Gauss Sphere Problem with Polynomials

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Gauss Sphere Problem with Polynomials"

Transcription

1 Gauss Sphere Problem with Polynomials Fan Zheng July 31, 013 Mathematics Subject Classifications: Primary 11P1; Secondary 11L03, 11L07 SPUR final paper, Summer 013 Mentor: Chenjie Fan Project suggested by David Jerison Abstract This paper provides an estimate of the sum of a homogeneous polynomial P over the lattice points inside a sphere of radius R. The polynomial P is assumed to be of degree ν and have zero mean over the sphere. It is proved that P (x) = O ɛ,p (R ν+83/64+ɛ ) x Z 3 x R MIT, supported by Summer Program of Undergraduate Research 1

2 Contents 1 Introduction: A Historical Review The Method of Chamizo and Iwaniec: Extensions, Improvements and Limitations 4 3 Converting the Long Sum to the Exponential Sum 9 4 Estimating the Exponential Sum 11 5 Estimating the Long Sum 0 6 Estimating the Short Sum 3 7 Appendix A: Modularity of the Theta Function 5 8 Appendix B: Better Results with New Exponent Pairs 30 9 Appendix C: Summary of Proved and Conjectured θ ν Acknowledgements 34 1 Introduction: A Historical Review The Gauss Sphere Problem is a generalization of the famous Gauss Circle Problem to three dimensions. It asks about the number of lattice points in a sphere of radius R. While it is easy to see that the leading term is asymptotically 4πR 3 /3, estimation of the error term is still open. Compared to the two dimensional case, the Gauss Sphere Problem has received less attention, yet it bears relation with a variety of topics in analytic number theory: average class numbers of negative discriminants, estimates of L- functions and Fourier coefficients of modular forms, to name a few. The trivial observation, already known to Gauss, that the error arises only from a shell of constant thickness on the surface of the ball, provides an error term of CR, where C is a calculable (effective) constant. The first breakthrough was made by Van der Corput, who used Poisson summation formula to transform the lattice in physical space to the one in frequency space. In this way, counting lattice points translates to summing

3 the Fourier transform of χ B(R), the characteristic function of the ball of radius R. Although the sum itself diverges, it can be brought convergent by smoothing χ B(R), or equivalently multiplication of a cut-off function in the frequency space. The width H of the range being smoothed, and correspondingly the radius H 1 in the frequency space to be summed, is a parameter that can be adjusted to obtain an optimal error bound. Using trivial estimates, Van der Corput obtained an error CRH 1 for the sum in the frequency space, and an error CR H from the smoothing in the physical space. Balancing these two errors by setting H = R 1/ Van der Corput got an error of CR 3/ for the Gauss Sphere Problem. The basic structure of Van der Corput s argument has remained unchanged ever since. Subsequent improvements came from a more careful analysis of both error terms. In the following discussions we shall use Vinogradov s notations: Definition 1. f = O(g) f g g f C > 0 such that f Cg. f g f g and g f. Subscript variables attached to these symbols mean that the implicit constant depend on the variables. In addition we introduce a short hand for the exponential function. Definition. For any z C. e(z) = exp(πiz). The frequency side was first exploited for improvements, because the Fourier transform of χ B(r) turns out to be (up to some constants) R e(r ξ ) ξ Since the denominator can be removed by Abel summation, the summation in the frequency space essentially involves the partial sum of the exponentials e(r ξ ) ξ Z 3 ξ N Of the numerous ways inverted to deal with such exponential sums, the two most important ones are Van der Corput A and B processes. The A process is also know as Weyl differencing. The B process is Poisson summation and stationary phase, much in the same spirit as discussed above. These two 3

4 processes have been abstracted to give the method of exponent pairs. For more details the reader is referred to Appendix B and [8]. Improvements in the frequency space culminated in the works of Chen [6] and Vinogradov [14], in which they independently improved the estimate of the exponential sum to O ɛ (RH 1/+ɛ ) in a sufficiently large range so that it can be balanced with O(R H) to give the improved error estimate O ɛ (R 4/3+ɛ ). The results of Chen and Vinogradov stood for another several decades before Chamizo and Iwaniec turned their attention to the physical space. Using character sums, they improved the trivial bound O(R H) on the error caused by smoothing to O ɛ (R 15/8+ɛ H 7/8 ), and thus lowered the 4/3 in the exponent further down to 9/ [5]. Currently, the world record on this problem is O ɛ (R 1/16+ɛ ), obtained by Heath-Brown [9] in much the same line as [5]. His step forward is an improved error bound O ɛ (R 11/6+ɛ H 5/6 ) of the character sum. Last but not least, it should be mensioned that current techniques are still insufficient to prove the famous conjecture of the true error bound (if true), which is believed to be O ɛ (R 1+ɛ ). The Method of Chamizo and Iwaniec: Extensions, Improvements and Limitations Chamizo and Iwaniec s method has gained popularity in a number of related problems in recent years. Aside from the above-mentioned [9], the reader is referred to [3] and [4] for more examples, and to [] for a non-technical account. This paper provides another application of this method, along with some improvements. The problem considered here is to sum a homogeneous polynomial of zero mean on the sphere. In other words, we provide an estimate for P (x) (1) x Z 3 x R where P is a homogeneous polynomial in three variables such that P (x) = 0 x =R It is well-known (see Corollary.50 of [7], for example) that P can be written as a linear combination of spherical harmonics times powers of x. The zero 4

5 mean of P ensures that the spherical harmonics involved are all non-constant. Therefore, it suffices to consider only harmonic homogeneous polynomial of degree ν > 0. Since P has zero mean, there is no main term of the form cr ν+3 as in the Gauss Sphere Problem. Thus the natural form of the estimate is P (x) = O ɛ,p (R ν+θν+ɛ ) x Z 3 x R where we have taken into account of the fact that sup x =R P (x) R ν. Normally, we would expect that (by some abuse of notation) θ ν = θ 0. In other words, the error in estimating the sum of a homogeneous polynomial of degree ν scales according to the degree of the polynomial. For example, Van der Corput s estimate x Z 3 x R 1 = 4πR3 3 + O(R 3/ ) easily generalizes to (see the proof of Lemma 3.5 (c) in [11]) P (x) = O P (R ν+3/ ) x Z 3 x R Naively, we would expect Heath-Brown s record-keeping result to give an error of O ɛ,p (R ν+1/16+ɛ ) for our problem. While this is true, the exponent 1/16 can actually be lowered (to 83/64) if we examine Heath-Brown s method more carefully and try to adapt it to the current case. As mentioned in the previous section, Heath-Brown decomposed the sum over lattice points into two parts, the long sum and the short sum. The long sum is essentially a smoothed version of the original sum: S f (R) = x Z 3 1 x f( x ) where f is a cutoff function growing like x when 0 x R and decreasing linearly to 0 when R x R + H, and H is the width of the cutoff function f, a parameter tunable for optimal bounds. 5

6 On the other hand, the short sum is S f (R, H) = x Z3 R x R+H 1 x f( x ) The actual sum (1), when P = 1, is then the difference between S f (R) and S f (R, H), which are estimated in different ways. The estimation of the long sum S f (R) essentially follows [5]. More precisely, in the notation in [8], the operation ABAB is performed to S f (R) to obtain the following estimate (thanks to [4] for pointing out a misprint in [5]) S f (R) = 4πR3 3 + πhr + O ɛ ((RH 1/ + R 11/8 H 1/8 + R 1/16 )H ɛ ) () The short sum, on the other hand, is converted to a character sum, an idea dated back to Gauss (see (1.) of [5]). Its estimation is made by Chamizo and Iwaniec, and improved by Heath-Brown to (see (3) in [9]) S f (R, H) = πr H + O ɛ ((R 11/6 H 5/6 + R 19/15 + R 7/6 H 1/6 )R ɛ ) Balancing S f (R) and S f (R, H) by setting H = R 5/8, we obtain Heath- Brown s bound x Z 3 x R 1 = 4πR3 3 + O ɛ (R 1/16+ɛ ) where, as Heath-Brown has remarked, the exponent 1/16 comes from trading the H factors in the term RH 1/ in S f (R) and the term R 11/6 H 5/6 in S f (R, H). The term R 1/16 in S f (R), on the other hand, is not optimal and has some room for improvement. The approach taken in this paper is essentially the same as that in [5] and [9], but two differences should be remarked, the first one more fundamental than the second. The first difference is in the estimate of the short sum. While character sum is used in the case P = 1, in our case when P is a harmonic homogeneous polynomial of degree ν > 0, modular forms come into play. In more detail, let a n = P (x) x Z3 x =n 6

7 and θ(z) = n N a n e(nz) It is known (Example, P 14 of [1]) that θ is a cusp form of weight k = ν + 3/ with respect to the congruence group {( ) } a b Γ 0 (4) = SL c d (Z), 4 c (when P = 1, θ is a modular form, but not a cusp form.) Therefore some powerful estimates of the Fourier coefficients a n of cusp forms are available, of which we use Blomer-Harcos bound (Corollary of [1]) a n ɛ,θ n k/ 5/16+ɛ (n, ) 5/8 R ν+7/8+ɛ (n, ) 5/8 and the triangle inequality to obtain Theorem : S f,p (R, H) = O ɛ,p (R ν+ɛ (R 15/8 H + R)) (3) which wins a factor of R 1/8 over the trivial short sum estimate O f (R ν+ H). (The GCD term, which means the largest power-of-two factor of n, is absorbed in the process of summation.) The second difference is a slight improvement of the estimate (), which is now necessary because the previously-mentioned improvements on the short sum pushes H up, making R 1/16 dominate RH 1/ in the long sum, so removing the R 1/16 is necessary (and possible, as already hinted by Heath- Brown). This is achieved by optimizing the two Weyl differencing steps (A processes). Specifically, in [5], the lengths of the first and the second Weyl differecing are set to N 1/ ɛ and U. This, however, is not optimal in all cases, especially when N R 6/5, for which the off-diagonal term actually dominates the diagonal term. This paper improves on this by tuning the lengths of the two Weyl differencing steps (Y in Lemma and T in Lemma 4), if possible, to balance the diagonal and off-diagonal terms. Our new estimate is Theorem 1, which for any homogeneous polynomial P of degree ν having zero mean on the sphere, says: S f,p (R) = O ɛ,p (R ν (RH 1/ + R 17/14 H 1/7 )H ɛ ) (4) Now let s see the effect of balancing (3) and (4). Let s ignore the common factor R ν and any factor of the form R ɛ or H ɛ. Moreover, let s assume 7

