hello, @iwriteonbananas

hey balarka

i wanna be on ignore by chris'sis, too

lol

yeah, you missed the great ignore party

dang it

what are you studying today?

i started doing some cellular homology stuff

i'll ignore you if you're upset about it, @iwriteonbananas :)

nah it's okay, i'll just tell chris'sis that i dont respect her next time she's online

@iwriteonbananas been trying to cope up with my schoolwork. done some more linear algebra, though.

excited about starting multivariable analysis (more than that, learning forms).

what schoolwork? i thought you were 14 yrs old and in high school

@iwriteonbananas oh, nice. but you have to do some degree theory before that.

im starting to learn about forms soon, too

it's a bit nasty at the moment

yes, i am, bananas. thus the "schoolwork"

yeah, learning a bit of degree theory

but i got alg top lecture and exercise class today, so whatever we do there will have priority over degree theory

ah, i see.

i don't like degree theory, the homology definition is ad-hoc. hopefully will get to know some of the geometric aspects when i do diff top.

wait i thought you guys just started singular homology?

we defined a general homology theory, i assume today we're going to start constructing singular homology

btw. do you know any other homology theory?

a homology theory not isomorphic to singular homology theory, you mean?

yes

well, i "know" bordism theory, but that is a generalized homology theory (dimension axiom fails horrifically)

so no, i don't think i know of anything else.

by dimension axiom you mean $H_0() = \Bbb{Z}$ and $H_i() = 0$?

yes

ok

i have always wanted to learn K-theory, though

what's K-theory?

dunno :P

lolol

@iwriteonbananas In general it is a way to assign $K$-groups to a category, one for each natural number

But I am only really familiar with $K_0$ (and mainly in the algebraic context)

@TobiasKildetoft what's a $K-$group?

since that is just the Grothendieck group

@iwriteonbananas A generalization of the Grothendieck group :)

I would not actually be able to write down the definition

okay :D

i wont ask what a grothendieck group is

grothendieck group is nothing so complicated

you turn a right cancellative monoid into a group by adjoining inverses

yeah, the Grothendieck group is fairly simple

okay

@BalarkaSen No, you turn a commutative monoid into a group

you don't want to try with an arbitrary cancelative monoid

whoops, yes.

Doesn't need to be cancellative

Greetings @robjohn @r9m

hi, @Ted

Hi @Balarka

@teadawg1337 I will never respect you or @BalarkaSen or the rest of the team that starred your messages.

Anyway, you're permanently ignored (and have some regrets I used my energy to talk to you).

i don't wish to be, @Chris'ssis

oh, googling K-theory tells me that K^0 is the grothendieck group of vector bundles over X under whitney sum, and K^-n(X) = K^0(S^nX). sounds fun.

@r9m the amazing thing is that I managed to develop the Knuth's problem!!! :-))))

I got the series with left-side in cubic variant!!! :-)

@AlexClark I'm not interested, really! I'm here to talk about math, not about people's life.

Maybe there should be a stackexchange where these guys can discuss about that.

@AlexClark I can communicate with them I think.

@AlexClark Both r9m and ronjohn are really great persons, never saw them talking about people's life, criticizing and so on.

@AlexClark I never do that either, but I'm often pushed in such discussions without my will.

@Rammus Yeah, but this is less important. It's a sign I'm around. :-)

someone can explain?

$x=0$

@Destinyfreedom When faced with such a question, you should start with "how could it fail to exist"

I know when $x=0$

But book say this integral exist,I think this author is wrong? do you argee me?

@Destinyfreedom what should go wrong at $0$?

By the way, people like robjohn and r9m are not unique, there is hope to meet here other such people - also note they are very good at mathematics- talking about mathematics, not about dust in the air. :-)

BBL (I have some research to do)

@AlexClark I was referring to non-mathematical discussions that aim people, and not necessary.

