THE CIRCLE METHOD, THE j FUNCTION, AND PARTITIONS

Transcription

1 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS JEREMY BOOHER. Introduction The circle method was first used by Hardy and Ramanujan [] in 98 to investigate the asymptotic growth of the partition function, which counts the number of ways to write n as a sum of positive integers. They exploited the fact that the generating function for the partition function is almost the Dedeind eta function, which is related to integral weight modular forms. The ey step is using the transformation law for the eta function to rewrite their integral over a circle into something more tractable. In 938, Rademacher published a strengthening of their method which allowed him to obtain a representation of pn as a convergent series [4]. He also generalized it so it could deal with the coefficients of the q expansions of other modular forms lie the j function [3]. In 943, he discovered a different contour of integration involving Ford circles that simplified the analysis of the partition function [5] and has become the standard account presented in textboos lie Apostol []. This paper will show two examples of using variants of the circle method to produce series representations for the partition function and for the coefficients cn of the j function. I will first review bacground about the eta and j functions and then construct the required contours of integration in terms of the Farey sequence and Ford circles. Then I will review some analytic facts involving Bessel functions and Kloosterman sums. After that I will derive the series representations for pn and cn. Throughout, a summation h, will denote a sum over relatively prime h and.. Preliminaries on the eta and j Functions First we define the Dedeind eta function. Bacground can be found in Apostol []. Definition. The eta function is defined on the upper half plane H = {z C : Imz > 0} to be. ηz := e πiz/ n= e πizn. It satisfies a transformational law very similar to that of an integral weight modular form. a b Theorem. For SL c d Z with c > 0 and z H, az + b. η = ɛa, b, c, d icz + d / ηz cz + d Date: May 6, 00.

2 JEREMY BOOHER a+d πi where ɛa, b, c, d := e c +s d,c and sh, is the Dedeind sum r= r hr The j function is defined in terms of Eisenstein series again, see Apostol []. [ ] hr. Definition 3. The j function also called the Klein j-invariant can be defined in terms of Eisenstein series E of weight to be.3 for z H. jz = 3 E 4 z 3 E 3 4z 7E 6z Remar 4. This is normalized using the convention that the coefficient of q in the q expansion will be. In Apostol s boo [] and Rademacher s paper [3] the 3 is omitted from the definition of the j function but added in later formula. It is a modular function for SL Z in the sense that jz = j and jz = jz+. This z follows from the fact that the Eisenstein series of weight is a modular forms for SL Z. Because E4z 3 7E6z is a cusp form while E 4 is not, jz has a simple pole at the cusp. In particular, letting q = e πiz, jz can be written as.4 jz = cnq n. n= Remar 5. The denominator E 3 4z 7E 6z is the modular discriminant z, which can also be expressed as π ηz 4. Combining this with the definition of the Eisenstein series shows the coefficients cn are integers. Rademacher represented the cn in terms of infinite series using a variant of the circle method. This will be proven in Section 5. Theorem 6. In the Fourier expansion for the modular function jz = c q + c0 + cnq n where q = e πiz, the coefficient cn for n is given by the convergent series.5 cn = π n = A n n= I 4π n where I is a Bessel function that will appear in Definition 6. A n is the series A n = h mod e πi nh+h with h defined to be an integer for which hh = mod. Remar 7. Although the convergent series does not apply for n =, 0, by manipulating power series in the definition of the j function it follows that c = and c0 = 744. One can represent the coefficients of ηz in terms of a similar series. This will be proven in Section 6.

