourier series\n\n\[ \n{t}^{2} = \frac{{\pi }^{2}}{3} + 4\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{{\left( -1\right) }^{n}}{{n}^{2}}\cos \left( {nt}\right) \n\]\n\ndeduce the value of\n\n\[ \n\mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{4}} \n\] | Solution Applying Parseval's theorem to this series, the left hand side becomes\n\n\[ \n{\int }_{-\pi }^{\pi }{\left( {t}^{2}\right) }^{2}{dt} = \frac{2}{5}{\pi }^{5} \n\]\n\nThe right hand side becomes\n\n\[ \n\pi {\left( \frac{{\pi }^{2}}{3}\right) }^{2} + \pi \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{16}{{n}^{4... | Yes |
Use the equation\n\n\[ \frac{d}{dt}\left\lbrack {t{e}^{-t}\frac{d{L}_{n}}{dt}}\right\rbrack + n{e}^{-t}{L}_{n} = 0 \]\n\nto show that \( {L}_{m}\left( t\right) ,{L}_{n}\left( t\right) ;m, n = 0,1,2,3\ldots \) obey the orthogonality relation\n\n\[ {\int }_{0}^{\infty }{e}^{-t}{L}_{m}\left( t\right) {L}_{n}\left( t\right... | Solution Start with the two equations\n\n\[ \frac{d}{dt}\left\lbrack {t{e}^{-t}\frac{d{L}_{n}}{dt}}\right\rbrack + n{e}^{-t}{L}_{n} = 0 \]\n\nand\n\[ \frac{d}{dt}\left\lbrack {t{e}^{-t}\frac{d{L}_{m}}{dt}}\right\rbrack + m{e}^{-t}{L}_{m} = 0 \]\n\ntake \( {L}_{m} \) times the first minus \( {L}_{n} \) times the second ... | Yes |
Example 5.1 Find all first order partial derivatives of the functions (a) \( {x}^{2}{yz} \), and (b) \( x\sin \left( {x + {yz}}\right) \) . | Solution (a) The partial derivatives are as follows:\n\n\[ \frac{\partial }{\partial x}\left( {{x}^{2}{yz}}\right) = {2xyz} \]\n\n\[ \frac{\partial }{\partial y}\left( {{x}^{2}{yz}}\right) = {x}^{2}z \]\n\nand\n\n\[ \frac{\partial }{\partial z}\left( {{x}^{2}{yz}}\right) = {x}^{2}y \]\n\n(b) The partial derivatives are... | Yes |
Example 5.2 Determine the solution to the boundary value problem:\n\n\[ \frac{\partial \phi }{\partial t} = \kappa \frac{{\partial }^{2}\phi }{\partial {x}^{2}},\;x \in \left\lbrack {0,2}\right\rbrack \]\n\n\[ \phi \left( {x,0}\right) = x + \left( {2 - {2x}}\right) H\left( {x - 1}\right) \text{ at time }t = 0 \]\n\n\[ ... | Solution The form of the function \( \phi \left( {x,0}\right) \) is displayed in Fig. 5.1. This function is expressed as a Fourier sine series by the methods outlined in Chap. 4. This is in order to make automatic the satisfying of the boundary conditions. The Fourier sine series is not derived in detail as this belong... | No |
Solve the heat conduction equation\n\n\\[ \n\\frac{\\partial \\phi }{\\partial t} = \\frac{{\\partial }^{2}\\phi }{\\\partial {x}^{2}} \n\\]\n\nin the region \\( t > 0, x > 0 \\) with boundary conditions \\( \\phi \\left( {x,0}\\right) = 0 x > 0 \\) (initial condition), \\( \\phi \\left( {0, t}\\right) = 1, t > 0 \\) (... | Solution Taking the Laplace transform (in \\( t \\) of course) gives\n\n\\[ \ns\\bar{\\phi } - \\phi \\left( 0\\right) = \\frac{{d}^{2}\\bar{\\phi }}{d{x}^{2}} \n\\]\n\n\\[ \ns\\bar{\\phi } = \\frac{{d}^{2}\\bar{\\phi }}{d{x}^{2}} \n\\]\nsince \\( \\phi \\left( {x,0}\\right) = 0 \\) . This is an ODE with constant coeff... | Yes |
Example 5.5 Starting with the linear Rossby wave equation\n\n\[ \n{\nabla }^{2}{\psi }_{t} + \beta {\psi }_{x} = 0 \n\]\n\ndetermine the long term response of an initially quiescent ocean if the variations in the \( y \) direction are much smaller than those in the \( x \) direction. | Solution The linear Rossby wave equation takes the form\n\n\[ \n{\nabla }^{2}{\psi }_{t} + \beta {\psi }_{x} = 0 \n\]\n\nTake Laplace transforms (in \( t \) ) using:\n\n\[ \n\mathcal{L}\{ \psi \} = \bar{\psi }\;\mathcal{L}\left\{ {\psi }_{t}\right\} = s\bar{\psi } - \psi \left( {x,0}\right) \n\]\n\ngiving:\n\n\[ \ns{\n... | Yes |
Example 5.6 Find an approximation to the solution of the partial differential equation\n\n\[ \frac{\partial \phi }{\partial t} = {c}^{2}\frac{{\partial }^{2}\phi }{\partial {x}^{2}} \]\n\nfor small times where \( \phi \left( {x,0}\right) = \cos \left( x\right) \), by using an asymptotic series. | Solution It is possible to solve this BVP exactly, but let us take Laplace transforms\n\nto obtain\n\[ s\bar{\phi } - \cos \left( x\right) = {c}^{2}\frac{{d}^{2}\bar{\phi }}{d{x}^{2}} \]\n\nthen try an asymptotic series of the form\n\n\[ \bar{\phi }\left( {x, s}\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{{... | Yes |
