1 56 -1- 4+L 52-i 56 2705 ~ 36 +i ( 2705 - 3/16) 56 2 of 10 Finally, in the last example we have 3) 21 = 14+1 = √17 > 1 => the series diverges. We have also studied Power series centered at Zo: ½ an (Z-Zo)", an EC. n=o (G4) Now we can adjust the previous example to include a variable ZEC, making the geometric series above into power series. Ex 15.2. For what values of z does the 2 (A+)" (2-3)" converge? n=0 power series Sol: In this case we are studying a power series centered at Zo = 3. We can apply the Ratio Test and obtain: anx (2-20)/h+1 = lim | (4zi ) (2-3)| = √17 lim 848 6 17-31 =L By Ratio Test: the series converges absolutely if L < 1 (diverges if L>+) => converges absolutely if 12-31 < 517 The radius of convergence of this power series is given by R = lim 1 an 1 = √17 17300 Qn+t Another type of question we may encounter is one in which We are asked to determine the power series of a given function. Ex 15.3: Find the power series expansion of the function f(z) = 22-3 about the point zo=0 and determine the radius of convergence. Sol: Observe that I can be written in the form of a convergent geometric series as follows: 22-3 n オー = f. 122 = -f 3 उं र ( 3²² ) " - - — — (³) "z" 4 8 =-2 3nti Z" n=o h=0 ทะ The series expansion is valid only if | 22 |< 1 >> the radius of IS R= 롤. What convergence of this series. Some natural questions to pose at this point are: happens if we asked to find the power series representation of a function of that cannot be expressed as a geometric series? Is there a general method by which we can find the power series representation of an arbitrary function? How do we even know that a function admits a power cocios representation? Now we will partially answer these questions. Ex 15.10: What is the laurent series of f(z) = (2-1) (7-2) on the annulus A = {ZEQ: 1<\2\<2} ? Sol: First, we write I a sum of 2 functions using a partial fraction expansion: ⑦ (2-1)(2-2) = 2-2-2-2-1 = 8 Σ ทะเ -1 글도 2 ( 1 − 3 ) 2 ( 1 − 1 ) = - — 22 ( 2 ) - j- z-n - the Laurent series 00 n=o Note: The first sum 1/200 (&)n exists only if 1 — k₁E) 12kg The second sum 2-n exists only 1½\<₁ => \Z/> 1 ทะเ 1 Thus, ⑦is valid only on the annular region 1<1212. Ex 15.11: Obtain the series expansion for f(z) = = /^2+4 valid in the region A 61 21 2i 1z-zil >4. The region here is a domain: the exterior of the circle centered at (0,2i) of radius 4. We want a series expansion about zo=di To do this we make a substitution Set W= z-ai and look for the expansion in W, where | W/> 4 To make the series expansion easier we manipulate again f(z) into a form similar to the series expansion for AZ. We 1 f(z) = 4i w (1-iw) 1 get: 4i w (1- iw) Now we use standard geometric series and 1 % (ix) n compute: _ 1wk4 4iw Aiw (1-in) nzo 1 - 4iw Win IWI > 4

1 56 -1- 4+L 52-i 56 2705 ~ 36 +i ( 2705 - 3/16) 56 2 of 10 Finally, in the last example we have 3) 21 = 14+1 = √17 > 1 => the series diverges. We have also studied Power series centered at Zo: ½ an (Z-Zo)", an EC. n=o (G4) Now we can adjust the previous example to include a variable ZEC, making the geometric series above into power series. Ex 15.2. For what values of z does the 2 (A+)" (2-3)" converge? n=0 power series Sol: In this case we are studying a power series centered at Zo = 3. We can apply the Ratio Test and obtain: anx (2-20)/h+1 = lim | (4zi ) (2-3)| = √17 lim 848 6 17-31 =L By Ratio Test: the series converges absolutely if L < 1 (diverges if L>+) => converges absolutely if 12-31 < 517 The radius of convergence of this power series is given by R = lim 1 an 1 = √17 17300 Qn+t Another type of question we may encounter is one in which We are asked to determine the power series of a given function. Ex 15.3: Find the power series expansion of the function f(z) = 22-3 about the point zo=0 and determine the radius of convergence. Sol: Observe that I can be written in the form of a convergent geometric series as follows: 22-3 n オー = f. 122 = -f 3 उं र ( 3²² ) " - - — — (³) "z" 4 8 =-2 3nti Z" n=o h=0 ทะ The series expansion is valid only if | 22 |< 1 >> the radius of IS R= 롤. What convergence of this series. Some natural questions to pose at this point are: happens if we asked to find the power series representation of a function of that cannot be expressed as a geometric series? Is there a general method by which we can find the power series representation of an arbitrary function? How do we even know that a function admits a power cocios representation? Now we will partially answer these questions. Ex 15.10: What is the laurent series of f(z) = (2-1) (7-2) on the annulus A = {ZEQ: 1<\2\<2} ? Sol: First, we write I a sum of 2 functions using a partial fraction expansion: ⑦ (2-1)(2-2) = 2-2-2-2-1 = 8 Σ ทะเ -1 글도 2 ( 1 − 3 ) 2 ( 1 − 1 ) = - — 22 ( 2 ) - j- z-n - the Laurent series 00 n=o Note: The first sum 1/200 (&)n exists only if 1 — k₁E) 12kg The second sum 2-n exists only 1½\<₁ => \Z/> 1 ทะเ 1 Thus, ⑦is valid only on the annular region 1<1212. Ex 15.11: Obtain the series expansion for f(z) = = /^2+4 valid in the region A 61 21 2i 1z-zil >4. The region here is a domain: the exterior of the circle centered at (0,2i) of radius 4. We want a series expansion about zo=di To do this we make a substitution Set W= z-ai and look for the expansion in W, where | W/> 4 To make the series expansion easier we manipulate again f(z) into a form similar to the series expansion for AZ. We 1 f(z) = 4i w (1-iw) 1 get: 4i w (1- iw) Now we use standard geometric series and 1 % (ix) n compute: _ 1wk4 4iw Aiw (1-in) nzo 1 - 4iw Win IWI > 4

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.2: Arithmetic Sequences

Problem 67E

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