Mathematics - 2014-05-30 (page 1 of 4)

Mike Miller

12:02 AM

hi

Hi @Fernando

hm

@FernandoMartin hai

sup @Pedro

btw @lolwut I was just kidding about ignoring you pal :-)

Pedro Tamaroff

12:08 AM

@skullpatrol pls he clearly a troll?

lol wut

@skullpatrol As I guessed when you said that you were ignoring me...

skullpatrol

Yep :D

lol wut

@PedroTamaroff What are you saying 'bout me?

skullpatrol

2 hours ago, by skullpatrol

It's all good in the hood :-)

Trolling serves a purpose.

lol wut

Pedro Tamaroff

12:13 AM

@lolwut you heard me homie

lol wut

@PedroTamaroff i dun herd u. i jus aint understandizing bro

skullpatrol

You need to learn your "street smarts" too...

...better here than out on the real street :-)

imo

lol wut

ya, imagines skullpatrol out on the streets

skullpatrol

it's a jungle out there pal

►

Mike Miller

@Ted yo

Studentmath

12:21 AM

Say, I have to find the enclosed space by $(x-2y+3)^2+(3x+4y-1)^2=100$, and I subsitute as recommended $u=x-2y , v=3x+4y$. I get a circle equation: $(u+3)^2+(v-1)^2=100$, why can't I immediately say the space enclosed is the space enclosed by the circle, which is only dependent on the radius?

Ted Shifrin

Howdy mr @Pedro

Studentmath

Which should be $\pi*r^2$ which is wrong here..

Ted Shifrin

Because you made a linear (not orthonormal) change of basis, @Studentmath. Compare our earlier discussion.

Pedro Tamaroff

@TedShifrin Hello Ted.

Studentmath

Hm, I see- I have to take into consideration the jacobian then.

With orthonormal change of basis, I wouldn't have to worry, correct? The axes simply rotate together

lol wut

12:23 AM

It's like saying, I am skinny if you draw me that way.

Ted Shifrin

Anything good today? @Pedro

Ayup @Studentmath

Mike Miller

@Ted I have a vector field $X$, and $\phi_t$ is its flow. Can you convince me that $$\mathscr L_X = \phi^\ast_1-\phi^\ast_0$$ as maps on closed $n$-forms?

Studentmath

Thanks Prof. @Ted :)

Mike Miller

Sorry - not closed forms, just forms

Pedro Tamaroff

@TedShifrin Continued with some Ramsey Theory in combinatorics. Looks cool.

Ted Shifrin

12:25 AM

Yes, quite cool, @Pedro.

That's wrong @Mike, so I shan't try.

Mike Miller

It didn't seem right. Is it close to something right?

Ted Shifrin

It seems like you're doing an integral of Lie derivative as $t$ varies from $0$ to $1$?

It's related to some of these chain homotopy formulas that show up in the Poincaré lemma.

Mike Miller

Ah, I see what you mean. That would probably not work.

Ted Shifrin

Shrugs unknowingly

Mike Miller

@Ted Actually I'm trying to construct a chain homotopy between $\phi_0^\ast$ and $\phi_1^\ast$. Using $\i_X$ as the chain homotopy seemed like the obvious choice!

Ted Shifrin

12:31 AM

You're doing this on $X\times I$?

Mike Miller

@Ted Whatddya mean? Where's the $\times I$ come in (other than that I'm assuming the flow generated is global, I guess)?

Ted Shifrin

You need to think about $$\int_0^1 \frac d{dt}\phi_t^*\omega dt.$$

Mike Miller

OK.

I'm convinced I want an $i_X$ in there too but I'll fiddle. Thanks!

Oh! I see the idea. Thanks!

Ted Shifrin

Or think about the homotopy formula in general (adjoint to $\partial(\sigma\times I$)).

Mike Miller

wat

eXtremiity

12:46 AM

Btw Mike I figured it out !

" To prove that two spaces are not homeomorphic, it is sufficient to find a topological property which is not shared by them."

Look at this beauty:
http://en.wikipedia.org/wiki/Topological_property#Compactness

I did not know this......But it makes sense now :))) So happy.

