Mathematics - 2015-05-06 (page 1 of 3)

because you can write it as $\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{\sqrt{\sqrt{3}}}$ which looks like $\frac{1}{\sqrt{2}}\frac{x}{\sqrt{x}}$ which by the trick becomes $\frac{1}{\sqrt{2}}\sqrt{x}$ which should be $\frac{1}{\sqrt{2}}\sqrt{\sqrt{3}}$ when we substitute back $x = \sqrt{3}$.

usukidoll

I heard differential geometry is hell

Ted Shifrin

Not really.

usukidoll

I wonder if anyone for Euclidean geometry is online... I've done a problem... I just need it checked

Ted Shifrin

Everything I Teach is hell. But I quit.

flakmonkey

How, in a proof, do you speak in a formal way of a particular element of a tuple?

usukidoll

12:12 AM

dear lord... even ^ doesn't work
http://math.stackexchange.com/questions/1269021/seperable-differential-equation-question#1269021

nsanger

The first/second/third/... element?

Mike Miller

can you give context? I don't know what you mean.

Samuel Yusim

"the $i$th coordinate" is what I might say

Mike Miller

oh, nevermind, I see

yeah $i$th term, $i$th element, $i$th coordinate, or even replace $i$ with your favorite other letter, all work

flakmonkey

If I'm trying to prove there is an element of set A whose first coordinate is the second coordinate of some element of set B... it seems slightly informal to speak of first, second, third, etc. coordinates if you think of coordinates as sets (using say Kuratowski definition)

Samuel Yusim

12:14 AM

solution: don't think that way

flakmonkey

yeah but I feel like at some point I should convince myself that in principle it can be done

Ted Shifrin

Ugh @ coordinates as sets

Better as maps

Samuel Yusim

better solution, I guess: come up with a definite operation to extract the ith coordinate of a set

nsanger

Maps are sets too!

Paul Plummer

Done with your stuff @AlexClark?

user147690

12:15 AM

@PaulPlummer Not yet

usukidoll

describe the rotation represented by the unit quaternion $ u =\frac{1}{6}(3+i+j+5k)$
but when I distribute the $\frac{1}{6}$ I get $ u = \frac{1}{2}+\frac{1}{6}i+\frac{1}{6}j+\frac{5}{6}k$ isn't that a 120 degree rotation because when I do $2 \theta = 120 \rightarrow \theta = 60$ $cos 60 = \frac{1}{2}$?

nsanger

Woops....

usukidoll

lol

user147690

@pjs36 Yes you were right, thanks. Seems to be working out now

user147690

1:24 AM

So if my eigenvalues $\alpha,\beta$ aren't in $\Bbb F_p$ they are in $\Bbb F_{p^2}$

user147690

1:49 AM

The following is what I can't follow(continuing from above): So they are distinct of the form $\alpha = a+b\sqrt{D}$ and $\beta = a-b\sqrt{D}$ where $a,b\in\Bbb F_p$

user147690

Where $D$ is described by saying:

$D\in F_p$ is not a perfect square in $F_p$ and thus we pick a square root of $D$ and call it $\sqrt{D}$

user147690

Then $\{1,\sqrt{D}\}$ is a basis of $\Bbb F_{p^@}$ over $\Bbb F_p$

user147690

So this is an adjoint field $F_p[\sqrt{D}]$ I guess, but I don't get where this $\sqrt{D}$ comes from

user147690

How do I choose this $\sqrt{D}$?

pjs36

I'm glad you got the other part sorted out! I guess the idea is that your characteristic polynomial, if the matrix is in $GL_2(\Bbb F_p)$, is quadratic, right?

user147690

2:03 AM

Yep $\lambda^2 - (a+d)\lambda + (ad-bc)=0$

user147690

And using the quadratic formula I obtain $\pm$ something outside the field, is that the idea?

user147690

$$\frac{a+d\pm\sqrt{(a+d)^2 - 4(ad-bc)}}{2}$$

pjs36

I'm terrible with field extensions, but I imagine that's the idea: the quadratic formula should still work, but we may not have a perfect square under the radical, exactly

Australia

3:41 AM

Let $A_{n\times n} $ be Normal matrix,and such $A^2=-I$,can we find this inverse of $(A'A+I)$?

wellington

3:52 AM

Hello, how's it going?

