Mathematics - 2015-12-05 (page 2 of 2)

r9m

14:00

@OFFSHARING I have a simple enough proof too :) but I need a bit of MZV algebra :| (nothing severe though)

Huy

@Khallil: being both open and closed is called clopen.

robjohn

@OFFSHARING Pretty good, but very busy at work lately. How about you?

Huy

@Khallil: so you know it's open, and the complement of any open set is ... ?

OFFSHARING

@r9m hehe, yeah, that is a good tool. Did you try the variant with $\zeta(3)$?

Khallil

Closed which means $V$ is closed.

Huy

14:01

exactly

Khallil

(If we're looking at the complement of $\varnothing$ in $V$.)

GBeau

putnam in 1 hour ^^

r9m

@OFFSHARING which variant?

OFFSHARING

@robjohn I see. Not bad, I'm working on my book and trying to do some research at the same time.

GBeau

I was part of my university's highest-ever scoring team last semester so I got pressured into taking it again (I only made a 2!)

but the department is buying us pizza during the break between sessions so it's okay

OFFSHARING

14:03

@r9m you have inside the brackets $\zeta(3)-1-1/2^3-1/3^3\cdots$ for the same problem. :-)

r9m

@OFFSHARING yea indeed :) :) won't be too different

OFFSHARING

@r9m These days I was working on something similar but very special, it required some kind of symmetry hard to note. However, the evaluation seems coming from another world. :D

r9m

@OFFSHARING okay! :)

OFFSHARING

@r9m Wait for my book. :-)

r9m

@OFFSHARING I patiently am :D

OFFSHARING

14:06

@r9m You won't imagine the stuff in it. ;)

No more saying.

r9m

@OFFSHARING as you wish! :) But seriously, what was the series again? I'm dying to remember :( I mean the one you showed me in chat .. the one with H_n^{3} thing

OFFSHARING

@r9m A long series with generalized harmonic number.

Precisely.

I'm out to buy some food for the dogs. Then I'm back to my work. I'm still connected until I'm back and then I'm off.

Huy

@OFFSHARING: my gf just got dogs. very cute =)

r9m

Okay .,. I'll lay more eggs in the mean time :(

OFFSHARING

@Huy Yeah, dogs are real friends, never fake ones. :-)

Huy

14:11

very much so. :)

Balarka Sen

14:30

@FrankScience I'll bookmark that message, I don't think I can seriously think about math while sick. Thanks.

Huy

@BalarkaSen: You're sick again?

Balarka Sen

I mentioned that yesterday.

Huy

I didn't read every single message you wrote yesterday.

Balarka Sen

Here you go.

Huy

Thanks for the reference.

Balarka Sen

14:32

:p

@Huy Have you learned the proof of Geometrization yet?

Huy

nope, trying to solve some exercises on Lie groups atm

as said, Lie groups first. :P

Balarka Sen

meh

Huy

what

they're cool too

Balarka Sen

I am not disagreeing, simply expressing frustration at "Lie groups first".

Huy

@BalarkaSen: what was the original motivation for outer semi-direct products?

Balarka Sen

14:37

what's the "outer" semidirect product? not familiar with that terminology.

Huy

@BalarkaSen: maybe you call it just the semi-direct

Balarka Sen

semidirect products are there to classify group extensions.

Huy

you take two groups G, H and some homomorphism $H \to Aut(G)$ and define a product on $G \times H$ using the homomorphism

Balarka Sen

yeah, I just call it the semidirect product.

Huy

ah, ok

Balarka Sen

14:38

I guess inner means $G$ and $H$ are subgroups of some bigger group.

Huy

splitting results in some isomorphism to a semidirect product, right?

Balarka Sen

do you know what a short exact sequence is?

Huy

yea

Balarka Sen

$0 \to N \to G \to H \to 0$ be short exact sequence of abelian groups.

if there is a map $s : H \to G$ such that composing with the 3rd map in the sequence gives me identity, $s$ is said to be a section, and the sequence is said to split.

splitting lemma says if you have a section (i.e., if the sequence splits) then $G \cong N \oplus H$.

