Mathematics - 2015-05-07 (page 3 of 6)

7:02 AM

@PaulPlummer we know that matrix multiplication is not commutative but i have found some matrices which when we multiply obeys commutativity is that fine?

Paul Plummer

@Rememberme Yes commutative group are groups where all $a,b$ we have $ab=ba$.

Balarka Sen

@SamuelYusim there is

Remember me

Hi@BalarkaSen

Paul Plummer

For example in every group we have $1a=a1$, even if the group is not commutative

Balarka Sen

the Galois short exact sequence can be generalized to any "good" category, I think. Graphs are good, I believe.

Remember me

7:05 AM

hmmm...thanks!!

Balarka Sen

for example, galois theory generalizes to $\mathbf{Grp}$, say.

Samuel Yusim

what is the galois short exact sequence?

Balarka Sen

$1 \to \mathbf{Gal}(K/E) \to \mathbf{Gal}(K/F) \to \mathbf{Gal}(F/E) \to 1$

Mike Miller

i remember talking about this way back when, and we had issues getting this to work for groups. did you end up getting that figured out?

Balarka Sen

I haven't done this for arbitrary categories in general, but this thing can be generalized to Grp (galois extensions get replaced by charecteristic subgroups).

@MikeMiller The issue was that the exact sequence was not short, but only left-exact.

no, i don't think i have thought about it much. i guess the right thing to do is to extend the exact sequence to the right by derived functors.

Mike Miller

7:09 AM

careful with that word: derived functors are something you get from left-exact functors between abelian categories. i'm not certain the thing you were doing was functorial, and Grp certainly isn't an abelian category

Balarka Sen

but then, all of this is essentially the same as group cohomology (the corresponding $\mathbf{Fix}$ functor and it's derived functors are cohomology groups of groups)

Samuel Yusim

this is getting more high-powered than I want it to

Balarka Sen

yeah, i don't even know if i can extend this to the right, but seeing that Fix can be extended, I guess we can.

yeah, we need AbGrp i guess

but i am not thinking about it anymore.

@MikeMiller it's most certainly a functor.

Mike Miller

since i've forgotten everything we were talking about, from what to what, and what's the functor

Balarka Sen

actually, no, i don't think that works.

Samuel Yusim

7:14 AM

I have no idea what you guys are talking about so I guess I'll just sit this one out

Mike Miller

@Samuel no worries, neither do we

Balarka Sen

i second that

Samuel Yusim

this is the worst answer to a question I've gotten in a while then

Balarka Sen

haha. well, i can give you the details if you want.

Mike Miller

i think balarka's original answer should have been 'i think you probably can, i dunno'

Samuel Yusim

7:16 AM

honestly that's probably satisfying for now

the only thing I care to ask for clarification on is what a "good" category is in the context of one of the messages from a few minutes ago

Balarka Sen

@MikeMiller whatever the functor is, the left-exact sequence isn't obtained from applying the functor on some short exact sequence, so it doesn't work.

Mike Miller

googling galois theory of graphs suggests one wants to do it like covering spaces rather than like fields ('deck transformations' is the keyword): see here

seems pretty readable

"good" category is not defined, except possibly as 'category i can do the thing i want to do in'

Samuel Yusim

this paper is in comic sans

6

Mike Miller

and you're going to enjoy it

Balarka Sen

@SamuelYusim categories where you can define automorphism group of objects (?) and where there exists enough objects such that if $A$ is a subthing of $B$, then $Aut(A)$ is also a subthing of $Aut(B)$ (<-- nonrigorous)

what's more fun is that there is a galois short exact sequence of deck transformation groups too =)

Samuel Yusim

7:20 AM

that doesn't necessarily happen with graphs

Mike Miller

every locally small category (i.e., Hom(A,B) is a set) has an automorphism group of objects

Samuel Yusim

e.g., consider the star and the "extension" of the star given by making each leaf into a path of distinct length

Balarka Sen

neither with groups : that's why you take characteristic subgroups

i.e., it mustn't happen that there are no such graphs. you have to take graphs where that happens.

these are all analogues of galois extensions.

@Mike covering spaces of a path connected, locally path connected and semilocally simply connected space forms a category, right?

i haven't checked, but probably it does

Mike Miller

you certainly don't mean to say simply connected

yeah they do; you don't need any conditions, those are just to ensure you have a universal cover

Balarka Sen

okay.