8 that in the short sum the term R 15/8 H dominates (the other term actually corresponds to θ ν = 1), and that in the long sum the term R 17/14 H 1/7 dominates (which is the very reason we want to optimize it). We obtain R 17/14 H 1/7 = R 15/8 H which gives H = R 37/64 and P (x) = S f,p (R) S f,p (R, H) = O ɛ,p (R ν+83/64+ɛ ) x Z 3 x R The second term in S f,p (R, H) is clearly dominated, while the first term in S f,p (R) is O ɛ,p (RH 1/ ) = O ɛ,p (R ν+165/18+ɛ ) which is also dominated. Therefore we have succeeded in showing θ ν /64, pushing Heath-Brown s exponent (for the purpose of our problem) down by 1/64, or 1/0 in relative terms, which is an improvement of 5%. In Appendix B we use a recent result of Huxley [10] to sketch a proof that θ ν can be further reduced to / Finally let s see how far we could possibly go from here. Assuming Lindelöf Exponent Pair Conjecture [3], we can substitute k = ɛ and l = 1/ + ɛ in (9) in Appendix B to reach θ ν 1 + 7/4, coming from the term RH 1/ in the long sum, which can be traced back to the diagonal term in the first Weyl differencing step. This is probably the best one can expect of the long sum, unless it is redone from scratch in a radically different way. On the other hand, improvements on the short sum may have a larger impact. If we assume Ramanujan s conjecture on modular forms of half integral weight, i.e. (ignoring all the ɛ s in the exponent) and use the triangle inequality we get a n f n k/ 1/ R ν+1/ (R+H) n=r a n f R ν+3/ H Balancing this result with Theorem 1 we obtain a significant better result θ ν 1 + 1/4. If we further assume Lindelöf Exponent Pair conjecture (i.e. k = ɛ and l = 1/ + ɛ) in Theorem 3 we obtain an error bound of O ɛ,p (R 7/6 3k+3l R 10k+10l+6 ) 8

9 Using [10] we can get θ ν /5790. Assuming Lindelöf conjecture we can get θ ν 1 + 5/. Finally we state a conjecture analogous to the famous Gauss Sphere Problem. Conjecture 1. θ ν = 1. In other words. Suppose P is a homogeneous polynomial of degree ν in three variables and P has zero mean on the sphere, then = O ɛ,p (R ν+1+ɛ ) x Z 3 x R A tabular summary of all the proved and conjectured exponents mentioned above can be found in Appendix C. 3 Converting the Long Sum to the Exponential Sum Suppose P is any homogeneous polynomial of degree ν > 0 in three variables, not necessarily having zero mean on the sphere. Definition 3. The long sum refers to the following sum: S f,p (R) = x Z 3 g P (x) where g P (x) = P (x)g(x), g(x) = f( x )/ x, and f is the cutoff function x, x [0, R] f(x) = R(R + H x)/h, x [R, R + H] 0, x R + H Lemma 1. S f,p (R) = P (x)g(x)dx + R 3 + ν 1 +ν +ν 3 =ν ν 1 +ν =ν H ν 1 (R + H) ν R ν 1 ξ Z 3 \0 ξ Z 3 \0 Q ν1 (ξ) sin(πr ξ + πν 1 ) ξ 3+ν 1+ν Q ν1,ν (ξ) sin(πh ξ + πν 1 ) cos(π(r + H) ξ + πν ) ξ 3+ν 1+ν +ν 3 9

10 where Q ν1 and Q ν1,ν are homogeneous polynomials of degree ν. Proof. By Poisson summation formula applied to g (whose validity will be justified later in Lemma 5) S f,p (R) = ξ Z 3 ĝ P (ξ) where the convention of the Fourier transform taken here is ˆf(ξ) = f(x)e( ξ x) R 3 and e(z) is defined to be exp(πiz) for all z C. If ξ = 0, then ĝ P (ξ) = g P (x)dx R 3 is the integral of the smoothed function g P over R 3, which contributes to the main term of the estimate of the long sum, and which actually vanishes if P has zero mean on the sphere. If ξ 0, then by (5.) in [5], we have so ĝ P (ξ) = P ĝ(ξ) = sin(πr ξ ) π ξ 3 R H sin(πh ξ ) π ξ 3 ( ) ( ξ sin(πr ξ ) R πi π ξ 3 H cos(π(r + H) ξ ) ) sin(πh ξ ) cos(π(r + H) ξ ) π ξ 3 By induction on ν, we can show that ( ) sin(πr ξ ) P ( ξ ) = R ν Q ν 1 1 (ξ) sin(πr ξ + πν 1 ) ξ 3 ξ 3+ν 1+ν ν 1 +ν =ν and ( ) sin(πh ξ ) cos(π(r + H) ξ ) P ( ξ ) ξ 3 = H ν 1 Q (R + H) ν ν1,ν (ξ) sin(πh ξ + πν 1 ) cos(π(r + H) ξ + πν ) ξ 3+ν 1+ν +ν 3 ν 1 +ν +ν 3 =ν where Q ν1 and Q ν1,ν are homogeneous polynomials of degree ν. 10

11 Now if we sum over ξ Z 3 \0 and use trigonometric identities, we get sums of the following type ξ Z 3 \0 Q(ξ)e(R ξ ), or ξ m ξ Z 3 \0 Q(ξ)e((R + H) ξ ) ξ m where Q is a homogeneous polynomial of degree ν. By Abel summation, it boils down to estimating sums of the following type Q(ξ)e(R ξ ) ξ Z 3 ξ N This is the sort of exponential sum that will play a key role in the estimation of the long sum. 4 Estimating the Exponential Sum In this section, Q denotes a homogeneous polynomial of degree ν 0 in three variables, not necessarily having zero mean on the sphere. For future convenience (in justifying the convergence of Poisson summation), we introduce a variable h Z 3, but all the estimates in this section are uniform in h. Definition 4. The exponential sum we are going to estimate is the following: Lemma. If R 1 then V N,Q,h (R) = ξ Z3 ξ N Q(ξ)e(R ξ + h ξ) V N,Q,h (R) ɛ,q N ν/+ɛ (min{n 3/, N 5/4 + N 15/14 R 3/14 }) Remark 1. The right hand side N 3/, 1 N R 1/ N ν/+ɛ N 15/14 R 3/14, R 1/ N R 6/5 N 5/4, N R 6/5 11

12 Proof. Clearly it suffices to prove Lemma for Q = a i b j c k with i+j +k = ν. The fact that V N,Q,h (R) N 3/ comes from the trivial bound V N,Q,h (R) ξ Z3 ξ N Q(ξ) #{ξ Z 3, ξ N} sup Q(ξ) N ν/+3/ ξ N To prove the second bound, we transform it into an expression that is the starting place of the estimates in [5]. By dyadic decomposition where so it suffices to show V N,Q,h (R) = V N,Q,h(R) = ξ Z3 ξ N log N k=1 V k,q,h (R) Q(ξ)e(R ξ + h ξ) V N,Q,h(R) ɛ,q N ν/+ɛ (N 5/4 + N 15/14 R 3/14 ) For a + b N, let χ a,b be a piecewise linear function that is equal to 1 on Z [ N a b, N a b ], and 0 on the other integers. Then by Theorem 7.3 in [8], ˆχ a,b (θ) is uniformly bounded by K(θ), where K L 1 (R) log N ɛ N ɛ. Therefore, up to the terms where two of a, b and 1

13 c are identical, which sum to ( N) O(N ν/ ) = O(N ν/+1 ), VN,Q,h(R) e(r a + b + c + ah 1 + bh + ch 3 )a i b j c k a +b +c N a,b c = a,b c a i b j e(ah 1 + bh ) c N e(r a + b + c + ch 3 )χ a,b (c)c k = a i b j e(ah 1 + bh ) e(r a + b + c + ch 3 )c k ˆχ a,b (θ)e(cθ)dθ a,b c c R N = a i b j e(ah 1 + bh ) ˆχ a,b (θ) e(r a + b + c )c k e(c(θ + h 3 ))dθ a,b c (a + b ) (i+j)/ a,b c 0<n=a +b N ɛ N (i+j)/+ɛ R R n (i+j)/ d(n) R R ɛ N (i+j)/+ɛ max θ K(θ) 0<n N c N ˆχ a,b (θ) e(r a + b + c )c k e(c(θ + h 3 ))dθ c N K(θ) e(r n + c )c k e(c(θ + h 3 )) c dθ N e(r n + c )c k e(c(θ + h 3 )) c dθ N e(r n + c )c k e(c(θ + h 3 )) 0<n N c N Now let s apply Weyl differencing: for 1 Y N we have Y e(r n + c )c k e(c(θ + h 3 )) c N d Y d = c N+Y c c+d N e(r n + (c + d) )(c + d) k e((c + d)(θ + h 3 )) d Y c+d N d e(r n + (c + d) )(c + d) k e((c + d)(θ + h 3 )) 13

14 By Cauchy Inequality, Y e(r n + c )c k e(c(θ + h 3 )) c N e(r n + (c + d) c )(c + d) k e((c + d)(θ + h 3 )) N+Y d Y c+d N d =N 1/ e(r( n + (c + d 1 ) n + (c + d ) )) N 1/ c N+Y d 1, Y c+d 1, N d 1, (c + d 1 ) k (c + d ) k e((d 1 d )(θ + h 3 )) =N 1/ (Y d)e(r( n + (m + d) n + (m d) )) d=(d 1 d )/ d Y m=c+(d 1 +d )/ m N d (m + d) k (m d) k e(d(θ + h 3 )) N k+1 Y + N 1/ (Y d)e(r( n + (m + d) n + (m d) )) 1 d Y m N d (m + d) k (m d) k e(d(θ + h 3 )) 14

15 Therefore e(r n + c )c k e(c(θ + h 3 )) 0<n N c N N max θ e(r n + c )c k e(c(θ + h 3 )) 0<n N c N N k N 3 Y 1 + N 3/ Y 1 e(r( n + (m + d) d Y m n + (m d) )) N 0<n N N k N 3 Y 1 + N 3/ Y 1 d(y) e(f(x, y)) y=4md y Y N ɛ N k+ɛ (N 3 Y 1 + N 3/ Y 1 V N,Y N (R)) where f(x, y) = R( x + y x) and V N,D (R) = y D 0<x=n+(m d) N 0<x N e(f(x, y)) Hence we conclude that, for all 1 Y N, V N,Q,h(R) ɛ,q N ν+ɛ (N 3 Y 1 + N 3/ Y 1 V N,Y N (R)) ɛ N ν+ɛ (N 3 Y 1 + R 1/ N Y 1/6 ) The last step follows from Lemma 3. As a result V N,Q,h(R) ɛ,q N ν/+ɛ (N 3/ Y 1/ + R 1/4 NY 1/1 ) If N R 6/5, then we let Y N to get V N,Q,h(R) ɛ,q N ν/+ɛ (N 5/4 + R 1/4 N 5/4 ) N ν/+ɛ N 5/4 If R 1/ N R 6/5, then we let 1 Y N 6/7 R 3/7 N to get V N,Q,h(R) ɛ,q N ν/+ɛ N 15/14 R 3/14 If N R 1/, then N 15/14 R 3/14 N 3/ and Lemma reduces to the trivial bound. 15

16 Lemma 3. If R 1 and 1 D N, then V N,D (R) (log N)(N 3/ + R 1/ D 7/6 N 1/1 ) Proof. By dyadic decomposition where V N,D (R) = log D k=1 If D D 0 = N 3/ R 1, then V N, (R) (log N) max V k N,D 1 D 1 D 1 (R) VN,D(R) = y D 0<x N e(f(x, y)) f x (x, y) = R ( 1 x 1 x + y ) = Ry x x + y( x + x + y) 1 Therefore by Theorem.1 of [8] e(f(x, y)) f x(x, y) 1 N 3/ R 1 D 1 0<x N so V N,D (R) N 3/ R 1 N 3/. Now suppose D 0 D N. This is possible only when N R. By applying the B process of the one-dimensional Van der Corput s method (Poisson summation and stationary phase, Lemma 3.6 of [8] with F = RDN 1/ ) we get 0<x N e(f(x, y)) = ξ f x(x,y) e(g(ξ, y) 1/8) f xx (α(ξ, y), y) 1/ + O(log(RDN 3/ )) + O(R 1/ D 1/ N 5/4 ) + O(1) = e(g(ξ, y) 1/8) f xx (α(ξ, y), y) + O(log R) + 1/ O(R 1/ D 1/ N 5/4 ) ξ f x(x,y) where α(ξ, y) satisfies f x (α(ξ, y), y) = ξ, and g(ξ, y) = f(α(ξ, y), y) α(ξ, y)ξ 16

