Dirichlet hyperbola method

A pdf version is Dirichlet hyperbola method.

1. Introduction

Theorem 1

\displaystyle \sum_{1\leq n\leq x}d(n)=\sum_{1\leq n\leq x}[\frac{x}{n}]=xlogx+(2\gamma-1) x+O(\sqrt{x}) \ \ \ \ \ (1)


Remark 1 I thought this problem initial 5 years ago, cost me several days to find a answer, I definitely get something without the argument of Dirchlet hyperbola method and which is weaker but morally the same camparable with the result get by Dirichlet hyperbola method.

Remark 2 How to get the formula:

\displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq n\leq x}[\frac{x}{n}]? \ \ \ \ \ (2)

In fact,

\displaystyle \sum_{1\leq n\leq x}d(x)=\sum_{1\leq ab\leq x}1=\sum_{1\leq n\leq x}[\frac{x}{n}]\ \ \ \ \ (3)

Which is the integer lattices under or lying on the hyperbola {\{(a,b)|ab=x\}}.

Remark 3 By trivial argument, we can bound the quantity as following way,

\displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}[\frac{x}{n}] & = & \sum_{1\leq ab\leq x}1\\ & = & x\sum_{i=1}^x\frac{1}{i}-\sum_{i=1}^x\{\frac{x}{i}\}\\ & = &xlnx+\gamma x+O(x) \end{array}

The error term is {O(x)}, which is too big. But fortunately we can use the symmetry of hyperbola to improve the error term.


\displaystyle \begin{array}{rcl} \sum_{1\leq n\leq x}d(n) & = & \sum_{ab\leq x}1\\ & = & \sum_{a\geq \sqrt{x}}[\frac{x}{b}]+\sum_{b\geq \sqrt{x}}[\frac{x}{a}]-\sum_{1\leq a,b\leq \sqrt{x}}1\\ & = & xlogx+(2\gamma-1)x+O(\sqrt{x}) \end{array}


Theorem 2

Given a natural number k, use the hyperbola method together
with induction and partial summation to show that

\displaystyle \sum_{n\leq x}d_k(n) = xP_k(log x) + O(x^{1-\frac{1}{k}+\epsilon}), n\leq x \ \ \ \ \ (4)

where {P_k(t)} denotes a polynomial of degree {k-1} with leading term {\frac{t^{k-1}}{(k-1)!}}.

Remark 4 {P_k(x)} is the residue of {\zeta(s)^kx^ss^{-1}} at {s=1}.


We can establish the dimension 3 case directly, which is the following asymptotic formula,

\displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=xP_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (5)

The approach is following, we first observe that

\displaystyle \sum_{1\leq xy\leq n}[\frac{n}{xy}]=\sum_{xyz\leq n}1 \ \ \ \ \ (6)

The problem transform to get a asymptotic formula for the lattices under 3 dimension hyperbola. The first key point is, morally {([n^{\frac{1}{3}}],[n^{\frac{1}{3}}],[n^{\frac{1}{3}}])} is the central point under the hyperbola.

Then we can divide the range into 3 parts, and try to get a asymptotic formula for each part then add them together. Assume we have:

  1. {A_x=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.
  2. {A_y=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.
  3. {A_z=\sum_{1\leq r\leq [n^{\frac{1}{3}}]}\sum_{1\leq xy\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]}.

Then the task transform to get a asymptotic formula,

\displaystyle A_x=A_y=A_z=xQ_2(logx)+O(x^{1-\frac{1}{3}+\epsilon}) \ \ \ \ \ (7)

But we can do the same thing for {\sum_{1\leq yz\leq [n^{\frac{2}{3}}]}[\frac{r}{yz}]} and then integral it. This end the proof. For general {k\in {\mathbb N}}, the story is the same, by induction.

Induction on {k} and use the Fubini theorem to calculate {\sum_{x_1...x_r\leq n}\frac{n}{x_1...x_r},\forall 1\leq r\leq k}. \Box

There is a major unsolved problem called Dirichlet divisor problem.

\displaystyle \sum_{n\leq x}d(n) \ \ \ \ \ (8)

What is the error term? The conjecture is the error term is {O(x^{\theta}), \forall \theta>\frac{1}{4}}, it is known that {\theta=\frac{1}{4}} is not right.

Remark 5

To beats this problem, need some tools in algebraic geometry.

2. Several problems

{\forall k\in {\mathbb N}}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{kn}{t}\}} ?

{\forall k\in {\mathbb N}}, {f(n)} is a polynomial with degree {k}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{f(n)}{t}\}} ?

{\forall k\in {\mathbb N}}, {g(n)} is a polynomial with degree {k}, is there a asymptotic formula for {\sum_{t=1}^n\{\frac{n}{g(t)}\}} ?