Mike Miller

7:27 AM

if $E \to X, E' \to X$ are covering spaces, the morphisms are maps $E \to E'$ that make the obvious diagram commute

Balarka Sen

yeah

Paul Plummer

math-fail.com/2010/07/latex-comic-sans-style.html

Balarka Sen

isomorphisms are deck transformations

Mike Miller

omfg installing that package right now

Balarka Sen

the fundamental theorem they mention is just a version of the fundamental theorem of covering spaces

[i didn't know there were zeta functions attached to graphs, though]

Paul Plummer

7:31 AM

Hmm, the wiki on comic sans says that there was a study done on fonts that are slightly harder to read, showing that people consistently retained more information. Maybe research papers need to move to comic sans...

Mike Miller

i find comic sans very easy to read, it's just ugly

Balarka Sen

lol, that'd be a nightmare, @Paul

Paul Plummer

And cranks can start using arial

Mike Miller

maybe people would retain information better if we wrote our papers in wingdings

5

Paul Plummer

Well they said they used comic sans italic

Mike Miller

7:33 AM

ooooh.

Paul Plummer

"A 2010 Princeton University study involving presenting students with text in a font slightly more difficult to read found that they consistently retained more information from material displayed in so-called disfluent or ugly fonts (Monotype Corsiva, Haettenschweiler, Comic Sans Italicized were used) than in a simple, more readable font such as Arial"

Balarka Sen

that's just because they tried hard to extract each word from the ugly text.

Paul Plummer

Maybe I will try that sometime, get some tex source off the arxiv and compile it with a comic sans italic font, or one of the others mentioned.

@BalarkaSen That was my first thought

lol

Remember me

hi@Julian

Mike Miller

lmfao i was googling around and someone was on the microsoft support forum because they turned their global font into wingdings and couldn't figure out how to turn it back

god bless this mess

Julian Rachman

7:36 AM

Hello @Rememberme

Haha

@PaulPlummer hi, btw

hiiiiiiiiii

Hello.

hello, @Julian.

7:44 AM

@BalarkaSen Hi, saw you are getting sick again

usukidoll

anyone out there know Euclidean Geometry?

Julian Rachman

@Balarka Hello. How are you going?

usukidoll

it's about circle inversion

Balarka Sen

@PaulPlummer yeah, but this time I guess it's just because of the heat.

Paul Plummer

That sucks. Are you dehydrated or is it just too hot

Balarka Sen

7:46 AM

way too hot. i think it hit 40 degrees today.

@JulianRachman so-so. what about you? have you been doing any math lately?

Paul Plummer

You live in hell? @BalarkaSen

usukidoll

XD

Balarka Sen

haha

Mike Miller

my hometown summers were 110-120 degrees during the daytime

Balarka Sen

that hell is called India, btw.

Remember me

7:47 AM

It really hot in india especially in west bengal @PaulPlummer

Mike Miller

the way people who work outside, e.g. construction workers, survive, was by working from 5-9 or 10 if they're lucky, packing up, coming back at 3 or 4 and working til sundown

lest they die; no point in paying the bills if you're dead

Balarka Sen

@MikeMiller that's just impossible

you're bluffing.

Mike Miller

fahrenheit

Balarka Sen

oh, haha

Julian Rachman

@Balarka Why "so-so?" Got some difficult Algebraic Topological concepts that have been irritating you? I have been a little slower lately due to exams but now I am working on compactness from Simmons

Balarka Sen

7:48 AM

i am talking about Celsius, btw.

Mike Miller

if it were 120 celcius you would be dead in minutes

i know

Remember me

@MikeMiller your blog is amazing

Julian Rachman

@Mike How's UCLA doing?

Balarka Sen

dead? you'll be dead after you turn in to vapor.

what, @Mike has a blog?

Remember me

assorteddetails.wordpress.com/author/assorteddetails @BalarkaSen have a look

Julian Rachman

7:50 AM

@Rememberme lol for my blog I was going to do something for the intersection between national poetry month and mathematics awareness month but it ended up still in my draft section.... :(

Balarka Sen

@JulianRachman No, it's just because I have been sick. altop has been fine.

