数学代考|素数分布理论MATH0083 Prime Numbers and their Distribution

这是一次UCL伦敦大学学院素数分布理论MATH0083 Prime Numbers and their Distribution课程的代写成功案例

这门课是一门本科生的复分析的入门课，质数一直令人类着迷
它们在数线上的不规则分布。数学中最伟大的定理之一是该定理给出了小于给定界限的素数的渐近估计值。小于给定界限的素数的渐近估计。简单而详细地证明素数定理是本课程的最终目标。本课程的最终目标。为了详细研究素数，我们需要建立一些工具。本课程的主要研究对象是本课程的主要研究对象是源于数论的函数，即算术函数。本课程的大部分时间将用于建立估计算术函数和的技术并展示它们与Dirichlet级数的等价性。后者允许我们后者使我们能够将实数和复数分析的强大技术用于算术函数的研究。这些工具将引导我们清楚地证明关于算术级数中素数的狄利克特定理素数定理。
推荐教材

1. 詹姆逊, 素数定理.
2. E. Stein and R. Shakarchi, Fourier Analysis. 导论, 普林斯顿分析讲座 I, (2003), 第7章和第8章.
   Analysis I, (2003), Chapters 7, and 8.
3. E. Stein and R. Shalarchi, Complex Analysis, Princeton Lectures in Analysis II, (2003)、
   第6章和第7章。

Solution. We first consider the case where $y$ is even. It follows that $x$ must also be even, which implies that $x^3 \equiv 0(\bmod 8)$. Now, $y$ is congruent to 0 or $2(\bmod 4)$. If $y \equiv 0(\bmod 4)$, then $y^2+4 \equiv 4(\bmod 8)$, so we can rule out this case. However, if $y \equiv 2(\bmod 4)$, then $y^2+4 \equiv 0(\bmod 8)$. Writing $y=2 Y$ with $Y$ odd, and $x=2 X$, we have $4 Y^2+4=8 X^3$, so that
$$
Y^2+1=2 X^3
$$
and
$$
(Y+i)(Y-i)=2 X^3=(1+i)(1-i) X^3 .
$$
We note that $Y^2+1 \equiv 2(\bmod 4)$ and so $X^3$ is odd. Now,
$$
\begin{aligned}
X^3 & =\frac{(Y+i)(Y-i)}{(1+i)(1-i)} \
& =\left(\frac{1+Y}{2}+\frac{1-Y}{2} i\right)\left(\frac{1+Y}{2}-\frac{1-Y}{2} i\right) \
& =\left(\frac{1+Y}{2}\right)^2+\left(\frac{1-Y}{2}\right)^2 .
\end{aligned}
$$
We shall write this last sum as $a^2+b^2$. Since $Y$ is odd, $a$ and $b$ are integers. Notice also that $a+b=1$ so that $\operatorname{gcd}(a, b)=1$. We now have that
$$
X^3=(a+b i)(a-b i)
$$
We would like to establish that $(a+b i)$ and $(a-b i)$ are relatively prime. We assume there exists some nonunit $d$ such that $d \mid(a+b i)$ and $d \mid(a-b i)$. But then $d \mid[(a+b i)+(a-b i)]=2 a$ and $d \mid(a+b i)-(a-b i)=2 b i$. Since $\operatorname{gcd}(a, b)=1$, then $d \mid 2$, and thus $d$ must have even norm. But then it is impossible that $d \mid(a+b i)$ since the norm of $(a+b i)$ is $a^2+b^2=X^3$ which is odd. Thus $(a+b i)$ and $(a-b i)$ are relatively prime, and each is therefore a cube, since $\mathbb{Z}[i]$ is a unique factorization domain. We write
$$
a+b i=(s+t i)^3=s^3-3 s t^2+\left(3 s^2 t-t^3\right) i .
$$

Comparing real and imaginary parts yields
$$
\begin{aligned}
& a=s^3-3 s t^2, \
& b=3 s^2 t-t^3 .
\end{aligned}
$$
Adding these two equations yields $a+b=s^3-3 s t^2+3 s^2 t-t^3$. But $a+b=1$, so we have
$$
\begin{aligned}
1 & =s^3-3 s t^2+3 s^2 t-t^3 \
& =(s-t)\left(s^2+4 s t+t^2\right) .
\end{aligned}
$$
Now, $s, t \in \mathbb{Z}$ so $(s-t)= \pm 1$ and $\left(s^2+4 s t+t^2\right)= \pm 1$. Subtracting the second equation from the square of the first we find that $-6 s t=0$ or 2 . Since $s$ and $t$ are integers, we rule out the case $-6 s t=2$ and deduce that either $s=0$ or $t=0$. Thus either $a=1, b=0$ or $a=0, b=1$. It follows that $Y= \pm 1$, so the only solutions in $\mathbb{Z}$ to the given equation with $y$ even are $x=2, y= \pm 2$.

