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00:00
well is k is in I then b isnt in I since I is a prime ideal
NO, that's incorrect.
It's not an exclusive or.
Keep your eye on the main ideas.
wierd
i thought it was exclusive
well is k is in I and b is in I then I isnt a proper subset of J
>
You could have $a=4 = 2\cdot 2$ and both $2$'s are in $I=\langle 2\rangle$.
AGH. I don't follow what you're saying and it's sidetracking us totally.
sorry im totally lost
It's NOT an exclusive "or." Just deal with cases.
If $b\in I$, we're done. If $k\in I$, what does this mean?
What did we say earlier (main point) about $I$? Elements look like what?
00:05
If $b\in I$ we are done so we can simply conside the case where If $k\in I$ and b is not in I?
$I=<a>=<bk> $
So if $k\in I$, what does that mean?
k=a ?
Noooo.
Just use the rules, step by step, instead of making insane hopeful leaps.
00:20
I don't know that anything in mathematics can be truly insane.
I'm all in favor of proving things by the "method of wishful thinking," but it has to be substantiated
$<a> \subset <k> $ is the only thing i can hink of
i dont really understand why b isnt a unit
No, we had $k\in\langle a\rangle$, so $k=a\ell$ for some $\ell$.
@TedShifrin do you have a favourite math problem?
We're going to prove that $b$ ends up a unit, @Faust. But you can't just leap there without justification.
00:24
For example, my one professor always shows his first year students the proof for sqrt(2) being irrational and tells them that if they don't think it is beautiful, it is not likely they will enjoy mathematics (jokingly of course).
I don't tend to make statements like that. But I do talk about things that I like a lot or are favorites, and my students sometimes joked about it.
@Faust: So putting it all together, we have $a=bk=b(a\ell)$. Now what? We're almost done.
can we multiply by $a^{-1} $?
I particularly have liked this one question about a game of tictactoe where players take turns filling a 3x3 matrix with 1's and 0's (first and second palyer resp.) until it is full. Then they take the determinant and player 1 wins if it is non-zero, otherwise player 2 wins. The question is which player has a winning strategy and what is it.
unintelligible noise
Do you have inverses in $\Bbb Z$, Faust?
00:28
no :(
But what can you do with that equation that's correct?
@anakhronizein Interesting
It is indeed interesting!
My initial guess at who had a winning strategy was wrong.
I have friends back at UGA who make up amazing challenging games like that. It's not my strength.
Another interesting question I have not yet been able to solve, but which a professor always gives as a fave problem is: "Find all positive rational solutions to a^b=b^a."
00:31
(x, x) and (2, 4) are the only ones IIRC
upto permutation
So a^1/a=b^1/b
so ln(a)/a = ln(b)/b
Oh, positive rational makes it interesting
I was thinking integers for some reason
So now plot $(\log x)/x$.
integers is ez
00:34
@Faust: I need to leave. Did you finish the argument?
@TedShifrin can we subtract? like a(bl-1)
so (ln(p)-ln(q))qr = (ln(r)-ln(s))ps where p,q,r,s are integer
Yes.
Now what do you say next, Faust?
i have no idea :(
We're in an integral domain!
00:35
qr ln(p) + ps ln(r) - qr ln(q) - ps ln(s) = 0
so $bl-1 =0$ as a cannot by defn
I'm not going anywhere
and im lost again
No, you're where you want to be.
$b\ell = 1$. What does that mean?
1 is in I?
00:37
@Leaky: Did you see my suggestion to graph $(\log x)/x$?
@TedShifrin sure
Well, $1$ is in $J$, actually, @Faust.
oh
Or $b$ is a unit (as you leapt to earlier).
thank you!
00:38
Either way, $\langle b\rangle = J = R$, so this means $I$ is maximal.
have a headache from that question but thank you understanding improved many fold
You need to force yourself to use definitions carefully and just take baby steps.
i need to understand the definitions better
i keep getting lost
Well, yes, but one way is to use them!
In math you must learn and understand definitions. Agreed.
@anakhronizein Is the right answer that player 1 has the winning strategy?
00:40
Diff geo is hard for that, too ... Lots of different definitions.