Julian Rachman

I need to post more. I just dont know what to post. @Rememberme

Remember me

I have no idea about your title@julian

Julian Rachman

@Balarka Oooooo... That sucks. I hope for only the best and that you will get better soon. :D

@Rememberme ? Title of my blog?

Remember me

yes

@BalarkaSen you saw mike's blog

Julian Rachman

7:52 AM

I just made it up. Have any suggestions?

I saw his blog. He started it like 3 days ago.

there is only one post so far

Balarka Sen

@Rememberme I am having a look.

Remember me

hmm.....@JulianRachman i am good at poetry and maths but i cant find an intersection

Julian Rachman

I have a draft, Ill be posting a belated-national poetry month one later in the week.

Remember me

Well i dont have a blog anyways how do ya create one

Julian Rachman

But do you have any suggests for the name of my blog? Any takers? @Mike @Balarka @Rememberme

You just go on the internet and search "how to start a blog" and poof!

Remember me

7:57 AM

@JulianRachman Will this title suffice ........ the beauty of mathematics is the essence of a poets lyrics...

Julian Rachman

That is a bit long

and my blog is not solely on poems and math. It was only for the month that I would post something on the current month

Remember me

okay

Balarka Sen

@MikeMiller haha, I like that "giving you more examples would be problematic". i wasn't familiar with the h-cobordism theorem (though i won't pretend i happen to see the full power of the theorem).

Remember me

let me think of something short

Julian Rachman

I am talking about a title for my entire blog. The MAIN thing

Short and Beautiful @Rememberme

Remember me

8:00 AM

Abstractness of maths and poetry...@Julian

Balarka Sen

ah, i just noticed the corollary. interesting.

Julian Rachman

Ooooooo...... Check out the new theme i got: jmrresearch.wordpress.com

@Rememberme again, my blog has NOTHING to do with poetry

Remember me

FIne

Julian Rachman

This is for the title of my blog. The thing that EVERYONE is going to see when they view it

Remember me

That is just like saying "the title of my poem is "MY poem""@Julian

Julian Rachman

8:03 AM

Wait. You understand my blog is about all math right?

Remember me

Yes just giving examples :p

usukidoll

@Rememberme do you know Euclidean Geometry

Julian Rachman

So like the name of my blog is "Beyond Abstraction". I want to change that to something else.

Any takers? Please

usukidoll

Beyond Logic?

Remember me

@usukidoll Its firsts time someone has asked me on chat that do i know something or not.... lol

usukidoll

8:06 AM

o-o

Julian Rachman

@usukidoll I am not into logic

Please. Suggestions?

Remember me

@JulianRachman reverberations in mathematics

Julian Rachman

nice. done. We will try it. :)

now what should be my url now???

Remember me

We will?

Julian Rachman

I want to change that too

Remember me

8:10 AM

who else is with you

Julian Rachman

I meant myself. sorry

Remember me

no worries Peace !!!!!@JulianRachman i have to go know

Julian Rachman

Ok. bye

and Thank you

ADG

8:25 AM

anyone for differential equations or integartion?

see math.stackexchange.com/questions/1271049/…

@MikeMiller I also wish to put a blog up, but could you guide me the steps to do so?

I tried in the past but failed

Balarka Sen

hi @iwriteonbananas

ADG

@BalarkaSen he's not here, but I am

Mike Miller

9:02 AM

@BalarkaSen There's just, like, no reason it should be true (except for the fact that it is). We haven't assumed much at all and we get for free everything we could possibly want. In addition, that it's true smoothly just as well as continuously is absurd, especially because here there be dragons (aka, exotic smooth structures). But well, there we have it.

If someone tells you they're willing to give you something for nothing, they're either trying to scam you or they're talking about the h-cobordism theorem.

3

iwriteonbananas

@BalarkaSen good morning

@MikeMiller hahah

Américo Tavares

11:17 AM

@robjohn Do you know if it is possible to generate with an adequate query a list of my own questions and answers that received downvotes, detailing the number of downvotes per question/answer?
I think I have not many, but I would like to know, without having to visit all of them and see.

robjohn

@AméricoTavares are you thinking about on this site or data.SE?

Américo Tavares

@robjohn This one.