17 For any n N, x n f(x, y) does not change sign, so (replacing n by n + 1) it is monotone in x. Thus α(ξ, y) and hence f xx (α(ξ, y), y) are monotone in ξ. By Abel summation, we obtain e(f(x, y)) V N,U (y; R) inf ξ U f xx (α(ξ, y), y) + 1/ R 1/ D 1/ N 5/4 + log R 0<x N R 1/ D 1/ N 5/4 V N,U (y; R) + R 1/ D 1/ N 5/4 + log R for U f x (x, y) = RDN 3/, where V N,U (y; R) = e(g(ξ, y)) Hence VN,D(R) = y D 0<x N e(f(x, y)) ξ U R 1/ D 1/ N 5/4 y D V N,U (y; R) + R 1/ D 1/ N 5/4 + D log R R 1/ N 5/4 ( y D V N,U (y; R) ) 1/ + R 1/ N 7/4 + N log R By Lemma 4 below we conclude that V N,D(R) R 1/ N 5/4 (R 1/ DN 3/4 + RD 7/6 N 4/3 ) + R 1/ N 7/4 + N log R N 3/ + R 1/ D 7/6 N 1/1 where we have used 1 D N R to dominate the first two terms over the last two. Lemma 4. If R N and 1 D N, then V N,U (y; R) RD N 3/ + R D 7/3 N 8/3 y D 17

18 Proof. Apply Weyl differencing to V N,U (y; R): for some T [1, U] to be chosen later, T V N,U (y; R) = T 1 e(g(ξ + t, y)) y D y D t=0 U ξ U = U ξ U T 1 t 1,t =0 y D t T(T t ) ξ U U T D + UT U T D + U T where G(ξ, y, t) = g(ξ + t, y) g(ξ, y), and e(g(ξ + t 1, y) g(ξ + t, y)) e(g(ξ, y, t)) y D 1 t T ξ U max 1 t T V N,D (t, ξ; R) = y D e(g(ξ, y, t)) e(g(ξ, y, t)) y D V N,D (t, ξ; R) From the computations in [5] we know that n y G(ξ, y, t) tnd n for n = 1,, 3, so estimates (.3.8) and (.3.9) in [8] are valid (with F replaced by tn and N replace by D). Note that in this case tn D, so the first term in both estimates dominates the second term, yielding V N,D (t, ξ; R) min{t 1/ N 1/, T 1/6 N 1/6 D 1/ } Therefore V N,U (y; R) U (T 1 D + min{t 1/ N 1/, T 1/6 N 1/6 D 1/ }) y D Let s record the numerology of some borderlines. Balancing the first and the second term yields T 1 = D /3 N 1/3. Balancing the first and the third yields T = D 3/7 N 1/7. Balancing the second and the third yields T 3 = D 3/ N 1. Balancing T 1 with U = RDN 3/ 1 yields D 1 = N 7/ R 3. Balancing T with U gives D = N 19/8 R 7/4. Balancing T 3 with U gives 18

19 D 3 = R N 1. Comparing D 0 = N 3/ R 1, D 1, D, D 3 and N gives the following chart: N ordering [R 6/5, R ] D 3 D 0 N, D D 1 [R, R 6/5 ] D 0 D 1,,3 N [1, R] D 1 D 0, N D 3 (1) R 6/5 N R. We always have D 3 D 0 and N D 1. Thus T 1,3 U, so we always take the first argument of the min, which is then dominated by T 1 D. Therefore by setting T = U we get V N,U (y; R) UD = RD N 3/ y D () R N R 6/5. (.1) D 1 D N. Then we have T 1 U, so we balance the first argument of the min with T 1 D by setting T = T 1 D /3 1 N 1/3 = N R 1 to get V N,U (y; R) U D 1/3 N 1/3 = R D 7/3 N 8/3 y D (Although the second argument in the min may be smaller, it does not happen for large D, so we have to balance the first argument with T 1 D in this case.) (.) D 0 D D 1. Then we have U T 1, so we take T = U to reach the same conclusion as in (1). (3) 1 N R. (3.1) N 1/ D N. In this case T 1 1, so we can set T = T 1 to reach the same conclusion as (.1). (3.) 1 D N 1/. In this case we let T = 1: V N,U (y; R) U (D + N 1/6 D 1/ ) = R D 3 N 3 + R D 5/ N 17/6 y D which is again dominated by (.1). Combining (1) through (3) we conclude that for all D N, V N,U (y; R) RD N 3/ + R D 7/3 N 8/3 y D 19

20 5 Estimating the Long Sum Lemma 5. Suppose Q is a homogeneous polynomial of degree ν 0 in three variables, not necessarily having zero mean on the sphere, and m ν + 3, then Q(ξ)e(R ξ + h ξ) lim N ξ m ɛ,q R ɛ 0< ξ N Moreover, the limit on the left hand side converges uniformly (although not absolutely) in h. Proof. By Abel summation where 0< ξ N Q(ξ)e(R ξ + h ξ) ξ m = N n=1 n m/ ξ =n N n m/ 1 V n,q,h (R) + N m/ V N,Q,h (R) n=1 V N,Q,h (R) = ξ N Q(ξ)e(R ξ + h ξ) Q(ξ)e(R ξ + h ξ) and we have ignored the discrepancy at ξ = 0, which is of order O Q (1) and is clearly dominated by other terms. By Lemma we have lim sup Q(ξ)e(R ξ + h ξ) N ξ m 0< ξ N ɛ,q n m/+ν/+1/+ɛ + R 3/14 n m/+ν/+1/14+ɛ 1 n R 1/ R 6/11 n R 6/5 + n m/+ν/+1/4+ɛ + lim N m/+ν/ (N 5/4 + N 15/14 R 3/14 ) N n R 6/5 n ɛ R1/ n=1 + R 3/14 n 3/7+ɛ R1/ n=r 6/5 + n 1/4+ɛ R6/5 n= R ɛ Moreover, for M > N R 6/5 we have Q(ξ)e(R ξ + h ξ) ξ m ɛ,q n 1/4+ɛ N n= + N m/+ν/ (N 5/4 + N 15/14 R 3/14 ) N< ξ M ɛ,q 0, as N 0

21 Theorem 1. Suppose P is a homogeneous polynomial of degree ν 0, not necessarily having zero mean on the sphere, R 1 and H 1, then S f,p (R) = P (x)g(x)dx + O ɛ,p R ν H ɛ (RH 1/ + R 17/14 H 1/7 ) R 3 Proof. Consider the function S f,p (h; R) = P (x)g(x)dx + R 3 ξ Z 3 \0 ĝ P (ξ)e(h ξ) where the summation over ξ in the second term on the right hand side is taken in the sense of Lemma 5, and ĝ P (ξ) (for ξ Z 3 \0) is given by ĝ P (ξ) = + ν 1 +ν =ν ν 1 +ν +ν 3 =ν R ν 1 Q ν 1 (ξ)e(r ξ ) ξ 3+ν 1+ν RH ν 1 1 (R + H) ν Q ν1,ν (ξ) sin(πh ξ + πν 1 ) cos(π(r + H) ξ + πν ) ξ 3+ν 1+ν +ν 3 Since 3 + ν 1 + ν ν + 3 deg Q ν1 + 3, Lemma 5 applies to show that ξ Z 3 \0 ν 1 +ν =ν R ν 1 Q ν 1 (ξ)e(r ξ + h ξ) ξ 3+ν 1+ν ɛ,q R ν+ɛ Similarly, since 3 + ν 1 + ν + ν 3 deg Q ν1,ν + 3, and R + H R, ξ Z 3 \0 ν 1 +ν +ν 3 =ν ν 1 1 ɛ, QR 1+ν +ɛ R ν+ɛ RH ν 1 1 (R + H) ν Q ν1,ν (ξ) sin(πh ξ + πν 1 ) cos(π(r + H) ξ + πν )e(h ξ) ξ 3+ν 1+ν +ν 3 (when ν = 0, i.e. P is constant, this sum actually vanishes), so we conclude that S f,p,h (R) = ξ Z 3 \0 ν +ν 3 =ν + O ɛ,p (R ν+ɛ ) RH 1 (R + H) ν 1 Q 0,ν (ξ) sin(πh ξ ) cos(π(r + H) ξ + πν )e(h ξ) ξ 3+ν +ν 3

22 By Abel summation and Lemma we have 0< ξ H 1 / H /4 sin(πh n) = n 3/+ν /+ν 3 n=1 H /4 = N=1 Q 0,ν (ξ) sin(πh ξ ) cos(π(r + H) ξ + πν )e(h ξ) ξ 3+ν +ν 3 ξ Z 3 ξ =n ( Q 0,ν (ξ) cos ( sin(πh ) N) N N 3/+ν /+ν 3 +O(H 3+ν +ν 3 V H /4,P,h(R)) H ɛ,p H H /4 N=1 ξ Z 3 0< ξ N π(r + H) ξ + πν ( Q 0,ν (ξ) cos N ν/ V N,P,h (R) + H 3+ν V H /4,P,h(R) ) e(h ξ) π(r + H) ξ + πν 1 N H N ɛ (N 3/4 + N 13/14 R 3/14 ) + H 1/ + H 6/7 R 3/14 HN ɛ (N 1/4 + N 1/14 R 3/14 ) H 1 + H 1/ + H 6/7 R 3/14 =H ɛ (H 1/ + H 6/7 R 3/14 ) and similarly ξ >H 1 / = N>H /4 Q 0,ν (ξ) sin(πh ξ ) cos(π(r + H) ξ + πν )e(h ξ) N ( 1 N 3/+ν /+ν 3 ξ 3+ν +ν 3 ) ξ Z 3 0< ξ N N H N 5/ ν/ V N,P,h (R) ɛ,p ( Q 0,ν (ξ) sin(πh ξ ) cos N H N ɛ (N 5/4 + N 10/7 R 3/14 ) N ɛ (N 1/4 + N 3/7 R 3/14 ) H = H ɛ (H 1/ + H 6/7 R 3/14 ) Combining the two parts we get S f,p (h; R) = P (x)g(x)dx + O ɛ,p R ν H ɛ (RH 1/ + R 17/14 H 1/7 ) R 3 ) e(h ξ) π(r + H) ξ + πν ) e(h ξ)

