@Alizter What do you mean?

@Mike

What was the code to avoid indentation throughout the document?

I don't know. Just don't hit enter between paragraphs, that works for me.

Ah, cool.

@Chris'ssis That is simpler than integrals?

oh long time no see

@PedroTamaroff \parindent=0cm

@TedShifrin I have something you might like to hear to say.

About 3 years ago I went on a differential-geometry spree, this was A-level times (intuition as in "well, what else would it converge to?" times) because I loved it and I was going through some old files

Differential Geometry: A First Course in Curves and Surfaces, Preliminary Version, Theodore Shifrin. University of Georgia, dedicated to the memory of Shiing-Shen Chern.

I'm chuffed to meet the author!

Could someone help me compile a TeX file?

I keep getting errors.

@PedroTamaroff maybe :)

@leo I am using TeXMaker.

When I hit the arrow that shows me the .pdf I get "file not found".

=/

Whoops.

It just worked.

LOL

@leo Do you know how to make apostrophes look good?

I mean, if I write "function" I get two " in one direction only.

@PedroTamaroff like which one?

I want them to "close" up well.

@PedroTamaroff I see

Well, " gives the one facing left.

@PedroTamaroff `` and ''

That is, 99

I need 66

that's the thing above tab and quotes

@AlecTeal Ah, OK.

quotes as in two apostrophies

``something"

@AlecTeal And how do I insert "dagger notes"?

As in ${\rm blah}^\dagger$ and then I want a line in the bottom with a comment.

hi can anyone tell me if the characteristic polynomial of the identity operator of an $n$ dimensional vector space is $x^{n-1}(x-n)$?

@Twink How did you get to that?

@PedroTamaroff I usually use \footnote

@PedroTamaroff use look at the foot\fotenote{whatever you want to be in the footnote}

wait

I made a mistake

If you google "TeX + thing I want to do"

Oh my god you're still alive -.-

Great

it's $(x-1)^n$ right?

yes I had made a mistake thanks

@Twink ;)

@AlecTeal But that gives me ${}^{1}$.

I just want a line and no $1$. Or a dagger instead.

D:

@leo How do I make indentations to margins smaller?

Thanks for the note above, @Alec:)

hey guys i was wonder I have to show $1+2\sqrt 2$ has infinite order in $R(\sqrt 2)$. Could I show this using the binomial theorem?

Do you know about norms?

nope havent learned about those

The binomial theorem should suffice for you, then

A simple proof, though: $1+2\sqrt{2}$ is a positive real number greater than one. Exponentiating it will only get you larger numbers...

yeah thats what I was thinking

thanks

@user60887 What is $R$?

I think I meant $Z(\sqrt2 2)$.

not $R$

Yeah, if that's the case, then what I said works

For an arbitrary ring (of characteristic zero) it's still true, but you can't do it that way

@user60887 Ah, $\Bbb Z[\sqrt 2]$.

yeah its part of the problem I asked yesterday. I'm just finishing it up

@user60887 Well, $1+2\sqrt 2>0$.

And $x,y>0\implies xy>0$.

You're inside $\Bbb R$ after all.

well yeah that is true.

@PedroTamaroff do you want to control the width of the margin?

O solo las sangrías?

@leo Ya lo encontre en un documneto viejo.

@PedroTamaroff como hiciste?

\usepackage[top=3cm, bottom=3cm, left=3cm, right=3cm]{geometry}

@PedroTamaroff exactly :-)

@PedroTamaroff in that case that's equivalent to \usepackage[margin=3cm]{geometry}

@leo Ah.

so i was wondering is it possible to have a ring where 0=1? My book states that in their definition of a ring that it is not required for $0\neq 1$ but Im guessing it matters

@user60887 Well, if $0=1$ your ring collapses to the trivial ring.

Boring stuff.

=)

oh the zero ring you mean?

@user60887: This is a matter of definition/convention. Personally, I require rings to have an identity, not equal to $0$.

