Mathematics - 2018-10-25 (page 1 of 2)

Adam

03:14

one can only consider the individual elements of the two sequences for which a hypergeometric function is based to be independent variables in such a function right? and the other arbitrary parameters, being the cardinality of each of the two sequences, must also both be considered independent variables, making the total number of variables of a hypergeometric function for any two defined multisets A and B to be equal to |A|+|B|+3

Unknown MathMan

How we can show a group of order 140 has a subgroup of order 2? Sylow theorem can only say it has a subgroup of order 4.

Pig

can you produce a subgroup of order 2 when you have one of order 4?

Secret

$\Bbb{Z}_2 \times \Bbb{Z}_2$

user282639

Hey guys, could use some hlep with this question: math.stackexchange.com/questions/2968638/…

Unknown MathMan

@Pig how?

I'm only thinking about abelian property of order 4 group.

I have another question, how to search for a general question in math.SE. Like I want to search for "Does the group" __ "acts on"__ type question.

BAYMAX

03:45

How can I show injectivity of a function which takes a matrix

Like $g(B) = BC$

but $B$ is a rectangular matrix and $C$ is a square matrix

Buddhini Angelika

Z3:=Group((1,2,3));
Group([ (1,2,3) ])
gap> Z3gen:=GeneratorsOfGroup(Z3)[1];
(1,2,3)
gap> Z5:=Group((1,2,3,4,5));
Group([ (1,2,3,4,5) ])
gap> g:=DirectProduct(Z5,Z5);
Group([ (1,2,3,4,5), (6,7,8,9,10) ])
gap> Auts:=AutomorphismGroup(g);
<group with 4 generators>
gap> AutsOfOrd3:=Filtered(Elements(Auts),elt->Order(elt)=3);;
gap> ExHom:=GroupHomomorphismByImages(Z3,Auts,[Z3gen],[AutsOfOrd3[1]]);
[ (1,2,3) ] -> [ [ (1,2,3,4,5), (6,7,8,9,10) ] -> [ (1,5,4,3,2)(6,10,9,8,7), (1,2,3,4,5) ] ]
gap> s:=SemidirectProduct(Z3,ExHom,g);

*Is there a way to identify the elements in that form?

The Great Duck

04:12

help

please

0

Q: How to convert the normal vector of a point in some polygonal surface transformed onto the surface of a triangle in space?

It's complicated to explain what the transformation I'm dealing with is but basically it is a transformation that takes a 3D model and maps it to the surface of one polygon on another model. The issue is that I'd prefer to not have to recompute the normal vectors of the post-transformation model ...

Adam

04:27

are the rules for SE regarding satire accounts the same as facebook?( ie admins wont auto delete after they ping your ip on the proviso you arnt upsetting zucc)

The Great Duck

@Adam I don't even know what you mean

Adam

ok well you've never gotten bored and made a fake facebook account and taken on a persona of a person you know your friends would be really annoyed by?

The Great Duck

@Adam define fake? SE has no real names policy.

Adam

lol ok that's fine I guess that's all that's necessary for me to know

The Great Duck

@Adam If you mean sockpuppetry that's not allowed. If you mean a satire account for the purpose of posting satire questions then obviously no. If you're just taking on some weird personality that isn't your own to have fun then really that's a question of whether said personality is allowed according to the site's policies.

However

the most important thing of all

is why are you posting this on math stack exchange when there is a main meta SE that you can actually ask questions like that on?

Adam

04:36

hmm I guess because opening another tab is always an inconvenience

but at least I'm asking that already proves im reaching adulthood maturity levels and seeings im becoming so mature its not likely that I will get in a silly of the magnitude that's needed to do something like that anyway. but I just wanted to ask in case I relapse

The Great Duck

@Adam that was a plural you in my comments. For all I know you've simply observed the existence of a satire account and don't know whether to report it.

Slereah

09:17

0

Q: Does a manifold with two disjoint compact boundaries have two disjoint collared neighbourhoods?

Take a Hausdorff manifold $M$, with a disjoint boundary $\partial M$ composed of $\partial M_1$ and $\partial M_2$, such that the boundary is compact (I think this doesn't hold for non-compact one, with $\mathbb{R}^2$ with $|y| > x^{-1}$ removed as a counterexample). Is there a collared neighbour...

differential-geometry

plz halp

I am bad at topology

Pretty sure it's true but I always run into the same issue

Kemono Chen

09:28

Is it possible to use Bernoulli numbers and Euler numbers to express the $n$th coefficient of the Taylor series of $\frac 1{1+2\cos x}$?