23 By Lemma 5 we know that the sum S f,p (h; R) converges uniformly in h. On the other hand, by Parseval identity we know that the sum converges in L (R 3 /Z 3 ) to g P (h) = x Z 3 P (x + h)g(x + h) Since this is a continuous function, it is identical (everywhere) to the uniform limit S f,p (h; R). In particular, taking h = 0 yields S f,p (R) = x Z 3 P (x)g(x) = S f,p (0; R) = P (x)g(x)dx + O ɛ,p R ν H ɛ (RH 1/ + R 17/14 H 1/7 ) R 3 6 Estimating the Short Sum Now we turn to the short sum. Suppose P is a homogeneous polynomial of degree ν > 0 in three variables, now having zero mean on the sphere. Note that deg P = 0 implies P = 0, which is trivial. Definition 5. By the short sum we mean the following: S f,p (R, H) = x Z3 R x R+H 1 x f( x ) Moreover, we let and define the theta-series a n = x Z3 x =n P (x) θ(z) = n N a n e(nz) = x Z 3 P (x)e( x z) where as usual e(z) = e πiz. First we suppose that P is harmonic, then by Lemma 9 in Appendix, θ is a cusp form of half integral weight k = ν + 3/ 5/ for Γ 0 (4). For all n N we have: 3

24 Lemma 6. ([1], Proposition 1.5.5) Lemma 7. ([1], Corollary ) a n ɛ,θ n k/ 1/4+ɛ a n ɛ,θ n k/ 5/16+ɛ (n, ) 5/8 Theorem. Suppose P is a homogeneou polynomial of degree ν 0 in three variables, having zero mean on the sphere. Then S f,p (R, H) ɛ,p R ν+ɛ (R 15/8 H + R) Proof. By Corollary.50 of [7], we can write [ν/] P (x) = x d P ν d (x) d=0 where P ν d is a homogeneous harmonic polynomial of degree ν d. If ν is odd, then there is no term P 0 in the sum (and actually a n vanishes identically by considering the symmetry x x). If ν is even, then integration over the sphere gives [ν/] 0 = P (x) = P ν d (x) = P 0 S S d=0 so the sum does not include P 0 either. Now by Lemma 6 and Lemma 7, (R+H) (R+H) S f,p (R, H) = f(n)a n n n=r a n n=r [ν/] (R+H) ɛ,p R d (n, ) 5/8 + R ν d+1+ɛ d=0 R ν+ɛ ( R 7/8 R ν d+3/ 5/8+ɛ k log RH R ν+ɛ (R 15/8 H + R) n=r (n, ) RH RH k 5k/8 + R ) 4

25 7 Appendix A: Modularity of the Theta Function Suppose n. Let C[x] denote a polynomial with complex coefficients in the variable x C n. Let C[x] ν be the subspace of C[x] consisting of homogeneous polynomials of degree ν. Let H[x] denote the subspace of harmonic polynomials. Lemma 8. C[x] ν H[x] = span C {( n j=1 a jx j ) ν, (a j ) n j=1 C n, n j=1 a j = 0}. Proof. Define a positive definite inner product on C[x] by letting x α, x β = α!δ αβ. The Laplacian and the multiplication operator M(f) = f x operate on C[x]. It is easy to check that they are adjoints under this inner product. Therefore H = ker = (ran M) We restrict the whole space to C[x] ν, where the inner product is still positive definite, and denote the right hand side by L. It suffices to show that L = (ran M) Since C[x] ν is finite dimensional, it is equivalent to L = ran M For f(x) = α =d c αx α C[x] ν we have ( n ) ν a j x j, f(x) = j=1 Therefore α =d f L f(a i ) = 0, (a i ) C n, α! α! aα x α, α =d c α x α = α! c α a α = α!f(a) α =d n a i = 0 x f(x) f ran M i=1 where we have used Hilbert s Nullstellensatz and the fact that ( x ) is a radical ideal in C[x] when n. (When n 3 it is actually irreducible, or equivalently, prime, in the UFD C[x]. When n = we have x = (x 1 + ix )(x 1 ix ) does not have repeated factors.) 5

26 Now suppose p ν C[x] ν H[x]. Let θ(z) = x Z n p ν (x)e( x z). Then θ is holomorphic on the upper half plane H. Lemma 9. θ is a cusp form of weight ν + n/ on Γ 0 (4)\H, where {( ) } a b Γ 0 (N) = SL c d (Z), N c More precisely, for every γ Γ 0 (4) we have θ(γz) = j(γ, z) ν+n θ(z) where ( c j(γ, z) = d) ɛ 1 d (cz + d)1/ where ( c d) is the Legendre symbol { as extended by Shimura ([13], point 3 in 1, d = 1 mod 4 Notation and Terminology), ɛ d =, and all square roots i, d = 3 mod 4 are taken in the principal branch ( 1 is taken to be e(+1/4).) Proof. If ν is odd then θ = 0 by considering the symmetry x x. Therefore we assume ν to be even. By Lemma 8, ( n ) ν n θ span C a j x j e( x z), a j = 0 x Z n j=1 Therefore we may assume that θ is one of the basis vectors on the right hand side. ( ) a b First we suppose SL c d (Z). If c = 0, then a = d = ±1. In either case we have j(γ, z) = 1, consistent with the fact that θ(z + b) = θ(z) for any b Z. j=1 6

27 Now we suppose c 0, then ( ) ( ) az + b a θ = θ cz + d c 1 c(cz + d) = ( n m (Z/cZ) n x Z n j=1 = ( n e(am /c) m (Z/cZ) n x Z n i=1 ν (( a a j (cx j + m j )) e By Poisson summation formula, ( n ν ( c a j (cx j + m j )) e (cz + d) x Z n j=1 = ( ) ( ) ν im ξ n (π) n/ exp a j ci ξj ( c 4πic ξ πz n j=1 cz+d ( ) ( ) ( n/ n im ξ cz + d =(ci) ν c ic ( n ξ πz n exp c 1 c(cz + d) ν ( a j (cx j + m j )) e x + m c ( ) n/ e j=1 ) 1 πi c cz + d ) cz + d 8πic ξ ) cx + m ) x + m c (5) ) ν ( ( 1 a j ξj e z + d ) ) ξ 16π c We compute the derivative on the rightmost exponential inductively. j=1 where ( n ) ( ( 1 a j ξj e z + d ) ) ( ( 1 ξ = C(z, c, d, a, ξ)e z + d ) ) ξ 16π c 16π c Suppose j=1 C(z, c, d, a, ξ) = i ( z + d ) ( ) n a j ξ j 4π c j=1 ) k ( ( 1 a j ξj e z + d ) ) ( ( 1 ξ = C(z, c, d, a, ξ) k e z + d ) ) ξ 16π c 16π c (6) ) 7

28 Then ( n ) k+1 ( ( 1 a j ξj e z + d ) ) ( ( 1 ξ = C(z, c, d, a, ξ) k+1 e z + d ) ) ξ 16π c 16π c j=1 n +kc(z, c, d, a, ξ) k 1 a j ξj C(z, c, d, a, ξ) Since n j=1 a j = 0, it is easy to see that the second term on the right hand side vanishes, so by induction we have shown that (6) holds for all k N. Therefore the inner sum of (5) is ( 1) ν ξ πz n exp = ( 1)ν (i) n/ ξ Z n e where ( ) ( ) [ ( n/ n im ξ cz + d cz + d c ic 4π j=1 ( ) ( m ξ z + d ) n/ ( ) ν cz + d P (ξ) ν e c c j=1 )] ν ( 1 a j ξ j e ( ( 1 z + d 4 c Now we supppose 4 c and carry out the summation over m in (5). S n (ξ, a, c) = S(ξ j, a, c) = m Z/cZ m (Z/cZ) n e ( ) am + m ξ = c ( ) am + mξ j e = c m Z/cZ n S(ξ j, a, c) j=1 ( ) d(m + mξ j ) e c 16π ) ) ξ We now show that if ξ j is odd, then S(ξ j, a, c) = 0. In fact, since 4 c and (c, d) = 1, d is odd, so S(ξ j, a, c) = [ ( ) ( )] d(m + mξ j ) d((m + c/) + (m + c/)ξ j ) e + e c c m Z/cZ = ( ) d(m + mξ j ) e [1 + e(d(m + c/4 + ξ j /))] = 0 c m Z/cZ ( z + d ) ) ξ c (7) 8

29 Therefore the sum over ξ in (5) is reduced to ( ( 1) ν n ) ( S(ξ (i) n/ j, a, c) z + d c ξ Z n j=1 where the sum over m is reduced to S(ξ j, a, c) = ( ) d(m + mξ j ) e c m Z/cZ ) n/ (cz + d) ν P (ξ) ν e (( z + d c ( ) = e dξ j e c m Z/cZ By the well-known Gauβ sum, (1 + i)ɛ 1 ( d c c ) d, c, d > 0 ( ) dm ( (1 i)ɛ d c c e = d), c > 0, d < 0 ( c (1 i)ɛ m Z/cZ d c c ) d, c < 0, d > 0 ( (1 + i)ɛ 1 d c c d), c, d < 0 the identities ɛ d = iɛ 1 d, 1 + i = i, z + d c c = cz + d, c > 0 z + d c c = i cz + d, c < 0 ) ξ ) ( ) dm and the extension of the Legendre symbol by Shimura, the Gauβ sum can be written uniformly as ( ) dm e = ( c ) ( iɛ 1 d cz + d z + d ) 1/ c d c Therefore, θ m Z/cZ ( ) az + b = ( 1) ν ɛ n d cz + d (cz + d)ν+n/ P (ξ) ν e(z ξ ) ξ Z n Recall that ν is even, so ( 1) ν = ɛ ν d = 1, and ( ) az + b θ = ɛ ν n d (cz + d) ν+n/ P (ξ) ν e(z ξ ) = j(γ, z) ν+n θ(z) cz + d ξ Z n It is clear that θ is a cusp form, because the sum in (7) is absolutely convergent and decays rapidly as Im z + for all γ SL (Z). c 9

30 8 Appendix B: Better Results with New Exponent Pairs Through personal communication with Chamizo, the author realized that the exponent 83/64 in the estimate P (x) = O ɛ,p (R ν+83/64+ɛ ) x Z 3 x R still has some room for improvement using the new exponent pairs recently obtained by Huxley [10]. Exponent pairs are a useful tool in estimating exponential sums of the following type e(f(n)) n N More precisely, suppose f behaves sufficiently close to a power function x s, in the sense that there is a constant c > 0 such that for all r N, d r dr f(x) c dxr dx r x s then for every ɛ > 0, P > 0, and N > N ɛ,p, f is in the class F(N, P, s, c, ɛ) as defined in P 30, [8]. Exponent pairs allow us to give estimates to the above exponential sum. Definition 6. (P 30, [8]) For 0 k 1/ l 1, (k, l) is an exponent pair if we have the following estimate e(f(n)) s (cn s 1 ) k N l + c 1 N s+1 n N Remark. If f cn s 1 1, then the second term is dominated by the first term. Example 1. By the triangle inequality, (k, l) = (0, 1) is an exponent pair. Exponent pairs are abstractions of the Van der Corput A and B processes as carried out in the main part of the paper. Specifically, if (k, l) is an 30

