# 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.

Proof:

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

$\Box$

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 ${s=1}$.

Proof:

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

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

Proof:

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

$\Box$

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}, ${-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, 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), 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

-1.Motivation-

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', 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. More precisely,

**problem 2**

What happen when $V((\alpha_1,...,\alpha_m))=\Pi_{1\leq i , 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 :

Theorem(Matomaki,Radziwill)
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:

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