At $x=0$

Akiva Weinberger

10:30

Just read that "the Riemann curvature tensor expresses the tidal force that a body feels when moving along a geodesic" and that "the Ricci curvature, or trace component of the Riemann tensor contains precisely the information about how volumes change in the presence of tidal forces"

How do I understand that?

In a sphere, for example, which has Ricci curvature 1 everywhere, what "tidal forces" are being felt?

Oh, I think I get it?

If I take an circular object living in a sphere and push it in a certain direction (and it continues moving under physics), its center of mass will travel in a geodesic

but most points won't

but they'll want to travel in a geodesic because of intertia

so they'll kinda just push inwards a little bit the whole time, and the object will fell a little compressive stress

This is assuming the object's not rotating, in which case it'll feel tensile stress outwards anyway 'cause of centrifugal force

@BalarkaSen Was that right or am I talking nonsense

And then I think this compressive stress will want to turn the object from a circle into an ellipse

Secret

11:03

I felt tidal forces is more complicated than sheer and stress

given the cube of the number of dimensions the riemannian curvature tensor has for a vector space of dimension n

It's as if placing a potato on the manifold, how it is squished depends on which coordinate direction it is travelling along

Mike Miller

12:45

@Akiva I think attempts to explain Ricci curvature physically are somewhat misguided. The bit about change of volume is more or less just them saying "Det is volume and trace is the derivative of determinant".

abenthy

12:55

I have a question about associative algebras. Let $A$ be an associative algebra over a field $K$ such that $A \otimes_K L$ is isomorphic to a matrix algebra $M_n(L)$ for some field extension $L/K$. If an element of $A$ is diagonalizeable in $M_n(L)$, are its eigenspaces defined over $K$, i.e. do they have a $K$-basis?

Akiva Weinberger

@MikeMiller Determinant of the Riemann curvature tensor?

(Also, it's so fortunate that Riemann and Ricci are both R. Or unfortunate. I'm not actually sure.)

(Is the R in $R_{ij}$ for Ricci or Riemann?)

Slereah

$R_{ij}$ is commonly called the Ricci tensor

Akiva Weinberger

Yeah but it's also the contraction of $R_{ijk\ell}$ so it's conceivable they just wanted to use the same notation for both

so like even if it were named the Sicci tensor or something it would still be R maybe

Moot point anyway

Slereah

fairly moot

Gentlemen write it as $\mathrm{Ricci}$

$$\mathrm{Ricci}(X,Y)$$

Mike Miller

13:13

@AkivaWeinberger No, just as operators on matrices, trace is the derivative of Det at the identity.

Slightly different at other points but basically still trace.

Ric is tr(R). That's the best intuition one can get, I think. But to make specific claims about what it "is, physically" I think is ill advised.

Read paragraph maybe 0.61 in Besse where they say something similar.

Akiva Weinberger

@MikeMiller Yes, but the trace of R is the derivative of the determinant of R at the identity? I don't understand, R isn't a matrix, how do you take its determinant

Mike Miller

...you plug in two tangent vectors and you get a matrix? Taking the trace of that is exactly the contraction you're talking about.

I am not taking the determinant of R at any point here.

abenthy

Okay, solved my own question. The matrix $\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$ is diagonalizable in $M_2(\mathbb{C})$ with eigenvalues $\pm i$ and the eigenspaces have no rational or real basis.

Akiva Weinberger

I see

Mike Miller

@AkivaWeinberger The precise reference I was looking for is Besse, Einstein manifolds, section 0.H and in particular 0.60

Unknown MathMan

14:41

I want to compute centralizer of $(13)(24)$ in $S_4$. Is there any better way than checking elements one by one?

Mike Miller

Not really as far as I know

You can save a little time by observing that if $x$ commutes with $y$ then so does $x^k$

The converse helps you see quickly that any 3-cycle eg cannot commute with that

Unknown MathMan

Ok

How you are saying powers of 2-cycles can't be a 3-cycle?