Next, we consider the case where $y$ is odd. We write $x^3=(y+2 i)(y-2 i)$. We can deduce that $(y+2 i)$ and $(y-2 i)$ are relatively prime since if $d$ divided both, $d$ would divide both their sum and their difference, i.e., we would have $d \mid 2 y$ and $d \mid 4 i$. But then $d$ would have even norm, and since $y$ is odd, $(y+2 i)$ has odd norm; thus $d$ does not divide $(y+2 i)$. Hence, $(y+2 i)$ is a cube; we write
$$
y+2 i=(q+r i)^3=q^3-3 q r^2+\left(3 q^2 r-r^3\right) i .
$$
Comparing real and imaginary parts we have that $2=3 q^2 r-r^3$ so that $r \mid 2$, and the only values $r$ could thus take are \pm 1 and \pm 2 . We get that the only possible pairs $(q, r)$ we can have are $(1,1),(-1,1),(1,-2)$, and $(-1,-2)$. Solving for $y$, and excluding the cases where $y$ is even, we find that $x=5, y= \pm 11$ is the only possible solution when $y$ is odd.

Solution. First we note that if $\alpha$ is algebraic, then so is $\bar{\alpha}$ (the complex conjugate of $\alpha$ ), since they satisfy the same minimal polynomial. Also, every element in an algebraic number field is algebraic, since if the field $\mathbb{Q}(\gamma)$ has degree $n$ over $\mathbb{Q}$, then for any $\beta \in \mathbb{Q}(\gamma)$ the elements $1, \beta, \ldots, \beta^n$ are surely linearly dependent. This implies that $\alpha+\bar{\alpha}=2 \operatorname{Re}(\alpha)$ and $\alpha-\bar{\alpha}=2 i \operatorname{Im}(\alpha)$ are algebraic, since they are both in the field $\mathbb{Q}(\alpha, \bar{\alpha})$.

We can apply Liouville’s theorem to $\operatorname{Re}(\alpha)$ to get a constant $c=$ $c(\operatorname{Re}(\alpha))$ such that
$$
\left|\operatorname{Re}(\alpha)-\frac{p}{q}\right| \geq \frac{c}{q^m}
$$
where $\operatorname{Re}(\alpha)$ has degree $m$. Now,
$$
\begin{aligned}
\left|\alpha-\frac{p}{q}\right| & =\sqrt{\left(\operatorname{Re}(\alpha)-\frac{p}{q}\right)^2+(\operatorname{Im}(\alpha))^2} \
& \geq\left|\operatorname{Re}(\alpha)-\frac{p}{q}\right| \
& \geq \frac{c}{q^m} .
\end{aligned}
$$
To prove the result, it remains only to show that if the degree of $\alpha$ is $n$, then the degree of $\operatorname{Re}(\alpha) \leq n$. Consider the polynomial
$$
g(x)=\prod_{i=1}^n\left[2 x-\left(\alpha^{(i)}+\bar{\alpha}^{(i)}\right)\right],
$$
where $\alpha=\alpha^{(1)}, \alpha^{(2)}, \ldots, \alpha^{(n)}$ are the algebraic conjugates of $\alpha$. Certainly $\operatorname{Re}(\alpha)$ satisfies this equation, so we must verify that its coefficients are in $\mathbb{Q}$.

To prove this, we need some Galois Theory. Let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$, and let $F$ be the splitting field of this polynomial (i.e., the normal closure of $\mathbb{Q}(\alpha)$ ). Recall that $f$ is also the minimal polynomial of $\bar{\alpha}$, and so $F$ contains $\alpha^{(i)}$ and $\bar{\alpha}^{(i)}$ for $i=1, \ldots, n$. Consider the Galois group of $F$, that is, all automorphisms of $F$ leaving $\mathbb{Q}$ fixed. These automorphisms permute the roots of $f$, which are simply the conjugates of $\alpha$. It is easy to see that the coefficients of $g(x)$ will remain unchanged under a permutation of the $\alpha^{(i)}$ ‘s, and so they must lie in the fixed field of the Galois group, which is $\mathbb{Q}$.