3 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 3 Theorem 8. Write the q series for eπiz/4 ηz q 4 ηz = with q = e πiz as n= pnq n. Let B n := 0 h< eπish, πinh/. Then the coefficient pn is given by the convergent series pn = B n d π sinh n 3 4 π dn = n 4 Remar 9. The argument will actually give an error term for summing terms of this series. As soon as the error term is less than, the nowledge that pn is an integer gives a way to compute pn. This is a significant improvement over trying to compute pn by counting partitions. In fact, because the function pn satisfies various congruence identities, it is usually possible to tolerate a much larger error and sum even fewer terms. 3. The Farey Sequence and Rademacher s Path of Integration The subdivision of the circle which Rademacher used in his early wor can be expressed in terms of the Farey sequence, a sequence of fractions arising in elementary number theory. Later, Rademacher constructed his refined contour of integration in terms of Ford circles, which are also defined in terms of the Farey sequence. 3.. The Farey Sequence. Definition 0. The Farey sequence of order n, denoted by F n, is the increasing sequence of reduced fractions in [0, ] with denominator less than or equal to n. Example. The Farey sequence of order 7 is 0, 7, 6, 5, 4, 7, 3, 5, 3 7,, 4 7, 3 5, 3, 5 7, 3 4, 4 5, 5 6, 6 7, When passing from F n to F n+m, new terms are added to the Farey sequence. If a < c, b d then their mediant is defined to be a+c. Elementary algebra shows it lies between a and c. b+d b d It is also easy to tell when certain fractions are consecutive. Proposition. Suppose 0 a < c with bc ad =. a and c are consecutive terms b d b d in F n when n satisfies maxb, d n b + d. Proof. Since bc ad =, the fractions are in lowest terms. If maxb, d n, then a and b c d are both in F n. We need to show they are consecutive if n b + d. Suppose there is a reduced fraction h lying strictly between them. But then we have 3. = bc ad = bc dh + dbh a and since a < h < c implies that c dh and bh a, we conclude b + d. Thus c d are consecutive. a c and b d

4 4 JEREMY BOOHER Also observe that if bd ad = and h is the mediant, it lies between them so bh a and c dh. But using 3., = b + d if and only if bh a = and c dh =. Following Apostol, say the fractions a < h satisfy the uni-modular relation if bh a =. b This gives enough information to mechanically construct F n+ from F n. Theorem 3. The sequence F n+ includes the sequence F n. Each fraction which is not in F n is the mediant of consecutive terms of F n. If a b and c d are consecutive in any F n, they satisfy the uni-modular relation. Proof. The proof proceeds by induction on n. For n =, the Farey sequence is 0, while F is 0,,. The statement is certainly satisfied. ow suppose that a and c satisfy the b d uni-modular relation and are consecutive in F n. They are consecutive whenever maxb, d m b + d by applying Proposition. Consider their mediant h. We now h satisfies the uni-modular relation with a and with c. This implies that h and are relatively prime. Furthermore, a b d b and c are not consecutive in F d = F b+d, but a and h are consecutive and liewise h and b c because maxb, =. This shows that in passing from F d n to F n+ each new fraction inserted is a mediant of a consecutive pair in F n and each consecutive pair satisfies the uni-modular relation. This completes the induction. This is enough theory to see how the Farey sequence of order allows the subdivision of the circle of radius e π as done in Rademacher s original wor [3] and [4]. Both and ηz jz tend to infinity as z approaches the cusps for SL Z. These correspond to q = e ir for r a rational number, points on the boundary of the unit circle. The cusps that turn out to contribute the most to the behavior of these functions are those with small denominators, which the Farey sequence represents. Definition 4. In the Farey sequence of order, let the fraction h lie between a and c. b d Let ξ h, be the image of a+h, c+h under the map θ b+ d+ e π +πiθ. Let ξ 0, be the image of,. Define + + ϑ h, = and b+ ϑ h, =. d+ π The arcs ξ h, are a division of the circle of radius e which will let us focus attention on the most important cusps. ote that h a+h = hb+ a h = ϑ b+ b+ h, and ϑ h, = c+h h. d+ Thus ϑ h, reflects the size of the arc between the mediant of a and h. b 3.. Ford Circles. The Ford circles can be defined in terms of the Farey sequence, and will lead to Rademacher s improved contour of integration for the partition function. Definition 5. Given a rational number h circle centered at h + i with radius. with h, =, the Ford circle Ch, is the Proposition 6. Two Ford circles Ca, b and Cc, d are either tangent or disjoint. They are tangent if and only if bc ad = ±. Corollary 7. Ford circles of consecutive Farey fractions are tangent. Proof. By the Pythagorean theorem, the distance between the centers of the circles is a b c + d b. d

