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# COMPLEX VARIABLES II MATH 618

Texas A&M

GPA 3.92

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This 3 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 618 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/226032/math-618-texas-a-m-university in Mathematics (M) at Texas A&M University.

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Date Created: 10/21/15

Math 618 Example April 227 2007 An entire function of intermediate growth Answering Jan Cameron7s question7 Dakota Blair suggested the following example of an entire function that grows faster than any polynomial but that has order zero 00 H 1 i 3 n n1 Since the function has in nitely many zeroes7 it is not a polynomial conse quently7 as John Paul Ward pointed out7 the function must grow faster than any polynomial7 by the version of Liouville7s theorem in Exercise 78 in the textbook On the other hand7 since Zjnmlf6 converges for every positive 67 the function has order zero1 If M0 denotes the maximum of the modulus of our function on a circle of radius r then evidently logMr iloglt1 n1 39 The preceding general considerations show that log M0quot must grow faster than k log for every positive constant k but slower than r5 for every positive constant 6 ln class7 I tried unsuccessfully to show that log M0 grows like log 7 27 and it turns out that the true growth rate of log M0 is very slightly slower In the following proposition7 the symbol means that the ratio of the two expressions has limit 1 Proposition We have the asymptotic relation Zloglt1igt GST OO TF1 n Qloglogr Proof The idea is to break the series at a suitable place and to make different estimates both from above and from below on the two sums To simplify the expression on the right hand side7 it is convenient to replace the variable r 1We did not actually prove in class that the order of a canonical in nite product equals the convergence exponent of the zeroes7 but this general property of in nite products follows from a small change in the argument that solves problem 4 on the third takehome examination Page 1 of 3 Dr Boas Math 618 Example April 227 2007 by exp Then we have to show that 00 6V 5 Zloglt1ix17 assaoo 1 n og 5 Let N be the unique integer such that NMltN1x where N depends on s This integer is a good place to break the series Since log1 s lt s when x gt 07 we have that 0 lt 2 log 1 6 V nN1 7139 00 lt Z 6m nN1 5 71 e n By the choice of N7 all the terms of the series on the right hand side are less than 17 and each term is less than half the preceding one Consequently7 the series is bounded above by 20 12quot7 which converges to 27 a bound that is independent of N and 5 Using Landau7s O and 0 notation see section 2B of the textbook7 we can say that 00 N e 10g1 m 01 21 1 If n S N7 then 1 exp nl 2exp nl7 so N N N 6 e Zloglt lt Zlog 1 n ltNlog2Zlogltm n1 n1 n1 Combining this inequality with the preceding equation shows that 00 e N Zlog 1 n ON N Z lognl 1 n1 Notice that the term N cannot be included in the term 0N7 because 5 is not a constant and N depends on 5 ln estimating the remaining sum7 I use that lognl nlogn 0717 which results from the easy part of Stirling7s formula section 34D in the textbook Since 2771 n 0N27 we then have that Zlog 1 TF1 n Page 2 of 3 Dr Boas N ONZN 7 anogn Math 618 Example April 227 2007 The function mlogz is a monotonically increasing positive function when x gt 17 so comparing our sum with the area under a graph shows that N N N1 zlogzdzltanognlt zlogxdx 1 1 2 Since leogzdm zz logx 7 i227 it follows that N anogn NZlogNnL 0N27 n1 and therefore 2 log 1 V 1 n Finally7 we need to make the relationship between N and s explicit The de nition of N says that logNl 3 lt logN 17 0N2 N i N2 log N 0 again by Stirling7s formula EleogN7 and N NNZlogN Combining this information with our preVious growth estimate shows that 2 log 1 n n1 To rewrite the right hand side in terms of 57 observe that N log N2 s 2 log N 2 log N39 Moreover7 since E N Nlog N7 it follows that 6N2 log N 1 01 NzlogN logsilogNiloglogNa07 so logs2logN Thus i1 16 5 1 1 1 0g 71 ilogs 0 7 as claimed D Page 3 of 3 Dr Boas

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