Oh and Mike, my question was could I transform that open unit disk into that open strip using a mobius transformation of some sort.

I don't know why I am asking that. I just thought that it may have

Mike Miller

Oh. Nope. Möbius transformations send circles to circles, so the boundary of the image of your disc under a möbius transformation will be some half-plane or interior/exterior of a disc.

eXtremiity

I see, thanks.

Mike Miller

I should clarify, they send circles in the Riemann sphere to circles in the Riemann sphere. Aka, they send (lines and circles) to (lines and circles). Circles can get sent to lines, lines to circles.

lol wut

2:09 AM

Call me an idiot, but I never knew that $f(f^{-1}(X))=X\cap\mathrm{im}(f)$ until I realized it just now.

robjohn

@lolwut it's a good thing to know.

lol wut

@robjohn It trivializes the statement $f$ is surjective if and only if $f\circ f^{-1}$ is the identity on the power set of the domain

robjohn

okay

Studentmath

2:45 AM

Gah, I really suck at getting the bounds right.

Sush

3:13 AM

@robjohn, please help me with these doubts.

I got a nice answer for the question I asked you on that day, just I am some small steps apart, so please help!

robjohn

@Sush what is the problem?

Sush

The problem I have written in this comment.

@robjohn

I don't get all those 4 doubts answered.

skullpatrol

@lolwut I would never call you an idiot :-)

Sush

@robjohn, ok, I got 4th doubt, it is because probability integral transformation, itself.

lol wut

@skullpatrol aww :D

Sush

3:47 AM

Please someone let me know why $G^{-1}[F(X)]\leq z$ implies that $[F(X)\leq G(z)] \cap [G^{-1}[F(X)] > z]$ is a subset of $F(X)=G(z)$?

skullpatrol

4:17 AM

@PedroTamaroff how are you finding the heavier course load?

Martin Sleziak

6:00 AM

Quote from a book by Aliprantis and Border: The "classical" proof uses some results from the theory of infinite continued fractions, which is not widely taught these days.

That makes me wonder whether once continued fractions were widely taught (whatever it could mean).

skullpatrol

Finite continued fractions are widely taught.

David Kirby

continued fractions have become quite popular in programming in recent years as a computationally efficient (and often more accurate) alternative to the traditional lancsoz/stirling approximations

stieltjes was the man

lol wut

6:42 AM

DAYUM! I be workin' hard ter find uh answer to a question and when I post it... no brothas givin' me no upvotes

David Kirby

i'd be hesitant to assign too much meaning to the votes here. HOWEVER, with a name like 'lol wut' you should get up votes purely on principle.

skullpatrol

What "principle" would that be?

David Kirby

lol wut?

skullpatrol

yep

lol wut

skullpatrol prefers to take things literally. On the other hand, who doesn't?

skullpatrol

6:49 AM

(note the username of the questioner)

David Kirby

(scanning through russell's 'principles of mathematics' looking for a joke)

(and failing)

skullpatrol

better to have tried and failed than never to have tried

David Kirby

oh here we go : § 499. Are propositional concepts individuals?

(failed)

Jasper Loy

8:21 AM

@BalarkaSen A

Balarka Sen

@JasperLoy Ahoy.

Jasper Loy

I am waiting for lhf.

lol wut

Here you go @JasperLoy, math.stackexchange.com/questions/814648/…

Jasper Loy

@lolwut That is too tedious, lol

Balarka Sen

8:40 AM

@JasperLoy Try for some high hangings sometimes.

@lolwut What does your username mean? "LOL WHAT" or "LOL WOOT"?

lol wut

First one, traditionally.

Balarka Sen

That makes things clearer.

lol wut

I am a non-traditionalist, myself.

skullpatrol

you need a question mark then

lol what?

Balarka Sen

@skullpatrol Yes, and in the latter case you need '!'. lol wut!

skullpatrol

8:56 AM

or both :-) lol! wut!?

Balarka Sen

Not yet, what about the order of ! and ?? ? after ! or ! after ?

Better, $? \!\! \text{!}$

Yes, this looks good.

skullpatrol

Are ? and ! commutative?

@skullpatrol No.

@skullpatrol No!?

?! $\neq$ !?

