33

There is no such function. Since $f$ would have to map $\mathbb R$ onto $\mathbb R$ for the equation to make sense at all $x\in\mathbb R$, it follows that $f^{-1}(x)\to -\infty$ also as $x\to -\infty$, so $f'\to 0$. Thus $f(x)\ge x$, say, for all small enough $x$, hence $f^{-1}(x)\le x$ eventually, but then the equation shows that $f'\le e^x$, which is ...

answered Jan 2 '17 at 1:45

Christian Remling

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29

Teaching/research job in any University (in a research oriented department or teaching-oriented
department) DOES NOT require a lot of travel. Invitations to conferences or seminar talks come
indeed but this does not mean you have to go, if you don't want to.
Traveling to two conferences per year (usually one week or less) and/or 1-2 seminar talks per year
(...

28

Edit: I've updated this answer to reflect the helpful comments made by Andres Koropecki and Ian Morris.
As the other answers mentioned, the first crucial distinction you must make is that some properties refer to a topological dynamical system $(X,T)$, while others refer to a measure-preserving dynamical system $(X,T,\mu)$. Thus there are two different ...

answered Nov 24 '12 at 22:33

Vaughn Climenhaga

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14

A clear statement is the following:
A compact 3-manifold $M$ is hyperbolic if and only if it has infinite fundamental group and does not contain any essential surface with $\chi \geqslant 0$.
You may remember that by saying that $M$ is hyperbolic unless there is some clear obstruction, and the obstrucion is $\pi_1$ finite or the existence of some surface ...

13

Yes, if $X$ is $\sigma$-compact: $X\smallsetminus E$ is $F_\sigma$, hence a countable union of compact sets. Since a continuous image of a compact set is compact (and therefore closed, in a Hausdorff space), $f(X\smallsetminus E)$ is also $F_\sigma$, hence $f(E)$ is $G_\delta$.
EDIT: While ljjpfx seems to be happy with this answer as his or her spaces are ...

answered Nov 13 '12 at 12:54

Emil Jeřábek

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13

One thing you can try is to select numerical values for $a$ and $b$ (say $a=b=1$ or $a=b^{-1}=2$) and then find a power series solution $y=1+\sum_{k>0}c_k(x-1)^k$ and inspect the coefficients $c_k$. With $a=b=1$ they are
$$ 3, 2, 4, 11, 35, \frac{721}{6}, \frac{18163}{42}, \frac{540391}{336},
\frac{98091}{16}, \frac{26684211}{1120},\dotsc $$
If there ...

answered May 9 '17 at 8:55

Neil Strickland

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12

Since you are interested in a divisor, you only need to know its degree, that is its intersection with a line. A generic line on $Gr(1,3)$ is given by the set of all lines contained in a plane $P$ and passing through a point $Q$. So, you want to know how many tangents to $S$ pass through $Q$ and lie in $P$.
Consider the intersection $S_P = S \cap P$.
Since ...

12

Dominique Manchon's lecture notes, which are very well-known amongst people working on Connes--Kreimer renormalisation, offer exactly the sort of detailed, accessible introduction to Hopf algebras and Connes--Kreimer renormalisation that you're looking for. However, you should first be thoroughly comfortable with abstract linear algebra and with the basics ...

11

You can't have $f^{-1}(x) = 1/f(x)$ with both sides defined for all $x \in \mathbb R$. Namely, if $t = f^{-1}(0)$ then $f^{-1}(t) = 1/f(t) = 1/0$ is undefined.
EDIT:
With the correction that $f$ maps $\mathbb R^* = \mathbb R \backslash \{0\}$ to itself, here is one class of solutions. Take any $f_0$ that maps $(0,1]$ one-to-one onto $(-\infty,-1]$ with $...

10

Yes, it is, on convenient locally convex vector spaces.
Convenient is a very weak completeness condition.
See 33.20 in:
Andreas Kriegl, Peter W. Michor: The Convenient Setting of Global Analysis. Mathematical Surveys and Monographs, Volume: 53, American Mathematical Society, Providence, 1997.(pdf)

10

The result does appear in Dunford/Schwartz, Linear Operators Part I (page 437), but is only stated as an exercise.
Edit after @JosephVanName' comment:
Conway's Functional Analysis has the result for completely regular spaces as Theorem 6.6 (page 140).

10

Let us consider $s$ with $2^s\leq n<2^{s+1}$. First we prove the conjecture when all the entries of $A$ are $1$'s. Then $\mathrm{perm}(A)=n!$, hence by Legendre's formula the exponent of $2$ in it equals $n-t$, where $t$ is the number of $1$'s in the binary expansion of $n$. For $n=2^{s+1}-1$ we have $t=s+1$, hence the exponent equals $n-s-1$. For $2^s\...

10

The answer is indeed no as David Cohen has pointed out, and more generally the answer is determined by the complements of the sets U and U'.
The complete solution is effectively due to R.L. Moore (1925), the key fact being every nondegenerate monotone upper semicontinuous decomposition of the 2-sphere yields a 2-sphere.
Thus to decide if two open connected ...

10

Since you are talking about rational maps, I assume you mean "open dense" in the Zariski topology, so that $X$ and $Y$ are algebraic varieties. Therefore we have a particular case of the following well-known statement in algebraic geometry.
Proposition 1. Let $k$ be an an algebraically closed field of characteristic zero and $f \colon X \to Y$ a ...

answered Sep 29 '14 at 9:17

Francesco Polizzi

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9

You may like to look into the following article.
Robert Anschuetz II and H. Sherwood, When Is a Function's Inverse Equal to Its Reciprocal? The College Mathematics Journal
Vol. 27, No. 5 (Nov., 1996), pp. 388-393.