By considering the Fourier cosine and sine transforms of the function \( f\left( t\right) = {e}^{-{at}},\; \) a a constant, evaluate the two integrals\n\n\[{\int }_{0}^{\infty }\frac{\cos \left( {kx}\right) }{{a}^{2} + {x}^{2}}{dx}\text{ and }{\int }_{0}^{\infty }\frac{x\sin \left( {kx}\right) }{{a}^{2} + {x}^{2}}{dx}.... | Solution First of all note that the cosine and sine transforms can be conveniently combined to give\n\n\[{F}_{c}\left( \omega \right) + i{F}_{s}\left( \omega \right) = {\int }_{0}^{\infty }{e}^{\left( {-a + {i\omega }}\right) t}{dt}\]\n\n\[= {\left\lbrack \frac{1}{-a + {i\omega }}{e}^{\left( {-a + {i\omega }}\right) t}... | Yes |
Theorem 6.3 Let \( f\left( x\right) \in G\{ \mathbb{R}\} \) and \( a, b \in \mathbb{R}, a \neq 0 \) and denote the Fourier transform of \( f \) by \( \mathbb{F}\left( f\right) \) so\n\n\[ \mathbb{F}\left( f\right) = {\int }_{-\infty }^{\infty }f\left( x\right) {e}^{-{i\omega x}}{dx}. \]\n\nLet \( g\left( x\right) = f\l... | Proof As is usual with this kind of proof, the technique is simply to evaluate the Fourier transform using its definition. Doing this, we obtain\n\n\[ \mathcal{F}\left( {g\left( \omega \right) }\right) = {\int }_{-\infty }^{\infty }f\left( {{ax} + b}\right) {e}^{-{i\omega x}}{dx}. \]\n\nNow simply substitute \( t = {ax... | Yes |
Find the solution to the two-dimensional Laplace equation\n\n\[ \n\\frac{{\\partial }^{2}\\phi }{\\partial {x}^{2}} + \\frac{{\\partial }^{2}\\phi }{\\partial {y}^{2}} = 0,\\;y > 0 \n\]\n\nwith\n\n\[ \n\\frac{\\partial \\phi }{\\partial x}\\text{ and }\\phi \\rightarrow 0\\text{ as }\\sqrt{{x}^{2} + {y}^{2}} \\rightarr... | Solution Let\n\n\[ \n\\bar{\\phi }\\left( {k, y}\\right) = {\\int }_{-\\infty }^{\\infty }\\phi \\left( {x, y}\\right) {e}^{-{ikx}}{dx} \n\]\n\nthen\n\n\[ \n{\\int }_{-\\infty }^{\\infty }\\frac{{\\partial }^{2}}{\\partial {y}^{2}}{e}^{-{ikx}}{dx} = \\frac{{\\partial }^{2}}{\\partial {y}^{2}}\\left( {{\\int }_{-\\infty... | No |
Theorem 6.4 (Parseval’s, for transforms) If \( f\left( t\right) \) has a Fourier transform \( F\left( \omega \right) \) and \[ {\int }_{-\infty }^{\infty }{\left| f\left( t\right) \right| }^{2}{dt} < \infty \] then \[ {\int }_{-\infty }^{\infty }{\left| f\left( t\right) \right| }^{2}{dt} = \frac{1}{2\pi }{\int }_{-\inf... | Proof The proof is straightforward:- \[ {\int }_{-\infty }^{\infty }f\left( t\right) {f}^{ * }\left( t\right) {dt} = {\int }_{-\infty }^{\infty }f\left( t\right) \frac{1}{2\pi }{\int }_{-\infty }^{\infty }F\left( \omega \right) {e}^{i\omega t}{d\omega dt} \] \[ = \frac{1}{2\pi }{\int }_{-\infty }^{\infty }F\left( \omeg... | Yes |
Determine the energy spectral densities for the following functions:\n\n(i)\n\n\\[ \nf\\left( t\\right) = \\left\\{ \\begin{array}{ll} A & \\left| t\\right| < T \\\\ 0 & \\text{ otherwise } \\end{array}\\right.\n\\]\n\nThis is the same function as in Example 6.2. | Solution The energy spectral density \\( {\\left| F\\left( \\omega \\right) \\right| }^{2} \\) is found from \\( f\\left( t\\right) \\) by first finding its Fourier transform. Both calculations are essentially routine.\n\n(i)\n\n\\[ \nF\\left( \\omega \\right) = {\\int }_{-\\infty }^{\\infty }f\\left( t\\right) {e}^{i\... | Yes |
Using the Father, Mother, Daughter1 and -Daughter1 wavelets defined above, is it possible to decompose the signals\n\n\\[ \n{g}_{1}\\left( t\\right) = \\left\\{ \\begin{matrix} 4, & 0 \\leq t < \\frac{1}{4} \\\\ - 6, & \\frac{1}{4} \\leq t < \\frac{1}{2} \\\\ 3, & \\frac{1}{2} \\leq t \\leq \\frac{3}{4} \\\\ 2, & \\fra... | Solution If we simply proceed to try and solve the vector equation:\n\n\\[ \n{a\\phi } + {b\\psi } + c{\\psi }_{1,1} - d{\\psi }_{1,1} = {\\left( 4, - 6,3,2\\right) }^{T}\n\\]\n\nwe get the four simultaneous equations:\n\n\\[ \na + b = 4\n\\]\n\n\\[ \na + b = - 6\n\\]\n\n\\[ \na - b + c - d = 3\n\\]\n\n\\[ \na - b - c ... | Yes |
Find the Fourier transform of the mother wavelet function \( \psi \left( t\right) \) . | Solution The definition of the mother wavelet \( \psi \left( t\right) \) was given earlier, and is:\n\n\[ \psi \left( t\right) = \begin{cases} 1, & 0 \leq t < \frac{1}{2} \\ - 1, & \frac{1}{2} \leq t < 1 \\ 0, & \text{ otherwise. } \end{cases} \]\n\nThus\n\n\[ {F}_{\psi }\left( \omega \right) = {\int }_{-\infty }^{\inf... | Yes |
Calculate \( {B}_{2}\left( t\right) = {B}_{1}\left( t\right) * {B}_{1}\left( t\right) \) . Use the convolution theorem to find its Fourier transform and hence generalise the result. | Solution Using direct integration, if \( 0 \leq t \leq 1 \) we have\n\n\[ \n{B}_{2}\left( t\right) = {B}_{1} * {B}_{1} = {\int }_{0}^{t}{B}_{1}\left( \tau \right) {B}_{1}\left( {t - \tau }\right) {d\tau } = {\int }_{0}^{t}{d\tau } = t \n\] \n\nwhereas if \( 0 \leq t - \tau \leq 1 \) (which is \( t - 1 \leq \tau \leq t ... | Yes |