31 exponent pair, then by applying Weyl differencing (and optimization of the length) we can get another exponent pair ( k A(k, l) = k +, k + l + 1 ) k + On the other hand, an application of Poisson summation and stationary phase (the B-process) will give as the pair B(k, l) = (l 1/, k + 1/) For details and proofs we refer the reader to [8]. Example. From (k, l) = (0, 1) we can get B(k, l) = (1/, 1/) and AB(k, l) = (1/6, /3), which are basically the pairs we used to bound V N,D (t, ξ, R). From Van der Corput A and B processes we can get a host of exponent pairs. This is, however, not the whole story. Bombieri and Iwaniec (BI,??), using large sieve inequalities, obtained a new exponent pair (k, l) = (9/56 + ɛ, 37/56 + ɛ) which is unreachable from A and B processes. Their results were subsequently improved by Huxley, giving one more exponent pair (3/05 + ɛ, 69/410 + ɛ) in [10]. New exponent pairs can offer further optimizations. In [3], Chamizo and Cristóbal applied the exponent pair (k, l) = BA (3/05 + ɛ, ɛ) to the following sum V N,D (t, ξ; R) = y D e(g(ξ, y, t)) which occurs in the proof of Lemma 4. The result is Proposition 3.6 of [3], which in our notation says Lemma 10. If (k, l) is an exponent pair, R N and 1 D N, then V N,U (y; R) ɛ RD 3k+l+3 k 3 k+ N k+ + R 1/+ɛ D 3/ N 3/4 y D Plugging Lemma 10 into the final lines of Lemma 3 we get V N,D(R) R 1/ D 1/ N 5/4 y D V N,U (y; R) + R 1/ D 1/ N 5/4 + N log R ɛ N 3/+ɛ + R 1/ D k+l+ k+ N k 1 4k+4 31

32 which implies the same estimate for V N,D (R). Finally, backtracing to Lemma, we get VN,Q,h(R) ɛ,q N ν+ɛ (N 3 Y 1 + R 1/ k+l 1 l + N 4k+4 Y k+ ) (8) Balancing the two term gives Comparing Y 0 with N gives Y 0 = N 3k l+5 4k+l+4 R k+1 k+l+ N 0 = R k+ k l+3 Therefore when N N 0, Y 0 N, so we take Y 0 N and the first term in (8) dominates. Otherwise, when N N 0, we can balance the two terms in (8) by setting Y = Y 0 to get V N,Q,h(R) ɛ,q N ν+ɛ R k+1 k+l+ N + k+3l+1 4k+l+4 Combining the two estimates and adding up dyadic intervals we conclude V N,Q,h (R) ɛ,q N ν/+ɛ (N 5/4 + R k+1 4k+l+4 N 1+ k+3l+1 8k+4l+8 ) which improves Theorem 1 to Theorem 3. If f, P and R are as in Theorem 1, then S f,p (R) = P (x)g(x)dx + O ɛ,p R ν H ɛ (RH 1/ + R 1+ R 3 k+1 4k+l+4 H k+3l 1 4k+l+4 ) Now we balance the error terms in Theorem 3 and Theorem using Lemma.4 of [8] to obtain P (x) = O ɛ,p (R ν+1+ɛ (R 7/4 + R 15k+1l+1 40k+40l+4 )) (9) x Z 3 x R Plugging in the exponent pair (k, l) = BA (3/05 + ɛ, ɛ) = (743/04 + ɛ, 69/506 + ɛ) we get the dominant exponent ν / ɛ which is a further improvement on the previous exponent ν + 83/64 + ɛ. The latter can in turn be recovered by taking the well known exponent pair (k, l) = (1/, 1/) = B(0, 1), 3

33 9 Appendix C: Summary of Proved and Conjectured θ ν This section summarizes all the proved and conjectured exponents mentioned in Section (and some more). For simplicity, we have omitted the normalizing factors R ν, the O symbol and all ɛ s in the exponents. References and decimal values are included in parentheses. The ellipsis refers to terms that are clearly dominated by the main term, and question marks indicate conjectures. The last column refers to the optimal value of α such that setting H = R α gives the desired error bound. The reader is referred to Remark 3 for meaning of the acronyms and to Section for a detailed account of relevant terminology. log H Long sum estimates Short sum estimates θ ν log R Applicability RH 1 (Van der Corput) R H (Trivial) 3/ 1/ P (1.5) (.5) RH 1/ +... R H (Trivial) 4/3 /3 P ([6] and [14]) ( ) (.66667) RH 1/ + R 1/16 R 15/8 H 7/ / 7/11 P = 1 +R 11/8 H 1/8 ([5]) ([5]) ( ) (.63636) RH 1/ + R 1/16 R 11/6 H 5/ /16 5/8 P = 1 +R 11/8 H 1/8 ([5]) ([9]) (1.315) (.65) RH 1/ + R 17/14 H 1/7 R 15/8 H + R 83/64 37/64 S P = 0 (Theorem 1) (Theorem ) (1.9688) (.57813) RH 1/ + R H R 15/ H + R S P = 0 (Theorem 3 with [10]) (Theorem ) (1.9476) (.5804) RH 1/ + R 6/5 H 1/10?? R 15/8 H + R 31/4?? 7/1?? S P = 0 (Theorem 3 with LEP) (Theorem ) (1.9167) (.58333) RH 1/ + R 17/14 H 1/7 R 3/ H 1/?? 3/18?? 4/9?? P = 1 (Theorem 1) (GLH) (1.7778) (.44444) RH 1/ + R 6/5 H 1/10?? R 3/ H 1/?? 5/4?? 1/?? P = 1 (Theorem 3 with LEP) (GLH) (1.5) (.5) RH 1/ + R 17/14 H 1/7 R 3/ H?? 5/4?? 1/4?? S P = 0 (Theorem 1) (RC) (1.5) (.5) RH 1/ + R H R 3/ H?? 5710?? 895?? S P = 0 (Theorem 3 with [10]) (RC) (1.466) (.5338) RH 1/ + R 6/5 H 1/10?? R 3/ H?? 7/?? 3/11?? S P = 0 (Theorem 3 with LEP) (RC) (1.77) (.773) Ultimate Conjecture i.e. Conjecture 1: 1???????? P 33

34 Remark 3. Acronyms: LEP = Lindelöf Exponent Pair Conjecutre [3]. GLH = Generalized Lindelöf Hypothesis [3]. RC = Ramanujan Conjecture [1]. 10 Acknowledgements The author wish to express his thanks to MIT for offering such an incredible program as SPUR, to Prof. David Jerison for suggesting this marvellous topic, to Prof. Etingof and Prof. Fox for overseeing the progress of the topic, to Chenjie Fan for his meticulous mentoring over the six weeks, and to Prof. Chamizo for enlightening and encouraging correspondence. Speaking of the technicalities, credits go to Prof. Etingof for opening the door of modular forms to me, to Prof. Fox for suggesting some connections to combinatorics, to Chenjie Fan for pointing the direction to the proof of Lemma 8 and justification of the convergence of the Poisson summation formula, and for helping the author with some L A TEXing problems. Thanks are also due to Prof. Chamizo for bringing to my attention the latest results in the theory of exponent pairs. References [1] Valentin Blomer and Gergely Harcos. Hybrid bounds for twisted L- functions. Journal für die reine und angewandte Mathematik, 61:53 79, 008. [] F. Chamizo. Lattice point counting and harmonic analysis [3] F. Chamizo and E. Cristóbal. The sphere problem and the L-functions. Acta. Math. Hungar., 01. [4] F. Chamizo, E. Cristóbal, and A. Ubis. Lattice points in rational ellipsoids. Journal of Mathematical Analysis and Applications, 350:83 89, 009. [5] Fernando Chamizo and Henryk Iwaniec. On the sphere problem. Revista Matemática Iberoamericana, 11():417 49,

35 [6] Jing-Run Chen. Improvement on the asymptotic formulas for the number of lattice points in a region of the three dimensions (ii). Sci. Sinica., 1: , [7] Gerald B. Folland. Introduction to Partial Differential Equations. Princeton University Press, [8] S. W. Graham and G. Kolesnik. Van der Corput s Method of Exponential Sums. Cambridge University Press, [9] D. R. Heath-Brown. Lattice points in the sphere. Number Theory in Progress, ():883 89, [10] M. N. Huxley. Exponential sums and the Riemann zeta function V. [11] David Jerison, Lionel Levine, and Scott Sheffield. Internal DLA and the Gaussian free field. Available at [1] Peter Sarnak. Some Applications of Modular Forms. Cambridge University Press, [13] Goro Shimura. On modular forms of half integral weight. Annals of Mathematics, 97(3): , [14] Vinogradov. On the number of integer points in a sphere (russian). Izv. Akad. Nauk. SSSR Ser. Mat., 7: ,

ON SUMS OF PRIMES FROM BEATTY SEQUENCES. Angel V. Kumchev 1 Department of Mathematics, Towson University, Towson, MD , U.S.A.

ON SUMS OF PRIMES FROM BEATTY SEQUENCES. Angel V. Kumchev 1 Department of Mathematics, Towson University, Towson, MD , U.S.A. INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A08 ON SUMS OF PRIMES FROM BEATTY SEQUENCES Angel V. Kumchev 1 Department of Mathematics, Towson University, Towson, MD 21252-0001,

More information

Solutions to Complex Analysis Prelims Ben Strasser

Solutions to Complex Analysis Prelims Ben Strasser Solutions to Complex Analysis Prelims Ben Strasser In preparation for the complex analysis prelim, I typed up solutions to some old exams. This document includes complete solutions to both exams in 23,

More information

Taylor and Laurent Series

Taylor and Laurent Series Chapter 4 Taylor and Laurent Series 4.. Taylor Series 4... Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor series of a given function f : R R is given by: f (x + f (x (x x + f (x

More information

17 The functional equation

17 The functional equation 18.785 Number theory I Fall 16 Lecture #17 11/8/16 17 The functional equation In the previous lecture we proved that the iemann zeta function ζ(s) has an Euler product and an analytic continuation to the

More information

Large Sieves and Exponential Sums. Liangyi Zhao Thesis Director: Henryk Iwaniec

Large Sieves and Exponential Sums. Liangyi Zhao Thesis Director: Henryk Iwaniec Large Sieves and Exponential Sums Liangyi Zhao Thesis Director: Henryk Iwaniec The large sieve was first intruded by Yuri Vladimirovich Linnik in 1941 and have been later refined by many, including Rényi,

More information

GAUSS CIRCLE PROBLEM

GAUSS CIRCLE PROBLEM GAUSS CIRCLE PROBLEM 1. Gauss circle problem We begin with a very classical problem: how many lattice points lie on or inside the circle centered at the origin and with radius r? (In keeping with the classical

More information

Mean square limit for lattice points in a sphere

Mean square limit for lattice points in a sphere ACTA ARITHMETICA LXVIII.4 (994) Mean square limit for lattice points in a sphere by Pavel M. Bleher (Indianapolis, Ind.) and Freeman J. Dyson (Princeton, N.J.). Main result. Let N(R) = #{n Z 3 : n R} be

More information

Course Description for Real Analysis, Math 156

Course Description for Real Analysis, Math 156 Course Description for Real Analysis, Math 156 In this class, we will study elliptic PDE, Fourier analysis, and dispersive PDE. Here is a quick summary of the topics will study study. They re described

More information

1.3.1 Definition and Basic Properties of Convolution

1.3.1 Definition and Basic Properties of Convolution 1.3 Convolution 15 1.3 Convolution Since L 1 (R) is a Banach space, we know that it has many useful properties. In particular the operations of addition and scalar multiplication are continuous. However,

More information

Roth s Theorem on Arithmetic Progressions

Roth s Theorem on Arithmetic Progressions September 25, 2014 The Theorema of Szemerédi and Roth For Λ N the (upper asymptotic) density of Λ is the number σ(λ) := lim sup N Λ [1, N] N [0, 1] The Theorema of Szemerédi and Roth For Λ N the (upper

More information

Real Analysis Problems

Real Analysis Problems Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1 1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals.

