Now we determine an by the following equations;

Transcription

1 8. On a Power Series which has only Algebraic Singularities on its Convergence Circle. U. By Masatsugu TSUJI. (Received April 1, 1927.) Let f(x)= ƒ n=0anxn have the convergence radius 1, then Jentzch has proved that every point on the unit circle is the limiting point of roots of sections f(x)=a0+a1x+ c c+anxn. In my previous paper (1), I have investigated a class of power series which has no limiting points of roots outside its convergence circle, for example, the series for which an `Ank, where k is any real number, or when f(x)= ƒ n=0anxn has only one algebraic singular point on the unit circle, then an `Ank, so that there are no limiting points of roots of s ections outside the unit circle(2). I will here give an example of a power series with the convergence radius 1, such that every point outside the unit circle is the limiting point of roots of sections. Let all rational points outside x =1 be numbered so that x1, x2 c c xn, c c where and so on. In general Now we determine an by the following equations; (1) M. Tsuji, this journal, 2 (1926), (2) M. Tsuji, this journal, 3 (1926), 82.

2 From these equations we get,

3 ON A POWER SERIES WHICH HAS ONLY ALGEBRAIC SINGULARITIES. 51 On the other hand therefore Hence the convergence radius r of f(x)= (a0=1) is 1 and since every point outside x =1 is the limiting point of roots of sections, and f(x)_??_0. Accordingly r=1. This power series f(x)= has the requi red properties Now we will show that if has only algebraic singu lar points on x =1, there exists a domain outside x =1 near a certain singular point on x =1 in which limn= fn(x)= uniformly and consequently free from limiting points of roots of sections. Let the singular points on x =1 be 1/ą1, c c 1/ąk, then f(x) can be expanded about 1/ąr in the form: f(x)=b0(r)(1-Ěrx)m/pr+b1(r)(1-Ěrx)mr+1/pr+ c c where mr, pr are integers and pr>0 (r=1, 2, c ck), k being the number of singular points on x =1. Let br(r)(1-Ěrx)mr+vr/pr be the first term in the above expansion such that br(r) 0 and mr+vr/pr is neither zero nor a positive integer, and Đ be the minimum of mr+vr/pr (r=1, 2, 3, c c k). We number Ě so that

4 52 M. TSUJI I have already proved (3) n 0 uniformly for x r0>1 where we write cr=bƒër(r) (r=1, c ck1) cr 0 and r0 is any number greater than 1. Hence if there are only poles on x =1, then ƒê is the maximum order of poles on x=1 and 1/ƒÖ1, c c 1/ƒÖk1 are the poles of maximum order. Now if x lies outside x =1 and sufficiently near to 1/ƒÖ1, for example, we can determine a neighbourhood of 1/ƒÖ1 outside x =1 so that Hence so that limn= f(x)= uniformly in that neighbourhood. Hence there are no limiting points of roots of sections. As a special case we have, Theorem T. If f(x)= ƒ n=0anxn has the convergence radius 1 and no other singularities as poles on x=1, then limn= fn(x)= uniformly in a certain domain outside x =1 near every pole of maximuon order Again let f(x)= ƒ n=0anxn has only algebraic singularities on x =1, then as I have mentioned above, Now putting (3) M. Tsuji, l. c. Foot note (2) p. 80.

5 ON A POWER SERIES WHICH HAS ONLY ALGEBRAIC SINGULARITIES. 53 we can prove(4) that for any k1 consecutive ƒðn, ƒðn+1, c c ƒðn+k-1, there exists at least one r (n r n+k1-1) such th t where ƒã0 is an absolute constant so far as we consider the domain x r0>1, r0 being any number greater than 1. Hence we can find I can not prove whether this lower limit 2 is the true limit for all power series with the convergence radius 1, but if f(x) has only algebraic singularities on x=1, then as above this limit 2 can be reduced to 1. (4) M. Tsuji, l. c. Foot note (2) p. 70. (5) M. Tsuji, this Journal 1 (1925) 31.

6 54 M. TSUJI 4 where k is the number of singular points on x=1. From this follows that in any k consecutive terms an there is at lust one non-vanishing term., an+1, c can+k-1 Mr. G. Polya (7) and Mr. Narumi (on the annual meeting April 1927 of the Physico-mathematical Society of Japan) extended this theorem for functions with algebrico-logarithmic singularities of more general type. Mr. Ostrowski (8) investigated the same problem and remarked that in the above form, k can be taken as the number of poles of maximum multiplicity on x=1 when f(x) has only poles on x=1. Mr. Mandelbrojt (9) proved that if aĕ+nq (n=0, 1 c c c) vanish, then on x=1. Mr. Ostrowski (8) extending this theorem, proved that if aĕ1+nq=0, aĕ2+nq=0 c c aĕk+nq=0 (n=0, 1. 2 c c) and if q is a prime, then there exist at least k+1 singular points on x=1. I will prove that when f(x) has only algebrico-logarithmic singulari ties on x=1, we can find a suitable ĕ independent of q such that every sequence of number aĕ+nq (n=0, 1, 2 c c c) contains infinitely many nonvanishing terms, where q is any integer. algebrico-logarithmic singularities on x=1, k in number, then there exists suck a number ĕ independent of q that q being any positive integet. And such ĕ can be found in any k consecutive integers. (6) M. Tsuji l. c. Foot note (2) (7) G. Polya C, R 184 (28 Feb. 1927). (8) Ostrowski: Uber Singularitaten gewiseer mit Lucoken behatteten Potenzreihen Jahresbericht der D. M. V. 35 (1926) (9) These

7 ON A POWER SERIES WHICH HAS ONLY ALGEBRAIC SIXGULARITIES. 55 Hence any k consecutive ƒ ƒï+nq, ƒ ƒï+(n+1)q, c cƒ ƒï+(n+k-1)q, for n n0 whose indices form an arithmetic progression, contain at least one which is not zero. Proof. As I have proved (10) if f(x)= åƒ n=0anxn has only algebrico logarithmic singularities on x =1 where ƒê, cƒë, ƒöƒë have the same meaning as the last paragraphs. Now we put then Among ƒö1q, c ƒöqk1 some may be equal, say, We put then We consider k1 equations in c1ƒö1ƒï, cck1ƒöƒïk1: From this it follows that c1ƒö1ƒï= c=ck1ƒök1ƒï=0 or c1= c= ck1=0 which contradicts the hypothesis c1 0, c ck1 0. Hence there exists one such that Such ƒð is independent of q. Some of the brackets is therefore not zero, and without loss of generality we can suppose (10) M. Tsuji, l. c, Foot note (2) p. 77.

8 56 ON A POWER SERIES WHICH HAS ONLY ALGEBRAIC SINGULARITIES. We write p for ƒï+ƒð and get From this follows easily that Mathematical Institute, Imperial University, Tokyo.