Search Results

The main observation: Let $f,g$ be analytic in the disk $\{|z|\le 2\delta\}$ and real on the interval $(-2\delta,2\delta)$. Assume that $f(0)=0$ and $f(x)<0$ for $0<|x|<2\delta$. Let $\psi$ be any $C …

The reason the asymptotics for the meromorphic functions works is that we know the exact coefficients for $c_k(z-z_0)^{-k}$ and can always remove the singularity by subtracting a few terms of this kind … The integral over the larger circle will decay exponentially faster than the typical coefficient, so all asymptotics will come from the cut. …

De Bruijn's "Asymptotic methods in analysis" is an excellent book for beginners. You'll need to work through it diligently to learn everything but no advanced a priori knowledge is required. Also, you …

OK, here is my argument (sorry for the delay). First of all, $Z$ is essentially the maximum of the absolute value of the polynomial $P(z)=\prod_j(1-z_jz)$ on the unit circumference (up to a factor of …

Power series decomposition does the job immediately: you can find $\int_0^1u^n\log^2 u du$ integrating by parts twice and the Euler identity for the sum of inverse squares finishes the story. As to di …

$$ \begin{aligned} &\int_0^\pi t^\beta e^{iyt}\frac{dt}{t}dt=y^{-\beta}\int_0^{\pi y} t^{\beta} e^{it}\frac{dt}{t}= \cr &=y^{-\beta}e^{i\pi\beta/2}\left[\int_0^{\pi y} t^\beta e^{-t}\frac{dt}{t}+O(t …

You can just play with the same truncation techniques you used already. $u\ge 0$ (replace by $\max(u,0)$ on $\mathbb R$) If $x\ge 0$ and $u(x)\ge \frac{1}{c(x)}$, then $u(y)\le u(x)$ for $y\ge x$ (r …

Unless I misunderstand the question, the answer is that $a(x,0)$ can be pretty much anything it wants. Take any smooth $f(x)$ supported on $[0,1-\delta]$. Put $$ a(x,y)=f(x)-[y(e^{1/y}-1)]^{-1}\int_ …

The set can be infinite (but only countable). For an example choose any $t_j$ linearly independent over $\mathbb Q$ and let $V$ be the set of all $v$ such that $vt_j\mod 1 \notin (\frac 12-a_j,\frac 1 …

Replacing $y$ with $y-\frac 12$, we get the integral $$ 2\int_0^\infty\left[\int_0^\infty z^{-\nu}e^{-\frac z2}e^{-x^2\frac{y^2+0.25}{2z^2}} \left(e^{\frac{x^2y}{2z^2}}-e^{-\frac{x^2y}{2z^2}}\right)dz …

We'll show that in the regime $q\ge p>(2+\delta)\log_2q$, $p,q\to\infty$, the portion of subsets of $A\times B$ with $|A|=p$, $|B|=q$ that have non-trivial automorphisms is at most $$ O(2^{-p}p^2q^2)\ …

In my opinion, that other question has been answered by michael completely in the very first comment. However, since the question arose again, let me just spell the details of michael's answer out. W …

If all you want is a compositional square root of something like $e^z-1$ analytic in some disk around the origin, I would go for $e^z-1-\frac 34 z=\frac z4+h(z)$. Then, putting $f(z)=\frac z2+g(z)$, w …

OK. It is going to be long and extremely boring, but, as promised, here goes. We'll get the precision $O(mL^2)$ where $m=\max(a,b,c), L=\max(\log\frac 1a,\log\frac 1b,\log\frac 1c)$, which should be e …