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 2 Part 1 pps

... imaginary part of some analytic function.Solution 8 .11 We write the real and imaginary parts of f(z) = u + ıv.u =x4/3y 5/ 3x2+y2 for z = 0,0 for z = 0., v =x 5/ 3y4/3x2+y2 for ... 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 .5 Take the logarithm of the equation to get a linear equation.Cauchy-Riemann EquationsHint 8.6Hint 8.7Hint 8.8 For the ﬁrst part...

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 Fourier ... 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, ... +z36+z 5 12 0+ ···= 1 −z46+ ···z3z2+ z6 1 36+ 1 60+ ···= 1 z 5 1 −z46+ ··· 1 −z490+ ···= 1 z 5  1 −z46+ ··· 1 +z490+ ···= 1 z 5  1 −7 45 z4+...

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).7 75 1 23448 12 16 1 23448 12 16 Figure 14 .1: The p opulation...

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 √ 5 2 1+ √ 5 2√ 5 = 1+ √ 5 2 1+ √ 5 2√ 5 = 1 √ 5 Substitute this result into the equation for c2.c2= 1 r2 1 ... c2.c2= 1 r2 1 − 1 √ 5 r 1 =2 1 −√ 5  1 − 1 √ 5 1 +√ 5 2= −2 1 −√ 5  1 −√ 5 2√ 5 = − 1 √ 5 Thus the nthterm in the Fibonacci sequence has the formulaan= 1 √ 5  1 +√ 5 2n− 1 √ 5  1 ... diﬀerence equation for bnthat is of order one less than theequation for an. 11 78 1 234 5 6 -1 -0 .5 0 .5 1 1 .5 1 234 5 6 -1 -0 .5 0 .5 1 1 .5 Figure 23.3: The graph of approximations and numerical...

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

... continuous. -1 -0 .5 0 .5 1 -1 -0 .5 0 .5 1 -1 -0 .5 0 .5 1 -1 -0 .5 0 .5 1 Figure 25. 1: Polynomial Approximations to cos(πx). 12 860 1 234 5 0. 25 0 .5 0. 75 1 1. 25 1 .5 1. 75 2Figure 24 .1: Plot of K0(x) and ... =2√π∞x 1 2t−2te−t2dt=2√π 1 2te−t2∞x−2√π∞x 1 2t−2e−t2dt= 1 √πx 1 e−x2− 1 √π∞xt−2e−t2dt. 12 64 5 10 15 20 25 -16 -14 -12 Figure 24 .5: The logarithm of the nthterm ... (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...

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

... 1 2t+z24t2t−n 1 et−z2/4tdt 16 29n Γ(n)√2πxx 1/ 2e−xrelative error 5 24 23.6038 0. 01 65 15 8. 717 83 · 10 10 8.66 954 · 10 10 0.0 055 25 6.20448 · 10 236 .18 384 · 10 230.0033 35 2. 952 33 · 10 382.9 453 1 · 10 380.0024 45 ... ( 1) nJn(z)Thus we see that J−n(z) and Jn(z) are not linearly independent for integer n. 16 25 246 8 10 12 14 -0.4-0.20.20.40.60.8 1 5 10 15 20-0.2-0 .1 0 .1 0.20.3Figure 34 .1: ... −− 1 22π 1/ 2z−3/2sin z −2π 1/ 2z 1/ 2cos z= 2 1/ 2π 1/ 2z−3/2sin z + 2 1/ 2π 1/ 2z−3/2sin z −2 1/ 2π 1/ 2cos z=2π 1/ 2z−3/2sin z −2π 1/ 2z 1/ 2cos...

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

... variablest → 1/ t in the integral representation of y 1 (t).y 1 (t) =Ctn 1 e 1 2z(t 1/ t)dt=C( 1/ t)n 1 e 1 2z( 1/ t+t) 1 t2dt=C( 1) nt−n 1 e 1 2z(t 1/ t)dt= ( 1) ny2(t)Thus ... equation.CLe 1 2z(t 1/ t)v(t) dt = 0Cz2 1 4t + 1 t2+ z 1 2t − 1 t2+z2− n2e 1 2z(t 1/ t)v(t) dt = 0 (34 .1) By consideringddtte 1 2z(t 1/ t)= 1 2xt + 1 t+ 1 e 1 2z(t 1/ t)d2dt2t2e 1 2z(t 1/ t)= 1 4x2t ... 1 e 1 2z(t 1/ t)d2dt2t2e 1 2z(t 1/ t)= 1 4x2t + 1 t2+ x2t + 1 t+ 2e 1 2z(t 1/ t)we see thatLe 1 2z(t 1/ t)=d2dt2t2− 3ddtt + 1 − n2e 1 2z(t 1/ t).Thus...

Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 5 ppt

... 1) 2Rn= 0α(α − 1) + α −(2n − 1) 2= 0α = ±(2n − 1) Rn= c 1 r2n 1 + c2r 1 2nThe solution which is bounded in 0 ≤ r ≤ 1 isRn= r2n 1 . 17 42We substitute this form into the wave equation ... obtain a problem with homogeneous boundary conditions.Hint 37 .13 Hint 37 .14 Hint 37. 15 Hint 37 .16 Hint 37 .17 17 35 Solution 37.4 1. ut= ν(uxx+ uyy)XY T= ν(XY T + XYT )TνT=XX+YY= ... independent solutions,T (1) n= cos(2n − 1) cπt2L, T(2)n= sin(2n − 1) cπt2L. 17 72Hint 37 .11 Hint 37 .12 There are two ways to solve the problem. For the ﬁrst method, expand the solution...

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