9

Maple does not find a closed-form solution. Note that the substitution $y = 1/u$ produces the Abel DE $$u' = - a u^3 - \frac{b+x}{x} u^2$$ but Maple can't find a closed-form solution for that either: it doesn't seem to fall into one of the known exactly solvable classes.
EDIT: You might also note that scaling the dependent and independent variables by $y = ...

8

Hurewicz theorem says that for a simply connected space $X$, $\pi_2(X)\cong H_2(X,\mathbb Z)$. So $\pi_2(K3)\cong H_2(K3,\mathbb Z)\cong \mathbb Z^{22}$. Here is a link:
http://en.wikipedia.org/wiki/Hurewicz_theorem

8

$\newcommand{\set}[1]{\lbrace #1 \rbrace}$I will assume that the notation $\Sigma X$ in the question denotes the unreduced suspension of the space $X$.
Quick answer: The notion of homotopy equivalence $\Sigma X\to I$ rel ends described in the question is actually equivalent to the contractibility of $\Sigma X$, since the inclusions of the "ends" $\set{0,1}$ ...

7

Have you ever considered the NSA or other government contractors? The standard way to get a job as a mathematician at the NSA is to apply for one of their development programs, which lets you tour around through several different groups within the NSA (but all would be in Maryland, probably) for 3 years and then settle down into one you like. They love math ...

7

I originally found the motivation for the definition of a Hopf algebra difficult to see. This article by Scott Carnahan makes the definition seem like a completely natural generalisation of the definition of a group. Think of that article as a "pre-introduction" that makes other introductions to Hopf algebras easier to grasp.

7

There are several methods, but let me describe (what is perhaps the simplest) one: On $\mathbb{R}^4$ with coordinates $(t,x,u,p)$, consider the pair of $2$-forms
\begin{aligned}
\Upsilon_0 &= (du-p\ dx)\wedge dt,\\\\
\Upsilon_1 &= du\wedge dx + d(u^{4/3}p)\wedge dt + \lambda v\ dx\wedge dt.
\end{aligned}
It is easy to see that a graphical surface $\...

answered Mar 19 '13 at 20:09

Robert Bryant

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7

A formal Taylor series (e.g.f.) solution about the origin can be obtained a few ways.
Let $f^{(-1)}(x) = e^{b.x}$ with $(b.)^n=b_n \;$ and $ \; b_0=0$.
Then A036040 (Bell polynomials) gives the e.g.f.
$$e^{f^{(-1)}(x)}= e^{e^{b.x}}= 1 + b_1 x + (b_2+b_1^2) \frac{x^2}{2!}+(b_3+3b_1b_2+b_1^3)\frac{x^3}{3!}+\cdots \; ,$$
and the Lagrange inversion / series ...

6

For any even $k>4$ there is a decomposition of $S^2$ into $k$ congruent triangles with angles $\pi/2,\pi/2, 4\pi/k$.
For $k=n+2$ in order to get a decomposition of $S^n$ into $k$ congruent simplexes you should just inscribe in $S^n$ the regular simplex and project its hyper-faces to the sphere from its centre.
In general, a sphere of arbitrary ...

6

Logically these are two different subjects.
Ergodic theory is about transformations of a space equipped with a measure
(and the transformation preserves this measure). The measure is given in advance.
The last 3 notions you mention are
related to this situation, and express the various degrees of "mixing".
Topological dynamics is about continuous ...

answered Nov 24 '12 at 14:42

Alexandre Eremenko

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6

The answer is no. Consider $B\subset A$ with $\mu(B)$ very small, $C_1=A\setminus B$ and $C_2=X\setminus C_1$. Then the r.h.s. equals $\mu(B)^2/\mu(A)\mu(C_2)$ which can be less than $\mu(B)$ since $\mu(A)$ and $\mu(C_2)$ can be grater than $1/2$ and $\mu(B)$ less than $1/10$.
Added later. Under the assumption that each $A$ an $B$ is a union of some $C_\...

6

It seems easier to integrate over the unit square, giving
$$ C^2(x) + S^2(x) = \int_0^x \int_0^x \cos(y^2-z^2) \;\mathrm{d}y\;\mathrm{d}z. $$
Using the Taylor series for cosine, and then normalizing the integral by setting $s = xy$, $t = xz$, gives
$$ C^2(x) + S^2(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{4n+2} \int_0^1 \int_0^1 (s^2-t^2)^{2n} \; \...

6

I do not know any name for it. However, note that the concept you are definining does not really depend on the function $f$, only on the equivalence relation induced by $f$ (which is sometimes called the kernel of $f$, unless you work with groups).
A set $S$ is called "saturated" with respect to an equivalence relation $\theta$ iff $S$ is a union of ...

6

The answer is no.
First, the ordering and density hypothesis are irrelevant (you do not use the ordering, and the density can be managed independently of the measure assumption we are trying to satisfy).
The Lebesgue measure of the set of $x\in(-1,1)$ such that $x\in B(x_,;r_n)$ for at least one
$n>N$ is at most $2\sum_{n>N} r_n$. Your set is the ...

answered Apr 17 '13 at 12:06

Benoît Kloeckner

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6

How about Eilenberg--Mac Lane spaces?
Let $G$ and $H$ be any groups. For pointed homotopy classes, $\langle K(G,1), K(H,1)\rangle $ is $\operatorname{Hom}(G,H)$, and for unpointed homotopy classes $[K(G,1), K(H,1)]$ is $\operatorname{Hom}(G,H)/H$, the orbits under conjugation by $H$.
When $n>1$ and $G$ and $H$ are abelian, we have $$\langle K(G,n),K(H,n)...

6

You can download a copy of my University of Central Florida thesis that covers this topic in even greater detail than the Journal article below:
http://stars.library.ucf.edu/rtd/3139/
Robert Anschuetz

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