MY-ASSIGNMENTEXPERT™可以为您提供cseweb.ucsd.edu Math204C Analytic Number Theory解析数论的代写代考和辅导服务！
Math204C课程简介
Course Webpage: http://cseweb. ucsd.edu/ dakane/Math204C/
Professor: Daniel Kane
Email: dakane “at” ucsd.edu
Office Hours: Mondays 12:00-1:00 in AP\&M 7131, or by appointment.
Course Description: We will study the basics of analytic number theory with an eye towards the recent proof of bounded gaps between prime numbers. The course will cover the Prime Number Theorem and Dirichlet’s Theorem, the basic theory of L-functions, sieve methods, the large sieve, and exponential sums.
Prerequisites
Prerequisites: In order to understand this class a student should have a solid background in elementary number theory including an understanding of the standard multiplicative functions, a knowledge of basic algebraic number theory will be important in a couple of places, and basic facts about complex analysis will be used throughout. Beyond this, it will be important to have a high degree of mathematical maturity.
Textbook: The textbook for the course will be Analytic Number Theory by Henryk Iwaniec and Emmanuel Kowalski.
Math204C Analytic Number Theory HELP（EXAM HELP， ONLINE TUTOR）
Question 1 (Riemann Hypothesis). The Riemann Hypothesis states that all of the nontrivial zeroes of $\zeta(s)$ lie on the line $\Re(s)=1 / 2$. Show that this would imply the much stronger bounds for the prime number theorem of
$$
\vartheta(x)=x+O\left(\sqrt{x} \log ^2(x)\right)
$$
Assuming the Riemann Hypothesis, we can obtain a stronger bound for the prime counting function $\pi(x)$. Recall that $\pi(x)$ counts the number of primes less than or equal to $x$, and is related to the Chebyshev functions $\psi(x)$ and $\vartheta(x)$ by the formulas
$$
\psi(x)=\sum_{p^k \leq x} \log (p), \quad \vartheta(x)=\sum_{p \leq x} \log (p)
$$
where the sums are taken over primes $\$ \mathrm{p} \$$ and prime powers $\$ \mathrm{p}^{\wedge} \mathrm{k} \$$.
We first use partial summation to relate $\psi(x)$ and $\vartheta(x)$. Let $f(x) = \log x$ and $g(x) = \sum_{p\leq x} 1$. Then, we have \begin{align*} \psi(x) &= \sum_{p^k\leq x} \log(p) \ &= \sum_{p\leq x} \sum_{k=1}^{\lfloor \log_p x \rfloor} \log(p) \ &= \sum_{p\leq x} (\lfloor \log_p x \rfloor) \log(p) \ &= \sum_{n\leq x} (\lfloor \log n \rfloor) \Lambda(n) \ &= \sum_{n\leq x} f(n) g(n), \end{align*} where $\Lambda(n)$ is the von Mangoldt function, which equals $\log p$ if $n=p^k$ for some prime $p$ and positive integer $k$, and equals $0$ otherwise. Similarly, we have
$$
\vartheta(x)=\sum_{n \leq x} f(n)
$$
Next, we use the Riemann Hypothesis to bound the error term in the integral. From the Riemann Hypothesis, we know that all of the nontrivial zeros of $\$ \backslash z e t a(s) \$$ lie on the line $\$ \backslash \operatorname{Re}(\mathrm{s})=\backslash f r a c{1}{2} \$$. In particular, we can write
$$
\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}=\prod_p \frac{1}{1-p^{-s}}=\sum_{n=1}^{\infty} \frac{\mu(n)}{n^s} \sum_{d \mid n} \log d,
$$
where $\$ \backslash m u(n) \$$ is the Möbius function. Using Perron’s formula, we obtain
$$
\psi(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty} \frac{\zeta^{\prime}(s)}{\zeta(s)} \frac{x^s}{s} d s+O(\sqrt{x})
$$
Question 2 (Even Zeta Values). Recall that we have
$$
\zeta(s) \Gamma(s)=\int_0^{\infty} t^s\left(\frac{e^{-t}}{1-e^{-t}}\right) \frac{d t}{t}
$$
Show how this integral can be related to an integral of the same quantity over a contour from $\infty$ to itself looping around 0 (Hint: for non-real s consider the difference of the two branches). Use this to relate the values of $\zeta$ at negative integers to the coefficients of $1 /\left(1-e^{-t}\right)$. Use this to get a formula for $\zeta(2 n)$ for positive integers $n$. You may want to use the fact that $\Gamma(s) \Gamma(s+1 / 2)=2^{1-2 s} \sqrt{\pi} \Gamma(2 s)$ (this can be proved by noting that the ratio is periodic $\bmod 1$ and has limit 1 as $s \rightarrow \infty$ ).
Let $\$ s \$$ be a complex number with $\$$ loperatorname ${R e}(s)>1 \$$. We can rewrite the integrand of the given integral as
$$
\frac{e^{-t}}{1-e^{-t}}=\sum_{k=0}^{\infty} e^{-k t}
$$
Substituting this into the integral gives
$$
\zeta(s) \Gamma(s)=\int_0^{\infty} t^s \sum_{k=0}^{\infty} e^{-k t} \frac{d t}{t}=\sum_{k=1}^{\infty} \int_0^{\infty} t^s e^{-k t} \frac{d t}{t}
$$
For any complex number $\$ \mathrm{~s} \$$, the integral
$$
\int_0^{\infty} t^s e^{-k t} \frac{d t}{t}
$$
converges absolutely and uniformly on compact subsets of the halfplane $\$ \backslash$ operatorname ${\operatorname{Re}}(\mathrm{s})>-1 \$$. We can therefore integrate term by term to obtain
$$
\zeta(s) \Gamma(s)=\sum_{k=1}^{\infty} k^{-s} \Gamma(s) .
$$
This expression is valid for all complex numbers $\$ \mathrm{~s} \$$ with \$\operatorname ${\operatorname{Re}}(\mathrm{s})>1 \$$.
To extend this expression to the whole complex plane, we use the functional equation for $\$ \backslash z e t a(\mathrm{~s}) \$$ :
$$
\pi^{-s / 2} \Gamma(s / 2) \zeta(s)=\pi^{-(1-s) / 2} \Gamma\left(\frac{1-s}{2}\right) \zeta(1-s)
$$
Setting $\$ s=1-s \$$ and solving for $\$ \backslash z e t a(s) \$$, we obtain
$$
\zeta(s)=2^s \pi^{s-1} \sin \left(\frac{\pi s}{2}\right) \Gamma(1-s) \zeta(1-s)
$$
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