Can you show your argument please :)

MatheinBoulomenos

15:00

@UnknownMathMan conjugation in $S_n$ has simple form. For $\sigma \in S_4$, you get that $\sigma^{-1} \circ (13)(24) \circ \sigma = (\sigma(1) \sigma(3)) ( \sigma(2) \sigma(4))$

So the centralizer of $(13)(24)$ will consist of those elements $\sigma \in S_4$ that stabilize the partition $\{1,2,3,4\} =\{1,3\} \sqcup \{2,4\}$

Daminark

Yo Mathein

MatheinBoulomenos

what I mean by this is that $\sigma$ centralizes $(13)(24)$ iff $\{\{\sigma(1),\sigma(3)\},\{\sigma(2),\sigma(4)\}\}=\{\{1,3\},\{2,4\}\}$

Hi @Daminark!

Unknown MathMan

@MatheinBoulomenos by stabilize do you mean anything related to stabilizer?

Daminark

How's everything going?

MatheinBoulomenos

@Daminark pretty well thanks

Liad

15:06

hi, how can i find how much subrings $F_{p^n}$ has?

MatheinBoulomenos

@UnknownMathMan $S_n$ acts on the set of all partitions of $\{1,\dots,n\}$, it's the stabilizer of a particular partition, yeah

Liad

@Daminark hi

Daminark

Hey Liad, how's it going?

Also yo Eric

MatheinBoulomenos

@Liad hint: try to show first that every subring is a field, after that you need to know some things about finite fields

Liad

good how are you ? @Daminark

MatheinBoulomenos

15:08

How's your application stuff going? @Daminark

Daminark

Doing alright, thanks! Dedicating today to the research statement, and then implementing some suggestions for the personal statement (luckily shouldn't be too difficult)

Prob gonna talk about elliptic curves for the research statement, since I have epsilon of a case that I know a little bit more than just "I saw a facebook meme about them and they seemed cool from that", and because it allows me to talk about cryptography a bit

MatheinBoulomenos

@Daminark you should mention the facebook meme as well

Kasmir Khaan

Mathein :D

MatheinBoulomenos

Hi @Kasmir!

Kasmir Khaan

Long time no see man !

How are you ?

MatheinBoulomenos

15:15

Doing well, thanks. And yourself?

Kasmir Khaan

doing good at your new courses i hope ? :D

good good ! finally starting to understand algebra :D

MatheinBoulomenos

yeah, ANT3 is really tough

Kasmir Khaan

took me a while but getting there =P

Unknown MathMan

@MatheinBoulomenos I knew $$\sigma^{-1} \circ (a_1a_2\cdots a_k) \circ \sigma = (\sigma(a_1) \sigma(a_2)\cdots \sigma(a_k))$$, i.e; for single $k$-cycle, but here we have two 2-cycles between $\sigma$, and $\sigma$ may contain something common hence they can't commute I thought. Can you help me to understand what you have said for two 2-cycles assuming I know it for one cycle?

Kasmir Khaan

I am gonna soon start with ANT1

Daminark

15:16

What are you guys doing in ANT3?

Kasmir Khaan

fazer book

Dami !_

;D

Daminark

How's it going Kasmir?

MatheinBoulomenos

@Daminark global CFT, Galois cohomology and Galois representations

Daminark

Damn, sounds fun

MatheinBoulomenos

currently working on the cohomological proofs for global CFT

Kasmir Khaan

15:18

am good ty ! :D

mathein that sounds way super hard ._.

MatheinBoulomenos

@UnknownMathMan if $G$ is a group, then conjugation is a homomorphism, i.e. $g^{-1}(ab)g=g^{-1}ag g^{-1}bg$

Kasmir Khaan

I planning to do galois theory in jan but what is those things u just mention ?

MatheinBoulomenos

you can apply that two $a$ and $b$ two disjoint 2-cycles and then get what I said

Kasmir Khaan

Galois cohomology and Galois representations

._.'

MatheinBoulomenos

that would take a while to explain

but Galois representations are not that hard, they're "just" representations of a Galois group, you know representations and Galois groups you will encounter in january

Kasmir Khaan

15:21

okay neat :D

I notice that algebra the beginning at least

is just about knwoning the defintions properly

like really understand them

but afterwards I assume one would need alot more than that =p

but kas will cross that bridge once he is ready ._.'