SOOO many
OK, I'm off to cook. Anyone who wants to come for dinner should show up in a few hours.
wth u cooking thats a few hours?
Something French probably!
I'm making fall weather comfort food.
French-y meat loaf and mac 'n cheese with 4 kinds of cheese.
A bunch of prep work and then oven time.
00:41
Sounds delicious, enjoy
numy
It's even better leftover, if you show up tomorrow.
Bye for now, all.
@KevinDriscoll that was my first guess. So no, that's not the case. ;)
This is really kind of crazy because the 1st player gets to go last!
Yeah it is pretty surprising.
I also found that playing it with friends led to P1 winning despite the winning strategy being P2's.
00:48
Ya that doesnt shock me because clearly player 2 has to plan very carefully
Because player 1's last move must not matter. So Player 2 has to have already won the game somehow on the 2 rows that are already filled in
Note that since P2 has a winning strategy (so the determinant can be forced to be 0), you can just find a winning strategy for P1 playing in a specific spot first.
I never considered the spot P1 plays first in to matte. Or do you mean the last move?
Well it doesn't matter.
But the point is that the determinant is always 0, even if you permute columns and rows.
(assuming you have a winning strategy)
Ya the 1st move's spot cant matter because all 9 spots are equivalent as far as the determinant goes
Yuppers.
So given that, it's not too hard to find a winning strategy by exhaustion.
00:59
Oh. I've been playing a completely different game!
I missed the part where you said the players could only use 0 and 1
I was playing with both players just inserting arbitrary numbers @anakhronizein
does the determinant live in Z or Z/2? does it matter for the answer?
Knowing the answer, if the 2nd player has a winnign strategy with the det in Z, he must also have one for Z/2, so I guess it doesnt matter
oops - silly me, i was thinking it was harder to be zero in Z/2
What do you mean by Z or Z/2, @MikeMiller
He means whether you fidn the determinant, or find the determinant mod 2
Z/2 means $\mathbb{Z}/2 \mathbb{Z}$
01:15
Oh I see. It's Z for the original question.
And hence it doesn't matter.
01:31
@MikeMiller concerning what you had said, how do we utilize $(d\alpha)^n|_\xi$ to orient $\xi$?
Is it because it is of top degree on xi?
I can see how we utilize the equivalent condition $\alpha\wedge(d\alpha)^n$ to orient $M$.
Hello fellow math lovers, I have a quick question. I have been laboriously looking for similar problems on the internet and I am using this place as a last resort. Does anyone here know where I can find a problem set covering questions about finding equation of line tangent to a curve? This type of problem appears in Chapter 3 Section 1 of Stewart Calculus (Single Variable) but I am not satisfied with my performance. I am primarily concerned with finding values of unknown variables.
Can you give an example of what you are having troubles with?
Like "find the equation of the line tangent to f(x)=x^2 at x=1", what trouble do you encounter?
That is a basic one. I have the basics down, no problem there. I am stuck at more advanced ones. For example: find the value of c such that the line y = 3x/2 + 6 is tangent to the curve y = c*sqrt(x). This problem is not so straight forward, and there are many like it that require finding value of missing variable. I am good to go once I have the equations and I can then use matrix algebra to get the values but getting all the equations correctly is where I get stumped at times.
@anakhronizein Quite right. In general, an orientation of a rank n vector bundle is equivalent to the data of a trivialization of $\Lambda^n E$, which is determined by a non vanishing section of $\Lambda^n E^*$. An n-form on a manifold restricts to an n-form in this sense on any subbundle of TM.
(Note that this is not the same as what I would normally call an E-valued form, but rather just an appropriately skew-symmetric and multilinear functional on E.)
01:46
Thanks, that makes sense!
I just realized that this gives a sorta pleasant fact: (4n-1)-dimensional contact manifolds are orientable, and (4n+1)-dimensional contact manifolds have orientable contact hyperplane field, then a contact manifold has trivial normal bundle to the hyper plane field (is co-orientable) if and only if it’s both orientable and has orientable contact field. So all of those orientations are given if and only if we have a global defining 1-form.
Straightforward but cute.
@MikeMiller I had heard a quote that I think was attributed to Arnold that said something along the lines of "all geometry is contact geometry". Do you know what was meant by this?