Some people don't require a ring to have $1$.

yeah it seems everyone has their own requirements for a ring but my book relaxes them a bit

they let 0 be the additive identity and 1 be the multiplicative identity

Yes, we all do that :)

for subrings my books states that if R is a subring of S iff 3) R must contain the same identity as S but my instructor states that is not always true. is that necessarily true what my instructor states?

@TedShifrin Ted, do you know where I may find du Bois Reymond's counterexample?

That is, a function continuous at a point whose Fourier series diverges there.

@MarianoSuárez-Alvarez Hola!

Um, what about Körner's beautiful book I once recommended to you ages ago?

@TedShifrin I have it yes. Maybe I should check there. =)

Hi @MarianoSuárez-Alvarez.

@TedShifrin Do you know Bonnet's mean value theorem for integrals?

Maybe. Remind me ... I don't remember who did what.

@TedShifrin The claim is that if $f\geqslant 0$ over $[a,b]$ and $B\geqslant f(b^{-})$ then we can find a point $c$ such that $$\int_a^b fg=B\int_{x_0}^b g$$

@user60887: If you don't require rings to have identity, it's false. What about $2\Bbb Z\subset\Bbb Z$?

Oh @Pedro, I've figured that out before. ($x_0=c$?)

@TedShifrin Ah, yes. =)

g night

Hi all

@TedShifrin I don't know what you mean. After all, a ring is a commutative ring with identity...

Hi T.

@Mike: To me, yes. Not to all authors/ teachers.

Commutative? Huh?

I'm being silly, of course.

That's how about half the books I've looked at lately start with.

(Because they have no need for other rings, of course)

Well, personally, I have taught/written as Michael Artin taught me. Rings have $1$ and ring homomorphisms take $1$ to $1$. Shrug.

I'm talking first undergrad course, not research level.

I think my course didn't define them that way, but then we started specializing because we had no interest in rings without identity

I wonder how Hungerford defines them.

I've always thought that rngs are silly. I had an algebra professor, though, who absolutely hated additional assumptions. He liked non-unitary modules.

@Mike I don't believe Hungerford requires 1.

You're correct, it seems

Well, some algebraists like formal symbol-pushing, too.

Also non-unitary modules are silly. Free modules are so ugly in that case.

Oh, no, I'm thinking of modules over an arbitrary ring (rather than CRing with identity)

Oops.

I think my students are going to hate me when I hand back my exams...

What are you teaching?

Survey calculus course

For non-math majors. Mostly just business people.

Ooh.

Mostly just drawing pictures of rectangles, and doing derivatives numerically. Lots of interpretation of what it all means.

I know the type of class, yeah.

I also know the type of effort that usually goes into it from the students :(

Most of my students are actually pretty decent, and do all the work. They just don't like the more mathy bits, like the derivative rules.

It's good to hear that they put in the work. Sorry about the exam grades then...

@robjohn Do you think it would be a good idea to transfer this question to MathOverflow? It had no answers here, but I'm afraid they'll close the question. What do you think?

@TedShifrin

Nevermind =P

He's left, I think

@anon This is peculiar.

The book by Korner on Harmonic Analysis addresses the reader as female.

that is a rhetorical device that happens here and there

Anyone here knows where I can find some examples of sprays?, specially I'm looking for examples on the sphere.

Good evening all

hello

@chris

$$ \int_{-\pi/4}^{\pi/4} \frac{ x^7 - 3x^5 + 7x^3 - x +1}{\cos^2x} $$

My favourite integral for the day

Greetings

@N3buchadnezzar Does one need pen and paper? It seems "No".

@Chris'ssis Deppens if you see the trick or not

@N3buchadnezzar $$\int_{-\pi/4}^{\pi/4} \frac{ x^7 - 3x^5 + 7x^3 - x +1}{\cos^2x} \ dx = \int_{-\pi/4}^{\pi/4} \frac{ 1}{\cos^2x} \ dx =2$$

qed

@Chris'ssis =) I like it

@N3buchadnezzar Yeah, it's cute.