9:00 AM

Of course ?! is not the same as !. Dunno why you're asking that!

skullpatrol

maybe ?=1

the identity element for questions

Balarka Sen

Well, topologically '?' and '!' are the same. Continually deforming both gives two non-connected points.

@N3buchadnezzar What are you studying nowadays?

N3buchadnezzar

Astrophysics and point set topology / geometric topology

lol wut

@BalarkaSen loll, loooloo:lolll ':' olo ':' olo llo lolo. loll:llolll oololl:lo ooll o:lol tlo lolllollolloo oo:lll.

Balarka Sen

@lolwut Free semigroup on 2 generators?

lol wut

9:09 AM

@BalarkaSen

monoid

Balarka Sen

Yeah, my bad.

lol wut

three generators

Balarka Sen

@lolwut Oh, I ignored ':'.

@N3buchadnezzar Yuck point-set topology.

Jasper Loy

@BalarkaSen What got you interested in math?

Balarka Sen

@JasperLoy To be frank, quadratic equations.

Jasper Loy

9:11 AM

@BalarkaSen LOL. And I still have not learnt how to solve the cubic and quartic

Balarka Sen

@JasperLoy Yes, what I asked myself is this : can you do the same with cubic, quartic and higher degrees?

I found out the answer was yes for the first two and no for the others. That was pretty confusing (without any intuitive explanation and all), so I learned galois theory.

Jasper Loy

I think all undergrad courses should teach the cubic and quartic

They teach galois theory but not that, sad

Balarka Sen

@JasperLoy Don't they?

Jasper Loy

@BalarkaSen They usually don't

Balarka Sen

@JasperLoy Solving cubic and quartics are a part of galois theory.

Jasper Loy

9:14 AM

Look in all the abstract algebra texts.

Balarka Sen

@JasperLoy Dummit-Foote gives it.

Jasper Loy

Really? Hmm

Ilan Aizelman WS

Hey all! Simple question math.stackexchange.com/questions/814672/…

Balarka Sen

@JasperLoy Yes. What book are you reading?

Jasper Loy

@BalarkaSen I looked through many algebra texts, maybe I did not look carefully enough. I will be using Cohn's 3 volumes for algebra.

Balarka Sen

9:22 AM

Never read it. Try standard texts like Dummit-Foote.

Jasper Loy

@BalarkaSen As for reading right now, I am not reading anything. I hope to start reading next year when I am better

Balarka Sen

@JasperLoy Haven't you said the exact same thing the year before this one?

Jasper Loy

@BalarkaSen Yes, I did, but I am still very unwell, so I will say this as many times as required

@BalarkaSen I have said this for many years

@BalarkaSen Solving my mental problems is as hard as proving FLT

Balarka Sen

@JasperLoy Ha, that gave Wiles some mental problems.

Jasper Loy

@BalarkaSen I want you to know that I am not delaying for fun. I want to get well more than anything in the world, but my wounds are too deep to mend easily

Balarka Sen

9:26 AM

I understand that. My tone was not snarky, just for the information.

Jasper Loy

My stress has caused half my hair to drop.

skullpatrol

9:42 AM

@JasperLoy I just created a lhf for you pal :-)

Patlatus

Hi all. First time in se chats

Do you discuss anything interesting in chats usually?

Welcome

Morning all.

hi

10:08 AM

Morning everyone!

I've got a problem with a fairly basic integral if you overlook a few things. However, I'm not willing to overlook the square root of a square.

$$ \begin{aligned} \int \dfrac{x^2}{\sqrt{x^2 - 1}} \text{ d}x \ \overset{x = \cosh u}= \int \dfrac{\cosh^2 u \cdot \sinh u}{\left| \sinh u \right|} \text{ d}u \end{aligned} $$

How would I deal with the $\left| \sinh u \right|$ in the denominator for the indefinite integral?

Jasper Loy

@skullpatrol I just answered.

skullpatrol

@JasperLoy icic

Balarka Sen

@JasperLoy You are forgetting something.

Jasper Loy

@BalarkaSen What thing?

Balarka Sen

"lol"

skullpatrol

10:18 AM

@JasperLoy if I edit my mistake there will be no question :-)

Jasper Loy

LOL

skullpatrol

should I edit?