MatheinBoulomenos

yeah, for the basics knowing your definitions can go a long way

Kasmir Khaan

that what I did not do last time !

anyway glad to see you back here even for a short while :D

wish u good luck on ANT3 I know you can do it! @MatheinBoulomenos

MatheinBoulomenos

thanks!

and good luck with your algebra! Galois theory is deep and has a lot of applications, although the proofs are not hard, but it brings together groups, group actions, rings and fields in a pleasing way, I'm sure you'll like it

Kasmir Khaan

thanks Mathein ! iam really looking foward for that :D

Liad

@MatheinBoulomenos thanks!

MatheinBoulomenos

15:29

@Liad you're welcome

Hi @TobiasKildetoft

Tobias Kildetoft

@MatheinBoulomenos Hi

@MatheinBoulomenos Did you see what I wrote about our centre director's reveal at the meeting with the funding body for the centre?

MatheinBoulomenos

@TobiasKildetoft no, I didn't

Liad

any ideas on how can i find a maximal proper ideal in $k[x] $ ? k is some field

Tobias Kildetoft

@MatheinBoulomenos After going through the various grants and awards people at the centre had gotten the past year, he had a slide stating that the following would have to not be repeated outside the room until noon. Then he revealed that he had just been awarded a 10M euro grant as PI.

MatheinBoulomenos

wow!

@Liad hint: an ideal $I\subset k[x]$ is maximal iff $k[x]/I$ is a field

Liad

15:36

i just thought about something

we have $m \subset k$ maximal proper ideal

we can take $m[x]$

what do you think ? @MatheinBoulomenos

Semiclassical

Hmm. Consider the dot product between two n-vectors. It’s a symmetric bilinear function, and it’s invariant under a rotation of both arguments.

Is it the only such function on n-vectors?

(Up to an overall multiplicative constant)

Mike Miller

@TobiasKildetoft Can I have some?

Tobias Kildetoft

@MikeMiller Well, do you do mathematical physics?

Semiclassical

(I think it just boils down to Schur’s lemma?)

MatheinBoulomenos

@Liad if $k$ is a field, then there no ideals except for $(0)$ and $k$

Liad

15:51

@MatheinBoulomenos ahh. right !

@MatheinBoulomenos well i already knew your hint , can i have another one ? ^^

MatheinBoulomenos

it's hard to give a hint without spoiling it

Tobias Kildetoft

@Liad So if you could do your idea, what would be two sets that generate that ideal?

Mike Miller

@TobiasKildetoft That depends who you ask, I think

loch

Put the word quantum in your work and pretend its physics :p

Tobias Kildetoft

@Liad It would not be if $k$ was not a field. But just drop the "maximal" part and think about my question.

Liad

15:54

yea i saw the "could" only in second reading

Mike Miller

Well it's already called gauge theory

Tobias Kildetoft

@MikeMiller Because there will probably be several postdoc positions opening up in mathematical physics related to this

@MikeMiller Gauge theory is definitely something they do

Mike Miller

On the other hand if you asked me I would only say that if I had a monetary incentive to :P

MatheinBoulomenos

I'm working on quantum Galois cohomology and quantum Galois representations and global quantum class field theory

Mike Miller

@MatheinBoulomenos Change that to global class quantum field theory and you'd do even better

Semiclassical

15:56

I guess my question reduces to: is the n-by-n identity matrix the only one which commutes with every n-by-n orthogonal matrix?

Liad

@TobiasKildetoft you are asking about two sets that generates $m[x]$ where $m$ is some ideal in $k$ ?

Tobias Kildetoft

@Liad Yes

Liad

hmm.

m and x ? @TobiasKildetoft

Tobias Kildetoft

@Liad Ok, and this gave you something either too small or too big. How can you fix that?

Unknown MathMan

@MatheinBoulomenos Thanks :) It needs only 8 computation and lot more easier. Can I know which book you have followed for Group theory(if you don't mind)? I haven't choosen any book yet, checking which one will fit for me.

Liad

16:01

what do you mean by "this gave you something too small" ? @TobiasKildetoft

Tobias Kildetoft

@Liad You can ignore that and focus on the case where it was too big.

Liad

like when $m$ was maximal?

MatheinBoulomenos

@UnknownMathMan I didn't learn it from any particular book, more from the lectures at uni

Tobias Kildetoft

well, it became too big when $m$ was $k$. But could you then make it smaller?