Calculate the centre and RMS value for the general window function \( {\chi }_{\lbrack - \tau ,\tau )} \) defined by\n\n\[ \n{\chi }_{\lbrack - \tau ,\tau )} = \left\{ \begin{array}{ll} 1, & - \tau \leq t < \tau \\ 0, & \text{ otherwise. } \end{array}\right.\n\]\n\nFind also the centre and RMS values of its Fourier tra... | Solution The norm of \( {\chi }_{\lbrack - \tau ,\tau )} \) is given by the square root of the integral\n\n\[ \n{\int }_{-\infty }^{\infty }{\left| {\chi }_{\lbrack - \tau ,\tau )}\right| }^{2}{dt} = {\int }_{-\tau }^{\tau }{dt} = {2\tau }\n\]\n\nSo \( \begin{Vmatrix}{\chi }_{\lbrack - \tau ,\tau )}\end{Vmatrix} = \sqr... | Yes |
Determine the Fourier transform of the function\n\n\[ f\\left( t\\right) = \\left\\{ \\begin{matrix} - 1, & - 1 < t < - \\frac{1}{2} \\\\ 1, & \\frac{1}{2} < t < 1 \\\\ 0, & \\text{ otherwise. } \\end{matrix}\\right. \] | Solution The Fourier transform is straightforward enough to find:\n\n\[ \\widehat{f}\\left( \\omega \\right) = \\int_{-\\infty}^{\\infty} f\\left( t\\right) e^{-i\\omega t} dt \]\n\n\[ = \\int_{-1}^{-\\frac{1}{2}} \\left( -1\\right) e^{-i\\omega t} dt + \\int_{\\frac{1}{2}}^{1} e^{-i\\omega t} dt \]\n\n\[ = 2\\int_{\\f... | Yes |
Example 7.8 Let \( f\left( t\right) = \sin \left( {\pi t}\right) \) and the window function \( b\left( t\right) \) be the simple symmetrical top hat function:\n\n\[ b\left( t\right) = \left\{ {\begin{array}{ll} 1, & - 1 \leq t \leq 1 \\ 0, & \text{ otherwise } \end{array}.}\right. \]\n\nDetermine the STFT \( {f}_{G} \) | Solution Using the given window function \( b\left( t\right) \) we see immediately that\n\n\[ {f}_{b}\left( t\right) = \left\{ {\begin{array}{ll} \sin \left( {\pi t}\right) , & - 1 \leq t \leq 1 \\ 0, & \text{ otherwise } \end{array}.}\right. \]\n\nwhence\n\n\[ {f}_{G} = {\int }_{-1}^{1}\sin \left( {\pi t}\right) {e}^{... | Yes |
Theorem 8.1 If \( f\left( z\right) \) is analytic in a domain \( D \) and on its (closed) boundary \( C \) then\n\n\[{\oint }_{C}f\left( z\right) {dz} = 0\]\n\nwhere the small circle within the integral sign denotes that the integral is around a closed loop. | Proof There are several proofs of Cauchy's theorem that place no reliance on the analyticity of \( f\left( z\right) \) on \( C \) and only use analyticity of \( f\left( z\right) \) inside \( C \) . However these proofs are rather involved and unenlightening except for those whose principal interest is pure mathematics.... | Yes |
Theorem 8.2 (Residue Theorem) If \( f\left( z\right) \) is analytic within and on \( C \) except at points \( {z}_{1},{z}_{2},\ldots ,{z}_{N} \) that lie inside \( C \) where \( f\left( z\right) \) has poles then\n\n\[{\oint }_{C}f\left( z\right) {dz} = {2\pi i}\mathop{\sum }\limits_{{k = 1}}^{N}\left( {\text{ sum of r... | Proof This result follows immediately on a straightforward generalisation of the case just considered to the case of \( f\left( z\right) \) possessing \( N \) poles inside \( C \) . No further elaboration is either necessary or given. | No |
Example 8.1 Use suitable contours \( C \) to evaluate the two real integrals\n\n(i) \( {\int }_{0}^{\infty }\frac{\cos \left( {\pi x}\right) }{{a}^{2} + {x}^{2}}{dx} \) | ## Solution\n\n(i) For the first part, we choose the contour \( C \) shown in Fig. 8.4, that is a semi-circular contour on the upper half plane. We consider the integral\n\n\[ \n{\oint }_{C}\frac{{e}^{i\pi x}}{{a}^{2} + {z}^{2}}{dz} \n\]\n\nNow,\n\n\[ \n{\oint }_{C}\frac{{e}^{i\pi z}}{{a}^{2} + {z}^{2}}{dz} = {\int }_{... | Yes |
Find the value of the real integral\n\n\[ \n{\int }_{0}^{\infty }\frac{{x}^{\alpha - 1}}{1 + x}{dx}\text{ where }0 < \alpha < {1\alpha }\text{ a real constant. }\n\] | Solution In order to find this integral, we evaluate the complex contour integral\n\n\[ \n{\int }_{C}\frac{{z}^{\alpha - 1}}{1 + z}{dz} \n\]\n\nwhere \( C \) is the keyhole contour shown in Fig. 8.7.\n\nSometimes there is trouble because there is a singularity where the cut is usually drawn. We shall meet this in the n... | Yes |
Use a semi-circular contour in the upper half plane to evaluate\n\n\[ \n{\int }_{C}\frac{\ln z}{{z}^{2} + {a}^{2}}{dz}\;\left( {a > 0}\right) \text{ real and positive } \n\]\n\nand deduce the values of two real integrals. | Solution Figure 8.8 shows the semi-circular contour. It is indented at the origin as \( z = 0 \) is an essential singularity of the integrand \( \frac{\ln z}{{z}^{2} + {a}^{2}} \) . Thus\n\n\[ \n{\int }_{C}\frac{\ln z}{{z}^{2} + {a}^{2}}{dz} = {2\pi i}\{ \text{ Residue at }z = {ia}\} \n\]\n\nprovided \( R \) is large e... | Yes |
