MTH3101 Spring 2017 HW Assignment 6: Chap. 5: Sec. 65, #6-8; Sec. 68, #5, 7; Sec. 72, #8; Sec. 73, #5, 6. The due date for this assignment is 4/06/17.
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1 MTH30 Spring 07 HW Assignment 6: Chp. 5: Sec. 65, #6-8; Sec. 68, #5, 7; Sec. 7, #8; Sec. 73, #5, 6. The due dte for this ssignment is 4/06/7. Sec. 65: #6. Wht is the lrgest circle within which the Mclurin series for the function tnh z converges to tnh z? Write the first two nonzero terms of tht series. Solution. We hve f z = tnh z = sinh z cosh z, where sinh z = n +! zn+ z <, cosh z = n! zn z < so tht the circle of convergence of f z, i.e., the lrgest circle within which the Mclurin series for the function tnh z converges to tnh z, is z = R, where R = min { z : z C nd cosh z = 0}. Now we hve so tht cosh z = ez + e z implying cosh z = 0 if nd only if e z = if nd only if z = log = i rg { z : z C nd cosh z = 0} = implying = [Arg + inπ] = π + inπ = π + in, n = 0, ±, ±,... { π + π R = min { z : z C nd cosh z = 0} π π = =. n : n = 0, ±, ±,... }
2 Next, we hve implying f z = tnh z = n z n z < π = n! zn n +! zn+ [ ] = +! z + z + 3! z3 + [ n ] = n! z + z + 3! z3 + =! z + + z + 3! z3 + = z + z + 3! z3 + = z + 3! z 3 + = z 3 z3 + 0 = f 0 = 0, = f 0 =, = 0, 3 = 3. Sec. 65: #7. Show tht if f z = sin z, then f n 0 = 0 nd f n+ 0 = n n = 0,,,.... Proof. The Mclurin series for the function f z = sin z is where f z = sin z = n z n = n n +! zn+ z < n = f n 0 n = 0,,,... n! implying tht for the even numbers we hve 0 = n = f n 0 n!
3 nd for the odd numbers we hve n n +! = n+ = f n+ 0 n +! so tht f n 0 = 0 nd f n+ 0 = n n = 0,,,.... This proves the result. Note: An lterntive proof is to use the fct tht implying tht d sin z = cos z, dz d sin z = sin z dz d n dz n sin z = n sin z, d n+ dz n+ sin z = n cos z from which it follows tht d n sin z dzn = n sin 0 = 0, z=0 d n+ sin z dzn+ = n cos 0 = n. z=0 Sec. 65: #8. Rederive the Mclurin series 4 in Sec. 64 for the function f z = cos z by using the definition cos z = eiz + e iz in Sec. 37 nd ppeling to the Mclurin series for e z in Sec. 64; b showing tht Proof. We hve so tht f n 0 = n nd f n+ 0 = 0 n = 0,,,.... e z = n! zn z < f z = cos z = eiz + e iz = [ ] n! izn + n! izn [ ] i n = + n i n z n n! = n z n z <, 3
4 where f n 0 n! f n+ 0 n +! = n = i n + n i n = n + n = n! n! n! n = n+ = for n = 0,,,..., tht is, i n+ + n+ i n+ i n+ i n+ = = 0 n +! n +! f z = cos z = n z n = n n! zn z <. This proves. An lterntive proof is to use the fct tht implying tht d cos z = sin z, dz d cos z = cos z dz d n dz n cos z = n cos z, d n+ dz n+ cos z = n sin z from which it follows tht d n cos z dzn = n cos 0 = n, z=0 d n+ cos z dzn+ = n sin 0 = 0 z=0 so tht from this nd the fct f z = cos z is entire we hve f z = cos z = n z n z < where n n! = f n 0 n! 0 = f n+ 0 n +! = n = n+ for n = 0,,,.... This proves b. 4
5 Sec. 68, #5. The function f z = z z = z z, which hs two singulr points z = nd z =, is nlytic in the domins D : z <, D : < z <, D 3 : < z <. Find the series representtion in powers of z for f z in ech of those domins. Solution. On D we cn use the convergence of the geometric series z = z n z < to get the convergent series representtion f z = z z = z + z = z n + z n = [ + n ] z n z <. Now for the geometric series we hve the convergent series representtion z = z z = z z = z n n z < < z nd hence on D we hve the convergent series representtion f z = z z = z n + n z = z n + n+ z n < z < 5
6 nd on D 3 we hve the convergent series representtion f z = z z = z + z = z n + n z = z n + n+ z n = n+ z n < z <. Sec. 68, #7. Let denote rel number, where < < nd derive the Lurent series z = n < z <. zn n= n= b After writing z = e iθ in, equte the rel prts nd then imginry prts on ech side of the result to derive the summtion formuls n cos θ cos nθ = n= cos θ +, n sin θ sin nθ = cos θ +, where < <. Compre with Exercise 4, Sec. 6. Proof. We hve the convergent series z = z n z < n= from which it follows tht we hve the convergent series z = z z = z n n= z z < = z n < z from which it follows tht we hve the convergent series z = z = z n < z = n < z <. zn n= 6