Theorem 3 {k\in {\mathbb N}}, then we have

\displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n}=k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma) \ \ \ \ \ (9)


\displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+\{\frac{kn}{2}\}+...+\{\frac{kn}{n}\}}{n} & = & \frac{\sum_{i=1}^k\frac{kn}{i}-\sum_{i=1}^n[\frac{kn}{i}]}{n}\\ & = & k(lnn+\gamma +\epsilon_n)-\frac{\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}]}{n} \end{array}


Now we try to estimate

\displaystyle S_k(n)=\sum_{i=1}^{kn}[\frac{kn}{i}]-\sum_{i=n+1}^{kn}[\frac{kn}{i}] \ \ \ \ \ (10)

In fact, we have,

\displaystyle \begin{array}{rcl} S_k(n) & = & (2\sum_{i=1}^{[\sqrt{kn}]}[\frac{kn}{i}]-[\sqrt{kn}]^2)-(\sum_{i=1}^k[\frac{kn}{i}]-kn)\\ & = & 2\sum_{i=1}^{[\sqrt{kn}]}\frac{kn}{i}-\sum_{i=1}^k\frac{kn}{i}+2\{\sqrt{kn}\}[\sqrt{kn}]+\{\sqrt{kn}\}^2-2\sum_{i=1}^{[\sqrt{kn}]}\{\frac{kn}{i}\}+\sum_{i=1}^k\{\frac{kn}{i}\}\\ & = & 2kn(ln[\sqrt{kn}]+\gamma+\epsilon_{[\sqrt{kn}]})-kn\sum_{i=1}^k\frac{1}{i}+r(n)\\ & = & knln(kn)+kn(2\gamma-\sum_{i=1}^k\frac{1}{i})+r'(n)\\ & = & knln n+kn(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+r'(n) \end{array}

Where {-3\sqrt{n}<r(n)<3\sqrt{n}}, {-3\sqrt{n}<r'(n)<3\sqrt{n}}.

So by 1 we know,

\displaystyle \begin{array}{rcl} \frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} & = & k(lnn+\gamma+\epsilon_n)-klnn-k(2\gamma+lnk-\sum_{i=1}^k\frac{1}{i})+\frac{r'(n)}{n}\\ & = & k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)+\frac{r'(n)}{n}+k\epsilon_n \end{array}

So we have,

\displaystyle \lim_{n\rightarrow \infty}\frac{\{\frac{kn}{1}\}+...+\{\frac{kn}{n}\}}{n} =k(\sum_{i=1}^k\frac{1}{i}-lnk-\gamma)=k\epsilon_k \ \ \ \ \ (11)

Remark 6 In fact we can get {0<k\epsilon_k<\frac{1}{2}, \forall k\in {\mathbb N}}, by combining the theorem 3 and 1.

3. Lattice points in ball

Gauss use the cube packing circle get a rough estimate,

\displaystyle \sum_{n\leq x}r_2(n)=\pi x+O(\sqrt{x}) \ \ \ \ \ (12)


In the same way one can obtain,

\displaystyle \sum_{n\leq x}r_k(n)=\rho_kx^{\frac{k}{2}}+O(x^{\frac{k-1}{2}}) \ \ \ \ \ (13)

Remark 7 Where {\rho_k=\frac{\pi^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}} is the volume of the unit ball in {k} dimension.

Dirchlet’s hyperbola method works nicely for the lattic points in a ball of dimension {k\geq 4}. Langrange proved that every natural number can be represented as the sum of four squares, i.e. {r_4(n)>0}, and Jacobi established the exact formula for the number of representations

\displaystyle r_4(n)=8(2+(-1)^n)\sum_{d|n,d\ odd}d. \ \ \ \ \ (14)

Hence we derive,

\displaystyle \begin{array}{rcl} \sum_{n\leq x}r_4(n) & = & 8\sum_{m\leq x}(2+(-1)^m)\sum_{dm\leq x, d\ odd}d\\ & = & 8\sum_{m\leq x}(2+(-1)^m)(\frac{x^2}{4m^2}+O(\frac{x}{m}))\\ & = & 2x^2\sum_1^{\infty}(2+(-1)^m)m^{-2}+O(xlogx)\\ & = & 3\zeta(2)x^2+O(xlogx) = \frac{1}{2}(\pi x)^2+O(xlogx) \end{array}

This result extend easily for any {k\geq 4}, write {r_k} as the additive convolution of {r_4} and {r_{k-4}}, i.e.

\displaystyle r_k(n)=\sum_{0\leq t\leq n}r_4(t)r_{k-4}(n-t) \ \ \ \ \ (15)

Apply the above result for {r_4} and execute the summation over the remaining {k-4} squares by integration.

\displaystyle \sum_{n\leq x}r_k(n)=\frac{(\pi x)^{\frac{k}{2}}}{\Gamma(\frac{k}{2}+1)}+O(x^{\frac{k}{2}-1}logx) \ \ \ \ \ (16)


Remark 8 Notice that this improve the formula 12 which was obtained by the method of packing with a unit square. The exponent {\frac{k}{2}-1} in 16 is the best possible because the individual terms of summation can be as large as the error term (apart from {logx}), indeed for {k=4} we have {r_4(n)\geq 16n} if {n} is odd by the Jacobi formula. The only case of the lattice point problem for a ball which is not yet solved (i.e. the best possible error terms are not yet established) are for the circle({k=2}) and the sphere ({k=3}).

