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Lemma 13.29. Let \( M \) and \( N \) be compact manifolds and let \( D \) be a compact manifold with boundary, with\n\n\[ \dim M + \dim D < \dim N\text{.} \]\n\nLet \( f : M \rightarrow N \) and \( g : D \rightarrow N \) be smooth maps. Suppose \( E \) is a closed subset of \( D \) such that \( g\left( E\right) \) is d... | Proof. Step 1: The local move. For \( x \in D \smallsetminus E \), let us choose a neighborhood \( U \) of \( g\left( x\right) \) diffeomorphic to \( {\mathbb{R}}^{n} \), where \( n = \dim N \) . We may then define a map\n\n\[ h : {f}^{-1}\left( U\right) \times {g}^{-1}\left( U\right) \rightarrow U \]\n\nby\n\n\[ h\lef... | Yes |
Proposition 13.31. The Stiefel diagram is invariant under the action of \( W \) and under translations by elements of \( \Gamma \) . | Proof. Since \( W \) permutes the roots, invariance of the Stiefel diagram is evident. Meanwhile, if \( \gamma \) is in the kernel \( \Gamma \) of the exponential map, the adjoint action of \( {e}^{2\pi \gamma } \) on \( \mathfrak{g} \) is trivial. Thus, for \( X \in {\mathfrak{g}}_{\alpha } \), we have\n\n\[ X = {\ope... | Yes |
Proposition 13.32. Let \( \Gamma \rtimes W \) denote the set of affine transformations of \( \mathfrak{t} \) that can be expressed in the form\n\n\[ H \mapsto w \cdot H + \gamma \]\n\nfor some \( \gamma \in \Gamma \) and some \( w \in W \). Then \( \Gamma \rtimes W \) forms a group under composition, with the group law... | Proof. We merely compute that\n\n\[ w \cdot \left( {{w}^{\prime } \cdot H + {\gamma }^{\prime }}\right) + \gamma = \left( {w{w}^{\prime }}\right) \cdot H + \gamma + w \cdot {\gamma }^{\prime },\]\n\nso that the composition of the affine transformations associated to \( \left( {\gamma, w}\right) \) and \( \left( {{\gamm... | Yes |
Proposition 13.34. 1. The extended Weyl group equals \( I \rtimes W \), the semidirect product of the ordinary Weyl group \( W \) and the coroot lattice \( I \) . | Proof. For Point 1, let \( {V}_{\alpha },\alpha \in R \), be the hyperplane through the origin orthogonal to \( \alpha \) . Since \( \left\langle {\alpha ,{H}_{\alpha }}\right\rangle = 2 \), we see that the hyperplane \( {L}_{\alpha, n} \) in Definition 13.30 is the translate of \( {V}_{\alpha } \) by \( n{H}_{\alpha }... | Yes |
Proposition 13.37. Every loop in \( K \) is homotopic to a loop in \( T \) . | Proof. By Proposition 13.8, the group \( K \) is a fiber bundle with base \( K/T \) and fiber \( T \) . Suppose now that \( l \) is a loop in \( K \) and that \( {l}^{\prime }\left( \tau \right) \mathrel{\text{:=}} \left\lbrack {l\left( \tau \right) }\right\rbrack \) is the corresponding loop in \( K/T \) . Since \( K/... | Yes |
Lemma 13.38. Suppose \( \gamma \in \Gamma \) but \( \gamma \notin I \) . Then there exist \( {\gamma }^{\prime } \) in \( I \) and \( w \) in \( W \) such that the affine transformation\n\n\[ H \mapsto w \cdot \left( {H + \gamma + {\gamma }^{\prime }}\right) \]\n\n(13.10)\n\nmaps \( A \) to itself but is not the identi... | Proof. By Proposition 13.31, translation by \( \gamma \) maps the alcove \( A \) to some other alcove \( {A}^{\prime } \) . Then by Proposition 13.34, there exists an element of the extended Weyl group that maps \( {A}^{\prime } \) back to \( A \) . Thus, there exist \( {\gamma }^{\prime } \in I \) and \( w \in W \) su... | Yes |
Proposition 13.40. If \( H \in \mathfrak{t} \), then \( {e}^{2\pi H} \) belongs to \( Z\left( K\right) \) if and only if\n\n\[ \langle \alpha, H\rangle \in \mathbb{Z} \]\n\nfor all \( \alpha \in R \) . | Proof. By Exercise 17 in Chapter 3, an element \( x \) of \( K \) is in \( Z\left( K\right) \) if and only if \( {\operatorname{Ad}}_{x}\left( X\right) = X \) for all \( X \in \mathfrak{k} \), or, equivalently, if and only if \( {\operatorname{Ad}}_{x}\left( X\right) = X \) for all \( X \in \mathfrak{g} = {\mathfrak{k}... | Yes |
Proposition 13.42. The map\n\n\[ \n\gamma \mapsto {e}^{2\pi \gamma },\;\gamma \in {\Lambda }^{ * }, \n\]\n\nis a homomorphism of \( {\Lambda }^{ * } \) onto \( Z\left( K\right) \) with kernel equal to \( \Gamma \) . Thus,\n\n\[ \nZ\left( K\right) \cong {\Lambda }^{ * }/\Gamma \n\]\n\nwhere \( {\Lambda }^{ * } \) is the... | Proof. As we have noted, Corollary 11.11 implies that \( Z\left( K\right) \subset T \) . Since the exponential map for \( T \) is surjective, Proposition 13.40 tells us that the map \( \gamma \mapsto \) \( {e}^{2\pi \gamma } \) maps \( {\Lambda }^{ * } \) onto \( Z\left( K\right) \) . This map is a homomorphism since \... | Yes |
Let \( {\Lambda }^{ * } \) denote the dual of the root lattice and let \( I \) denote the coroot lattice. If \( K \) is simply connected, then\n\n\[ Z\left( K\right) \cong {\Lambda }^{ * }/I \] | Proof. If \( {\pi }_{1}\left( K\right) \) is trivial, the kernel \( \Gamma \) of the exponential map must equal the coroot lattice \( I \), which means that\n\n\[ Z\left( K\right) \cong {\Lambda }^{ * }/\Gamma = {\Lambda }^{ * }/I \] | Yes |
If \( K = \mathrm{{SO}}\left( 4\right) \), then the lattices \( {\Lambda }^{ * },\Gamma \), and \( I \) are as in Figure 13.7 and both \( {\pi }_{1}\left( K\right) \) and \( Z\left( K\right) \) are isomorphic to \( \mathbb{Z}/2 \) . Explicitly, \( Z\left( {\mathrm{{SO}}\left( 4\right) }\right) = \{ I, - I\} \) . | Proof. If we compute as in Sect. 7.7.2, but adjusting for a factor of \( i \) to obtain the real roots and coroots, we find that the coroots are the matrices\n\n\[ \left( \begin{array}{rrr} 0 & a & b \\ - a & 0 & b \\ & - b & 0 \end{array}\right) \]\n\n(13.14)\n\nwhere \( a = \pm 1 \) and \( b = \pm 1 \) . We identify ... | Yes |
Example 13.45. Suppose that \( K \) is a connected, compact matrix Lie group with Lie algebra \( \mathfrak{k} \), that \( \mathfrak{t} \) is a maximal commutative subalgebra of \( \mathfrak{k} \), and that the root system of \( \mathfrak{k} \) relative to \( \mathfrak{t} \) is isomorphic to \( {G}_{2} \) . Then both \(... | Proof. As we can see from Figure 8.11, each of the fundamental weights for \( {G}_{2} \) is a root. Thus, every algebraically integral element for \( {G}_{2} \) (i.e., every integer linear combination of the fundamental weights) is in the root lattice. Thus, all three of the lattices in (13.12) must be equal. By dualiz... | No |