Theorem 8.3 If the Laplace transform of \( F\left( t\right) \) exists, that is \( F\left( t\right) \) is of exponential order and\n\n\[ f\left( s\right) = {\int }_{0}^{\infty }{e}^{-{st}}F\left( t\right) {dt} \]\n\nthen\n\n\[ F\left( t\right) = \mathop{\lim }\limits_{{k \rightarrow \infty }}\left\{ {\frac{1}{2\pi i}{\i... | Proof The proof of this has already been outlined in Sect. 6.2 of Chap. 6. However, we have now done enough formal complex variable theory to give a more complete proof. The outline remains the same in that we define \( {F}_{k}\left( \omega \right) \) as in that chapter, namely\n\n\[ {F}_{k}\left( \omega \right) = {\in... | Yes |
Use the Bromwich contour to find the value of\n\n\[ \n{\mathcal{L}}^{-1}\left\{ \frac{1}{\left( {s + 1}\right) {\left( s - 2\right) }^{2}}\right\} \n\] | Solution It is quite easy to find this particular inverse Laplace transform using partial fractions as in Chap. 2; however it serves as an illustration of the use of the contour integral method. In the next example, there are no alternative direct methods.\n\nNow,\n\n\[ \n{\int }_{C}\frac{{e}^{st}}{\left( {s + 1}\right... | Yes |
Example 8.6 Find the asymptotic behaviour as \( y \rightarrow \infty \) for fixed \( x \) of the solution of the partial differential equation\n\n\[ \n\\frac{{\\partial }^{2}\\phi }{\\partial {y}^{2}} = \\frac{{\\partial }^{2}\\phi }{\\partial {x}^{2}} - \\phi ,\\;x > 0,\\;y > 0,\n\]\n\nsuch that\n\[ \n\\phi \\left( {x... | Solution Taking the Laplace transform of the given equation with respect to \( y \) using the by now familiar notation leads to the following ODE for \( \\bar{\\phi } \) :\n\n\[ \n\\left( {1 + {s}^{2}}\\right) \\bar{\\phi } = \\frac{{d}^{2}\\bar{\\phi }}{d{x}^{2}}\n\]\n\nwhere the boundary conditions at \( y = 0 \) hav... | Yes |
Example 1.3. Take \( S = \mathbb{Z} \), and let \( \sim \) be the relation defined by\n\n\[ a \sim b \Leftrightarrow a - b\\text{is even.} \]\n\nThen \( \mathbb{Z}/ \sim \) consists of two equivalence classes: | Indeed, every integer \( b \) is either even (and hence \( b - 0 \) is even, so \( b \sim 0 \), and \( b \in {\\left\\lbrack 0\\right\\rbrack }_{ \\sim } \) ) or odd (and hence \( b - 1 \) is even, so \( b \sim 1 \), and \( b \in {\\left\\lbrack 1\\right\\rbrack }_{ \\sim } \) ). | Yes |
Proposition 2.1. Assume \( A \neq \varnothing \), and let \( f : A \rightarrow B \) be a function. Then\n\n(1) \( f \) has a left-inverse if and only if it is injective. | Proof. Let's prove (1).\n\n\( \left( \Rightarrow \right) \) If \( f : A \rightarrow B \) has a left-inverse, then there exists a \( g : B \rightarrow A \) such that \( g \circ f = {\operatorname{id}}_{A} \) . Now assume that \( {a}^{\prime } \neq {a}^{\prime \prime } \) are arbitrary different elements in \( A \) ; the... | Yes |
Proposition 2.3. A function is injective if and only if it is a monomorphism. | Proof. \( \left( \Rightarrow \right) \) By Proposition 2.1, if a function \( f : A \rightarrow B \) is injective, then it has a left-inverse \( g : B \rightarrow A \) . Now assume that \( {\alpha }^{\prime },{\alpha }^{\prime \prime } \) are arbitrary functions from another set \( Z \) to \( A \) and that\n\n\[ f \circ... | Yes |
Let \( A, B \) be sets. Then there are natural projections \( {\pi }_{A},{\pi }_{B} \) | defined by\n\n\[ \n{\pi }_{A}\left( \left( {a, b}\right) \right) \mathrel{\text{:=}} a,\;{\pi }_{B}\left( \left( {a, b}\right) \right) \mathrel{\text{:=}} b \]\n\nfor all \( \left( {a, b}\right) \in A \times B \) . Both of these maps are (clearly) surjective. | Yes |
Similarly, there are natural injections from \( A \) and \( B \) to the disjoint union: | obtained by sending \( a \in A \) (resp., \( b \in B \) ) to the corresponding element in the isomorphic copy \( {A}^{\prime } \) of \( A \) (resp., \( {B}^{\prime } \) of \( B \) ) in \( A \coprod B \) . | Yes |
If \( \sim \) is an equivalence relation on a set \( A \), there is a (clearly surjective) canonical projection | \[ A \rightarrow A/ \sim \] obtained by sending every \( a \in A \) to its equivalence class \( {\left\lbrack a\right\rbrack }_{ \sim } \) | Yes |
Theorem 2.7. Let \( f : A \rightarrow B \) be any function, and define \( \sim \) as above. Then \( f \) decomposes as follows:\n\n\n\nwhere the first function is the canonical projection \( A \rightarrow A/ \sim \) (a... | Proof. Spelling out the first item discussed above, we have to verify that, for all \( {a}^{\prime },{a}^{\prime \prime } \) in \( A \), \n\n\[ {\left\lbrack {a}^{\prime }\right\rbrack }_{ \sim } = {\left\lbrack {a}^{\prime \prime }\right\rbrack }_{ \sim } \Rightarrow f\left( {a}^{\prime }\right) = f\left( {a}^{\prime ... | Yes |
