advanced mathematical methods for scientists and engineers bender pdf

... 1566 32.7.3 Cosine and Sine Transform in Terms of the Fourier Transform . . . . . . . . . . . . . . . . . . 1568 32.8 Solving Differential Equations with the Fourier Cosine and Sine Transforms . . . ... = p(x)/q(x) where p(x) and q(x) are rational quadratic polynomials. Give possible formulas for p(x) and q(x). Hint, Solution 12 33 The Gamma Function 1605 33.1 Euler’s Formul a . . . . . . . ... also contains the word Scientists or Engineers ” the advanced book may be quite suitable for actually learning the material. xxvii 43.3 The Method of Characteristics and the Wave Equation ....

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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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... δ − √ x is a decreasing function of x and an increasing function of δ for positive x and δ. Bound this function for fixed δ. Consider any positive δ and . For what values of x is 1 x − 1 x + δ > ... Consider y = x 3 and the point x = 0. The function is differentiable. The derivative, y  = 3x 2 is positive for x < 0 and positive for 0 < x. Since y  is not identically zero and the sign ... a n > 0 for all n > 200, and lim n→∞ a n = L, then L > 0. 4. If f : R → R is continuous and lim x→∞ f(x) = L, then for n ∈ Z, lim n→∞ f(n) = L. 5. If f : R → R is continuous and lim n→∞ f(n)...

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... of x and an increasing function of δ for positive x and δ. Thus for any fixed δ, the maximum value of √ x + δ − √ x is bounded by √ δ. Therefore on the interval (0, 1), a sufficient condition for ... −2) −1/3 The first derivative exists and is nonzero for x = 2. At x = 2, the derivative does not exist and thus x = 2 is a critical point. For x < 2, f  (x) < 0 and for x > 2, f  (x) > 0. ... satisfying, e(x, δ) ≤ (δ), for all x in the closed interval. Since (δ) is continuous and increasing, it has an inverse δ(). Now note that |f(x) − f(ξ)| <  for all x and ξ in the closed interval...

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... − 1 4 cos  t 2  i − 1 4 sin  t 2  j. See Figure 5.8 for plots of position, ve locity and acceleration. Figure 5.8: A Graph of Position and Velocity and of Position and Acceleration Solution 5.2 If r(t) has ... Figure 5.12.) We find the volume obtained by rotating the 172 Set x = 2 and x = −2 to solve for a and b. Hint 4.16 Expanding the integral in partial fractions, x + 1 x 3 + x 2 − 6x = x + 1 x(x ... |r  (t)|. Differentiation Formulas. Let f(t) and g(t) be vector functions and a(t) be a scalar function. By writing out components you can verify the differentiation formulas: d dt (f · g) = f  ·...

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... y 2 e ı arctan(x,y) . Cartesian form is convenient for addition. Polar form is convenient for multiplication and division. Example 6.3.1 We write 5 + ı7 in polar form. 5 + ı7 = √ 74 e ı arctan(5,7) We ... u 0 , u 1 , u 2 and u 3 are real numbers and ı,  and k are objects which satisfy ı 2 =  2 = k 2 = −1, ı = k, ı = −k and the usual associative and distributive laws. Show that for any quaternions ... 4. 6.3 Polar Form Polar form. A complex number written in Cartesian form, z = x + ıy, can be converted polar form, z = r(cos θ + ı sin θ), using trigonometry. Here r = |z| is the modulus and θ = arctan(x,...

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... real-variable counterparts. 7.1 Curves and Regions In this section we introduce curves and regions in the complex plane. This material is necessary for the study of branch points in this chapter and later for contour integration. Curves. ... function, f(z) = z. In Cartesian coordinates and Cartesian form, the function is f(z) = x + ıy. The real and imaginary components are u(x, y) = x and v(x, y) = y. (See Figure 7.9.) In modulus -2 -1 0 1 2 x -2 -1 0 1 2 y -2 -1 0 1 2 -2 -1 0 1 2 x -2 -1 0 1 2 x -2 -1 0 1 2 y -2 -1 0 1 2 -2 -1 0 1 2 x Figure ... arctangent that is between 0 and π. The domain and a plot of the selected values of the arctangent are shown in Figure 7.8. CONTINUE. 7.4 Cartesian and Modulus-Argument Form We can write a function...

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... in Cartesian form and z = r e ıθ in polar form. e u+ıv = r e ıθ We equate the modulus and argument of this expression. e u = r v = θ + 2πn u = ln r v = θ + 2πn With log z = u + ıv, we have a formula for ... See Figure 7.18 and Figure 7.19 for plots of the real and imaginary parts of the cosine and sine, respectively. Figure 7.20 shows the modulus of the cosine and the sine. The hyperbolic sine and cosine. ... infinity and its only singularity is at z = 1, the only possi bili ties for branch points are at z = 1 and z = ∞. Since log  1 z −1  = −log(z −1) and log w has branch points at zero and infinity,...

