Arkasas Tech Uiversity MATH 94: Calculus II Dr Marcel B Fia 85 Power Series Let {a } =0 be a sequece of umbers The a power series about x = a is a series of the form a (x a) = a 0 + a (x a) + a (x a) + =0 Note that a power series about x = a always coverges at x = a with sum equals to a 0 Example 85 Which of the followig is a power series? (a) A polyomial of degree m (b) The geometric series + x + x + (c) The series x + x + x 3 + (d) The series + x + (x ) + (x ) 3 + (x 3) 4 + (a) A polyomial of degree m is a power series about x = 0 sice p(x) = a 0 + a x + a x + + a m x m Note that a = 0 for m + (b) The geometric series + x + x + is a power series about x = 0 with a = for all (c) The series x + x + x 3 + is ot a power series sice it has egative powers of x (d) The series + x + (x ) + (x ) 3 + (x 3) 4 + is ot a power series sice each term is a power of a differet quatity Covergece of Power Series To study the covergece of a power series about x = a oe starts by fixig x ad the costructig the partial sums S 0 (x) =a 0 S (x) =a 0 + a (x a) S (x) =a 0 + a (x a) + a (x a) S (x) =a 0 + a (x a) + a (x a) + + a (x a)
Thus obtaiig the sequece {S (x)} =0 If this sequece coverges to a umber L, ie lim S (x) = L, the we say that the power series coverges to L for the specific value of x Otherwise, we say that the power series diverges Power series may coverge for some values of x ad diverge for other values The followig theorems provide a tool for determiig the values of x for which a power series coverges ad those for which it diverges Theorem 85 Suppose that =0 a (x a) is a power series that coverges for x = c ad diverges for x = d The (i) =0 a (x a) coverges absolutely for x a < c a ; ad (ii) =0 a (x a) diverges for x a > d a Proof (i) If c = a the the series coverges to a 0 So we assume that =0 a (c a) coverges with c a By the th term test, lim a (c a) = 0 Thus, there exists a positive iteger N such that a (c a) < for all N Let M = N =0 a (c a) + The, a (c a) M for all 0 Suppose that x a < c a The, for 0 we have a (x a) = a (c a) x a c a M x a c a But the series =0 M x a c a is a coverget geometric series sice x a c a < Hece, by the compariso test =0 a (x a) coverges ad cosequetly =0 a (x a) coverges absolutely (ii) Suppose that x a > d a If =0 a (x a) is coverget at x the by (i), =0 a (d a) is absolutely coverget sice d a < x a This cotradicts the fact that =0 a (d a) is diverget Hece, =0 a (x a) diverges Theorem 85 Give a power series a (x a) = a 0 + a (x a) + a (x a) + =0 The oe of followig is true: (i) The series coverges oly at x = a;
(ii) the series coverges for all x; (iii) there is some positive umber R such that the series coverges absolutely for x a < R ad diverges for x a > R The series may or may ot coverge for x a = R That is for the values x = a R ad x = a + R Proof Let C be the collectio of all real umbers at which the series =0 a (x a) coverges If C = {a} the series coverges oly at x = a ad diverges for all x a This establishes (i) If C = (, ) the the series coverges for all values of x This establishes (ii) To prove (iii), we assume that C {a} ad C (, ) These coditios guaratee the existece of a real umber d a such that =0 a (d a) diverges Hece, by applyig (ii) of the previous theorem, x a < d a wheever x is i C This shows that C is bouded Let R > 0 be the smallest umber such that x a < R for all x i C Hece, (A) if x a > R the x is ot i C ad so =0 a (x a) diverges (B) If x a < R the we ca fid a x 0 is C such that x a < x 0 a R, by the way R is defied For otherwise y a x a for all y i C ad therefore R < x a, a cotradictio Sice =0 a (x 0 a) coverges, by (i) of the previous theorem =0 a (x a) coverges absolutely We caot asserts what happes at either x = a R or x = a + R See Example 853 below The largest iterval for which the power series coverges is called the iterval of covergece If a power series coverges at oly the poit x = a the we defie R = 0 If a power series coverges for all values of x the we defie R = We call R the radius of covergece Fidig the radius of covergece The ext theorem gives a method for computig the radius of covergece of may series Theorem 853 Suppose that =0 a (x a) is a power series with a 0 for all 0 = 0 the R = ad this meas that the series co- a (a) If lim + a verges for all x a (b) If lim + a = L > 0 the R = a (c) If lim + a for all x a L = the R = 0 ad this meas that the series diverges 3
Proof Cosider the ratio a + (x a) + a (x a) = a + x a a { } a Suppose that lim + a = L 0 The the sequece a + a x a =0 coverges to L x a (a) If L = 0 the x a L = 0 < ad by the ratio test the series coverges for all real umbers x Thus, R = (b) Suppose that L > 0 ad fiite If L x a < or x a < L the by the ratio test, the series =0 a (x a) is coverget If x a > L the by the ratio test, the series is diverget Hece, R = L (c) If L = the the series diverges for all x a Thus, R = 0 Remark 85 It follows from the previous theorem that the radius of covergece satisfies R = lim a + a Example 85 Fid the radius of covergece of the series =0 x! The radius of covergece of the power series =0 x! is R = sice R = lim a +! a = lim (+)! = lim + = 0 Thus, the series coverges everywhere Example 853 Fid the iterval of covergece of the series = ( ) (x ) By the previous theorem R = lim + = so that R = Hece, by Theorem 85, the series coverges for x < (ie, 0 < x < ) ad diverges for x > What about the edpoits x = 0 ad x =? If we replace x by 0 we obtai the series = which is diverget (Harmoic series) If we replace x by we obtai the alteratig series = ( ) which coverges by the alteratig series test Thus, the iterval of covergece is 0 < x 4
Example 854 What is the radius of covergece of the series =0 x? Usig the ratio test we fid (+) x lim + (+) x = lim + x 4 = x 4 Thus, the series coverges if x 4 < or x < Hece, the radius of covergece is R = 5