Theorem 4

\displaystyle \sum_{n\leq x}\tau(n^2+1)=\frac{3}{\pi}xlogx+O(x) \ \ \ \ \ (17)

4. Application in finite fields

Suppose {f(x)\in {\mathbb Z}[x]} is a irreducible polynomial. And for each prime {p}, let

\displaystyle \rho_f(p)=\# \ of \ solutions\ of f(x)\equiv 0(mod\ p) \ \ \ \ \ (18)

By Langrange theorem we know {\rho_f(p)\leq deg(f)}. Is there a asymptotic formula for

\displaystyle \sum_{p\leq x}\rho_f(p)? \ \ \ \ \ (19)

A general version, we can naturally generated it to algebraic variety.

\displaystyle \rho_{f_1,...,f_k}(p)=\#\ of \ solutions\ of f_i(x)\equiv 0(mod\ p) ,\ \forall 1\leq i\leq k \ \ \ \ \ (20)

Is there a asymptotic formula for

\displaystyle \sum_{p\leq x}\rho_{f_1,...,f_k}(p)? \ \ \ \ \ (21)

Example 1 We give an example to observe what is involved. {f(x)=x^2+1}. We know {x^2+1\equiv 0 (mod \ p)} is solvable iff {p\equiv 1 (mod\ 4)} or {p=2}. One side is easy, just by Fermat little theorem, the other hand need Fermat descent procedure, which of course could be done by Willson theorem. In this case,

\displaystyle \sum_{p\leq n}\rho_f(p)=\# \ of \{primes \ of \ type\ 4k+1 \ in \ 1,2,...,n\} \ \ \ \ \ (22)

Which is a special case of Dirichlet prime theorem.

Let {K} be an algebraic number field, i.e. the finite field extension of rational numbers, let

\displaystyle \mathcal{O}_K=\{\alpha\in K, \alpha \ satisfied \ a\ monic \ polynomial\ in\ {\mathbb Z}[x]\} \ \ \ \ \ (23)


Dedekind proved that,

Theorem 5

  1. {\mathcal{O}_K} is a ring, we call it the ring of integer of {K}.
  2. He showed further every non-zero ideal of {\mathcal{O}_K} could write as the product of prime ideal in {\mathcal{O}_k} uniquely.
  3. the index of every non-zero ideal {I} in {\mathcal{O}_K} is finite, i.e. {[\mathcal{O}_K:I]<\infty}, and we can define the norm induce by index.

    \displaystyle N(I):=[\mathcal{O}_K:I] \ \ \ \ \ (24)

    Then the norm is a multiplication function in the space of ideal, i.e. {N(IJ)=N(I)N(J), \forall I,J \in \ ideal\ class\ group\ of\ \mathcal{O}_K}.

  4. Now he construct the Dedekind Riemann zeta function,

    \displaystyle \zeta_K(s)=\sum_{N(I)\neq 0}\frac{1}{N(I)^s}=\prod_{J\ prime \ ideal\ }\frac{1}{1-\frac{1}{N(J)^s}},\ \forall Re(s)>1 \ \ \ \ \ (25)


Now we consider the analog of the prime number theorem. Let {\pi_K(x)=\{I,N(I)<x\}}, does the exist a asymptotic formula,

\displaystyle \pi_K(x)\sim \frac{x}{ln x}\ as\ x\rightarrow \infty? \ \ \ \ \ (26)

Given a prime {p}, we may consider the prime ideal

\displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (27)

Where {\mathfrak{P}_i } is different prime ideal in {\mathcal{O}_K}. But the question is how to find these {\mathfrak{P}_i}? For the question, there is a satisfied answer.

Lemma 6 (existence of primitive element) There always exist a primetive elements in {K}, such that,

\displaystyle K={\mathbb Q}(\theta) \ \ \ \ \ (28)

Where {\theta} is some algebraic number, which’s minor polynomial {f(x)\in {\mathbb Z}[x]}.

Theorem 7 (Dedekind recipe) Take the polynomial {f(x)}, factorize it in the polynomial ring {{\mathbb Z}_p[x]},

\displaystyle f(x)\equiv f_1(x)^{e_1}...f_{r}(x)^{e_r}(mod \ p) \ \ \ \ \ (29)

Consider {\mathfrak{P}_i=(p, f_i(\theta)) \subset \mathcal{O}_K}. Then apart from finite many primes, we have,

\displaystyle p\mathcal{O}_K=\mathfrak{P}_1^{e_1}\mathfrak{P}_2^{e_2}...\mathfrak{P}_k^{e_k} \ \ \ \ \ (30)

Where {N(\mathfrak{P}_i)=p^{deg{f_i}}}.