Given that I'm in the following situation: $\int{\frac{sec^{2}\theta}{(tan^{2}\theta +1)^{2}}}d\theta$, should I expand the denominator?

Wait, isn't that an identity. Hold on

@Owatch Yep there's an identity there

Am I correct to say that: $\frac{1}{sec^{2}\theta}$ = $cos^{2}\theta$ ?

@Owatch Yeh

Why not to simply write $$ \int{\frac{(tan \theta)'}{(tan^{2}\theta +1)^{2}}}d\theta$$?

Because I normally don't express the numerator the way you do.

I guess $sec^{2}\theta$ is the derivative of $tan\theta$, but I never usually express anything like that in my work.

What will it allow me to do?

@Owatch It makes you consider the variable change $\tan \theta=x$.

tan theta is already x.

@Owatch No it is not

The problem is written in terms of x, but I am supposed to use Trigonometric substitutions.

Then I don't understand what your saying.

@Owatch You have not mentioned an $x$ anywhere

I've already changed it to tan\theta.

@Owatch If you feel comfortable with other variable, just use another variable, say $y$ instead of $x$.

@Owatch If you need help with a problem, it is a good idea to state the original problem, or at least that what you have is something derived from it

@Chris'ssis to which form? I mean in what way? :)

@r9m Getting another cool series in the right side. I'll let you know when I put things on paper. :-)

@Chris'ssis okay! :D

$\int{ \frac{ 1 } { (x^{2} + 1)^{2} } } dx$ was the original problem. This then became: $\int{ \frac{sec^{2}\theta}{(tan^{2}\theta + 1)^{2}}}d\theta$

@Owatch You complicated things as I can see at first sight.

It's what I am supposed to do in the section.

It says to use the table to make substitutions.

@Owatch Use that $$\frac{ 1 } { (x^{2} + 1)^{2} }=\frac{ 1 +x^2-x^2} { (x^{2} + 1)^{2} }$$

@Owatch what table?

In any case, I have it down to $\int{ cos^{2}\theta d\theta}$

@Owatch that one can be done by parts

@Owatch yes .. after that $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$ .. which can be done easily

or what r9m said

(which is a lot easier, I just always forget those identities)

@TobiasKildetoft imgur.com/rl8tyb0

@Owatch Indeed. It seems trigonometric substitution is a better option (than what I believed at first sight).

@AlexClark I guess it would be when I am using the material outside of the textbook enviroment. Like using theorems and ideas on projects, or coming up with problems and questions that I end up answering ideas that would be in a textbook.

Anyway, one can approach such questions in many ways.

This is a nice elementary integral $$\int \frac{1}{(x+2)^7-x^7-128} \ dx$$

Yah, I am definelty not say textbooks are useless for learning, it is just I don't really understand something full untill I am using it or can come up with it answering questions outside of the pristine textbook environment. I guess in the end that was maybe what I was trying to get at, when I have my nose in a book too long, I tend to realize that I don't understand things as well as I should, or want, even though I technically answered questions on the subject and solved problems, and read it

@AlexClark

So basically "finishing a book" won't necessarily be the best way to learn, a textbook is good place to start though

Sounds cool, I meant to have a look at the paper you posted.

How should I deal with $sin2\theta$, if I have $tan\theta = x$

I drew the triangle, and I know what sin is, but I'm not sure how to write $sin2\theta$

@Owatch If you just have $sin(2\theta)$ then that is easily handled

Yeah, I need to express it in terms of $sin2\theta$

What do you need to express in terms of that?

I have $sin\theta = \frac{x}{\sqrt{1+x^{2}}}$

Thanks

I don't need to integrate it, I want to express $sin2\theta$ in terms of x, since I substituted $x$ for $tan\theta$.

@Owatch $\sin(2\theta) = \frac{2\tan(\theta)}{1+\tan^2(\theta)}$

@r9m I wish I could send proposals to AMM without giving the proofs, and see if I meet situations like Ramanujan when no one answers for many months (or years). :-)

(or to other important mathematicals magazines)

@Chris'ssis They are probably afraid that people would submit incorrect things (and have no easy way to check it)

@TobiasKildetoft Yeah, that's sure.