Huy

ah

Balarka Sen

14:41

but this fails horrendously for groups in general. what holds is that if you have a section, then $G$ is semdirect product of $N$ and $H$.

Huy

I remember vaguely

aha

Balarka Sen

so, $G$ has a nice presentation in terms of the presentations of $N$ and $H$. that's the point.

Huy

yea

I think my prof used that to compute certain fundamental groups

with a similar sequence

Krijn

You could try this with $Q_8$ and the center of $Q_8$ and then look for $H$. It's the example that made it clear to me

Balarka Sen

fundamental group of what?

Krijn

14:43

Because $Q_8 / Z(Q_8) \cong D_4 / Z(D_4)$ and $Z(Q_8) \cong Z(D_4)$ but of course not $Q_8 \cong D_4$

Huy

@BalarkaSen: $SL_n(R)$ and $SL_n(C)$

Balarka Sen

good example, @Krijn. but unfortunately, $Q_8$ is not a semidirect product of anything.

Krijn

That was my point @BalarkaSen

Balarka Sen

no exact sequence with middle term $Q_8$ splits.

ok, I see.

@Huy gross. but you can do this by the Gram-Schmidt process.

Krijn

Btw, this is why $D_4$ and $Q_8$ have the same character table

Balarka Sen

14:45

ah.

@Huy curious: what kind of exact sequence did he use?

Huy

@BalarkaSen: $\pi_2(G/H) \to \pi_1(H) \to \pi_1(G) \to \pi_1(G/H) \to \pi_0(H)$

Balarka Sen

oh, the fibration long exact sequence

well, that's not very short

Huy

in our case the first and latter are trivial

Balarka Sen

ok.

r9m

15:03

that's how people are cutting trees these days =P

Huy

15:15

haha, standard solution of some of my exercises:

In order to show that $\Phi$ is a homomorphism of rings, it suffices to show that $\Phi$ obeys the product on the generators $\{1,i,j,k\}$. This works and is supertedious to tex.

Alec Teal

Okay, I have one of those "what is the name of this property" questions, it's left the main page now (so I probably wont get an answer) forgive the shameless bump: http://math.stackexchange.com/questions/1561000/is-there-a-name-for-this-property-on-sequences
(property is: $$\exists x\in X\forall\epsilon>0[|B_\epsilon(x)\cap (x_n)_{n=1}^\infty|=\aleph_0]$$)

robjohn

15:32

@AlecTeal $\left(x_n\right)_{n=1}^\infty$ has a limit point

Alec Teal

Thanks, I'll keep that as a fallback.

(The context isn't topological and I really want the language of hovering of lingering)

Paradox 101

16:55

@Huy when proving whether a space is a metric space we just have to prove that the associated metric is positive, symmetric and follows the triangle inequality right?

Huy

yeah, and that $d(x,y) = 0 \iff x = y$

Paradox 101

Ok and what does the '#' mean? What does it refer to exactly?

Huy

number of elements

Paradox 101

so that means that $d(x,y)$ is the number of elements such that the sequences $x_n$ and $y_n$ are not equal?

Huy

no

$x_n$ and $y_n$ are elements of the sequences $x$ and $y$

you take two sequences $x$ and $y$ and then you count how many elements differ

Paradox 101

17:03

sorry i still don't get it

what exactly will be the metric here?

Huy

$d(x,y)$ is the metric

say you have two sequences $x = (1,2,3,4,5,5,5,5,\dots)$ and $y = (5,5,5,5,5,5,5,5,\dots)$

then $d(x,y) = 4$

(don't use this notation)

Paradox 101

Ok. Given that $x_n$ and $y_n$ are not equal then wouldn't that make the metric infinite?

