advertisement

I ntrnat. J. Math. Math. Sci. Vol. 2 No. 2 (1979) 239-250 239 REMAINDERS OF POWER SERIES J.D. McCALL Department of Mathematics LeMoyne-Owen College Memphis, Tennessee 38126 U.S.A. G.H. FRICK:E Department of Mathematics Wright State University U.S.A. 45431 Dayton, Ohio W.A. BEYER Department of Mathematics Los Alamos Scientific Laboratory Los Alamos, New Mexico 87545 U.S.A. (Received February 21, 1978 and in Revised form March 20, 1979) n 17’n=N anZ I" z 2. questions. z 2 (b) n < Suppose {N[ ON(Z I) Put S be cofinite? for T En= 0 anZ Suppose ABSTRACT. > has radius of convergence R and z21 ON(Z) < R, and T is either z 2 or a neighborhood of for z e T}. can S be infinite? Two questions are asked: (a) an a neighborhood of z 2. a # O. can S This paper provides some answers to these The answer to (a) is no, even if T if lim ON(Z) z 2. The answer to (b) is no, Examples show (b) is possible if T z2 and for T J.D. McCALL, G.H. FRICKE AND W.A. BEYER 240 KEY WORDS AND PHRASES. Power-Series, Remainders, Radius of Convergence. AMS (MOS) SUBJECT CLASSIFICATION (1970) CODES. 1. 30AI0. INTRODUCTION. This paper originated in a question of approximation by power series raised in Query 51 in the American Mathematical Society Notices Ill. (The query originated in considerations of analytically continuing a polynomial series from the interval mn=O anZn has Suppose f(z) Suppose Zll > {nlOn(Z I) [-1,1] < radius of convergence R and < R and T is either z z21 On(Z) to the region of convergence of the series.) for z g T}. ON(Z) or a neighborhood of z 2 m IXn=N anZ n I. Put S 2. S is cofinite if its complement is finite. Two questions are asked- (a) can S be cofinite? (b) can S be infinite? One might expect the answer to both questions to be no since one expects the approximation to f by partial sums of its power series to be worse, closer to the circle of convergence. Section 2 shows This paper provides some answers to these questions. is impossible for any T. Section 3 shows lim an (b) is possible for T a neighborhood of z a # 0. Section 4 shows (b) is impossible if T (b) is possible for T 2 and Section 5 shows 2 and 2. Section 5 suggests the conjecture that if S must be "thin." z z (a) T is a neighborhood of z2, then The S which appears in Section 5 is lacunary. These questions can also be raised about other series of orthonormal polynomials with elliptic domains of convergence. (cf. Szeg [5], pp. 309-10). REMAINDERS OF POWER SERIES 2. 241 S CANNOT BE COFINITE. The following theorem was suggested by P. Lax [3]. TEOREM 1 [z2[/[Zl[ If lira a n 11/n 1/R < , {n[[k=n akz 2 then the set S 0 < k [Zl[, [z2[ < R and 0 < 6 < k akZl [} < 6 n [k=n cannot be cofinite. PROOF. I. Suppose S contains Then for O n a nonempty tail set I; i.e. n_l implies n+l n, > (z) 1 > 1 n+l (z) [a n [[z 1 in > 6-(n+1) Crn+l (z2) 6-(n+l)[[a n ][z2 ]n n(Z2)] [6-(n+l)[z2 In > [a n [an] [z 1 [n [an[[Zl In [Zl [n] -6-1 0 n (Zl) Hence (1+6 -1) Suppose 1/R so large that I/R e,. lakl On(Z1) 0. > lan[ Choose g [6-(n+1)[z2 In- [zl[n] > 0 so that 1/k < (1/R + ) for k _> (R -1 + 1-(R-I+:) Iz > ’k=n lakl IZl Ik > n(Zl) [an[ [6-( n+l)lz2 In 1+6-1 < 1 and choose n z Also choose n n. Then [(R-l+)[Zl[] n > e)[Zl[ IZl In] so, that lan I1/n > J.D. McCALL, G.H. FRICKE AND W.A. BEYER 242 > n (R- l-g) n [6- (n+l) Iz21 -1 1+6 Izl n] Now in addition to the other conditions on n, choose n large enough so that Then, since (R-l+g) [Zl[ [z2[ R-l_g > 6 [l_(R-l+)lZll]l/n- (l+6-1)l/n one obtains upon letting g + 0 and n + contradicting 6 < [z Suppose R set, {la n -1 l/nln > 2]/[z 1[. Then [a O. 1 lanlll/nl. iargest element lan211/n2 obtain a sequence for n _> n..1 ni, Also lim n [1/n converges to zero. If we zero to the add the new set is closed and bounded and thus compact with the Iargest eiement is a m" Deleting lall la211/2,..., lanlll/nl, Thus we in the remaining set and so forth. i 1,2,..., with i g’1 O. [an i[ I/ni g 1 0 and [a Thus for i large enough that there n [1/n gi[Zl[ < . i < I" 243 REMAINDERS OF POWER SERIES no > 1-ilZl[ [k=n. [ak[ [Zl[ k -(ni+l) i 1+5 -1 nil Iz2l 1+5-1 6 Now choose n.x so that ei IZll (61Zll/Iz21)ni lan i > 1/n.- (1-e ilzl) > o (Zl) 1 n [z2[ n [Zl[ l 6[z 1 -1 Iz2l < 5 -1 Iz2l Then -1 6 -1 1/n. l- 61z 1 Iz21 (1+5 or Izll > 1/n. Letting :.1 1 1/n 5(1+5 (1-ilzll) + 0 and n. + 1 m, / Iz21 i one obtains Izll>contradicting 6 < lz2[/[z I [. Iz21 6 This completes the proof of Theorem 1. The following observation about general series was made by a referee. O Ap be convergent If 0 p lbpl E Ap p>N < < E >N then Apbp Let 244 J.D. McCALL, G.H. FRICKE AND W.A. BEYER is not cofinite. For let R p>n Ap n Then A Rn-Rn+l" If S were co- finite, then for n > n o >_n Rn+l p or p>N If N o p>N p>Np>n is selected so large that plb < 1/2, then for N > N o 1 p>N p>N which is a contradiction. A p>N If one puts a P z2 bp= zl 22 )P then under the hypothesis of Theorem I, one obtains the weaker result that the set Z akz2 k=n k akZl I k=n cannot be cofinite. 3. CASE OF LIMN AN A 0. In this section it is shown that (b) if limn an and ]z] < R a 1 # 0. is impossible for even a single point The proof is as follows. For > 0, N large enough, 245 REMAINDERS OF POWER SERIES nN anzn ON(Z) < lal Izln ll-zl + a n=N n=N / g (a n a)z n [z[ n Also N Izl ll-zl Z zn a a n=N + n n-N n (a-a)z n N < o (z) + e n Izl 1-1zl Thus lal Suppose IzlN e I-zl oN(z2) lal < IzllN <- N(Z) <- i’ zl ON(ZI) Iz2l N I1_z21 lal (1) Then (I) gives N Iz2l < 1_lz21 <-ON(Z 2) O’N(Z 1) IZll N i1_zll + e IZll [z 1[ < [z 2[. _< IZll N 1_1Zll Taking Nth roots, letting N Iz21 a contradiction of I-zl N + e for infinitely many N. <_ lal for infinitely many N. IzlN + m, and g + 0, yields 246 4. J.D. McCALL, G.H. FRICKE AND W.A. BEYER {z2} FOR T- (b) IS POSSIBLE. {z2}. The following example shows (b) is possible if T Let 2 -1 (1-2z)(1-z) F(z) l-2z + z 2 2z 3 + 4 z 2z 5 + One has Iz 2k O2k(Z) ]z] 2z 2k ]I 2k+l + z 2k+2 2z + z 2 =""[z[ 2k [1- 2z[[1and thus O2k(I/2) 0. So for any z 2z 2z z 2k+3 + 3 + 2 -1 1/2 and 0 < I Izll < I, O2k(Z I) > O2k(1/2). Note that for an e-neighborhood of 1/2" 0 < g < I/2 and for any z (2k(Z) faster than with I ]z I] < I/2 {zl N g, Iz O2k(Z I) at any point z in N except I/2. 1/21 < converges to zero So we cannot extend the result to a neighborhood of 1/2. 