A natural problem for the heat equation (in one spatial dimension) is this one:\n\n\[ \n{u}_{xx}\left( {x, t}\right) = {u}_{t}\left( {x, t}\right) ,\;x > 0,\;t > 0;\;u\left( {x,0}\right) = 0,\;x > 0;\;u\left( {0, t}\right) = 0,\;t > 0.\n\] | The obvious and intuitive solution is, of course, that the rod will remain at temperature 0, i.e., \( u\left( {x, t}\right) = 0 \) for all \( x > 0, t > 0 \) . But the mathematical problem has additional solutions: let\n\n\[ \nu\left( {x, t}\right) = \frac{x}{{t}^{3/2}}{e}^{-{x}^{2}/\left( {4t}\right) },\;x > 0, t > 0.... | No |
A simple example of instability is exhibited by an ordinary differential equation such as \( {y}^{\prime \prime }\left( t\right) + y\left( t\right) = f\left( t\right) \) with initial conditions \( y\left( 0\right) = 1,{y}^{\prime }\left( 0\right) = 0 \). | If, for example, we take \( f\left( t\right) = 1 \), the solution is \( y\left( t\right) = 1 \). If we introduce a small perturbation in the right-hand member by taking \( f\left( t\right) = 1 + \varepsilon \cos t \), where \( \varepsilon \neq 0 \), the solution is given by \( y\left( t\right) = 1 + \frac{\mathrm{i}}{2... | No |
Find all solutions \( u\left( {x, t}\right) \) of \( {u}_{xx} = {u}_{tt} \) for \( x > 0, t > 0 \), that satisfy \( u\left( {x,0}\right) = {2x} \) and \( {u}_{t}\left( {x,0}\right) = 1 \) for \( x > 0 \) and, in addition, \( u\left( {0, t}\right) = {2t} \) for \( t > 0 \) . | Since the first quadrant of the \( {xt} \) -plane is convex, all solutions of the equation must have the appearance\n\n\[ u\left( {x, t}\right) = \varphi \left( {x - t}\right) + \psi \left( {x + t}\right) ,\;x > 0, t > 0. \]\n\nOur task is to determine what the functions \( \varphi \) and \( \psi \) look like. We need ... | Yes |
The modulus of a complex number \( z = x + {iy} \) is defined as \( \left| z\right| = \sqrt{z\bar{z}} = \sqrt{{x}^{2} + {y}^{2}} \). As a consequence, | \[ \left| {e}^{z}\right| = \left| {e}^{x + {iy}}\right| = \left| {{e}^{x} \cdot {e}^{iy}}\right| = {e}^{x}\left| {\cos y + i\sin y}\right| = {e}^{x}\sqrt{{\cos }^{2}y + {\sin }^{2}y} = {e}^{x}. \] | Yes |
Let us start from the formula \( {e}^{ix}{e}^{iy} = {e}^{i\left( {x + y}\right) } \) and rewrite both sides of this, using (2.1). | On the one hand we have\n\n\[ \n{e}^{ix}{e}^{iy} = \left( {\cos x + i\sin x}\right) \left( {\cos y + i\sin y}\right) \n\]\n\n\[ \n= \cos x\cos y - \sin x\sin y + i\left( {\cos x\sin y + \sin x\cos y}\right) , \n\]\n\nand on the other hand,\n\n\[ \n{e}^{i\left( {x + y}\right) } = \cos \left( {x + y}\right) + i\sin \left... | Yes |
If \( f\left( t\right) = {e}^{ct} \) with a complex coefficient \( c = \alpha + {i\beta } \), we can find the derivative, according to (2.3), like this: | \[ {f}^{\prime }\left( t\right) = \frac{d}{dt}\left( {{e}^{\alpha t}\left( {\cos {\beta t} + i\sin {\beta t}}\right) }\right) = \frac{d}{dt}\left( {{e}^{\alpha t}\cos {\beta t}}\right) + i\frac{d}{dt}\left( {{e}^{\alpha t}\sin {\beta t}}\right) \] \[ = \alpha {e}^{\alpha t}\cos {\beta t} - {e}^{\alpha t}\beta \sin {\be... | Yes |
Let \( c \) be a non-zero real number. To compute the integral of \( {e}^{ct} \) over an interval \( \left\lbrack {a, b}\right\rbrack \), we can use the fact that \( {e}^{ct} \) is the derivative of a known function, by Example 2.4: | \[ {\int }_{a}^{b}{e}^{ct}{dt} = {\left\lbrack \frac{{e}^{ct}}{c}\right\rbrack }_{t = a}^{t = b} = \frac{{e}^{cb} - {e}^{ca}}{c}. \] | Yes |
Lemma 2.1 Suppose that the series \( \mathop{\sum }\limits_{{k = 1}}^{\infty }{a}_{k} \) is convergent with the sum s. Then also\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\sigma }_{n} = s \] | Proof. Let \( \varepsilon > 0 \) be given. The assumption is that \( {s}_{n} \rightarrow s \) as \( n \rightarrow \infty \) . This means that there exists an integer \( N \) such that \( \left| {{s}_{n} - s}\right| < \varepsilon /2 \) for all \( n > N \) . For these \( n \) we can write\n\n\[ \left| {{\sigma }_{n} - s}... | Yes |
Example 2.6. Let \( {a}_{k} = {\left( -1\right) }^{k - 1}, k = 1,2,3,\ldots \), which means that we have the series \( 1 - 1 + 1 - 1 + 1 - 1 + \cdots \) . Then \( {s}_{n} = 0 \) if \( n \) is even and \( {s}_{n} = 1 \) if \( n \) is odd. The means \( {\sigma }_{n} \) are | \[ {\sigma }_{n} = \frac{1}{2}\;\text{ if }n\text{ is even,}\;{\sigma }_{n} = \frac{\frac{1}{2}\left( {n + 1}\right) }{n} = \frac{n + 1}{2n}\;\text{ if }n\text{ is odd. } \] Thus we have \( {\sigma }_{n} \rightarrow \frac{1}{2} \) as \( n \rightarrow \infty \) . This divergent series is indeed summable \( \left( {C,1}\... | Yes |
Theorem 2.1 Let \( I = \left( {-a, a}\right) \) be an interval (finite or infinite). Suppose that \( {\left\{ {K}_{n}\right\} }_{n = 1}^{\infty } \) is a sequence of real-valued, Riemann-integrable functions defined on \( I \), with the following properties:\n\n(1) \( {K}_{n}\left( s\right) \geq 0 \) .\n\n(2) \( {\int ... | Proof. Let \( \varepsilon \) be a positive number. Since \( f \) is continous at the origin there exists a number \( \delta > 0 \) such that\n\n\[ \left| s\right| \leq \delta \; \Rightarrow \;\left| {f\left( s\right) - f\left( 0\right) }\right| < \varepsilon . \]\n\nFurthermore, \( f \) is bounded on \( I \), i.e., the... | Yes |
Define \( {K}_{n} : \mathbf{R} \rightarrow \mathbf{R} \) by\n\n\[ \n{K}_{n}\left( s\right) = \left\{ \begin{matrix} n, & \left| s\right| < 1/\left( {2n}\right) \\ 0, & \left| s\right| > 1/\left( {2n}\right) \end{matrix}\right.\n\] | It is obvious that the conditions 1-3 are fullfilled. | No |
Corollary 2.1 If \( {\left\{ {K}_{n}\right\} }_{n = 1}^{\infty } \) is a positive summation kernel on the interval \( I,{s}_{0} \) is an interior point of \( I \), and \( f \) is continuous at \( s = {s}_{0} \), then\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{\int }_{I}{K}_{n}\left( s\right) f\left( {{s}_{0}... | The proof is left as an exercise (do the change of variable \( {s}_{0} - s = u \) ). | No |
Can we invent a mathematical object that can be interpreted as the limit of such a sequence? | The problems in Examples 2.11 and 2.12 above have been addressed by many physicists ever since the later years of the nineteenth century by using the following trick. Let us assume that the independent variable is \( t \) . Introduce a \ | No |
