@MattN probably on Chrome... Anyway: my only point was that it is very painful to type a period after every word if you're too lazy to write a script for that.
Postman Matt, Postman Matt, Postman Matt and his red and white hat, All the birds are singing, and the day is just beginning. Matt feels he's a really happy man.
Everybody knows his bright red avatar, All his friends will smile as he waves to greet them, Maybe... You can never be sure, There'll be knock...[knock knock] Ring [ring ring] Theorems through your door.
@HenningMakholm FYI: I have wondered about that question for years. You could say it's the reason I decided to go back to school and get a math degree.
@Henning My questions are always too simple to post for the geniuses on this site (plus, I like to interact and ask questions about the answer). Anyway, here goes:
@Jeff Working modulo p we have (-1)^r r! = (-1)^r·1·2·3···r = (-1)(-2)(-3)···(-r). Then change each factor to a different representative of the residue class to get the product on the left.
Bleh, I have only one reference for the induced coverage on a subcategory and the author writes that it is straightforward to verify that it is a Grothendieck topology if the original coverage was...
@Jeff You don't need to think in terms of canceling. Canceling is inferring from ab=ac to b=c. Here you need the simpler rule: a=b infers ac=bc, which does not depend on coprimality.
@Jeff One says that if you know an equation between products, you can derive an equality between the factors, under certain conditions. In the other you start with knowing an equality between some factors and derive the fact that two products are equal. There are no extra conditions in the latter direction.
@tb Just as well; I only have a functioning sound card setup at work.
Has there been an increase of flags recently or are the moderators just a bit absent? Every other time I visit the main site I see this yellow bullet light up...
@Jeff I type AltGr+u followed by the letter it goes over, or AltGr+[ for ä and AltGr+] for ö. But that only works on my own customized keyboard layout :-)
@Jeff On most OSes you can install a "US International" keyboard layout that will give you dead-key combinations for most diacritics used with Latin scripts.
@Jeff AltGr is the key that replaces the right Alt key on US keyboards. But as I said, AltGr+u works in my personal customized keyboard only. (And on other custom layouts that derive from the one the University of Copenhagen's CS department used on their US keyboards 15 years ago).
@HenningMakholm i have a BS in comp. sci. But I never made too much of it and eventually that career petered out. I'm almost done with my math graduate degree (yet still feel, not dumb, but not quite up to speed on math)
@robjohn haha. i know you are :D. It's a good thing I'm an American, cuz I can never figure out those keyboard things for making diacriticals.
@robjohn By the way, I looked at your solution for the integral this morning. Very beautiful (as expected). Somehow it works out but I don't see what is really going on.
Hmm, it seems that Microsoft's "US International" keyboard actually usurps the regular quote/apostrophe key to use as dead keys. Sounds like it would be hell to code on.
@Jeff Some newer languages (eg Java) support non-English letters in identifiers natively, so it certainly can be done. However I don't think it is common. Certainly I code in English -- but I don't want to switch keyboard between coding and typing text (which is sometimes in Danish).
@tb Yeah -- needing AltGr for backslash and braces makes TeX (and many other code languages) very painful. That's why I'm not using a Danish keyboard too.
@robjohn My (personal) keyboard has US-English layout and so does have a backslash. But in the standard Danish layout one must press AltGr plus the 102th key that sits between left shift and Z.
@tb The idea is to swap the $r(1+t)$ and $\frac1r(1-s)$ for $\frac1r(1+s)$ and $r(1-s)$ with leftover $\sqrt{\frac{1-t^2}{1-s^2}}$. The formulas will look similar with small differences and you can sometimes exploit those differences (as in the problem at hand)
@Jeff My information is that they mostly type using romanization (pinyin/romaji) and then select the appropriate character from a menu after each word.
@robjohn No -- it's the one that is on 102-key keyboards but not on 101-key ones. (And also on 105 but not 104 key keyboards, for ones with Windows keys added to the layout).
Question (advanced calc 1): $p(x), q(x)$ are polynomials, $r(x)=\frac{p(x)}{q(x)}$ is a rational function. Prove that $\lim_{x \rightarrow a} r(x)=r(a)$ (for all $r \in Reals$ such that $r(a)$ is defined)....
isn't that just saying $\lim_{x \rightarrow a} r(x)=p(a)/q(a)$
@Jeff Not sure what they think. Apparently there's a movement inside China for using pinyin in everyday written text, because the traditional characters is just too cumbersome to use. (This may be a biased impression, though; I have it primarily from Victor Mair's posts on Language Log).
@robjohn I see. It still has a bit of a ... magical feel to it. I was trying to get some intuition for why this works. I was trying to translate it into something more geometrical because it looks awfully similar to some calculations I encountered in hyperbolic geometry but I couldn't figure out exactly what those were. Maybe I'm also tempted to think so because your substitution looks like the reciprocal of the Cayley transform. Thanks for the explanations so far.
@Jeff if you had started your post with \rm {\bf Hint :} \rm \ \ \ then you wouldn't have had that kind of trouble... Anyway: I think it's always good to look at a graph and have some feel of what is going on.
@Jeff Well, just say that it isn't intended to be a rigorous argument. But it upsets me that some doesn't seem see that a graph not touching the $x$-axis means that the function has no roots...