Huy

seems like it

so maybe they allow $\infty$

or maybe there's something about $X$ that I don't know

Paradox 101

since they are not equal then this won't be a metric space as it violates the $d(x,y)=0$ if $x=y$ condition?

Huy

no, $d(x,y) = 0 \iff x = y$ is clearly satisfied

Paradox 101

17:09

But here the metric is defined so that $x$ and $y$ are not equal

i should check the other conditions too

Huy

no?

are we reading the same part?

Paradox 101

I mean in the definition of the metric it's said that $x_n$ and $y_n$ are not equal

Huy

no

look at my example above again

you didn't understand it yet

Paradox 101

In your example there are 4 elements which are different and the rest are the same i.e. the 5's

so the metric is 4

but since no element of sequence $x$ equals $y$ here then the metric would go on and on. But since the set $X$ is a sequence of bounded sequences, the metric should also be bounded for it to be a metric space. But here the metric is not bounded

Huy

why is $X$ a sequence???

why should the metric be bounded?

Paradox 101

17:16

Sorry I meant to write $X$ is a set of bounded sequences

but if the set is composed of bounded sequences, shouldn't the metric also be bounded?

Huy

why?

Paradox 101

I'm not basing this on any actual definitions, just wondering

I mean when we generally speak of metric spaces we just include a metric with a set. For them to be well a metric space shouldn't they share a set of properties?

Mike Miller

Morning.

Balarka Sen

Morning, @MikeMiller.

Ted Shifrin

17:44

@Huy @Paradox: Did you actually decide if $d(x,y)$ is a real number for all sequences $x$, $y$? :)

goodnight @MikeM

oh, @Huy already asked my same question ...

Paradox 101

@TedShifrin given that the sequences belong to natural numbers, wouldn't $d(x,y)$ be a real number?

Mike Miller

Morning @Ted.

Long day today.

Mike Miller

18:06

...but short conversation.

Clarinetist

Morning everyone. It's weird to experience finals week again

Huy

hey @Clarinetist

we meet at last

Frank Science

18:34

Today I learnt a useful characterization of UFD.

Suppose $A$ is a Noetherian domain, then $A$ is a UFD if and only if for each $a,b\in A$, $(a:b)$ is principal.

Anubhav.K

@MikeMiller can you tell me any proof of this fact, if $p:N→M$ is a covering, N orientable, M nonorientable, then$ \pi_M∘\pi_1(p)=\pi_N $ where $\pi_M$ is the morphism $\pi_1(M)→Z/2$, sending loops to the orientation of the basepoint according to lifts in the orientation double cover.

Frank Science

As a corollary, if $(A,\mathfrak m)$ is a Noetherian local domain and its completion $\hat A$ is a UFD, then so is $A$.

Mike Miller

18:55

That notation is terrible. First note that the RHS is zero. So the point is that a loop in N goes to an orientation-preserving loop in M. I don't see why this is true off the top of my head, though it's intuitively clear.

Surely it's bene done somewhere on MSE.

Frank Science

Personally I think we can only assume that $p$ is a local homeo and without any assumption on orientable.

L33ter

20:01

I want to ask something about this proof

how did we get that line ?

L ?

I don't understand so here the sketched line is the plane

that we constructed

i.e the one that is orthagonal to u

and now we measure the area -->

is L the line that intersects with the plane $R^2x1$?

Huy

L33ter

I spit on your theorem

hahaha

@Huy

do you want to check my problem above

I just want to make sure I am correct

Huy

@L33ter: what do you not understand?

L33ter

so here how we understand we construct $f_i

$f_i(u)$

first we get a point $u$ in $S^1$ and then we construct that plane as we did above

and then what we do afterwards we consider the line L that intersect the plane $R^2x1$ and that plane P

and then we measure the area that is on the same side as u

i.e the line L is the one that we just begin in order to intersect the plane P and the plane $R^2x1$

i.e we look at the first point of intersection and then we draw a line L

is that correct?