I have a question about tracking the path of the Sun, specifically the hour angle.

I'm trying to calculate the zenith angle, but my value is changing too rapidly. And I don't know how to correct it. ---this is for a JAVA program, but the math is where my problem is.

I've been using this page for reference: en.wikipedia.org/wiki/Solar_zenith_angle

Is anybody able to assist?

Remember me

@Australia if i am not wrong you can always find an inverse if the determinant is nonzero

Hi professor @BalarkaSen

:p

user147690

@Balarka Could you tell me what you would cover in a 5 minute presentation that introduces central extensions?

Remember me

4:08 AM

@AlexClark i think balarka is not here....

user147690

@Balarka Currently my order is:

1) General terminologies, just defining mono/epi morphisms
2) Defining an exact sequence
3) Defining an extension and what makes it a central extension
4) Haven't done yet, but possibly straight onto Discrete Heisenberg group

Mike Miller

5 minutes is not very long. What you have is probably more than enough.

user147690

Well 1-3 are less than a page, so 5 minutes is probably 1.5 pages right?

Mike Miller

Why not practice giving your talk? It's only five minutes.

That'll tell you how long it is. Try to get someone to watch

user147690

I will when I do 4

Semiclassical

4:09 AM

practicing is the essential thing, yeah

otherwise you're just working in the dark as far as how the talk actually functions as such

pjs36

Personally, @AlexClark, I think a definition is only as good as why I should care about it. I've made the mistake of going too technical, and ignoring examples that tell people why they should think what I'm talking about is cool :) So I'd weight interesting examples more heavily that being super-precise and covering every inch of the terminology

user147690

Yes that is true, but he wanted us to follow roughly:

3 minutes define,
1 minute motivation
1 minute example

pjs36

Oh wow! OK, fair enough... weird

user147690

And my define isn't 3 min

Mike Miller

@AlexClark: You're underestimating how long it will take you to do this.

pjs36

4:12 AM

Then I guess you should follow your (or his) heart :P

Remember me

What are you defining@AlexClark

user147690

@Rememberme Central extensions

Mike Miller

Especially if you've never given a talk before, it's very, very difficult to guess the timing.

Remember me

you are giving a talk??

wow....

user147690

@MikeMiller Yes true, never done it

Australia

4:14 AM

@Rememberme,I have prove this determinant is nonzero,Now I can't find this (A'A+I) inverse

user147690

@Rememberme Just for class lol

user147690

Is this a good start?

user147690

imgur.com/vLjRpft

user147690

Have I missed anything important in defining or gotten anything wrong?

Mike Miller

If you're not using anything categorical about epis or monos, why bother naming them?

Just talk about injections and surjections.

user147690

4:16 AM

Just a shorter way of saying injective group homomorphism

user147690

So I don't have to keep saying it pretty much

Mike Miller

I think it will only serve to confuse if people haven't seen it before. You gain nothing from defining it.

user147690

Okay fair point

Remember me

Well i think my professor is again wrong here ....
Whats the domain of $\sqrt{x^2+|x|-2}+\sqrt{x^2+16}$....@pjs36 i found that the domain is $[-4,-2]\cup[2,4]$ but my professor says it is an open interval on -4 i dont understand how?

Mike Miller

When you're giving your definitions, give simple examples, and explain why they are examples. For instance, for exact sequence, $1 \to \Bbb Z \to \Bbb Z \to \Bbb Z/n\Bbb Z \to 1$ is a good example.

The audience will be lost without something to ground what you're saying.

user147690

4:18 AM

Okay, so I should throw that in while still defining

Remember me

i tried plugging 4 and i dont get an undefined value...god knows how he is getting it??

pjs36

Good advice, @MikeMiller, I was thinking the same thing. Definitions are scant on their own; you can fly through them in 2 seconds of nervousness. A good example of everything will be enlightening, and give you good pacing

Australia

@Rememberme,I think is [ wolframalpha.com/input/…

this domain is $\ge\sqrt{17}$ when $x\in R$

Remember me

@Australia my domain is absolutely right then

@Australia am i right

Christopher

Hey. Does anyone find 2.48105... familiar?