Jasper Loy

@skullpatrol Not necessary

Shisui

@BalarkaSen @skullpatrol @JasperLoy Anyone? :-)

Balarka Sen

I'd rather sub $x = \cos(u)$.

Martin Sleziak

10:25 AM

@Shisui $\cosh u$ is not a bijection. So you are probably taking $x=\cosh u$ for $u\ge 0$.

Shisui

What's a bijection in simple terms?

@MartinSleziak

Martin Sleziak

For such u's you have $\sinh u\ge 0$ and you do not have problems with absolute value.

:-)

one-to-one

and onto

10:26 AM

If you want to make a substitution, you want to have for each $x$ exactly one $u$.

Shisui

So substitutions should be one-to-one functions or complications occur?

@MartinSleziak

Ah I see!

Thanks!

Martin Sleziak

Or for each u exactly one x. Depending on the direction you're going.

Shisui

What kinds of complications occur if our substitution is not one-to-one?

@MartinSleziak

Martin Sleziak

Well, at least for definite integrals you have one problem: How to define range after the transformation.

For example if $x=\cosh u$ and $x$ is in the interval $[1,2]$, you have two possible choices for endpoints after the transformation.

You are right that you do not really need the function to be bijective.

Jasper Loy

Answered another lhf, but no votes...

Hippalectryon

10:31 AM

@JasperLoy what's a lhf ?

Jasper Loy

@Hippalectryon Low hanging fruit = easy question

Martin Sleziak

But if there are two (or more) possibilities you can simply choose one of them.

And it seems I was wrong in what I was telling you about your integral. I should be more careful.

David Kirby

Unless it's tedious I usually just try both-- usually it's very clear which direction you want to go.

Martin Sleziak

Domain for $1/\sqrt{x^2-1}$ is $[-\infty,-1]\cup[1,\infty]$.

What I wrote only works on $[1,\infty]$. Since here you can use $x=\cosh u$.

So if we are working only on the interval $[1,\infty]$, then you can use $x=\cosh u$ and choose $u\ge 0$, which simplifies things @Shisui

But since the function is even, dealing with it on $[-\infty,-1]$ should not be much of a problem.

Shisui

That makes sense!

@MartinSleziak

Thanks

G.T.R

10:44 AM

@Shisui are you Extremity?

Shisui

@G.T.R Perhaps ...

Balarka Sen

@G.T.R. Can you recall the book in which you found the problem regarding quintics?

i.e., the problem of inversion of the system of elementary symmetric polynomials?

Shisui

So would that mean that the integral is defined over $|x|>1$?
@MartinSleziak

Jasper Loy

I think I will keep my accounts for life this time, lol

Balarka Sen

That's a good thought.

skullpatrol

10:54 AM

Keep the good thoughts.

Shisui

I'm still really shaky on understanding this

For the indefinite integral of $$ \dfrac{x^2}{\sqrt{x^2 -1}}$$ we can use the substitution of $x = \cosh u$ where $u \geq 0$ for the interval $x \in \left[ 1, + \infty \right)$ but which substitution would we use for the interval $x \in \left( -\infty, \ -1 \right]$ ?
@MartinSleziak

Balarka Sen

Here's a problem for everyone : Given an arbitrary function $f : \Bbb R \to \mathcal{A} \subseteq \Bbb{R}$, and $f'$ defined as the derivative of the function $f$ as $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ Is it always possible to compute $f'$ without going through the computation of the limit in the definition above (first principles)?

An example at hand is $f = \sin : \Bbb R \to [-1, 1]$.

G.T.R

@BalarkaSen the author was Ulam Something

@BalarkaSen this books.google.fr/books/about/…

Balarka Sen

@G.T.R Ulam? Not possible the one related to Ulam spiral?

G.T.R

Yes he himself wrote the book

And he also made breakthroughs developing the H-bomb

Balarka Sen

11:11 AM

@G.T.R Stanislaw Ulam! My god!

Martin Sleziak

@Shisui Let us denote your function by $f(x)$. Suppose that we are able to solve the problem for $[1,\infty)$, i.e. that we can find the primitive function $F$ such that $F'(x)=f(x)$ for each $x\in[1,\infty)$.