Remark 9 The apart primes are those divide the discriminant.

Now we can argue that 4 is morally the same as counting the ideals whose norm is divide by {p} in a certain algebraic number theory.

And we have following, which is just the version in algebraic number fields of 2.

Theorem 8 (Weber) {\#} of ideals of {\mathcal{O}_K} with norm {\leq x} equal to,

\displaystyle \rho_k(X)+O(x^{1-\frac{1}{d}}), where \ d=[K:Q] \ \ \ \ \ (31)



The large sieve and the Bombieri-Vinogradov theorem


Large sieve a philosophy reflect as a large group of inequalities which is very effective on controlling some linear sum or square sum of some correlation of arithmetic function, some idea of which could have originated in harmonic analysis, merely rely on almost orthogonality.

One fundamental example is the estimate of the quality,

\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|

One naive idea of control this quality is using Cauchy-schwarz inequality. But stupid use this we gain something even worse than trivial estimate. In fact by triangle inequality and trivial estimate we gain trivial bound: \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq x. But by stupid use Cauchy we get following,

\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|\leq ((\sum_{n\leq x}|\Lambda(n)|^2)(\sum_{n\leq x}|\chi(n)|^2))^{\frac{1}{2}}\leq xlog^{\frac{1}{2}}x

But this does not mean Cauchy-Schwarz is useless on charge this quality, we careful look at the inequality and try to understand why the bound will be even worse. Every time we successful use Cauchy-Schwarz there are two main phenomenon, first, we lower down the complexity of the quantity we wish to bound, second we almost do not loss any thing at all. So we just reformulate the quantity and find it lower down the complexity and the change is compatible with the equivalent condition of Cauchy-Schwarz. For example we have following identity,

\sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}|=\sqrt{ \sum_{n\leq x}|\Lambda(n)\overline{\chi(n)}| \sum_{m\leq x}|\Lambda(m)\overline{\chi(m)}|}=\sqrt{ \sum_{k_1,k_2\in \mathbb F_p^{\times}}\sum_{n',m'\leq \frac{x}{p}}|\Lambda(n')\Lambda(m')\overline{\chi(k_1)\chi(k_2)}| }

So we could understand this quality as the Variation of primes in arithmetic profession constructed by \{pn+b| b\in\{1,2,...,p-1\}\}. But this is still difficult to estimate, merely because of we need to control the variation of convolution of \Lambda with itself on \mathbb F_p^{\times}\simeq \{pn+b| b\in\{1,2,...,p-1\}\}.

Now we change our perspective, recall a variant of Cauchy-Schwarz inequality, which called Bessel inequality, as following,

Bessel inequality

Let {g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}} be finitely supported functions obeying the orthonormality relationship,

\displaystyle \sum_n g_j(n) \overline{g_{j'}(n)} = 1_{j=j'}

for all {1 \leq j,j' \leq J}. Then for any function {f: {\bf N} \rightarrow {\bf C}}, we have,

\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2)^{1/2}.

Pf: The proof is not very difficult, we just need to keep an orthogonal picture in our mind, consider \{g_{j}(n)\}, 1\leq j\leq J to be a orthogonal basis on l^2(\mathbb N), then this inequality is a natural corollary.

Have this inequality in mind, by the standard argument given by transform from version of orthogonal to almost orthogonal which was merely explained in the previous note.  We could image the following corresponding almost orthogonal variate of “Bessel inequality” is true:

Generalised Bessel inequality

Let {g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}} be finitely supported functions, and let {\nu: {\bf N} \rightarrow {\bf R}^+} be a non-negative function. Let {f: {\bf N} \rightarrow {\bf C}} be such that {f} vanishes whenever {\nu} vanishes, we have

\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2} \times ( \sum_{j=1}^J \sum_{j'=1}^J c_j \overline{c_{j'}} \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)} )^{1/2}

for some sequence {c_1,\dots,c_J} of complex numbers with {\sum_{j=1}^J |c_j|^2 = 1}, with the convention that {|f(n)|^2/\nu(n)} vanishes whenever {f(n), \nu(n)} both vanish.


The correlation of Mobius function and nil-sequences in short interval

I wish to establish the following estimate:

Conjecture :(correlation of Mobius function and nil-sequences in short interval)

\lambda(n) is the liouville function we wish the following estimate is true.

\int_{0\leq x\leq X}|\sup_{f\in \Omega^m}\sum_{x\leq n\leq x+H}\lambda(n)e^{2\pi if(x)}|dx =o(XH).

Where we have H\to \infty as x\to \infty, \Omega^m=\{a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0 | a_m,...,a_1,a_0\in [0,1]\} is a compact space.

I do not know how to prove this but this is result is valuable to consider, because by a Fourier identity we could transform the difficulty of (log average) Chowla conjecture to this type of result.