Huy

20:20

@L33ter: use \times not x, it's really confusing

L33ter

alright

yeah I think my intuition is correct

Huy

@L33ter the most important thing is just Borsuk Ulam imo

L33ter

yeah

I understand that perfectly the proof of it

Huy

Borsuk Ulam gives you existance of a point $u$ such that the areas are halved

but not a construction

L33ter

yeah

but you see

we want to see that such L exist

Huy

20:23

yeah

I know

I'm just pointing out that even though the proof looks like we're constructing something, we don't

L33ter

yeah

we are proving

that there exists such L

Huy

it's a bit confusing that they in the beginning call such a line $L$ and then later $L$ is the intersection of the plane with $\mathbb{R} \times \{1\}$ which as seen in the construction in general is far away from being $L$

L33ter

but we need L as $f_i$ depends on L

as we measure the area that is in the same direction as u

so in the picture above it is -->

Huy

yeah, but it's in general a different $L$ than the one we are trying to find

L33ter

yeah

so for example

because each $f_i$ depends

Huy

20:27

even though in the very first sentence of the proof they call the line we are trying to find $L$ too

L33ter

on the point $u$

no, we prove such L exist we don't have like a explicit formula to find it

Huy

I think we're talking past each other

I'm not objecting to the proof, just to the use of the same name for two different things, which can be very confusing.

L33ter

yeah

I agree with you

MGA

What are good books providing a rigorous treatment of multivariable and vector calculus? Is this equivalent to differential geometry?

Huy

from my experience, the two you mention are usually done exclusively on $\mathbb{R}^n$

Frank Science

20:52

@MGA Henri Cartan, Differential Forms and Differential Calculus.

Mike Miller

21:04

You have a very strange choice of books.

@MGA One of the regulars here would probably suggest his own book, Ted Shifrin's "Multivariable Mathematics". I would not say that what you're talking about is "equivalent" to differential geometry.

Balarka Sen

Shifrin's book is great, I second that.

Frank Science

@MGA and Dieudonné, Foundations of modern analysis.

and the second volume of Zorich's Mathematical Analysis also covers this.

Mike Miller

21:29

anyone have access to this?

nevermind

Ted Shifrin

21:54

@MikeM @Balarka: Thanks for the plug. People definitely have a misconception about what differential geometry is. A lot of people think a first graduate course on differentiable manifolds is differential geometry; it is not, IMHO. But a rigorous treatment of linear algebra/multivariable calculus is a prerequisite for anything about manifolds.

@MGA — if you want to learn an undergraduate differential geometry course, studying curves and surfaces in $\Bbb R^3$, a solid understanding of linear algebra (linear maps, eigenvalues/eigenvectors) and basic multivariable calculus (directional derivatives, line integrals) is good enough to get started.

@Frank: Dieudonné, much as I love it, is very much a graduate text for everything in it. Super super sophisticated and terse.

skill patrol

Hello Professor @TedShifrin

How are you?

:-)

L33ter

22:12

@TedShifrin I saw couple of your videos yesterday

its perfect

user174558

Hi @Nickolas!

Ted Shifrin

22:44

hi @skull @skill

skill patrol

How's the bridge coming along? @TedShifrin

Clarinetist

G'afternoon everyone

Mike Miller

Proctoring for the next three hours. Wonder if I can finish a paper.

SAWblade

Probably. :0

skill patrol

G'day mate :-) @Clarinetist

Frank Science

23:06

@TedShifrin Thanks for the comment. Here there's a course for third year undergrads, named after Topologie et Calcul Différentiel, covers similar to that of Dieudonné's book, including ODEs in Banach spaces. I don't know, however, whether this is the right way to get with these things. Apparently in China we didn't learn things at this abstract level.

Daeto

hey there guys

anybody here?

evinda

Hello!!!

Could I ask you something?

If we have the function $f(x)=e^{-150 \left( x-\frac{1}{2} \right)^2}$, how can we find its 1-periodic extension?

