@infinitesimal "Avoid Excess. Live life in harmony and balance. Avoid excesses. Even good things, pursued or attained without moderation, can become a source of misery and suffering. " - The Ten Golden Rules on Living the Good Life
I humbly suggest to make some kinds of progress you will need some ERP (exposure and response prevention) therapy. In other words, working on your need to do certain things to reduce your anxiety.
Ah, I have a lot to say about that. Well, there are some things I need to check and some things I don't, so there is ERP already.
So I will just cut down to check the things I need to check, and check them once and for all, and move on with life, and prevent future occurrences. Again, very vague.
To make a crude analogy-the sound of a tiger roaring nearby understandable creates a natural response to flee. With OCD, the innate alarm signal gets triggered by similar things that aren't actually an imminent danger.
Some people will say I am disputing science about SSRIs, that's because they have not read enough of the literature. I am not saying it doesn't work, just saying there is no solid proof they do work.
GABA, for example helps regulate serotonin in the brain, but only the GABA our body manufactures in the brain "works" because of the blood-brain barrier.
And a lot of things we do affect GABA production in our bodies-diet, sleep cycles, etc.
So it's not just "here, take this pill, and all better"
SSRI stands for selective serotonin reuptake inhibitor-it merely prevents you from depleting your existing serotonin levels-if these are already low, they won't do much good.
ecstacy (or MMDA) is a serotonin agonist, it makes you feel "happy", but at the cost of below-normal serotonin levels for days, sometimes weeks, afterwards.
I think in your case, Jasper, they will be of some help, but you're a special case, and I really think you need someone who will "stay with you long enough" to see the big picture.
I felt alienated from people in general. In large crowds, I would find myself getting angry at the ease at which people moved through life, and the utter banality of what they were doing.
Mostly Chinese, with a minority of Malay, and a slighly smaller Indian minority. Former British colony (hence the official language). Politics is dominated by the PAP, laws are fairly strict.
The problem is the following. I have a two sheeted covering map $p:\widetilde X\to X$; and I want to show it admits a unique $\Bbb Z/2\Bbb Z$-covering structure.
So, I have a pretty good idea (I think) of what I should be doing.
@MarianoSuárez-Alvarez I mean I want to make $C_2$ act on $\widetilde X$ such that each multiplication is an homeo and such that for each $\tilde x\in\widetilde X$ there is a nbhd $V$ of $\tilde x$ such that $g\neq h$ implies $gV\cap hV$ is empty.
you have a map $f:(Z,z_0)\to(X,x_0)$ and a covering $p:(\tilde X,\tilde x_0)\to(X,x_0)$, and you want a map $g:(Z,z_0)\to(\tilde X,\tilde z_0)$ such that $pg=f$.
you know how to construct lifting of paths along a covering
Guys, please avoid posting only your questions like that, and avoid that style. You can use the [text](link)code to get smaller links. I would rather you post here if you had a specific doubt about your question (perhaps a user posted something and didn't explain a detail, and such user is away).
Let A, B two C*algebras and f:A->B is a isometric homomorphism of algebras (which means: f is linear and multiplicative), then it should not follow that f is involutive
My condition is very unstable these days. I can feel OK now but a few hours later I feel terrible. Need to try to calm down and stabilise my thoughts. Maybe the meds will help.
Hey @DanielFischer!!! I am given the recurrence relation $T(n)=T(n-1)+n$. Couldn't we solve it as follows? $$T(n)=T(n-1)+n\\T(n-1)=T(n-2)+(n-1)\\T(n-2)=T(n-3)+(n-2)\\ \dots \\ \dots \\ \dots \\ T(2)=T(1)+2$$
At the step $i$ we will have the formula $T(n-(i-1))=T(n-i)+(n-(i-1))$.
Does the recursion end when $n-i=1$ or when $n-i=0$ ?
@DanielFischer It is given that Τ(n) is constant for $n \leq 2$. So does the recursion end when $n-i=2$? So is the last relation this one: T(3)=T(2)+3 ?
That formula holds for $n \geqslant 2$, but not for $n \in \{0,1\}$. So to be fully correct, you need to distinguish the cases if you give an explicit formula. Doesn't matter for the complexity of course.
@DanielFischer Great :) The exercise asks the following: Give asymptotic upper and lower bounds for $T(n)$. Suppose that $Τ(n)$ is constant for $n \leq 2$. Make your bounds as tight as possible, and justify your answers. Doing it as above, are our bounds as tight as possible or do we have to prove that $T(n) \in \Theta(n^2)$.
@evinda "As tight as possible" means you need the tightest possible upper and lower bounds. Most of the time that gives you a $\Theta$ [here we can do even better, $T(n) \sim \frac{n^2}{2}$], but sometimes you may get something like $T(n) \in \Omega(n\log n)$ and $T(n) \in o(n^2)$ as tightest bounds.
Could someone verify this http://upload.wikimedia.org/wikipedia/en/math/7/2/a/72a1058ad2087aec467af24bddcf9479.png ? Also, good morning/evening to all here!
@Kugelblitz It seems fine at the first 1 second looking (it must be viewed in the spirit of the Riemann sums).
user134177
hi
user134177
2:52 PM
i will repeat my question a last time. Let A, B two C*algebras. do you know an example of a map f:A->B wich is a isometric homomorphism of algebras (which means: f is linear and multiplicative), but does not preserve involution?
@DanielFischer Have you ever seen this problem ? Consider $2n$ points in $\mathbb R^2$. Prove the existence of a line that splits the points into exactly two groups of $n$ points (no point should lie on the line)
@LeGrandDODOM Never seen it, but: There are only finitely many slopes of lines passing through more than one of the points. Pick a slope such that any line with that slope passes through at most one point. Move it.