5 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 5 The square of the sum of the radii is quantities is a b c + d b d b + = d b +. The difference between these two d ad bc bd b d = ad bc 0. b d Equality holds if and only if ad bc = ±, and equality corresponds to the circles being tangent. Before proceeding, we need a lemma from plane geometry. Lemma 8. A point C lies on the circle with diameter AB if and only if CD = AD DB, where D is the intersection of AB with the perpendicular to AB through C. Proof. Pic coordinates so the center of the circle is 0, 0 and A = r, 0 and B = r, 0. Then C = x, y lies on the circle if and only if x + y = r. But D = x, 0, so CD = y, AD = x + r and DB = r x, so CD = AD DB if and only if y = r x ext we consider how the Ford circles corresponding to three consecutive Farey fractions fit together. Theorem 9. Let a < h < c be three consecutive Farey fractions. The points of tangency b d of Ch, with Ca, b and Cc, d are given the points and α h, = h α h, = h + b + b + i + b d + d + i + d. Furthermore, α h, lies on the semicircle whose diameter is the interval [ a, h ] on the real line. b Proof. This is an exercise in plane geometry. α h, divides the line segment L joining the center z of Ca, b with the center z of Ch, into segments of length and. Thus b α h, = + b z + By definition z = a + i and z b b = h + i. Thus a b α h, = + i b + = = + b + b b + z b h + i b ab + i + h + i b ab + h + i b + = h b + b + i b +

6 6 JEREMY BOOHER where the last step uses that bh a =. The formula for α h, is obtained in a similar manner. To chec that α h, is on the semicircle, we simply use the previous lemma. The length of the perpendicular is the imaginary part of α h,,. It divides the segment from a to h b + b b into two segments of length and h a b. Then we have +b b +b b h + b a b b b = + b + b b b + b b + b b = + b b + b = b + as desired to show α h, lies on the semicircle Rademacher s Contour of Integration. Given this description of the Ford circles, we can construct a path of integration for every integer. It will be a path connecting the points i and i +. Definition 0. Consider the Ford cycles of the Farey sequence F. If a b, h, and c d are consecutive the point of tangency between Ca, b and Ch, and between Ch, and Cc, d divide Ch, into two arcs. P is the union of the arc with larger imaginary parts. For the circles C0, and C, use only the part of the upper arc with real part between 0 and, and consider them as part of the same arc associated to the point 0. Remar. Lie the Farey arcs on the circle of radius e π, this contour is sirting the points h in the Farey sequence of order. They correspond to cusps on the boundary of the unit circle under the map x = e πiτ. While Rademacher s contour is more complicated to describe, it greatly simplifies the analysis for the partition function. It is also useful to understand this path under the transformation z = i τ h. Proposition. This sends the circle Ch, to a circle K of radius around the point z =. The points of contact αh, and α h, from Proposition 9 are send to z h, := + b + i b + b z h, := + d i d + d. The upper arc joining αh, and α h, corresponds to the arc not touching the imaginary z axis. Proof. The translation τ h i translates Ch, so its center is at. Multiplication by i expands the radius to and rotates the z plane 90 degrees clocwise. This results in a circle of radius centered at. The formulas for the z i are just substitution. It will be easier to analyze integrals over the upper arcs in Ch, by converting them to integrals over this circle. To this end, the following bounds will be useful. There is an illuminating picture of the contour in the Wiipedia article on the Hardy-Littlewood circle method.

7 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 7 Proposition 3. With the notation of the previous proposition, z h, = + b and z h, = + d Moreover, all points z on the chord joining z h, and z h, satisfy z < provided a b, h, and c d are consecutive in F. The length of the chord is at most. Proof. Given the formulas for z h, and z h, in Proposition, the first assertion is clear. To prove the second, note that any point on the chord is z = sz h, + tz h, for s and t non-negative real numbers with s + t =. Thus z max z h,, z h,. However, as 0 + b + b we now that + b + b + > using Proposition. Combined with the formula for z h, and the same argument for z h, this gives that z as desired. The length of the chord is z h, z h, z h, + z h,. 4. Bessel Functions and Kloosterman Sums The first order of business is to state the relevant facts about Bessel functions. Definition 4. The Bessel function of the first ind J a is defined by the power series m x m+a J a z :=. m!γm + a + m=0 Remar 5. If a is a positive integer, Γm + a + is just m + a!. This function arises as a solution to the differential equation x d y dx + xdy dx + x a y = 0. The Bessel functions appearing later are the imaginary or modified Bessel functions of the first ind. Definition 6. The imaginary Bessel function of the first ind I a is defined to be i a J a iz. This has series expansion 4. I a z = m=0 x m+a. m!γm + a + It can also be represented in various ways as integrals. The relevant one is that 4. I v z = vv πi c+ i c i t v e t+z /4t dt