We have that $x \in [0,1]$.

@anon @robjohn Do you maybe have an idea?

Michael

23:30

hello!

How would you do integral of lnx/x^2

Indefinite integral*

Danu

Hi guys

Mike Miller

Morning.

Michael

Hello guys

I would like some help with the $\int (lnx)/(x^2)$

Mike Miller

Parts.

Michael

Can you work it out infront of me please?

anon

23:35

try your hand first

Michael

I know how to do it by parts

how about u substituion?

anon

if you know how to do it then why do you need help?

Michael

I need help with a different technique

Owatch

$u = ln(x)$, $du = \frac{1}{x}dx$, $dv = x^{2}$. $v = \frac{1}{3}x^{3}$

Michael

Another person said u substitution was much faster and easier. He didn't explain how to do it htough

anon

23:37

that seems like something you should have said to begin with

Owatch

Maybe try that?

Michael

@Owatch Ok thanks! Is it possible with U substition?

Owatch

dunno

Maybe, not focused on it.

Michael

$\int \frac {lnx}{x^2}$

Owatch

I don't see a good u substitution.

Michael

23:42

"@Michael I'm going to presume that the natural log contains only x. Yes, this integral is incredibly simple. Do a u-substitution of 1/x. du will then equal ln(x). If you wish to challenge yourself and do it a really long way, you could alternatively integrate by parts, which is essentially doing a double substitution of u and dv. Then, the integral of udv = uv - the integral of vdu. This comes from the multiplication rule for derivatives."

Owatch

$\frac{ln(x)*x^{3}}{3} - \int{\frac{x^{3}}{3x}}$dx

Easy

Michael

$\int \frac {1}{u^2}$ ? Would that be a good way to look at it? Or is that wrong?

Mike Miller

If u=1/x then du=-1/x^2 dx. Whoever you're talking to should be more careful b

anon

@Michael if u=1/x then du will be -dx/x^2, not ln(x). so we get int ln(1/u)(-du), which means you have to integrate ln(u)

Ted Shifrin

@Frank: I probably should have specified standard undergraduate in the US. Harvard has an infamous course (in which a handful of students spend 30-40 hours a week on the class) that is very much at the level of Dieudonné.

Michael

23:46

in Calculus and analysis, Nov 26 at 4:36, by TheGreatDuck

@Michael I'm going to presume that the natural log contains only x. Yes, this integral is incredibly simple. Do a u-substitution of 1/x. du will then equal ln(x). If you wish to challenge yourself and do it a really long way, you could alternatively integrate by parts, which is essentially doing a double substitution of u and dv. Then, the integral of udv = uv - the integral of vdu. This comes from the multiplication rule for derivatives.

Mike Miller

The source doesn't change my response.

Ted Shifrin

@Michael: If it had been $\int (\ln x)/x\,dx$, that person would have been correct.

Michael

@TedShifrin oh ok. So it is impossible with u substition?

Ted Shifrin

That person you just quoted wrote crap, by the way. Read it. If you substitute $u=1/x$, he's telling you $du=????$. What do you think?

evinda

@TedShifrin Hi!!! Could I ask you something?
If we have the function $f(x)=e^{-150 \left( x-\frac{1}{2} \right)^2}$, how can we find its 1-periodic extension?
We have that $x \in [0,1]$.

Michael

23:48

@TedShifrin Yes it makes sense now. So only integration by parts then

Ted Shifrin

@Michael: You can substitute $u=\ln x$, but you'll still have to do an integration by parts in the end.

anon

@Michael at some point in the integral computation you must replace int(blah) with just blah, so it requires having memorized some integrals. none of the basic ones I have memorized allow a direct u-substitution solution.

Michael

@anon Thank you for your assistance sir! I have another question. Is there a benefit of hyperbolic substitution in an integral instead of a trigonometric substitution?