Now we can notice that $f(x)=f(-x)$.

So if we take $G(x)=-F(-x)$ for $x\le-1$,, then $G'(x)=F'(-x)=f(-x)=f(x)$.

So from the primitive function on $[1,\infty)$ you can get primitive function on $(-\infty,-1]$ simply by using the fact that you want the primitive function to be odd. (The original function was even.)

Of course, then you can add constants, so you can get $G(x)+C_1$ and $F(x)+C_2$.

@Shisui But if you insist that you want to try to do this by substitution, you could use $x=-\cosh u$.

Clearly, $-\cosh u$ attains exactly the values in the interval $(-\infty,1]$.

Shisui

The primitive function definition is much better than sledge-hammering another substitution.
@MartinSleziak

What do you mean by adding constants?

Martin Sleziak

Primitive function is determined up to a constant.

If $F(x)$ is primitive to $f(x)$, then so is $F(x)+C$.

Shisui

What's the exact definition of primitive?

Martin Sleziak

Of course, you can use one constant on $(-\infty,-1]$ and another one on $[1,\infty)$.

$F(x)$ is a primitive function (antiderivative) of $f(x)$ iff $F'(x)=f(x)$.

I guess antiderivative is a more common name in English speaking countries than primitive function...?

Shisui

11:26 AM

Yep

Martin Sleziak

ok, I'm off to lunch. See you later! (And have fun with your integrals!)

Shisui

So what we've done is found a primitive function for which our integral is satisfied over a certain interval and used the fact that the function is even to find another primitive function for our original function over the other interval as a result of symmetry?

Enjoy your lunch! I'll be on later too :3

Will do!

@MartinSleziak

Martin Sleziak

Yes, that's more or less what I was trying to explain above.

Shisui

Nice!

:-)

Balarka Sen

@G.T.R. It is truly amazing to see how many people were involved in theory of equations, isn't it?

Charlie

11:53 AM

hi @skull

skullpatrol

@Charlie hi

Charlie

@skullpatrol :) how are you?

G.T.R

I didn't know grumpy cat was a girl

Charlie

yup

skullpatrol

@Charlie fine thanks, how are you?

Charlie

11:55 AM

@skullpatrol fine

skullpatrol

:-)

Jasper Loy

@Charlie Are you getting married soon? lol

Sush

Please please someone help me with these doubts, no. 1. & no. 2.

Charlie

@JasperLoy I don't know

@skullpatrol (-:

Sush

:-)(-:

Charlie

11:57 AM

hehe

Sush

@Charlie, what is רק אהרון מבין, אולי אחרים?

@Sush it's hebrew

Ok! @Charlie

@Sush :D

)8< >8(

11:59 AM

haha

Chris's sis

Greetings

Charlie

hi chrissy

Chris's sis

@robjohn are you around?

@Charlie Cat!!! How are you doing? :-)

Charlie

@Chris'ssis fine fine, and you?

Sush

I think main has gone slower :(

Chris's sis

12:00 PM

@Charlie Not that bad. Trying to compute something here ...

Charlie

@Chris'ssis hmmmm.....

G.T.R

Is Chris'ssis also a girl ?

Chris's sis

@G.T.R You never know ...

skullpatrol

how many sis = sisters are boys?

Chris's sis

;)

G.T.R

12:02 PM

Come on, it's the second evasive answer in the chat today

Sush

@Chris'ssis, will you please help me here only with doubt no. 1 & 2? Please! I have been stuck there for two days!

Charlie

@skullpatrol twisted sisters....are they girls?

Chris's sis

@Sush I'll look at it a bit later. Now I need to finish something.

skullpatrol

@Charlie no, they are twisted :)

Sush

@Chris'ssis, ok.

Charlie

12:04 PM

@skullpatrol then just because it has the name doesn't mean it follows the definition

skullpatrol

true

&

false

at the same time :D

Sush

I think main is very slow now-a-days! When I first came to SE, only 5 minutes were taken to be answered. Now, I have to wait for at least half-a-day!

Chris's sis

Oh, I had to regulate a polylog identity because of the sign ...