There is some clue to show this type of result could be true, the first one is the result established by Matomaki and Raziwill in 2015:

Theorem (multiplication function in short interval)

f(n): \mathbb N\to \mathbb C is a multiplicative function, i.e. f(mn)=f(n)f(m), \forall m,n\in \mathbb N. H\to \infty as x\to infty, then we have the following result,

\int_{1\leq x\leq X}|\sum_{x\leq n\leq x+H}f(n)|=o(XH).

And there also exists the result which could be established by Vinagrodov estimate and B-S-Z critation :

Theorem(correlation of multiplication function and nil-sequences in long interval)

f(n): \mathbb N\to \mathbb C is a multiplicative function, i.e. f(mn)=f(n)f(m), \forall m,n\in \mathbb N. g(n)=a_n^m+...+a_1n+a_0 is a polynomial function then we have the following result,

\int_{1\leq n \leq X}|f(n)e^{2\pi i g(n)}|=o(X).

Diophantine approximation of algebraic number

An important theorem in Diophantine approximation is the theorem of Liuoville:

**Liuoville Theorem** If x is a algebraic number of degree n over the rational number then there exists a constant c(x) > 0 such that:\left|x-{\frac {p}{q}}\right|>{\frac {c(x)}{q^{{n}}}}

holds for every integer p,q\in N^* where q>0.

This theorem explain a phenomenon, the approximation of algebraic number by rational number could not be very well. Which was generated later to **Thue–Siegel–Roth theorem**, them could be used to proof a lots of constant is not algebraic, i.e. transcendentals .

My questions is in another direction, now let us not just consider one root  \alpha_1 of a integer polynomial P(x)=a_mx^m+...+a_1x+a_0 but consider all roots of it, i.e. \{\alpha_1,...,\alpha_m\}, which is based on a observation : If we define

\sigma_k(P(x))=\sum_{1\leq \alpha_{i_1}<\alpha_{i_2}<...<\alpha_{i_k}\leq m}\alpha_{i_1}\alpha_{i_2}...\alpha_{i_k}

By **Vieta theorem** we know \sigma_k(n)\in \mathbb Q for all k\in N^*, this will lead to some restriction and in fact destroy the uniformly distribution of (\alpha_1,...,\alpha_m)\in [0,1]^m. In fact the most important one is the determination of Vandermon Determinant:
V(P(x))=\Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j).

We know \Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\in \mathbb Q so when \Pi_{1\leq \alpha_i<\alpha_j\leq m}(\alpha_i-\alpha_j)\neq 0 we could use this to proof a nontrivial estimate for \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}.
\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).

by combine the A-G inequality and \Pi_{1\leq \alpha_i<\alpha_j\leq n}(\alpha_i-\alpha_j)=\lambda\neq 0.While by continue fractional expansion we only know a trivial estimate of type \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n}).

my question is the following:
Is there still have a nontrivial estimate for \sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z} (which could be slight weaker), if we don’t have the whole power of **Vieta theorem**? more precisely:

**problem 1**

if we have \sigma_k((\alpha_1,...,\alpha_m))=\lambda_k\in \mathbb Q for all k\in \{1,2,...,m'\} where m'<m, is there still some nontrivial estimate of,

\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}

hold for all n\in N^*?

One reason to consider this could be true is that although \{\alpha_1,...,\alpha_m\} is not roots of a integer polynomial but we could image in some suitable metric space X the gromov-hausdorff distance of tuple (\alpha_1,...,\alpha_m) and a tuple come form roots of integer polynomial is small . And it seems reasonable to image this type of asymptotic quality is continue with the G-H distance on X.

Another problem is what happen when V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0. More precisely,

**problem 2**

What happen when V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i<j\leq n}(\alpha_i-\alpha_j)=0 , is this result,

\sum_{1\leq k\leq m}||\alpha_kn||_{\mathbb R/\mathbb Z}= O(\frac{1}{n^{\frac{1}{m-1}}}).

still true?

Let us go a litter further, if these two problem both have a satisfied answer, what is the higher dimensional case?

**problem 3**

Given m\in \mathbb N^*. If tuple (y_1,...,y_k) is very closed to the zero set of a variety in \mathbb Z[x_1,...,x_m] in \mathbb (Z^{m})^k in the sense a lots of symmetric sum of y_1,...,y_k belong to \mathbb Q^m, will this lead to some good estimate for

\sum_{1\leq s\leq k}||y_sn||_{\mathbb R^m/\mathbb Z^m}?

I think these type of result should be investigated very well, Iappreciate to any pointer with useful comments and answer, both on given some strategy to solve these problems or given some reference about these problems.

Multiplication function on short interval

The most important beakgrouth of analytic number theory is the new understanding of multiplication function on share interval, this result is established by Kaisa Matomäki & Maksym Radziwill. Two very young and intelligent superstars.

The main theorem in them article is :

As soon as H\to \infty when x\to \infty, one has:

                    \sum_{x\leq n\leq x+H}\lambda(n)= o(H)

for almost all x\sim X .


