MTH 122 Calculus II Essex County College Division of Mathematics and Physics 1 Lecture Notes #20 Sakai Web Project Material

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1 MTH 1 Calculus II Essex Couty College Divisio of Mathematics ad Physics 1 Lecture Notes #0 Sakai Web Project Material 1 Power Series 1 A power series is a series of the form a x = a 0 + a 1 x + a x + a x + where x is a variable ad the a s are costats called the coefficiets of the series The domai of this fuctio is the set of all x for which this series is coverget A power series i (x b) is a power series cetered at b, where b is a costat a (x b) = a 0 + a 1 (x b) + a (x b) + a (x b) + where x is a variable ad the a s are costats called the coefficiets of the series The domai of this fuctio is the set of all x for which this series is coverget, ad you should otice that this series always coverges for x = a Theorem: For a give power series a (x b) = a 0 + a 1 (x b) + a (x b) + a (x b) + there are oly three possibilities: (a) The series coverges oly whe x = b The radius of covergece is defied to be r = 0 (b) The series coverges for all x The radius of covergece is defied to be r = (c) There is a positive umber R such that the series coverges if x b < R ad diverges for x b > R What happes at x b = R should also be examied The radius of covergece is betwee b r ad b + r, icludig ay edpoits where the series coverges 4 Theorem: If the power series a (x b) = a 0 + a 1 (x b) + a (x b) + a (x b) + 1 This documet was prepared by Ro Bao (robao@mathographyorg) usig L A TEX ε Last revised Jauary 10, 009 1

2 has a radius of covergece R > 0, the the fuctio defied by a 0 + a 1 (x b) + a (x b) + a (x b) + = a (x b) is differetiable (ad therefore cotiuous) o the iterval (b R, b + R) 5 Method for Computig the Radius of Covergece: To calculate the radius of covergece, r, for the power series use the ratio test a (x b), (a) If is ifiite, the r = 0 (b) If the r = (c) If a a = 0, a = k x b, where k is a costat, the r = 1 k Now let s look at some key examples The geometric series x = 1 + x + x + is certaily recogizable, but let s do the Ratio Test to see where it coverges = x +1 x = x < 1 So 1 < x < 1, but what if x = ±1? If x = 1 we get f (1) = 1 = , which is clearly diverget Now if x = 1 we get f ( 1) = 1 = , which is also clearly diverget So the Radius of Covergece is R = 1, ad the Iterval of Covergece is ( 1, 1)

3 Here s aother (urecogizable) series!x = 1 + x + x + 6x + ad we will do the Ratio Test to see where it coverges = ( + 1)!x +1!x = ( + 1) x =, however, if x = 0 this it will be zero ad the series will be coverget Here we eed to adopt a covetio, that (x a) 0 = 1 eve if x = a, so f (0) = 1 So the Radius of Covergece is R = 0, ad the Iterval of Covergece is [0, 0] or {0} idicatig a set with oe elemet Here s aother (urecogizable) series (x 1) = (x 1) + (x 1) + (x 1) + ad we will do the Ratio Test to see where it coverges = (x 1) +1 x 1 ( + 1) (x 1) = = x 1 < 1, + 1 so the series will be coverget whe 0 < x < We will still eed to look at x = 0 ad x =, because the Ratio Test is icoclusive here Whe x = 0 we get ( 1) f (0) = = , which is a alteratig harmoic series ad hece coverget Now we eed to look at x = where we get ( 1) f () = = a diverget harmoic series So the Radius of Covergece is R = 1, ad the Iterval of Covergece is [0, ) Here s aother (urecogizable) series (!) = 1 x 4 + x4 64 ad we will do the Ratio Test to see where it coverges = x + + [( + 1)!] (!) x = x 4 ( + 1) = 0 so the series will be coverget whe < x < So the Radius of Covergece is R =, ad the Iterval of Covergece is (, )