Chris's sis

12:29 PM

►

This song inspires me ...

Charlie

@Chris'ssis :D

Chris's sis

@Charlie :D

skullpatrol

Old John inspires me :D

15 hours ago, by Old John

Just thought I would drop in to let people know I am still alive :)

Charlie

haha

@skullpatrol >.<

skullpatrol

@Charlie the man still does hill walking

Charlie

12:43 PM

@skullpatrol yes

Charlie

1:02 PM

@skullpatrol bye skull

skullpatrol

@Charlie bye chucky :-)

Jasper Loy

1:28 PM

Wow my lhf are getting me plenty votes

Studentmath

2:06 PM

Gah. I want to find the Area of the face of some ball, between the linear spaces $z=h_1$ and $z=h_2$, I have no idea how to bound it. I know each gets me a circle when it meets the ball, but.. no idea.

I -think- each gets me a circle.

Sush

2:21 PM

Please someone help me!!!

@MartinSleziak, please help me with this

@DanielFischer, please let me know why (not necessarily strict) monotonicity of $G$, we have $G^{-1}[F(X)]\leq z \Rightarrow F(X) \leq G(z)$ here?

Daniel Fischer

2:36 PM

@Sush By monotonicity of $G$, we have $$G^{-1}[F(X)] \leqslant z \implies G\left(G^{-1}[F(X)]\right) \leqslant G(z).$$

Studentmath

Hrmpf. http://www.wolframalpha.com/input/?i=integral+of+1%2F%E2%88%9A%28%28%E2%88%9A%28a%5E2-y%5E2+%29%29%5E2-x%5E2+%29
Shouldn't it be $\arcsin(\frac{x}{\sqrt{a^2-y^2}})$

Sush

@DanielFischer, sir, still not getting! How monotonicity implies that?

@DanielFischer, sorry, got it!

Daniel Fischer

@Sush Monotonicity is $a \leqslant b \implies G(a) \leqslant G(b)$. (Of course we're talking about monotonically nondecreasing functions here, for monotonically non-increasing functions, the second inequality is reversed.)

Sush

@DanielFischer, thank you so much.

Daniel Fischer

You're welcome.

Sush

2:43 PM

Will you please solve one more doubt? I have been stuck there for two days! @DanielFischer

Daniel Fischer

I can try. What?

Sush

I have asked it here

@DanielFischer

Studentmath

I think I should switch to polar coordinates.. gah, I need more experience

Sush

@DanielFischer, Why due to monotonicity of $G$, $F(X)<G(z)\implies G^{-1}[F(X)]\leq z$?

Studentmath

I need to find the surface area of a ball, $x^2+y^2+z^2=a^2$, blocked by the planes $z=h_1$, $z=h_2$, $0\le h_1 \le h_2 \le a$. Should I switch to polar co-ordinates? Going with cartesian, I have no idea how to get the right bounded area R to do it over

I can do the first integral and it got me out nicely, but Then I am stuck as I have no idea how to bound the area correctly, so no idea what to put in for the second integral to solve..

Daniel Fischer

2:54 PM

@Sush $[F(X)\leqslant G(z)]$ is the union of two disjoint parts, $[F(X) < G(z)]$ and $[F(X) = G(z)]$. For the former, we have $G^{-1}[F(X)] \leqslant z$ as a premise, so $$\begin{align}[F(X)\leqslant G(z)] \cap [G^{-1}[F(X)] > z] &= \left([F(X) < G(z)]\cap [G^{-1}[F(X)] > z]\right) \cup \left([F(X) = G(z)]\cap [G^{-1}[F(X)] > z]\right)\\ &=[F(X) = G(z)]\cap [G^{-1}[F(X)] > z]\\ &\subset [F(X) = G(z)].\end{align}$$

robjohn

@Chris'ssis I am now. What's up?

Chris's sis

@robjohn I forgot what I wanted to ask you. :-)

Daniel Fischer

@Sush If $y > z$, then by monotonicity we have $G(y) \geqslant G(z)$. Therefore $G(y) \neq F(X)$, and that means $y \neq G^{-1}[F(X)]$.

Transcript for

$Mathematics$ Mathematics