8 8 JEREMY BOOHER for c > 0, Rev > 0. Finally, there are sometimes elementary expressions for Bessel functions when a is a half integer. The relevant fact is that 4.3 I 3 z = z π d dz sinh z All of these facts were culled from the literature on Bessel functions. Apostol used Watson s boo on Bessel functions [6]. The only obscure fact is the integral formula, which Apostol found on page 8 with a slightly different but equivalent path of integration. The other necessary preliminary is nontrivial bounds on Kloosterman sums. Definition 7. For integers a, b and m, define K m a, b := z x Z/mZ e πi. ax+bx m where x denotes a lift to Z of the inverse of x in Z/mZ. There are different ways to obtain estimates on K m a, b. The following special case is sufficient and was nown before the Weil bound was proven. Rademacher attributes it to Salié and Davenport. Define A n := K n, : then we have the bound 4.4 A n C 3 +ɛ, n 3. Rademacher also uses an extension of this to incomplete Kloosterman sums. The footnote on page 507 [3] explains where to loo to derive the same estimate over shorter intervals. 5. The Convergent Series for the j Function The proof of Theorem 6 will need a number of lemmas. They will be proven after they are used to deduce the main theorem. The starting point is to express the coefficients of the j function in terms of integrals. Lemma 8. Let fe πiτ = jτ. For n > 0, we have 5. cn = 0 h< e πinh ϑ h, ϑ h, fe πih π iφ e πn iφ dφ where ϑ h, and ϑ h, are the boundary points of the Farey arcs defined in Definition 4. Also, recall that denotes the sum over relatively prime h and. This can be split up into two pieces corresponding to the q and n=0 cnqn parts of the q-expansion of jz. Lemma 9. Let w denote iφ, and define Dx = cmx m := fx x. We have that cn = Qn + Rn with Qn := Rn := 0 h< 0 h< ϑ e πi nh+h h, e πinh ϑ h, ϑ h, ϑ h, m=0 e π w +πnw dφ De πih π w e πnw dφ

9 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 9 where h denotes an integer with hh = mod. ow divide the contour of integration into three parts between the points ϑ h, = b + + < + d + = ϑ h, where the a b and c d are the adjacent fractions in the Farey sequence of order. Define 5.4 Q 0 n := Q n := Q n := = h mod = h mod = h mod e πi nh+h + + e πi nh+h + b+ e πi nh+h d+ + e π w +πnw dφ e π w +πnw dφ e π w +πnw dφ Then Qn = Q 0 n + Q n + Q n. The first step is to estimate Q 0 n. Lemma 30. Letting A n = h mod order with purely imaginary argument equation 4.. Then 5.5 Q 0 n = π n = e πi nh+h and I denote the Bessel function of the first A n 4π n I + Oe πn n 3 3 +ɛ. ext both Q n and Q n can be analyzed together. Lemma 3. We have that Q n and Q n are Oe πn n 3 3 +ɛ. The last step is to analyze Rn. Lemma 3. We have that Rn = Oe πn n 3 3 +ɛ. These estimates give the proof of Theorem 6. Proof. Using Lemmas 30, 3, and 3 in the equation from Lemma 9 gives 5.6 cn = Qn + Rn = Q 0 n + Q n + Q n + Rn cn = π n = A n 4π n I + Oe πn n 3 3 +ɛ. For a fixed n > 0, letting go to infinity maes the error term goes to zero. This establishes Theorem 6.

10 0 JEREMY BOOHER 5.. Proof of Lemma 8. Cauchy s formula says that cn = fx dx πi xn+ where C is the circle of radius e π < centered at the origin. The union of the disjoint Farey arcs ξ h, for h in the Farey sequence of order is exactly this circle. Thus we have fx cn = dx πi xn+ 0 h< ξ h, In terms of the variable φ on ξ h, defined by x = e π + πih +πiφ arc length centered at e πi h, the integral becomes 5.7 cn = as dx = x πi dφ. 0 h< e πinh ϑ h, ϑ h, f C e πih π iφ e πn iφ dφ 5.. Proof of Lemma 9. Since j aτ + b = jτ for τ in the upper half plane and cτ + d a b SL c d Z, taing τ = iz + h a b h and = hh + c d h gives j iz + h = j i z + h and hence f e πz 0 h< + πih = f ϑ h, e π z + πih. ow letting w = iφ applying this to the representation 5.7 gives ϑ cn = e πinh h, f e πih π w e πnw dφ. This has the advantage of moving the contour in relation to the cusps to facilitate the analysis. Writing fx = x + Dx with Dx = m=0 cmxm we can split the integral into two integrals. They are precisely the Rn and Qn listed in Lemma 9, with Rn arising from x and Qn from Dx Proof of Lemma 30. Splitting up the integral Qn as in 5.4, we will first analyze Q 0 n. The integral is independent of h, so + Q 0 n = A n e π +πnw w dφ where we define = A n = h mod + e πi nh+h