The LAPLACE transform of a function \( f \) is defined to be another function \( \widetilde{f} \), given by\n\n\[ \widetilde{f}\left( s\right) = {\int }_{0}^{\infty }f\left( t\right) {e}^{-{st}}{dt} \] \n\nfor all \( s \) such that the integral is convergent (see Chapter 3). The Laplace transform of \( \delta \) cannot... | It is indeed customary to write\n\n\[ \widetilde{f}\left( s\right) = {\int }_{0 - }^{\infty }f\left( t\right) {e}^{-{st}}{dt} = \mathop{\lim }\limits_{{k \nearrow 0}}{\int }_{k}^{\infty }f\left( t\right) {e}^{-{st}}{dt}. \] \n\nWith this definition one finds that \( \widetilde{\delta }\left( s\right) = 1 \) for all \( ... | Yes |
Consider the function \( f : \mathbf{R} \rightarrow \mathbf{R} \) that is given by\n\n\[ f\left( t\right) = \left\{ \begin{array}{l} 1 - {t}^{2}\text{ for }t < - 2 \\ t + 2\text{ for } - 2 < t < 1 \\ 1 - t\text{ for }t > 1 \end{array}\right. \] | This can now be compressed into one formula:\n\n\( f\left( t\right) \)\n\n\[ = \left( {1 - {t}^{2}}\right) \left( {1 - H\left( {t + 2}\right) }\right) + \left( {t + 2}\right) \left( {H\left( {t + 2}\right) - H\left( {t - 1}\right) }\right) + \left( {1 - t}\right) H\left( {t - 1}\right) \]\n\n\[ = \left( {1 - {t}^{2}}\r... | Yes |
If \( \chi \) is a function that is continuous at \( a \), what should be meant by the product \( \chi \left( t\right) {\delta }_{a}\left( t\right) \) ? | Since \( {\delta }_{a}\left( t\right) \) is \ | No |
Find the first two derivatives of \( f\\left( t\\right) = \\left| t\\right| \) . | Solution. Rewrite the function without modulus signs, using Heaviside windows:\n\n\\[ \nf\\left( t\\right) = \\left| t\\right| = - t\\left( {1 - H\\left( t\\right) }\\right) + {tH}\\left( t\\right) = {2tH}\\left( t\\right) - t.\n\\]\n\nDifferentiation then gives\n\n\\[ \n{f}^{\\prime }\\left( t\\right) = {2H}\\left( t\... | Yes |
Another example of the same type, though more complicated. The function \( f\left( x\right) = \left| {{x}^{2} - 1}\right| \) can be rewritten as | \[ f\left( x\right) = \left( {{x}^{2} - 1}\right) H\left( {x - 1}\right) + \left( {1 - {x}^{2}}\right) \left( {H\left( {x + 1}\right) - H\left( {x - 1}\right) }\right) + \left( {{x}^{2} - 1}\right) \left( {1 - H\left( {x + 1}\right) }\right) = \left( {{x}^{2} - 1}\right) \left( {{2H}\left( {x - 1}\right) - {2H}\left( {... | Yes |
Solve the differential equation \( {y}^{\prime } + {2y} = \delta \left( {t - 1}\right) \) for \( t > 0 \) with the initial value \( y\left( 0\right) = 1 \) . | Solution. The method of integrating factor can be used. An integrating factor is \( {e}^{2t} \) :\n\n\[ \n{e}^{2t}{y}^{\prime } + 2{e}^{2t}y = {e}^{2t}\delta \left( {t - 1}\right) \; \Leftrightarrow \;\frac{d}{dt}\left( {{e}^{2t}y}\right) = {e}^{2}\delta \left( {t - 1}\right) .\n\]\n\nIn rewriting the right-hand side w... | Yes |
Find all solutions of the differential equation \( {y}^{\prime \prime } + {4y} = \delta \) . | Solution. The classical method for this sort of problem amounts to first finding the general solution of the corresponding homogeneous equation, which is \( {y}_{H} = {C}_{1}\cos {2t} + {C}_{2}\sin {2t} \), where \( {C}_{1} \) and \( {C}_{2} \) are arbitrary constants. Then we should find some particular solution of th... | Yes |
In Sec. 1.3, on the wave equation, the final example turned out to have a solution that was not really a differentiable function. Now we can put this right, by allowing the generalized derivatives introduced in this section. The solution involved the function \( \varphi \), defined by\n\n\[ \varphi \left( s\right) = \f... | The two first derivatives are\n\n\[ {\varphi }^{\prime }\left( s\right) = - \frac{1}{2} + H\left( s\right) + {s\delta }\left( s\right) = - \frac{1}{2} + H\left( s\right) ,\;{\varphi }^{\prime \prime }\left( s\right) = \delta \left( s\right) . \]\n\nThe complete solution of the problem in Sec. 1.3 can be written\n\n\[ u... | Yes |
Let \( f\left( t\right) = {e}^{at}, t \geq 0 \). Then, | \[ {\int }_{0}^{\infty }f\left( t\right) {e}^{-{st}}{dt} = {\int }_{0}^{\infty }{e}^{{at} - {st}}{dt} = {\left\lbrack \frac{{e}^{\left( {a - s}\right) t}}{a - s}\right\rbrack }_{t = 0}^{\infty } = \frac{1}{s - a}, \] provided that \( a - s < 0 \) so that the evaluation at infinity yields zero. Thus we have \( \widetild... | Yes |
Let \( f\left( t\right) = t, t > 0 \). Then, integrating by parts, we get | \[\n\widetilde{f}\left( s\right) = {\int }_{0}^{\infty }t{e}^{-{st}}{dt} = {\left\lbrack t \cdot \frac{{e}^{-{st}}}{-s}\right\rbrack }_{t = 0}^{\infty } + \frac{1}{s}{\int }_{0}^{\infty }1 \cdot {e}^{-{st}}{dt}\n\]\n\n\[= 0 + \frac{1}{s}\mathcal{L}\left\lbrack 1\right\rbrack \left( s\right) = \frac{1}{{s}^{2}}\]\n\nThi... | Yes |
Let \( f\left( t\right) = {e}^{ibt} \). Then we can imitate Example 3.1 above and write\n\n\[ \n{\int }_{0}^{\infty }f\left( t\right) {e}^{-{st}}{dt} = {\int }_{0}^{\infty }{e}^{\left( {{ib} - s}\right) t}{dt} = {\left\lbrack \frac{{e}^{\left( {{ib} - s}\right) t}}{{ib} - s}\right\rbrack }_{t = 0}^{\infty }\n\]\n\n\[ \... | For \( s > 0 \) the substitution as \( t \rightarrow \infty \) will tend to zero, because the factor \( {e}^{-{st}} \) tends to zero and the rest of the expression is bounded. The result is thus that \( \mathcal{L}\left\lbrack {e}^{ibt}\right\rbrack \left( s\right) = 1/\left( {s - {ib}}\right) \), which means that the ... | No |
Theorem 3.1 If \( f \in {\mathcal{E}}_{k} \), then \( \widetilde{f}\left( s\right) \) exists for all \( s > k \) . | Proof. We begin by observing that condition (i) for the class \( {\mathcal{E}}_{k} \) implies that the integral\n\n\[ \n{\int }_{0}^{T}f\left( t\right) {e}^{-{st}}{dt} \n\]\n\nexists finitely for all \( s \) and all \( T > 0 \) . Now assume \( s > k \) . Thus there exists a number \( M \) and a number \( {t}_{0} \) so ... | Yes |
Using rule 1 and the result of Example 3.3 in the preceding section, we can find the Laplace transforms of cos and sin: | \n\[ \mathcal{L}\left\lbrack {\cos {bt}}\right\rbrack \left( s\right) = \frac{1}{2}\mathcal{L}\left\lbrack {{e}^{ibt} + {e}^{-{ibt}}}\right\rbrack \left( s\right) = \frac{1}{2}\left( {\frac{1}{s - {ib}} + \frac{1}{s + {ib}}}\right) = \frac{s}{{s}^{2} + {b}^{2}}, \]\n\n\[ \mathcal{L}\left\lbrack {\sin {bt}}\right\rbrack... | Yes |
