Would any one be so kind as to construct a function $f: \Bbb R \to [0..\infty)$ with $\displaystyle \limsup_{x\to\pm\infty} f(x) = \infty$ and $\int_\Bbb R f(x) \ \mathrm dx = 1$?
i got good scores in my english exam. examiner told me that he liked my analysis of one of Blake's poems that I don't think any of our textbooks do justice
I'll try something. I once started walking but then soon got bored.
"Besides, one afternoon I did step into the street; If I returned before night, I did so because of the fear that the faces of the common people inspired in me, faces as discolored and pale as the palm of one's hand."
@BalarkaSen Do you know what $p$ means in this formulation of the axiom of subsets? $\forall X \forall p \exists Y \forall u (u \in Y \equiv (u \in X \land \phi(u,p)))$
Just curious, for which rational $c_i$ can the algebraic number $\sqrt{c_1+c_2\sqrt{c_3}}$ be "flattened," i.e., written such that there are no nested radicals?
It's possible for some $c_i$, such as if $c_1=d_1^2+d_2^2d_3,c_2=2d_1d_2,c_3=d_3$ for some $d_i$
Then, $\sqrt{c_1+c_2\sqrt{c_3}}=d_1+d_2\sqrt{d_3}$ (excepting negative things, etc., they can be dealt with separately)
@Sha If there were more than two components, you could connect two (add an edge) and still have a disconnected graph, so it would not be with maximum edges
@Astyx tu peux lire "axiom of subset" ici est la propiete a deux parametres... mais si tu cliques le lien "axiom of subset", la formulation est differente...
A preuver que $A \in A$ n'est pas possible: Soit $A$ un ensemble tel que $A \in A$. Considere l'ensemble $\{A\}$ (axiome de la paire, axiome d'extensionnalite). Nous avons que $x \in A \equiv x = A$. Alors, $x \cap A = A \ne \varnothing$, qui contredit l'axiome de fondation: $\forall S[S \ne \varnothing \implies (\exists x[x \in S \land x \cap S = \varnothing])]$
Soit $f: X \to \mathscr P(X)$ une fonction surjective. Nous savons que $\forall a \in X: f(a) \subseteq X$. Soit $B = \{a \notin f(a)|a \in X\}$. Nous savons que $B \subseteq X$. Soit $f(b) = B$. Si $b \in B$, donc $b \notin B$... Si $b \notin B$, donc $b \in B$...
Introduction: Consider the orbit $Gx$ where $G=(\{0,\pm 1\},+)$ and $x\in (\mathbb{Z},+)$. This is path independent because given any fixed $x,y\in (\mathbb{Z},+)$ and a word $w$ formed by a sequence of elements in $G$, $w \in (\mathbb{Z},+)$
Alternative explanation in terms of vectors. The above description is equivalent to a 1 dimensional vector space where the allowed vectors are of integer length and the allowed elements of addition are the unit vector, its additive inverse and the zero vector. The path independence is then straightforward as an…
Now, consider a finite subset of size n of the previous structure and impose extra conditions as follows: Given any starting point x and a word w. To compute wx, do the following: 0. Evaluate the operations from the leftmost to the rightmost 1. For every +1 in w, increase the size of the set by 1 in both directions 2. For every -1 in w, decrease the size of the set by 1 in both directions
Now consider some starting point x and some word w. w forms a path that link x to wx. Notes what happens:
in semigroup theory, word has a special meaning. It means any expression formed by elements of semigroups. Basically, it is an ordered sequence of operations https://en.wikipedia.org/wiki/Word_(group_theory) The term is more frequently used in groups however
@DHMO An orbit is all transformation made by a group on some set X, i.e. the set $Gx=\{gx|g \in G, x \in X\}$. This terminology is used in the context of group actions. For semigroups, you have an analogous concept called semigroup actions and these are done by transformation semigroups
In a book I've been reading (focused on combinatorial optimization), the following definition is given
The dimension $\text{dim} \; X$ of a non empty subset $X \in \mathbb{R}^2$ is defined to be
$$
n - \max \left\{rank(A) : A \in \mathbb{R}^{n \times n}, Ax = Ay \; \forall x,y \in X\right\}...
When solving equations like
$$\begin{align} 4x-4 &=\frac{(2x)^2}{x} \\
-4 &= \frac{4x^2}{x} -4x \\
-4 &= 4x -4x \\[0.2em]
-4 &= 0\end{align}$$
using the equality-symbol feels like abuse of notation, since you'll end up with $-4=0$, which is not an equality. For instance I feel it would be bette...
[Random] Prove that $\oint f(x)dx=0$ for a vector field $f$ (1) implies path independence on $f$ (2). Prove: (1)=>(2) Suppose $\oint f(x)dx=0$. Then $\int_{C}f(x)dx-\int_{D}f(x)dx$, $\int_C f(x)dx=\int_{D}f(x)dx$ (2)=>(1) reverse the above proof.
[Random] Consider $S_{2n+1}=\{-n,...,0,...,n\}\subset (\mathbb{Z},+)$ with the following additional rule: 1. If the operation is an integer $k < 0$, reduce the size of $S_{2n+1}$ by $k$ in both the negative and positive ends starting from the largest and smallest element
2. If the operation k will result in the element k+x to fall outside of the current $S_m$, then k+x=x
Now the following can be demonstrated: Noncommutativity of components of k: Consider $S_{2n+1}$, $x=2n$, $kx=2-1+x=2+(2n-1)=2n+1,S_{2n}$, but $k'x=-1+2+x=-1+2n+1=2n,S_{2n}\neq kx,S_{2n}$
Nonassociativity Consider $S_{2n+1}$, $x=0$, $kx=-1+1+(-1+x)=2n,S_{2n-1}$ but $kx=(-1+1)-1+x=0-1+x=2n,S_{2n}$ and $kx=-1+(-1+1)+x=-1+0+x=2n,S_{2n}$
Ok this is not screwed up enough. The ideal case is a semigroup such that given any word of any length and constitutent, they are all unique elements in that it is not power associative, alternative, lie or any other notion of asosciativity
->I am trying to use thise to build some kind of path dependent geometry, where every single path including the start and endpoints matter in the navigation within such space
@DHMO $C\subseteq\Bbb R^n$ is closed. $F$ is the set of points $f$ such that there is a unique element of $C$ closest to $f$; in other words, if $c_1,c_2\in C$, and $d(f,C)=d(f,c_1)=d(f,c_2)$, then $c_1=c_2$.
The claim, which I'm now sure is false in general, is that $F$ is open.
@Semiclassical Yeah, I like doing it that way because I get to keep the variable name. -- I've seen people write it like "Define $n'=n-1$" and then, after the last step, say something like, "Dropping the primes, we get…" to get back the original variable name.
@AkivaWeinberger Let us transform (riskily) to "$G$ is the set of points $g$ such that there are more than one element in $C$ closest to $g$, then $G$ is closed."
Actually, that's sorta obvious isn't it. To be on the hypersurfaces requires p,q constraints respectively, so in general being on both of them requires all p+q constraints.
In a book I've been reading (focused on combinatorial optimization), the following definition is given
The dimension $\text{dim} \; X$ of a non empty subset $X \in \mathbb{R}^2$ is defined to be
$$
n - \max \left\{rank(A) : A \in \mathbb{R}^{n \times n}, Ax = Ay \; \forall x,y \in X\right\}...