Mike Miller

4:24 AM

no

pjs36

@Rememberme The second square root doesn't affect anything; you just need to solve $x^2 + |x| - 2 \geq 0$, and WA says it's that. By hand you'd just need to consider two cases to find the zeros and behavior; it's a bit of a pain, but not undoable

Remember me

@pjs36 so am i right

user147690

So after saying what an exact sequence is you think I should say that?

Mike Miller

think back to how you began to understand exact sequences

it certainly wasn't by looking at the definition until you 'got' what they were about, right? it was by looking at examples

user147690

Actually it was from memory

user147690

4:27 AM

Or maybe not, I think Paul taught me them on reflection

pjs36

@Rememberme Nobody looks right to me, which makes me nervous. It looks like the domain is $|x| \geq 1$, according to this input from Wolfram Alpha.

user147690

Oh yes, now I recall!

user147690

It was learning about semi-direct products

user147690

Balarka randomly explained them to me

Remember me

@pjs36 i missed a detail in the second part it is $-x^2+16$

pjs36

4:33 AM

Well that definitely makes a difference. Then it looks like the answer would be the intersection of two domains, leading to $1 \leq |x| \leq 4$.

Remember me

@pjs36 i really feel like kicking myself now another mistake it is -|x| sorry!!!!!!!!!!!!

pjs36

OK, then yes, I agree with what you have

pjs36

4:55 AM

Wow, I had no idea central extensions had anything to do with cohomology (a word I have seen, but know nothing about). So that's why we care...

user147690

an extension is central if the normal subgroup $N$ of $G$ lies in the center of $G$ for:

$$1\to N \to G \to H \to 1$$

user147690

When it says lies in the center does it mean it can be the center AND/OR any normal subgroup that is a subgroup of the center

Mike Miller

the latter

the point is that $H$ needs to act trivially on it (which is why we want it in the center)

@pjs36: Cohomology is a word meaning a huge number of things. Saying that something relates to cohomology is almost analagous to saying something is related to groups; in the sense that everything is (to both!)

user147690

the latter meaning it can't be the center?

user147690

[Sorry I was confused if you meant latter as in latter to the options AND/OR, or if you meant the latter statement is the only true one]

Mike Miller

4:59 AM

no? we want $N$ to be in the center; it can be a proper subgroup or it can be the center itself

oh, I never saw the AND/OR

sorry

pjs36

Got it, @MikeMiller. Pretend I said something intelligent then :P

Mike Miller

I wasn't scolding you, if that's what it sounded like! I was just pointing out that it shows up everywhere.

At the level of definitions, cohomology is just an operation you do to something called a chain complex. And chain complexes show up when you're thinking about just about anything; topological spaces, smooth manifolds, group actions, schemes (and various things about schemes), dynamical systems, knots, the intersection of Lagrangian submanifolds of symplectic manifolds...

and so you see this thing everywhere

(homology is the covariant version; cohomology is contravariant; sometimes one seems to be more natural than the other)

pjs36

haha, no, but I didn't realize just how vague my statement was. But the wiki tells me group cohomology is vaguely analogous to group representations, and I do understand the importance of those, but have never looked into any of the various cohomologies

Mike Miller

Whenever you've got a G-module you can take its group cohomology, is where that comes from, I think

pjs36

Interesting, I see. I'd never seen chain complexes, but now I know they exist! And in many places

Mike Miller

5:09 AM

absolutely everywhere

pjs36

Tell me they have nothing to do with simplicial complexes; I've seen those, and they look terribly unfun to work with.

user147690

Hmm I don't understand quotient maps for matrices:

I want my $N=\begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix}$ to have:

$$I\to N \to \begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix}\to \begin{bmatrix}1&x&0\\0&1&y\\0&0&1 \end{bmatrix} \to I$$

What is the the map explicitely from $H_3\to H_3/ N$?

pjs36

OK, I take that back, but they've always seemed that way (simplicial complexes, I mean)

Mike Miller

The first ones constructed were associated to a simplicial complex :P But no, a chain complex is just a long sequence $$\dots \to C_{n+1} \xrightarrow{\partial_{n+1}} C_n \xrightarrow{\partial_n} C_{n-1} \to \dots$$ such that $\partial_n\partial_{n+1}=0$

(sequence of groups, or more commonly modules)

user147690

$N$ above is the center of $H_3$

Mike Miller

5:11 AM

one usually just writes $\partial^2 = 0$

and simplicial complexes aren't so bad.

pjs36

Very interesting; there's always more to learn, always much more to learn...

