@Hippalectryon this question I showed you some hours before it's exceptionally easy for some reason

Even the form of the answer says all about that.

Hi, again. Quick easy question, just to make sure: math.stackexchange.com/questions/1354132/…

@user183297 yeah, projections of a continuous function are continuous, Henning is right.

I didn't check whether the projection he hinted for you to check is continuous or not though.

@dREaM there was nothing else to say.

I think he thinks it is not, which would settle the problem.

I doubt.

he doesn't know what a short exact sequence is.

the thing I wrote after you comment was meant for user183297

Are short exact sequences the exact sequences that are not long?

oh, ok.

@dREaM no.

it's an exact sequence of the form $0 \to A \to B \to C \to 0$

oh yes, that looks short

@VibhavPant yes you would, map each set to its complement

in the sense that anything shorter than this would be a fancy way to write down an isomorphism, yes

@VibhavPant It works coz the sum of all elements is odd.

The key is that it is essentially $A \hookrightarrow B \twoheadrightarrow C$ because of the trivial objects at either end.

@dREaM so is the complement of an odd sum set == an even sum set?

hello @Fargle

Hi @Balarka, rest of chat

@Hippalectryon I can't add to my book such stuff. Not that $-\csc^2(x)$ is the derivative of $\cot(x)$ and then you immediately get a well-known integral.

@Fargle what kind of mathematics have you been thinking about, then?

@Chris'ssistheartist Even for the more complex versions ?

dReEaM, what the hell you thinking about in a math forum ) I saw what you wrote, it wasnt my intention )

"she goes up and down at very high rate"

yeah, if set $A$ has sum $n$ the complement has sum $\frac{50\cdot49}{2}-n$ which is an odd number minus $n$.

it has a different parity than $n$.

@Hippalectryon I was only referring to the version I previously showed you, that with $\log(\tan(x))$.

@BalarkaSen Lately, ring theory. A bit of analysis as well, picking up my self-study where it stopped a while ago.

cool. what about rings?

@dREaM ah, thanks!

@user183297 that is why you shouldn't talk about the cosine function as a she.

Just the basic stuff, for right now. I've got some gaps to fill, holes to plug. Dummit and Foote is helping with that.

D-F has good exercises.

wait wait

Definitely! Very cerebral. I liked proving that $R[x]$ has zero divisors iff $R$ has zero divisors.

@VibhavPant Sure, no problem. Although there are other bijcetive correspondances you could establish if the sum of all elements turned out to be even, but in this case this works.

@VibhavPant In fact all you need for the claim to be true is for the initial set to have at least one odd number. so it doesn't have to be set $\{1,2,3,4\dots 50\}$, it can be any set with at least one odd number, try to find a bijective correspondance between the sets of odd sum and the sets of even sum.

I personally wanted to establish one between the odd sum subsets and $\mathcal{P}(\{1,2...,49\})$, since they're the same number

no, that would be harder. It is better to establish it between the sets with odd sum and the sets of even sum

@dREaM is it possible to find one though?

@dREaM thats you right about it

(just curious)

@VibhavPant No it isn't, because you don't know which parity the elements have beforehand.

Well, technically it is possible since the sets have the same cardinality, but there is no sensible way to do so.

yeah :P

@Fargle Isn't that trivial?

I'm sure there is a way of constructing a function

maybe I'll ask it MSE

@dREaM Er, I think I meant the result that comes after that.

which one?

Where if $f(x)g(x) = 0$ for $f,g \in R[x]$, then $rf(x) = 0$ for some $r \in R$.

($R$ being assumed commutative, of course. And unital, but I don't think that matters for this proof.)

@dREaM "trivial" is a rather inappropriate word for any kind of mathematics. it's not hard, sure, but not trivial.

@dREaM was trolling a while back.

@Rememberme I think so, I don't remember.

Everything in math is "trivial" or "by definition" if you want to use that term loosely enough.

@VibhavPant Look at this article.

Similar.

@Soham I think I owe an apology for the super-quick explanation of the galois theory-covering space phenomenon.

@BalarkaSen heh, don't overdo it, man.

it was cool.

and a lot to ponder.