In my understanding of the result, the main strategy is:

Step 1:Parseval indetity, monotonically inequality

Parseval indetity, monotonically inequality, this is something about the L^2 norms of the quality we wish to charge. It is just trying to understanding

\frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx

as a fuzzy thing by a more chargeable quality:

  \frac{1}{X^2}\int_{0}^{\infty}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt

In fact we do a cutoff, the quality we really consider is just:

\frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt

established the monotonically inequality:

\frac{1}{X}\int_{X}^{2X}|\frac{1}{H}\sum_{x\leq n\leq x+H}\lambda(n)|^2dx << \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt

In my understanding, This is a perspective of the quality, due to the quality is a multiplicative function integral on a domain (\mathbb N^*) with additive structure, it could be looked as a lots of wave with the periodic given by primes, so we could do a orthogonal decomposition in the fractional space, try to prove the cutoff is a error term and we get such a monotonically inequality.

But at once we get the monotonically inequality, we could look it as a compactification process and this process still carry most of the information so lead to the inequality.

It seems something similar occur in the attack of the moments estimate of zeta function by the second author. And it is also could be looked as something similar to the  spectral decomposition with some basis come from multiplication unclear, i.e. primes.


Step 2: Involved by multiplication property, spectral decomposition 

I called it is “spectral decomposition”, but this is not very exact. Anyway, the thing I want to say is that for multiplication function \lambda(n), we have Euler-product formula:

Euler-product formula:
                      \Pi_{p,prime}(\frac{1}{1-\frac{\lambda(p)}{p^s}})=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s}

But anyway, we do not use the whole power of multiplication just use it on primes, i.e. \lambda(pn)=\lambda(p)\lambda(n) leads to following result:

\lambda(n)=\sum_{n=pm,p\in I}\frac{\lambda(p)\lambda(m)}{\# \{p|n, p\in I\}+1}+\lambda(n)1_{p|n;p\notin I}

This is a identity about the function \lambda(n), the point is it is not just use the multiplication at a point,i.e. \lambda(mn)=\lambda(m)\lambda(n), but take average at a area which is natural generated and compatible with multiplication, this identity carry a lot of information of the multiplicative property. Which is crucial to get a good estimate for the quality we consider about.


Step 3:from linear to multilinear , Cauchy schwarz

Now, we do not use one sets I, but use several sets I_1,...,I_n which is carefully chosen. And we do not consider [X,2X] with linear structure anymore , instead reconsider the decomposition:

[X,2X]=\amalg_{i=1}^n (I_i\times J_i) \amalg U

On every I_i\times J_i it equipped with a bilinear structure. And U is a very small set, $|U|=o(X)$ which is in fact have much better estimate.

\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt =\sum_{i=1}^n\int_{I_i\times J_i}  \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt

Now we just use a Cauchy-Schwarz:

\sum_{i=1}^n\int_{I_i\times J_i}  \frac{1}{X^2}\int_{|log(X)|^{100}}^{\frac{X}{H}}|\sum_{n\leq X}\lambda(n)n^{it}|^2dt +\int_N |\sum_{n\leq X}\lambda(n)n^{it}|^2dt$


Step 4: major term estimate


step 5:minor term estimate


step 6: estimate the contribution of area which is not filled


An approach to Vinogradov estimate

Vinogradov estimate is:

|\sum_{n=1}^{N}e^{2\pi i\alpha P(n)}|\leq c_A\frac{N}{log^A N}

For fix \alpha is irrational and \forall A>0 ... (*).

Assume deg(P)=n, this could view as a effective uniformly distribute result of dynamic system:  ([0,1]^n,T), where T: x\to (A+B)x, b is a nilpotent matrix, matrix A is identity but with a irrational number \alpha in the (n, n) elements.

First approach

we could easily to get a “uniform distribute on fiber” result without very much tough estimate to attach the theorem. That is just a application by my  “rigid trick” that is describe in my early note. But this approach is according to the understanding of the result as a uniformly distribute result on Torus T^n, we could do this approach with the last S^1, which will corresponding to \partial^{n-1}x_k, i.e. we could apply the “rigid trick” to prove sequences (x_k,\partial^1 x_k,..., \partial^{n-1} x_k) is uniformly distribute according to \partial^{n-1} x_k\in S^1 .


But this approach seems difficult to generate. The difficulty is come from both there is no  similar uniformly distribute of the other perimeter use the rigid trick (At least as I know, I try to prove there could be one but I failed) and if in the best case we have the similar uniformly distribute result for other perimeter there is still some thing more need to be established. See this graph for a counterexample that the uniformly distribute for all fiberation could not derive a uniformly distribute for the original space.


Second approach 

In this approach we need use the information of continue fractional to get some information (Which is of course critical to get some information about the estimate). But I do not know if it is necessary, maybe this could be a interesting question weather the information come from continue fractional must involve to get such a estimate in the future, but not today.