4 Ca You Do These? 1 Problems 1 For what values of x is the series coverget? ( ) x + 1 Show that the power series for coverges for all x e x = 1 + x + x! + x! + x! + = The Bessel fuctio order zero is defied by the series (!) Fid the iterval of covergece, which is also the domai 4 Determie the radius of covergece for x! (x 1) (x 1) + (x 1) (x 1)4 4 ( 1) 1 (x 1) + = 5 Determie the radius of covergece for x x! + x5 5! x7 7! + = ( 1) 1 x 1 ( 1)! 6 Fid the radius ad iterval of covergece of the series 1 + x + 4 x x + 7 Fid the radius ad iterval of covergece of the series (x + ) +1 Daiel Beroulli the Swiss mathematicia defied defied ad it was later geeralized by the Germa Friedrich Bessel 4

5 Solutios 1 For what values of x is the series ( ) x + 1 coverget? a = = x x Usig the Ratio Test we have x or 1/ < x < 1/ Sice the Ratio Test is icoclusive at x = ±1/, we eed to see what happes at these values Lookig at x = 1/ we have ( ) 1 ( 1) f = + 1 which is a alteratig series ad coverges by the Alteratig Series Test For x = 1/ we have ( f 1 ) 1 = + 1 which is diverget by the Itegral Test You could have also use the p-series for compariso to show that this is diverget So the Radius of Covergece is R = 1/, ad the Iterval of Covergece is ( 1/, 1/] Show that the power series for coverges for all x e x = 1 + x + x! + x! + x! + = x! QED = x +1 ( + 1)! x! = x +1 ( + 1)!! x = x = x + 1 = 0 5

6 The Bessel fuctio order zero is defied by the series (!) Fid the iterval of covergece, which is also the domai ( 1) +1 x + = + [( + 1)!] (!) = ( 1) x 4 ( + 1) = x ( + 1) = 0 < 1 So the radius of covergece is ifiite (all x), thus the domai of the Bessel fuctio order zero is R, that is the iterval of covergece is R 4 Determie the radius of covergece for (x 1) (x 1) + (x 1) (x 1)4 4 ( 1) 1 (x 1) + = ( 1) (x 1) +1 = ( 1) 1 (x 1) + 1 ( 1) (x 1) +1 = + 1 ( 1) 1 (x 1) = ( 1) (x 1) + 1 = x = x 1 < 1 This power series coverges for x 1 < 1 (ie 0 < x < ) ad diverges for x 1 > 1 If we check the edpoits we will see that the series coverges for 0 < x The radius of covergece is 1 You may recall that this is the series (cetered at 1) for l x 5 Determie the radius of covergece for x x! + x5 5! x7 7! + = ( 1) 1 x 1 ( 1)! Daiel Beroulli the Swiss mathematicia defied defied ad it was later geeralized by the Germa Friedrich Bessel 6

7 +1 = ( 1) 1 x 1 ( + 1)! ( 1)! = +1 ( 1)! ( + 1)! ( 1) 1 x 1 = ( 1) x ( + 1) = x 1 ( + 1) = 0 < 1 This power series coverges for all x The radius of covergece is ifiite You may recall that this is the series for si x 6 Fid the radius ad iterval of covergece of the series 1 + x + 4 x x + (+1) x (+1) = x = 4x = 4x 1 = 4x < 1 Hopefully you recall how to solve quadratic iequalities from MTH-119 That is, solvig 4x < 1 ad you ll get 1/ < x < 1/ At both edpoits you will get a diverget series, so the radius of covergece is 1/, a the iterval of covergece is 1/ < x < 1/ 7 Fid the radius ad iterval of covergece of the series (x + ) +1 ( + 1) (x + ) = + (x + ) = ( + 1) (x + ) x x + = = < 1 Hopefully you recall how to solve absolute valued iequalities from MTH-119 That is, solvig x + < ad you ll get 5 < x < 1 At both edpoits you will get a diverget series, so the radius of covergece is, a the iterval of covergece is 5 < x < 1 7

1 Introduction to Sequences and Series, Part V

1 Introduction to Sequences and Series, Part V MTH 22 Calculus II Essex Couty College Divisio of Mathematics ad Physics Lecture Notes #8 Sakai Web Project Material Itroductio to Sequeces ad Series, Part V. The compariso test that we used prior, relies