62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

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1 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of the series I geeral, the series will coverge or diverge, depedig o the choice of x. The power series always coverges for x = 0 to the umber c 0. Example 98. For what values of x does the power series series x / coverge? Solutio: By the root test, x = x x as

2 62. POWER SERIES 125 So, the series coverges for all 1 < x < 1 ad diverges as x > 1 or x < 1. The root test is icoclusive for x = ±1. These values have to ivestigated by differet meas. For x = 1, the power series becomes the harmoic series 1/ which is diverget. For x = 1, the power series becomes the alteratig harmoic series ( 1) / which is coverget. Thus, the power series coverges if x [ 1, 1) ad diverges otherwise. Give a umber a, cosider a power series i the variable y = x a c y = c (x a) It is also called a power series cetered at a or a power series about a. Let S be the set of all values of x for which a power series i x coverges ad let S a be the set of all values of x for which the correspodig power series i (x a) coverges. What is the relatio betwee S ad S a? Sice the series are obtaied from oe aother by merely shiftig the value of the variable by a umber a, x x a, the set S a is therefore obtaied by addig the umber a to every elemet of S: (60) x S a x a S = S a = {x x a S} For example, the series (x 2) / coverges if x 2 [ 1, 1) or x [1, 3) ad diverges otherwise by Example 98. Thus, the problem of fidig the set S a is equivalet to the problem of fidig the set S Power series as a fuctio. Suppose that a power series i x coverges o a set S. The it defies a fuctio o S: f(x) = c x, x S The set S is called the domai of such a fuctio. Fuctios defied by power series are most commo i applicatios. May of them have special otatios (like elemetary fuctios si, cos, exp, etc). Their properties are well studied. I what follows it will be show that familiar elemetary fuctios si(x), cos(x), exp(x), etc ca also be represeted as power series. Example 99. Fid the domai of the Bessel fuctio of order 0 that is defied by the power series J 0 (x) = ( 1) 2 2 (!) 2 x2

3 SEQUENCES AND SERIES where, by commo covetio, 0! = 1. Solutio: Sice a = c x 2 cotais the factorial, the ratio test is more coveiet: a +1 a = x 2 c +1 c = x (!) 2 2 2(+1) (( + 1)!) 2 = x ( + 1) 2 0 as. So, the series coverges for all x. Values of a fuctio defied by a power series ca be estimated by partial sums which are polyomials i the variable x: f(x) f (x) = c k x k = c 0 + c 1 x + c 2 x c x k=0 Thus, partial sums defie a sequece of polyomials that coverges to the fuctio o S, f (x) f(x) for all x S. The accuracy of the approximatio is determied by the remaider R (x) = f(x) f (x). The accuracy assessmet is discussed i Sectio Sice the remider R (x) is a fuctio o S, the error of the approximatio is ot geerally uiform, i.e. it depeds o x Radius of covergece. The set S o which a power series is coverget is a importat characteristic ad its properties have to be studied. Lemma 2. (Properties of a power series) (i). If a power series c x coverges whe x = b 0, the it coverges wheever x b. (ii). If a power series c x diverges whe x = d 0, the it diverges wheever x d. Proof: If c b coverges, the by the ecessary coditio for covergece, c b 0 as. This meas, i particular, that for ε = 1 there exists a iteger N such that c b < ε = 1 for all > N. Thus, for > N, c x = c b x = c b x < x b b b which shows that the series c x coverges by compariso with the geometric series q where q = x/b ad x/b < 1 or x < b. Suppose that c d diverges. If x is ay umber such that x > d, the c x caot coverge because, by part (i) of the lemma, the covergece of c x implies the covergece of c d. Therefore c x diverges. This lemma allows us to establish the followig descriptio of the set S.

4 62. POWER SERIES 127 Theorem 45. (Covergece properties of a power series) For a power series c x there are oly three possibilities: 1. The series coverges oly whe x = 0 2. The series coverges for all x 3. There is a positive umber R such that the series coverges if x < R ad diverges if x > R. Proof: Suppose that either case 1 or case 2 is true. There there are umbers b 0 ad d 0 such that the power series coverges for x = b ad diverges for x = d. By Lemma 2 the set of covergece S lies i the iterval: x d for all x S. This shows that d is a upper boud for the set S. By the completeess axiom, S has a least upper boud R = sup S. If x > R, the x S ad c x diverges. If x < R, the x is ot a upper boud for S ad there exist a umber b S such that b > x. Sice b S, c x coverges by Lemma 2. Theorem 45 shows that a power series coverges i a sigle ope iterval ( R, R) ad diverges outside this iterval. The set S may or may ot iclude the poits x = ±R. This questio requires a special ivestigatio just like i Example 98. So, the umber R is characteristic for covergece properties of a power series. Defiitio 17. (Radius of covergece) The radius of covergece of a power series c x is a positive umber R > 0 such that the series coverges i the ope iterval ( R, R) ad diverges outside it. A power series is said to have a zero radius of covergece, R = 0, if it coverges oly whe x = 0. A power series is said to have a ifiite radius of covergece, R =, if it coverges for all values of x. The ratio or root test ca be used to determie the radius of covergece. Corollary 3. (Radius of covergece of a power series) Give a power series c x, if if c +1 lim c lim = α = R = 1 α c = α = R = 1 α where R = 0 if α = ad R = if α = 0 Proof: Put a = c x i the ratio test (Theorem 60.4). The a +1 / a = x c +1 / c x α. The series coverges if x α < 1 which shows that R = 1/α. Similarly, usig the root test (Theorem 40), a = x c x α < 1, which shows that R = 1/α.

5 SEQUENCES AND SERIES Remark. If the sequeces i Corollary 3 do ot coverge, the Theorem 41 should be used where a = c x. Oce the radius of covergece has bee foud ad 0 < R <, the cases x = ±R have to be ivestigated by some other meas (as the root or ratio test is icoclusive i this case) to determie the iterval of covergece S of a power series. Example 100. Fid the radius of covergece ad the iterval of covergece of the power series c x where c = ( q) / + 1 ad q > 0 Solutio: c +1 c = q q = q = q 1 + 1/ 1 + 2/ q = α Therefore R = 1/α = 1/q. If x = 1/q, the c x = ( 1) / + 1 = ( 1) b. The sequece b coverges mootoically to zero so that ( 1) b coverges by the alteratig series test. If x = 1/q, the c x = 1/ + 1 > 1/ 2, 1. The p series 1/ 1/2 diverges (p = 1/2 < 1) so that 1/ + 1 diverges by the compariso test. Thus, the iterval of covergece is S = [ 1/q, 1/q). Example 101. Fid the radius of covergece ad the iterval of covergece of the power series 2 (x + 1) /q where q > 0. Solutio: Put y = x +1. If S is the iterval of covergece of c y where c = 2 /q the the iterval of covergece i questio is obtaied by addig 1 to all umbers i S accordig to the rule (60). c = 1 q 2 = 1 q ( ) 2 1 q = α So R = 1/α = q. If y = q, c y = 2 ad the series 2 diverges (a = 2 does ot coverge to 0). If y = q, the c x = ( 1) 2 ad the series diverges because a = ( 1) 2 does ot coverge to 0. The series coverges oly if y = x + 1 < q ad, hece, the iterval of covergece is x ( q 1, q 1) (the iterval ( q, q) shifted by 1).

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