But it could be on data.SE as well.

robjohn

@AméricoTavares I don't think you can sort the list of your questions by the number of downvotes. The only way might be to use data.SE and write the appropriate SQL query.

Américo Tavares

@robjohn So I would have to learn how to write such a query.

robjohn

@AméricoTavares or find one that is already written.

Américo Tavares

11:22 AM

Of course. Where can I search, in data.SE?

@robjohn Of course. Where and how can I search for? in data.SE, I assume. I'll try. Thanks!

robjohn

@AméricoTavares Try this one data.stackexchange.com/math/query/79428/downvotes

@AméricoTavares It shows you have 11 posts downvoted

user96977

how do i solve y = ax (mod m) for x, if i can compute the multiplicative inverse u such that: yu = axu = 1 (mod m)

user96977

i'm assuming it is possible, i could be wrong...

Américo Tavares

@robjohn I think not, but I am going to check.

robjohn

@AméricoTavares doesn't that cover questions and answers?

Américo Tavares

11:38 AM

@robjohn @robjohn I think not, but I am going to check.

robjohn

@AméricoTavares I see questions and answers in that list

Américo Tavares

@robjohn Yes, you are right, 7 answers are listed. Thank you for your help!

Balarka Sen

@iwriteonbananas g'morning

robjohn

@TruthSerum You can use the Extended Euclidean Algorithm. I describe a version in this answer.

user147690

@BalarkaSen Good morning m8

Balarka Sen

11:53 AM

@MikeMiller yeah, I guess there's no intuitive reason it should be true, although it is. but I don't feel surprised since I don't know if there is an intuitive reason it should not be true.

it is powerful, for sure, now that I have seen the corollary in your blog which proves Poincare conjecture for dim $\geq$ 6

user147690

What is the explicit reason why I can say $Z(H_3(\Bbb Z))\cong \Bbb Z$

Balarka Sen

are you going to post an outline of a proof in your blog? that'd be pretty cool.

@AlexClark morning

user147690

Because the centre is based solely on one integer?

Balarka Sen

it's generated by a single element. compute the center explicitly to see it.

user147690

It is generated via $z=\begin{bmatrix} 1&0&1\\0&1&0\\0&0&1\end{bmatrix}$

Balarka Sen

11:57 AM

right, so it's an infinite cyclic group on a single generator, i.e., $\langle a \rangle$. it has to be isomorphic to $\Bbb Z$.

the explicit isomorphism is obtained from sending $a$ to $1$ and extending linearly.

user147690

And my $(\Bbb Z^2 \cong)\{(x,y)\}$ comes from the quotient map how?

Chris's sis

Greetings

user147690

When I quotient map: $H_3(\Bbb Z) \to H_3(\Bbb Z) / Z(H_3(\Bbb Z))$

Balarka Sen

@AlexClark send $\begin{bmatrix}1&a&b\\0&1&c\\0&0&1 \end{bmatrix}$ to $(a, c)$. What's the kernel?

user147690

Well it must be $\langle b\rangle$, but I guess I am in the mindset of linear operators still

Balarka Sen

12:00 PM

since you're leaving out $b$, it's the group generated by the matrix $z$ above, isn't it?

@AlexClark $b$ is not even an element of $Z(H_3(\Bbb Z))$

user147690

@BalarkaSen Indeed

Balarka Sen

so there you have the exact sequence $1 \to \Bbb Z \to H_3 \to \Bbb Z^2 \to 1$

Chris's sis

@robjohn this is one of the most beautiful integrals I ever met $$\int _0^1\int _0^1\frac{1}{(x+y) \sqrt{(1-x) (1-y)}}dydx$$

user147690

So I should explicitly state my generators at the start :\

robjohn

@Chris'ssis looks interesting...

Balarka Sen

12:02 PM

@AlexClark generators of what?