Theorem 3.2 If \( f \in {\mathcal{E}}_{{k}_{0}} \), then \( \left( {t \mapsto {tf}\left( t\right) }\right) \in {\mathcal{E}}_{{k}_{1}} \) for \( {k}_{1} > {k}_{0} \) and\n\n\[ \mathcal{L}\left\lbrack {{tf}\left( t\right) }\right\rbrack \left( s\right) = - \frac{d}{ds}\widetilde{f}\left( s\right) \] | Proof. We shall use a theorem on differentiation of integrals. In order to keep it lucid, we assume that \( f \) is continuous on the whole of \( {\mathbf{R}}_{ + } \) ; otherwise we would have to split into integrals over subintervals where \( f \) is continuous, and this introduces certain purely technical complicati... | Yes |
We know that \( \mathcal{L}\left\lbrack 1\right\rbrack \left( s\right) = 1/s \) for \( s > 0 \). Then we can say that | \[ \mathcal{L}\left\lbrack t\right\rbrack \left( s\right) = \mathcal{L}\left\lbrack {t \cdot 1}\right\rbrack \left( s\right) = - \frac{d}{ds}\frac{1}{s} = - \left( {-\frac{1}{{s}^{2}}}\right) = \frac{1}{{s}^{2}},\;s > 0. \] | Yes |
Theorem 3.3 Assume that \( f \in \mathcal{E} \) is continuous on \( {\mathbf{R}}_{ + } \) . Also assume that the derivative \( {f}^{\prime }\left( t\right) \) exists for all \( t \geq 0 \) (with \( {f}^{\prime }\left( 0\right) \) interpreted as the righthand derivative) and that \( {f}^{\prime } \in \mathcal{E} \) . Th... | Proof. Suppose that \( f \in {\mathcal{E}}_{{k}_{0}} \) and \( {f}^{\prime } \in {\mathcal{E}}_{{k}_{1}} \), and take \( s \) to be larger than both \( {k}_{0} \) and \( {k}_{1} \) . Let \( T \) be a positive number. Integration by parts gives\n\n\[ {\int }_{0}^{T}{f}^{\prime }\left( t\right) {e}^{-{st}}{dt} = f\left( ... | Yes |
Theorem 3.4 (a) If \( f \in \mathcal{E} \), then\n\n\[ \mathop{\lim }\limits_{{s \rightarrow \infty }}\widetilde{f}\left( s\right) = 0 \] | Proof. (a) Let \( \varepsilon > 0 \) be given and choose \( \delta > 0 \) so small that\n\n\[ {\int }_{0}^{\delta }\left| {f\left( t\right) }\right| {dt} < \varepsilon \]\n\nLet \( k > 0 \) be such that \( f \in {\mathcal{E}}_{k} \) and let \( {s}_{0} > k \) . Then for \( s > {s}_{0} \) we get\n\n\[ \left| {\widetilde{... | Yes |
Let us try to solve the initial value problem\n\n\\[ \n{y}^{\prime \prime } - 4{y}^{\prime } + {3y} = t,\;t > 0;\;y\left( 0\right) = 3,\;{y}^{\prime }\left( 0\right) = 2.\n\\] | We assume that \\( y = y\\left( t\\right) \\) is a solution such that \\( y \\), as well as \\( {y}^{\prime } \\) and \\( {y}^{\prime \prime } \\), has a Laplace transform. By Theorem 3.3 we have then\n\n\\[ \n\\mathcal{L}\\left\\lbrack {y}^{\prime }\\right\\rbrack \\left( s\\right) = s\\widetilde{y} - y\\left( 0\\righ... | Yes |
Find \( f\left( t\right) \), when \( \widetilde{f}\left( s\right) = \frac{{2s} + 3}{{s}^{2} + {4s} + {13}} \) . | Solution. Complete the square in the denominator: \( {s}^{2} + {4s} + {13} = {\left( s + 2\right) }^{2} + 9 \) . Then split the numerator to enable us to recognize transforms of cosines and sines:\n\n\[ \frac{{2s} + 3}{{s}^{2} + {4s} + {13}} = \frac{2\left( {s + 2}\right) - 1}{{\left( s + 2\right) }^{2} + {3}^{2}} = 2 ... | Yes |
Find \( g\\left( t\\right) \), if \( \\widetilde{g}\\left( s\\right) = \\frac{2s}{{\\left( {s}^{2} + 1\\right) }^{2}} \) . | Solution. We recognize the transform as a derivative:\n\n\[ \n\\widetilde{g}\\left( s\\right) = - \\frac{d}{ds}\\frac{1}{{s}^{2} + 1}.\n\]\n\nBy Theorem 3.2 and the known transform of the sine we get \( g\\left( t\\right) = t\\sin t \) . | Yes |
Example 3.11. Solve the initial value problem\n\n\\[ \n{y}^{\\prime \\prime } + 4{y}^{\\prime } + {13y} = {13},\\;y\\left( 0\\right) = {y}^{\\prime }\\left( 0\\right) = 0.\n\\] | Solution. Transformation gives\n\n\\[ \n\\left( {{s}^{2} + {4s} + {13}}\\right) \\widetilde{y} = \\frac{13}{s} \\Leftrightarrow \\widetilde{y} = \\frac{13}{s\\left( {{\\left( s + 2\\right) }^{2} + 9}\\right) }.\n\\]\nExpand into partial fractions:\n\n\\[ \n\\widetilde{y} = \\frac{1}{s} - \\frac{s + 4}{{\\left( s + 2\\r... | Yes |
Example 3.12. Solve the initial value problem\n\n\\[ \n\\left\\{ {\\begin{array}{l} {x}^{\\prime } = x + {3y}, \\\\ {y}^{\\prime } = {3x} + y; \\end{array}\\;x\\left( 0\\right) = 5,\\;y\\left( 0\\right) = 1.}\\right.\n\\] | Solution. Laplace transformation gives\n\n\\[ \n\\left\\{ {\\begin{array}{l} s\\widetilde{x} - 5 = \\widetilde{x} + 3\\widetilde{y} \\\\ s\\widetilde{y} - 1 = 3\\widetilde{x} + \\widetilde{y} \\end{array}\\; \\Leftrightarrow \\; \\left\\{ \\begin{array}{l} \\left( {1 - s}\\right) \\widetilde{x} + 3\\widetilde{y} = - 5 ... | Yes |
Find a solution of the problem\n\n\\[ \n\\frac{{\\partial }^{2}u}{\\partial {x}^{2}} = \\frac{\\partial u}{\\partial t},\\;0 < x < 1, t > 0 \n\\]\n\n\\[ \nu\\left( {0, t}\\right) = 1, u\\left( {1, t}\\right) = 1,\\;t > 0 \n\\]\n\n\\[ \nu\\left( {x,0}\\right) = 1 + \\sin {\\pi x},\\;0 < x < 1. \n\\] | Solution. We introduce the Laplace transform \\( U\\left( {x, s}\\right) \\) of \\( u\\left( {x, t}\\right) \\), i.e.,\n\n\\[ \nU\\left( {x, s}\\right) = \\mathcal{L}\\left\\lbrack {t \\mapsto u\\left( {x, t}\\right) }\\right\\rbrack \\left( s\\right) = {\\int }_{0}^{\\infty }u\\left( {x, t}\\right) {e}^{-{st}}{dt}. \n... | Yes |
Let \( f\left( t\right) = {e}^{t}, g\left( t\right) = {e}^{-{2t}} \) . Then | \[ f * g\left( t\right) = {\int }_{0}^{t}{e}^{u}{e}^{-2\left( {t - u}\right) }{du} = {\int }_{0}^{t}{e}^{u - {2t} + {2u}}{du} = {e}^{-{2t}}{\int }_{0}^{t}{e}^{3u}{du} \] \[ = {e}^{-{2t}}{\left\lbrack \frac{1}{3}{e}^{3u}\right\rbrack }_{u = 0}^{u = t} = \frac{1}{3}{e}^{-{2t}}\left( {{e}^{3t} - 1}\right) = \frac{{e}^{t} ... | Yes |
If \( g\left( t\right) = 1 \), then \( f * g\left( t\right) = {\int }_{0}^{t}f\left( u\right) \;{du} \) | Thus,“integration” can be considered to be convolution with the function 1 | No |
Theorem 3.6 The Laplace transform of a convolution is the product of the Laplace transforms of the two convolution factors: | Proof. Let \( s \) be so large that both \( \widetilde{f}\left( s\right) \) and \( \widetilde{g}\left( s\right) \) exist. We have agreed in section 3.1 that this means that the corresponding integrals converge absolutely. Now consider the improper double integral\n\n\[ {\iint }_{Q}\left| {f\left( u\right) g\left( v\rig... | Yes |