Unknown MathMan

@MatheinBoulomenos Ok, thanks for sharing :)

MatheinBoulomenos

16:04

@UnknownMathMan but there are many good books. I personally really like "Aluffi - Algebra Chapter 0", but it's a bit unorthodox with its emphasis on categories

Tobias Kildetoft

Anyone who knows if it is possible to specify the name given to the citation when using \nocite with bibtex?

Daminark

Inb4 logicians create "Quantum model theory" and get all the funding because it's quantum and the NSF thinks model means mathematical modeling

Eric Silva

the NSF deserves to be tricked

Will Njundong

I have been given a definite integral and told to approximate it using left reimann sum. Is there a simpler way to get $+ C$ than first evaluating the integral directly, then using the answer to figure out $C$?

This question comes right before another that says "Evaluate the integral exactly", so I suspect there is a method to figure out what $C$ is without having evaluated the integral exactly, but I cant figure it out

Daminark

@EricSilva can't argue with the truth

Mike Miller

16:22

The math postdoc application specifically had a huge glitch where you couldn't check the "I am a citizen" box - but applications without that box checked would be desk rejected. So you had to call the helpline and have them do it for you. And once you did, you couldn't edit anything or it would break again

No other fields, just math

Eric Silva

our government is broken on literally every conceivable level

Secret

WWWIII when?

Mike Miller

After the impending climate catastrophe m8

Secret

that's pretty soon, given we only have 12 years left

Eric Silva

hot take we’ve been doomed since the deformed beast that is modernity first crawled out of its leathery amniotic sac

Daminark

16:29

D:

Will Njundong

yea well the human race doesnt deserve to get far anyways :P

Mike Miller

Yeah but you don't quit the struggle

Rithaniel

Does a set in a cover have to be a contiguous interval? Or would something like "every even number in one set and every odd number in the other set" be a valid cover for $\mathbb{N}$? (lets say discrete topology)

Eric Silva

@MikeMiller the beast has burrowed deep and so the struggle is within and without

Alessandro Codenotti

@TobiasKildetoft what do you mean? It pulls the info from the .bib to decide how to cite it

Sharath Zotis

16:37

How to find all homomorphisms from $\mathbb{Z}_8 \text{to} D_4$

Mike Miller

@Rithaniel A cover is comprised of open sets, and every point is contained in at least one of the open sets in the cover. That is all you know.

Sharath Zotis

$\mathbb { Z } _ { 8 } \text { to } D_ { 4 }$

Mike Miller

Every set is open in the discrete topology.

Sharath Zotis

so I know that $<1>$ is generator in $\mathbb{Z}_8$

and we know that $|1| = 8$ in z8

Alessandro Codenotti

Doesn't that depend on the author you're reading? Some talk about open covers when they want open sets in the cover

Sharath Zotis

16:39

so $\theta(1) | 8$

Mike Miller

@AlessandroCodenotti maybe but I think open is meant here.

@SharathZotis Go on!

Eric Silva

ive never said "cover" and without meaning open cover but that's me

Rithaniel

Alright, so it's not necessary for them to be contiguous. Danke.

Alessandro Codenotti

My professor always specified open cover. He also didn't assume nbhds to be open though

Sharath Zotis

hmm Im kind of stuck here @MikeMiller

I think I have to find the order of the group $D_4$

Eric Silva

16:41

@AlessandroCodenotti what fresh hell is this

Sharath Zotis

and then $\theta(1)$ has to divide that also right?

well the order of $D_4$ is 8 as well so it doesnt give us new info

Mike Miller

Contiguous doesn't even make sense in an arbitrary topological space!

Sharath Zotis

so the order of $\theta(1)$ has to be either 1,2,4

Mike Miller

@SharathZotis That gives you all of the information. What you know so far is that a homomorphism from $\Bbb Z/n$ is determined by where it sends 1, and the place it sends 1 must have order dividing n.

So you just need to know what elements have order dividing 8.

Adam

It's a legit first step right?

Sharath Zotis

16:46

so if u map 1 to $\rho$ we get a homomorphism

Adam

Sharath Zotis

as $|\rho| = 4$

Mike Miller

I dunno what $\rho$ is. But you don't need to work this hard.

Liad

@TobiasKildetoft @MatheinBoulomenos (x) is a maximal ideal in $k[x]$ donesn't it ?

