Advanced Mathematical Methods for Scientists and Engineers Episode 1 Part 1 pdf

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

... 8 .12 Consider the complex functionf(z) = u + ıv =x3 (1+ ı)−y3 (1 ı)x2+y2 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 ... 8 .1. 3 and use Result 8 .1. 1.Hint 8.4Use Result 8 .1. 1.Hint 8.5Take the logarithm of the equation to get a linear equation.Cauchy-Riemann EquationsHint 8.6Hint 8.7Hint 8.8 For the ﬁrst part ... y2)2=−(x − 1) 2+ y2((x − 1) 2+ y2)2 and 2(x − 1) y((x − 1) 2+ y2)2=2(x − 1) y((x − 1) 2+ y2)2The Cauchy-Riemann equations are each identities. The ﬁrst partial derivatives...

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

... negativedirection.648(b) 1 z (1 −z)= 1 z+ 1 1 − z= 1 z− 1 z 1 1 − 1/ z= 1 z− 1 z∞n=0 1 zn, for |z| > 1 = − 1 z∞n =1 z−n, for |z| > 1 = −−∞n=−2zn, for |z| > 1 624Result 13 .5.2 ... Exercise 13 .10 .)642The series about z = ∞ for 1/ (z + 2) is 1 2 + z= 1/ z 1 + 2/z= 1 z∞n=0(−2/z)n, for |2/z| < 1 =∞n=0( 1) n2nz−n 1 , for |z| > 2= 1 n=−∞( 1) n +1 2n +1 zn, ... z=∞n=0( 1) nzn−∞n=0( 1) n2n +1 zn, for |z| < 1 =∞n=0( 1) n 1 − 1 2n +1 zn, for |z| < 1 f(z) =∞n=0( 1) n2n +1 − 1 2n +1 zn, for |z| < 1 614 In general, if f(x) is an odd function with...

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 1 potx

... =y(x)x.P (1, u) + Q (1, u)u + xdudx= 0This equation is separable.P (1, u) + uQ (1, u) + xQ (1, u)dudx= 0 1 x+Q (1, u)P (1, u) + uQ (1, u)dudx= 0ln |x| + 1 u + P (1, u)/Q (1, u)du ... divide it into two equations on separate domains.y 1 − y 1 = 0, y 1 (0) = 1, for x < 1 y2− y2= 1, y2 (1) = y 1 (1) , for x > 1 797• y+ 3xy+ 2y = x2• y= yyThe ... doubles everyhour. For the continuous problem, we assume that this i s true for y(t). We write this as an equation:y(t) = αy(t).775 1 23448 12 16 1 23448 12 16 Figure 14 .1: The p opulation...

Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 5 pdf

... EquationsHint 17 .11 Hint 17 .12 Hint 17 .13 Hint 17 .14 Substitute y = xλinto the diﬀerential equation. Consider the three cases: a2> b, a2< b and a2= b.Hint 17 .15 Hint 17 .16 Exact EquationsHint ... conditions, y 1 (0) = 1 and y 1 (0) = 0, for the ﬁrst solution become,c 1 = 1, −ac 1 +√a2− b c2= 0,c 1 = 1, c2=a√a2− b.The conditions, y2(0) = 0 and y2(0) = 1, for the second ... solution become,c 1 = 1, −ac 1 + c2= 0,c 1 = 1, c2= a.The conditions, y2(0) = 0 and y2(0) = 1, for the second solution become,c 1 = 0, −ac 1 + c2= 1, c 1 = 0, c2= 1. The fundamental...

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

... thatan=n/2j=0−4j2−2j +1 (2j+2)(2j +1)  for even n,0 for odd n. 11 93c 1 = 1 − r2r2 1 − r 1 r2= 1 − 1 √52 1+ √52√5= 1+ √52 1+ √52√5= 1 √5Substitute this result into the equation for c2.c2= 1 r2 1 ... second equation.c 1 r2 1 + 1 r2 (1 − c 1 r 1 )r22= 1 c 1 (r2 1 − r 1 r2) = 1 − r2 11 81 0.2 0.4 0.6 0.8 1 1.20.30.40.50.60.70.80.9Figure 23.2: Plot of the solution and approximations.Recall ... arer = 1 ± (1 − 2p)2pr 1 = 1 − pp=qp, r2= 1. If p > 1/ 2 the roots arer = 1 ± (2p − 1) 2pr 1 = 1, r2=−p + 1 p=qp.Thus the general solution for p = 1/ 2 isan= c 1 + c2qpn.The...