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... Hints Cartesian and Modulus-Argument Form Hint 7.1 Hint 7.2 Trigonometric Functions Hint 7.3 Recall that sin(z) = 1 ı2 ( e ız − e −ız ). Use Result 6.3.1 to convert between Cartesian and modulus-argument form. Hint ... on which f(0) = ı √ 6. Write out an explicit formula for the value of the function on this branch. Figure 7.33: Four candidate sets of branch cuts for ((z − 1)(z − 2)(z − 3)) 1/2 . Hint, Solution 294 ... 4 . Figure 7.48 first shows the branch cuts and their s tereographic projections and then shows the stereographic projections alone. Solution 7.21 1. For each value of z, f(z) = z 1/3 has three...

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... condition for the analyticity of f(z). Let φ(x, y) = u(x, y) + ıv(x, y) where u and v are real-valued functions. We equate the real and imaginary parts of Equation 8.1 to obtain another form for the ... ı) −1/3 = 3 √ r e ıθ/3 1 3 √ s e −ıφ/3 1 3 √ t e −ıψ/3 = 3  r st e ı(θ−φ−ψ)/3 we have an explicit formula for computing the value of the function for this branch. Now we compute f (1) to see if we chose the correct ranges for the angles. (If not, we’ll just ... . . 2π), (2π . . . 4π), . . .}. Now we choose ranges for θ and φ and see if we get the desired branch. If not, we choose a different range for one of the angles. First we choose the ranges θ ∈...

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... equations for à and are satised if and only if the Cauchy-Riemann equations for u and v are satisfied. The continuity of the first partial derivatives of u and v implies the same of à and . Thus ... =  x 3 (1+ı)−y 3 (1−ı) x 2 +y 2 for z = 0, 0 for z = 0. Show that the partial derivatives of u and v with respect to x and y exist at z = 0 and that u x = v y and u y = −v x there: the Cauchy-Riemann ... function. Solution 8.11 We write the real and imaginary parts of f(z) = u + ıv. u =  x 4/3 y 5/3 x 2 +y 2 for z = 0, 0 for z = 0. , v =  x 5/3 y 4/3 x 2 +y 2 for z = 0, 0 for z = 0. The Cauchy-Riemann...

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... Consider analytic functions f 1 (z) and f 2 (z) defined on the domains D 1 and D 2 , respectively. Suppose that D 1 ∩ D 2 is a region or an arc and that f 1 (z) = f 2 (z) for all z ∈ D 1 ∩ D 2 . (See ... converges uniformly for D 1 = |z| ≤ r < 1. Since the derivative also converges in this domain, the function is analytic there. 440 Figure 8.7: The velocity potential φ and stream function ψ for Φ(z) ... Substitute this expression for v into the equation for ∂v/∂x. −y e −x sin y − x e −x cos y + e −x cos y + F  (x) = −y e −x sin y − x e −x cos y + e −x cos y Thus F  (x) = 0 and F (x) = c. v = e −x (y...

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... contour and do the integration. z − z 0 = e ıθ , θ ∈ [0 . . . 2π)  C (z − z 0 ) n dz =  2π 0 e ınθ ı e ıθ dθ =     e ı(n+1)θ n+1  2π 0 for n = −1 [ıθ] 2π 0 for n = −1 =  0 for n = −1 ı2π for ... u y ) dx dy + ı  D (u x − v y ) dx dy = 0 Since the two integrands are continuous and vanish for all C in Ω, we conclude that the integrands are identically zero. This implies that the Cauchy-Riemann ... ıφ dy) =  D (ıφ x − φ y ) dx dy = 0 Since the integrand, ıφ x − φ y is continuous and vanishes for all C in Ω, we conclude that the integrand is identically zero. This implies that the Cauchy-Riemann...

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... converges if and only i f for any  > 0 there exists an N such that |a n − a m | <  for all n, m > N. The Cauchy convergence criterion is equivalent to the definition we had before. For some ... integrals along C 1 and C 2 . (We could also see this by deforming C onto C 1 and C 2 .)  C =  C 1 +  C 2 We use the Cauchy Integral Formula to evaluate the integrals along C 1 and C 2 .  C (z 3 + ... integral  C e zt z 2 (z + 1) dz. There are singularities at z = 0 and z = −1. Let C 1 and C 2 be contours around z = 0 and z = −1. See Figure 11.6. We deform C onto C 1 and C 2 .  C =  C 1 +  C 2 520 11.4 Exercises Exercise...

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... , converges for α > 1 and diverges for α ≤ 1. Hint, Solution 564 Example 12.3.2 Convergence and Uniform Convergence. Consider the series log(1 − z) = − ∞  n=1 z n n . This series converges for |z| ... necessary condition for uniform convergence. The Weierstrass M-test can succeed only if the series is uniformly and absolutely convergent. Example 12.2.1 The series f(x) = ∞  n=1 sin x n(n + 1) is uniformly and ... Thus this series is not uniformly convergent in the domain |z| ≤ 1, z = 1. The series is uniformly convergent for |z| ≤ r < 1. 545 12.2.2 Uniform Convergence and Continuous Functions. Consider...

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Ngày tải lên: 06/08/2014, 01:21

Ngày tải lên: 06/08/2014, 01:21

Ngày tải lên: 06/08/2014, 01:21

Ngày tải lên: 06/08/2014, 01:21

Ngày tải lên: 06/08/2014, 01:21