introducing euler product / zeta function
Theorem. The Euler Product: The birth of AnalyticNT. For all \(s\in \mathbb{C}\) with \(\mathrm{Re}[s]>1\) we have \[\zeta(s) = \sum_{n=1}^{\infty} n^{-s} = \prod_p \frac{1}{1-p^{-s}}. \]
Proof. Thus sum on the left converges absolutely, so you can re-arrange terms however you want.
If we expand out the sum \[ \prod_p \sum_{i=1}^{\infty} p^{-i} \] We get a terms \(n^{-s}\) for every \(n\in \mathbb{N}\).
This is by the fundamental theorem of numbers: you can decompose them into prime factors.
Proposition. An interesting consequence of this Euler Product thing is that the prime counting function \(\pi\) (how many primes are there up to a certain point) satisfies \(\pi(n) > \Omega(n^{1-\varepsilon})\) for any \(\varepsilon>0\).
Proof. Suppose that \(\pi(n) \le O(n^{1-\varepsilon})\). Fix \(s = 1-\varepsilon/2\).
Fix some \(n_0\) and let \(A\) denote the set of primes between \(n_0,2n_0\). Then, \[ \prod_{p\in A} \frac{1}{1-p^{-s}} \approx (1+\frac{1}{n_0^{1-\varepsilon/2}})^{n_0^{1-\varepsilon/2-\varepsilon/2}} \approx e^{n_0^{-\varepsilon/2}}. \]
So, \[ \prod_p \frac{1}{1-p^{-s}} \approx \exp\left(\sum_{n=1}^{\infty} (2^{-\varepsilon})^{n}\right) <\infty.\] But this is ridiculuous because \(\zeta(1-\varepsilon/2) = \infty\).
Proposition. A sometimes useful estimate: \[ \frac{1}{s-1} \le \zeta(s) \le \frac{1}{s-1}+1.\]
Follows from basic calculus
Theorem. For any \(s\in (1,2)\), \[ \left|\sum_{p} p^{-s} - \log \frac{1}{s-1}\right| \le O(1). \]
Proof. By taylor series / desmos: \[ \log\zeta(s) = \sum_p -\log(1-p^{-s}) \approx \sum_p p^{-s}. \]
From earlier (calculus): \[ \log\zeta(s) \in ( \frac{1}{s-1}, \frac{1}{s-1} + 10). \]
Theorem. \[ \sum_p p^{-s} = s\int_1^{\infty} \pi(y) y^{-s-1} dy. \]
Proof.
Well, \(p^{-s} = s\int_p^{\infty}y^{-s-1}dy.\) If you do \[ \sum_p \int_p^{\infty} y^{-s-1}\] you indeed pick up \(\pi(y)\) terms \(y^{-s-1}\).
This is sometimes called the mellin transform. It’s maybe a sort of laplace transform?
Proposition. Meme: The fact that \(\zeta(2)= \pi^2/6\notin \mathbb{Q}\) implies that there are infinitely many primes.
In fact there are some theorems that show that \(\pi\) is “very far from rational”. But these still only allow you to prove really weak things like \(\pi(n) > \log\log n\).
Example. Problem 7 on the first pset is to get some understanding of \(\phi(n)\). I think I should do this one before trying the next two.
Example. There are \(n/\zeta(2) + O(\sqrt{n})\) squarefree numbers in \([n]\).
Hmm, so I kinda get that approximately the pr of a number being square free is kinda like
\[ \prod_p \left(1-\frac{1}{p^2}\right) = \frac{1}{\zeta(2)}. \] But I really am not yet sure how to reason about the error terms.
Example. There are \(n^2/\zeta(2) + O(n\log n)\) ordered pairs of relatively prime numbers in \([n]^2\).
dirichlet characters
Towards Dirichlet’s theorem.
Define \(\chi_4(x)\) as \(+1\) if \(x\equiv 1\mod 4\), \(-1\) if \(x\equiv -1 \mod 4\) and \(0\) else.
Then we define an \(L\)-series. \[ L(s, \chi_4) = \sum_{n=1}^{\infty} \chi_4(n) n^{-s}. \]
We get another product formula.
Let \(\pi(x) = \sum_{p\le x}\chi_4(p)\).
We can show: For all \(s\in (1,2)\), \[ s\int_1^{\infty} \pi(y, \chi_4) y^{-1-s}dy = O(1) \]
This allows us to conclude that there are infinitely many primes congruent to both \(1 \mod 4\) and \(-1 \mod 4\). It also shows the “logarithmic density” of \(1\mod 4\) primes is \(1/2\). Logarithmic density is somehthing which is apparently not necessarily equivalent to actual density. not writing it down.
Definition. Fix positive integer \(q\). Dirichlet character modulo \(q\) is a function \(\chi:\mathbb{Z}\to \mathbb{C}\) that is
- \(q\)-periodic: \(n\equiv n' \mod q \implies \chi(n)=\chi(n')\)
- For \(n\in \mathbb{Z}\), \(\chi(n)= 0 \iff n\not\perp q\).
- multiplicative: \(\chi(mn)=\chi(m)\chi(n)\) for all integers \(m,n\)
Associated Dirichlet \(L\)-series: For \(s>1\) \[ L(s, \chi) = \sum_{n=1}^{\infty} \chi(n)n^{-s} = \prod_p \frac{1}{1-\chi(p)p^{-s}}. \]
Lemma. The characters mod \(q\) form a group under pointwise multiplication with identity the trivial character \(\chi_0\) and \(\chi^{-1}= \overline{\chi}.\)
This group is the Pontrjagin dual of \((\mathbb{Z}/q\mathbb{Z})^{*}\). It is homomorphisms to the unit circle extended by zero lifted to Z. Apparently this means it is a discrete fourier transform or something.
Definition. For a finite abelian group its Pontrjagin dual is homomorhpisms to unit circle in \(\mathbb{C}\).
at this point it got a bit facny and we actually proved dirichlet’s theorem afaict. probably a story for a different day though.
von Mangoldt
\[\nu_p(x!) = \sum_{k\ge 1} \left\lfloor x/p^{k} \right\rfloor.\] Let \(\Lambda(n)\) be \(\log p\) if \(n\) is a power of prime \(p\), and \(0\) if \(n\) is not a prime power. Then we have \[ x! = \prod_p p^{\nu_p(x!)} \] \[ \log x! = \sum_{n\ge 1} \left\lfloor x/n \right\rfloor \Lambda(n) \]
\[ \psi(x) = \sum_{n\in [x]} \Lambda(n).\]
We can find some stuff out about \(\Lambda\)
TODO: write more here.
entering complex analysis land
It seems understandable, but it will definitely take a bit of reminding myself how contour integrals work.