advanced mathematical methods for scientists and engineers solutions

... 4.13 Let u = x, and dv = sin x dx. Hint 4.14 Perform integration by parts three succes sive times. For the first one let u = x 3 and dv = e 2x dx. Hint 4.15 Expanding the integrand in partial fractions, ... − 2)(x + 2) = a (x − 2) + b (x + 2) 1 = a(x + 2) + b(x −2) 139 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 ...  n=0 (n∆x)∆x Hint 4.7 Let u = sin x and dv = sin x dx. Integration by parts will give you an equation for  π 0 sin 2 x dx. Hint 4.8 Let H  (x) = h(x) and evaluate the integral in terms of

Ngày tải lên: 06/08/2014, 01:21

... cut beteen z = 1 and z = 13/12. This puts a branch cut between w = ∞ and w = 0 and thus separates the branches of the logarithm. Figure 7.54 shows the branch cuts in the positive and negative sheets ... ı(θ−φ−ψ)/3 e = 3 st 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 ... z = ±1 and each go to infinity. We can also make the function single-valued with a branch cut that connects the points z = ±1. This is because log(z + 1) and −log(z − 1) change by ı2π and −ı2π,

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... the analytic function is of the form, az 3 + bz 2 + cz + ıd, 457 with a, b and c complex-valued constants and d a real constant. Substituting z = x + ıy and expanding products yields, a  x 3 ... the contour and do the integration z − z0 = eıθ , θ ∈ [0 2 ) 2 eınθ ı eıθ dθ (z − z0 )n dz = C = 0   2 eı(n+1)θ n+1 0 [ıθ ]2 0 for n = −1 = for n = −1 0 2 for n = −1 for n = −1 ... (z, −ı log z). We integrate to obtain an expression for f (z 2 ). 1 2 f  z 2  = u(z, −ı log z) + const We make a change of variables and solve for f(z). f(z) = 2u  z 1/2 , − ı 2 log z  + const.

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

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... |z + 1| > 2 for |z + 1| > 2 for |z + 1| > 2 1 − 2 n−1 (z + 1)n , = n=0 2n , (z + 1)n n 1 2 , (z + 1)n 1 − 2n , (z + 1)n+1 ∞ for |z + 1| > 1 and |z + 1| > 2 for |z + 1| > ... 2 for r < 1, for r = 1, for r > 1 In the above example we evaluated the contour integral by parameterizing the contour This approach is only feasible when the integrand is... integrand ... n=−∞  ı 2  −n−1 z n , for |z| < 2 = −1  n=−∞ (−ı2) n+1 z n , for |z| < 2 619 − 1 z − 2 = 1/2 1 − z/2 = 1 2 ∞  n=0  z 2  n , for |z/2| < 1 = ∞  n=0 z n 2 n+1 , for |z| < 2 − 1 z

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... integrand below the branch cut is a constant times the value of the integrand above the branch cut. After demonstrating that the integrals along C  and C R vanish in the limits as  → 0 and R ... 2 for n ∈ Z+ for n = 0 Now we consider... at z = ±1 ± 2 The poles at z = −1 + 2 and z = 1 − 2 are inside the path of integration We evaluate the integral with Cauchy’s Residue Formula ... → 0 for some α > −1 then the integral on C  will vanish as  → 0. f(z)  z β as z → ∞ for some β < −1 then the integral on C R will vanish as R → ∞. Below the branch cut the integrand

Ngày tải lên: 06/08/2014, 01:21

... (x), for x ≥ 0, for x ≤ 0 The initial condition for y− demands that the solution be continuous Solving the two problems for positive and negative x, we obtain y(x) = e1−x , e1+x , for ... for the reader... 0 for x = 0 for x > 0 Since sign x is piecewise defined, we solve the two problems, y+ + y+ = 0, y− − y− = 0, y+ (1) = 1, y− (0) = y+ (0), for x > 0 for x < 0, and ... y  y −y 2 = 1 We expand in partial fractions and integrate.  1 y − 1 y −1  y  = 1 ln |y| − ln |y −1| = x + c 781 We have an implicit equation for y(x). Now we solve for y(x). ln     y

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... 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 ... homogeneous solutions. The characteristic equation for the matrix is χ(λ) =     4 − λ −2 8 −4 − λ     = λ 2 = 0 λ = 0 is an eigenvalue of multiplicity 2. Thus the Jordan canonical form of

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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 ... multiplicity 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, ... coefficients of tα−1 log t and tα−1 to determine xi and η (A − αI)xi = 0, 896 (A − αI)η = xi These equations have solutions because λ = α has generalized eigenvectors of first and second order Note

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... fundamental set of solutions at the point x = ξ, subs titute (x − ξ) for x in the above solutions. 17.1.3 Higher Order Equations The constant coefficient equation of order n has the form L[y] = y (n) ... equation for y, but note that it is a first order equation for y We can solve directly for y d dx 2 3/ 2 x y =0 3 2 y = c1 exp − x3/2 3 exp Now we just integrate to get the solution for ... solution... solutions that are real valued when x is real and positive 944 17 .3 Exact Equations Exact equations have the form d F (x, y, y , y , ) = f (x) dx If you can write an equation in the form

