Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 5 pps

... n)n11.∞n =2 (−1)nln1n 12. ∞n =2 (n!) 2 (2n)!13.∞n =2 3n+ 4n+ 5 5n− 4n− 3 5 62 Im(z)Re(z)RR 2 1Im(z)Re(z)RR 2 1Cr1r 2 zCCC1 2 zFigure 12. 5: Contours for a Laurent ... 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 ... + (−1) 2 z 2 + (−1)3z3+ ···= 1 − z + z 2 − z3+ ···Example 12. 5. 4 Find the ﬁrst few terms in the Taylor series expansion of1√z 2 + 5z + 6about the origin. 55 4-4-3 -2 -1-3 -2 -1Figure...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 1 pps

... 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 )= 1379 for ... function.Solution 8.11We write the real and imaginary parts of f(z) = u + ıv.u =x4/3y 5/ 3x 2 +y 2 for z = 0,0 for z = 0., v =x 5/ 3y4/3x 2 +y 2 for z = 0,0 for z = 0.The Cauchy-Riemann...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 4 ppsx

... equation p(z) = z6−5z 2 +10 = 0 lie in the annulus 1 < |z| < 2. Exercise 11 .5 Evaluate as a function of tω =1 2 Ceztz 2 (z 2 + a 2 )dz, 50 5Integral Test.Result 12. 1 .2 If the coeﬃcients ... 1)dz.There are singularities at z = 0 and z = −1.Let C1 and C 2 be contours around z = 0 and z = −1. See Figure 11.6. We deform C onto C1 and C 2 .C=C1+C 2 52 0 11.4 ExercisesExercise 11.1What ... 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...

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 5 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 ... region about the y axis, only the-3 -2 -1 1 2 3-3 -2 -11 2 3 -2 0 2 -2 0 2 -2 0 2 -2 0 2 -2 0 2 Figure 5. 11: The curve x 2 + xy + y 2 = 9.portions in the second and fourth quadrants make a contribution....

Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 6 pps

... 4x 2 + y 2 +(x − 2) 2 + y 2 = 4x 2 + y 2 = 16 − 8(x − 2) 2 + y 2 + x 2 − 4x + 4 + y 2 x − 5 = 2 (x − 2) 2 + y 2 x 2 − 10x + 25 = 4x 2 − 16x + 16 + 4y 2 14(x − 1) 2 +13y 2 = 1Thus ... (1 2 )1 /2 =11 /2 = ±1 and 11 /2  2 = (±1) 2 = 1.Example 6.6 .2 Consider 2 1 /5 , (1 + ı)1/3 and (2 + ı) 5/ 6. 2 1 /5 = 5 √ 2 e 2 k /5 , for k = 0, 1, 2, 3, 4199Example 6 .5. 1 Suppose that we ... modulus-argumentform.√3 + ı 20 =√3 2 + 1 2 eı arctan(√3,1) 20 = 2 eıπ/6 20 = 2 20eı4π/3= 104 857 6−1 2 − ı√3 2 = − 52 4 288 − ı 52 4 288√36No, I have no idea why...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 2 pptx

... =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...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 3 ppt

... = 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...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 6 doc

... π)Hint 12. 23CONTINUEHint 12. 24CONTINUEHint 12. 25 Hint 12. 26Hint 12. 27Hint 12. 28Hint 12. 29Hint 12. 30CONTINUE 58 1Solution 12. 22 cos z = −cos(z − π)= −∞n=0(−1)n(z −π)2n(2n)!=∞n=0(−1)n+1(z ... 5| 2 limk→∞(k + 2) 2 (k + 1) 2 < 1|z + 5| 2 limk→∞ 2( k + 2) 2( k + 1)< 1|z + 5| 2 limk→∞ 2 2< 1|z + 5| 2 < 1 5. ∞k=0(k + 2 k)zkWe d etermine the ... <e.4.∞k=0(z + 5) 2k(k + 1) 2 We u se the ratio formula to determine the domain of convergence.limk→∞(z + 5) 2( k+1)(k + 2) 2 (z + 5) 2k(k + 1) 2 < 1|z + 5| 2 limk→∞(k...

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

... 0.|z|=3z1z − ı /2 −1z − 2 +c(z − 2) 2 + ddz = 0|z|=3(z − ı /2) + ı /2 z − ı /2 −(z − 2) + 2 z − 2 +c(z − 2) + 2c(z − 2) 2 dz = 0 2 ı 2 − 2 + c= 0c = 2 −ı 2 Thus we see that ... − 2/ z= −1z∞n=0 2 zn, for |2/ z| < 1= −∞n=0 2 nz−n−1, for |z| > 2 = −−1n=−∞ 2 −n−1zn, for |z| > 2 620 1 /2 < |z| < 2 and 2 < |z|. For |z| < 1 /2, ... 1)zn+ d, for 1 /2 < |z| < 2 For 2 < |z|, we havef(z) =−1n=−∞(− 2) n+1zn−−1n=−∞ 2 −n−1zn− (2 − ı /2) 2 n=−∞n + 1 2 n +2 zn+ df(z) = 2 n=−∞(− 2) n+1−1 2 n+1(1...

Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot

... theresidue by expanding the function in a Laurent series.(1 − cos z) 2 z7= z−71 −1 −z 2 2+z4 24 + Oz6 2 = z−7z 2 2−z4 24 + Oz6 2 = z−7z44−z6 24 + Oz8=14z3−1 24 z+ ... −n−1k=0limz→eıπ(1+2k)/nlog z + (z −eıπ(1+2k)/n)/znzn−1= −n−1k=0ıπ(1 + 2k)/nneıπ(1+2k)(n−1)/n= −ıπn 2 eıπ(n−1)/nn−1k=0(1 + 2k)e 2 k/n= 2 eıπ/nn 2 n−1k=1ke 2 k/n= 2 eıπ/nn 2 ne 2 /n−1=πn ... =C 2/ az 2 + ( 2/ a)z − 1dz= 2 2/ az1− z 2 = 2 1ı√1 − a 2 660Hint, SolutionFourier Cosine and Sine IntegralsExercise 13.19Evaluate∞−∞sin xxdx.Hint, SolutionExercise 13 .20 Evaluate∞−∞1...

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