@N3buchadnezzar maybe it helps to write the denominator as $(x - 1)^4 + x^4 + 2$

It also helps to use $u = 1 - x$ :p

@N3buchadnezzar this is what my form of the denominator suggests because of the symmetry.

@Chris'ssis Let $f(x) = x^4 + 1$ then the integral is of the form

$$ \int_0^a \frac{f(x)\,\mathrm{d}x}{f(x) + f(x-a)} $$ By letting $u = x-a$, and taking the avreage one obtains $a/2$.

@Chris'ssis Indeed!

:-)

Me gusta

Very nice. Why did you post three things?

@N3buchadnezzar this is the way all my proofs look like. I work a lot on them, I love them.

I only looked at the first one =)

@N3buchadnezzar the other 2 were messages.

Ah, okay ;)

@N3buchadnezzar I asked some students to compute it elementarily. No one did it so far.

@Chris'ssis This should be doable for a clever student. I like the problems where you continously meet small challenges, but never are totally stuck. Or have to use some extremely clever crux move to obtain the solution (although they are nice as well to showoff to others).

@N3buchadnezzar You mean you created that question?

I saw the idea posted in a problem on this site =)

@N3buchadnezzar you mean that $E\ge0$ for $t \in [0,1)$.

@Chris'ssis Yes?

@N3buchadnezzar To be honest, I'd like to ponder over this in terms of elementary tools like Cauchy-Schwarz inequality.

Well integrating the inequality solves the whole shabang :p

@N3buchadnezzar or you can simply ponder over it in terms of polynomials of degree 2.

@N3buchadnezzar Note that the roots are $r_1=0, r_2=1$ and the coefficient of $t^2$ is negative.

Thus $$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2\ge 0 \ \ \forall \, t \in [0,1]$$

@GustavoBandeira will they close the question even though it has a score of 3?

@Chris'ssis Like the problem? Easy = yes, fun = ?

@Chris'ssis Nice, did not think about it that way. Very nice.

@N3buchadnezzar Yeah, it's funny. I like it! :-)

But does this actually prove the final inequality then ?

That $ \log 2 < (1/2)^{1/2}$ and not that $\log 2 \leq (1/2)^{1/2}$

@N3buchadnezzar well we may write that $$(1+\sqrt{2})^2(t+1)-(t+1+\sqrt{2})^2>0 \ \ \forall \, t \in (0,1)$$

@N3buchadnezzar $0$ is reached for $t=0=1$

But the integration goes from $0$ to $1$, does this not mean that $0$ and $1$ is included?

@N3buchadnezzar no problem at the integration ends, you can integrate from $0$ to $1$.

@Chris'ssis I feel a tad stupid but if you do $\int_0^1 x dx$, then $0$ and $1$ are not included? Scratches head

@N3buchadnezzar look at it in terms of Riemann sums and see what happens.

Ah! Clever. AS you said it a bunch of rectangles poped up in my head, funny. Thanks

@robjohn how about your geometry result? Is it a known result?

@Chris'ssis I have not found any mention of it yet. However, that doesn't mean it has not been found before.

@robjohn true. I always think like that when I find something.

@Chris'ssis I've been looking quite hard and it is certainly not easy to find an answer, that is for certain.

@robjohn hehe, I've just remembered this answer of yours here math.stackexchange.com/questions/156930/geometry-related-limit

(related to geometry)

@Chris'ssis I had forgotten about that one. I'm glad you liked it.

@robjohn Yeah, very nice.

> This question has been deleted. No more answers will be accepted.

How I love that :(

@DanielFischer had the question been answered?

No, no answers before. The OP probably figured it out themselves. No action needed, I'm just whining to relieve the frustration ;)

@robjohn I'm not sure. But mathematical hipsters are everywhere, they always do something like that.