Adam

in reference to the image cbf with typing latex tonight and maples translator is gutter ball

Sharath Zotis

16:48

$/rho$ is just a rotation

$/rho$

a rotation in $D_4$ has order 4, so mapping 1 to $D_4$ will be a homomorphism correct?

Mike Miller

Yes, but I am trying to help you avoid unnecessary work.

If you want to check all the elements by hand I won't stop you.

Adam

great another night of homomorphic isomorphic might morphic power rangers that ignore me

Sharath Zotis

how to go the other way

find homomorophisms from $D_4 to \mathbb{Z}_4$

*$\mathbb{Z}_4$

Mike Miller

What do you think you should try?

Sharath Zotis

well I dont think we can do what we did before as $D_4$ is not cyclic right?

we can't simply map 1 to all elements which divide the order

should we just map elements that have the same order?

Will Njundong

17:03

why does u substitution not work with $$\int(4-x^2)^{1/2}dx$$ bounds 0 - 2

Mike Miller

No, that's not all of the homomorphisms, and you don't know that it would be a homomorphism.

Last time you worked with a generating set (one element) and decided what to map it to based on the relation it satisfied (order 8). Can you try something like that here?

Sharath Zotis

right, so would the generating set be the elements which have order which divide 8?

Mike Miller

That would be every element. Not a very small generating set.

You want to minimize the amount of work you have to do.

Rithaniel

Is a cover a subcover of itself?

Mike Miller

@Adam You don't have a right to help. That entitlement makes it strictly less likely someone will be interested in spending time on you.

Alessandro Codenotti

17:07

@Liad who's $k[x]/(x)$?

Liad

k

Alessandro Codenotti

Which is a field

Liad

yes. just wanted to check im not missing something :)

Alessandro Codenotti

Nope, it is maximal indeed

Can you tell me other maximal ideals of $k[x]$?

Liad

(x^k) ?

Alessandro Codenotti

17:11

That doesn't work, think about why

Can you find a proper ideal $I$ such that $(x^k)\subset I$?

Liad

Ah. yes

the previous one

Alessandro Codenotti

Indeed

Tobias Kildetoft

@AlessandroCodenotti I mean the name it uses for the reference (like ABC12 for a paper from 2012, but I have a series of papers I would like it to number from 1 and up instead, as ABC1, ABC2,...)

Alessandro Codenotti

Ah, I'm not sure, there is the style parameter for global changes but I don't know how to change it for only some of them

Tobias Kildetoft

I was hoping I could just pass some parameter to the \nocite command and then manually write the names (a bit of a hack, but I would like to cite the series as ABC1-11 rather than as 11 separate papers).

Rithaniel

17:27

Hmmm, need a direct argument for the real numbers being Lindelof

Sharath Zotis

$\theta : \mathbb { Z } _ { n } \rightarrow \mathbb { Z } _ { m }$ What pairs of nums gives us a surjective homomorphism, what are the homomorphism and its kernel?

Is it if n and m have common factors?

Mike Miller

you already have all the information you need to answer this problem, and you demonstrated it while doing the first problem about homomorphisms from $\Bbb Z/8$ to $D_4$

Sharath Zotis

I believe if m does not divide n

Adam

@MikeMiller inferring a sense of entitlement from my comment comes across as inept you really shouldn't be so presumptuous about a person's character with such little information about them

Sharath Zotis

so m must divide n in order to get a surjective homomorphism

is this correct @MikeMiller

Mike Miller

17:31

Yup! You should write down a proof, and calculate the kernel.

Sharath Zotis

Thanks @MikeMiller

IM still struggling with the $D_4 \text{ to } \mathbb{Z}_4$

using ur idea of a generating set

where $t$ is a flip and $\rho$ is a rotation I have

$t^2 = \rho^ = (\rho t)^2 = e$

sorry $D_4 \text{ to } \mathbb{Z}_8$

Tobias Kildetoft

@AlessandroCodenotti Ahh well, I just decided to cite as ABC01-16 since these were the relevant years. And now I also managed to get it down to exactly 2 pages without the bibliography as required.

Rithaniel

So, take $\mathbb{R}$ in the discrete topology, and the cover where every number comprises one set in the cover. This cover does not have a countable subcover, so this space is not Lindelof, correct?

