advanced mathematical methods for scientists and engineers pdf

... 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

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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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... 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 ... = 0, has the solution y = cx a . Thus for the second order equation we will try a solution of the form y = x λ . The substitution 940 y = x λ will transform the differential equation into an algebraic ... 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 of an exact equation,

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...  −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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... formula for the Gamma function ∞ Γ(z) = z n=1 1+ n z 1+ z n −1 We derived this formula from Euler’s formula which is valid only in the left half-plane However, the product formula is valid for ... [f (x) − f (−x)]) where F, Fc and Fs are respectively the Fourier transform, Fourier cosine transform and Fourier sine transform Hint, Solution F[f (x)] = 1576 Exercise 32.21 Find u(x) as the solution ... Hankel’s formula converges for all complex z For non-positive, integral z the integral does not vanish Thus because of the sine term the Gamma function has simple poles at z = 0, −1, −2, For positive,

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... 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

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... 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.

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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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... 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

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... (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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... For positive A, the solution is bounded at the origin only for c = 0 For A = 0, there are no bounded solutions For negative A, the solution is bounded there for any value of c and ... function However, the integral form above is as nice as any other and we leave the answer in that form 2 dy − 2xy(x) = 1, dx y(0) = 1 We determine the integrating factor and then integrate the ... Figure 14.10: The Solution as α → 0 and α → ∞ In the limit as α → ∞ we have, 1 e−x +(α − 2) e−αx α→∞ α − 1 α − 2 −αx e = lim α→∞ α − 1 1 for x = 0, = 0 for x > 0 lim y(x) = lim α→∞ This

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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 ... 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...

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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

Ngày tải lên: 06/08/2014, 01:21

... 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 of e−ax in this linear combination is non-zero, it is linearly independent to y1 For ... 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 is thus a solution For a = 0, y2 is one half the derivative of xa evaluated ... solution for u has two constants of integration However, the solution for y should only have one constant of integration as it satisfies a first order equation Write y in terms of the solution for u and

Ngày tải lên: 06/08/2014, 01:21

... 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 ... that satisfy the left and right boundary conditions are c1 (x − a) and c2 (x − b) Thus the Green’s function has the form G(x|ξ) = c1 (x − a), c2 (x − b), for x ≤ ξ for x ≥ ξ Imposing continuity

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 ... (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

Ngày tải lên: 06/08/2014, 01:21

... 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 ... u 1 (x)u 2 (ξ) W (ξ) for x < ξ, u 1 (ξ)u 2 (x) W (ξ) for x > ξ. The solution for u is u =  b a G(x|ξ)f(ξ) dξ. Thus if there is a unique solution for v, the solution for y is y = v +  b ... (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

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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 = ... 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 = ... cosh(aL) Thus the Green function is g(x; ξ) = − sinh(ax) cosh(a(ξ−L)) a cosh(aL) for. .. know a closed-form formula for the an we can calculate the an in order by substituting into the difference

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