It is hopefully crystal clear by now that sets (as objects), together with set-functions (as morphisms), form a category; if not, the reader must stop here and go no further until this assertion sheds any residual mystery 17. | - \( \operatorname{Obj}\left( \operatorname{Set}\right) = \) the class of all sets;\n\n- for \( A, B \) in \( \operatorname{Obj}\left( \operatorname{Set}\right) \) (that is, for \( A, B \) sets) \( {\operatorname{Hom}}_{\operatorname{Set}}\left( {A, B}\right) = {B}^{A} \) . | No |
Suppose \( S \) is a set and \( \sim \) is a relation on \( S \) satisfying the reflexive and transitive properties. Then we can encode this data into a category:\n\n- objects: the elements of \( S \) ;\n\n- morphisms: if \( a, b \) are objects (that is, if \( a, b \in S \) ), then let \( \operatorname{Hom}\left( {a, b... | We have to define 'composition of morphisms' and verify that the conditions specified in [3.1] are satisfied. First all, do we have 'identities'? If \( a \) is an object (that is, if \( a \in S \) ), we need to find an element\n\n\[ \n{1}_{a} \in \operatorname{Hom}\left( {a, a}\right) \text{.}\n\]\n\nThis is precisely ... | No |
Let \( S \) again be a set. Define a category \( \widehat{\mathrm{S}} \) by setting\n\n- \( \operatorname{Obj}\left( \widehat{\mathrm{S}}\right) = \mathcal{P}\left( S\right) \), the power set \( S \) (cf. \( \$ \underline{1.2} \) and Exercise 2.11);\n\n- for \( A, B \) objects of \( \widehat{\mathrm{S}} \) (that is, \(... | Checking the axioms specified in [3.1] should be routine (make sure this is the case!). | No |
Let \( \mathrm{C} \) be a category, and let \( A \) be an object of \( \mathrm{C} \). We are going to define a category \( {\mathrm{C}}_{A} \) whose objects are certain morphisms in \( \mathrm{C} \) and whose morphisms are certain diagrams of \( \mathrm{C} \) (surprise!). | - \( \operatorname{Obj}\left( {\mathrm{C}}_{A}\right) = \) all morphisms from any object of \( \mathrm{C} \) to \( A \) ; thus, an object of \( {\mathrm{C}}_{A} \) is a morphism \( f \in {\operatorname{Hom}}_{\mathrm{C}}\left( {Z, A}\right) \) for some object \( Z \) of \( \mathrm{C} \). Pictorially, an object of \( {\... | Yes |
For the sake of concreteness, let's apply the construction given in Example 3.5 to the category constructed in Example 3.3, say for \( S = \mathbb{Z} \) and \( \sim \) the relation \( \leq \) . Call \( \mathrm{C} \) this category, and choose an object \( A \) of \( \mathrm{C} \) -that is, an integer, for example, \( A ... | \[ \left( {m,3}\right) \rightarrow \left( {n,3}\right) \] if and only if \( m \leq n \) . In this case \( {\mathrm{C}}_{A} \) may be harmlessly identified with the ’subcategory’ of integers \( \leq 3 \), with ’the same’ morphisms as in \( \mathrm{C} \) . | Yes |
Example 3.8. As a ’concrete’ instance of a category as in Example 3.7, let \( \mathrm{C} = \) Set and \( A = \) a fixed singleton \( \{ * \} \) . Call the resulting category \( {\operatorname{Set}}^{ * } \) .\n\nAn object in \( {\operatorname{Set}}^{ * } \) is then a morphism \( f : \{ * \} \rightarrow S \) in Set, whe... | Objects of \( {\operatorname{Set}}^{ * } \) are called ’pointed sets’. Many of the structures we will study in this book will be pointed sets. For example (as we will see) a ’group’ is a set \( G \) with, among other requirements, a distinguished element \( {e}_{G} \) (its ’identity’); ’group homomorphisms' will be fun... | No |
Example 3.9. It is useful to contemplate a few more 'abstract' examples in the style of Examples 3.5 and 3.7 These will be essential ingredients in the promised revisitation of some of the operations mentioned in [1.3] Their definition will appear disappointedly simple-minded to the reader who has mastered Examples 3.5... | I will leave to the reader the task of formalizing this rough description. This example is really nothing more than a mixture of \( {\mathrm{C}}_{A} \) and \( {\mathrm{C}}_{B} \), where the two structures interact because of the stringent requirement that the same \( \sigma \) must make both sides of the diagram commut... | No |
Proposition 4.2. The inverse of an isomorphism is unique. | Proof. We have to verify that if both \( {g}_{1} \) and \( {g}_{2} : B \rightarrow A \) act as inverses of a given isomorphism \( f : A \rightarrow B \), then \( {g}_{1} = {g}_{2} \) . The standard trick for this kind of verification is to compose \( f \) on the left by one of the morphisms, and on the right by the oth... | Yes |
