POWER SERIES R. E. SHOWALTER

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1 POWER SERIES R. E. SHOWALTER. sequeces We deote by lim = tht the limit of the sequece { } is the umber. By this we me tht for y ε > 0 there is iteger N such tht < ε for ll itegers N. This mkes precise the sttemet tht s gets lrge, the error gets smll. This is lso deoted by, d we sy the sequece is coverget. Exmple.. For the sequece = we hve lim = 0. Tht is, the sequece { } is coverget d lim = 0. Limits c be tke ito sums, multiples d quotiets (whe the deomitor hs o-zero limit), such s i the two followig exmples. Exmple.2. For the sequece =, we hve + lim + = lim + = =. + lim Or directly we c compute + = + 0. Exmple.3. For the sequece = = , we hve so we hve lim = 3 0 = 3. Sice we kow the limit, we c check +0+0 directly for covergece from the crude estimte 3 < I fct, for y fuctio F ( ) which is cotiuous t the poit, If, the F ( ) F (). The two precedig exmples follow from the respective cses F (x) = 3 x2 d F (x) = d +x +x+x 2 =.

2 2 R. E. SHOWALTER Note tht for y umber h 0 d iteger we hve ( + h) + h. This estimte is useful to idetify limits of powers. Exmple.4. To evlute the limit of the sequece = p, if p <, we set p = for some h > 0 to see tht (+h) = ( + h) < + h 0, d if p >, we set p = h + with h > 0 to get p = ( + h) + h s. I summry, we hve { 0 if p <, lim p = if p =, d the sequece does ot coverge otherwise. Exmple.5. For the sequece of frctiol powers, = p, we proceed similrly. If p >, the set p = + h so tht p = ( + h ) > + h d 0 < h < p. This shows tht lim p =. If 0 < p <, the p =, q >, so lim p = = =. q I summry, we hve lim p = for ll 0 < p. lim q Here s more delicte oe tht will rise i pplictios. Exmple.6. For the sequece =, we write = ( ) = + h, d the = ( + h ) + h, so tht we hve h 0. This shows tht < + 2h + h , d we see tht. Tht is, lim =. Eve though the bse is growig i this exmple, the frctiol power still brigs the sequece to. Ad this hppes eve with higher powers of the bse: for y fixed iteger M we hve lim (M ) = lim M ( ) = lim M =.

3 POWER SERIES 3 Cuchy Criterio. If the series { } is coverget with lim =, the for y pir of itegers m, we get m m +, so for both of m, sufficietly lrge, it follows tht m is rbitrrily smll. The sequece is Cuchy if for y ε > 0 there is iteger N such tht for ll pirs of itegers m, N we hve m < ε. I prticulr, we just oted tht every coverget sequece is Cuchy. A fudmetl property of the rel umbers is tht every Cuchy sequece is coverget. This provides useful test for covergece tht does ot deped o kowig the limit of the sequece. 2. series Let { } be sequece. The defie ew sequece {s } by s = m = ,. m= This is the sequece of prtil sums of { } or the series, d is the -th term of the series. If the sequece {s } is coverget, we sy the series coverges d deote its limit lso by = lim s. It follows from the Cuchy test for covergece of the sequece {s } tht the series is coverget if d oly if for y ε > 0 there is N such tht s m s = m < ε for ll m N. Filly, we ote tht if the series coverges, the we ecessrily hve = s s s s = 0, so the sequece of terms { } coverges to 0. The most importt exmple is the geometric series obtied from the terms = p. Exmple 2.. The sequece of prtil sums is s m = + p + p 2 + p p m, m. The we compute s m ps m = p m+ to get { m + if p =, s m = p m+ if p. p This shows tht the series coverges to the limit d it is ot coverget otherwise. =0 p = p if p <,