Example 3.16. As an illustration of the theorem we can take the situation in Example 3.14. There we have\n\n\[ \widetilde{f}\left( s\right) = \frac{1}{s - 1},\;\widetilde{g}\left( s\right) = \frac{1}{s + 2}, \] | \[ \widetilde{f}\left( s\right) \widetilde{g}\left( s\right) = \frac{1}{\left( {s - 1}\right) \left( {s + 2}\right) } = \frac{\frac{1}{3}}{s - 1} - \frac{\frac{1}{3}}{s + 2} = \mathcal{L}\left\lbrack {f * g}\right\rbrack \left( s\right) . \] | Yes |
Find a function \( f \) that satisfies the integral equation\n\n\[ f\left( t\right) = 1 + {\int }_{0}^{\infty }f\left( {t - u}\right) \sin {udu},\;t \geq 0. \] | Solution. Suppose that \( f \in \mathcal{E} \). Then we can transform the equation to get\n\n\[ \widetilde{f}\left( s\right) = \frac{1}{s} + \widetilde{f}\left( s\right) \cdot \frac{1}{{s}^{2} + 1} \]\n\nfrom which we solve\n\n\[ \widetilde{f}\left( s\right) = \frac{{s}^{2} + 1}{{s}^{2}} \cdot \frac{1}{s} = \frac{{s}^{... | Yes |
Solve the initial value problem\n\n\\[ \n{y}^{\prime \prime } + 4{y}^{\prime } + {13y} = {\delta }^{\prime }\left( t\right) ,\;y\left( {0 - }\right) = {y}^{\prime }\left( {0 - }\right) = 0.\n\\] | Transformation gives\n\n\\[ \n\left( {{s}^{2} + {4s} + {13}}\right) \widetilde{y} = s \Leftrightarrow \widetilde{y} = \frac{s}{{\left( s + 2\right) }^{2} + 9} = \frac{s + 2}{{\left( s + 2\right) }^{2} + 9} - \frac{2}{3} \cdot \frac{3}{{\left( s + 2\right) }^{2} + 9}.\n\\]\nThe solution is found to be\n\n\\[ \ny\left( t... | Yes |
Find the general solution of the differential equation \( {y}^{\prime \prime } + 3{y}^{\prime } + {2y} = \delta \) . | Solution. It should be wellknown that the solution can be written as the sum of the general solution \( {y}_{H} \) of the corresponding homogeneous equation \( {y}^{\prime \prime } + 3{y}^{\prime } + {2y} = 0 \), and one particular solution \( {y}_{P} \) of the given equation. We easily find \( {y}_{H} = {C}_{1}{e}^{-t... | Yes |
If \( {a}_{n} = 1 \) for all \( n \geq 0 \), the \( Z \) transform is | \[ \mathop{\sum }\limits_{{n = 0}}^{\infty }\frac{1}{{z}^{n}} = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\left( \frac{1}{z}\right) }^{n} = \frac{1}{1 - \frac{1}{z}} = \frac{z}{z - 1} \] which is convergent for all \( z \) such that \( \left| z\right| > 1 \) | Yes |
If we know that \( {a}_{0} = 1,{a}_{1} = 2 \) and\n\n\[ \n{a}_{n + 2} = 3{a}_{n + 1} - 2{a}_{n},\;n = 0,1,2,\ldots ,\n\]\n\nfind a formula for \( {a}_{n} \) . | To solve the problem in Example 3.21, we multiply the formula (3.9) by \( {z}^{-n} \) and add up for \( n = 0,1,2,\ldots \) :\n\n\[ \n\mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n + 2}{z}^{-n} = 3\mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n + 1}{z}^{-n} - 2\mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{z}^{-n}.\... | Yes |
Example 3.22. We have already seen that if \( a = {\left\{ {\lambda }^{n}\right\} }_{0}^{\infty } \), then | \[ \mathcal{Z}\left\lbrack a\right\rbrack \left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{\lambda }^{n}{z}^{-n} = \frac{z}{z - \lambda },\;\left| z\right| > \left| \lambda \right| . \] | Yes |
The sequence \( a = \{ n!{\} }_{0}^{\infty } \) has no \( Z \) transform, because the series \( \mathop{\sum }\limits_{{n = 0}}^{\infty }n!{z}^{-n} \) diverges for all \( z \) . | As stated at the beginning of this section, a sufficient (and actually necessary) condition for \( A\left( z\right) \) to exist is that the numbers \( {a}_{n} \) grow at most exponentially: \( \left| {a}_{n}\right| \leq M{R}^{n} \) for some numbers \( M \) and \( R \) . It is easy to see that this condition implies the... | No |
Theorem 3.7 (i) The transformation \( \mathcal{Z} \) is linear, i.e., \[ \mathcal{Z}\left\lbrack {\lambda a}\right\rbrack \left( z\right) = \lambda \mathcal{Z}\left\lbrack a\right\rbrack \left( z\right) ,\;\left| z\right| > {\sigma }_{a}, \] \[ \mathcal{Z}\left\lbrack {a + b}\right\rbrack \left( z\right) = \mathcal{Z}\... | Proof. The assertions follow rather immediately from the definitions. | No |
Example 3.25. Example 3.23 and rule (ii) give us the transform of the sequence \( {\left\{ {\lambda }^{n}/n!\right\} }_{0}^{\infty } \) | \[ \Lambda \left( z\right) = {e}^{1/\left( {z/\lambda }\right) } = {e}^{\lambda /z}. \] | Yes |
Find a formula for the so-called FIBONACCI numbers, which are defined by \( {f}_{0} = {f}_{1} = 1,{f}_{n + 2} = {f}_{n + 1} + {f}_{n} \) for \( n \geq 0 \) . | Solution. Let \( F = \mathcal{Z}\left\lbrack f\right\rbrack \) . If we \( Z \)-transform the recursion formula, using (iii) from the theorem, we get\n\n\[ \n{z}^{2}F\left( z\right) - {z}^{2} - z = \left( {{zF}\left( z\right) - z}\right) + F\left( z\right) \n\]\n\nwhence \( \left( {{z}^{2} - z - 1}\right) F\left( z\righ... | Yes |
Find \( x\left( t\right), t = 0,1,2,\ldots \), from the equation \[ \mathop{\sum }\limits_{{k = 0}}^{t}{3}^{-k}x\left( {t - k}\right) = {2}^{-t},\;t = 0,1,2,\ldots \] | Solution. The left-hand side is the convolution of \( x \) and the function \( t \mapsto \) \( {\left( 1/3\right) }^{t} \), so that taking \( Z \) transforms of both members gives \[ \frac{z}{z - \frac{1}{3}} \cdot X\left( z\right) = \frac{z}{z - \frac{1}{2}}. \] (We have used the result of Example 3.22.) We get \[ z) ... | Yes |
Let \( {\left\{ {a}_{n}\right\} }_{n = 0}^{\infty } \) be a sequence having a \( Z \) transform \( A\left( z\right) \), and define a function \( f \) by\n\n\[ f\left( t\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\delta }_{n}\left( t\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}\delta \left( {... | Thus, via a change of variable \( z = {e}^{s} \), the two transforms are more or less the same thing. | Yes |
The polynomial \( {P}_{1}\left( s\right) = {s}^{2} + {2s} + 2 \) has zeroes \( s = \) \( - 1 \pm i \) . Both have real part -1, so that the device described by the equation \( {y}^{\prime \prime } + 2{y}^{\prime } + {2y} = x\left( t\right) \) is stable. | In contrast, the polynomial \( {P}_{2}\left( s\right) = {s}^{2} + {2s} - 1 \) has zeroes \( s = - 1 \pm \sqrt{2} \) . One of these is positive, which implies that the corresponding black box is unstable. Finally, the polynomial \( {P}_{3}\left( s\right) = {s}^{2} + 1 \) has zeroes \( s = \pm i \) . These have real part... | Yes |