Hmm
There's loads of stuff on converting parametric equations to implicit ones, but how would I go about doing the reverse?
Or is this an X/Y problem? (Currently trying to trisect a certain implicitly defined arc segment.)
02:02
@Iamlearningmath well how do you personally approach this one?
@anakhronizein Well I have already solved this one. But When I first solved it, I found out my answer was incorrect then I spent half an hour finding the correct one. First thing I do is write down what is required of me, what I need to get there and I start plugging away. But in general, my go to steps are, find the slope, find x value, find value...not necessarily in that order.
@anakhronizein Not really. I could ask someone who would know.
Well when you got it wrong initially, what was the problem with your answer? Where did you go wrong?
maybe you can pass from dynamics on a manifold to Reeb dynamics on its unit tangent bundle.
@MikeMiller If you could, that would be appreciated, though it's not that important. Let me double check where it came from.
02:12
@AkivaWeinberger Actually it kinda makes sense because spam is basically the opposite of information
:P
I think you can do something similar in symplectic geometry, lifting vector fields on M to Hamiltonian vector fields on T^*M.
And then I’d buy you can restrict.
Geiges states in his preface: "One of the most eloquent of modern panegyrists of contact geometry is Vladimir Arnold, who proclaimed on several occasions since 1989 that ‘contact geometry is all geometry’"
Therefore $f(information)=spam$, $f^{-1}(spam)=information$
where $f=spam = (maps)^{-1}$
He also tended to be grandiose.
Arnold? Or Geiges? :P
02:17
In addition $spam$ is idempotent because spamspamspamspam is still spam
actually no, if a function is idempotent, it is its own inverse, bleh
Idempotent is where E^2=E.
Like a projection.
So you had it right initially.
Hmmm, so apparently a characteristic foliation of S is orientable if the normal bundle of S is isomorphic to the quotient bundle $(TM/\xi)|_S$
@anakhronizein Arnold. The worst I’ve heard of Geiges is that his book isn’t great.
Do you know why his book might not be great?
I find he is a good writer.
As in, he is not boring.
Hmm, more specifically, is it possible to convert an implicit polynomial relation into a parametric equation with constant "velocity?"
Wait a second
facepalm
I should just solve for arc length
02:36
@anakhronizein Nah, just some grumbly grad students. If it works for you keep using it.
I am a grumbly grad student sometimes.
I grumble about un-fun writers. ;)
Do you have a particular book you really like to advocate for in terms of geometry, @MikeMiller ?
I am open to suggestions for new reading material no matter the topic or difficulty.
I probably do but I have a fever and a broken brain right now. I’ll name some another time.
Sure. Hope you feel better!
...
Definite integrals are weird
02:52
@anakhronizein Oh, Besse’s Einstein manifolds is a fantastic Riemannian text and I love Joyce’s book on compact manifolds with special holonomy.
Broadly speaking I’m more a topologist than geometer.
Topology is appreciated. :)
I tend to lump topology and geometry into one. topometry.
Speaking of grumbly grad students
Oh then I definitely have a few less well-known must-reads. Let me look at my references.
I want to announce to all the US-based grad students that the current version of the tax bill in the US congress would make tuition waivers taxable income. For almost all of us, I imagine, that would substantially increase our tax burden, even after accounting for the larger standard deduction. So just a heads up to everyone to be aware, and prepare if necessary.
3
1) Kreck, “differential algebraic topology”, reframes homology as bordism of a kind of singular smooth manifold (the fun to say stratifold), thereby allowing one to use smooth techniques from transversalitt theory in a pleasantly direct way, and reinterpret a lot of standard algebra in this way
It’s often how I like to think about these things (also useful is the brief paper of Lipyanskiy, “geometric homology”, which uses manifolds with corners instead)
02:57
Its kind of amazing to me how Ive taken 1 class and my understanding of chat messages increased by at least 100%
Is Kreck fairly introductory?
Quite - if you know a first course in algebraic topology and have heard of the Thom transversality theorem you should be able to read it
in fact I don’t think either of those are strictly necessary since I think he uses his book to teach a first alg top course
First course in algebraic topology constituting fundamental group, singular, cellular, and axiomatic homology, and de Rham cohomology?
