CHAPTER 6 Series and Residues In Problems 27 and 28 without

CHAPTER 6 Series and Residues In Problems 27 and 28, without a the Tay 27 and 28, without actually expanding, determine lor series of the given function centered at the indi anding, determine the radiu determine the radius of tered at the indicated point. indicated point. 4 + 52 29. What is the radius of convergence R of the Macl 30. What is the radius of convergence R of the Maclau R of the Maclaurin series in Problem Rof the Maclaurin series in Problem In Problems 31 and 32, expand the given function in Tayl indicated points. Give the radius of convergence R of ea which both series converge. n function in Taylor series centered us of convergence R of each series. Sketch t 242 32. f(z) =-, zo = 1 + i, 4s3 d in this section to find the sum of the on to find the sum of the giv In Problems 33 and 34, use results obtained in this section 33. 3tzk 35. Find the Maclaurin series (14) by differentiating the Maclaurin series(13) 36. The error function erf(z) is defined by the integral erf(z) = series k-0 k! 0 2[a. Find a Maclaurin series for erf z) by integrating the Maclaurin series fore Find a Maclaurin series for erf(z) by integrating the Maclaurin series for e In Problems 37 and 38, approximate the value of the given expression using the i

Solution

Solution: Given function f(z) = (4+5z)/(1+z² )

= 2i{ 1/(z+i) - 1/(z-i) } + 5/2 { 1/(z+i) + 1/(z-i) }

= (2i + 5/2){1/(z+i) } - (2i - 5/2){1/(z-i)}

= (4i+5)/2.{ 1/(z+i)} - (4i-5)/2.{ 1/(z-i)}

={(4i+5)/2}g(z) - {(4i-5)/2}.h(z)    ................................. ................. (1)

where g(z) = { 1/(z+i)}, h(z) = {1/(z-i)}

Now Taylor\'s expansion for g(z),

g(z) =\\sigma_{0}^{\\infty} a_n (z-2-5i)ⁿ , a_n = {g⁽ⁿ⁾(2+5i)}/n!

g(z) = { 1/(z+i)} = (z+i)^-1

implies that g⁽¹⁾(z)= (-1) (z+i)^-2 , g⁽²⁾(z)= (-1) (-2)(z+i)^-3 ,...., g⁽ⁿ⁾(z)= (-1)(-2).....(-n) (z+i)^-(n+1) = (-1)ⁿn! (z+i)^-(n+1)

g⁽ⁿ⁾(z)/ n! = (-1)ⁿ (z+i)^-(n+1)

So, a_{n} = g⁽ⁿ⁾(2+5i)/ n! = (-1)ⁿ (2+5i+i)^-(n+1)= (-1)ⁿ (2+6i)^-(n+1)

Similarly h⁽ⁿ⁾(2+5i)/ n! = (-1)ⁿ (2+5i-i)^-(n+1)= (-1)ⁿ (2+4i)^-(n+1)

Hence Taylor\'s series becomes

f(z) = {(4i+5)/2}g(z) - {(4i-5)/2}.h(z)

Now, for radius of convergence of g(z) , a_n = (-1)ⁿ (2+6i)^-(n+1)

or |a_n |= | (-1)ⁿ (2+6i)^-(n+1)|= (4+36)^-(n+1)/2 = 40^-(n+1)/2

or 1/R = lim_{n \\to \\infty} |a_n |^1/n = lim_{n \\to \\infty} 40^-(n+1)/2n = 40^(-1/2)

So R = 40^(1/2) = 2(10)(^1/2)

Similarly radius of convergence of h(z) is square root of (4+16) = square root of 20

Hence radius of convergence R of the Taylor\'s series of f(z) is minimum of

{square root of 20, square root of 40} = square root of 20.