4

user147690

Hmmm $\begin{bmatrix}1&0&-z\\0&1&0\\0&0&1 \end{bmatrix}\begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix}= \begin{bmatrix}1&x&0\\0&1&y\\0&0&1\end{bmatrix}$

Remember me

math.stackexchange.com/questions/1266587/… hmm any takes

user147690

5:31 AM

0

Q: Central extension of the Discrete Heisenberg group $H_3(\Bbb Z)$

I want to use the Discrete Heisenberg group $H_3(\Bbb Z)$ as an example for a presentation on central extensions. $H_3(\Bbb Z) = \begin{bmatrix}1&x&z\\0&1&y\\0&0&1 \end{bmatrix}x,y,z\in \Bbb Z$ Now I have shown that the center of $H_3(\Bbb Z)$ is $Z = \begin{bmatrix}1&0&z\\0&1&0\\0&0&1 \end{bm...

user147690

:-)

pjs36

@AlexClark If you're parameterizing your matrices as triples $(x, y, z)$, could you just write the your map as something like $(x, y, z) \mapsto (x, y, 0)$? I realize it's not the most elegant thing in the world, and quite possibly wrong, but maybe helpful.

user147690

Can you see my explicit maps at the bottom of that post @Pjs?

pjs36

I can, but I haven't thought about your group too much, and I don't really understand what $A \mapsto -zA$ means. Is that not multiplying each entry by $-z$?

user147690

Oh woops I wrote that wrong

user147690

5:38 AM

I edited it, is that better now?

pjs36

Now there's something I'm good at, proof reading! :P

user147690

@pjs36 Haha thanks for that catch

user147690

What does $H_3(\Bbb Z)/N$ even mean for matrices?

pjs36

I still don't know what the right answer is. But I'm a little leery about a map like that, because (again, remember, I haven't been paying full attention), I thought we were trying to map into some kind of quotient group. I would expect a quotient group to look like cosets; matrices where "we don't care" what's in one (or several) of the entries

So I would sort of expect something more along the lines of $\begin{bmatrix}1&x&*\\0&1&y\\0&0&1\end{bmatrix}$, where we really don't care what's in that upper right entry.

user147690

Well $\phi(A)=Z^{-1}A$ just rips out the that $z$ variable

user147690

5:41 AM

and makes it $0$

pjs36

OK, which may indeed be isomorphic to what I have in mind

Vrouvrou

@robjohn hello, please how you find the last condition? please

user147690

@pjs36 Is $ZX = H_3(\Bbb Z)$ a good indicator for me?

robjohn

@Vrouvrou what last condition?

user147690

Yes it should be

Vrouvrou

5:45 AM

@robjohn this condition : $\frac{N-p\theta}{N-p}\gt-1$.

@robjohn ?

robjohn

@Vrouvrou $$\left(\frac Np-\theta\right)\frac{Np}{N-p}+N-1\gt-1$$

That is the exponent of $|x|$ in the integral

pjs36

Well, I do think there are a few things that need ironed out, @AlexClark, but you're getting there. I can't put my finger on the specifics, but for one: Is $\phi : A \mapsto Z^{-1}A$ well defined? If $Z$ is a specific matrix, it seems like you really have a collections of maps there. If it's meant to be a group, then what does a group to the $-1$ power mean?