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... x−a For a = 0, we have y1 = [xa ]a=0 = 1, y2 = Consider the solutions d a x da = ln x a=0 xa − x−a a Clearly y1 is a solution for all a For a = 0, y2 is a linear combination of xa and x−a and ... eax − e−ax y2 = lim a→0 a Consider the solutions eαx − e−αx α→a α Clearly y1 is a solution for all a For a = 0, y2 is a linear combination of eax and e−ax and is thus a solution Since the coefficient ... independent solutions for u(ξ) is {eıξ , e−ıξ } Since eıξ + e−ıξ eıξ − e−ıξ and sin ξ = , ı2 another linearly independent set of solutions is {cos ξ, sin ξ} cos ξ = The general solution for y(x)

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... Equation Normal Form Hint 19.1 Transform the equation to normal form Transformations of the Independent Variable Integral Equations Hint 19.2 Transform the equation to normal form and then apply ... form and then apply the scale transformation... transformation x = λξ + µ Hint 19 .3 Transform the equation to normal form and then apply the scale transformation x = λξ Hint 19.4 Make the ... x = dx dt Hint 19.5 Transform the equation to normal form 1 034 19 .7 Solutions The Constant Coefficient Equation Normal Form Solution 19.1 1 4... pn−1 (x) dx u(x) transforms the differential equation

Ngày tải lên: 06/08/2014, 01:21

... 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 ... Likewise for hyperbolic sines and hyperbolic cosines. Example 21.2.1 Consider y  − 2y  + y = t 2 . The homogeneous solutions are y 1 = e t and y 2 = t e t . We guess a particular solution of the form ... (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

Ngày tải lên: 06/08/2014, 01:21

... (ξ) for a ≤ x ≤ ξ, y 2 (x)y 1 (ξ) p(ξ)W (ξ) for ξ ≤ x ≤ b, where y 1 and y 2 are non-trivial homogeneous solutions that satisf y B 1 [y 1 ] = B 2 [y 2 ] = 0, and W (x) is the Wronskian of y 1 and ... (ξ) for x < ξ, u 1 (ξ)u 2 (x) W (ξ) for x > ξ, u 1 and u 2 are solutions of the homogeneous differential equation that satisfy the left and right boundary conditions, respectively, and W ... How does the solution for λ = 1 differ from that for λ = 1? The λ = 1 case provides an example of resonant forcing Plot the solution for resonant and non-resonant forcing Hint, Solution

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... p(ξ)W (ξ) for a ≤ x ≤ ξ, for ξ ≤ x ≤ b Here v1 and v2... satisfy the left and right homogeneous boundary conditions 1157 Since g(x; ξ) is a solution of the homogeneous equation for x = ... G(x|ξ) = for 0 ≤ x ≤ ξ, for ξ ≤ x ≤ 1 Solution 21.14 The differential equation for the Green function is G − G = δ(x − ξ), G(0|ξ) = G(∞|ξ) = 0 Note that sinh(x) and e−x are homogeneous solutions ... functions of period 2π and 2π/λ. This solution is plotted in Figure 21.5 on the interval t ∈ [0, 16π] for the values λ = 1/4, 7/8, 5/2. 1140 Figure 21.6: Resonant Forcing For λ = 1, we have y =

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... integer and n ≥ 0, the series for one of the solutions reduces to an even polynomial of degree 2n. 4. Show that if α = 2n + 1, with n an integer and n ≥ 0, the series for one of the solutions ... Legendre polynomials P n (x) and P m (x) must satisfy this relation for α = n and α = m respectively. By multiplying the first relation by P m (x) and the second by P n (x) and integrating by parts ... independent solutions of the equation as Taylor series about x = 0. For what values of x do the series converge? Show that for certain values of λ, called eigenvalues, one of the solutions is

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... eigenvalues before you determine the eigenfunctions Thus this formula could not be used to compute the eigenvalues However, we can often use the formula to obtain information about the eigenvalues before ... eigenvalues as λn and the eigenfunctions as φn for n ∈ Z+ For the moment we assume that λ = 0 is not an eigenvalue and that the eigenfunctions are real-valued We expand the function f ... We will find the series expansion of the Heaviside function first by expanding directly and then by integrating the expansion for the delta function. Finding the series expansion of H(x) directly.

Ngày tải lên: 06/08/2014, 01:21

...  −x for − π ≤ x < 0 π −2x for 0 ≤ x < π. The Fourier series converges to the function defined by ˆ f(x) =          0 for x = −π −x for − π < x < 0 π/2 for x = 0 π −2x for ... converge for a fairly general class of functions. Let f(x − ) denote the left 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 ... Note that this formula is valid for m = 0, 1, 2, . . Similarly, we can multiply by sin(mx) and integrate to solve for b m . The result is b m = 1 π  π −π f(x) sin(mx) dx. a n and b n are called

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... (−1)2 nπ(2 − π n2 ) nπ(2−π n2 ) for odd n for even n 1382 sin(nπx) dx which has the solutions, λn = nπ , b−a nπ(x − a) b−a φn = sin , n ∈ N We expand the solution and the inhomogeneity in the eigenfunctions ... there is no value of a for which both cos a and sin a vanish, the system is not orthogonal for any interval of length π First note that π cos nx dx = for n ∈ N If n = m, n ≥ and m ≥ then π cos nx ... problem is φ + λ2 φ = 0, φ(a) = φ(b) = 0, 1398 which has the solutions, nπ nπ(x − a) , φn = sin b−a b−a We expand the solution and the inhomogeneity in the eigenfunctions λn = ∞ yn sin y(x)

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