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title: Questions

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# Questions

## Bijection in 2.2
The bijection is
$$\{(a,b) \in \Bbb N^2\} \xrightarrow{\sim} \{(d,a,b) | d \in \Bbb N, \gcd(a,b)=1\},$$
the map is given by $(a,b) \mapsto (\gcd(a,b),a/\gcd(a,b), b/\gcd(a,b))$, with inverse $(d,a,b) \mapsto (da, db)$. 

Look at prime decomposition to validate this. 

## What happens in 8.2a?
I don't think details are that important. The part you ask foris simply the Taylor expansion of $-\frac{\pi}{(2\pi)^s}$ at 1. 

## 8.2b
Ouh, I think $q$ should have been a $t$. Idk how that happened. 
The inequality uses that $2\sigma \geq 1$ and $\sigma^2 \leq 1$ as $\frac 12 \leq \sigma \leq 1$. 

## 9.1
Cutting the integral is not necessary. It's just an easy way to deal with logs in integrals I think.

For the error term, just substitut $y = \frac{\log t}{\log x}$. 
Now $y \leq \frac 12$ (as $t \leq \sqrt x$), and we can make use of the Taylor expansion for
$\frac{1}{1-y}$, which is given by $\sum_{n =0}^\infty y^k
= 1 + O(y)$. But you have a point! This really needs that $|y|$ doesnot get arbitrarily close to $1$.

## 9.4
Not a stupid question at all! We have shown that for two consecutive exceptional moduli $q<q'$, we have $q' > q^2$. 
If we let $q_0 < q_1 < \dots$ be the sequence of exceptional modulus, we find that $q_i \geq q_0^{(2^i)} \geq 2^{(2^i)}$,
i.e. $\log_2(q_i) \geq 2^i$ i.e. $\log_2 \log_2(q_i) \geq i$. 
Also $\log_2\log_2(x) = O(\log \log x)$. 
If we have some $x > 0$ with $q_i < x < q_{i+1}$, we have 
$\#\{ex.\ moduli\ up\ to\ x\} = i \leq \log \log x$. 


## 10.1
I don't know what you are referring to. But yeah, I found a few 
small errors in my solutions on [the website with solutions.](https://mvconsbruch.github.io/AnaNT) (for example, there is
one $\bar \chi(a)$ missing.)

## Abosulute convergence of $\frac {\zeta'}{\zeta}(s)$ in $\Re s > 1$. 
We do have absolute convergence, as a direct consequence of the 
fact that we have the following bounds for the corresponding Dirichlet series:
$$ -\frac{\zeta'}{\zeta}(s) = \sum_{n=1}^\infty \Lambda(n) n^{-s}
\leq \sum_{n=1}^\infty \log(n) n^{-s},$$
and the series on the RHS is absolutely convergent in $\Re s > 1$. 

## 11.1 // Residue of $G(s)$ at $1$
Everything involving terms of the form $\log x$ comes from the 
taylor expansion of $x^s = e^{(\log x) s}$ at $0$, which is given by 
$$x^s = 1 + (\log x)s + \frac 12 (\log x)^2 s^2 + ...$$
Also see [Wolframalpha](https://www.wolframalpha.com/input?i=taylor+expansion+of+x%5Es+at+s%3D0).
In particular, if $G$ has a pole of order 2 at $0$, we get a term involving $(\log x)^2$. 
(Remember that we integrate over $$G(s) \frac{x^s}{s}$$, so in total we have a degree-3-singularity at 0, hence the -1st term of the Laurent expansion (which is the residue) of $G(s)\frac{x^s}s$ has a $(\log x)^2$.)

## 12.1 
This is just the cauchy product of the euler factors. We have 
$$\zeta(s) = \prod_{p} \left( 1+ p^{-s} + p^{-2s} + \dots \right),$$
Hence the euler factor at $p$ of $\zeta(s)^2$ is given by
$$(1 + 2p^{-s} + 3p^{-2s} + \dots) = (1 + p^{-s} + p^{-2s} + \dots)^2$$

