@KannappanSampath No. I'm not sure how to describe it but I don't think I like it. OTOH, I don't want to decide just yet because I haven't read very much (because progress is slow because it's no fun so far)
@MattN strange. Might be JSTOR's abysmal cookie system, try to clear them while you're still using webvpn. Anyway, here's a freely available text doing something very similar.
Well, apparently because "The notation ∇×F has its origins in the similarities to the 3 dimensional cross product, and it is useful as a mnemonic in Cartesian coordinates if we take ∇ as a vector differential operator del. "
Thanks :) I'm just curious because people are very eager to suggest edits doing that (well, I'm exaggerating, but it happens from time to time). I never liked the notation people use in vector analysis, I always found it more confusing than helpful, but apparently, I'm part of a minority...
@MattN I don't know what I would write. But I would not suggest an edit to replace $\operatorname{rot}{F}$ or $\operatorname{curl}{F}$ by $\nabla \times F$.
@MattN I haven't tried very hard and I'm pretty sure iTunes sells it. Let me put it this way: I already have enough methods to waste my time effectively so I don't pursue other approaches actively...
That's true. This basically reduces to the series representation of $e^x$, as follows:
For $x \geq 0$ you have $e^x = \sum_{n = 0}^\infty \frac{x^n}{n!} \geq \frac{x^{n+1}}{(n+1)!}$.
Now a polynomial of degree $n$ grows at most as $C x^n$ for $C$ large enough, that is $C x^n \geq p(x)$. Thus for $x \gt C (n+1)!$ you have $e^x \geq x \cdot \frac{x^{n}}{(n+1)!} \gt C x^n \geq p(x)$.
Now for $y \gt e$ you have $y^x \geq e^x$... Piece those things together and done.
Well, I've shown that for every polynomial $p$ you eventually have $e^x \geq p(x)$. Indeed, it shouldn't be hard to modify this argument to show that $\frac{p(x)}{e^x} \,\xrightarrow{x \to \infty}\,0$. Does this answer your question?
@MattN by having two different people voting at the same time? Anyway, the paradox seems to be resolved now...
@MattN Oh, you meant me there? I hadn't interpreted it that way. I was just pointing out that some people don't like pity upvoting. And I haven't voted on that question, btw.
@MattN It feels like it is impossible to react to that in a sensible way. I wouldn't put it in those general terms... But I do have the feeling that oftentimes feelings aren't actually helpful. :)
@MattN exactly. Right after I mentioned that I'm terribly disorganized and lousy at staying focused, especially on things I'm not particularly interested in...
@tb Its not just you I am there to give you company in being terribly disorganized and lousy at staying focused, especially on things I'm not particularly interested
@MattN I heard it's very good. I read his Undergraduate Algebraic Geometry a long time ago and it was excellent, so I guess his commutative algebra will be of similar quality.
@SivaramAmbikasaran I'm in no position to judge. I heard all kinds of opinions on it. It seems quite terse and lots of important information seems to be in the exercises. I guess that means it's not for the faint-hearted. I tried to read it a long time ago but I wasn't terribly excited, but commutative algebra is not really my cup of tea anyway.
@BenjaminLim You should be aware of the fact that you already know more algebra than I do...
Because you keep asking me questions on algebra with the expectation that I can answer them... Well, maybe I could, but I never learned algebra properly.
Yes, but you're learning it right now, so you're much more into it than I. You apparently haven't made the experience of how much you can forget in a few years of not using something at all.
@tb. Yes. I have a good understanding of what a accumulation point is. I am at a loss though of how to go about determining them. Especially proving that those are the only ones.