@SohamChowdhury damn, I'm not versed in Linear Algebra :(

@SohamChowdhury that's what bothers me.

@VibhavPant ask around, neither am I ;)

@BalarkaSen that's a good thing, if you ask me.

not for youngsters.

a similar thing happened when I learned what an aut is.

:P

hmm, I'll try

@BalarkaSen haha

@Hippalectryon I like this quote so much 'The two most important days in your life are the day you are born and the day you find out why.' - Mark Twain

If that's so, I'll probably only have one important day in my life :c

:-))))))))))

@Hippalectryon Maybe to do mathematics and excel in it? :D

I just do that as a hobby really. Other than that I'm pretty bored.

I don't have anything motivating me that much.

OK

@anon: I can write every element of $SL_2(\mathbb R)$ as a product of three commutators. can I do it in fewer?

@TedShifrin I read when the Hessian of $f-\lambda g$ is negative definite, we are at a local maximum. I'm not quite following why. Does $L'' < 0$ tell me I am at a max? How does this give me different information than first order derivatives of $L$?

@MikeMiller the iwasawa decomposition says everything is of the form $abc$ where $a,b,c$ are certain canonical representatives of the three nontrivial conjugacy classes, so maybe consider that.

(well, that's a bad way to put it, but bleh)

that's how i can write things as a product of three commutators

ah

wait, you only need two. conjugacy increases commutator length by one; so just write each of these canonical representatives as a commutator

i guess the real question is: do you need two?

(the relevant formula is $xbx^{-1} = [x,bx]b$)

@anon: that was dumb. $[axa^{-1},aya^{-1}] = a[x,y]a^{-1}$.

yeah

@Hippalectryon it's very easy to get that.

Ok.

@Mike I have my own question over in the MO room if you want to think about combinatorics.

@anon that is combinatorics?

sure, it's a combinatorial problem

@Chris'ssistheartist: since you are looking at integrals involving the exponential of $-\cot^2(x)$, you might be interested in this question. The accepted answer was wrong until I pointed out what was wrong and that the question was about all $x\gt0$, not just $0\lt x\lt\pi/2 $. Now his answer is the same as mine.

@robjohn ooo, that's an interesting limit.

@Chris'ssistheartist It is interesting. A stream of delta approximations.

@Chris'ssistheartist Didn't we look at something like that a while ago?

@robjohn You smashed that nicely. :-)

@robjohn You referred to that limit with many spikes where all reduces to summing all the values gathered around the spikes?

@Chris'ssistheartist I don't remember exactly. The question just seemed familiar

@dREaM yeah, look up "combinatorial species"

@robjohn Yeah, it's about that one.

@anon isn't that the name of a rock band?

I doubt there's any band with "combinatorial" in their name

OTOH:

lol

I love how it says (group)

SL2 (group), as opposed to $SL_2$ (group)

@Chris'ssistheartist It might be this answer that I am remembering.

@robjohn yeah :-))))) How might I forget the question where I fell into spikes metaphorically? :-)

afternoon @ted

Hi @Semiclassic

@anon what d'you think is the correct analog for neighborhood in galois theory? $k \hookrightarrow \overline{k}$ be a point of $k$. is there a notion of "open neighborhood" of this point in $k$?

dunno

drats

@Stan: It's not the full Hessian that we're talking about being negative definite, only what it does to tangent vectors to the constraint set. However, one of the last exercises in my book gives a test in terms of the bordered Hessian matrix.

hi @Ted

hi @Balarka

Hi @Ted

heya @Fargle

@Ted bananas rediscovered de Rham cohomology, apparently.

astonishing to hear ...

@TedShifrin: it is still June gloomy here. It will be hot later, but humid.

he explained what forms are, and I pointed him out that there is a codifferential map $C_0 \to C_1$ from 0-forms to 1-forms that sends $a$ to $D_a(-)$. he generalized it to $k$-forms in general.

OT: The limit $\displaystyle\lim_{(x,y)\to(0,1)}\frac{x(y-1)}{e^x-y}$ doesn't exist. To prove that I can approach it along the curve $y=e^x$, correct?

