mathematical methods for scientists and engineers mcquarrie ebook

... Uniform Convergence 536 12.2.1 Tests for Uniform Convergence 537 12.2.2 Uniform Convergence and Continuous Functions 539 12.3 Uniformly Convergent Power Series 539 12.4 Integration and ... Transform 1560 32.6 The Fourier Cosine and Sine Transform 1562 32.6.1 The Fourier Cosine Transform 1562 32.6.2 The Fourier Sine Transform 1563 32.7 Properties of the Fourier Cosine and ... Cosine and Sine Transform 1564 32.7.1 Transforms of Derivatives 1564 32.7.2 Convolution Theorems 1566 32.7.3 Cosine and Sine Transform in Terms of the Fourier Transform 1568 32.8 Solving

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... with u = f(n+1)(x − ξ) and dv = n!1ξn Hint 4.11 The arc length from 0 to x is Z x 0 a(x − 2) + b(x + 2) 1 = a(x + 2) + b(x − 2) Trang 8Set x = 2 and x = −2 to solve for a and b.Hint 4.17 Z 4 0 ... cos x + sin x + CSolution 4.14 Let u = x3 and dv = e2x dx Then du = 3x2dx and v = 12e2x x2e2x dx Trang 15Let u = x2 and dv = e2x dx Then du = 2x dx and v = 12 e2x.Z  1 2x e 2x−12 A(x − 2) + ... define ∆xi = xi+1− xi and ∆x = maxi∆xi and choose ξi ∈ [xi xi+1] Z b a √ x dx = Z 1 0 √ x dx + Z 2 0 + 2 3x 3/2 2 0 ddx Trang 20Solution 4.4First we expand the integrand in partial fractions

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... curves and regions in the complex plane This material is necessary for the study ofbranch points in this chapter and later for contour integration Curves Consider two continuous functions x(t) and ... function of r and θ with the substitutions z = x + ıy and z = r eıθ, respectively Then we can separate the real and imaginary components or write the function in modulus-argument form, f (z) = ... 7.4.2 Consider the functions f (z) = z, f (z) = z3 and f (z) = 11−z We write the functions in terms of rand θ and write them in modulus-argument form f (z) = z = r eıθ Trang 371 − 2r cos θ + r2

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... the function w = z2− 2
(z + 2)
1/3.Construct and define a branch so that the resulting cut is one line of finite extent and w(2) = 2 What is w(−3) forthis branch? What are the limiting values ... SolutionsCartesian and Modulus-Argument Form Solution 7.1 Let w = u + ıv We consider the strip 2 < x < 3 as composed of vertical lines Consider the vertical line: z = c + ıy, y ∈ R for constant ... 4c2v2, v ∈ RThe boundaries of the region, x = 2 and x = 3, are respectively mapped to the parabolas:  See Figure 7.35 for depictions of the strip and its image under the mapping The mapping is

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... cut beteen z = 1 and z = 13/12 Thisputs a branch cut between w = ∞ and w = 0 and thus separates the branches of the logarithm Figure7.54 shows the branch cuts in the positive and negative sheets ... z1/2 z 7→ z + 5For the positive branch of g(z), the branch cut is mapped to the line x = 5 and the z plane is mapped to thehalf-plane x > 5 log(w) has branch points at w = 0 and w = ∞ It is ... 1 and each go to infinity We can also make the function single-valued with a branch cut that connects two of the points z = −1, 0, 1 andanother branch cut that starts at the remaining point and

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... functions, f (x, y) and g(x, y) They are said to be functionally dependent if there is a an h(g) such that f (x, y) = h(g(x, y)) f and g will be functionally dependent if and only if their Jacobian ... = h(g(x, y)) f and g will be functionally dependent if and only if their Jacobian vanishes If f and g are functionally dependent, then the derivatives of f are fx = h0(g)gx fy = h0(g)gy.Thus we ... ıvy ≤ Z b a |f (x)| |dx| ≤ (b − a) max a≤x≤b|f (x)| Trang 15Now we prove the analogous result for the modulus of a contour integral. Z C f (z) dz =

