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

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Unformatted text preview: Physics 212A Homework #4 Name: 1.Using the complex analysis technique discussed in class, calculate n = 1 4 n 4 + 1 . Verify with Sum. Consider 1 4 z 2 + 1 Cot @ z D dz = = sum of enclosed residues. The sum we want has terms Table B 1 4 n 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 Cot @ z D 4 z 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 = 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 4 n 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 Cot @ z D 4 z 4 + 1 , 8 z, <F &, zeros F :- 1 8 + 8 Tanh B 2 F ,- 1 8 + 8 Tanh B 2 F ,- 1 8- 8 Tanh B 2 F ,- 1 8- 8 Tanh B 2 F> and their sum is sumfres = Plus res1- 1 2 Tanh B 2 F solving for sum yields : sum . Solve @ 2 sum- 1 + sumfres 0, sum D : 1 4 2 + Tanh B 2 F > Verify : Sum B 1 4 n 4 + 1 , 8 n, 0, <F 1 4 2 + Tanh B 2 F 2. The function [x] = n = 1 1 n x is a function of a complex variable x which is defined for Re[x]>1 . For real values, [x] is a sum of inverse powers of integers. For example, [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. [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 z 3 z- 1 z must be calculated. a) For large z, the integrand goes like z 3 - z . Express the integrand as a product of the asympotic behavior and a series in - 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 = - z z 3 1- - z - z z 3 1- - z z goes from 0 to , so - 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- - z . - z fi r 1 1- r 2 junk4sol.nb 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 = - z z 3 H srl L . r fi - z 9 - z z 3 , - 2 z z 3 , - 3 z z 3 , - 4 z z 3 , - 5 z z 3 , - 6 z z 3 , - 7 z z 3 , - 8 z 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...
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212HomeWork4sol - Physics 212A Homework #4 Name: 1.Using...

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