@robjohn: The last thing I want to hear is hot and humid. I might just as well stay in GA, where we also have water to drink.

he's pondering on the cochain complex right now

although we just talked about the complex for $\Bbb R^n$

I don't like your notation, @Balarka. If you recall from the beginning of your multivariable work, that means directional derivative in the direction of $a$.

well, yes, that's what I mean.

@TedShifrin It's foggy in the morning (June gloom) and hot in the afternoon. Can't help but be muggy.

1-forms are just linear functionals

the directional derivative at the direct of $a$ is a linear functional

you can generalize this to $k$-forms : for example, for $k = 2$, you can send $f dx \wedge dy + g dy \wedge dz + h dx \wedge dz$ to $df \wedge dx \wedge dy + g \wedge dy \wedge dz + dh \wedge dx \wedge dz$

oh, very confusing notation. $a$ is usually the point, and $v$ is a vector you feed in. Remember that when you differentiate, you're differentiating as the point varies, and then feeding in a direction.

right, sorry. should have written $\vec{a}$

@Balarka: That isn't my point. At any rate, you hardly need to lecture me on how to define the exterior derivative.

@BalarkaSen No, you're still confused.

no, lecturing you wasn't my intention. i didn't know it was called the exterior derivative

bananas told me he could do that

When you get to Chapter 8, you'll learn this.

ok, i am abandoning this discussion since it's making me confuzzled.

bananas is doing the work, you should talk to him

he apparently doesn't understand what's the geometric interpretation of wedge product (neither do I), I guess you can give him some intuition

@robjohn what's the temp. in California?

That's also in chapter 8. It's all about determinants.

i have to confess that i entirely forget what the geometric interpretation of the wedge product is :/

@TedShifrin ok. interesting.

just looked. 22 degrees. you call that hot?!

@TedShifrin It's not the full hessian?

@Stan: Reread the exercise. Also, check your email.

@Semiclassic: It's generally not discussed :)

it's 26 degrees in here, after much rain. we're enjoying how cold it is.

pshh

well, okay

@TedShifrin will do. gracias señor :D

@BalarkaSen California is a big place; it probably is between 45° and 95°, depending on where you are.

whoa.

Huh... I was unable to post to chat for a while. I could see what people said, but got an error when I tried to send.

@TedShifrin off to read! this is going to be so much fun :)

Me too, @robjohn. Several times. But I reloaded the page and it was ok.

hasta luego

me too. it said "unknown error"

i was just having that myself a minute ago, but reloading it fixed it

see ya @Stan

I reloaded the chat too.

@BalarkaSen It might be colder on top of Mt Whitney and such

LOL, might be? @robjohn

what about Death Valley?

Death Valley is hot, I suppose.

@TedShifrin that's what I did, too. Odd. My browser said it was a server error.

@BalarkaSen It is, but at this time of the morning, I doubt it is above 95°

I could be wrong

@TedShifrin well, in the sun, it could be 45°

Tennessee has been pretty damn hot, but that's usual. Summers over 90, winters below 20.

@Fargle Summers here usually reach 110-115° (occasionally higher). Winters where I live only get below 30° fewer than 7 days a year.

Why d'you people measure things in F?

Old habits die hard. Also, weather reports give temperatures in Fahrenheit.

@BalarkaSen because that is what is used (on most thermometers and weather stations)

ah. we commonly use C over here.

1) tradition 2) because it's nice to have a system where 100 degrees = "my god its hot" and 0 degrees = "i'm freezing my ass off"

@BalarkaSen yeah, and you use those funny centimeter things.

well, what d'you use instead of centimeter?

@Semiclassical rather than 0° = it's chilly and 100° = I've been dead for a while now, but germ free?

@BalarkaSen inches

@Semiclassical it's 79 in F here, then. I think it's pretty chilly...

ehhh, my scale of chilly is a bit different having lived in minnesota my whole life

@robjohn weird, I never got used to using inches.

