@Matt with people from Caucus it's very dangerous to say some blames in the address of their family members, especially mother. They will find you by ip and ride on horses to your house, with sabers in their teeth
@Matt I'm not sure I understand you. I never heard this convention that simple functions are positive. For me a simple function is simply a function taking only finitely many values. Using this we have that $s$ is simple and close to $f$ in $L^1$. So you can without loss of generality assume that $f \geq 0$ in the sequel which makes the argument a bit easier.
@robjohn when we were talking to one Armenian guy, he told me some obscene spells ;) some of them are really strange. I mean, very strange blames. I would never say that it is a blame
I'll bake the potatoes for 40 minutes, slightly cook the tomatoes and then put it all together for a nice three hours vacation in a pot alongside the beef and the onions and whatnot.
Actually, when I read about the technical foundations of Windows, it feels like a building of Jell-O constructed on a foundation of steel and concrete...
They could have created a modern, secure OS. But they didn't.
I'm not sure how well UAC works, really. Given that all you have to do is press "Yes", users might just train themselves to always click that button to make the annoying dialog box go away...
@tb I've always relied on the crutch of knowing how great circles behave; I suppose that's what's impeding me in imagining three-dimensional elliptic geometry, to give one example.
@JM you should like that. Lots of special functions involved in the determination of their volume, for example. Polylogarithms and the like pop up out of seemingly nowhere.
Why is $\displaystyle \sum\limits_{n=0}^{\infty} 0.7^n$ equal $1/(1-0.7) = 10/3$ ?
Can we generalize the above to
$\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?
Are there some values of $x$ for which the above formula is invalid?
What about if we take only a finite number of t...
@Srivatsan Well, a geometry synthetically consists of a collection of points and relations satisfying some axioms. Now look at a subcollection and restriction of the relations that satisfy the same axioms. Then look at subsubcollections and restrictions of the relations, and one notion of dimension builds on a maximal chain of such subcollections.
@Srivatsan $\mathbb{R}^n \supset \mathbb{R}^{n-1} \supset \cdots \supset \mathbb{R}^{2}$. You have collections of points and of lines, and various axioms that relate them together (incidence, parallelism, congruence, etc.) Restricting those to $\mathbb{R}^k$ gives you a subgeometry. Now you'd have to prove that this chain is maximal.
Essentially what you said. You can probably set up an induction. Note that you start with $\mathbb{R}^n$, so you have a concrete model. Then you should try to prove that every proper subgeometry can be embedded into $\mathbb{R}^{n-1}$.
@Srivatsan Ah, but I thought about it and did feel I did a significant bit. Enough to make it worthwhile. Ah, but the last remark was about the NE question...
@Gigili, that "They are if you say!" is really rude because you're basically just disagreeing with me by sarcastically pointing out that I don't have absolute authority - I never claimed to anyway - without making any definite statement, which means you have no responsibility to back up anything up.
@Gigili, the only reason I responded to your objections to this was for your own benefit, so I think you should try to be a bit nicer about it. If you really think there's something to this website (which you haven't read) maybe point it out?
I guess you're exaggerating, I said you cannot judge when you're not sure about it. Calling a website garbage or rubbish like that is not rude? consider - one percent - there's some efforts behind that. I'm certainly not going to continue this discussion with you @QED.
@Gigili There are many problems with that site, but for a start, it doesn't actually answer the question. Section (5) talks about finding the $k$th prime, but if you look carefully, you can see that the function needs two variables, a k and an n, and it will tell you ifn is the $k$th prime. So you need to find the $k$th prime, n, in order to find the $k$th prime!
Well, suppose I am given a smooth function $h$ on $U \cap V$. How do I find functions $f : U \to \mathbb{R}$, $g : V \to \mathbb{R}$ such that $h = f - g$?