On Taylor Series for which liman+1/an=1,

Transcription

1 On Taylor Series for which liman+1/an=1, by Richard H. COOKE, London. very restricted, but interesting class ; it has been considered (in particular) by the authors cited in the list of references at the end of this note. Two small, and disconnected questions concerning stich series are here considered. (i) We have the well-known theorem of Fabry (2) : Theorems of converse typo have been given, by H ad a m a r d (1), when the singularities of f(z) on z =1 are all isolated and of finite order, and there is only one singularity of the highest order, situated at eta(2), and by Ju nge n (3), who has shown that if f(z) has only alg(braico-logarithmic singularities on z =1, of which that at z=1 is of greater weight than the others, then lim an'+1/an'=1, where (n') is a sequence of density 1(8 ). But Hadamard(4) gave an example to show that even when there is only one singularity on z =1, lim an+1/an does not exist necessarily. Our first question is: if z=1 is an essential sigularity, and the sole singularity of f(z), in the entire plane, can any Simple conditions be given so that liman+1/an=1? (ii) It was proved by Narumi and Izumi, in the papers (6), (7) cited at the end of this note, that (1) A simple transformation will replace 1 by ela; so that without loss of generality, we may take a=0 when convenient. (2) Hadamard (1), 148. (3) Juiigen (3), 276. (4) Hadamard (1), 109.

2 320 RICHARD H. COOKE : of tho first robult by Shimizu (8) runs as follows : then(2) Our second question is: what are the corresponding result for points inside and on the circle of convergence? 2. In answer to the first question, the following result is proved. Let be a uniform function having z=1 as an isolated essential singularity, and its sole singularity in Since z = 1 is the sole, singular point of f (z), f(z) is an integral function of 1/(I-z), and therefore of Thus f(z)=en(z/(1-z))n, where lim en=0 ; so that mo can be found so that for m>m0, and an arbitrary positive c, (2.1) (2.2) For a sufficiently. largo n, lot Rn, be the sum of the above series from the (m+1)-th term to the n-th term, inclusivo. Let be an arbitrarily small positive number, then if n is so large that n8 > mo, wo have by (2.1), (1) The restriction that I +k(n) ( Is non-decreasing for nn0 was removed by Izum1, (7), (2) I have replaced i4, zn-lm, as given by Shimizu, In,the formuls for am(z), by zm+1, which appears to me to be the correct expression.

3 ON TAYLOR SERIES FOR WHICE lim an+1/an= Now which is true if i. e., if But Cmem+1/n-1 Cmem+1=n/(n-m) is greatest When m=[n], (m=0, 1,..., [n]). Therefore (2.3) Since 8 is arbitrarily small, the result follows by (2.2) and (2.3). Simple examples of the theorem are given by taking f (z) =e/(1-s), cosh (z/(1-z)). 3. We now consider the second question. We first prove that If Rn(z) is the remainder after the term anzn in Z anzn, and lim an+1/an=1, then Rn(z)anzn+1/(1-z) uniformly for z sr<1. We have But so that

4 322 RICHARD H COOKE: Hence and so Rn(z)-anzn+1/(1-z) uniformly for z < r<1. The corresponding result of Narumi-Izumiquoted in 1, when z r> 1, may be restated as follows : If an/an+1-1, then at all regular points of f (z) =anzn, Rn (z) ~anzn+1/(1-z) as n-oo uniformly for z r<1, where Rn(z) f(z) -s (Z). For, at regular points, f(z) M, and M/a,zn+10 as n-oo when I z r> 1. Thus the results for points inside and outside the circle of convergence are the same, at all regular points of f(z). We next prove that If, in T anzn, line an_,/an(n) =1, where lira k(n + 1)/4'(n) -1, and (n) I is non-decreasing for n no, then Rn(z) anzn+1/{, (n)-z} uniformly for I z I s (1--5)1 *(n) (, 8 being an arbitrarily small positive number. When I z j s (1-8) I *(n) j, the binomial series for is absolutely convergent; there are then integers m and N, such that for nznaino, and (i) (ii) {iii) (iv) (iv) follows since I fi(n) I is non-decreasing for n;a;no, and (i), (ii), (iii) are obvious from the hypothesis ; s, e', 8 are given arbitrarily small positive numbers.