Vrouvrou

why + N-1 ? @robjohn

Australia

Let $A_{n\times n} $ be Normal matrix,and such $A^2=-I$,can we find this inverse of $(A'A+I)$? @robjohn,can you help ?

pjs36

So for one @AlexClark, I propose using calligraphic font (or script; something) for groups here, because it's very tempting to confuse individual matrices with groups, when you have subgroups $Z, H$ and matrices $A$.

user147690

5:55 AM

@pjs36 Yes true, good call

user147690

I also screwed up as anon commented

user147690

Hopefully this is reparable

pjs36

I think it is, if you instead think of things in $\mathcal{H}$ as cosets $\begin{bmatrix}1&x&*\\0&1&y\\0&0&1\end{bmatrix}$, where we just don't care what's in the upper right.

Vrouvrou

@robjohn we want to see what is the condition to have that $\int_{\Omega}|x|^{(\frac{N}{p}-\theta){p^*}}dx$ converge with $0\in \Omega$

pjs36

So you're "modding out" by that entry; you just stop caring about it :)

user147690

5:58 AM

I don't know how to write that properly though

robjohn

@Vrouvrou yes. in $\mathbb{R}^N$, $\mathrm{d}x=\omega_{N-1}r^{N-1}\,\mathrm{d}r$ in polar coordinates.

Mike Miller

I mean, what pjs said is literally fine

The quotient is isomorphic to $\Bbb Z^2$, so you could write it like that, if you wanted

you could write the surjection in your sequence as $\begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix} \mapsto (x,y)$

Vrouvrou

@robjohn i really don't understand why we do polar coordinates ?

user147690

So $$I \to N \to H_3(\Bbb Z) \to \{(x,y,0)\} \to I$$

robjohn

@Vrouvrou We are integrating a function of $r=|x|$, why wouldn't we?

Vrouvrou

6:01 AM

@robjohn you change the variable $|x|=r$ ?

pjs36

@AlexClark I've secretly been thinking of your group as triples $(x, y, z)$ with a funny multiplication: $(x, y, z) \star (\alpha, \beta, \gamma) = (x + \alpha, y + \beta, z + \gamma + x\beta)$, if I can TeX blind...

If it helps

robjohn

@Vrouvrou how else would you compute that integral?

user147690

That does

Mike Miller

really the way most people would write this extension is $1 \to \Bbb Z \to H \to \Bbb Z^2 \to 1$

where the first map is $z \mapsto \begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix}$, and the second is as above

pjs36

Yeah, well we're not most people, Mike! :P

user147690

6:04 AM

Okay thanks @Mike that helps a heap

Vrouvrou

@robjohn so $\int_{\Omega} |x|^{(\frac{N}{p}-\theta)p^*}dx=\int_{\Omega} r^{(\frac{N}{p}-\theta)p^*} dr$ but i don't understand how you find dr , what is $\omega_{N-1}$ ?

Mike Miller

having a terrible time deciding what to name my blog

robjohn

@Vrouvrou that only works in $\mathbb{R}^1$

Mike Miller

my friends already shot down "Concerned about dying in the fire" and "Very good at dodging bullets"

user147690

lmao

pjs36

6:06 AM

Then are they really your friends? Those were some quality titles.

Vrouvrou

@robjohn what only works in $\mathbb{R}^1$ ?? please

Mike Miller

"That's not about math", they said, "People will think it's weird", they said

I was pretty sold on "Sordid details following" but I got a frowny face for that one

I might go with "Assorted details following"

Vrouvrou

@robjohn where are you ?

@robjohn you mean that $\int_{\Omega} |x|^{(\frac{N}{p}-\theta)p^*}dx=\int_{\mathbb{R}} r^{(\frac{N}{p}-\theta)p^*} dr$ but how to find dr ?

pjs36

Well, @MikeMiller, I would have blindly clicked on any of those. I have faith in your abilities

And I'm off to bed! Everyone upvote this question

robjohn

@Vrouvrou $$\int_{|x|\lt R}|x|^{\left(\frac Np-\theta\right)p^\ast}\,\mathrm{d}x =\frac{2\pi^{N/2}}{\Gamma(N/2)}\int_0^Rr^{\left(\frac Np-\theta\right)p^\ast+N-1}\,\mathrm{d}r$$

user147690

6:14 AM

@pjs36 Hahaha thanks pjs

user147690

@pjs36 Goodnight

Vrouvrou

@robjohn from where $\frac{2\pi^{N/2}}{\Gamma(N/2)}$ comes ?