@Chris'ssis there are other problems from AMM (past) which have no solution given by anyone yet .. you could mail them I suppose, asking them to publish the (possibly updated) list from time to time ..

@r9m btw, is there a list with all AMM proposed problems somewhere available (with problems only)?

@Chris'ssis not that I know of! :( there is a list from back in 1970's or sth .. with all past problems categorized either by year or by topic ..

@Rammus How did you come to this?

@r9m It would be very nice to have a list with all AMM problems since the beginning of AMM birth. :-)

It's a treasure eventually ...

@Chris'ssis perhaps you could ask them to publish such a list .. most likely they have sth like that in their database ..

@r9m I'm sure many others already did it before.

Is looks just like $\frac{x}{\sqrt{1+x^{2}}}$ multiplied by 2, and with x represented as $tan\theta$, but the missing square root means it must be more complicated than that.

So, if $sin\theta$ is $\frac{x}{\sqrt{1+x^{2}}}$ and $cos\theta$ is $\frac{1}{\sqrt{1+x^{2}}}$

I will have a large answer.

Should be fine though. It's all back in terms of 'x'.

@AlexClark I am now.

$\int{ \frac{1}{x(x^{2}+4)^{2}}}dx$

I want to solve this using partial fractions.

I tried expanding the $(x^{2}+4)^{2}$, but I feel that isn't the right approach.

Try $x = 2 \tan \theta$

Would that not be a trigonometric substitution way of approaching the problem?

Oh you want it specifically with, partial fraction sorry D:

Well you can do it like that too

You need two factors for $x^2+4$

It has no roots.

yes, it has

$-2i, 2i$.

I can rewrite it as $x^{2} + 2^{2}$.

$\frac{a}{x}+\frac{(Bx+c)}{x^2+4}$ and $\frac{ex^3+fx^2+cx+D}{(x^2+4)^2}$ or something if not complex aswell

I've never worked with i before.

Trignometric seems best anyway

But you can try complex root @BalarkaSen has point

This way would be too long

@Owatch Do you mean to say that you don't know what complex numbers are or do you mean to say that you just haven't done a lot of fiddling with them?

Both.

Then you should really do trig substitution .-.

what Mann says

Alright. .

Partial fraction gonna be nice massacre

feel free to go through a bunch of pains, though.

I don't want to, It just says I should be doing these by partial functions.

But I suppose I can do it another way. . .

Then use the partial i mentioned

Maybe I should just pick another problem.

or that ^^

I have a message for all you that starred that message in the right panel: I will never respect you! Besides that, I'd appreciate if you never read me, see your business and your math. Avoid me totally.

Thanks.

Give pieces of advices to your families, your friends, not to me, there in your groups. If you wanna talk about respect, find on google some forums about that and write there such things.

@BalarkaSen Why don't you talk instead about elementary mathematics, and not about me? I have many middle school questions to ask. Would you like to attend them?

Anyway, I don't wanna get annoyed more.

I have important stuff to do.

$\int{\frac{1}{x^{2}+3}dx}$

Should I multiply by $(x^{2}-3)$ numerator and denominator?

Ah, I appear to have made an error earlier.

Resolved.

Heya

@Chris'ssis I thought you were supposed to ignore me?

btw, it is not for you to decide what I am going to study about or not. I have merely told you to stop making false universal statements (and many apparently do agree with me).

I couldn't care less about your middle-school questions : they might be interesting or they mightn't be, but those are not the kind of mathematics I care about.

morning

morning, @Mike

i'm wondering whether i should read some bordism theory from tom Dieck.

Read what makes you happy.

yes, i do, but i am unsure whether some basic bordism theory would make me happy.

Ah, that's a hard question for all of us to answer.