I was looking or a refresher on algebraic topology so this might be a good way of doing it without getting bored!
oh yeah you’re well-prepared
Balarka and I were looking at Candel and Conlon’s foliation books a while back but we got bogged down at some point and then busy with work
those were fun
I was an undergraduate when I took that algebraic topology course. Also it was a category theorist so I feel like there was a lot of unnecessary categorical background that took up time. Hence the need for a refresher.
03:05
Haha. I sympathize.
@Kevin: Yup, gotta love those Repugnicans. Anything to make money for the multi-billionaires who won't send them money otherwise.
The fancy language can be useful and interesting but it’s not how I think about the basic things.
I shortly after was reading Milnor's Morse Theory book and had to ask a professor why there were always rings included in the homology notation.
I find category theory is useful after the fact, but definitely not for learning.
I am glad that I have had the opportunity to see it in my classes though. Had a course on proof/type/category theory.
That was fun.
Sounds interesting. I only know a little logic.
What classifies as a 'little'? Like basic model theory?
03:24
@TedShifrin Yeah, I'm salty about it. There doesn't seem to be any logic to raising taxes of grad students to pay for an estate tax cut other than they have to placate their donors. If there's a bright center to the tax-reform universe, this bill is on the planet furthest from
Somehow 'salty' seems the wrong word for that @KevinDriscoll
When you lose a video game and get frustrated afterwards, you're salty. When the GOP in Washington is planning to change the tax code so that the rich pay less and you, personally, pay more...
I think unironic contempt/anger are quite appropriate.
Salt is an ironic version of anger, and taxes are not ironic.
I thought we paid taxes ironically!
Really though I see what you're saying
03:40
Ya @Semiclassical might be right. I'm not 100% sure how to describe how I feel about it.
03:58
Yeah
Regarding my own feelings I wonder if there’s an element of privilege built in
I’m used to feeling disgust at how GOP policies tend to run over other people. I’m not used to being personally affected by it myself
@KevinDriscoll so yea, privilege
Ya thats part of the reason I'm not viscerally angry. Its annoying, but I'll live.
04:19
Hey has anybody ever heard that the rigidity of a graph is equivalent to the definability of equality on vertices of the graph in first order logic? I just proved it and I keep looking it up online and cannot find it shown anywhere else
04:47
Hey @Ted!
05:03
I don't know the right terminology for this, so it will look naive. If I have a function y = f(x) (y and x are typically vectors), and I have some set of "smallest" basis vectors {a_i} such that f(x+a_i) = f(x) for any x (from which a whole vector space of such terms could be built), is there a standard way to refer to an individual a_i, and to the smallest set of them that could generate all possible cases where f is unchanged?
This is some kind of symmetry, so I presume this is related to Group theory (of which I have none) -- I don't really know what I am doing here.

I've been calling these minimal a_i things as "atoms" (they're essentially vectors that are almost all 0, but with a few elements that are +1 or -1 in particular patterns) but it's better if I find out what they're more conventionally called.
There's nothing with the same property that can have fewer non-zero elements than they have, so they're the simplest units to work with.
05:18
The underlying point of all this is to demonstrate to practioners in a particular application area who are constructing predictions using a particular f that their favorite f is completely blind to particular kinds of structure in x (i.e. f(x')=f(x) where x'-x is not something they would wish to be blind to). Being able to talk about how to generate a collection of interesting (i.e. surprising to them) x' more conventionally would be handy.
I can do the required calculations readily enough but better notation and terminology will help (not just in describing it but in looking up stuff in order to understand more about what I am doing rather than just building it all from scratch)
I'm not even sure where to start looking really.
Any suggestions of terms to use for an a_i or the smallest collection of them? Theorems relating to how to show I really do have the smallest set that generates all x' with the same output? Useful things to read about this stuff?
05:36
@Glen_b Sorry I'm not sure I understand. You have a vector-valued function $f$. And then what you find is that there is a single set of basis vectors $a_i$ such that $f( x + \sum_i c_i a_i) = f(x)$ for any $x$
Is that right?