More information

SPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS

SPRING 2006 PRELIMINARY EXAMINATION SOLUTIONS SPRING 006 PRELIMINARY EXAMINATION SOLUTIONS 1A. Let G be the subgroup of the free abelian group Z 4 consisting of all integer vectors (x, y, z, w) such that x + 3y + 5z + 7w = 0. (a) Determine a linearly

More information

One of the Eight Numbers ζ(5),ζ(7),...,ζ(17),ζ(19) Is Irrational

One of the Eight Numbers ζ(5),ζ(7),...,ζ(17),ζ(19) Is Irrational Mathematical Notes, vol. 70, no. 3, 2001, pp. 426 431. Translated from Matematicheskie Zametki, vol. 70, no. 3, 2001, pp. 472 476. Original Russian Text Copyright c 2001 by V. V. Zudilin. One of the Eight

More information

Finite-dimensional spaces. C n is the space of n-tuples x = (x 1,..., x n ) of complex numbers. It is a Hilbert space with the inner product

Finite-dimensional spaces. C n is the space of n-tuples x = (x 1,..., x n ) of complex numbers. It is a Hilbert space with the inner product Chapter 4 Hilbert Spaces 4.1 Inner Product Spaces Inner Product Space. A complex vector space E is called an inner product space (or a pre-hilbert space, or a unitary space) if there is a mapping (, )

More information

Quantitative Oppenheim theorem. Eigenvalue spacing for rectangular billiards. Pair correlation for higher degree diagonal forms

Quantitative Oppenheim theorem. Eigenvalue spacing for rectangular billiards. Pair correlation for higher degree diagonal forms QUANTITATIVE DISTRIBUTIONAL ASPECTS OF GENERIC DIAGONAL FORMS Quantitative Oppenheim theorem Eigenvalue spacing for rectangular billiards Pair correlation for higher degree diagonal forms EFFECTIVE ESTIMATES

More information

PUTNAM TRAINING PROBLEMS

PUTNAM TRAINING PROBLEMS PUTNAM TRAINING PROBLEMS (Last updated: December 3, 2003) Remark This is a list of Math problems for the NU Putnam team to be discussed during the training sessions Miguel A Lerma 1 Bag of candies In a

More information

Department of Mathematics, University of California, Berkeley. GRADUATE PRELIMINARY EXAMINATION, Part A Fall Semester 2016

Department of Mathematics, University of California, Berkeley. GRADUATE PRELIMINARY EXAMINATION, Part A Fall Semester 2016 Department of Mathematics, University of California, Berkeley YOUR 1 OR 2 DIGIT EXAM NUMBER GRADUATE PRELIMINARY EXAMINATION, Part A Fall Semester 2016 1. Please write your 1- or 2-digit exam number on

More information

Mid-term Exam Answers and Final Exam Study Guide CIS 675 Summer 2010

Mid-term Exam Answers and Final Exam Study Guide CIS 675 Summer 2010 Mid-term Exam Answers and Final Exam Study Guide CIS 675 Summer 2010 Midterm Problem 1: Recall that for two functions g : N N + and h : N N +, h = Θ(g) iff for some positive integer N and positive real

More information

HASSE-MINKOWSKI THEOREM

HASSE-MINKOWSKI THEOREM HASSE-MINKOWSKI THEOREM KIM, SUNGJIN 1. Introduction In rough terms, a local-global principle is a statement that asserts that a certain property is true globally if and only if it is true everywhere locally.

More information

A LeVeque-type lower bound for discrepancy

A LeVeque-type lower bound for discrepancy reprinted from Monte Carlo and Quasi-Monte Carlo Methods 998, H. Niederreiter and J. Spanier, eds., Springer-Verlag, 000, pp. 448-458. A LeVeque-type lower bound for discrepancy Francis Edward Su Department

More information

Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi

Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.

More information

McGill University Math 354: Honors Analysis 3

McGill University Math 354: Honors Analysis 3 Practice problems McGill University Math 354: Honors Analysis 3 not for credit Problem 1. Determine whether the family of F = {f n } functions f n (x) = x n is uniformly equicontinuous. 1st Solution: The

More information

INDEFINITE THETA FUNCTIONS OF TYPE (n, 1) I: DEFINITIONS AND EXAMPLES

INDEFINITE THETA FUNCTIONS OF TYPE (n, 1) I: DEFINITIONS AND EXAMPLES INDEFINITE THETA FUNCTIONS OF TYPE (n, ) I: DEFINITIONS AND EXAMPLES LARRY ROLEN. Classical theta functions Theta functions are classical examples of modular forms which play many roles in number theory

More information

PRIME NUMBER THEOREM

PRIME NUMBER THEOREM PRIME NUMBER THEOREM RYAN LIU Abstract. Prime numbers have always been seen as the building blocks of all integers, but their behavior and distribution are often puzzling. The prime number theorem gives

More information

Cycle lengths in sparse graphs

Cycle lengths in sparse graphs Cycle lengths in sparse graphs Benny Sudakov Jacques Verstraëte Abstract Let C(G) denote the set of lengths of cycles in a graph G. In the first part of this paper, we study the minimum possible value

More information

HARMONIC ANALYSIS TERENCE TAO

HARMONIC ANALYSIS TERENCE TAO HARMONIC ANALYSIS TERENCE TAO Analysis in general tends to revolve around the study of general classes of functions (often real-valued or complex-valued) and operators (which take one or more functions

More information

On Rankin-Cohen Brackets of Eigenforms

On Rankin-Cohen Brackets of Eigenforms On Rankin-Cohen Brackets of Eigenforms Dominic Lanphier and Ramin Takloo-Bighash July 2, 2003 1 Introduction Let f and g be two modular forms of weights k and l on a congruence subgroup Γ. The n th Rankin-Cohen

More information

Rational Points on the Unit Sphere

Rational Points on the Unit Sphere Rational oints on the Unit Sphere Eric Schmutz Mathematics Department, Drexel University hiladelphia, ennsylvania, 19104 Eric.Jonathan.Schmutz@drexel.edu May 3, 008 MSC: 11J99,14G05; Keywords: Diophantine

More information

An Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010

An Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010 An Introduction to Complex Analysis and Geometry John P. D Angelo, Pure and Applied Undergraduate Texts Volume 12, American Mathematical Society, 2010 John P. D Angelo, Univ. of Illinois, Urbana IL 61801.

More information

Math 172 Problem Set 8 Solutions

Math 172 Problem Set 8 Solutions Math 72 Problem Set 8 Solutions Problem. (i We have (Fχ [ a,a] (ξ = χ [ a,a] e ixξ dx = a a e ixξ dx = iξ (e iax e iax = 2 sin aξ. ξ (ii We have (Fχ [, e ax (ξ = e ax e ixξ dx = e x(a+iξ dx = a + iξ where

More information

Exercise Solutions to Functional Analysis

Exercise Solutions to Functional Analysis Exercise Solutions to Functional Analysis Note: References refer to M. Schechter, Principles of Functional Analysis Exersize that. Let φ,..., φ n be an orthonormal set in a Hilbert space H. Show n f n

More information

REAL AND COMPLEX ANALYSIS

REAL AND COMPLEX ANALYSIS REAL AND COMPLE ANALYSIS Third Edition Walter Rudin Professor of Mathematics University of Wisconsin, Madison Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any

More information

We start with a simple result from Fourier analysis. Given a function f : [0, 1] C, we define the Fourier coefficients of f by

We start with a simple result from Fourier analysis. Given a function f : [0, 1] C, we define the Fourier coefficients of f by Chapter 9 The functional equation for the Riemann zeta function We will eventually deduce a functional equation, relating ζ(s to ζ( s. There are various methods to derive this functional equation, see

More information

L p Functions. Given a measure space (X, µ) and a real number p [1, ), recall that the L p -norm of a measurable function f : X R is defined by

L p Functions. Given a measure space (X, µ) and a real number p [1, ), recall that the L p -norm of a measurable function f : X R is defined by L p Functions Given a measure space (, µ) and a real number p [, ), recall that the L p -norm of a measurable function f : R is defined by f p = ( ) /p f p dµ Note that the L p -norm of a function f may

More information

We are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero

We are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero Chapter Limits of Sequences Calculus Student: lim s n = 0 means the s n are getting closer and closer to zero but never gets there. Instructor: ARGHHHHH! Exercise. Think of a better response for the instructor.

More information

ANALYSIS QUALIFYING EXAM FALL 2017: SOLUTIONS. 1 cos(nx) lim. n 2 x 2. g n (x) = 1 cos(nx) n 2 x 2. x 2.

ANALYSIS QUALIFYING EXAM FALL 2017: SOLUTIONS. 1 cos(nx) lim. n 2 x 2. g n (x) = 1 cos(nx) n 2 x 2. x 2. ANALYSIS QUALIFYING EXAM FALL 27: SOLUTIONS Problem. Determine, with justification, the it cos(nx) n 2 x 2 dx. Solution. For an integer n >, define g n : (, ) R by Also define g : (, ) R by g(x) = g n

More information

Chapter 8. P-adic numbers. 8.1 Absolute values

Chapter 8. P-adic numbers. 8.1 Absolute values Chapter 8 P-adic numbers Literature: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, 2nd edition, Graduate Texts in Mathematics 58, Springer Verlag 1984, corrected 2nd printing 1996, Chap.

More information

PARTIAL DIFFERENTIAL EQUATIONS. Lecturer: D.M.A. Stuart MT 2007

PARTIAL DIFFERENTIAL EQUATIONS. Lecturer: D.M.A. Stuart MT 2007 PARTIAL DIFFERENTIAL EQUATIONS Lecturer: D.M.A. Stuart MT 2007 In addition to the sets of lecture notes written by previous lecturers ([1, 2]) the books [4, 7] are very good for the PDE topics in the course.

More information

4 Uniform convergence

4 Uniform convergence 4 Uniform convergence In the last few sections we have seen several functions which have been defined via series or integrals. We now want to develop tools that will allow us to show that these functions

More information

THE GENERALIZED ARTIN CONJECTURE AND ARITHMETIC ORBIFOLDS

THE GENERALIZED ARTIN CONJECTURE AND ARITHMETIC ORBIFOLDS THE GENERALIZED ARTIN CONJECTURE AND ARITHMETIC ORBIFOLDS M. RAM MURTY AND KATHLEEN L. PETERSEN Abstract. Let K be a number field with positive unit rank, and let O K denote the ring of integers of K.