11 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS ote that this is an example of a Kloosterman sum. The strategy to evaluate the integral is to view it as one of the sides of a rectangle in the w plane with vertices ± ± i. + Since w = iφ, Q 0 n = A n i = + i + i + e π w +πnw dw. Q 0 n is the integral over the right side of the rectangle. Let L n denote the integral over the entire rectangle, J the integral from + i to + i, J + + the integral from + i to i, and J the integral from i to i. + + Then we have 5.8 Q 0 n = A n i L i = A n J + J + J 3. It is easy to estimate J and J 3. On the paths of integration w = u ± i, u +, and Re w = u u < Thus the integrand is less than e 8π+πn, so = 5.9 J and J 3 e 8π+πn. For J, the path of integration is w = + iv with v. The real part of w is + always < 0 while Re = < 0. Thus the integrand is O note this doesn t w 4 +v wor on the right side, which is good. The path has length, so J < + <. Combining 5.9 with 5.0 and the bounds on the Kloosterman sum A n from 4.4 we get πn A n J + J + J 3 = O e 3 +ɛ n, 3. = As long as n which we are assuming, n, n. Furthermore, Thus we can estimate 5. = = 3 +ɛ = O 3 +ɛ. A n J + J + J 3 = Oe πn n 3 3 +ɛ. =

12 JEREMY BOOHER The last step is to deal with L n. ow if R is the rectangle it had a positive orientation then using the power series for e x we get π L n = e π πi +πnw w dw = πi By the residue theorem, the integral πi R R µ=0 R π µ w µ! πnw ν dw. ν! ν=0 π µ w πnwν µ! ν! is zero unless there is a simple pole at 0, which requires ν µ =. Thus we have π L n = n π n ν+ ν!ν +! ν=0 = n I 4π n where I z is the Bessel function defined in we obtain the desired Q 0 n = π n = A n Putting this together with 5.8 and I 4π n + O e πn n 3 3 +ɛ 5.4. Proof of Lemma 3. The next step is to bound Q n and Q n. I will do the case of Q n: the argument for Q n is nearly identical and is the one Rademacher chooses to do. The definition of Q n is Q n := = h mod e πi nh+h + b+ e π w +πnw dφ. Splitting the interval [, ] up into the intervals [, ] for l from b + b+ + l l+ to +, it follows that 5. Q n = + = l=+ l+ l e π w +πnw dφ h mod <b+ l e πi nh+h where the extra condition on the last sum allows the extension of the second sum from + to +. Because b h mod which follows from the uni-modular relation on the Farey sequence, the restriction < b + l does in fact constrain the choice of h. This maes the last sum an incomplete Kloosterman sum, which using 4.4 tells me that h mod <b l e πi nh+h = O 3 n, 3 = O 3 n 3

13 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 3 π But now in the integral of 5., on the intervals the real part of + πnw is w Re π w + πnw = Re π iφ + πn iφ = π + n 4 + φ π π n 8π + πn. + n Combining this with the rest of the integral gives + Q n = O e πn n 3 l 3 +ɛ l + = l=+ = O e πn n 3 = 3 ɛ = O e πn n 3 3 +ɛ. This and the analogous result for Q n establish Lemma Proof of Lemma 3. By the definition of Rn and D in Lemma 9, Rn = = h mod ote that for any m, on the interval Re πm = w e πinh ϑ h, πm 4 + φ ϑ h, m=0 [ ], + + cme πih m πm w e πnw dφ πm + φ πm = πm. I decompose the interval [ ϑ h,, ] [ ] [ ] ϑ h, into,,,, and +b [ ],, and then further decompose the first and last intervals. The first decomposes + b+ [ ] [ ] into, for l from b + to +, the last into,. Let S l l+ l+ l be the integral from the middle part: S := = m=0 cm + + e πm w +πnw dφ h mod e πi nh mh.