06:11
It's right, but there's no unique set of basis vectors - you could make a different basis easily; my collection of a's just happen to be made up of vectors with the fewest possible number of non-zero terms in them (and by restricting them to only have +1 or -1 they'll then be unique up to an overall change of sign)
Ya I didnt mean unique
I mean is it true that there exists a single set of basis vector for any $x$?
sorry if I misconstrued your use of "single" there
That is, can you use the same $a_i$ for every $x$?
Ah. Yes, the same a's work for every x
The a's are like zero-elements.
f(a_i) = 0
Okay, so then what you have is that your function is invariant on some linear subspace of whatever space $x$ lives in
Well but I presume $f$ is not linear
06:14
COrrect, f is not linear.
Although if you condition on a bunch of stuff, parts of it are. That's not very helpful.
So then teh fact that $f(a_i) = 0 $ doesnt tell us much about $f(x+a_i)$
No. only that the 0-vector is another possible x, I guess. Oh, actually, ... I have to be careful, f(a_i) might strictly be undefined, more correctly for any case where a linear combination of a's is defined, the value would be zero
even "invariant on some linear subspace" is helpful at the moment.
So you're certainly correct that your function has a symmetry. And there's some group thats associated to that symmetry.
Okay, good. But this is stuff I don't know how to even talk about. I need to learn a little.
Not sure where to start
I think teh first thing to do is to look at your $a_i$
and figure out how to write $x + c_i a_i$ as some matrix $A$ acting on $x$
So for example, there will be $1$s along the diagonal but then many other entries
06:21
Written as a vector typical a_i will have two +1's and two -1's and the rest 0's but the nonzero values have to be in particular places
Actually, I believe all the a_i will be like that,
Actually what I said isnt quite right
You could flip all the signs but I have been just taking the leading term to be +1.
So one thing to figure out is exactly what linear subspace your $a_i$ span
Ah right. And so you can think of $x + c_i a_i$ as being $x + v$ for some vector $v$. And then think of $v$ as being $A w$ for some matrix $A$ and vector $w$. That vector will be arbitrary. But the matrix will have to have a very particular structure
related to the $a_i$ and where they ahve 0s and where they don't
And knowing that matrix $A$ should give you a better idea of what kind of symmetry this is
Thinking about what you're saying there...
SO let me write out a trivial example
06:30
Okay; these vectors are stacked up from a rectangular array that is cut off on the lower right (it most often will be triangular) in some convenient order. The ai's are such that all rows and columns add to 0.
in the original array.
The cut off parts don't exist
Suppose $x \in \mathbb{R}^3$ and $a_1 = \{1, 0, 0 \}$ and $a_2 = \{0, -1, 0 \}$. Then we can say $f(x + A w) = f(x)$ for arbitrary $w$ provided that we make $A$ the matrix $$\begin{array}{lcr} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}$$
My contention is firstly that every such set of values (v in your notation) where all rows and columns sum to 0 are buildable from a's that in the original array have a 2x2 matrix +1,-1 and -1,+1
Hang on, while I read the link on how to see the LaTeX markup in chat.
Still trying to figure that out.
Its not too bad, just drag the chatjax bookmark from the link to your bookmark bar
and then click the resulting button
while your on this page of course
Got it
Thanks. Your few words were more helpful than the link page
Didn't help that I was dragging to the wrong place
RIght, yes, you can make a big matrix like A with an arbitrary diagonal.
So then the structure of this $A$ should tell you what group you're working with
For example its not the entire general linear group, becuase than $A$ could be any invertible matrix. So thats not it
06:44
Okay. So the particular A's then denote the group -- or is there some characteristic of A that does? This is probably the point where I should stop using up your valuable time and go read something about exactly how to do what you're saying to do.
The $A$s should tell you what group is hidden in there. They aren't themselves the group, they're actually something like a sum of representations of elements of the group
But thats just a vocabulary point
That's still useful.
Since vocab is one thing I was seeking -- how to talk about things.
The stuff I have tried to read on group theory doesn't seem to relate directly to what I am trying to do but everything you have said makes sense. This suggests that I am just looking in the wrong places
What have you been looking at?
Most of the math stuff is quite abstract
Elementary treatments of group theory on line
But my objects are vectors which makes things structured in a particular way that's probably why I am not finding what I need
I don't know what to even search for.