Any way, there is two different type of continue fractional:



Anyway, these could be understand as a same thing more or less (if fact we can calculate some quantitive with a_i,q_i which is roughly the same). That is just the orbits \{e^{2\pi i\alpha}\} have quasi-period property, that is to say, under certain norms, it could be understand as the limits of periodic sequences. So it is natural to approximation \{e^{2\pi i\alpha}\} by periodic sequences and will lead to a very good point-wise coverage result:

T_k^{n}(x) \longrightarrow T^{n}(x)

Where T_k^{n}(x)=e^{2\pi i\sum_{i=1}^k\frac{1}{p_1...p_i}} is just the periodic approximation sequence which come from the best approximation (critical point of ||\frac{q}{p}-\alpha||), which natural occur in continue fractional. And by this we already arrive a non qualitative form result of (*) with deg(P)=1.

But unfortunately this approximation is too good to be true for deg(P)\geq 2 case. The reason of this result could be true is just because the natural estimate for the best approximation of \alpha; i.e. Dirichlet approximation theorem.

But for higher degree case, although we could not expect this thing to be true, we still could image a weaker but enough result to be true:

\{T_k^{n}(x)\} \longrightarrow \{T^{n}(x)\}

in the Gromov Hausdorff metric sense, and the $T_k^n(x)$ is carefully chose, which have a finite torsion structure(which could be view as a multilinear structure which will play a central role in the estimate). Here is a graph for deg(P)=2:


Roughly speaking,  in general deg(P)=n case, there is a cube structure in the orbits e^{2\pi iP(n\alpha)} and is critical to observe that the progression of difference structure in it. The goal of this approach is to establish some result from the finite torsion structure(multilinear structure). That is to say, the boundary is high order thing in all direction but there is only one direction attend to infinity the other is just a finite torsion, and we wish to get more information from the extra structure.

This also have a physics explaining, for which see the graph:


Third approach

For P(n)=an^2+bn+c case:


Covering a non-closed interval by disjoint closed intervals

this note will talk about the Ostrowski representation and approximation by continue fraction.

As well-known,by the Weyl criterion,\{n\alpha\} is uniformly distribution in [0,1] iff \alpha\in R-Q.

i.e. we have:\forall 0\leq a\leq b\leq 1,we have:

\lim_{N\to \infty}|\{1\leq n\leq N|\{n\alpha\}\in [a,b]\}|=(b-a)N+o(N).

but this will not give the effective version.i.e. we do not the the more information about the decay of o(N).

we will give a approach of effective version of \alpha with smooth condition by give another proof of the uniformly distribution (in fact to to decomposition the interval [a,b] in to a finite sums of special intervals).and get the result:

D_N=\int_{M}D_N(\theta)d\mu=\int_Msup_{0<a<b<1}|\sum_{n=1}^{N}\chi_{(a,b)}(\{\theta n \})-N(b-a)|d\mu\sim O(log N)

if the term in the continuous fraction of \alpha have a up bound.this is so called \alpha is smooth.

Heat flow and the zero of polynomial-a approach to Riemann Hypesis

this is a note after reading the blog:Heat flow and the zero of polynomial.

1.instead of consider the original version:


consider the corresponding “equidistribution version” is also interesting:

\partial_{zz}f(z,t)=\theta(z,t)\partial_tf(z,t),especially \theta(z,t)=e^{2\pi i\alpha t},\alpha\in R-Q.


where f(z)=z^n+a_{n-1}z^{n-1}+...+a_1z+a_0.

f(z,t)=\sum_{k=1}^n\sum_{0\leq m\leq k-2,2|k-m}\frac{k!}{m!(k-m)!}z^mt^{k-m}.

=\sum_{k=1}^m\sum_{0\leq m\leq k-2,2|k-m}C_k^mt^{k-m})z^mt^{k-m}



F_t:(z_1(t),...,z_n(t))\longrightarrow (\frac{z_1(t)}{t},...,\frac{z_n(t)}{t}).

F_t\cdot f(z,t)=\sum_{m=0}^{n-2}(\sum_{k=m,2|k-m}^nC_{k}^mt^{k-n})z^m.

\lim_{t\to \infty}F_t\cdot f(z,t)=\sum_{m=0,2|n-m}^{n-2}C_n^mz^m.(*)

even term \longrightarrow constant.(after renormelization)

odd term \longrightarrow 0(invariant).so at least the sum zeros of is invarient.

by the algebraic fundamental theorem,we have n zero \{z_1,...,z_n\}of (*).

until now,we already now if the n zeros is distinct,then because the energy is the energy is the same and the entropy is increase so \exists T>>0,\forall t_i,t_j>T,\{t>T|z_i(t)\} \cap \{t>T|z_j(t)\}=\emptyset.\lim_{t\to \infty}|z_i(t)|=\infty and \lim_{t\to \infty}arg(z_i(t))=z_i.

but how to know the information of the change of direction at “blow up” time?

1.change direction only at blow up.

2.energy invariant \sum_{1\leq i\neq j\leq n}\frac{1}{|x_i-x_j|^2}.

