212HomeWork4sol - Physics 212A Homework#4 Name 1.Using the...

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Physics 212A Homework #4 Name: 1.Using the complex analysis technique discussed in class, calculate n = 0 ¥ 1 4 n 4 + 1 . Verify with Sum. Consider 1 4 z 2 + 1 Π Cot @ Π z D dz = 0 = sum of enclosed residues. The sum we want has terms Table B 1 4n 4 + 1 , 8 n,0,5 <F : 1, 1 5 , 1 65 , 1 325 , 1 1025 , 1 2501 > The residues at the integers are Table B Residue B Π Cot @ Π z D 4z 4 + 1 , 8 z,n <F , 8 n, - 5,5 <F : 1 2501 , 1 1025 , 1 325 , 1 65 , 1 5 ,1, 1 5 , 1 65 , 1 325 , 1 1025 , 1 2501 > If sum= n = 0 ¥ 1 4 n 4 + 1 , then the sum of residues at the integers is 2 sum-1. The poles of f[z] are at zeros = n .Solve A 4n 4 + 1 0,n E ComplexExpand : - 1 2 - 2 , 1 2 + 2 , 1 2 - 2 , - 1 2 + 2 > The residues at each pole are res1 = Map B Residue B Π Cot @ Π z D 4z 4 + 1 , 8 z, <F &,zeros F : - 1 8 + 8 Π Tanh B Π 2 F , - 1 8 + 8 Π Tanh B Π 2 F , - 1 8 - 8 Π Tanh B Π 2 F , - 1 8 - 8 Π Tanh B Π 2 F> and their sum is sumfres = Plus res1 - 1 2 Π Tanh B Π 2 F solving for sum yields : sum .Solve @ 2sum - 1 + sumfres 0,sum D
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: 1 4 2 + Π Tanh B Π 2 F > Verify : Sum B 1 4n 4 + 1 , 8 n,0, ¥ <F 1 4 2 + Π Tanh B Π 2 F 2. The function Ζ [x] = n = 1 ¥ 1 n x is a function of a complex variable x which is defined for Re[x]>1 . For real values, Ζ [x] is a sum of inverse powers of integers. For example, Ζ [4]=1+ 1 2 4 + 1 3 4 + 1 4 4 + .. . As discussed in class, complex variable techniques provide a method for exactly evaluating this type of sum for x= even integer. In Mathematica , the function is called Zeta, which works for arbitrary complex x. Ζ [x] is also related to a number of important integrals. For example, to calculate the free energy of the photons in a cavity at a fixed temperature, the integral 0 ¥ z 3 ª z - 1 z must be calculated. a) For large z, the integrand goes like z 3 ª - z . Express the integrand as a product of the asympotic behavior and a series in ª - z . Integrate the series term by term to generate a sum. Evaluate the sum by complex variable techniques. Verify the sum using Sum and Zeta and verify the integral using Integrate. Write the integrand as int2 = ª - z z 3 1 - ª - z ª - z z 3 1 - ª - z z goes from 0 to ¥ , so ª - z goes from 1 to 0. Expand the denominator in powers of e - z . Series only works for expansions of a simple symbol, so replace e - z with r denom = 1 1 - ª - z . ª - z fi r 1 1 - r 2 junk4sol.nb
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sr = Series @ denom, 8 r,0,7 <D Normal 1 + r + r 2 + r 3 + r 4 + r 5 + r 6 + r 7 for clarity, convert this into a list : srl = List sr 9 1,r,r 2 ,r 3 ,r 4 ,r 5 ,r 6 ,r 7 = so the terms of the sum are : int21 = ª - z z 3 H srl L .r fi ª - z 9 ª - z z 3 , ª - 2z z 3 , ª - 3z z 3 , ª - 4z z 3 , ª - 5z z 3 , ª - 6z z 3 , ª - 7z z 3 , ª - 8z z 3 = Now integrate them term by term : int22 = Map @ Integrate @ , 8 z,0, ¥ <D &,int21 D : 6, 3 8 , 2 27 , 3 128 , 6 625 , 1 216 , 6 2401 , 3 2048 > pull out a factor of 6 int22 6 Expand : 1, 1 16 , 1 81 , 1 256 , 1 625 , 1 1296 , 1 2401 , 1 4096 > and the series is recognizable as 1 n 4 . So the value of the integral is 6 Ζ [4] . Lets check this: Integrate B z 3 ª z - 1 , 8 z,0, ¥ <F Π 4 15 junk4sol.nb 3
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6Zeta @ 4 D Π 4 15 To evaluate the sum= n = 1 ¥ 1 n 4 using the technique discussed in class, consider 1 z 4 Π Cot @ Π z D dz = 0 = 2 sum + Res @ 0 D Residue B 1 z 4 Π Cot @ Π z D , 8 z,0 <F - Π 4 45 so sum = - H 1 2 L Residue B 1 z 4 Π Cot @ Π z D , 8 z,0 <F Π 4 90 We can also do it using Sum Sum @ 1 n^4, 8 n,1, ¥ <D Π 4 90
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