5. CASE OF T A NEIGHBORHOOD OF z 2. THEOREM 2. IZll < < R and Iz21 _< m, For each R, 0 < R PROOF. where b k with 2. Suppose R max O<j<n2 k I. Put n 2 k 4 k and Pk(Z) (]/b k) The power series j wili be shown to satisfy the Theorem for R Note that and z 2 2mn=0 anZ n with radius of convergence R such N, ON(Zl)/3 ON(Z) for all z in some neigh- a power series that for infinitely many values of borhood of z there exist points z with z zn2k-I I/2)n2k, (z 2k=l Pk (z) 2n=0 -1/4 and z 2 a z n 1/2. 247 REMAINDERS OF POWER SERIES + n2k n2k-1 < (2) n2k+l and n2k_l (log 4/log for all k. (2) implies that each a n 3 + I) < (3) n2k is either zero or appears exactly once as a coefficient in the expansion of some Pk(Z). max0<j<n2k Then Let Jk be the integer for which v\ 1/(j aj+n2k_l (n2.O +n2k k 1/(J+n2k -1 2- j (0 < j < This is less than or equal to one for all j and equal to one for j n2k) Jk’ which implies the radius of convergence is one. For all z with 1/21 < 1/4" z 1 IPk+ l(z)[ bk+l Next, for Iz- Izl < k < IPk(Z) Izl < (1/4) 1/21 < 1/4, IPk(Z)l IPk(-1/4) Izl n2k+l Iz n2kol Iz-1/21 1/21 n2k+2 n2k (1/4) n2k+2 Iz-1/21 n2k+2 n2k IPk(Z)l n2k_ 1 Iz-1/21 n2k n2k (4 4n2k 1 (4/3) n2k 248 J.D. McCALL, G.H. FRICKE AND W.A. BEYER < 4-n2k 4n2k-l(4/3)n2k 4 by (3). Hence, o n2k- for n2k_ 1 -n2k < 1/4 3 Iz- 1/2) Z J=n2k_ (z) Z _< < 1/4, a .Z 4k-j n2k- 1 and z 2 n.’s are even. Z IPj(z)[ < (1/3) by (4) IPk(-I/4)I by (5) Z b 1 (-1/4)n2j-I (-3/4)n2j{ (1/3) o since all <_ ]Pk(Z)[ IPk(Z)l < (1/3) j 1 j=k (4/3) () (-1/4) This shows that the assertion holds for z I -I/4 1/2. For the case 0 < R < m, use the power series result holds for z I For the case R -R/2, z 2 R/2, and the neighborhood m, let n2k-l n2k 2 -Jk bk (n2k-I) Jk For 0 < j < n2k- =0 an(z/R)n" Iz Then the R/21 < R/4. 249 REMAINDERS OF POWER SERIES [aj+n2k_l 1/(J+n2k_ < (n2k- < (n2k) and hence lim R I. 6. AVERAGE REMAINDER Suppose in a z n n a I/n n /(J+n2k- 1) -j n.2k! k n2k-1 (nak_l (n2k) as k 2 n2k -j 2 -n2k_i/(J+n2k_ 1) -n2k- / (n2k -1/5 0. + n2k- 0 The rest of the proof follows the case has a radius of convergence R. It follows from results Plya and Szeg [4, Part III, problems 307-310] that the geometric GN(r) exp -o log oN(re10)d0 mean" (r < R) and the p th mean, p > O- o(rei0)d 0 (r < R) are both monotone increasing functions of r for each N and log IN(r) P are convex functions of log r. mean sense, oN(z) GN(r)- and log Thus in the geometric mean sense and pth become larger as one approaches the circle of convergence. J.D. McCALL, G.H. FRICKE AND W.A. BEYER 250 ACKNOWLEDGMENT The authors thank Dr. Leon Heller of the Los Alamos Scientific Laboratory, who proposed this problem, for many discussions and contributions. REFERENCES I. Beyer, W. A. and L. Heller, Query 51, Amer. Math. Soc. Notices 21 (1974), 280. 2. Fricke, G. H., Answer to Query 51, Amer. Math. Soc. Notices 22 (1975), 72. 3. Lax, P., private communication, June 14, 1974. 4. P61ya, G. and G. SzegS, Problems and Theorems in Analysis, Vol. I, Springer-Verlag, New York, 1972. 5. SzegS, G., Orthogonal Polynomials, Amer. Math. Soc. Colloquium Publications 23, (1959).