5 ON TAYLO1t BERIES FOR WHICH lim an+i/an= Then for iam, nn, we have by (ii) and (iv), so that since z _ (1 8). I *(n) 1, and (n) IsI+(n+1), (nn0). s" being an arbitrary positive number, if m is taken sufficiently large. Hence j being an arbitrarily small positive number,, when m and n are taken sufficiently large. The result then follows, since am+1(n)/an+1. The corresponding result of Na ru m i-i zu m i can also be given as at all regular points of f(z), where An(z) f(z)-sn(z). We now prove that i, e., am+k]am-1 as moo, then uniformly for z r<1, Let

6 if 0<a<2 and when a 324 RICHARD H. COOKE : Then But, replacing zk in the above extension of the Naruml-Izumi theorem to the case where z r<1, we have from the hyposhesis, Hence, uniformly for we have or which is the result stated. When k-1, this reduces to agreeing with the previous result; and the (corrected) result of Shimizu, as given in 1, then reduces to sm(z) amzm+1/(z -1), agreeing with theorem of Narumi-Izumi. Finally, we consider the corresponding result for points on the circle of convergence., The assumption that an/a+1-1 only seems insufficient to arrive at any conclusions, but if we assume that cn/an+1=1+cn-1+0(n-1), where c> 1, we obtain results precisely analogous to those already obtained for points inside and outside the circle of convergence. We prove, in fact, that If, in anzn, an/an+1=1+cn-1+o(n-1) where c>1,,then =0 we have Rn(1) (c-1) Under, the conditions of the theorem, 2:" a.z" converges absolutely at all points of z =1, and (first taking the case o =O) Thus where

7 ON TAYLOR SERIF 3 FOR WHICH lim an+1/an= and for n>n0, there is a positive constant K such that (3.1) Also (3.2) Thus from (3.1) and (3.2), we see that where r=n+s. Hence and consequently so that, with the previous notation, we obtain as before, (3.3) Now by Wilton's transformation of series(1)) (1) Wilton, (10), 83, formula (1.11).

8 326 RICHARD H. COOKE: (3.4) But, by partial integration, we see that and Hence and Consequently, by (3.3) and (3.4), whence This proves the theorem.

9 ON TAYLOR SERIES FOR WHICH liman+1/an= References. (1) J. Hadamard, "Essai sur letude des fonctions donnies par leur developements do Taylor," Journal do Math, (4), 8 (1892), (2) E. Fabry, "Sur les points singuliers d'une fonction donnee par son developeanent en eerie, et sur l'impossibilit6 du prolongement analytique dans los cas tres g6n6raux," Annales scientifiques do L'lvcole Norniale Sup., 3e serie, 13 (1896). (3) R. Jungen, "Sur lee series de Taylor n'ayant quo des singularit6s alggbrico-logarithmiques sur leur cerele de convergence," Commentarli 'Math. ielvetici, 3 (1931), (4) J. Rey Pastor, "Sobre ]as singularida des algebricologaritmica de las flpclone analiticas," Bol. del sem. Matematico institute mat. Hispano Americano, 4 (1935), (5) V. Bernstein, " Aleune osservazione sopra un teorema di Fabry," Atti Accad. naz. Lincei Rend. (6), 21 (1935), (6) S. Narumi, "On the distribution of the zero points of sections of a power series," Japanese Journal of Math, 4 (1927), and (7) S. Izumi, (same title as in (6)), ibid, and (8) T. Shimizu, "'On some power series and their sctions," ibid, (9) J. Hadamard and S. Mandelbrojt, La seriede Taylor et son prolongement analytique, (Paris, 1926). (10) J. R. Wilton "'Some applications of a transformation of a series," Proe. London Math. Soc, (2), 27 (1928), Birkbeek College, University of London, England. (Received July 24, 1939).