Ya, you're looking at something that isn't part of the mathematicians core group theory education
Because you're not dealing with abstract groups
You're dealing with some concrete representation
06:50
Indeed. I don't mind a little abstractness where necessary, particularly if I can see how it relates.
And for that you need to know something about whats called linear representation theory
Thank you.
Which is all about how we represent elements of groups as matricies
Let me check something real quick
Ah. I see even from the wikipedia page that that is quite probably the kind of thing I need, but I will need something like a set of notes with some exercises or a reasonable text on it.
Yes definitely. Or someone nearby who already knows what they're doing!
06:56
There's a long list of references on the "representation theory" wikipedia page. I don't know even where to start there
But at least now I have "linear representation theory" and some terms and thoughts about what I need to work with
I will be afk a few minutes but still "around"
Testing...

https://www.desmos.com/calculator/7xsolk4qug

$$n,m \in \Bbb{N} : f(n,m) = \{ \text{truncate}(\sqrt{2},k)(x_n\pm\frac{1+m}{2+n+m}\}, x_n = \{\}) $$
@Glen_b So check out this lecture by William Harter and particularly the first 15 minutes or so. There's a lot of physics mumbo-jumbo in there that isn't important for you. But the thing I think is important is that he shows this matrix $H$
$$f(\frac{p}{q}) = \text{Truncate}(p\sqrt{2},q-1)$$
and that matrix doesn't have arbitrary entires. for exmaple, all its diagonal entries have to be the same.
and there are similar relations among the other entries
And what he shows here is the very basics of how you break this $6 \times 6$ matrix that he has into 6 pieces. And each piece corresponds to the element of a group. In this case its $C_6$. The matrix $A$ that you will find should play the same role as his matrix $H$. And of course who knows what, if anything, the group you'll have will be.
@Kevin thanks very much for your help.
07:06
And here's his textbook where he goes over all this stuff: amazon.com/Principles-Symmetry-Dynamics-Spectroscopy-William/dp/…
Again, theres a lot of physics mumbo jumbo soo I dont know how intelligible it'll all be to you
But at least if I were working on your problem, this is the direction I would start out in
Probably not too intelligible given my high school physics from ... nearly 40 years ago.
But that's okay, it may still help
At least I can start looking at things with a basis for generating actual questions.
@Glen_b Indeed. And if all this works out, at the end of the day you'll be able to say something like. This $f$ is invariant under the transformation generated by the group $G$, so if you have 2 inputs that are related to each other by some $g \in G$ you're gonna get the same result. And so if you don't want to treat those 2 inputs the same, then you gotta come up with a different $f$
Or something to that effect
yeah, something like that is just what I want to be able to say.
And depending on what your $x$ represent physically, maybe you can give some natural interpretation of what the elements of $g$ mean in this case
every rectangular array with the lower right corner cut off is different in size but all these problems have the same structure
The x's are usually either monetary amounts or counts of things
07:14
When you say lower right corner cut off, what does that mean exactly?
Like I have an $m \times n$ matrix and I do what to it?
slice a diagonal line across it at some place from top right to bottom left. You don't have any values below that diagonal.
They don't exist (well they will in the future)
So you might have say:
 23   47   18
 15   21
 22
Oh I see. So now we have some weird thing thats not even a matrix anymore. But I presume now we do something to the numbers that are left?
if you started with a 3x3 and cut off the last 2 diagonals
Yes, so there's a bunch of calculations performed on them of a particular form that produce predictions for the lower part.
Which you will have in the future
I see so we're using the values that are left over to then try and predict what values should go in those lower right corner spots
Yes.
07:20
and presumably there's some model related to the real world and what these numbers represent that doesn that predicting
Sure.
But the most commonly used approach is very mechanical, and has some very odd features.
It's those odd features I want to be able to more clearly characterize.
Ok yea. Sounds interesting and potentially doable.
I can show all manner of interesting properties but I don't currently have the tools I need to do a proper job of saying everything important there is to say; I didn't even have the words to go looking.