Ted Shifrin

@evinda, what do you mean by finding its $1$-period extension? That function satisfies $f(0)=f(1)$, so it makes sense that it has a $1$-periodic extension. You want a formula?

howdy @anon

anon

yo

Ted Shifrin

23:51

ponders @anon as a "Sir." :)

user174558

I had a weird dream. In it, I was a woman sleeping with a man.

anon

@Michael I imagine there are certainly situations where hyperbolic does what trig does not, but from the perspective of complex variables they're the same thing

Ted Shifrin

no comment @Jasper

Michael

@anon What is a complex variable? Sorry I am fresh of Highschool math.

user174558

@TedShifrin How many turkeys did you eat on Thanksgiving?

anon

23:52

a variable that is assumed to represent a complex number

Michael

@anon what field of math are complex numbers really common in?

anon

if you look up the hyperbolic trig functions you will see they are just the usual trig functions but "rotated" by multiplication-by-i. complex numbers are common in, of course, complex analysis, and also anything that could use it. also relevant in e.g. linear algebra.

Ted Shifrin

@anon @Michael For something like $\displaystyle\int\frac {dx}{x\sqrt{x^2-1}}$, hyperbolic trig is a lot nicer than regular trig. And it shows up in physics/geometry applications very naturally. Look up catenary.

barely epsilon, @Jasper.

@Michael: Complex numbers show up in every part of mathematics, pretty much.

GBeau

just got out of the putnam

Ted Shifrin

And calculus with complex numbers/functions, too.

How was it @GBeau?

GBeau

23:54

I thought A1 was the easiest Putnam problem I've ever seen, but maybe I was just primed to solve it

evinda

@TedShifrin It does not hold that $f(x+1)=f(x), \forall x \in [0,1] $. This will hold for the period extension, right? How can we find this?

Ted Shifrin

I haven't seen this year's exam, since I'm no longer teaching. I guess I'll google for it.

GBeau

it might not be online yet

I think the west coast is still taking the exam right now

Mike Miller

It won't be online for a few days.

anon

which one was A1 again?

Ted Shifrin

23:55

What do you mean by find, @evinda?

Oh, @MikeM, usually there are places that post 'em immediately, just not MAA.

user174558

@anon Rotman's Advanced Modern Algebra, Third Edition, Part 1 has just been published.

GBeau

@anon did you take it?

anon

just define $f(x)$ to be $f(r)$ where $r$ is $x$'s fractional part, no? @evinda

@Gbeau yeah

Michael

Is there a 4th dimensional coordinate system in math?

GBeau

@anon the triangle bounded by the hyperbola

Ted Shifrin

23:56

As many dimensions as you want, yes, @Michael, including infinitely many.

anon

there are coordinate systems in any number of dimensions

@GBeau oh yeah. I actually went back and did that one after I did a different one.

Ted Shifrin

<--- goes to bake an apple tart and leaves anon and Mike in charge

speaking of hyperbolic trig functions ... :)

anon

the one I wanted to solve was the matrix with arithmetic-progression-entries one

GBeau

@anon yeah I thought A1 was unusually easy

evinda

@TedShifrin f(x-floor(x)) satisfies this condition. But how do we deduce that this is a 1-periodic extension of f?

Michael

23:57

What happens when you try to take the limit of an equation with more than one variable? Ak.a multiavariable equation. Is it even possible?

GBeau

it's practically a simple calculation relative to the rest of the putnam problems I've gone through

Ted Shifrin

Just check directly @evinda. Plug in $x+1$ for $x$.

evinda

@anon Could you explain it further to me how we extend like that f to be 1-periodic?

GBeau

but like I said, maybe it's just because I've done so many minimization/maximization problems

so I was primed to solve that one quickly

anon

@evinda geometrically, it should be obvious. the act of moving x left or right by the period 1 until you get it into [0,1) is just the act of taking the fractional part.

user174558

23:59

I feel that Michael is trolling, like last week.

Michael

Im not

Transcript for

$Mathematics$ Mathematics