The difference equation \( y\left( {t + 2}\right) + \frac{1}{2}y\left( {t + 1}\right) + \frac{1}{4}y\left( t\right) = x\left( t\right) \) has \( P\left( z\right) = {z}^{2} + \frac{1}{2}z + \frac{1}{4} \) with zeroes \( z = - \frac{1}{4} \pm \frac{\sqrt{3}}{4}i \) . These satisfy \( \left| z\right| = \frac{1}{2} < 1 \) ... | The equation\n\n\[ y\left( {t + 3}\right) + {2y}\left( {t + 2}\right) - y\left( {t + 1}\right) + {2y}\left( t\right) = x\left( t\right) \]\n\nis unstable. This can be seen from the constant term \( \left( { = 2}\right) \) of the characteristic polynomial; as is well known, this term is (plus or minus) the product of th... | Yes |
Define \( f \) by saying that \( f\left( t\right) = {e}^{t} \) for \( - \pi < t < \pi \) and \( f\left( {t + {2\pi }}\right) = f\left( t\right) \) for all \( t \) . (This leaves \( f\left( t\right) \) undefined for \( t = \left( {{2n} + 1}\right) \pi \), but this does not matter. The value of a function at one point or... | \[ {c}_{n} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{t}{e}^{-{int}}{dt} = \frac{1}{2\pi }{\int }_{-\pi }^{\pi }{e}^{\left( {1 - {in}}\right) t}{dt} = \frac{1}{2\pi }{\left\lbrack \frac{{e}^{\left( {1 - {in}}\right) t}}{1 - {in}}\right\rbrack }_{t = - \pi }^{\pi } \] \[ = \frac{{e}^{\pi - {in\pi }} - {e}^{-\pi + {in\p... | Yes |
Example 4.2. Let \( f \) be an odd function with period \( {2\pi } \), that satisfies \( f\left( t\right) = \left( {\pi - t}\right) /2 \) for \( 0 < t < \pi \) . Find its Fourier series! (See Figure 4.2.) | Solution. Notice that the description as given actually determines the function completely (except for its value at one point in each period, which does not matter). Because the function is odd we have \( {a}_{n} = 0 \) and\n\n\[ \n{b}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }\frac{\pi - t}{2}\sin {ntdt} \n\]\n\n\[ \n= \f... | Yes |
Example 4.3. Let \( f\left( t\right) = {t}^{2} \) for \( \left| t\right| \leq \pi \) and define \( f \) outside of this interval by proclaiming it to have period \( {2\pi } \) (draw a picture!). Find the Fourier series of this function. | Solution. Now the function is even, and so \( {b}_{n} = 0 \) and\n\n\[ \n{a}_{n} = \frac{2}{\pi }{\int }_{0}^{\pi }{t}^{2}\cos {nt}\;{dt}\overset{\begin{matrix} {n \neq 0} \\ \downarrow \\ \downarrow \end{matrix}}{ = }\frac{2}{\pi }{\left\lbrack {t}^{2}\frac{\sin {nt}}{n}\right\rbrack }_{0}^{\pi } - \frac{2}{n\pi }{\in... | Yes |
Lemma 4.1 Suppose that \( f \) is as in the definition of Fourier series. Then\n\n1. The sequence of Fourier coefficients is bounded; more precisely,\n\n\[ \left| {c}_{n}\right| \leq \frac{1}{2\pi }{\int }_{\mathbf{T}}\left| {f\left( t\right) }\right| {dt}\;\text{ for all }n. \]\n\n2. The Fourier coefficients tend to z... | Proof. For the \( {c}_{n} \) we have\n\n\[ \left| {c}_{n}\right| = \frac{1}{2\pi }\left| {{\int }_{\mathbf{T}}f\left( t\right) {e}^{-{int}}{dt}}\right| \leq \frac{1}{2\pi }{\int }_{\mathbf{T}}\left| {f\left( t\right) }\right| \left| {e}^{-{int}}\right| {dt} = \frac{1}{2\pi }{\int }_{\mathbf{T}}\left| {f\left( t\right) ... | Yes |
Lemma 4.2\n\n\[ \n{D}_{N}\left( u\right) \mathrel{\text{:=}} \frac{1}{\pi }\left( {\frac{1}{2} + \mathop{\sum }\limits_{{n = 1}}^{N}\cos {nu}}\right) = \frac{1}{2\pi }\mathop{\sum }\limits_{{n = - N}}^{N}{e}^{inu} = \frac{\sin \left( {N + \frac{1}{2}}\right) u}{{2\pi }\sin \frac{1}{2}u}. \n\] | Proof. The equality of the two sums follows easily from Euler's formulae. Let us then start from the \ | No |
Lemma 4.3\n\n\[ \n{F}_{N}\left( u\right) \mathrel{\text{:=}} \frac{1}{N + 1}\mathop{\sum }\limits_{{n = 0}}^{N}{D}_{n}\left( u\right) = \frac{1}{{2\pi }\left( {N + 1}\right) }{\left( \frac{\sin \frac{1}{2}\left( {N + 1}\right) u}{\sin \frac{1}{2}u}\right) }^{2}. \n\] | The proof can be done in a way similar to Lemma 4.2 (or in some other way). It is left as an exercise. | No |
Lemma 4.4 The Fejér kernel \( {F}_{N}\left( u\right) \) has the following properties:\n\n1. \( {F}_{N} \) is an even function, and \( {F}_{N}\left( u\right) \geq 0 \) .\n\n2. \( {\int }_{-\pi }^{\pi }{F}_{N}\left( u\right) {du} = 1 \) .\n\n3. If \( \delta > 0 \), then \( \mathop{\lim }\limits_{{N \rightarrow \infty }}{... | Proof. Property 1 is obvious. Number 2 follows from\n\n\[{\int }_{-\pi }^{\pi }{D}_{n}\left( u\right) {du} = \frac{1}{\pi }{\int }_{-\pi }^{\pi }\left( {\frac{1}{2} + \cos u + \cdots + \cos {nu}}\right) {du} = 1,\;n = 0,1,2,\ldots, N,\]\n\nand the fact that \( {F}_{N} \) is the mean of these Dirichlet kernels. Finally,... | Yes |
Theorem 4.2 If \( f \) is continuous on \( \mathbf{T} \) and its Fourier coefficients \( {c}_{n} \) are such that \( \sum \left| {c}_{n}\right| \) is convergent, then the Fourier series is convergent with sum \( f\left( t\right) \) for all \( t \in \mathbf{T} \), and the convergence is even uniform on \( \mathbf{T} \). | The uniform convergence follows using the Weierstrass \( M \) -test just as at the beginning of this chapter. | No |
Theorem 4.3 Suppose that \( f \) is piecewise continuous and that all its Fourier coefficients are 0 . Then \( f\left( t\right) = 0 \) at all points where \( f \) is continuous. | In fact, all the partial sums are zero and the series is trivially convergent, and by Theorem 4.2 it must then converge to the function from which it is formed. | No |
Corollary 4.1 If two continuous functions \( f \) and \( g \) have the same Fourier coefficients, then \( f = g \) . | Proof. Apply Theorem 4.3 to the function \( h = f - g \) . | No |
Example 4.4. Find a solution \( y\left( t\right) \) with period \( {2\pi } \) of the differential-difference equation \( {y}^{\prime }\left( t\right) + {2y}\left( {t - \pi }\right) = \sin t, - \infty < t < \infty \) . | Solution. Assume the solution to be the sum of a \ | No |
Theorem 4.5 Suppose that \( f \) has period \( {2\pi } \), and suppose that \( {t}_{0} \) is a point where \( f \) has one-sided limiting values and (generalized) one-sided derivatives. Then the Fourier series of \( f \) converges for \( t = {t}_{0} \) to the mean value \( \frac{1}{2}\left( {f\left( {{t}_{0} + }\right)... | We emphasize that if \( f \) is continuous at \( {t}_{0} \), the sum of the series is simply \( f\left( {t}_{0}\right) \) . At a point where the function has a jump discontinuity, the sum is instead the mean value of the right-hand and left-hand limits. | Yes |
