I'm currently trying to get to grips with the Hardy Littlewood circle method so I'm working through Vaughan's book.
In the past I've been very bad for leaving a point behind if I don't follow it so I'm not trying to get rid of that habit by asking more questions. For example on page 3 of his book (see page here) Vaughan tells us that for
$F(z)= \sum_{m=1}^{\infty} z^{a_m}$
if $a_m=m^2,\ \rho=1- \frac{1}{n}$, with n large and $e(\alpha)= e^{2 \pi i \alpha} $ then the function F has 'peaks' when $z= \rho e(\alpha)$ is 'close' to the point $e(a/q)$ with $q$ 'not too large'.
So (although I accept that this sounds stupid) I'm not sure about what he means by peaks, does he simply mean F has a peak or? and also why do the values of $z$ close the point $e(a/q)$ cause this?
He then goes on to tell us that F has an asymptotic expansion in the neighbourhood of such points, roughly speaking valid when $|\alpha - a/q| \leq 1/(q \sqrt{n})$ and $ q \leq \sqrt{n}$.
Here I'm not sure what he means by asymptotic expansion and why that neighbourhood is valid.
Any help with any of this would be greatly appreciated.
$F$ is a complex-valued function, so you can think of "peaks" as local maxima of $|F(\rho e(\alpha))|$, where $0 < \rho < 1$ is fixed. You can imagine that these peaks happen around certain rational values $\alpha = a/q$ ($q$ not too large) because the oscillations of $e(\alpha)$ are "in phase" with each other at rational points of the $[0,1)$ interval. As $\rho$ gets closer to 1, the local peaks become more frequent and pronounced.
Vaughan mentions the asympotic expansion: $$F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) \sim \frac{C}{q} S(q,a)(1 - \rho(\beta))^{-1/2}$$ where $n$ is large, $\rho = 1 - 1/n$ and $$S(q,a) = \sum_{m=1}^q e(am^2/q).$$ Vaughan says the asymptotic expansion works for denominator $q \leq \sqrt{n}$ and $\beta$ small, roughly $\beta \leq 1/(q\sqrt{n})$. You can interpret this to mean that $F(\rho e(a/q+\beta))$ is approximately equal to $\frac{C}{q} S(q,a)(1 - \rho(\beta))^{-1/2}$ for $\rho$ close to 1 and $q, \beta$ in the ranges given, where $C$ here is actually $\sqrt{\pi}/2$.
In fact the asymptotic estimate does not seem to quite work on the full range of $q$ and $\beta$ that Vaughan gives, and I think we need to be somewhat more restrictive. I give more explicit estimates below. Perhaps somewhat else can give better estimates that allow the asymptotic to work for a larger range.
First, we'll need an application of partial summation. Suppose that $f:\mathbb{R}_{\geq 0} \rightarrow \mathbb{C}$ is continuous and $f(x) \rightarrow 0$ as $x \rightarrow \infty$. Then $$\sum_{n=1}^\infty f(n) = \int_0^\infty f(x) \, dx + \int_0^\infty \{x\} f'(x) \,dx$$ where $\{x\}$ is the fractional part of $x$. More generally, if we have a congruence condition for our summation, we have the estimate $$\sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty f(n) = \frac{1}{q} \int_0^\infty f(x) \, dx - \int_0^\infty f(x) \,d\left\{\frac{x-m}{q}\right\}$$ $$= \frac{1}{q} \int_0^\infty f(x) \, dx - f(0)\left(1 - \frac{m}{q}\right) + \int_0^\infty \left\{\frac{x-m}{q}\right\} f'(x) \,dx.$$ Therefore $$\left| \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty f(n) - \frac{1}{q} \int_0^\infty f(x) \, dx \right| \leq |f(0)| + \int_0^\infty |f'(x)| \, dx.$$
Let's apply this to the function $$f(x) = \left(\rho e(\beta) \right)^{x^2}.$$ We have $$\left| \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty \left(\rho e(\beta) \right)^{n^2} - \frac{1}{q} \int_0^\infty \left(\rho e(\beta) \right)^{x^2} \, dx\right| \leq 1 + \int_0^\infty \left|\left(\left(\rho e(\beta) \right)^{x^2}\right)'\right| \, dx.$$ We evaluate the derivatives and integrals to get $$\left| \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty \left(\rho e(\beta) \right)^{n^2} - \frac{\sqrt{\pi}}{2q\sqrt{-\log{\rho e(\beta)}}}\right| \leq 1 + \int_0^\infty \left| 2x\left(\left(\rho e(\beta) \right)^{x^2}\right)(\log \rho + 2 \pi i \beta)\right| \, dx$$ $$\leq 1 + (-\log \rho + 2 \pi \beta)\int_0^\infty 2x\rho^{x^2} \, dx = 2 + \frac{2 \pi \beta}{(-\log{\rho})}.$$
Now we can estimate the full sum. We split into residue classes to get $$\sum_{n=1}^\infty \left(\rho e(a/q + \beta) \right)^{n^2} = \sum_{m (\bmod q)} e(am^2/q) \sum_{\substack{n=1\\ n \equiv m (\bmod q)}}^\infty \left(\rho e(\beta) \right)^{n^2}.$$
Therefore we have an estimate for the full sum $$\left|F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) - \frac{\sqrt{\pi}/2}{q} S(q,a) \frac{1}{\sqrt{-\log \rho e(\beta)}}\right| \leq \left(2 + \frac{2 \pi \beta }{(-\log \rho)}\right) q.$$
The asymptotic estimate will be with fixed modulus $q$ and sending $\rho \rightarrow 1^-$. Specifically, we fix $q$ and a small $\epsilon > 0$, and put $\rho = 1 - 1/n$. We send $n \rightarrow \infty$, so that $\rho \rightarrow 1^-$. At the same time, for each $n$ we need to choose $\beta$ such that $\beta = O(n^{-2/3-\epsilon})$, so $\beta$ goes to zero faster than Vaughan indicated.
Note that for small $\rho$ and $\beta$, we have $$-\log \rho e(\beta) \sim 1- \rho e(\beta) = O\left(\frac{1}{n^{2/3+\epsilon}}\right).$$ We thereby obtain an estimate $$F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) = \frac{\sqrt{\pi}/2}{q} S(q,a) \frac{1}{\sqrt{1- \rho e(\beta)}} + O(n^{1/3 - \epsilon}).$$ Since $(1- \rho e(\beta))^{-1/2} \gg n^{1/3 + \epsilon/2} \gg n^{1/3 - \epsilon}$, we get the asymptotic $$F\left(\rho e\left(\frac{a}{q}+\beta\right)\right) \sim \frac{\sqrt{\pi}/2}{q} S(q,a) \frac{1}{\sqrt{1- \rho e(\beta)}}$$ as $n \rightarrow \infty$.