4 4 R. E. SHOWALTER A series is bsolutely coverget if the series of bsolute vlues is coverget. The Cuchy criterio for covergece together with the iequlity m m shows tht bsolute covergece implies covergece of the series. The series is coditiolly coverget if it is coverget but ot bsolutely coverget. Exmples will be give below. Covergece Tests. Theorem 2.. Compriso Test: If there is costt C 0 for which Cb for ll sufficietly lrge, d if b is coverget, the is bsolutely coverget. Proof. Use the estimte bove with the Cuchy test for covergece of the two series. For exmple, by comprig with the geometric series, it follows tht if there is costt C 0 d iteger N for which Cp for ll N for some 0 p <, the is bsolutely coverget. Sufficiet coditios re give by the followig. Corollry 2.2. Rtio Test: If + p <, N, or if lim p <, the is bsolutely coverget. + Corollry 2.3. Root Test: If p <, N, or if lim p <, the is bsolutely coverget. Itegrl Test. Our lst criterio for covergece of series with oegtive terms is obtied by comprig with improper itegrl. Suppose the fuctio f( ) is cotiuous, decresig, d f(x) 0 for ll x 0. The for x + we hve f() + f(x)dx f(+) so we obti m m+ m f() f(x)dx f( + ) This shows tht the improper itegrl f(x)dx coverges if d oly if the series f() = f()+ f(+) coverges. This criterio is the itegrl test. Exmple 2.2. By tkig the fuctio f(x) =, we fid the series x α coverges if d oly if α >. α

5 POWER SERIES 5 Altertig Series. The series ( ) + b with b 0 is ltertig series. Tht is, successive terms = ( ) + b lterte sig. If lso the terms re decresig, b + b, the we c rrge the terms i two wys ( ) + b = (b b 2 ) + (b 3 b 4 ) + (b 5 b 6 ) +... to see tht = b (b 2 b 3 ) (b 4 b 5 ) (b 6 b 7 )... s 2 s 4 s 6... d s s 3 s These show tht the odd terms re decresig d they lie bove the eve terms which re icresig. Filly we ote tht s 2 + s 2 = b 2 +, so if lim b = 0, the these two sequeces coverge to the commo vlue which is lim s. We summrize this s Theorem 2.4. A ltertig series ( ) + b with b 0, b + b d lim b = 0 is coverget. Exmple 2.3. The hrmoic series ( ) + = is coverget (ltertig series) but ot bsolutely coverget (by the itegrl test). 3. Sequeces d Series of Fuctios Let {f ( )} be sequece of fuctios o set of umbers S. This sequece is poitwise coverget to fuctio f( ) o S if lim f (x) = f(x) for every x S. Tht is, for every x S d ε > 0, there is iteger N for which f (x) f(x) < ε for ll N. (The iteger N depeds o ε d o x.) The sequece {f ( )} is uiformly coverget to f( ) o S if for every ε > 0 there is iteger N for which f (x) f(x) < ε for ll N d for ll x S. (The iteger N depeds o ε.) Exmple 3.. The sequece f (x) = x coverges poitwise to f(x) = 0 o the set S = (, ). The covergece is uiform o S = ( p, p) for y p with 0 < p <. See Exmple.4.