Although the easiest way to go in this group theory direction is to somehow wrangle someone who does this for a living, and sadly Im not at that level yet
To find someone in a math department who works with linear representation theory and try to get them to look at this?
07:24
Yes, although I'm not exactly sure what the math people who do this actually do all day. Like what their specific expertise is. You might actually have more luck in a physics department.
For example let me give you the analogy I have in mind
Prof. HArter in that lecture looks as the case of 6 atoms or balls or whatever arranged in the shape of a hexagon and each one is connected by a spring to its nearest neighbor
Possible. However, I'd want to go in with enough background that I come off as serious-but-ignorant rather than a crank. Physicists see even more of them than mathematicians.
Thats what that matrix $H$ represents. And he investigates the structure and whatnot.
okay...
and its osme $6 \times 6$matrix
I ahven't watched the video yet, btw. It's in another tab waiting nearf the start
07:28
and then you look at a different problem which has little particles arranged in the shape of a 12-gon but htis time there are 2 kinds of spring. And 1 spring goes between each pair of balls and you alternate which one you use
So its like A-type, B-type, A-type, B-type etc. til you get all teh wya round
And again you do this matrix thing
And its some $ 12 \times 12$ mess, so not at all teh $6 \times 6$ thing you had earlier
but hten you find out by analyzing it that they both contain representation of the same group
... Looking with hindsight that doesn't sound helpful to you. But there's some connection there about guys of different sizes nevertheless hiding the same structure
oh, cool.
thanks for all your help
No problem, hope it ends up working out
 
5 hours later…
12:47
1
Q: Is $\Bbb{R}^\omega$ (with uniform topology) locally path connected?

user193319Okay. Let $A \subseteq \Bbb{R}^\omega$ be the set of all bounded sequences in $\Bbb{R}$. The problem I am working on is trying to show that $x \in \Bbb{R}^\omega$ is lies in the same component of $0$ if and only if $x$ is a bounded sequence, where $\Bbb{R}^\omega$ is endowed with the uniform topo...

13:42
i need to compute $\int_{\gamma} z \ ^ 3 dz$ over a square ( it is a complex line integral) it's supposed to be zero, right?
@MathematicsAminPhysics A room for experiments with MathJax was unfrozen. Just in case it's useful for you. (I've seen that you've occasionally do some experiments with formatting in various rooms. And in the past, you have similar room of your own.)
@Liad Should be, yeah
Note the difference between $dz$ and $|dz|$
(With Riemann sums and real integrals, $dz$ represents the width of each of those tiny rectangles. But here $dz$ can be complex, so that picture gets confused.)
14:06
0
Q: complex line integral, calculation mistake

LiadI need to calculate $\int_{\gamma} z \ ^ 3 dz$ where $\gamma = \gamma_1 *\gamma_2*\gamma_3*\gamma_4$ . $\gamma_1(t) = -2t +1+i$ $\gamma_2(t) = -2it +i-1$ $\gamma_3(t) = 2t -i-1$ $\gamma_4(t) = 2it +1-i$ $t \in [0,1]$ That is, $\gamma$ is a square that starts at 1+i , with positive directio...

@AkivaWeinberger can you take a look ?
14:32
$$\left(\frac{b+a}{m+a}+\frac{\text{round}\left(10^{a+2}\left(2s-\frac{\text{rou‌​nd}\left(10^{a+1}s\right)}{10^{a+1}}\right)\right)}{10^{a+2}},\frac{b+a}{m+a} \frac{\text{round}\left(10^{a+2}\right)}{10^{a+2}}\right)$$
$$\left(-\frac{b+a}{m+a}+\frac{\operatorname{round}\left(10^{a+2}s\right)}{10^{a‌​+2}},-\frac{b+a}{m+a}\cdot\frac{\operatorname{round}\left(10^{a+2}\right)}{10^{a+‌​2}}\right)$$
$m \in \Bbb{N}, m > 0$
$a , b \in \Bbb{N}$
$s \in \Bbb{I}$
Plus one more rule: Ignore any duplicates for sequences with higher $a$ s
i.e. If $a=1$, get some y value for $\frac{1}{2}$
Then for $a=2$, all y values by $\frac{2}{4}$ are ignored
and so on
@AkivaWeinberger

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