@robjohn semi is talking about F, you don't die in 100 F

i think he's making a comparison with how it'd be in C?

that was the point of the message.

note that "rather than"

oh, nvm

@anon he was talking about Fahrenheit; I was saying "rather than Centigrade", where 100° means...

for some reason MSE just switched me to my other account without me telling it to

I had to switch back

and my point is just that 0 degrees C (freezing point) isn't so bad some days in winter :)

@anon I think we've all had to restart browsers. Something wiggy with the servers.

@anon which other? ;)

"The coldest winter I ever lived was a summer in San Francisco" Mark Twain

Wait, I didn't say all, and I'd give the maximum mark for a powerful idea, not necessarily the whole solution.

would you remind them of the definitions of those sinh and cosh integrals? :)

@Semiclassical :-))))))))

@Chris'ssistheartist We can be thankful you are not then.

:-)))))

@Semiclassical would it matter?

depending on the mechanism by which it's solved, yes

Yeah, the idea explained in a few words would be enough to me to give the maximum score.

Sometimes there is also a lot of boring calculation that I may understand, so the useful idea is the most important thing.

@Chris'ssistheartist without doing any work, I'd say the proper substitution and changing the order of integration should work

@anon I guess neighborhoods of $*$ are commutative diagrams consisting of $* : k \hookrightarrow k^{alg}$, $\widetilde{*} : L \hookrightarrow k^{alg}$ and inclusion $k \hookrightarrow L$, where $L/k$ is an extension.

@robjohn Indeed, proper substitution is required.

this is motivated from this geometric picture : neighborhoods of $s : * \hookrightarrow X$ are commutative diagrams consisting of $s$, $\widetilde{s} : * \hookrightarrow U$ and inclusion $U \hookrightarrow X$.

@Chris'ssistheartist First would be $u=\cot(x)$ as that makes the $\csc^2(x)e^{-\csc^2(x)}\,\mathrm{d}x$ look a bit nicer and works well with the arguments to Chi and Shi

hmm, I guess the algebraic analogy is more specialized because $k \hookrightarrow L$ is actually an analog for the covering map

After that, I'd need to see exactly how things looked

@robjohn It might work.

this indicates there might be different structures than fields at work

doh

for reference: Shi, Chi

@Semiclassical That integral works nicely with the Godfather song while given to a class. :-))))

main thing i notice immediately is that you end up getting an integrand of $(\cosh t+\sinh t-1)/t=(e^t-1)/t$

when you do add Chi and Shi of the same argument

at which point everything becomes pretty transparent, i suspect, since $1+\cot^2 x = \csc^2 x$

though the presence of $\gamma$ and $\log x$ make me a bit paranoid

@Semiclassical :-)

hmm, I think I want a $k$-algebra instead of $L$.

@robjohn The limit $\displaystyle\lim_{(x,y)\to(0,1)}\frac{x(y-1)}{e^x-y}$ doesn't exist. To prove that I can approach along the curve $y=e^x$, correct?

@Cristopher That works. Also, if you look at it near $(0,1)$, it is $\frac{x(y-1)}{x-(y-1)}$

@Semiclassical It's a good idea eventually.

@robjohn Oh, I see. Thank you

@Semiclassical a question just arouse to my mind. When it is more useful to return from $(e^t-1)/t$ to Shi and Chi, if there are such integrals. :-)

@Cristopher Since $\frac{uv}{u-v}=\frac1{\frac1v-\frac1u}$ doesn't have a limit at $(0,0)$, the original function doesn't have a limit at $(0,1)$

@Cristopher However, Mathematica might choke on that one, too.

@robjohn Haha, thanks for editing to make it clear. And yes, wolfram alpha says the limit is 0....

@Cristopher In most directions it approaches $0$, but if there is a path along which the limit doesn't go to $0$, the limit doesn't exist

@robjohn Yes, I understand. Mathematica still isn't completely reliable when it comes to multivariable limits, I see. I hope they improve that

@Cristopher using the stuff I've said above, we have that along the path $\left(u,\frac{2u+1}{u+1}\right)$, the limit is $1$

Ah, I see

@robjohn Looking at the plot in Mathematica all seems fine.