Tobias Kildetoft

Which is great, because there will be wine and cheese in the library in less than half an hour.

Alessandro Codenotti

Nice, what's being celebrated?

Tobias Kildetoft

17:42

@AlessandroCodenotti Nothing in particular. We are on "retreat" here, and each evening there is cheese and wine in the library

(though of course we do still celebrate that huge grant)

Alessandro Codenotti

@TobiasKildetoft remind me where "here" is, I'll need to find a place for a phd in a couple of years :P

Tobias Kildetoft

@AlessandroCodenotti "here" is precisely an estate in southern Jutland. But the university is Aarhus, and the grant is for mathematical physics and geometry, the use thereof

Alessandro Codenotti

I see

Tobias Kildetoft

The mathematicians on the grant are J.E. Andersen and M. Kontsevich.

(not familiar with the physicists)

Mike Miller

Ah. That would do it

Sharath Zotis

17:46

How to find all homomorphism $\text {from } D _ { 4 } \text { to } \mathbb { Z } _ { 8 }$

Im having some trouble and am stuck

Will Njundong

Cant figure out how to start on this one, please help: "Suppose $\int f(t)dt=3$ (bound 0 to 1). Determine the value of $\int f(2t)-4 dt$ (bound 0 to .5)

How do i work with that 2t??

Tobias Kildetoft

@WillNjundong Substitution

Will Njundong

so.. 2t = u?

Liad

i need to find a condition on $k:[0,1] \to [0,\infty)$ (k is a measurable function) s.t $\int_0^1 |f(x)|^2 k(x)dx = 0$ iff $f=0$ (that is , $(f,g) = \int_0^1 f \overline{g} k(x) dx$ defines inner product ) . someone can help ?

@WillNjundong yes

$f\in X$ where $X$ is the space of functions modulo equality a.e.

i think $k(x)$ should satisfy that if $A$ has a posstive measure, then $k(A)$ also has. is that work?

Will Njundong

so i've got ${1/2}\int f(u)du - {{1}/{2}}\int 4 du$ (bound 0 to .5)

Tobias Kildetoft

17:55

@WillNjundong Wrong bound

Will Njundong

so after that substitution i use bounds: 0 to 1?

Liad

you use $u = 2t$ so bounds are $2*0$ to $2*5$

Will Njundong

aahh, so I can now put 2t back where i have u?

Liad

if u=2t then $du = 2dt$ so $dt = 1/2 du$ so the integral you wrote is $\int (f(u) -4 )1/2 du$ and the bounds are $0,10$

Tobias Kildetoft

@Liad It was .5, not 5

Will Njundong

17:59

shouldve corrected him the first time, sorry

Liad

ah , why not writing 0.5 ^^

but that's the same just write 2*bound

@TobiasKildetoft any thoughts about my question ? ^^

Tobias Kildetoft

@Liad No

Will Njundong

so will i now be replacing $f(u)$ with 3? I'm trying to figure out why its this way

Liad

you have $\int_0^1 f(t) dt =3 $ and you want to find $\int_0^5 f(2t)-4 dt$

tatan

Given $\vec C$ is a constant vector, while evaluating $\int \vec{C} x \vec{dl} $, can we take out $\vec C$ out of the integral and evaluate $\int \vec{dl}$ and then cross it with C? Note that 'x' in the middle is for cross product between the vectors

Will Njundong

18:04

@Liad how to show those bounds in mathjax?

Liad

$\int_0^5 f(2t)-4 dt = \int_0^5 f(2t)dt - \int_0^5 4 dt$

ok so far @WillNjundong ?

Will Njundong

yes

Liad

now $\int_0^{1/2} 4 dt$ is easy.

$\int_0^{1/2} f(2t)dt = \int f(u) \dfrac{1}{2} du$ where $u = 2t$. now what are the bounds of the second integral ?

Will Njundong

should be .5 :)

Liad

good thing i didn't wrote the bounds on the second one :-)

Will Njundong

18:08

0 to 1

Liad

nice

so it is $1/2\int_0^1 f(u)du$

what's $\int_0^1 f(u) du$ ?

Will Njundong

thats... precisely where im stuck. :/

it doesnt make sense to me to just plug the 2t back in and proceed

Liad

so change back u to t if that help

@WillNjundong dont

use $u = t$

Will Njundong

in thats case thats 3

Liad

ok great.

