Mathematical Methods for Physics and Engineering - Matematica.NET

25.10 HINTS AND ANSWERSt = −i and then approaches the origin in the fourth quadrant in a curve thatis ultimately antiparallel to the positive real axis. The other contour, C 1 ,isthemirror image of this in the real axis; it is confined to the upper half-plane, passesthrough t = i and is antiparallel to the real t-axis at both of its extremities. Thecontribution to J ν (z) from the curve C k is 1 2 H(k) ν , the function H ν(k) being knownas a Hankel function.Using the method of steepest descents, establish the leading term in an asymptoticexpansion for H ν(1) for z real, large and positive. Deduce, without detailedcalculation, the corresponding result for H ν(2) . Hence establish the asymptoticform of J ν (z) for the same range of z.25.23 Use the method of steepest descents to find an asymptotic approximation, validfor z large, real and positive, to the function defined by∫F ν (z) = exp(−iz sin t + iνt) dt,Cwhere ν is real and non-negative and C is a contour that starts at t = −π + i∞and ends at t = −i∞.25.10 Hints and answers25.1 Apply Kirchhoff’s laws to three independent loops, say ADBA, ADEA andDBED. Eliminate other currents from the equations to obtain I R = ω 0 CV 0 [(ω0 2 −ω 2 − 2iωω 0 )/(ω0 2 + ω2 )], where ω0 2 =(LC)−1 ; |I R | = ω 0 CV 0 ; the phase of I R istan −1 [(−2ωω 0 )/(ω0 2 − ω2 )].25.3 Set c coth u 1 = −d, c coth u 2 =+d, |c cosech u| = a and note that the capacitanceis proportional to (u 2 − u 1 ) −1 .25.5 ξ = constant, ellipses x 2 (a+1) −2 +y 2 (a−1) −2 = c 2 /(4a 2 ); η = constant, hyperbolaex 2 (cos α) −2 − y 2 (sin α) −2 = c 2 . The curves are the cuts −c ≤ x ≤ c, y =0and|x| ≥c, y =0.Thecurvesforη =2π are the same as those for η =0.25.7 (a) For a quarter-circular contour enclosing the first quadrant, the change in theargument of the function is 0 + 8(π/2) + 0 (since y 8 + 5 = 0 has no real roots);(b) one negative real zero; a conjugate pair in the second and third quadrants,− 3 , −1 ± i.225.9 Evaluate∫π cot πz( 1 + z)( dz1+ z)2 4around a large circle centred on the origin; residue at z = −1/2 is 0; residue atz = −1/4 is4π cot(−π/4).25.11 The behaviour of the integrand for large |z| is |z| −2 exp [ (2α − π)|z| ]. The residueat z = ±m, foreachintegerm, issin 2 (mα)(−1) m /(mα) 2 . The contour contributesnothing.Required summation = [ total sum − (m = 0 term) ]/2.25.13 Note that ¯f(s) has no pole at s =0.Fort1 in the left half-plane. For 0