Math 341 Lecture #31 6.5: Power Series

Math 341 Lecture #31 6.5: Power Series We ow tur our attetio to a particular kid of series of fuctios, amely, power series, f(x = a x = a 0 + a 1 x + a 2 x 2 + where a R for all N. I terms of a series of fuctios, we have f (x = a x which is ifiitely differetiable (ad cotiuous for each = 0, 1, 2, 3,.... Theorem 6.5.1. If a power series a x coverges at some poit R, the it coverges absolutely for ay x satisfyig x <. See the Appedix for a proof. The mai implicatio of Theorem 6.5.1 is that the subset of R o which a power series coverges is either {0}, a bouded iterval cetered at 0, or R. I the bouded iterval case, there is some ambiguity about the edpoits because of the strict iequality i Theorem 6.5.1. The bouded iterval of covergece ca be oe of four possible forms: ( R, R, [ R, R, ( R, R], or [ R, R] for some R > 0. We call R the radius of covergece i the bouded iterval case. We assig R = 0 whe the set of covergece is {0}, ad R = whe the set of covergece is R. Recall that a stadard way to fid the radius of covergece R for a power series is the Ratio Test. We ow tur our attetio to properties of a power series that coverges absolutely at some poit. Theorem 6.5.2. If a power series a x coverges absolutely at a poit, the the power series coverges uiformly o the compact iterval [ c, c] where c =. Proof. Suppose for some we have that coverges. a x 0 With M = a x 0 for each = 0, 1, 2, 3,..., we have for all x satisfyig x that Sice we have the covergece of a x a x 0 = M. M = a x 0,

the Weierstrass M-Test implies that a x coverges uiformly o A = [ c, c] for c =. We ca ow prove that whe a power series coverges o a ope iterval ( R, R with R > 0 or R =, the power series is a cotiuous fuctio o ( R, R. For a fixed x 1 satisfyig 0 < x 1 < R we pick such that x 1 < < R. The covergece of a x 0 implies by Theorem 6.5.1 that a x coverges absolutely o the ope iterval x <. Sice 0 < x 1 <, we have that a x coverges absolutely at x 1. By Theorem 6.5.2, we have that a x coverges uiformly o [ x 1, x 1 ]. Sice each f (x = a x is cotiuous o [ x 1, x 1 ], we have by Theorem 6.4.2 that the power series is cotiuous o [ x 1, x 1 ]. Sice x 1 satisfyig 0 < x 1 < R is arbitrary, we have that a x is cotiuous o ( R, R. Example. Determie where the power series is cotiuous. ( 1 x Without usig the Ratio Test, we ca determie the radius of covergece for this series. The i the deomiator suggests that we evaluate the power series at x = 1: we have which diverges by Corollary 2.4.7. ( 1 ( 1 = This says that the radius of covergece satisfies R 1. For ay 0 < x 1, we have that x / is a decreasig sequece covergig to 0, so by the Alteratig Series Test (Theorem 2.7.7, we have covergece of the power series o (0, 1]. This says that the radius of covergece satisfies R 1. Thus we have that R = 1, ad so the power series is cotiuous o ( 1, 1. The power series coverges at x = 1, but it is cotiuous at x = 1? Could the coditioal covergece at x = 1 prevet cotiuity at x = 1? We see i this example that the iterval of covergece could be of the form ( R, R], ad the questio is the is the power series cotiuous o ( R, R]. 1

We aswer this i the affirmative. Theorem 6.5.4 (Abel s Theorem. If g(x = a x coverges at x = R > 0, the the power series coverges uiformly o [0, R]. A similar result holds whe the power series coverges at x = R < 0. See the Appedix for a proof. This aswers the questio that whe the iterval of covergece of a power series icludes or oe or both edpoits we have cotiuity of the power series o that iterval. We ca say somethig eve stroger. Theorem 6.5.5. If a power series coverges poitwise o a set A R, the it coverges uiformly o ay compact subset K of A. Proof. Let K be a compact subset of A. The the poits a = if K ad b = sup K both belog to K, ad we have that K [a, b]. Sice a, b A, we have the covergece of the power series at a ad b. If a < 0 ad b > 0, the by Abel s Theorem we have uiform covergece of the power series o [a, 0] ad [0, b], hece uiform covergece o [a, b], ad thus o K as well. If a 0 the b a 0, givig uiform covergece o [0, b] ad hece o K. If b 0 the a b 0, givig uiform covergece o [a, 0] ad hece o K. How about the differetiability of a coverget power series? Accordig to Theorem 6.4.3, we eed uiform covergece of the term by term differetiated series. By Theorem 6.5.5, it suffices to get poitwise covergece of the term by term differetiated series. Theorem 6.5.6. If a x coverges for all x ( R, R for R > 0, the the term-wise differetiated series a x 1 coverges for all x ( R, R as well. See the Appedix for a proof. We ca ow summarize the totality of the theory of covergece power series we have prove i this sectio. Theorem 6.5.7. Suppose that g(x = a x coverges o a iterval A R. The g is cotiuous o A ad differetiable o ay ope subiterval of A, where g (x = a x 1. Moreover, g(x is ifiitely differetiable o ay ope subiterval of A, ad successive derivatives of g are obtaied through term by term differetiatio of g (x, g (x, etc.