Proposition 4.3. With notation as above:\n\n- Each identity \( {1}_{A} \) is an isomorphism and is its own inverse.\n\n- If \( f \) is an isomorphism, then \( {f}^{-1} \) is an isomorphism and further \( {\left( {f}^{-1}\right) }^{-1} = f \) .\n\n- If \( f \in {\operatorname{Hom}}_{\mathrm{C}}\left( {A, B}\right), g \i... | Proof. These all 'prove themselves'. For example, it is immediate to verify that \( {f}^{-1}{g}^{-1} \) is a left-inverse of \( {gf} \) : indeed 20,\n\n\[ \left( {{f}^{-1}{g}^{-1}}\right) \left( {gf}\right) = {f}^{-1}\left( {\left( {{g}^{-1}g}\right) f}\right) = {f}^{-1}\left( {{1}_{B}f}\right) = {f}^{-1}f = {1}_{A}. \... | No |
As noted in Proposition 4.3, identities are isomorphisms. They may be the only isomorphisms in a category: for example, this is the case in the category \( \mathrm{C} \) obtained from the relation \( \leq \) on \( \mathbb{Z} \), as in Example 3.3 Indeed, for \( a, b \) objects of \( \mathrm{C} \) (that is, \( a, b \in ... | So an isomorphism in \( \mathrm{C} \) necessarily acts from an object \( a \) to itself; but in \( \mathrm{C} \) there is only one such morphism, that is, \( {1}_{a} \) . | Yes |
An automorphism of an object \( A \) of a category \( \mathrm{C} \) is an isomorphism from \( A \) to itself. The set of automorphisms of \( A \) is denoted \( {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) ; it is a subset of \( {\operatorname{End}}_{\mathrm{C}}\left( A\right) \) . | By Proposition 4.3, composition confers on \( {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) a remarkable structure:\n\n- the composition of two elements \( f, g \in {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) is an element \( {gf} \in {\operatorname{Aut}}_{\mathrm{C}}\left( A\right) \) ;\n\n- composition ... | Yes |
In the categories of Example 3.3, every morphism is both a monomorphism and an epimorphism. | Indeed, recall that there is at most one morphism between any two objects in these categories; hence the conditions defining monomorphisms and epimorphisms are vacuous. | Yes |
The category obtained by endowing \( \mathbb{Z} \) with the relation \( \leq \) (see Example 3.3) has no initial or final object. | Indeed, an initial object in this category would be an integer \( i \) such that \( i \leq a \) for all integers \( a \) ; there is no such integer. Similarly, a final object would be an integer \( f \) larger than every integer, and there is no such thing. | Yes |
I claim that if initial/final objects exist, then they are unique up to a unique isomorphism. | I will invoke this fact frequently, so here is its official statement and its (immediate) proof:\n\nProposition 5.4. Let | No |
If \( {I}_{1},{I}_{2} \) are both initial objects in \( \mathrm{C} \), then \( {I}_{1} \cong {I}_{2} \) . | Proof. Recall that (by definition of category!) for every object \( A \) of \( \mathrm{C} \) there is at least one element in \( {\operatorname{Hom}}_{\mathrm{C}}\left( {A, A}\right) \), namely the identity \( {1}_{A} \) . If \( I \) is initial, then there is a unique morphism \( I \rightarrow I \), which therefore mus... | Yes |
Claim 5.5. Denoting by \( \pi \) the ’canonical projection’ defined in Example 2.6, the pair \( \left( {\pi, A/ \sim }\right) \) is an initial object of this category. | Proof. Consider any \( \left( {\varphi, Z}\right) \) as above. We have to prove that there exists a unique morphism \( \left( {\pi, A/ \sim }\right) \rightarrow \left( {\varphi, Z}\right). | No |
Proposition 5.6. The disjoint union is a coproduct in Set. | Proof. Recall (§1.4) that the disjoint union \( A \coprod B \) is defined as the union of two disjoint isomorphic copies \( {A}^{\prime },{B}^{\prime } \) of \( A, B \), respectively; for example, we may let \( {A}^{\prime } = \{ 0\} \times A,{B}^{\prime } = \{ 1\} \times B \) . The functions \( {i}_{A},{i}_{B} \) are ... | Yes |
Since we explicitly require \( G \) to be nonempty, the most economical way to concoct a group is by letting \( G = \{ e\} \) be a singleton. There is only one function \( G \times G \rightarrow G \) in this case, so there is only one possible binary operation on \( G \), defined by\n\n\[ e \bullet e \mathrel{\text{:=}... | The three axioms trivially hold for this example, so \( \{ e\} \) is equipped with a unique group structure.\n\nThis is usually called the trivial group; purists should call any such group \( a \) trivial group, since every singleton gives rise to one. | Yes |
Proposition 1.6. If \( h \in G \) is an identity of \( G \), then \( h = {e}_{G} \) . | Proof. Using first that \( {e}_{G} \) is an identity and then that \( h \) is an identity, one get.\n\n\[ h = {e}_{G}h = {e}_{G} \]\n\n(Amusingly, this argument only uses that \( {e}_{G} \) is a ’left’ identity and \( h \) is a ’right’ identity.) | No |