7 Hence for ny C we hve z = n= n < z <. zn This proves. Now in let be such tht < < nd let z = e iθ, θ R so tht < = z to get the convergent series e iθ cos θ + = e iθ cos θ + cos θ + sin θ = e iθ cos θ + sin θ = e iθ e iθ = e iθ e iθ e iθ = e iθ = n e inθ = n e inθ n= which is good for ll θ R so tht by tking rel nd imginry prts we get cos θ cos θ + = Re e iθ = Re n e inθ cos θ + n= = n Re e inθ = n cos nθ, n= n= n= sin θ cos θ + = Im e iθ = Im n e inθ cos θ + n= = n Im e inθ = n sin nθ n= = n sin nθ. n= n= This proves b. Sec. 7, #8. Prove tht if f is nlytic t z 0 nd f z 0 = = f m z 0 = 0, then the function g defined by mens of the equtions g z = { fz z z 0 m+ when z z 0 f m+ z 0 m+! when z = z 0 is nlytic t z 0. Proof. By the hypothesis there exists n ε > 0 such tht the Tylor series of f is given by f z = f n z 0 z n z z 0 < ε n! n=m+ 7
8 so tht g z = n=m+ = f n z 0 z n m+ z z 0 < ε n! f n+m+ z 0 n + m +! zn z z 0 < ε is the Tylor series of g. Therefore, this implies tht g is nlytic t z = z 0. Sec. 73, #5. Note how the expnsion z sinh z = z 3 6 z + 7 z + 0 < z < π, 360 is n immedite consequence of the Lurent series 8 in Sec. 73. Then use the method illustrted in Exmple 4, Sec. 68, to show tht C z dz = πi sinh z 3, when C is positively oriented unit circle z = with prmeteriztion winding once round the origin. Proof. The Lurent series 8 in Sec. 73 is so tht sinh z z = + z 3! + z4 5! + = 6 z z4 + 0 < z < π z sinh z = z 3 sinh z z which is the Lurent series of tht z sinh z = = z 3 6 z z4 + = z 3 6 z + 7 z + 0 < z < π 360 z sinh z expnded in powers of z. This implies n z n= c n = z 3 6 z + 7 z + 0 < z < π 360 hs its coeffi cients given by c n = πi z dz, n = 0, ±, ±,... sinh z zn+ nd, in prticulr, for n =, we hve C πi z sinh z dz = c = 6 C 8
9 implying tht C z dz = πi sinh z 6 = πi 3. This proves the result. Sec. 73, #6. Follow these steps, which illustrte n lterntive to strightforwrd division, to obtin eqution 8 in Exmple, Sec. 73. Write + z 3! + z4 5! + = d 0 + d z + d z + d 3 z 3 + d 4 z 4 +, where the coeffi cients in the power series on the right re to be determined by multiplying the two series in the eqution = + z 3! + z4 d0 5! + + d z + d z + d 3 z 3 + d 4 z 4 +. Perform this multipliction to show tht 0 = d 0 + d z + d + 3! d 0 z + d 3 + 3! d z 3 + d 4 + 3! d + 5! d 0 z 4 + when z < π. b By setting the coeffi cients in the lst series in prt equl to zero, find the vlues of d 0, d, d, d 3, nd d 4. With these vlues, the first eqution in prt becomes eqution 8, Sec. 73. Proof. We hve for the Lurent series sinh z z sinh z z = + z 3! + z4 5! + = d 0 + d z + d z + d 3 z 3 + d 4 z < z < π = + z 3! + z4 + 0 < z < π 5! implying convergent of the series for z < π of the series on the right nd side nd hence convergent of the series = + z 3! + z4 d0 5! + + d z + d z + d 3 z 3 + d 4 z 4 + = d 0 + d z + d z + d 3 z 3 + d 4 z 4 + z + d 0 3! + d z 3 3! + d z 4 3! + z 4 + d 0 5! + = d 0 + d z + + d 4 + d 3! + d 0 5! d + d 0 3! z 4 + z + d 3 + d z 3 3! 9
10 for z < π nd thus convergent of the series 0 = d 0 + d z + d + d 0 3! + d 4 + d 3! + d 0 5! z 4 + z + d 3 + d z 3 3! for z < π. This proves. Now for prt b we use the fct tht the coeffi cients of convergent power series re unique to imply by tht d 0 = 0, d = 0, d + d 0 3! = 0, d 3 + d 3! = 0, d 4 + d 3! + d 0 5! = 0, nd solving these equtions we find tht d 0 =, d = 0, d = 3! = 6, d 3 = d 3! = 0, d 4 = d 3! d 0 5! = 6 5! = = = = =
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