Additional Notes on Power Series

Transcription

1 Additioal Notes o Power Series Mauela Girotti MATH 37-0 Advaced Calculus of oe variable Cotets Quick recall 2 Abel s Theorem 2 3 Differetiatio ad Itegratio of Power series 4 Quick recall We recall here the mai facts about power series. A detailed descriptio of the topic has bee covered a few weeks ago ad it refers to the Sectio of the textbook. Defiitio. Let {a } be a sequece of real umbers ad c R. The power series cetered at c with coefficiets {a } is the fuctio + a (x c). Every power series is a fuctio defied o a certai domai (called iterval of covergece) that is idetified by the radius of covergece R: Theorem 2. Let f(x) := + a (x c) be a power series. There exists a costat R R {+ } such that the series coverges absolutely for x c < R (i.e. for ay x (c R, c + R)) ad diverges for x c > R. Theorem 3 (Hadamard s Theorem). The radius of covergece R of the power series f(x) := + a (x c) is give by R = lim sup + a. Remark 4. The radius of covergece of a power series ca also be calculated as R = a lim +. + a

2 2 Abel s Theorem From the theorems above we have that give a power series, the series coverges absolutely for ay poit x i the ope iterval (c R, c + R), however o iformatio is give about what happes at the edpoits x = c ± R ad i geeral the series may coverges or diverges there. O the other ed, suppose that we have a power series (say, cetered at c = 0) f(x) := + with radius of covergece R (therefore the series coverges for ay x ( R, R)) ad suppose that for example the series a R < +. How do we kow that f(r) = a R? (i.e. we just plug the value R ito the defiitio of the power series) To state the problem i a more correct way, we eed to prove that assumig that a R < +, the lim f(x) = a R. x R The followig theorem has a positive aswers to this problem. We are cosiderig here the simple(r) case where the ceter of the series is c = 0, the radius of covergece is R = ad we ll be checkig oly the right ed poit of the iterval of covergece, amely x =. The result ca be easily geeralized to ay power series cetered at a poit c R with radius of covergece R > 0 ad with x equal either oe of the edpoits of the iterval of covergece. Theorem 5 (Abel s Theorem). Let f(x) := + a x a power series with radius of covergece R =. If a < +, the lim f(x) = a. x Proof. The mai tool used here will be the summatio by parts formula for two fiite sequeces {u,..., u } ad {v 0,..., v }: k= a x u k (v k v k ) = u v u v 0 v k (u k+ u k ). We kow that a < +, meaig that the sequece of partial sums {s := a } coverges (s S). We recall also that a = s s for all N ad a 0 = s 0. Now, we will use the summatio by parts formula for u k = x k ad v k = s k a k x k = a 0 + (s k s k )x k = a 0 + s x s 0 x s k (x k+ x k ) k= k= k= = a 0 ( x) + s x + s k ( x)x k = s x + s k ( x)x k k= 2

3 This chai of equality is always valid ad additioally we kow that for ay x < the left had side of the idetity is (absolutely) coverget. Also, the term s x 0 as +, because x 0 ( x < ) ad the sequece of partial sums {s } of the (umeric series) a is coverget, therefore it is bouded. Takig the limit as k + o both sides, we have a k x k = ( x) s k x k. 2. If S = a = lim + s is the sum of the (umeric) series, we wat to show that a x S as x : first of all a x S = ( x) s x S = ( x) (s S)x (the secod equality comes from the fact that ( x) x = ( x) series). x =, geometric 3. The, sice s S, this meas that ɛ > 0 M ɛ N such that s S < ɛ for M ɛ ; so, we fix ɛ > 0 ad we split the series (the first term will be a just a fiite sum, therefore bouded) ( x) M ɛ (s S)x ( x) s S x + ( x) s S x =M ɛ ( x)k + ( x)ɛ x = ( x)k + ( x)ɛ x where we do t write x because x < ; also, sice we wat to take the limit as x, we ca assume that 0 < x <. I coclusio, ( x) (s S)x K( x) + ɛ 0. Example. The series ( ) + f(x) = (x ) = has radius of covergece R = ( ) lim sup + + = lim sup + = 3