:P

I can't think of any interesting results that don't assume you know more than you do, other than the Thom-Pontryagin theorem, which is good. You need to learn singular cohomology.

I haven't read the book you're referring to so can't say whether or not it's accessible.

ah, ok, i see. well, thanks for letting me know.

so you mean i need singular cohomology to understand Thom-Pontryagin theorem?

No, that's the one thing I can think of that you don't need singular cohomology to understand.

here is the link if you want to have a look, but if you are busy, it's ok.

ahh, i see.

ok, the relevant sections are not very long, so worst case it's too much for you and you come back another day.

thom spectra are good

ok, nice, thanks again!

sure

Hello@BalarkaSen

@Chris'ssis what do you think you are ??? You keep on making false statements and disrespect maths you should be the one who should be disrespected its good that people star messages as such so that people like you get the right treatment

Show that every member of the family of functions: $y = \frac{lnx + C}{x}$ is a solution of the differential equation{ $x^{2}y\ + xy = 1$

Okay, so the first thing I do is find: $y'$

And using the Quotient Rule, I get: $\frac{-lnx+c}{x^{2}}$

Then, I substitute that into the differential equation, along with the original y. This gives me: $\frac{x^{2}}{1} * (\frac{-lnx+C}{x^{2}}) + \frac{x}{1} * (\frac{lnx+C}{x}) = 1$

If you simplify that, I get: $-lnx+C + lnx + C = 1$

Or, $-lnx + lnx + C = 1$

I think the constant shouldn't be there, so it would be: $-lnx + lnx = 1$

However, this doesn't look right to me. Shouldn't I get $1 = 1$?

Well you forgot a x in xy@Owatch

I did?

Oh right.

Wait.

Doesn't it cancel with the x it is being multiplied by?

@Rememberme Who am I? I'm someone that works extremely hard on mathematics, with great ambitions in terms of mathematics, one that wants to become as soon as possible like Ramanujan. What are your ambitions in terms of mathematics?

This is me, to be clear once and for all.

Busy with research now.

Might your research incorporate finding if $y = \frac{lnx+C}{x}$ is a solution of the differential equation...

Aw.w

Be successful in most of mathematics and respect every part of maths@Chris'ssis

I can check if I did the quotient rule right, but I've checked already. I don't know why I'm not getting 1=1.

$$\frac{x^{2}}{1} * (\frac{-lnx+C}{x^{2}}) + \frac{x}{1} * (\frac{lnx+C}{x^2}) = 1$$@Owatch

The second one is wromg

It's x * y, not y'

y = $\frac{lnx+C}{x}$

Not over $x^{2}$, that is y'.

You haven't written that in your question

Its here I think: chat.stackexchange.com/transcript/message/21702036#21702036

Oh I madea mistake

The first y, should be y'. However, the second is correct.

(In mine)

$y = \frac{lnx+C}{x}$, and $x^{2}y' + xy = 1$.

So: $ x^{2} * (\frac{-lnx + C}{x^{2}}) + x * (\frac{lnx + C}{x}) = 1$

Given that $y'$ is $\frac{-lnx+C}{x^{2}}$

@Rememberme This is a false situation that never appeared, that is I never stoned any area of mathematics. Show me only one message when I stoned a certain area of mathematics. Maybe you don't understand the English I use, anything is possible.

How could someone that likes mathematics talk against mathematics? Don't you see how much nonsense is in what you and @BalarkaSen say?

Atrocious are your thoughts @Chris'ssis

And you are more than a perfect example of such a person@Chris'ssis

BBL

@Rememberme This might help you to relax some more $$\int_0^{\pi/4} \frac{\cos (2 x) }{1+\sin ^2(2 x)}\log (\cos (x)) \, dx$$

Pfft, mere childs play!

@Chris'ssis Do you like maths puzzles ?