More information

Representation theory and quantum mechanics tutorial Spin and the hydrogen atom

Representation theory and quantum mechanics tutorial Spin and the hydrogen atom Representation theory and quantum mechanics tutorial Spin and the hydrogen atom Justin Campbell August 3, 2017 1 Representations of SU 2 and SO 3 (R) 1.1 The following observation is long overdue. Proposition

More information

Math Linear Algebra II. 1. Inner Products and Norms

Math Linear Algebra II. 1. Inner Products and Norms Math 342 - Linear Algebra II Notes 1. Inner Products and Norms One knows from a basic introduction to vectors in R n Math 254 at OSU) that the length of a vector x = x 1 x 2... x n ) T R n, denoted x,

More information

On distribution functions of ξ(3/2) n mod 1

On distribution functions of ξ(3/2) n mod 1 ACTA ARITHMETICA LXXXI. (997) On distribution functions of ξ(3/2) n mod by Oto Strauch (Bratislava). Preliminary remarks. The question about distribution of (3/2) n mod is most difficult. We present a

More information

An introduction to some aspects of functional analysis

An introduction to some aspects of functional analysis An introduction to some aspects of functional analysis Stephen Semmes Rice University Abstract These informal notes deal with some very basic objects in functional analysis, including norms and seminorms

More information

CALCULUS JIA-MING (FRANK) LIOU

CALCULUS JIA-MING (FRANK) LIOU CALCULUS JIA-MING (FRANK) LIOU Abstract. Contents. Power Series.. Polynomials and Formal Power Series.2. Radius of Convergence 2.3. Derivative and Antiderivative of Power Series 4.4. Power Series Expansion

More information

On the Error Term for the Mean Value Associated With Dedekind Zeta-function of a Non-normal Cubic Field

On the Error Term for the Mean Value Associated With Dedekind Zeta-function of a Non-normal Cubic Field Λ44ΩΛ6fi Ψ ο Vol.44, No.6 05ffμ ADVANCES IN MAHEMAICS(CHINA) Nov., 05 doi: 0.845/sxjz.04038b On the Error erm for the Mean Value Associated With Dedekind Zeta-function of a Non-normal Cubic Field SHI Sanying

More information

Bernstein s inequality and Nikolsky s inequality for R d

Bernstein s inequality and Nikolsky s inequality for R d Bernstein s inequality and Nikolsky s inequality for d Jordan Bell jordan.bell@gmail.com Department of athematics University of Toronto February 6 25 Complex Borel measures and the Fourier transform Let

More information

TOOLS FROM HARMONIC ANALYSIS

TOOLS FROM HARMONIC ANALYSIS TOOLS FROM HARMONIC ANALYSIS BRADLY STADIE Abstract. The Fourier transform can be thought of as a map that decomposes a function into oscillatory functions. In this paper, we will apply this decomposition

More information

Remarks on the Pólya Vinogradov inequality

Remarks on the Pólya Vinogradov inequality Remarks on the Pólya Vinogradov ineuality Carl Pomerance Dedicated to Mel Nathanson on his 65th birthday Abstract: We establish a numerically explicit version of the Pólya Vinogradov ineuality for the

More information

. A NOTE ON THE RESTRICTION THEOREM AND GEOMETRY OF HYPERSURFACES

. A NOTE ON THE RESTRICTION THEOREM AND GEOMETRY OF HYPERSURFACES . A NOTE ON THE RESTRICTION THEOREM AND GEOMETRY OF HYPERSURFACES FABIO NICOLA Abstract. A necessary condition is established for the optimal (L p, L 2 ) restriction theorem to hold on a hypersurface S,

More information

Primes in arithmetic progressions

Primes in arithmetic progressions (September 26, 205) Primes in arithmetic progressions Paul Garrett garrett@math.umn.edu http://www.math.umn.edu/ garrett/ [This document is http://www.math.umn.edu/ garrett/m/mfms/notes 205-6/06 Dirichlet.pdf].

More information

ON THE GAPS BETWEEN ZEROS OF TRIGONOMETRIC POLYNOMIALS

ON THE GAPS BETWEEN ZEROS OF TRIGONOMETRIC POLYNOMIALS Real Analysis Exchange Vol. 8(), 00/003, pp. 447 454 Gady Kozma, Faculty of Mathematics and Computer Science, Weizmann Institute of Science, POB 6, Rehovot 76100, Israel. e-mail: gadyk@wisdom.weizmann.ac.il,

More information

Problem 1A. Suppose that f is a continuous real function on [0, 1]. Prove that

Problem 1A. Suppose that f is a continuous real function on [0, 1]. Prove that Problem 1A. Suppose that f is a continuous real function on [, 1]. Prove that lim α α + x α 1 f(x)dx = f(). Solution: This is obvious for f a constant, so by subtracting f() from both sides we can assume

More information

Real Analysis Notes. Thomas Goller

Real Analysis Notes. Thomas Goller Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions........................... 2 1.2 Measurable Functions........................ 2 1.3 Integration..............................

More information

This ODE arises in many physical systems that we shall investigate. + ( + 1)u = 0. (λ + s)x λ + s + ( + 1) a λ. (s + 1)(s + 2) a 0

This ODE arises in many physical systems that we shall investigate. + ( + 1)u = 0. (λ + s)x λ + s + ( + 1) a λ. (s + 1)(s + 2) a 0 Legendre equation This ODE arises in many physical systems that we shall investigate We choose We then have Substitution gives ( x 2 ) d 2 u du 2x 2 dx dx + ( + )u u x s a λ x λ a du dx λ a λ (λ + s)x

More information

arxiv: v1 [math.co] 3 Nov 2014

arxiv: v1 [math.co] 3 Nov 2014 SPARSE MATRICES DESCRIBING ITERATIONS OF INTEGER-VALUED FUNCTIONS BERND C. KELLNER arxiv:1411.0590v1 [math.co] 3 Nov 014 Abstract. We consider iterations of integer-valued functions φ, which have no fixed

More information

Write on one side of the paper only and begin each answer on a separate sheet. Write legibly; otherwise, you place yourself at a grave disadvantage.

Write on one side of the paper only and begin each answer on a separate sheet. Write legibly; otherwise, you place yourself at a grave disadvantage. MATHEMATICAL TRIPOS Part IB Wednesday 5 June 2002 1.30 to 4.30 PAPER 1 Before you begin read these instructions carefully. Each question in Section II carries twice the credit of each question in Section

More information

HARMONIC ANALYSIS. Date:

HARMONIC ANALYSIS. Date: HARMONIC ANALYSIS Contents. Introduction 2. Hardy-Littlewood maximal function 3. Approximation by convolution 4. Muckenhaupt weights 4.. Calderón-Zygmund decomposition 5. Fourier transform 6. BMO (bounded

More information

ADJOINTS, ABSOLUTE VALUES AND POLAR DECOMPOSITIONS

ADJOINTS, ABSOLUTE VALUES AND POLAR DECOMPOSITIONS J. OPERATOR THEORY 44(2000), 243 254 c Copyright by Theta, 2000 ADJOINTS, ABSOLUTE VALUES AND POLAR DECOMPOSITIONS DOUGLAS BRIDGES, FRED RICHMAN and PETER SCHUSTER Communicated by William B. Arveson Abstract.

More information

Part 3.3 Differentiation Taylor Polynomials

Part 3.3 Differentiation Taylor Polynomials Part 3.3 Differentiation 3..3.1 Taylor Polynomials Definition 3.3.1 Taylor 1715 and Maclaurin 1742) If a is a fixed number, and f is a function whose first n derivatives exist at a then the Taylor polynomial

More information

Lecture 5 - Hausdorff and Gromov-Hausdorff Distance

Lecture 5 - Hausdorff and Gromov-Hausdorff Distance Lecture 5 - Hausdorff and Gromov-Hausdorff Distance August 1, 2011 1 Definition and Basic Properties Given a metric space X, the set of closed sets of X supports a metric, the Hausdorff metric. If A is

More information

Decoupling Lecture 1

Decoupling Lecture 1 18.118 Decoupling Lecture 1 Instructor: Larry Guth Trans. : Sarah Tammen September 7, 2017 Decoupling theory is a branch of Fourier analysis that is recent in origin and that has many applications to problems

More information

MATHS 730 FC Lecture Notes March 5, Introduction

MATHS 730 FC Lecture Notes March 5, Introduction 1 INTRODUCTION MATHS 730 FC Lecture Notes March 5, 2014 1 Introduction Definition. If A, B are sets and there exists a bijection A B, they have the same cardinality, which we write as A, #A. If there exists

More information

Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is. e ikx f(x) dx. (1.

Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is. e ikx f(x) dx. (1. Chapter 1 Fourier transforms 1.1 Introduction Let R be the line parameterized by x. Let f be a complex function on R that is integrable. The Fourier transform ˆf = F f is ˆf(k) = e ikx f(x) dx. (1.1) It

More information

Kernel Method: Data Analysis with Positive Definite Kernels

Kernel Method: Data Analysis with Positive Definite Kernels Kernel Method: Data Analysis with Positive Definite Kernels 2. Positive Definite Kernel and Reproducing Kernel Hilbert Space Kenji Fukumizu The Institute of Statistical Mathematics. Graduate University

More information

X n D X lim n F n (x) = F (x) for all x C F. lim n F n(u) = F (u) for all u C F. (2)

X n D X lim n F n (x) = F (x) for all x C F. lim n F n(u) = F (u) for all u C F. (2) 14:17 11/16/2 TOPIC. Convergence in distribution and related notions. This section studies the notion of the so-called convergence in distribution of real random variables. This is the kind of convergence

More information

MASTERS EXAMINATION IN MATHEMATICS

MASTERS EXAMINATION IN MATHEMATICS MASTERS EXAMINATION IN MATHEMATICS PURE MATHEMATICS OPTION FALL 2007 Full points can be obtained for correct answers to 8 questions. Each numbered question (which may have several parts) is worth the same

More information

DECOUPLING INEQUALITIES IN HARMONIC ANALYSIS AND APPLICATIONS

DECOUPLING INEQUALITIES IN HARMONIC ANALYSIS AND APPLICATIONS DECOUPLING INEQUALITIES IN HARMONIC ANALYSIS AND APPLICATIONS 1 NOTATION AND STATEMENT S compact smooth hypersurface in R n with positive definite second fundamental form ( Examples: sphere S d 1, truncated

More information

YOUNG TABLEAUX AND THE REPRESENTATIONS OF THE SYMMETRIC GROUP

YOUNG TABLEAUX AND THE REPRESENTATIONS OF THE SYMMETRIC GROUP YOUNG TABLEAUX AND THE REPRESENTATIONS OF THE SYMMETRIC GROUP YUFEI ZHAO ABSTRACT We explore an intimate connection between Young tableaux and representations of the symmetric group We describe the construction

More information

1, for s = σ + it where σ, t R and σ > 1

1, for s = σ + it where σ, t R and σ > 1 DIRICHLET L-FUNCTIONS AND DEDEKIND ζ-functions FRIMPONG A. BAIDOO Abstract. We begin by introducing Dirichlet L-functions which we use to prove Dirichlet s theorem on arithmetic progressions. From there,

More information

A LOWER BOUND ON BLOWUP RATES FOR THE 3D INCOMPRESSIBLE EULER EQUATION AND A SINGLE EXPONENTIAL BEALE-KATO-MAJDA ESTIMATE. 1.