Find the Fourier series of the odd function of period 2 that is described by \( f\left( t\right) = t\left( {1 - t}\right) \) for \( 0 \leq t \leq 1 \) . Using the result, find the value of the sum\n\n\[ \n{s}_{1} = \mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{\left( -1\right) }^{k}}{{\left( 2k + 1\right) }^{3}} \n\] | Solution. Since the function is odd, we compute a sine series. The coefficients are\n\n\[ \n{b}_{n} = 2{\int }_{0}^{1}t\left( {1 - t}\right) \sin {n\pi tdt} = \text{ (integrations by parts) } = \frac{4\left( {1 - {\left( -1\right) }^{n}}\right) }{{n}^{3}{\pi }^{3}}, \n\]\n\nwhich is zero for all even values of \( n \) ... | Yes |
Consider the function of Example 4.2 on page 78. Its Fourier series can be written\n\n\\[ \nf\\left( t\\right) \\sim \\mathop{\\sum }\\limits_{{n = 1}}^{\\infty }\\frac{\\sin {nt}}{n} = \\mathop{\\sum }\\limits_{{n \\neq 0}}\\frac{1}{2in}{e}^{int}.\n\\]\n\n(Notice that the last version is correct - the minus sign in th... | The derivative of \\( f \\) consists of an \ | No |
Consider the vector \( \mathbf{z} = \left( {1, i}\right) \in {\mathbf{C}}^{2} \). With the usual formula for the scalar product we would get \( {\left| \mathbf{z}\right| }^{2} = 1 \cdot 1 + i \cdot i = 1 - 1 = 0 \). The length of the vector would be 0, which is no good. Still more strange would be the case for the vect... | By considering the one-dimensional space \( \mathbf{C} \), we can get an idea for a more suitable definition. The vector, or number, \( z = x + {iy} \) is normally identified with the vector \( \left( {x, y}\right) \) in \( {\mathbf{R}}^{2} \). The length of this vector is the number \( \left| z\right| = \sqrt{{x}^{2} ... | Yes |
In \( {\mathbf{C}}^{n} \), we can define an inner product (using natural notation) by the formula | \[ \langle \mathbf{z},\mathbf{w}\rangle = {z}_{1}{\bar{w}}_{1} + {z}_{2}{\bar{w}}_{2} + \cdots + {z}_{n}{\bar{w}}_{n} \] | Yes |
Let \( C\left( {a, b}\right) \) be the set of continuous, complex-valued functions defined on the compact interval \( \left\lbrack {a, b}\right\rbrack \), and put\n\n\[ \langle f, g\rangle = {\int }_{a}^{b}f\left( x\right) \overline{g\left( x\right) }{dx} \] | The fact that this is an inner product is almost trivial, except possibly for condition 4. That implication follows from the fact that if a continuous function is non-negative on an interval and its integral is 0 , then the function must indeed be identically 0 (see textbooks on calculus). | Yes |
\[ \left| {\langle u, v\rangle }\right| \leq \left\| u\right\| \left\| v\right\| \;\text{(Cauchy–Schwarz inequality)} \] | Proof. (a) If \( u = 0 \), then both members are 0 and the statement is true. Thus, let us assume that \( u \neq 0 \) . Put \( \alpha = - \langle v, u\rangle /\langle u, u\rangle \) . Then,\n\n\[ 0 \leq \parallel {\alpha u} + v{\parallel }^{2} = \langle {\alpha u} + v,{\alpha u} + v\rangle = \alpha \overline{\alpha }\l... | Yes |
Let \( C\left( \mathbf{T}\right) \) consist of the continuous complex-valued functions on the unit circle \( \mathbf{T} \), and let the inner product be defined by\n\n\[ \langle f, g\rangle = \frac{1}{2\pi }{\int }_{\mathbf{T}}f\left( x\right) \overline{g\left( x\right) }{dx} \]\n\nLet \( {\varphi }_{k}\left( x\right) ... | The sequence \( {\left\{ {\varphi }_{k}\right\} }_{k = - \infty }^{\infty } \) is thus an orthonormal set in \( C\left( \mathbf{T}\right) \) . | Yes |
Theorem 5.3 (Least squares approximation) Let \( {\left\{ {\varphi }_{k}\right\} }_{k = 1}^{N} \) be an orthonormal set in an inner product space \( V \) and let \( u \) be a vector in \( V \) . Among all the linear combinations \( \Phi = \mathop{\sum }\limits_{{k = 1}}^{N}{\gamma }_{k}{\varphi }_{k} \), the one that m... | Proof. For brevity, we write \( \left\langle {u,{\varphi }_{k}}\right\rangle = {c}_{k} \) . We get\n\n\[ \parallel u - \Phi {\parallel }^{2} = \langle u - \Phi, u - \Phi \rangle = \left\langle {u - \mathop{\sum }\limits_{{k = 1}}^{N}{\gamma }_{k}{\varphi }_{k}, u - \mathop{\sum }\limits_{{k = 1}}^{N}{\gamma }_{k}{\varp... | Yes |
Theorem 5.4 The ON system \( {\left\{ {\varphi }_{j}\right\} }_{j = 1}^{\infty } \) is complete in \( V \) if and only if for every \( u \in V \) it holds that\n\n\[ \parallel u{\parallel }^{2} = \mathop{\sum }\limits_{{j = 1}}^{\infty }{\left| \left\langle u,{\varphi }_{j}\right\rangle \right| }^{2} \]\n\n(the PARSEVA... | The proof follows from (5.1). | No |
Theorem 5.6 If the \( {ON} \) system \( {\left\{ {\varphi }_{j}\right\} }_{j = 1}^{\infty } \) is complete in \( V \), then\n\n\[ \langle u, v\rangle = \mathop{\sum }\limits_{{j = 1}}^{\infty }\left\langle {u,{\varphi }_{j}}\right\rangle \overline{\left\langle v,{\varphi }_{j}\right\rangle } \]\n\nfor all \( u, v \in V... | Proof. Let \( {P}_{n}\left( u\right) \) be the projection of \( u \) on to the subspace spanned by the \( n \) first \( \varphi \) ’s:\n\n\[ {P}_{n}\left( u\right) = \mathop{\sum }\limits_{{j = 1}}^{n}\left\langle {u,{\varphi }_{j}}\right\rangle {\varphi }_{j} \]\n\nBy Theorem 5.2 we have\n\n\[ \left\langle {{P}_{n}\le... | Yes |
Let \( I = \left( {-\pi ,\pi }\right) \) or, equivalently, interpret \( I \) as the unit circle \( \mathbf{T} \), and take \( w\left( x\right) = 1 \) . The inner product here is almost the same as in Example 5.5, page 108. There we showed that the functions \( {\varphi }_{n}\left( x\right) = {e}^{inx} \) are orthogonal... | The formula for projections on to an orthogonal set of vectors has the form\n\n\[ \n{P}_{N}\left( f\right) = \mathop{\sum }\limits_{{n = 1}}^{N}\frac{\left\langle f,{\varphi }_{n}\right\rangle }{\left\langle {\varphi }_{n},{\varphi }_{n}\right\rangle }{\varphi }_{n} = \mathop{\sum }\limits_{{n = 1}}^{N}\frac{\left\lang... | Yes |
In the same space as in the preceding example we study the system consisting of firstly all the functions \( {\varphi }_{n}\left( x\right) = \cos {nx}, n = 0,1,2,\ldots \) , secondly all the functions \( {\psi }_{n}\left( x\right) = \sin {nx}, n = 1,2,3,\ldots \) | Using some suitable trigonometric formula (or Euler's formulae), it is easily proved that all these functions are mutually orthogonal, and furthermore \( \left\langle {{\varphi }_{0},{\varphi }_{0}}\right\rangle = {2\pi } \) , while all the other members of the system have the square of the norm equal to \( \pi \) . | No |