Proposition 1.7. The inverse is also unique: if \( {h}_{1},{h}_{2} \) are both inverses of \( g \) in \( G \) , then \( {h}_{1} = {h}_{2} \) . | Proof. This actually follows from Proposition 1.4.2 (by viewing \( G \) as the set of isomorphisms of a groupoid with a single object). The reader should construct a stand-alone proof, using the same trick, but carefully hiding any reference to morphisms. | No |
Proposition 1.8. Let \( G \) be a group. Then \( \forall a, g, h \in G \)\n\n\[ \n{ga} = {ha} \Rightarrow g = h,\;{ag} = {ah} \Rightarrow g = h.\n\] | Proof. Both statements are proven by multiplying (on the appropriate side) by \( {a}^{-1} \) and applying associativity. For example,\n\n\[ \n{ga} = {ha} \Rightarrow \left( {ga}\right) {a}^{-1} = \left( {ha}\right) {a}^{-1} \Rightarrow g\left( {a{a}^{-1}}\right) = h\left( {a{a}^{-1}}\right) \Rightarrow g{e}_{G} = h{e}_... | Yes |
Lemma 1.10. If \( {g}^{n} = e \) for some positive integer \( n \), then \( \left| g\right| \) is a divisor of \( n \) . | Proof. As observed, \( n \geq \left| g\right| \) by definition of order, that is, \( n - \left| g\right| \geq 0 \) . There must then exist a positive integer \( m \) such that\n\n\[ r = n - \left| g\right| \cdot m \geq 0\;\text{ and }\;n - \left| g\right| \cdot \left( {m + 1}\right) < 0, \]\n\nthat is, \( r < \left| g\... | Yes |
Proposition 1.13. Let \( q \in G \) be an element of finite order. Then \( {g}^{m} \) has finite order \( \forall m \geq 0 \), and in fact\n\n\[ \left| {g}^{m}\right| = \frac{\operatorname{lcm}\left( {m,\left| g\right| }\right) }{m} = \frac{\left| g\right| }{\gcd \left( {m,\left| g\right| }\right) }.\] | Proof. The equality of the two numbers \( \frac{\operatorname{lcm}\left( {m,\left| g\right| }\right) }{m} \) and \( \frac{\left| g\right| }{\gcd \left( {m,\left| g\right| }\right) } \) follows from elementary properties of \( \gcd \) and \( \operatorname{lcm} : \operatorname{lcm}\left( {a, b}\right) = {ab}/\gcd \left( ... | Yes |
Proposition 1.14. If \( {gh} = {hg} \), then \( \left| {gh}\right| \) divides \( \operatorname{lcm}\left( {\left| g\right| ,\left| h\right| }\right) \) . | Proof. Let \( \left| g\right| = m,\left| h\right| = n \) . If \( N \) is any common multiple of \( m \) and \( n \), then \( {g}^{N} = {h}^{N} = e \) by Corollary 1.11 Since \( g \) and \( h \) commute,\n\n\[ \n{\left( gh\right) }^{N} = \underset{N\text{ times }}{\underbrace{\left( {gh}\right) \left( {gh}\right) \cdots... | Yes |
Lemma 2.2. If \( a \equiv {a}^{\prime }{\;\operatorname{mod}\;n} \) and \( b \equiv {b}^{\prime }{\;\operatorname{mod}\;n} \), then\n\n\[ \left( {a + b}\right) \equiv \left( {{a}^{\prime } + {b}^{\prime }}\right) {\;\operatorname{mod}\;n}. \]\n | Proof. By hypothesis \( n \mid \left( {{a}^{\prime } - a}\right) \) and \( n \mid \left( {{b}^{\prime } - b}\right) \) ; therefore \( \exists k,\ell \in \mathbb{Z} \) such that\n\n\[ \left( {{a}^{\prime } - a}\right) = {kn},\;\left( {{b}^{\prime } - b}\right) = \ell n. \]\n\nThen\n\n\[ \left( {{a}^{\prime } + {b}^{\pri... | Yes |
Proposition 2.3. The order of \( {\left\lbrack m\right\rbrack }_{n} \) in \( \mathbb{Z}/n\mathbb{Z} \) is 1 if \( n \mid m \), and more generally\n\n\[ \left| {\left\lbrack m\right\rbrack }_{n}\right| = \frac{n}{\gcd \left( {m, n}\right) }.\] | Proof. If \( n \mid m \), then \( {\left\lbrack m\right\rbrack }_{n} = {\left\lbrack 0\right\rbrack }_{n} \). If \( n \) does not divide \( m \), observe again that \( {\left\lbrack m\right\rbrack }_{n} = m{\left\lbrack 1\right\rbrack }_{n} \) and apply Proposition 1.13. | Yes |
Proposition 2.6. Multiplication makes \( {\left( \mathbb{Z}/n\mathbb{Z}\right) }^{ * } \) into a group. | Proof. Simple properties of gcd’s show that if \( \gcd \left( {{m}_{1}, n}\right) = \gcd \left( {{m}_{2}, n}\right) = 1 \), then \( \gcd \left( {{m}_{1}{m}_{2}, n}\right) = 1 \) . (For example, if a prime integer divided both \( n \) and \( {m}_{1}{m}_{2} \), then it would necessarily divide \( {m}_{1} \) or \( {m}_{2}... | Yes |
Proposition 3.2. Let \( \varphi : G \rightarrow H \) be a group homomorphism. Then\n\n- \( \varphi \left( {e}_{G}\right) = {e}_{H} \) ;\n\n- \( \forall g \in G,\varphi \left( {g}^{-1}\right) = \varphi {\left( g\right) }^{-1} \) . | Proof. The first item follows from the definition of homomorphism and cancellation: since \( {e}_{H} = {e}_{H} \cdot {e}_{H} \), \n\n\[ \n{e}_{H} \cdot \varphi \left( {e}_{G}\right) = \varphi \left( {e}_{G}\right) = \varphi \left( {{e}_{G} \cdot {e}_{G}}\right) = \varphi \left( {e}_{G}\right) \cdot \varphi \left( {e}_{... | Yes |
