advanced mathematical techniques for scientists and engineers pdf

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... term in the sum is the leading order behavior For a few examples, • For sin x ∼ x − x3/6 + x5/120 − · · · as x → 0, the leading order behavior is x • For f (x) ∼ ex(1 − 1/x + 1/x2− · · · ) as x ... the differential equation for y.us an approximate solution If one of the terms in the equation for s is much smaller than the others then we say thatthe remaining terms form a dominant balance ... equation for y Trang 8We make the assumption that s00  (s0)2 as x → ∞ and we know that ν2/x2  1 as x → ∞ Thus we drop thesetwo terms from the equation to obtain an approximate equation for s.(s0)2+

Ngày tải lên: 06/08/2014, 01:21

... (x−) denote theleft limit of f (x) and f (x+) denote the right limit Example 28.2.1 For the function defined f (x) = ( 0 for x < 0, x + 1 for x ≥ 0,the left and right limits at x = 0 are Trang ... by sin(mx) and integrate to solve for bm The result is bm = 1π Z π −π f (x) sin(mx) dx an and bn are called Fourier coefficients Although we will not show it, Fourier series converge for a fairly ... dxMultiplying by cos(mx) and integrating will enable us to solve for am Z π −π f (x) cos(mx) dx = πam am = 1π Z π −π f (x) cos(mx) dx Trang 3Note that this formula is valid for m = 0, 1, 2, Similarly,

Ngày tải lên: 06/08/2014, 01:21

... transforms and sine transforms.ˆG(ω; ξ) = −π f (t) g(|x − t|) − g(x + t)
dt G(x; ξ) = −π a 12π Z ∞ 0 Trang 24the integrand is analytic there, the integral is zero G(x; ξ) = 0 for x < ξ For ... transform?Hint, Solution Exercise 32.20 Show that F [f (x)] = 1 2(Fc[f (x) + f (−x)] − ıFs[f (x) − f (−x)])where F , Fc and Fs are respectively the Fourier transform, Fourier cosine transform and ... We derived this formula from Euler’s formula which is valid only in the left half-plane However, the product formula is valid for all z except z = 0, −1, −2, 33.4 Weierstrass’ Formula Trang 40The

Ngày tải lên: 06/08/2014, 01:21

... turn by cos(kt) and sin(kt) and integrating from 0 to 1. This would yields a set of two linear equations for c 1 and c 2 . 2. u(x) = λ  π 0 ∞  n=1 sin nx sin ns n u(s) ds We expand u(x) in a ... 2141 For even n, su bstituting φ(x) = 1 yields λ = 1 πa 0 . For n and m both even or odd, sub stituting φ(x) = cos(mx) or φ(x) = sin(mx) yields λ = 1 πa m . For even n we have the eigenvalues and ... differential equation is φ(x) =  a + bx   √ for λ = 0 √ a cos λx + b sin λx   a cosh √−λx + b sinh √−λx for λ > 0 for λ < 0 We see that for λ = 0 and λ < 0 only the trivial solution satisfies

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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 =

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 0 0 0 ··· 0 λ           856 Jordan Canonical Form. A matrix J is in Jordan canonical form if all the elements are zero except for Jordan blocks J k along the diagonal. J =      ...    The Jordan canonical form of a matrix is obtained with the similarity transformation: J = S −1 AS, where S is the matrix of the generalized eigenvectors of A and the generalized eigenvectors ... eigenvalues and associated eigenvectors of A [HINT: notice that λ = −1 is a root of the characteristic polynomial of A.] 2 Use the results from part (a) to construct eAt and therefore the...

Ngày tải lên: 06/08/2014, 01:21

... 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) ... greater than one, we will have solutions of the form, xitα , xitα log t + ηtα , xitα (log t)2 + ηtα log t + ζtα , ., analogous to the form of the solutions for a constant coefficient system, xi eατ ,

Ngày tải lên: 06/08/2014, 01:21

... 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

Ngày tải lên: 06/08/2014, 01:21

... 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...

Ngày tải lên: 06/08/2014, 01:21