4 so it coverges for x < ad it diverges for x >. We still eed to check the edpoits, i.e. x = 0 ad x = 2. For x = 2 the power series becomes a alteratig harmoic series f(2) = ( ) + = which coverges thaks to Leibiz s rule. For x = 0, the series become ( ) + f(0) = ( ) ( ) 2 = = = which is a harmoic series ad therefore diverges. I coclusio, the iterval of covergece is (0, 2]. Example 2. has coefficiets The series f(x) = = = ( ) x 2 = x x 2 + x 4 x 8 + x 6 x ad its radius of covergece is equal to R = a = { if = 2 k 0 if 2 k lim sup + a = so it coverges for x < ad it diverges for x >. At the edpoits we have that f() = f( ) = ( ) which is clearly a diverget series. Therefore the iterval of covergece is (, ). Note 6. This series is called lacuary power series because there are successively loger gaps ( lacua ) betwee the powers with o-zero coefficiets. 3 Differetiatio ad Itegratio of Power series We will ow tackle the problem of differetiatig ad itegratig power series. Theorem 7 (Term-by-term differetiatio of a power series). Suppose that the power series f(x) = a (x c) has radius of covergece R. The the power series g(x) = a (x c) = is well defied ad it also has radius of covergece R. Moreover, the power series f is differetiable x (c R, c + R) ad we have the idetity f (x) = g(x) = a (x c). = 4

5 We will see oly the proof of the first part of the theorem here. To prove that the power series g is ideed the derivative of the power series f some extra tools from Aalysis are required, that we do t possess at the momet. Proof. Assume without loss of geerality that c = 0. Let x < R: this implies (least upper boud properties of the set R) that there exists ρ > 0 such that x < ρ < R ad let r := x ρ (0, ). To estimate the terms i the differetiated power series by the terms i the origial series, we write a x = ( ) x a ρ = r a ρ ρ ρ ρ The series = r coverges by ratio test, sice ( + )r lim r = lim r ( + ) = r < therefore the sequece {r } is bouded (it actually goes to zero): M r < M, N. Therefore, a x = r a ρ M ρ ρ a ρ N ad the series a ρ coverges sice ρ < R (remember that for all x < R the power series coverges absolutely). Fially, the compariso test implies that a x coverges absolutely. Coversely, suppose that x > R, the a x diverges (sice a x diverges) ad a x x a x N, so the compariso test implies that a x diverges. I coclusio, the power series g(x) = = a (x c) has the same radius of covergece R as the power series f. Similarly, usig the theorem above ad the fudametal theorem of calculus, the followig theorem for itegratio of power series ca be prove. Theorem 8 (Term-by-term itegratio of a power series). Suppose that the power series f(x) = a (x c) has radius of covergece R. The the power series (x c) + g(x) = a + C + is well defied ad it also has radius of covergece R. Moreover, the power series g is oe ati-derivative of the power series f. 5

6 Example. Fid a power series represetatio for the fuctio (wherever it makes sese) f(x) = ( x) 2. We have that [ ] ( x) 2 = x ad for x < (geometric series; its radius of covergece is R = ) x = x. Therefore, for x < (the radius of covergece is also equal ) [ ] ( x) 2 = x = x = ( + )x. Example 2. Fid a power series represetatio for the fuctio (wherever it makes sese) f(x) = arcta(x). We kow that arcta(x) = dx + x 2 ad for x < (geometric series; its radius of covergece is R = ) + x 2 = ( x 2 ) = ( ) x 2. Therefore, for x < (the radius of covergece is also equal ) arcta(x) = ( ) x 2 = C + ( ) ad sice arcta(0) = 0, the C = 0. I coclusio, with radius of covergece equal. arcta(x) = ( ) x2+ x 2 dx = C ( ) x