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On Analytic Number Theory Lectures on Analytic Number Theory By H. Rademacher Tata Institute of Fundamental Research, Bombay 1954-55 Lectures on Analytic Number Theory By H. Rademacher Notes by K. Balagangadharan and V. Venugopal Rao Tata Institute of Fundamental Research, Bombay 1954-1955 Contents I Formal Power Series 1 1 Lecture 2 2 Lecture 11 3 Lecture 17 4 Lecture 23 5 Lecture 30 6 Lecture 39 7 Lecture 46 8 Lecture 55 II Analysis 59 9 Lecture 60 10 Lecture 67 11 Lecture 74 12 Lecture 82 13 Lecture 89 iii CONTENTS iv 14 Lecture 95 15 Lecture 100 III Analytic theory of partitions 108 16 Lecture 109 17 Lecture 118 18 Lecture 124 19 Lecture 129 20 Lecture 136 21 Lecture 143 22 Lecture 150 23 Lecture 155 24 Lecture 160 25 Lecture 165 26 Lecture 169 27 Lecture 174 28 Lecture 179 29 Lecture 183 30 Lecture 188 31 Lecture 194 32 Lecture 200 CONTENTS v IV Representation by squares 207 33 Lecture 208 34 Lecture 214 35 Lecture 219 36 Lecture 225 37 Lecture 232 38 Lecture 237 39 Lecture 242 40 Lecture 246 41 Lecture 251 42 Lecture 256 43 Lecture 261 44 Lecture 264 45 Lecture 266 46 Lecture 272 Part I Formal Power Series 1 Lecture 1 Introduction In additive number theory we make reference to facts about addition in 1 contradistinction to multiplicative number theory, the foundations of which were laid by Euclid at about 300 B.C. Whereas one of the principal concerns of the latter theory is the deconposition of numbers into prime factors, addi- tive number theory deals with the decomposition of numbers into summands. It asks such questions as: in how many ways can a given natural number be ecpressed as the sum of other natural numbers? Of course the decompostion into primary summands is trivial; it is therefore of interest to restrict in some way the nature of the summands(such as odd numbersor even numbers or per- fect squares) or the number of summands allowed. These are questions typical of those which will arise in this course. We shall have occasion to study the properties of -functions and their numerous applications to number theory, in particular theV theory of quadratic residues. Formal Power Series Additivenumber theory starts with Euler (1742). His tool was power series. His starting point was the simple relation xm. xn = xm+n by which multiplica- tion of powers of x is pictured in the addition of exponents. He therefore found it expedient to use power series. Compare the situation in multiplicative num- ber theory; to deal with the product n.m, one uses the equation nsms = (nm)s, thus paving the way for utilising Dirichlet series. While dealing with power series in modern mathematics one asks ques- 2 tions about the domain of convergence. Euler was intelligent enough not to ask this question. In the context of additive number theory power series are purely formal; thus the series 0! + 1! x + 2! x2 + is a perfectly good series in our ··· 2 1. Lecture 3 theory. We have to introduce the algebra of formal power series in order to vindicate what Euler did with great tact and insight. 2 A formal power series is an expression a0 + a1 x + a2x + . Where the symbol x is an indeterminatesymbol i.e., it is never assigned a numer··· ical value. Consequently, all questions of convergence are irrelevant. Formal power series are manipulated in the same way as ordinary power series. We build an algebra with these by defining addition and multiplication in the following way. If ∞ n ∞ n A = an x , B = bnx , = = Xn 0 Xn 0 ∞ n ∞ n we define A + B = C where C = cn x and AB = D where D = dn x , n=0 n=0 with the stipulation that we performP these operations in such a way thatP these equations are true modulo xN , whatever be N. (This reauirement stems from the fact that we can assign a valuation in the set of power series by defining ∞ n the order of A = anx to be k where ak is the first non-zero coefficient). n=0 Therefore cn and dnPmay be computed as for finite polynomials; then cn = an + bn, dn = a0bn + a1bn 1 + + an 1b1 + anb0. − ··· − A = B means that the two series are equal term by term, A = 0 means that all the coefficiants of A are zero. It is easy to verify that the following relations 3 hold: A + B = B + a AB = BA A + (B + C) = (A + B) + C A(BC) = (AB)C A(B + C) = AB + AC We summarise these facts by saying that the formal power series form a commutative ring. This will be the case when the coefficients are taken from such a ring, eg. the integres, real numbers, complex numbers. The ring of power series has the additional property that there are no divi- sors of zero (in case the ring of coefficients is itself an integrity domain), ie. if A, B = 0, either A = 0 or B = 0. We see this as follows: Suppose A = 0, B = 0. Let ak be the first non-zero coefficient in A, and b j the first non-zero coefficient ∞ n in B. Let AB = dnx ; then n=0 P dk+ j = a bk+ j + + ak 1b j+1 + akb j + ak+1b j 1 + + ak+ jb0 . ◦ ··· − − ··· 1. Lecture 4 In this expression the middle term is not zero while all the other terms are zero. Therefore dk+ j , 0 and so A.B , 0, which is a contradiction. From this property follows the cancellation law: If A , and A.B = A.C, then B = C. For, AB AC = A(B C). Since A , 0, B C◦ = or B = C. − − If the− ring of◦ coefficients has a unit element so has the ring of power series. As an example of multiplication of formal power series, let, 4 A = 1 x and B = 1 + x + x2 + − ··· ∞ n A = anx , where a0 = 1, a1 = 1, and an = 0 for n 2, = − ≥ Xn 0 ∞ n B = bnx , where bn = 1, n = 0, 1, 2, 3,... = Xn 0 ∞ n C = cn x , where cn = a0bn + a1bn 1 + + anb0; = − ··· Xn 0 then c0 = a0b0 = 1, cn = bn bn 1 = 1 1 = 0, n = 1, 2, 3,...; − − − so (1 x)(1 + x + x2 + ) = 1. − ··· We can very well give a meaning to infinite sums and products in certain cases. Thus A + A + = B, 1 2 ··· C C = D, 1 2 ··· both equations understood in the sense module xN , so that only a finite number N of A′ s or (C 1)′s can contribute as far as x . Let us apply− our methods to prove the identity: 1 + x + x2 + x3 + = (1 + x)(1 + x2)(1 + x4)(1 + x8) ··· ··· Let C = (1 + x)(1 + x2)(1 + x4) ... (1 x)C = (1 x)(1 + x)(1 + x2)(1 + x4) ... − − = (1 x2)(1 + x2)(1 + x4) ... − = (1 x4)(1 + x4) ... − 1. Lecture 5 Continuing in this way, all powers of x on the right eventually disappear, and we have(1 x)C = 1. However we have shown that (1 x)(1+x+x2+ ) = 1, therefore (1 −x)C = (1 x)(1 + x + x2 + ), and by the− law of cancellation,··· C = 1 + x + x2−+ which− we were to prove.··· ··· This identity easily lends itself to an interpretation which gives an example 5 of the application of Euler’s idea. Once again we stress the simple fact that xn xm = xn+m. We have · 1 + x + x2 + x3 + = (1 + x)(1 + x2)(1 + x4)(1 + x8) ··· ··· This is an equality between two formal power series (one represented as a product). The coefficients must then be identical. The coefficient of xn on the right hand side is the number of ways in which n can be written as the sum of powers of 2. But the coefficient of xn on the left side is 1. We therefore conclude: every natural number can be expressed in one and only one way as the sum of powers of 2. We have proved that 1 + x + x2 + x3 + = (1 + x)(1 + x2)(1 + x4) ··· ··· If we replace x by x3 and repeat the whole story, modulo x3N , the coeffi- cients of these formal power series will still be equal: 1 + x3 + x6 + x9 + = (1 + x3)(1 + x2.3)(1 + x4.3) ··· ··· Similarly 1 + x5 + x2.5 + x3.5 + = (1 + x5)(1 + x2.5)(1 + x4.5) ··· ··· We continue indefinitely, replacing x by odd powers of x. It is permissible to multiply these infinitely many equations together, because any given power of x comes from only a finite number of factors. On the left appears (1 + xk + x2k + x3k + ). ··· kYodd On the right side will occur factors of the form (1 + xN ). But N can be written uniquely as xλ.m where m is odd. That means for each N,1 + xN will occur once and only once on the right side. We would like to rearrange the factors to obtain (1 + x)(1 + x2)(1 + x3) ··· This may be done for the following reason. For any N, that part of the 6 formal power series up to xN is a polymial derived from a finite number of 1. Lecture 6 factors. Rearranging the factors will not change the polynomial. But since this is true for any N, the entire series will be unchanged by the rearrangement of factors. We have thus proved the identity ∞ (1 + xk + x2k + x3k + ) = (1 + xn) (1) ··· = kYodd Yn 1 ∞ n This is an equality of two formal power series and could be written an x n=0 ∞ n = bn x .
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