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 3 pdf

... continuous. -1 -0.5 0.5 1 -1 -0.50.5 1 -1 -0.5 0.5 1 -1 -0.50.5 1 Figure 25 .1: Polynomial Approximations to cos(πx). 12 860 1 23450.250.50.75 1 1.25 1. 5 1. 752Figure 24 .1: Plot of K0(x) and ... (t)2− 2x 1/ 2t− 1 2x 1/ 2= 0 1 4x−2+ u+− 1 4x 1 + u2− 2x 1/ 2− 1 4x 1 + u− 1 2x 1/ 2= 0u+ (u)2+− 1 2x 1 − 2x 1/ 2u+5 16 x−2= 0Assume that ... = 1 √πx 1 e−x2− 1 √π∞xt−2e−t2dt= 1 √πx 1 e−x2− 1 √π− 1 2t−3e−t2∞x+ 1 √π∞x32t−4e−t2dt= 1 √πe−x2x 1 − 1 2x−3+ 1 √π∞x32t−4e−t2dt= 1 √πe−x2x 1 − 1 2x−3+ 1 √π−34t−5e−t2∞x− 1 √π∞x 15 4t−6e−t2dt= 1 √πe−x2x 1 − 1 2x−3+34x−5− 1 √π∞x 15 4t−6e−t2dt 12 65The...

Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

... ∼ 4∞n =1 oddncos(nx). For x = nπ, this implies0 = 4∞n =1 oddncos(nx), 13 61 -1 -0.5 0.5 1 -0.2-0 .1 0 .1 0.2 1 0.250 .1 00 .1 1 1 10.5Figure 28.8: Three Term Approximation for a Function ... =−x − 1 for − 1 < x < 1/ 2x for − 1/ 2 < x < 1/ 2−x + 1 for 1/ 2 < x < 1. 13 55 1. Let SNbe the sum of the ﬁrst N terms in the Fourier series. Show thatdSNdx= 1 − ( 1) NcosN ... expansion.bn= 1/ 2 1 (−x − 1) sin(nπx) dx + 1/ 2 1/ 2x sin(nπx) dx + 1 1/2(−x + 1) sin(nπx) dx= 2 1/ 20x sin(nπx) dx + 2 1 1/2 (1 − x) sin(nπx) dx=4(nπ)2sin(nπ/2)=4(nπ)2( 1) (n 1) /2for...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 1 pdf

... integration by parts we can evaluate the integral. 1 0 (1 − τ)nτz 1 dτ = (1 − τ)nτzz 1 0− 1 0−n (1 − τ)n 1 τzzdτ=nz 1 0 (1 − τ)n 1 τzdτ=n(n − 1) z(z + 1)  1 0 (1 − τ)n−2τz +1 dτ=n(n ... 1)  1 0 (1 − τ)n−2τz +1 dτ=n(n − 1) ··· (1) z(z + 1) ···(z + n 1)  1 0τz+n 1 dτ=n(n − 1) ··· (1) z(z + 1) ···(z + n 1) τz+nz + n 1 0=n!z(z + 1) ···(z + n) 16 10 For x > 0 we close ... limn→∞nzn!z(z + 1) ···(z + n)= 1 zlimn→∞ (1) (2) ···(n)(z + 1) (z + 2) ···(z + n)nz= 1 zlimn→∞ 1 (1 + z) (1 + z/2) ··· (1 + z/n)nz= 1 zlimn→∞ 1 (1 + z) (1 + z/2) ··· (1 + z/n)2z3z···nz 1 z2z···(n...

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

... case 1 < t.u(x, t) =0, x + t < 1 12x+t 1 (1 − |ξ|) dξ, 1 < x + t < 1 12 1 1 (1 − |ξ|) dξ, x − t < 1, 1 < x + t 1 2 1 x−t (1 − |ξ|) ... =0, x + t < 1 12x+t 1 (1 − |ξ|) dξ, x − t < 1 < x + t 1 2x+tx−t (1 − |ξ|) dξ, 1 < x − t, x + t < 1 12 1 x−t (1 − |ξ|) dξ, x − t < 1 < x + t0, 1 < x − tu(x, ... < 1 12x − t < 1, 1 < x + t 1 4 (1 − (t − x − 2)(t − x)) 1 < x − t < 0 1 4 (1 + t − x)20 < x − t < 1 0, 1 < x − tFigure 45 .1 shows the solution at t = 1/ 2 and t...

Advanced Mathematical Methods for Scientists and Engineers Episode 6 Part 5 pdf

... withu = α (1 − v) for special values of α and r. Find v(x) for this special case. 216 8L 1 L2(L 1 x) = L 1 (L 1 l2− I)x= L 1 (λx − x)= (λ − 1) (L 1 x)L 1 L2(L2x) = (L2L 1 + I)L2x= ... 2√ 1 − t2log (1 + t), 1 2G+(t) + G−(t)= ıt log (1 + t). For t ∈ (−∞, 1) ,G+(t) = ı√ 1 − t2+ t(log(−t − 1) + ıπ) ,G−(t) = ı−√ 1 − t2+ t(log(−t − 1) − ıπ) , 215 3 Part ... u∂v∂nds 214 6F (z) =z 1 z 1/ 2−ıγ/(2π)zk(z 1) mF (z) = 1 z(z − 1) z 1 z−ıγ/(2π)F±(x) =e±ıπ(−ıγ/(2π))x (1 − x) 1 − xx−ıγ/(2π)F±(x) =e±γ/2x (1 − x) 1 − xx−ıγ/(2π)Deﬁnef(x)...

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