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... Let {an } and {bn } be the positive and negative terms in the sum, respectively, ordered in decreasing magnitude Note that both ∞ an and ∞ bn are divergent Devise a n=1 n=1 method for alternately ... Integrate the series for 1/z Differentiate the series for 1/z Integrate the series for Log z 580 Hint 12.21 Evaluate the derivatives of ez at z = Use Taylor’s Theorem Write the cosine and sine in terms ... criterion for series In particular, consider |SN +1 − SN | Hint 12.2 CONTINUE Hint 12.3 ∞ n ln(n) n=2 Use the integral test ∞ n=2 ln (nn ) Simplify the summand ∞ ln √ n ln n n=2 Simplify the summand

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... cosine and sine integrals have the form, Z ∞ 0 f (x) cos(ωx) dx and Z ∞ 0 f (x) sin(ωx) dx If f (x) is even/odd then we can evaluate the cosine/sine integral with the method we developed for Fourier ... rational function we expand it in partial fractions to obtain a form that is convenient to integrate We use the value of the integrals of f (z) to determine the constants, a, b, c and d I |z|=1 a z ... ı/2)14(1 − z/2) −2 = 4 − ı8 ∞ X n=0 −2n   −z2 n , for |z/2| < 1 = 4 − ı8 z2  1 −2z   −2z Trang 101/2 < |z| < 2 and 2 < |z| For |z| < 1/2, we havek3/3k (k + 1)3/3k+1 = 3 lim

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... integrand is bounded, the integral exists α x2+ α2 dx = ı2 Z 1/α 0 Trang 9Figure 13.8: The real and imaginary part of the integrand for several values of α.Note that the integral exists for all ... starts and ends a distance of  from z = 1 Let C be the negative, circular arc of radius  with center at z = 1 that joins the endpoints of Cp Let Ci, be the union of Cp and C (Cp stands for Principal ... integral exists for all nonzero real α and that < 1 Trang 10Now note that when α = 0, the integrand is real Of course the integral doesn’t converge for this case, but if wecould assign some

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... the integrand below the branch cut is a constant timesthe value of the integrand above the branch cut After demonstrating that the integrals along C and CR vanish in thelimits as  → 0 and R → ... R → ∞ The value of the integrand below thebranch cut, z = x eı2π is f (x)(log x + ı2π) Taking the limit as  → 0 and R → ∞, we have Z ∞ 0 If f (z)  zα as z → 0 for some α > −1 then the integral ... On the circle of radius , the integrand is o(−1) Since the length of C is 2π, the integral on C vanishes as  → 0 On the circle of radius R, the integrand is o(R−1) Since the length of CR

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... separate the dependent and independent variables Trang 9and integrate to find the solution.dy dx = xy 2 y−2dy = x dxZ Trang 10We have an implicit equation for y(x) Now we solve for y(x).ln y y − ... one-parameter family of functions y(x; c) Here y is a function of the variable x and theparameter c For simplicity, we will write y(x) and not explicitly show the parameter dependence Example 14.3.1 The equation ... dependent variable y(x) and its derivative and the parameter c If we algebraically eliminate c between the two equations, the eliminant will be a first order differential equation for y(x) Thus we see

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... function However, the integral form above is as nice as any otherand we leave the answer in that form.2 erf(x)  Trang 16Solution 14.6We determine the integrating factor and then integrate the equation ... α−1e−x+c e−αx for α 6= 1,(c + x) e−x for α = 1 Trang 174 8 12 16 1Figure 14.9: The Solution for a Range of α The solution which satisfies the initial condition is y = ( 1 α−1(e−x+(α − 2) e−αx) for α ... we make the transformation z = 1ζ and study the behavior of f (1/ζ) at ζ = 0.Example 14.8.8 Let’s examine the behavior of sin z at infinity We make the substitution z = 1/ζ and find theLaurent

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... a constant times xi Thus we look for the solution whose second component vanishes tosimplify the algebra.  e2t (d) To find a third solution we substutite the form x = xi(t2/2) e2t+ηt e2t+ζ e2t ... looping through the following three steps until all thethe Nk are zero: 1 Select the largest k for which Nk is positive Find a generalized eigenvector xk of rank k which is linearlyindependent ... generalized eigenvector ζ Note that ζ is onlydetermined up to a constant times xi Thus we look for the solution whose second component vanishes to Trang 35simplify the algebra. Trang 36Thus