@DanielFischer Hey

thanks for all your help

@fppf You're welcome. Anything new?

I guess that counts as not too bad ;)

haha thanks

I was spamming math.se with all the questions because this is the course at uni that I've felt the least comfortable in

@N3buchadnezzar I think you'd love this one $$\lim_{n\to\infty} n \left(1-\prod_{k=n}^{\infty} \frac{k^2-1}{k^2+1}\right)$$

@Chris'ssis That's two too easy :-)

@robjohn True. What tool did you employ?

$\frac{k^2-1}{k^2+1}=1-\frac2{k^2+1}$

That's it :-)

@robjohn hehe, ok! :-)

@Chris'ssis A concave function is super-linear... the line connecting the endpoints is $0$

I just scrolled back to see that question

@robjohn True

@robjohn this evening I put in LaTeX a mind-blowing proof to math.stackexchange.com/questions/389991/…. I mean it's more than "a mind-blowing proof", it's hard to describe.

@robjohn I'll show you later. :-)

@Chris'ssis okay

@Chris'ssis Let $f(x)=\prod\limits_{k-0}^\infty(1-x^{2^n})^{1/2^n}$ and $g(x)=\prod\limits_{k-0}^\infty(1+x^{2^n})^{1/2^n}$. Then, $f(x)/g(x)=(1-x)^2$, and for the problem given, $x=e^{-2}$

@Chris'ssis The question there leaves out the $k=0$ term and that gives $1-x^2$

Off to the park. BBL

@Chris'ssis I've added my answer to the question.

@robjohn Nice (+1). I had in mind a similar way.

@Chris'ssis 1 vs 22. I have a ways to go to catch up :-)

@robjohn :-)))

@PedroTamaroff good day!

@robjohn Indeed!

@PedroTamaroff Why does the typo by Jack M keep getting stars so that it stays in the sidebar?

I bet he meant $\pi r^2$

@robjohn Because people are mean!

@PedroTamaroff I guess I could cancel the stars...

@robjohn "Lead thy life as thou seest fit."

@anon what do you think?

doesn't really mattress to me

@anon sorry, were you napping on that mattress?

all the time (:

@Chris'ssis Yes, you could get by with just $f(x)$; is that what you did? I thought mentioning $g(x)$ simplified the explanation.

Hey guys, I am making a little presentation for my sequences and series class on Fourier theory, in particular, I presented two criteria for pointwise convergence: Dini's and Jordan's. Do you know any other relevant criteria that is not too advanced?

I tried to look at Korner but it is a bit advanced in general. I mean, it is readable, just not what I am looking for to present.

Is the set of antisymmetric matrices a ring?

I say no because it doesn't contain the multiplicative identity

...

@sonicboom That depends on your definition of a ring. Often a ring is defined without an identity.

In that case, consider $\begin{bmatrix}0&-1\\1&0\end{bmatrix} \begin{bmatrix}0&-1\\1&0\end{bmatrix} =\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$

on the other hand, the answer does not depend on one's definition of ring. consider closure under multiplication.

@robjohn my calc is rusty. what's a good way to compute the volume of the region $|x_1|+\cdots+|x_r|+2|z_1|+\cdots+2|z_s|\le c$ where $x_1,\cdots,x_r\in\Bbb R$, $z_1,\cdots,z_s\in\Bbb C$, where we consider $\Bbb R^r\times\Bbb C^s$ to be a $r+2s=n$-dimensional real vector space in the obvious way? I obtain integrals that I can't compute with mathematica, but the answer is supposed to be $2^r(\pi/2)^sc^n/n!$.

unfortunately I have to leave, so I'll just leave that there

?

Are the units in Z/(n) all the elements that are coprime to n? I.e. in Z/(15) the units are {1, 2, 4, 7, 8, 11, 13, 14}...is this always the case

yes

try proving it

i will! right now

@anon

hi @Chris'ssis

@robjohn I'll show you my way a bit later. I'm preparing things.