@anon I think I'm closing in at a formal analogy of why a field extension is similar to a covering map, by my defn. of a neighborhood (/ a slightly deformed version of it I am using)

excited, hoping it won't lead it to a dead end.

hello @KarlKronenfeld

hi

@robjohn What about this one? $\displaystyle\lim_{(x,y)\to(-1,0)}\frac{y^4(x+1)}{|x+1|^3+2|y|^3}$ Using the Squeeze theorem I get: $0\leq\dfrac{y^4|x+1|}{|x+1|^3+2|y|^3}\leq\dfrac{y^4|x+1|}{|x+1|^3}=\dfrac{y^4}{‌​|x+1|^2}$

@Karl due to this geometric picture, the correct analog of a neighrborhood of a point $*: k \hookrightarrow k^{alg}$ should be a commutative diagram consisting of $*$, $\tilde{*} : A \hookrightarrow k^{alg}$ and a hom $k \to A$, where $A$ is some algebraic structure lying over $k$. d'you have any idea what this algebraic structure should be?

Yeah, the neighborhood is really the kind of structure $A$, rather than the fluff you have listed so far.

No idea what it should be.

I don't know if I should take field. If I take a field, a lot of things are simplified because $A \otimes_k k^{alg}$ is a direct product of $k^{alg}$.

I think that says something about the evenly covered condition, but I haven't thrashed out the details yet.

field is too simple

@KarlKronenfeld well, I am particularly fond of my discovery of a point and fiber (which is either $\mathsf{Hom}_k(L, k^{alg})$ or $L \otimes k^{alg}$ or probably both, depending on what I want) among the fluff :P

You should be looking for an interesting property of the extension of k-alg over A.

@KarlKronenfeld can $A \otimes_k k^{alg} \cong \prod k^{alg}$ hold for something other than fields?

@KarlKronenfeld I am not sure if you have the hom definition of fibers in mind when you say that, but a tensor product definition seems like the right thing here. fibers are pullbacks in Top, so they should be pushouts in this analogy. fields don't have particularly interesting pushouts, so taking pushout in Ring makes sense (which is precisely tensor)

I'm quite literally working from the analogy with topological spaces, ignoring the details here.

@BalarkaSen Yeah, you want a certain kind of field (extension) methinks, I initially read that as "all fields" for some reason.

hm

well, I'm gonna sleep on it, I think. g'night!

Looks bad near $(0,1)$

@robjohn Yes, but there is a small part, if rotating, where one might believe all is fine.

it looks particularly badly behaved as you approach $(0,1)$ from the bottom side of that figure

@Cristopher Hölder implies that $$\frac{|x+1|y^4}{|x+1|^3+2|y|^3} \le\sqrt[3]{\frac{16}{3125}}\left(|x|^3+2|y|^3\right)^{2/3}$$

@Cristopher we have no control of $\frac{y^4}{(x+1)^2}$

can you help me reopen this question please? I closed it but it is not really a duplicate and wish to reopen it.

@robjohn Holder's theorem? I haven't heard of it before

oops

oh no

@Cristopher Hölder's Inequality?

That is the question, I want to reopen it because I closed it but someone wants to add a solution that doesn't fit the "duplicate"

@robjohn Thanks for the link. I never heard of that either. It's not entirely clear to me how you arrived at that result. Which section of the article should I look at?

thanks guys, it is open now.

Is the set of positive integers $n$ such that $n!+1$ divides $(2012n)!$ finite or infinite?

The answer is:

Yes

since $n!+1$ is less than $(2012n)!$, $n!+1$ must show up as a factor in $(2012n)!$. So every positive integer is an example.

lol wut?

$n!+1$ is not less than $2012n$

anyways, good night folks

...oh, bugger. brain fart

hi guys

@DanielFischer How is this not a duplicate?

@MikeMiller Well ...

Okay, so it is.

I see, Brian is objecting because these spaces aren't connected. Great.

@MikeMiller Not sure, to be honest, the connectivity requirement at the other question may make a difference.

You get examples in both questions. The other restricts the set of answers. In any case, not worth investing energy.

@Chris'ssistheartist: I thought this answer was nice since it only used pre-calc, but it seems to have gone unnoticed. Is it perhaps a bit hard to understand?