Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 7 pdf

... dish.1 62 Let u = x 2 and dv =e2xdx. Then du = 2x dx and v =1 2 e2x.x3e2xdx =1 2 x3e2x−3 2 1 2 x 2 e2x−xe2xdxx3e2xdx =1 2 x3e2x−34x 2 e2x+3 2 xe2xdxLet ... = x and dv =e2xdx. Then du = dx and v =1 2 e2x.x3e2xdx =1 2 x3e2x−34x 2 e2x+3 2 1 2 xe2x−1 2 e2xdxx3e2xdx =1 2 x3e2x−34x 2 e2x+34xe2x−38e2x+CSolution ... =0−1√−x dx + 2 0√x dx=10√x dx + 2 0√x dx= 2 3x3 /2 10+ 2 3x3 /2  2 0= 2 3+ 2 3 2 3 /2 = 2 3(1 + 2 √ 2) Solution 4.3ddxx 2 xf(ξ) dξ = f(x 2 )ddx(x 2 ) − f(x)ddx(x)=...

... mapping. 24 3 2. z + 2 2 − ız=x + ıy + 2 2 − ı(x − ıy)=x + ı(y + 2) 2 − y −ıx=x + ı(y + 2) 2 − y −ıx 2 − y + ıx 2 − y + ıx=x (2 − y) − (y + 2) x (2 − y) 2 + x 2 + ıx 2 + (y + 2) (2 − y) (2 − ... of z 2 is|z 2 | =z 2 z 2 = zz = (x + ıy)(x −ıy) = x 2 + y 2 . 25 0 -2 -101 2 x -2 -101 2 y-505 -2 -101 2 x -2 -101 2 yFigure 7. 10: A few branches of arg(z). -2 -101 2 x -2 -101 2 y01 2 -2 -101 2 x -2 -101 2 x -2 -101 2 y -2 0 2 -2 -101 2 xFigure ... arg(z). -2 -101 2 x -2 -101 2 y01 2 -2 -101 2 x -2 -101 2 x -2 -101 2 y -2 0 2 -2 -101 2 xFigure 7. 11: Plots of |z| and Arg(z). 25 1-1 1-0.4-0 .2 0 .2 0.4Figure 6 .20 : |(z)| + 5|(z)| =...

... 1) 2 + y 2 ) 2 =−(x − 1) 2 + y 2 ((x − 1) 2 + y 2 ) 2 and 2( x − 1)y((x − 1) 2 + y 2 ) 2 = 2( x − 1)y((x − 1) 2 + y 2 ) 2 The Cauchy-Riemann equations are each identities. The first partial derivatives ... (z 2 ) has a second order pole at z = 0 and first order poles at z = (nπ)1 /2 , n ∈ Z±.limz→0z 2 sin (z 2 )= limz→02z2z cos (z 2 )= limz→0 2 2 cos (z 2 ) − 4z 2 sin (z 2 )= 1 379 for ... cosh y −ı cos x sinh y 2. x 2 − y 2 + x + ı(2xy − y)Hint, SolutionExercise 8.5f(z) is analytic for all z, (|z| < ∞). f (z1+ z 2 ) = f (z1) f (z 2 ) for all z1 and z 2 . (This is known...

... =1 2 Logx 2 + y 2 + ı Arctan(x, y).4 52 2. We calculate the first partial derivatives of u and v.ux= 2 ex 2 −y 2 (x cos(2xy) − y sin(2xy))uy= 2 ex 2 −y 2 (y cos(2xy) + x sin(2xy))vx= ... xdirection.f(z) = ux+ ıvxf(z) = 2 ex 2 −y 2 (x cos(2xy) − y sin(2xy)) + 2 ex 2 −y 2 (y cos(2xy) + x sin(2xy))f(z) = 2 ex 2 −y 2 ((x + ıy) cos(2xy) + (−y + ıx) sin(2xy))Finding the derivative ... y − y cos y).f(z) = 2uz 2 , −ız 2 = 2 e−z /2 z 2 sin−ız 2 + ız 2 cos−ız 2 + c= ıze−z /2 ı sinız 2 + cos−ız 2 + c= ıze−z /2 e−z /2 + c= ıze−z+cExample...