Chris's sis

@robjohn It seems very hard without using the proper approaching way.

user147690

@BalarkaSen I mean $x=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$

user147690

and the one for $y$ and the above one for $z$

Balarka Sen

oh, sure. but that follows from the definition of the heisenberg group, doesn't it?

user147690

I guess if you were doing it correctly from the start, now I know that how I was going to start it is bad:

The Heisenberg group,$H_3(\Bbb Z)$ is the group of matrices of the form:

$$\begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix}$$ Where $x,y,z\in\Bbb Z$

Balarka Sen

12:06 PM

yes, that's the definition. i am not sure with what you're having problems.

user147690

I dunno, I just haven't had enough time to become comfortable with all of the concepts :\

user147690

I'll be fine though

ᴇʏᴇs

@TedShifrin Could you elaborate on what you mean by taking the derivative using a sequence here? math.stackexchange.com/questions/913582/…

user147690

I'm still pretty terrible, but I am sure you can recall how much more terrible I was 8 weeks ago @Balarka :P

Balarka Sen

It takes time to grasp all of these new concepts, @AlexClark. If I were you, I'd have just done classification of finitely generated abelian groups (which was all linear algebra over a ring).

user147690

12:09 PM

@BalarkaSen Only just starting to learn rings now

Australia

user147690

@BalarkaSen That wasn't an option actually

Balarka Sen

oh, alright. it'd have been easier, though.

user147690

Unless it is in a cryptic name I only had:

Braid groups, Free groups, Simple groups, Solvable groups, Sylow subgroups, Central extensions, Lattice of subgroups, Jordan-Holder theorem, Feit THompson theorem, Tensor product of finite groups

Balarka Sen

well, of course you should have chosen Sylow theory.

robjohn

12:18 PM

@Australia That is true

@Chris'ssis I have a double sum, but I don't see immediately how to sum it.

Australia

z=0 is clear

But how to prove?

robjohn

@Australia it is very easy using contour integration.

Australia

Hello, is e^{-pi x^2\cdot e^{2\pi i xz}} not e^{-\pi x^2}\cdot e^{2\pi i xz}

Chris's sis

@robjohn What kind of sums?

user147690

I'm not really comfortable with the whole cyclic generator thing.

If I understand this correctly, I can write:
$$x=\begin{bmatrix}1&1&0\\0&1&0\\0&0&1\end{bmatrix}$$
$$y=\begin{bmatrix}1&0&0\\0&1&1\\0&0&1\end{bmatrix}$$
And thus, perhaps this is wrong:
$$\langle x \rangle =\begin{bmatrix}1&x&0\\0&1&0\\0&0&1\end{bmatrix}$$
$$\langle y \rangle =\begin{bmatrix}1&0&0\\0&1&y\\0&0&1\end{bmatrix}$$

user147690

12:32 PM

and therefore, pretty sure this one isss wrong:
$$<xy>=\begin{bmatrix}1&x&0\\0&1&y\\0&0&1\end{bmatrix}$$

Balarka Sen

$\langle x_1, x_2, \cdots, x_n \rangle$ is used to denote a group, with elements being combination of $x_1, x_2, ..., x_n$ (called "the group generated by $x_i$s"). Additionally, $\langle x_1, x_2, ..., x_n | \prod_i {x_i}^{k_i} = 1 \rangle$ is used to denote the group generated by $x_1, x_2, ..., x_n$ with elements having relations $\prod_i {x_i}^{k_i} = 1$

This is called presentation of a group.

$\langle a \rangle$ means the group generated by $a$ with no additional relation. That means, the underlying set of the group is $\{\cdots, a^{-2}, a^{-1}, a, a^2, \cdots \}$

And multiplication is obvious. You can see that this is isomorphic to $\Bbb Z$ (how?).

robjohn

@AlexClark what is $\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}^2$?

Balarka Sen

What you're writing, on the other hand, is a matrix, not a group, @AlexClark.

user147690

@robjohn It is sadness :P

Balarka Sen

Some more examples, if you prefer : $\langle a | a^n = 1 \rangle$ is isomorphic to the cyclic group $\Bbb Z/n\Bbb Z$. $\langle a, b\rangle$ is the free group on two generators, with no additional relations. $\langle a, b | ab = ba\rangle$ is isomorphic to $\Bbb Z \times \Bbb Z$ (prove all of this)

robjohn

12:37 PM

@AlexClark indeed

user147690

So what I want to write is:

$$\langle x,y,z|z=xyx^{-1}y^{-1},xz=zx,yz=zy\rangle$$

Balarka Sen

@Alex I presume you haven't done much about presentation of groups?

user147690

@BalarkaSen Precisely one question on the past assignment

Balarka Sen

Yes, well, that's the Heisenberg group. I have no idea why you think $\langle x \rangle$ is that matrix.