@Charlie Hi. How are you? :-)

@Chris'ssis I'm fine, and you? :)

@Charlie Philosophizing about surrounding things and doing some math.

@Chris'ssis hmmm.....interesting, what kind surrounding things?

@Chris'ssis sounds good.

@Charlie Sense of life, God, various spiritual matters.

@Chris'ssis fascinating

Oh!

:-O

:-$\huge\text O$

$\huge \text {:-O}$

@sonicboom Try using Bezout's Identity

@anon does this proof look alright - math.stackexchange.com/questions/573513/…

@Charlie :-$\Huge\text D$

@leo Como estás?

@Charlie todo bien :)

@leo que bueno :)

@Chris'ssis I usually have those thoughts before I sleep

and it usually makes me stay awake, cry, those stuff.

I gave up

thinking about god

@Charlie For instance, what is free will? What if this doesn't really exist? Why I love math and I don't hate it? Who put this love in me? What if we're programmed to live in a certain way?

oh

@Chris'ssis yes. I think of it a lot

@leo oh

what if you are a brain in a vat?

let's get philosophical

hmm....

haha

I'm very existencialist

but more in a social way, not more in a cosmic way, I'd say

“Pain and suffering are always inevitable for a large intelligence and a deep heart. The really great men must, I think, have great sadness on earth.”

Me done good?

@Charlie last time I was in the library, they put new acquisitions in some place where everybody can see them. There was a book about philosophy, about the search for the sense of life. The author said that there always a reason to live. And the reason he gave is: look for some reason to live.

@leo indeed

@leo I feel a huge emptyness and powerlessness, sometimes

as if nothing had taste nor purpose

I inevitably converge to a state of frustration, no matter what I do

@Charlie Then I should be the greatest one! :-)

@Chris'ssis hehe

@Charlie hehe :-)

@Chris'ssis I think i've never been as sad as I am lately. which is weird, because I've never been as glad as I was when I went to uni. it's weird, i can't understand.

I think it's more thankfulness, not happiness itself

@Charlie The best moment to be happy is right now, not a bit later or tomorrow. I do my best to get that in every moment.

@Chris'ssis I can't focus on the right now

@Charlie :D

In Z/(n) the units and zero divisors completely partition the elements. However this is not the case in general. There may be elements in a ring that are not units but are also not zero divisors?

@Chris'ssis I'm having a nervous breakdown

I woke up today feeling so awful

huge anxiety

@Charlie Sometimes some relaxing moments are miraculously effective. Maybe to go out, do some physical exercise, go jogging (something like that).

@Chris'ssis yes...rare moments those are...

How does one get a bigger "slash" character for this sort of formula in MathJax: $$\mathbb F_q[x]/\left<\pi(x)\right>$$

@Chris'ssis :/ maybe...

@Charlie It works for me.

@Chris'ssis good :)

@Chris'ssis couple days ago, my mom asked me to stop studying for a bit

@Charlie I think you're mom was right. You need some breaks once in a while. :-)

@Chris'ssis I have breaks, but i feel guilty for it.

@ThomasAndrews do you mean $\mathbb{F}_q[x] \Big / \left\langle\pi(x)\right\rangle$

Something like that, yeah.

@ThomasAndrews \mathbb{F}_q[x] \Big / \left\langle\pi(x)\right\rangle

Yeah, seems now a bit bigger than I'd like, but better than before :)

@ThomasAndrews \mathbb{F}_q[x] \big / \left\langle\pi(x)\right\rangle

@Chris'ssis math.stackexchange.com/questions/364452/… ;)

@Charlie do you? why?

@Chris'ssis I should be working instead of resting

@leo Thanks.

@N3buchadnezzar nice, I upvoted! :-)

@Charlie People need some rest sometimes, we're not robots. We have to accept that! :-)

@Chris'ssis :///

@Chris'ssis I should be working, I should be working