Will Njundong

18:10

so i would then add 3 and 4 to get 7?

Liad

wait why?

Will Njundong

sorry thats wrong

Liad

you have $1/2 \int_0^1 f(u) - \int_0^5 4 dt$

Will Njundong

wrong bounds

Liad

0.5 * 3 - 4*5 :)

Rahul Jain

18:11

Hi, I want to know how many 4-digit numbers have no same digit, how many have two same, how many have 3 same and how many have all 4 same. leading zeroes are allowed. please help

you can give answers only, instead of explaining

Will Njundong

"0.5 * 3 - 4*5" Ive stared at that too long trying to figure out what u meant :/

Liad

3 mins ago, by Liad

you have $1/2 \int_0^1 f(u) - \int_0^5 4 dt$

i forgot the du

Will Njundong

where are 3 and 4 from ??

Liad

what's $\int_0^1 f(u)du $ ?

Will Njundong

or rather

Why are they factored in there? I know where theyre from

Semiclassical

18:16

for the second term, you've got $\int_0^5 4\,dt = 4\int_0^5 dt$

Liad

@WillNjundong what do you mean?

Will Njundong

I dont understand at all why that upper bound is 5 now

I understand how u got 5.

Liad

ahh wait it was $1$ at the first place?

or $.5$ ?

it's my mistake if so

Will Njundong

.5

Liad

ok so my bad ^^

its $1/2 *3 -4 *1/2$

Semiclassical

18:18

@tatan 1) For future purposes, you can use \times to write the cross product nicely

Liad

now that's clear ? @WillNjundong

Will Njundong

yes

Liad

ok , nice! sorry for the confusion..

Will Njundong

yes yes makes much more sense now haha

Liad

:)

Will Njundong

18:21

thank you so much for the help, good day

Semiclassical

2) One way to express the components of the cross product between two vectors A,B is as $(\vec{A}\times \vec{B})_i =\sum_{j,k=1}^3 \epsilon_{ijk}A_j B_k$

Liad

no problem, good day to you too

Semiclassical

where $\epsilon_{ijk}$ is the Levi-Civita symbol; this is +1 if ijk=123,231,312, -1 if ijk=321,213,132, and 0 otherwise

ninja hatori

0

Q: About Witt Cancellation in Modules over rings

I understand lemma 2.17 but in Theorem 2.18 I don't understand why $\beta(x,x)$ = $\beta(y,y)$ = $u$ and steps after that. Could somebody explain me this steps? I know that if we show $Rx$ isomorphic to $Ry$ then we are done since then its complementry subspaces are also isometric. How to show ...

linear-algebra commutative-algebra bilinear-form

Semiclassical

With that in hand, you can write the $i$th component of that integral as $$\int \epsilon_{ijk}C_j\, dl_k=\epsilon_{ijk}C_j\int dl_k=\epsilon_{ijk}C_j \left(\int d\vec{l}\right)_k$$

and then it's clear that yes, you can indeed write $\int \vec{C}\times d\vec{l} = \vec{C}\times \int d\vec{l}$

(the point is basically that the cross product is a linear function of both arguments, e.g. $(\vec{a}+\vec{b})\times \vec{c}=\vec{a}\times \vec{c}+\vec{b}\times \vec{c}$. That's enough to ensure that you get $\int \vec{C}\times d\vec{l}=\vec{C}\times \int d\vec{l}$)

Rahul Jain

18:30

Hi, I want to know how many 4-digit numbers have no same digit, how many have two same, how many have 3 same and how many have all 4 same. leading zeroes are allowed. please help
you can give answers only, instead of explaining

would any1 plz help

Semiclassical

Considering you're basically asking us to do the work of the problems for you...no, I think not.

Rahul Jain

nooo

I got this:

for all digits different = 10*9*8*7 = 5040,
for two digits same = 6*10*9*8 = 4320,
for three digits same = 4*10*9 = 360,
for all digits same = 10
total = 9730 but total should be 10000
leading zeros are allowed

can u please tell me which one is wrong and where am I wrong

Semiclassical

18:56

@RahulJain the 10000 consists of all 4 digit numbers, with no restrictions on repeated digits. Hence it'll contain numbers like 4466. Where does it fit into your classification?

(Note that your second calculation correctly counts the number of 4-digit numbers with exactly 2 repeated digits.)

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