It's a group, not a matrix.

user147690

Well I thought $\langle x\rangle $ generated that matrix

Balarka Sen

12:40 PM

$\langle x \rangle$ is a group, so I don't get what you mean when you say "it generates something". Also, what do you mean by "generating a matrix"?

user147690

@BalarkaSen nevermind I misunderstood it

Balarka Sen

so what'd be the correct statement?

user147690

For which thing?

Balarka Sen

about $\langle x \rangle$?

Write down $x$ and the underlying set of the group $\langle x \rangle$

user147690

Well it's a single element that is non-cyclic in it's field, so by fundamental theorem of finitely generated abelian groups it has no Torsion and thus is isomorphic to $\Bbb Z$ or something close to that lol

Balarka Sen

12:44 PM

No tricks, do it the hard way. You don't know what a field is, nor the fundamental theorem.

user147690

I do know what a field is lol and the fundemental theorem is all I know on this

user147690

But okay haha, I'll do it soon

Balarka Sen

also, i don't know what you mean by "single element that is noncyclic in it's field"

underlying set of $\langle x \rangle$ is not the integers either.

it's just isomorphic as a group to $\Bbb Z$.

user147690

Okay I won't say anything more until I do more reading, thanks heaps

user96977

1:16 PM

@robjohn thanks

user96977

but it still does not explain how to /solve/ the equation, given that i can compute the multiplicative inverse

Ted Shifrin

1:34 PM

hi @AlexC, @Balarka, @robjohn

Balarka Sen

hello, @Ted.

Ted Shifrin

mr eyeglasses, suppose $x_n\to a$ and $f(x_n)=x_n$ for all $n$. What is $f(a)$? What is $f'(a)$?

I won't ask you how quantum chemistry is going, @Balarka.

Balarka Sen

I am not fussed about quantum chemistry, really.

Ted Shifrin

ok ... glad to hear you're at peace. :)

Balarka Sen

not particularly... I am sick, once again.

Ted Shifrin

1:37 PM

oh damn

Balarka Sen

It's the heat, I think.

It hit 40 degrees (in celsius) today.

Ted Shifrin

I was in Europe one time when it was something like 42º day after day after day, and I was walking around in it.

Balarka Sen

yeesh.

Did you get back ok?

Ted Shifrin

LOL ... I'm still here ... and that was more than 15 years ago (yikes).

Make sure you drink lots of water, @Balarka.

Balarka Sen

Yeah, I am gonna take care of that. Been diluting glucose and pouring them in glasses all day.

How's your day, @Ted?

Ted Shifrin

1:43 PM

Just starting ... I have to agonize over final grades in my diff geo class. Not pretty.

Balarka Sen

aw, that's always a pain. how did they do at it?

Ted Shifrin

Not impressively. Or impressively poorly. What galls me is that two people in particular, who needed to pass with a C to graduate, couldn't even be bothered to come to the review session where I worked half the problems on the exam for them.

Balarka Sen

This batch in your uni seems to be particularly bad, seeing that you talk about their poor grades a lot in here.

Ted Shifrin

I didn't follow that.

Oh.

I can no longer motivate many students to work their butts off ... One of the reasons I decided to retire.

Balarka Sen

probably it's just the class, but I respect your decision.

Ted Shifrin

1:49 PM

No, it's not an isolated phenomenon. Lots of people are observing this.

Balarka Sen

I see. Well, I guess people do get old.

Ted Shifrin

Never mind.

Balarka Sen

I am sure you'll have lots of fun after retirement :)

ADG

@TedShifrin integration. can you help?

Ted Shifrin

Depends, @ADG.

ADG

1:52 PM

@TedShifrin here math.stackexchange.com/questions/1271049/…

@TedShifrin what do you say?

Ted Shifrin

I'm looking. I don't believe your physics is correct.

ADG

@pjs36 hi, would you laso take a look at math.stackexchange.com/questions/1271049/…?

Ted Shifrin

@ADG: It's bad form to ping everyone and ask them to do this ...

ADG

@TedShifrin hmm

I can't disagree

@TedShifrin BTW it matches somewhat with the options too.