In Sec. 4.1, we saw that the odd function \( f \) with period \( {2\pi } \) that is described by \( f\left( t\right) = \left( {\pi - t}\right) /2 \) for \( 0 < t < \pi \) has Fourier series \( \mathop{\sum }\limits_{1}^{\infty }\left( {\sin {nt}}\right) /n \) . Parseval’s formula looks like this | \[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{2}} = \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n}^{2} = \frac{1}{\pi }{\int }_{\mathbf{T}}{\left( f\left( t\right) \right) }^{2}{dt} = \frac{2}{\pi }{\int }_{0}^{\pi }\frac{{\left( \pi - t\right) }^{2}}{4}{dt} = \frac{1}{2\pi }{\left\lbrack \frac{{\left( \p... | Yes |
Theorem 5.9 Let \( f \in {L}^{2}\left( \mathbf{T}\right) \) . If the Fourier series of \( f \) is integrated term by term over a finite interval \( \left( {a, b}\right) \), the series obtained is convergent with the sum \( {\int }_{a}^{b}f\left( t\right) {dt} \) . | Proof. Let us first assume that the interval \( \left( {a, b}\right) \) is contained within a period, for instance, \( - \pi \leq a < b \leq \pi \) . We define a function \( g \) on \( \mathbf{T} \) by letting \( g\left( t\right) = 1 \) for \( a < t < b \) and 0 otherwise. We compute the Fourier coefficients of \( g \)... | Yes |
Theorem 5.10 (The Weierstrass approximation theorem) An arbitrary continuous function \( f \) on a compact interval \( K \) can be approximated uniformly arbitrarily well by polynomials. | In greater detail, the assertion is the following: If \( K \) is compact, \( f \) : \( K \rightarrow \mathbf{C} \) is continuous and \( \varepsilon \) is any positive number, then there exists a polynomial \( P\left( x\right) \) such that \( \left| {f\left( x\right) - P\left( x\right) }\right| < \varepsilon \) for all ... | Yes |
Find the polynomial \( p\\left( x\\right) \) of degree at most 4 that minimizes the value of\n\n\[ \n{\\int }_{-1}^{1}{\\left| \\sin \\pi x - p\\left( x\\right) \\right| }^{2}{dx} \n\] | Solution. By the general theory, the required polynomial can be obtained as the orthogonal projection of \( f\\left( x\\right) = \\sin {\\pi x} \) onto the first five Legendre polynomials:\n\n\[ \np\\left( x\\right) = \\mathop{\\sum }\\limits_{{k = 0}}^{4}\\frac{\\left\\langle f,{P}_{k}\\right\\rangle }{\\left\\langle ... | Yes |
Find the polynomial \( p\\left( x\\right) \) of degree at most 2, that minimizes the value of the integral\n\n\[ \n{\\int }_{0}^{\\infty }{\\left| {x}^{3} - p\\left( x\\right) \\right| }^{2}{e}^{-x}{dx} \n\] | The norm occurring in the problem belongs together with the Laguerre polynomials. These even happen to be orthonormal (not merely orthogonal), which means that the wanted polynomial must be\n\n\[ \np\\left( x\\right) = \\left\\langle {f,{L}_{0}}\\right\\rangle {L}_{0}\\left( x\\right) + \\left\\langle {f,{L}_{1}}\\righ... | Yes |
Let \( f\left( x\right) = \sqrt{1 - {x}^{2}} \) for \( \left| x\right| \leq 1 \) . Find the polynomial \( p\left( x\right) \), of degree at most 3, that minimizes the value of the integral\n\n\[ \n{\int }_{-1}^{1}{\left| f\left( x\right) - p\left( x\right) \right| }^{2}\frac{dx}{\sqrt{1 - {x}^{2}}}.\n\] | Solution. This inner product belongs with the Chebyshev polynomials. Because of \ | No |
Let us study the problem of heat conduction in a completely isolated rod, where there is no exchange of heat with the surroundings, not even at the end points. As before, the rod is represented by the interval \( \\left\\lbrack {0,\\pi }\\right\\rbrack \), and the temperature at the point \( x \) at time \( t \) is den... | This problem is largely similar to the previous one, and we attack it by the same means (cf. Sec. 1.4). Thus we start by looking for nontrivial solutions of the homogeneous sub-problem (E)+(B), and we try to find solutions having the form \( u\\left( {x, t}\\right) = X\\left( x\\right) T\\left( t\\right) \). Substituti... | No |
Example 6.2. Let us now modify the original problem in a few different ways. We let the rod be the interval \( \\left( {0,2}\\right) \), and the end points are kept each at a constant temperature, but these are different at the two ends. To be specific, say that \( u\\left( {0, t}\\right) = 2 \) and \( u\\left( {2, t}\... | Here, separation of variables cannot be applied directly; an important feature of that method is making use of the homogeneity of the conditions, enabling us to add solutions to each other to obtain other solutions. For this reason, we now start by homogenizing the problem in the following way. Since the boundary value... | Yes |
In our next variation we consider a rod with a built-in source of heat. The length of the rod is again \( \pi \), and we assume that at the point with coordinate \( x \) there is generated an amount of heat per unit of time and unit of length along the rod, described by the function \( \sin \left( {x/2}\right) \) . It ... | Here there is an inhomogeneity in the equation itself. We try to amend this by using the same trick as in Example 2: put \( u\left( {x, t}\right) = v\left( {x, t}\right) + \varphi \left( x\right) \) and substitute into (E) and (B). (Do it!) We conclude that it would be very\n\nnice to have\n\[ \n{\varphi }^{\prime \pri... | No |
We want to find \( u\left( {x, t}\right) \) for \( t > 0 \) and all the interesting values of \( x \) . This is collected into a problem of the following appearance:\n\n(E) \( \;{u}_{xx} = {u}_{tt},\;0 < x < \pi ,\;t > 0 \) ;\n\n(B) \( \;u\left( {0, t}\right) = u\left( {\pi, t}\right) = 0,\;t > 0 \) ;\n\n(6.11)\n\n\[ \... | Again,(E) and (B) are homogeneous conditions. The usual attempt \( u\left( {x, t}\right) \) \( = X\left( x\right) T\left( t\right) \) this time leads up to this set of coupled problems:\n\n\[ \left\{ {\begin{array}{l} {X}^{\prime \prime }\left( x\right) + {\lambda X}\left( x\right) = 0, \\ X\left( 0\right) = X\left( \p... | Yes |
Find \( u\left( {x, y}\right) \) that solves \( {u}_{xx} + {u}_{yy} = 0,0 < x < \pi ,0 < y < \pi \), with boundary conditions \( u\left( {x,0}\right) = \sin {3x} - 3\sin {2x} \) for \( 0 < x < \pi \), \( u\left( {x,\pi }\right) = u\left( {0, y}\right) = u\left( {\pi, y}\right) = 0,0 < x, y < \pi \). | Solution. Draw a picture! We have a homogeneous equation,\n\n(E)\n\n\[ \n{u}_{xx} + {u}_{yy} = 0 \n\] \n\ntogether with three homogeneous boundary conditions,\n\n\[ \n\left( {\mathrm{B}}_{1,2,3}\right) \;u\left( {0, y}\right) = u\left( {\pi, y}\right) = 0,\;u\left( {x,\pi }\right) = 0, \n\] \n\nand one non-homogeneous ... | Yes |
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