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.

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}<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)

 

广告

发表评论

Fill in your details below or click an icon to log in:

WordPress.com 徽标

您正在使用您的 WordPress.com 账号评论。 注销 /  更改 )

Facebook photo

您正在使用您的 Facebook 账号评论。 注销 /  更改 )

Connecting to %s