Ted Shifrin

So, you solved for the friction force. And then used $F=m dv/dt$. But the friction force is not the net horizontal force, is it?

You should be doing this in polar coordinates, anyhow, shouldn't you?

ADG

1:59 PM

@TedShifrin actually it is the only force that will do some work since all rest forces are perpendicular and will not do any work and I am using the expression/definition for tangential acceleration, i.e. $a_t=d|\bar v|/dt$ and only frictional force is tangential to the path.

@TedShifrin that'll be overkill anyways and I haven't been taught about them (but I know them.)

Ted Shifrin

Oh, I was taught to do such problems by looking at radial and tangential components in polar coordinates.

ADG

seems like the've used the integral $$\int\frac1{\sqrt{x^2+a^2}}=\ln|x+\sqrt{x^2+a^2}|+{\rm constant}$$

Ted Shifrin

That's not right. It's an arctan.

ADG

@TedShifrin typo

Ted Shifrin

Oh.

ADG

2:02 PM

ain't it?

Chris's sis

@ADG I have a nice integral for you.

Mats Granvik

Has anyone ever done any significant mathematics using the Floor, Round, Ceiling or Modulo function?

Ted Shifrin

Yeah, that looks right.

ADG

@Chris'ssis not now.

@Chris'ssis I instead have one for you.

Chris's sis

@ADG $$\int _0^1\int _0^1\frac{1}{(x+y) \sqrt{(1-x) (1-y)}}dydx$$

ADG

2:02 PM

sorry BTW.

@Chris'ssis you know na that I only study at highschool

Chris's sis

@ADG high school knowledge is enough :-)

ADG

@MatsGranvik yes

Mats Granvik

@ADG Convince me.

ADG

@Chris'ssis $x=\sin^2\theta; y=\sin^2\phi$

Chris's sis

@ADG Awesome ... (not sure if it works though) :-)

ADG

2:04 PM

@MatsGranvik $x=\lfloor x\rfloor + \langle x\rangle$

Chris's sis

@ADG Not much difference I think.

Ted Shifrin

Oh, @ADG, I get it.

They're asking for the distance travelled. Can you get to the distance without first solving for $v(t)$?

ADG

@TedShifrin stupid me. stupid,stupid.

I'm solving for time :(

Ted Shifrin

Not stupid.

ADG

@TedShifrin should've used : $$F=\frac{mv{\rm d}v}{{\rm d}x}$$

Ted Shifrin

2:07 PM

Right. The old chain rule trick.

ADG

@TedShifrin I never studied 'bout it in any maths books. thanks to physics btw :D

Ted Shifrin

I call it the "energy trick" in my differential geometry class, because, yes, it shows up a lot in physics.

ADG

@TedShifrin since you helped me, why don't you frame an answer for it and get it accepted.

@TedShifrin you're a physicist too I guess?

Ted Shifrin

No, I'm a mathematician, but I like physics :)

I try to teach a fair amount of basic physics intuition in my geometry and multivariable classes.

ADG

@TedShifrin good

Do you also have some familarity with matrices and determinants?

Balarka Sen

2:10 PM

Can't think of a mathematician who doesn't know matrices and determinants, @ADG :P

ADG

@MatsGranvik better you tell me what you want to solve or need help with.

Balarka Sen

Of course he does.

ADG

@BalarkaSen you too maybe.

(Irony)

Anyways the help is needed for math.stackexchange.com/questions/1270925/…

@robjohn isn't $\rm I^2=I$?

Balarka Sen

@ADG the only way to do it is to explicitly write out the determinant.

horrendous.

ADG

no plz. I beg.

Mats Granvik

2:12 PM

@ADG No I just wanted to know in general, but more specifically I have many times started with the Floor or Modulo function and tried to mimic it with Fourier series, only to conclude that it is not good mathematics or that it will tell me nothing about the problem I am looking at.

ADG

I'm feared of solving determinants by explicitly expanding them

Ted Shifrin

OK, @ADG, answer written.

@ADG: You're making up rules as you go along. You're telling us that $\det(A+B)=\det A+\det B$. Very false. :)

ADG

@TedShifrin thanks

Ted Shifrin

I would suggest using row operations.

Balarka Sen

@ADG Well, expanding isn't that hard. Use the cofactor formula, or whatever it's called.

ADG