please

robjohn

@Vrouvrou that is $\omega_{N-1}$. It is the constant needed to convert to polar coordinates. It is the surface area of the unit sphere in $\mathbb{R}^N$

Vrouvrou

@robjohn is this result is known ? where i can find it ?

user147690

@Mike I actually don't have an answer why anyone would care about my central extension for the heisenberg group, do you have any motivation for it?

user147690

6:19 AM

I have motivation for the heisenberg group, but none for extensions :\

robjohn

@Vrouvrou Look here en.wikipedia.org/wiki/Sphere#Generalization_to_other_dimensions

Mike Miller

it's an extension of $\Bbb Z$ by $\Bbb Z^2$ that's not split. finding group extensions is a hard but fundamental part of group theory; split extensions are quite easy to classify; extensions are general are not much better than impossible

central extensions are about as good as we can get in terms of extensions we can understand without going crazy. this is an example of one that's not split.

Vrouvrou

@robjohn why it is not sufficient to say that the condition is $\frac{N}{p}-\theta>0$ ? why we do ths step of change variable ?

user147690

@MikeMiller So split extensions are for semi-direct products right? And central extensions never split?

robjohn

@Vrouvrou Because you want to integrate that function in $\mathbb{R}^N$. What I give above is the formula for the integral over a sphere of radius $R$ in $\mathbb{R}^N$.

Mike Miller

6:28 AM

split extensions are semi-direct products. central extensions can split, but they don't usually

they're the 'next harder thing'

robjohn

@Vrouvrou If it can be integrated there, it can be integrated over $\Omega$

Vrouvrou

and over $\Omega$ it is the same thing ?@robjohn

robjohn

@Vrouvrou the only questionable region is near the origin. Everywhere else is a bounded function integrated over a bounded region.

user147690

@MikeMiller So with this, the maps are $(0,0,1)\mapsto(0,0,1)$ then $(0,0,z) \mapsto \begin{bmatrix}1&0&z\\0&1&0\\0&0&1\end{bmatrix}$ then $\begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix} \mapsto (x,y,0)$ then $(x,y,0)\mapsto (1,1,0)$ I am confused with the first and last map

user147690

I am confused by what the identity $1$ means at the start and end

Mike Miller

6:34 AM

$1$ is the trivial group. The first and last map is the only map to and from the trivial group.

user147690

Which is $(0,0,0)$ here?

Mike Miller

The trivial group is the trivial group is the trivial group. The identity element of your group is $(0,0,0)$.

You need to stop thinking about this as subgroups of the Heisenberg group. After all, the fourth term in your sequence is not a subgroup. It's a quotient group.

user147690

Ok

Mike Miller

(but it's emphatically not a subgroup!)

I'm heading to bed, sorry to do so as we're chatting

But 'm tired

user147690

@MikeMiller Okay goodnight, thanks for your help

robjohn

6:43 AM

@Vrouvrou do you understand?

robjohn

7:07 AM

. o O ( zzzz )

Julian Rachman

7:32 AM

Hello everyone. I have a question. If anyone is up.

Remember me

yes........

i am

Julian Rachman

Lol. How do you write notes from a textbook without copying everything word-for -word?

Remember me

What i do i first when the professor is teaching i highlight the imp points @JulianRachman Then when i have to write notes i first read the content under the highlighted points after doing this what i had learnt from the content i write it down in points Now when the points are over if you had read it correctly you can trace stuff from the paragraph which contain my points

And you dont even have to write everything....

@JulianRachman which subjects notes

Mew

hello

is anyone smart on?

Remember me

We are all idiots@Mew

Mew

7:46 AM

that's offensive

CAn someone please help with my question here: earthscience.stackexchange.com/questions/4816/…

0

Q: Why is the colour gradient in the sky at sunset on an angle?

Below is a picture of a typical sunset that I have been observing lately at an undisclosed location in the Southern Hemisphere: Given that the base of the picture is parallel to sea level, you can see that the line where the dark blue meets the red in the sky, is on an angle of elevation of ap...