# 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,

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