Appedix. Proof of Theorem 6.5.1. Suppose there is R such that a x 0 coverges. By Theorem 2.7.3 we have that the terms a x 0 0 as. Beig coverget, the sequece (a x 0 is bouded: there is a real M > 0 such that we have a x 0 M for all = 0, 1, 2, 3,.... For ay x R that satisfies x <, we have a x = a x M x. Sice x/ < 1, we have that the geometric series x coverges, ad so by the Algebraic Limit Theorem for Series (Theorem 2.7.1 we have that M x coverges as well. By the Compariso Theorem (Theorem 2.7.4 we have that a x coverges, ad so by the Absolute Covergece Test (Theorem 2.7.6 we have that coverges absolutely as well. a x Lemma 6.5.3 (Abel s Lemma. Let b satisfy b 1 b 2 b 3 0. Let a be a series whose sequece of partial sums s m = m a is bouded: there is A > 0 such that s m A for all m N. The for itegers > m 0 there holds a j b j 2Ab m+1. Proof. Each s m is defied for m 1. We will eed a s 0 which we set to 0. We first show the summatio by parts formula a j b j = s b +1 s m b m+1 + s j (b j b j+1

holds for all 1 ad all m 0. [Thik of this like itegratio by parts. ] Startig with the right had side of this formula, we have for m 1 that s b +1 s m b m+1 + s j (b j b j+1 = (a 1 + + a b +1 (a 1 + + a m b m+1. + (a 1 + + a m + a m+1 b m+1 (a 1 + + a m + a m+1 b m+2 + (a 1 + + a m+1 + a m+2 b m+2 (a 1 + + a m+1 + a m+2 b m+3 + (a 1 + + a 2 + a 1 b 1 (a 1 + + a 2 + a 1 b + (a 1 + + a 1 + a b (a 1 + + a b +1 = a m+1 b m+1 + a m+2 b m+2 + + a b = a j b j, which is the left had side of the formula. The oly differece i the case of m = 0 are the secod ad third terms after the first equal sig above, which result i a term of a 1 b 1. By this summatio by parts formula, for m 0, we have a j b j = s b +1 s m b m+1 + s j (b j b j+1 s b +1 + s m b m+1 + s j (b j b j+1 = s b +1 + s m b m+1 + s m (b m+1 b m+2 + + s (b b +1 Ab +1 + Ab m+1 + A(b m+1 b m+2 + + A(b b +1 = Ab m+1 + Ab m+1 = 2Ab m+1, which establishes the desired iequality. Proof of Abel s Theorem. Applyig the Cauchy Criterio to the coverget series a R we have for each ɛ > 0 the existece of N N such that for all > m N we have a m+1 R m+1 + a m+2 R m+2 + + a R < ɛ 3. We are drivig for the Cauchy Criterio for Uiform Covergece o the iterval [0, R]. To this ed, we have for ay x [0, R] ad > m N that a m+1 x m+1 + a m+2 x m+2 + + a x ( x m+1 ( x m+2 ( x = a m+1 R m+1 + am+2 R m+2 + + a R. R R R

We the apply Lemma 6.5.3 to the sequeces (a R ad (x/r to get for all > m N ad all x [0, R] that ( x m+1 ( x m+2 ( x a m+1r R m+1 + am+2 R R m+2 + + a R R ( ɛ ( x m+1 2 3 R < ɛ. By the Cauchy Criterio for Uiform Covergece o the iterval [0, R] (Theorem 6.2.5, we have the uiform covergece of the power series o [0, R]. Proof of Theorem 6.5.6. We work towards the Weierstrass M-Test to get uiform covergece of the term by term differetiated power series a x 1 o compact subitervals of ( R, R. To this ed, let x 1 (0, R ad pick t satisfyig x 1 < t < R. The for all x satisfyig x x 1, we have a x 1 = 1 ( x 1 t t 1 a t 1 ( t x1 1 t 1 a t. Sice a t coverges we have that the terms a t are bouded: there is L > 0 such that a t L for all = 0, 1, 2, 3,..., ad so for all x x 1 ad all N. a x 1 L t ( x 1 1 t 1 = M For s = x 1 /t ad b = s 1 we have that M = (L/tb where b +1 lim b = lim ( + 1s s 1 = lim ( + 1s = s. Sice 0 < s < 1, there exists N {0} N such that for all N we have that Pick 0 < r < 1 that for all N satisfies b +1 b < 1. b +1 b < r. The we have that b N+1 < b N r, ad by iductio that b N+k < b N r k for all k 1. The geometric series b N r k k=1

coverges because 0 < r < 1, ad so by the Compariso Test, the series coverges as well. Thus the series =N+1 M = b = =N+1 L t ( s 1 x 1 t 1 coverges, so that by the Weierstrass M-Test, the series a x 1 coverges uiformly o the iterval [ x 1, x 1 ], ad hece poitwise o [ x 1, x 1 ]. Sice x 1 satisfyig 0 < x 1 < R is arbitrary, we have covergece of the term by term differetiated series a x 1 at every poit x i ( R, R.