A LOWER BOUND ON BLOWUP RATES FOR THE 3D INCOMPRESSIBLE EULER EQUATION AND A SINGLE EXPONENTIAL BEALE-KATO-MAJDA ESTIMATE. 1. A LOWER BOUND ON BLOWUP RATES FOR THE 3D INCOMPRESSIBLE EULER EQUATION AND A SINGLE EXPONENTIAL BEALE-KATO-MAJDA ESTIMATE THOMAS CHEN AND NATAŠA PAVLOVIĆ Abstract. We prove a Beale-Kato-Majda criterion

More information

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

More information

SCALE INVARIANT FOURIER RESTRICTION TO A HYPERBOLIC SURFACE

SCALE INVARIANT FOURIER RESTRICTION TO A HYPERBOLIC SURFACE SCALE INVARIANT FOURIER RESTRICTION TO A HYPERBOLIC SURFACE BETSY STOVALL Abstract. This result sharpens the bilinear to linear deduction of Lee and Vargas for extension estimates on the hyperbolic paraboloid

More information

u( x) = g( y) ds y ( 1 ) U solves u = 0 in U; u = 0 on U. ( 3)

u( x) = g( y) ds y ( 1 ) U solves u = 0 in U; u = 0 on U. ( 3) M ath 5 2 7 Fall 2 0 0 9 L ecture 4 ( S ep. 6, 2 0 0 9 ) Properties and Estimates of Laplace s and Poisson s Equations In our last lecture we derived the formulas for the solutions of Poisson s equation

More information

Remarks on Bronštein s root theorem

Remarks on Bronštein s root theorem Remarks on Bronštein s root theorem Guy Métivier January 23, 2017 1 Introduction In [Br1], M.D.Bronštein proved that the roots of hyperbolic polynomials (1.1) p(t, τ) = τ m + m p k (t)τ m k. which depend

More information

C.7. Numerical series. Pag. 147 Proof of the converging criteria for series. Theorem 5.29 (Comparison test) Let a k and b k be positive-term series

C.7. Numerical series. Pag. 147 Proof of the converging criteria for series. Theorem 5.29 (Comparison test) Let a k and b k be positive-term series C.7 Numerical series Pag. 147 Proof of the converging criteria for series Theorem 5.29 (Comparison test) Let and be positive-term series such that 0, for any k 0. i) If the series converges, then also

More information

An introduction to Birkhoff normal form

An introduction to Birkhoff normal form An introduction to Birkhoff normal form Dario Bambusi Dipartimento di Matematica, Universitá di Milano via Saldini 50, 0133 Milano (Italy) 19.11.14 1 Introduction The aim of this note is to present an

More information

GATE Engineering Mathematics SAMPLE STUDY MATERIAL. Postal Correspondence Course GATE. Engineering. Mathematics GATE ENGINEERING MATHEMATICS

GATE Engineering Mathematics SAMPLE STUDY MATERIAL. Postal Correspondence Course GATE. Engineering. Mathematics GATE ENGINEERING MATHEMATICS SAMPLE STUDY MATERIAL Postal Correspondence Course GATE Engineering Mathematics GATE ENGINEERING MATHEMATICS ENGINEERING MATHEMATICS GATE Syllabus CIVIL ENGINEERING CE CHEMICAL ENGINEERING CH MECHANICAL

More information

ELEMENTARY LINEAR ALGEBRA

ELEMENTARY LINEAR ALGEBRA ELEMENTARY LINEAR ALGEBRA K. R. MATTHEWS DEPARTMENT OF MATHEMATICS UNIVERSITY OF QUEENSLAND Corrected Version, 7th April 013 Comments to the author at keithmatt@gmail.com Chapter 1 LINEAR EQUATIONS 1.1

More information

RANKIN-COHEN BRACKETS AND VAN DER POL-TYPE IDENTITIES FOR THE RAMANUJAN S TAU FUNCTION

RANKIN-COHEN BRACKETS AND VAN DER POL-TYPE IDENTITIES FOR THE RAMANUJAN S TAU FUNCTION RANKIN-COHEN BRACKETS AND VAN DER POL-TYPE IDENTITIES FOR THE RAMANUJAN S TAU FUNCTION B. RAMAKRISHNAN AND BRUNDABAN SAHU Abstract. We use Rankin-Cohen brackets for modular forms and quasimodular forms

More information

means is a subset of. So we say A B for sets A and B if x A we have x B holds. BY CONTRAST, a S means that a is a member of S.

means is a subset of. So we say A B for sets A and B if x A we have x B holds. BY CONTRAST, a S means that a is a member of S. 1 Notation For those unfamiliar, we have := means equal by definition, N := {0, 1,... } or {1, 2,... } depending on context. (i.e. N is the set or collection of counting numbers.) In addition, means for

More information

Measure Theory and Lebesgue Integration. Joshua H. Lifton

Measure Theory and Lebesgue Integration. Joshua H. Lifton Measure Theory and Lebesgue Integration Joshua H. Lifton Originally published 31 March 1999 Revised 5 September 2004 bstract This paper originally came out of my 1999 Swarthmore College Mathematics Senior

More information

2.3. VECTOR SPACES 25

2.3. VECTOR SPACES 25 2.3. VECTOR SPACES 25 2.3 Vector Spaces MATH 294 FALL 982 PRELIM # 3a 2.3. Let C[, ] denote the space of continuous functions defined on the interval [,] (i.e. f(x) is a member of C[, ] if f(x) is continuous

More information

g(x) = P (y) Proof. This is true for n = 0. Assume by the inductive hypothesis that g (n) (0) = 0 for some n. Compute g (n) (h) g (n) (0)

g(x) = P (y) Proof. This is true for n = 0. Assume by the inductive hypothesis that g (n) (0) = 0 for some n. Compute g (n) (h) g (n) (0) Mollifiers and Smooth Functions We say a function f from C is C (or simply smooth) if all its derivatives to every order exist at every point of. For f : C, we say f is C if all partial derivatives to

More information

Lecture 9 Metric spaces. The contraction fixed point theorem. The implicit function theorem. The existence of solutions to differenti. equations.

Lecture 9 Metric spaces. The contraction fixed point theorem. The implicit function theorem. The existence of solutions to differenti. equations. Lecture 9 Metric spaces. The contraction fixed point theorem. The implicit function theorem. The existence of solutions to differential equations. 1 Metric spaces 2 Completeness and completion. 3 The contraction

More information

Continued Fraction Digit Averages and Maclaurin s Inequalities

Continued Fraction Digit Averages and Maclaurin s Inequalities Continued Fraction Digit Averages and Maclaurin s Inequalities Steven J. Miller, Williams College sjm1@williams.edu, Steven.Miller.MC.96@aya.yale.edu Joint with Francesco Cellarosi, Doug Hensley and Jake

More information

NOTE ON A REMARKABLE SUPERPOSITION FOR A NONLINEAR EQUATION. 1. Introduction. ρ(y)dy, ρ 0, x y

NOTE ON A REMARKABLE SUPERPOSITION FOR A NONLINEAR EQUATION. 1. Introduction. ρ(y)dy, ρ 0, x y NOTE ON A REMARKABLE SUPERPOSITION FOR A NONLINEAR EQUATION PETER LINDQVIST AND JUAN J. MANFREDI Abstract. We give a simple proof of - and extend - a superposition principle for the equation div( u p 2

More information

A PROOF OF ROTH S THEOREM ON ARITHMETIC PROGRESSIONS

A PROOF OF ROTH S THEOREM ON ARITHMETIC PROGRESSIONS A PROOF OF ROTH S THEOREM ON ARITHMETIC PROGRESSIONS ERNIE CROOT AND OLOF SISASK Abstract. We present a proof of Roth s theorem that follows a slightly different structure to the usual proofs, in that

More information

Normed & Inner Product Vector Spaces

Normed & Inner Product Vector Spaces Normed & Inner Product Vector Spaces ECE 174 Introduction to Linear & Nonlinear Optimization Ken Kreutz-Delgado ECE Department, UC San Diego Ken Kreutz-Delgado (UC San Diego) ECE 174 Fall 2016 1 / 27 Normed

More information

Doubly Indexed Infinite Series

Doubly Indexed Infinite Series The Islamic University of Gaza Deanery of Higher studies Faculty of Science Department of Mathematics Doubly Indexed Infinite Series Presented By Ahed Khaleel Abu ALees Supervisor Professor Eissa D. Habil

More information

Continued fractions for complex numbers and values of binary quadratic forms

Continued fractions for complex numbers and values of binary quadratic forms arxiv:110.3754v1 [math.nt] 18 Feb 011 Continued fractions for complex numbers and values of binary quadratic forms S.G. Dani and Arnaldo Nogueira February 1, 011 Abstract We describe various properties

More information

Mathematical Research Letters 10, (2003) THE FUGLEDE SPECTRAL CONJECTURE HOLDS FOR CONVEX PLANAR DOMAINS

Mathematical Research Letters 10, (2003) THE FUGLEDE SPECTRAL CONJECTURE HOLDS FOR CONVEX PLANAR DOMAINS Mathematical Research Letters 0, 559 569 (2003) THE FUGLEDE SPECTRAL CONJECTURE HOLDS FOR CONVEX PLANAR DOMAINS Alex Iosevich, Nets Katz, and Terence Tao Abstract. Let Ω be a compact convex domain in the

More information

Heisenberg s inequality

Heisenberg s inequality Heisenberg s inequality Michael Walter 1 Introduction 1 Definition. Let P be a probability measure on R. Then its mean or expectation is defined as E(P) := x dp(x) Its variance is the expected square deviation

More information

A class of non-convex polytopes that admit no orthonormal basis of exponentials

A class of non-convex polytopes that admit no orthonormal basis of exponentials A class of non-convex polytopes that admit no orthonormal basis of exponentials Mihail N. Kolountzakis and Michael Papadimitrakis 1 Abstract A conjecture of Fuglede states that a bounded measurable set

More information

Solving a linear equation in a set of integers II

Solving a linear equation in a set of integers II ACTA ARITHMETICA LXXII.4 (1995) Solving a linear equation in a set of integers II by Imre Z. Ruzsa (Budapest) 1. Introduction. We continue the study of linear equations started in Part I of this paper.

More information

Joshua N. Cooper Department of Mathematics, Courant Institute of Mathematics, New York University, New York, NY 10012, USA.

Joshua N. Cooper Department of Mathematics, Courant Institute of Mathematics, New York University, New York, NY 10012, USA. CONTINUED FRACTIONS WITH PARTIAL QUOTIENTS BOUNDED IN AVERAGE Joshua N. Cooper Department of Mathematics, Courant Institute of Mathematics, New York University, New York, NY 10012, USA cooper@cims.nyu.edu

More information

A SHORT PROOF OF THE COIFMAN-MEYER MULTILINEAR THEOREM

A SHORT PROOF OF THE COIFMAN-MEYER MULTILINEAR THEOREM A SHORT PROOF OF THE COIFMAN-MEYER MULTILINEAR THEOREM CAMIL MUSCALU, JILL PIPHER, TERENCE TAO, AND CHRISTOPH THIELE Abstract. We give a short proof of the well known Coifman-Meyer theorem on multilinear

More information

5. Atoms and the periodic table of chemical elements. Definition of the geometrical structure of a molecule

5. Atoms and the periodic table of chemical elements. Definition of the geometrical structure of a molecule Historical introduction The Schrödinger equation for one-particle problems Mathematical tools for quantum chemistry 4 The postulates of quantum mechanics 5 Atoms and the periodic table of chemical elements

More information

INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES

INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES INTRODUCTION TO REAL ANALYSIS II MATH 4332 BLECHER NOTES You will be expected to reread and digest these typed notes after class, line by line, trying to follow why the line is true, for example how it

More information

Before we prove this result, we first recall the construction ( of) Suppose that λ is an integer, and that k := λ+ 1 αβ

Before we prove this result, we first recall the construction ( of) Suppose that λ is an integer, and that k := λ+ 1 αβ 600 K. Bringmann, K. Ono Before we prove this result, we first recall the construction ( of) these forms. Suppose that λ is an integer, and that k := λ+ 1 αβ. For each A = Ɣ γ δ 0 (4),let j(a, z) := (

More information