Proposition 3.3. Trivial groups are both initial and final in Grp. | Proof. It should be clear that trivial groups are final: there is only one function from a set to a singleton, that is, the constant function; this is vacuously a group homomorphism.\n\nTo see that trivial groups are initial, let \( T = \{ e\} \) be a trivial group; for any group \( G \), define \( \varphi : T \rightar... | Yes |
Proposition 3.4. With operation defined componentwise, \( G \times H \) is a product in Grp. | Proof. Recall (§ 15.4) that this means that \( G \times H \) satisfies the following universal property: for any group \( A \) and any choice of group homomorphisms \( {\varphi }_{G} : A \rightarrow G \), \( {\varphi }_{H} : A \rightarrow H \), there exists a unique group homomorphism \( {\varphi }_{G} \times {\varphi ... | Yes |
Proposition 4.1. Let \( \varphi : G \rightarrow H \) be a group homomorphism, and let \( g \in G \) be an element of finite order. Then \( \left| {\varphi \left( g\right) }\right| \) divides \( \left| g\right| \) . | Proof. As observed, \( \varphi {\left( g\right) }^{\left| g\right| } = {e}_{H} \) ; applying Lemma 1.10 gives the statement. | No |
There are no nontrivial homomorphisms \( \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z} \) : indeed, the image of every element of \( \mathbb{Z}/n\mathbb{Z} \) must have finite order, and the only element with finite order in \( \left( {\mathbb{Z}, + }\right) \) is 0 . | Indeed, the image of every element of \( \mathbb{Z}/n\mathbb{Z} \) must have finite order, and the only element with finite order in \( \left( {\mathbb{Z}, + }\right) \) is 0. | Yes |
Proposition 4.3. Let \( \varphi : G \rightarrow H \) be a group homomorphism. Then \( \varphi \) is an isomorphism of groups if and only if it is a bijection. | Proof. One implication is immediate, as pointed out above. For the other implication, assume \( \varphi : G \rightarrow H \) is a bijective group homomorphism. As a bijection, \( \varphi \) has an inverse in Set:\n\n\[{\varphi }^{-1} : H \rightarrow G\]\n\nwe simply need to check that this is a group homomorphism. Let ... | Yes |
Proposition 4.8. Let \( \varphi : G \rightarrow H \) be an isomorphism.\n\n\[ \text{-}\left( {\forall g \in G}\right) : \left| {\varphi \left( g\right) }\right| = \left| g\right| \text{;} \] | Proof. The first assertion follows from Proposition 4.1 the order of \( \varphi \left( g\right) \) divides the order of \( g \), and on the other hand the order of \( g = {\varphi }^{-1}\left( {\varphi \left( g\right) }\right) \) must divide the order of \( \varphi \left( g\right) \) ; thus the two orders must be equal... | Yes |
Lemma 5.1. If \( w \in W\left( A\right) \) has length \( n \), then \( 2{2}^{n\left\lfloor \frac{n}{2}\right\rfloor }\left( w\right) \) is a reduced word. | Proof. Indeed, either \( r\left( w\right) = w \) or the length of \( r\left( w\right) \) is less than the length of \( w \) ; but one cannot decrease the length of \( w \) more than \( n/2 \) times, since each nonidentity application of \( r \) decreases the length by two. | No |
Proposition 5.2. The pair \( \left( {j, F\left( A\right) }\right) \) satisfies the universal property for free groups on \( A \) . | Proof. This is also essentially evident, once one has absorbed all the notation. Any function \( f : A \rightarrow G \) to a group extends uniquely to a map \( \varphi : F\left( A\right) \rightarrow \) \( G \), determined by the homomorphism condition and by the requirement that the diagram commutes, which fixes its va... | Yes |
Example 5.3. It is easy to ’visualize’ \( F\left( {\{ a\} }\right) \cong \mathbb{Z} \) ; but it is already somewhat challenging for the free group on two generators, \( F\left( {\{ x, y\} }\right) \) . The best I can do is the following: behold the infinite graph \( {}^{23} \) | This is an example of the Cayley graph of a group (cf. Exercise 8.6): a graph whose vertices correspond to the elements of the group and whose edges connect vertices according to the action of generators.\n\nobtained by starting at a point (the center of the picture), then branching out in four directions by a length o... | No |
Proposition 5.6. For every set \( A,{F}^{ab}\left( A\right) \cong {\mathbb{Z}}^{\oplus A} \) . | Proof. The key point is again that every element of \( {\mathbb{Z}}^{\oplus A} \) may be written uniquely as a finite sum\n\n\[\n\mathop{\sum }\limits_{{a \in A}}{m}_{a}j\left( a\right) ,\;{m}_{a} \neq 0\text{ for only finitely many }a;\n\]\n\nonce this is understood, the argument is precisely the same as for Claim 5.4... | No |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.