3.general philosophy

deformation some function under some evolution equation, such like heat equation,wave equation,shrodinger equation.and there is some conversion thing under the equation,and some quantity that could calculate directly such like the trace of spectral.


this philosophy could generate to the analytic function case,but to make the limit case(I only know how ti deal with this now)coverage.we need very good control on the coefficient.

and to investigate the change of direction at blow up point maybe we need some knowledge about the burid group.




Sarnak猜想在skew product上的情形。

Cylinder map:
Cylender map:这是一个动力系统\Theta=(T,T^2),T:T^2\longrightarrow T^2 满足:\\
来自动力系统\Theta中的可观测量是指\xi(n)=f(T^n(x)),其中x\in T^2,$f\in C(T^2)$.
由于Cylender map是零熵的,这个情形下Sarnak猜想成立等价于:
S(N)=\sum_{n=1}^N\mu(n)\xi(n)=\sum{n=1}^N \mu(n)f(T^nx)
满足S(N)=o(N),由于f_{\lambda_1\lambda_2}=e^{2\pi i(\lambda_1 x+\lambda_2 y)}C(T^2)的一组基,只需对f_{\lambda_1\lambda_2}证明S(N)=o(N)\\
S(N)=\sum_{n=1}^N\mu(n)\xi(n)=\sum_{n=1}^N \mu(n)f(T^nx)\\

=\sum_{n=1}^N\mu(n)e^{2\pi ik(\lambda_1(x+n\alpha)+\lambda_2(nx+\frac{n(n-1)}{2}+y\sum_{i=1}^{n-1}h(x+i\alpha)))}\\

=\sum_{n=1}^N\mu(n)e^{2\pi i(\phi(n)+\sum_{i=1}^{n-1}h(x+i\alpha))}\\

=\sum_{n=1}^N\mu(n)e^{2\pi i(\phi(n)+\sum_{i=1}^{n-1}\sum_{m\in Z}\hat h(m)e^{2\pi im(x+i\alpha)})}\\

=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}} \\
其中我们暂时假定h是解析的,实际上我们要求对h的fourior级数有下界控制,总的来说就是\exists \tau_1,\tau_2:
e^{\tau_1 m}<<\hat h(m)<<e^{\tau_2 m}

\forall A>0,\forall \phi(n) 为多项式函数,我们有指数和估计:

此引理来自解析数论指数和理论, 那么\alpha \in Q情形是引理的直接推论。接下来处理\alpha \in R-Q情形,这种情形下,我们定义\alpha的连分数展开为:

如果\alpha的连分数展开有一致的上界,即存在C\in N^*,\forall n\in N^*,1\leq q_n\leq C那么:
sup_{0\leq a<b\leq 1}|\sum_{k=0}^{N-1}\chi_{(a,b)}(\{k\alpha\})-N(b-a)|=O(log N)



S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}

S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)

S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)
令:A_n=\mu(n)e(\phi(n)),B_n=e(\sum_{m\in Z}e(mx)\hat H(m))
\forall A>0,\forall \phi(n) 为多项式函数,我们有指数和估计:

对于P是一个质数,对于P<<N_1<<N:\\定义\chi_{p}(n)=e^{\frac{2\pi in}{p}}=e_p(n), 定义f:N^*\to Im(\chi_p)满足:\\
对于任何长度为N_1的一段区间$I$,对任意k\in \{0,1,...,p-1\},
\sharp\{n\in I|f(n)=e_p(k)\}=\frac{N_1}{p}+O(1)

其中C\sim P,A\\


cut-off 估计
在式子S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=o(N)
中,我们希望对m\in Z1\leq n \leq N做cut off来简化问题。\\
后者是简单的, 我们待定一个常数c,有:
S(N)=\hat S(n)+\sum_{n=1}^{cN}\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}=\hat S(n)+O(cN)
对m做cut off会稍微复杂一些,根据Fourior分析我们知道:\\
1如果h\in C^{\omega}(T),则
\hat h(m)=O(e^-\tau m).
2.若h\in C^{d}(T),则根据分部积分公式\hat h(m)=O(m^{-d}).\\
e(\sum_{|m|>\delta}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1})\sim \sum_{|m|>\delta}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}
=O(\sum_{|m|>\delta}m \cdot m^{-d})=O(\delta^{d-2})
所以至少当d>2时,我们可以找到\delta \to \infty当$N\to \infty$,使得|m|>\delta的部分可以被cut off.

此时我们定义正整数集N^*关于\alpha的Ostrowoski表示(wangzhiren 2):
其中r_i \in [0,q_{i}-1]



\forall k\in N^*

对任意k\in N^*,对任意长度为l_k=(a,b)的区间I_k\subset (0,1),\forall N\in N^*我们有:


S(N)=\sum_{n=1}^N\mu(n)e^{\phi(n)+\sum_{m\in Z}e(mx)\hat H(m)\frac{e(nm\alpha)-1}{e(m\alpha)-1}}