Since $\operatorname{Re}(\alpha)$ satisfies a polynomial with coefficients in $\mathbb{Q}$ of degree $n$, it follows that the minimal polynomial of $\operatorname{Re}(\alpha)$ must divide this polynomial, and so have degree less than or equal to $n$. This proves Liouville’s theorem for complex algebraic numbers.

Prove that each member of the set of $n-1$ consecutive integers
$$
n !+2, n !+3, \ldots, n !+n
$$
is divisible by a prime which does not divide any other member of the set.
Proof. For each $2 \leq k \leq n$, consider the three cases.

• Suppose $k$ is prime and $k>n / 2$. Then $k \mid n !+k$ and $k \nmid n !+j$ for $j \neq k$ since $n !+2 k>n !+j$. In this case we are done.
• Suppose $k$ is prime and $k \leq n / 2$. Then since $2 k \leq n$, for any prime $p \leq n, p \mid n ! / k$ and hence $p \nmid n ! / k+1$. It follows that $n ! / k+1$ has a prime factor larger than $n$. This implies $n !+k$ does as well.
• Suppose $k$ is composite. Similar to the above case, for any prime $p \leq n, p \mid n ! / k$. Thus $p \nmid n ! / k+1$ and so $n !+k$ has a prime factor larger than $n$.

Now suppose $k$ is not a prime larger than $n / 2$. Let $p_k>n$ be a prime dividing $n !+k$. Since $|(n !+j)-(n !+k)|<n$ we have
$$
n !+j \not \equiv n !+k \bmod p_k \quad \text { for } j \neq k .
$$
Therefore $p_k \mid n !+k$ and $p_k \nmid n !+j$ for $j \neq k$.

E Let $\lambda_a(n)=\sum_{d \mid n} d^a \lambda(d)$ where $\lambda(n)$ is Liouville’s function. Prove that if $\sigma>\max {1, \operatorname{Re}(a)+1}$, we have
$$
\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\frac{\zeta(s) \zeta(2 s-2 a)}{\zeta(s-a)}
$$
and
$$
\sum_{n=1}^{\infty} \frac{\lambda(n) \lambda_a(n)}{n^s}=\frac{\zeta(2 s) \zeta(s-a)}{\zeta(s)} .
$$
Proof. Since $\lambda_a=u * N^a \lambda$, for $\sigma>\max {1, \operatorname{Re}(a)+1}$, by Theorem 11.5,
$$
\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\left(\sum_{n=1}^{\infty} \frac{1}{n^s}\right)\left(\sum_{n=1}^{\infty} \frac{n^a \lambda(n)}{n^s}\right)=\zeta(s) \sum_{n=1}^{\infty} \frac{\lambda(n)}{n^{s-a}} .
$$
then shows
$$
\sum_{n=1}^{\infty} \frac{\lambda_a(n)}{n^s}=\frac{\zeta(s) \zeta(2 s-2 a)}{\zeta(s-a)} .
$$
Looking at the second sum, observe $\lambda$ is completely multiplicative and $\lambda\left(d^2\right)=1$ for all $d$. Hence if $d \mid n$,
$$
\lambda\left(\frac{n}{d}\right)=\lambda\left(\frac{n}{d}\right) \lambda\left(d^2\right)=\lambda(n d) .
$$
From here we find
$$
\lambda(n) \lambda_a(n)=\sum_{d \mid n} d^a \lambda(n d)=\sum_{d \mid n} d^a \lambda\left(\frac{n}{d}\right)=\left(N^a * \lambda\right)(n) .
$$
Thus for $\sigma>\max {1, \operatorname{Re}(a)+1}$, by Theorem 11.5,
$$
\sum_{n=1}^{\infty} \frac{\lambda(n) \lambda_a(n)}{n^s}=\left(\sum_{n=1}^{\infty} \frac{n^a}{n^s}\right)\left(\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^s}\right)=\frac{\zeta(2 s) \zeta(s-a)}{\zeta(s)}
$$

UprivateTA™为您的留学生涯保驾护航。

Fourier analysis代写

微分几何代写

离散数学代写

Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析

时间序列分析代写

Math代写| linear algebra代写| 线性代数代写| Advance linear algebra代写