@BalarkaSen hey

sup

dumb question incoming

throw it at me

@Chris'ssis blog.tanyakhovanova.com/2015/05/kvantiks-problems

^ for anyone else who likes riddles :P

actually nvm, hold on...

sorry, i think i got it lol

k

problem 1 is easy.

@BalarkaSen You have considered that you can't put one glass between two others right ?

yes

(actually the problem is unclear on that)

@BalarkaSen it's true that if the sequence of $\Bbb{Z}$-modules $0\to M \to N \to \Bbb{Z} \to 0$ is exact, then $N \approx M \oplus \Bbb{Z}$, right?

my solution is to pick the 5th glass and pour the contains into the 2nd one, and put it back

@BalarkaSen Yeah exactly.. wait you mean 5th ?

@iwriteonbananas yes

@iwriteonbananas: Yes. $\Bbb Z$ is free, hence projective.

what do you mean by projective?

@Hippalectryon The 5th into the 2nd! :-)

typo

The others are more difficult/funnier

the parrot one is funny

@iwriteonbananas Projective module. Usually stated a different way, but $P$ projective literally means $0 \to M \to N \to P \to 0$ splits.

don't care anymore

i dont know, what split means but i assume it means exactly what i claimed above

it is true that having a free thing at the 3rd term in you sequence makes it split

why does freeness imply projectiveness?

I'm giving you the keywords; these are proofs you can find if you google.

No point in reproducing a standard argument.

@Hippalectryon Yeah, I like them, but when I'm not doing anything else like writing a book, or preparing some paper to publish in some magazine. These days I was pretty involved in such things.

ok

@Chris'ssis How the book going ? :P

Split is stronger than the thing you said above; it does imply the thing you said above. It means the exact sequence is isomorphic to $0 \to M \to M \oplus P \to P \to 0$, where the first map is inclusion and the last is projection.

You can have exact sequences $0 \to M \to M \oplus P \to P \to 0$ where the maps aren't inclusion and projection. That's why it's stronger.

@Hippalectryon I wanted to say something like: "It will be mind-blowing", but it's up to the readers say that eventually. :-)

@Hippalectryon I also add to my book 11832 here jstor.org/discover/10.4169/…. Immediately I send my solution to the guys from AMM. It's a very nice proof, indeed, clear, easy and fast.

okay

@Chris'ssis nice :-)

@Hippalectryon The thing there to point out is that if you don't go the proper way you get stuck into some horrible expressions in terms of hypergeometric functions.

for an exact sequence $0\to M \to N \to \Bbb{Z}/2\Bbb{Z}\to 0$ it does not hold that $N\approx M\oplus \Bbb{Z}/2\Bbb{Z}$ right?

no

@iwriteonbananas It might and it might not

exercise : find such a nonsplit exact sequence

@BalarkaSen someone's grumpy

Hi@TobiasKildetoft

@Rememberme Hi

@TobiasKildetoft Are you a logician by any chance ? :P

@TobiasKildetoft can you please elaborate?

@iwriteonbananas Well, there are such sequences that split and ones that do not

@Hippalectryon No

@MikeMiller :P no, it's just that i am eating and typing with a single finger of my left hand at the same time.

oh

@iwriteonbananas You can of course always easily find a split one

Eating and typing, both with a single finger :3 we have a winner /s

@BalarkaSen are you free?

:D

@Rememberme Maybe he is just projective

@Hippalectryon Just give it a try.

Or maybe flat

@Chris'ssis Well I'm still not free yet, but as always I saved it for later.

@TobiasKildetoft Well i have a doubt can you help me pls

@Rememberme Maybe

@TobiasKildetoft lol

@Hippalectryon Sure, when you have time. It's that kind of amazing problem that you simply don't wanna miss.

haha at "projective"

three people left because they were so offended by that joke

lol

@Chris'ssis By the way since you are familiar (?) with Mathematica, do you know how to adimension equations (in Mathematica)?

@Hippalectryon Not that familiar.

Ok