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... eigenvectors, a and b then atα and btα are linearly independent solutions If λ = α has only one linearly independent eigenvector, a, then atα is a solution We look for a second solution of the form x ... − Q + R = 0 is a necessary and sufficient condition for this equation to be exact Hint, Solution Exercise 16.2 Determine an equation for the integrating factor µ(x) for Equation 16.1 900 (16.1) ... of independent variable τ = log t, y(τ ) = x(t), will transform (15.4) into a constant coefficient system dy = Ay dτ Thus all the methods for solving constant coefficient systems carry over directly

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... equation in the formExample 17.4.1 Consider the equation y00+√ xy0 = 0 This is a second order equation for y, but note that it is a first order equation for y0 We can solve directly for y0 ddx exp ... sines and cosines y = c1cos x + c2sin x + c3x cos x + c4x sin x Trang 9y = xλ will transform the differential equation into an algebraic equation.(λ − λ1)(λ − λ2) = 0 If the two roots, λ1 and λ2, ... ξ) for x in the above solutions. The constant coefficient equation of order n has the form L[y] = y(n)+ an−1y(n−1)+ · · · + a1y0+ a0y = 0 (17.3) Trang 6The substitution y = eλx will transform

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... the following canonical forms: y = 0 Hint, Solution Trang 2219.6 HintsThe Constant Coefficient Equation Normal Form Hint 19.1 Transform the equation to normal form Transformations of the Independent ... becomes (f0)2u00+ (f00+ pf0)u0+ qu = 0 For this to be a constant coefficient equation we must have (f0)2 = c1q, and f00+ pf0 = c2q, Trang 14for some constants c1 and c2 Solving the first condition,f0 ... = y1/2d Trang 6Chapter 19Transformations and Canonical Forms Prize intensity more than extent Excellence resides in quality not in quantity The best is always few and rare abundance lowers value

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... present general methods which work for any linear differential equation and any inhogeneity. Thus one might wonder why I would present a method that works only for some simple problems. (And why it ... (x), Bj [y] = bj , for j = 1, , n has the solution... the form G(x|ξ) = c1 + c2 x d1 + d2 x for x < ξ for x > ξ Applying the two boundary conditions, we see that c1 = 0 and d1 = −d2 The ... 1 for x < 0 for x > 0 The integral of the Heaviside function is the ramp function, r(x) x H(t) dt = r(x) = −∞ 0 x for x < 0 for x > 0 The derivative of the delta function is zero for

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... constant How does the solution for λ = 1 differ from that for λ 6= 1? The λ = 1 case provides an example of resonant forcing Plot the solution for resonant and non-resonant forcing Hint, Solution Exercise ... p(ξ)W (ξ) for a ≤ x ≤ ξ, y2(x)y1(ξ) p(ξ)W (ξ) for ξ ≤ x ≤ b, where y1 and y2 are non-trivial homogeneous solutions that satisfy B1[y1] = B2[y2] = 0, and W (x) is the Wronskian of y1 and y2. Trang ... function for (b) approaches that for (c) as a → ∞.Hint, Solution Exercise 21.16 1 For what values of λ does the problem have a unique solution? Find the Green functions for these cases 2 For what

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... 21.5: Non-resonant ForcingFor λ 6= 1, this is y = 12 Z t 0 cos(t − τ − λτ ) − cos(t − τ + λτ )
dτ = 12 = 12 The solution is the sum of two periodic functions of period 2π and 2π/λ This solution ... solution is plotted in Figure 21.5 on the interval t ∈ [0, 16π] for the values λ = 1/4, 7/8, 5/2 Trang 9Figure 21.6: Resonant ForcingFor λ = 1, we have y = 12 Z t 0 cos(t − 2τ ) − cos(tau)
dτ = ... 1 1 + log x = x + x log x − x log x = x Trang 3Writing the inhomogeneous equation in the standard form:Trang 4The Wronskian of the homogeneous solutions isW [cos x, sin x] = = cos2t + sin2t

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... is the angle from a to b and n is a unit vector that is orthogonal to a and b and in the direction s uch that the ordered triple of vectors a, b and n form a right-handed system. 29 a b b θ b Figure ... x z yj i k z k j i y x Figure 2.7: Right and left handed coordinate systems. You can visualize the direction of a ì b by applying the right hand rule. Curl the fingers of your right hand in the direction from ... arbitrary vectors a and b. We can write b = b ⊥ + b  where b ⊥ is orthogonal to a and b  is parallel to a. Show that a ì b = a ì b . Finally prove the distributive law for arbitrary b and c. Hint...

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