14 4 JEREMY BOOHER The last sum is a Kloosterman sum, so it is O 3 +ɛ n 3 uniformly in m. Thus S = O cm + e πm+πn 3 +ɛ n 3 = m=0 = O e πn n 3 cm e πm 3 +ɛ m=0 = O e πn n 3 3 +ɛ where the last step uses that m=0 cm e πm is a finite constant because jz is absolutely convergent at i. [ ] The other two integrals are similar, so I will only deal with the one over,. +b + Write S = = m=0 cm + l=+ l+ l e πm w +πnw dφ = h mod <+b l e πi nh mh. Again, the last sum is a Kloosterman sum, which we can bound by O 3 +ɛ n 3. Using the estimate on the real part, S = O cm 3 +ɛ n 3 = O = m=0 e πn n 3 3 +ɛ. e πm+πn Combining this with a similar statement for S 3 and the bound for S, we get the statement of Lemma 3. This completes the proof of Theorem The Convergent Series for the Partition Function 6.. Outline of the Proof. The same sort of argument wors to prove the series representation for pn in Theorem 8. The generating function for the partition function can be expressed as F q = pnq n = n= qn n= by expanding as a geometric series. This differs from by a factor of q q n 4. The first ηz order of business is to convert the transformation law for the eta function into one for the function F. Lemma 33. Let F t = n=, and let tm x = e πih πz and x = e πih where Rez > 0, h and are relatively prime, and hh mod. Then the transformation law becomes z / F x = e πish, π e z πz F x. π z

15 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 5 a b Proof. For SL c d Z with c > 0, the functional equation in Theorem implies ητ = ητ icτ + d a+d e πi c +s d,c where τ = aτ+b. Rewriting this in terms of F cτ+d eπiτ = eπiτ/ ητ gives F e πiτ = F e πiτ e πiτ τ icτ + d e πi a+d c +s d,c. Tae a = H, c =, d = h, b = hh+ equation becomes πz F e πih Replacing z by z/ gives the desired formula., and τ = iz+h. Then τ = iz + H πih = F e π π z z e z πz +πish,. and the ow fix n, and allow to vary. As with the case of the j function, we will extract coefficients using a Cauchy s theorem. This in turn can be written in terms of an integral along Rademacher s contour followed by another change of variables onto the circle K of radius centered at z =. Lemma 34. With the points z h, and z h, as in Section 3.3, z h, pn = F e πih πz i e πinh/ e nπz/ dz. 0 h< z h, The next step is to use the transformation law for the eta function and split the integral up to deal with the elementary factor 6. separately. To do so, define 6. I h, := I h, := z h, z h, z h, z h, Ψ z := z e π z πz Ψ ze πnz/ dz Ψ z F e πih π z e πnz/ dz where H is defined as in Lemma 33. Denote e πish, by ωh,. Lemma 35. With the notation above, pn = 0 h< i 5/ ωh, e πinh/ I h, + I h, The next step is of course to analyze the two integrals. I will contribute the main term. Lemma 36. We have that I h, = Ψ ze πnz/ dz + O 3 3 K where K is the circle with radius centered at z = I becomes an error term. with negative orientation.

16 6 JEREMY BOOHER Lemma 37. We have that I h, = O 3 3. ext, we put these together. ote that ωh, = as sh, is a real number. Thus we can conclude that i 5 ωh, e πinh/ C 3 3 C 3/ h< n= C 3 0 h< h,= = O. Combining this with Lemmas 36 and 37 gives pn = i 5/ ωh, e πinh/ Ψ ze πnz/ dz + O. K 0 h< Letting tend to infinity gives us that pn = i B n 5 z π 6.4 e z + πz n 4 dz K = where B n is the exponential sum B n := 0 h< = e πish, πinh/. The final step is to evaluate the integral in terms of Bessel functions and the hyperbolic sine function. Mae the change of variable w =, dz = dw. This sends the circle K z w to the the line with real part. Then 6.4 becomes pn = + i B n 5/ w 5/ e πw + π i n 4 w dw = ow substitute t = π w. This gives π pn = π B n 3 5 = which loos lie 4. with v = 3 and Rewriting in terms of I 3 π pn = z = π n gives 3 4 π i πi 6 n 4 π + i t π i. B n π I 3/ 5 e t+ π 6 n 4 t dt 3 n 4. n = Using the special value of Bessel functions of half odd-order from 4.3 gives pn = π = B n d dn sinh π n 3 4 n 4