... = 2 0eınθıeıθdθ=eı(n+1)θn+1 2 0 for n = −1[ıθ] 2 0 for n = −1=0 for n = −1 2 for n = −1 2. We parameterize the contour and do the integration.z − z0= 2 +eıθ, θ ∈ [0 . . . 2 )C(z − z0)ndz = 2 0 2 ... axis and is defined continuously on the real axis.)Hint, Solution481CLog z dz≤C|Log z||dz|=π /2 −π /2 |ln 2 + ıθ |2 dθ≤ 2 π /2 −π /2 (ln 2 + |θ|) dθ= 4π /2 0(ln 2 ... areux= 3x 2 − 3y 2 − 2y,uy= −6xy − 2x + 1.The derivative of f(z) isf(z) = ux− ıuy= 3x 2 − 2y 2 − 2y + ı(6xy − 2x + 1).On the real axis we havef(z = x) = 3x 2 − ı2x + ı.Using...

... zz3z=−ı+ 2 2! d 2 dz 2 (z3+ z + ı) sin zz + ız=0= 2 (−ı sinh(1)) + ıπ 2 3z 2 + 1z + ı−z3+ z + ı(z + ı) 2 cos z+6zz + ı− 2( 3z 2 + 1)(z + ı) 2 + 2( z3+ z + ı)(z ... formula.Czz 2 + 1dz =C1 /2 z −ıdz +C1 /2 z + ıdz=1 2 2 +1 2 2 = 2 3.Cz 2 + 1zdz =Cz +1zdz=Cz dz +C1zdz= 0 + 2 = 2 Solution 11.3Let C be the ... cosine and sine to simplify the integral.Ceazzdz = 2  2 0eaeıθeıθıeıθdθ = 2  2 0ea(cos θ+ı sin θ)dθ = 2  2 0ea cos θ(cos(sin θ) + ı sin(sin θ)) dθ = 2  2 0ea...

... closed form. (See Exercise 12. 9.)N−1n=1sin(nx) =0 for x = 2 kcos(x /2) −cos((N−1 /2) x) 2 sin(x /2) for x = 2 kThe partial sums have infinite discontinuities at x = 2 k, k ∈ Z. The partial ... n))5.∞n=1ln (2 n)ln (3n) + 16.∞n=01ln(n + 20 ) 7. ∞n=04n+ 13n− 2 8.∞n=0(Logπ 2) n9.∞n =2 n 2 − 1n4− 110.∞n =2 n 2 (ln n)n11.∞n =2 (−1)nln1n 12. ∞n =2 (n!) 2 (2n)!13.∞n =2 3n+ ... n)n11.∞n =2 (−1)nln1n 12. ∞n =2 (n!) 2 (2n)!13.∞n =2 3n+ 4n+ 55n− 4n− 35 62 Im(z)Re(z)RR 2 1Im(z)Re(z)RR 2 1Cr1r 2 zCCC1 2 zFigure 12. 5: Contours for a Laurent Expansion in an Annulus.Example 12. 6.1...

... π)Hint 12. 23CONTINUEHint 12. 24CONTINUEHint 12. 25Hint 12. 26Hint 12. 27 Hint 12. 28Hint 12. 29Hint 12. 30CONTINUE581Solution 12. 22 cos z = −cos(z − π)= −∞n=0(−1)n(z −π)2n(2n)!=∞n=0(−1)n+1(z ... series.Solution 12. 4∞n=1(−1)n+1n=∞n=112n − 1−12n=∞n=11(2n − 1)(2n)<∞n=11(2n − 1) 2 <1 2 ∞n=11n 2 =π 2 12 Thus the series is convergent.5 87 The series ... integrand. 20 .∞n =2 ln nn11/10Use the integral test.Hint 12. 4Group the terms.1 −1 2 =1 2 13−14=1 12 15−16=130··· 577 (b)f(z) =1 + z1 − z, f(ı) = ıf(z) = 2 (1 − z) 2 ,...