6 6 R. E. SHOWALTER Exmple 3.2. The sequece f (x) = x coverges poitwise to f(x) = o (0, ), but the covergece is ot uiform. The covergece is uiform o y set of the form S = (α, ) with 0 < α < or S = [α, ]. See Exmple.5. Theorem 3.. If the sequece {f ( )} is uiformly coverget to f( ) o S d if ech f ( ) is cotiuous, the the limit f( ) is cotiuous o S. Proof. Let lim f ( ) = f( ) uiformly o S d x 0 S. Let ε > 0. Uiform covergece implies there is N for which f N (x) f(x) < ε 3 for ll x S. Cotiuity of f N implies tht there is δ > 0 such tht f N (x) f N (x 0 ) < ε 3 for ll x S with x x 0 < δ. But the we hve f(x) f(x 0 ) f(x) f N (x) + f N (x) f N (x 0 ) + f N (x 0 ) f(x 0 ) < ε for ll x S with x x 0 < δ. Theorem 3.2. If the sequece of cotiuous fuctios {f ( )} is uiformly coverget to the (cotiuous) fuctio f( ) o S = [, b], the we hve lim f (x) dx = f(x) dx. Proof. Let ε > 0. There is N for which N implies f (x) f(x) < ε for ll x [, b]. The b for N. f(x) dx f (x) dx f(x) f (x) dx < ε Corollry 3.3. If the sequece of cotiuous derivtives {f ( )} is uiformly coverget to the (cotiuous) fuctio g( ) o S = [, b], d if the sequece {f ( )} coverges poitwise to f( ) o S = [, b], the f( ) is differetible d f = g. Proof. For ech x (, b] we hve x f (s) ds = f (x) f (), d tkig limits yields x g(s) ds = f(x) f() Let {f ( )} be sequece of fuctios o the set S. As before, we defie ew sequece {s ( )} by s ( ) = f m ( ) = f ( ) + f 2 ( ) + + f ( ),, x S. m= This is the sequece of prtil sums of {f ( )} or the series If the sequece {s } is poitwise (or uiformly) coverget, we sy the series f coverges poitwise (or uiformly, respectively) d deote its limit lso by f (x) = lim s (x). f.

7 POWER SERIES 7 Exmple 3.3. Tkig the sequece f (x) = x, we obti the geometric series x = o (, ). The series is bsolutely coverget, x =0 poitwise o (, ), d uiformly o y subitervl [, b] (, ). 4. Power Series Defiitio 4.. A ifiite series of the form (4.) (x x 0 ) = 0 + (x x 0 )+ 2 (x x 0 ) (x x 0 ) =0 is power series i x bout the poit x 0. This is series of fuctios costructed from the terms f (x) = (x x 0 ) for 0. The geometric series resulted from the prticulr choice of coefficiets = d x 0 = 0. Theorem 4.2. If the power series (4.) coverges t the poit x = x 0 + r, the it coverges bsolutely t y poit x with x x 0 < r, d for y p with 0 < p < r the covergece is uiform o those x with x x 0 p. Proof. Sice the series =0 r is coverget, we hve lim r = 0, so there is iteger N such tht r < for ll N. Thus, for ll N we hve (x x 0 ) = r x x 0 < x x 0, so by the r r compriso test we see tht the series (4.) coverges bsolutely for ll x with x x 0 < r. Moreover, these estimtes show the covergece is uiform for x x 0 p for y p < r. It follows tht the set of poits t which the series coverges is either the sigle poit 0, itervl (x 0 R, x 0 +R), possibly cotiig either edpoit, or the etire umber lie IR = (, ). The umber R is the rdius of covergece, d we set R = 0 i the first cse d R = i the lst. Theorem 4.3. Let R > 0 be the rdius of covergece of the power series (4.). The the fuctio f(x) = =0 (x x 0 ) is ifiitely differetible d its derivtive is give by the power series f (x) = (x x 0 ) with the sme rdius of covergece. Proof. Let x x 0 < R d choose ξ i the itervl of covergece, ie., ξ x 0 < R, with R 2 < ξ x 0 d x x 0 = p ξ x 0 with 0 p <. The the differetited series is bouded by (x x 0 ) = (ξ x 0 ) p 2 R Cp sice the coverget series =0 (ξ x 0 ) hs bouded terms. The series =0 p coverges by the limit rtio test, so the differetited series coverges for x x 0 < R by the compriso test.

8 8 R. E. SHOWALTER Corollry 4.4 (Tylor s formul). If the power series (4.) hs rdius of covergece R > 0 d its limit is the fuctio f(x), tht is, (4.2) f(x) = (x x 0 ) =0 = 0 + (x x 0 ) + 2 (x x 0 ) (x x 0 ) , x x 0 < R, the the -th derivtive of the sum f(x) t x 0 is give by f () (x 0 ) =!, 0. Deprtmet of Mthemtics, Orego Stte Uiversity, Corvllis, OR E-mil ddress: show@mth.oregostte.edu

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