17 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 7 This finishes the proof of Theorem Proof of Lemma 34. The starting point is Cauchy s integral formula, which combines with the Power series for F x to imply pn = F x dx πi xn+ where C is any positively oriented closed contour surrounding 0 lying inside the unit circle where F is holomorphic. Maing the change of variable x = e πiτ, a circle of radius e π centered at 0 in the x-plane is sent to the line joining i and + i in the τ-plane. Thus pn = i+ i C F e πiτ e πinτ dτ as dx = e πiτ dτ. Replace this line with the Rademacher contour P from Section 3.3, and let γh, denote the arc on the circle Ch, connecting αh, and α h,. Then pn = F e πiτ e πinτ dτ. γ h, 0 h< Mae the change of variable z = i τ h which sends Ch, onto a circle of radius with z = as its center. The arc γh, maps onto an arc joining the points z h, and z h,. Furthermore, we have dz = i dτ and τ = iz + h. Thus the integral becomes the desired z h, pn = F e πih πz i e πinh/ e nπz/ dz. h, z h, 6.3. Proof of Lemma 35. Lemma 33 tells me that F e πih πz z = ωh, e π z πz F e πih π z where H is chosen so hh = mod and ωh, was defined to be e πish,. Substituting this into Lemma 34 which gives a formula for pn in terms of a contour integral gives pn = 0 h< i e πinh/ ωh, z h, z h, Rearranging terms to match the definitions in 6. and 6. gives pn = 0 h< z e π z πz +πnz/ F e πih π z dz i 5/ ωh, e πinh/ I h, + I h, Proof of Lemma 36. Instead of integrating around the circle from z h, to z h,, we will integrate around the entire circle and analyze the error. ote that the circle has a a negative orientation. Let K denote the negatively oriented circle, J the arc from 0 to z h, and J the arc from z h, to 0. Then I h, can be written as the integral over K minus the integrals over J and J.

18 8 JEREMY BOOHER To estimate J, note that the length of the arc joining the point 0 and z h, is less than π z h, < π. On the circle Re/z = while 0 Rez. Thus Ψ ze πnz/ = e πn Rez/ z e π Re z π Rez eπn 4 e π/ since z is maximized at z h,. Thus the integral over J is O 3 3. The same argument wors for J. This shows that I h, = Ψ ze πnz/ dz + O 3 3. K 6.5. Proof of Lemma 37. Instead of integrating around the circle from z h, to z h,, we will integrate along the chord joining them. ote that 0 < Rez and Re z anywhere in the circle. Then we can estimate the integrand as follows: Ψ z F e πih π z e πnz/ = z e π Re z π Rez e πn Rez/ z e π Re z e πn/ < z e πn z e πn z e πn = z e πn = c z m= m= pme πm Re z pme πm 4 Re z m= pme πm 4 m= p4m e π4m /4 m= p4m y 4m m= pme πihm/ e πm/z where the second to last step uses pm < p4m for m and y := e π/4. The constant c equals e πn m= p4m y 4m which bounded because it is a sub-sequence of the q series for F evaluated at y which is inside the unit circle. ote it is independent of z or it depends on n, but n is fixed.

19 THE CIRCLE METHOD, THE j FUCTIO, AD PARTITIOS 9 ow the length of the chord is less than by Proposition 3. Furthermore, z on the chord. This implies that the integral is 6.5 I h, = O 3 3. Remar 38. One of the reasons Rademacher s improved contour of integration does not immediately apply to the j function is that, in contrast to Lemma 33, there is no extra factors in the transformation law in the j function because it is better behaved that the eta function. In particular, the z is necessary for this lemma to wor. Attempting to directly copy this argument naively tells me that fq = je πiz is bounded on the chord and gives me a bound of O for the integral. After summing over h, with 0 h < and h, = as in 6.3, this is not obviously bounded by anything better than O. A possible salvage would be to divide the segment up into smaller pieces and bound each separately. This loos lie the situation in Lemma 3, where the integral is divided up carefully into pieces to get a good enough bound: it is not clear whether using Rademacher s contour instead of Farey arcs would mae this division any easier. References. Tom Apostol, Modular functions and dirichlet series in number theory, Springer-Verlag, G. H. Hardy and S. Ramanujan, Asymptotic formula and combinatory analysis, Proc. of the London Math. Society 98, Hans Rademacher, The fourier coefficients of the modular invariant jτ, American Journal of Mathematics 938, , On the partition function pn, Proc. of the London Math. Society 938, , On the expansion of the partition function in a series, The Annals of Mathematics 943, George Watson, A treatise on the theory of bessel functions, second edition ed., Cambridge University Press, 995.