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2:00 PM
@DHMO but we use use that formula instead of section formula
 
@Ramanujan so?
 
We never get undefined while using section formula
 
but both areas are zero
 
So?
 
0/0 is undefined
 
2:03 PM
Are you considering ratio of areas of triangle??!?!
 
@SamuelYusim Why would one think of constructing such a polyhedron? I suppose it must be useful in some way - I'd like to understand the construction at a non-technical level.
 
@Ramanujan what is your question?
 
oh, sure.
 
@Ramanujan when it is vertical, you cannot get the ratio only through the x-coordinates.
 
2:07 PM
@DHMO by analogue of section formula we can get ration with either only x coordinate or y coordinate
 
How do I find the value of D?

http://math.stackexchange.com/questions/1977186/find-partial-fraction-2x2-1-x2-12-x23
 
@Ramanujan x-coordinate, not when the line is vertical; y-coordinate, not when the line is horizontal
@MithleshUpadhyay isn't D 0?
 
OK , so there is an exception which was no one said me
 
@Balarka so I left out any mention of optimization in order to hype you up with geometry first, but you need to delve into the depths of linear programming to see why we would think of constructing a polyhedron.
So, a linear program (or LP for short) is just a problem where you need to optimize a linear function $c^Tx$ (the objective function) subject to a set of linear inequalities, which usually gets expressed as $Ax \leq b$. The set of linear inequalities defines an intersection of half spaces, which is exactly what a polyhedron is. So if you just find a way to rephrase your problem as an LP you're in a really good position.
A key fact about LP's is duality, which says that for any maximization problem, there's a natural minimization problem for which the objective function will give the same value when optimized, and vice versa. Duality is also constructive. Now from duality you have another free set of inequalities which define your object, and you can use this information to prove validity of algorithms, bounds, etc.
 
@SamuelYusim Ah, true
 
2:16 PM
excitingly, linear programming in general can be solved in polynomial time in the size of the constraint matrix
 
Let the Möbius transformation $T(z) = \frac{z-i}{z+i}$ be given. Find the image under the mapping of $T$ over the area $G = \{z: |z|<1 \} \cap \{z: |z-1|<1 \}$
 
however, in combinatorial optimization you usually (e.g., min-cost spanning tree) have to add the constraint that a solution vector has only integer values. this is called integer programming, and it's NP-hard :(
 
oh right
 
@DHMO how can we solve these type of problems?
Iam really down when it say when n is even and when n is odd😣
 
2:20 PM
@Ramanujan which country uses a decimal point to indicate multiplication?
 
Iam not sure about other countries but in this case india uses in mathematical induction
 
@Ramanujan Then what is five over two?
 
5/2 ?
 
in decimal?
 
I will chat to you on Sunday and answer all you questions,pls help me now ( be a good water molecule)
 
2:24 PM
when n is odd, what is the (n+1)-th term?
 
Odd
 
?
 
You mean term, so it will be (it depends)
 
what is the term in terms of n?
 
n . n^2
 
2:27 PM
no
 
Why?
 
because the 4-th term is not 4.4^2
 
Sorry,i just saw 1st two terms, nth term is n^2 when n is odd and 2n^2 when n is even?
 
yes
 
@DHMO , That's correct. You are right.
 
2:33 PM
@DHMO one request, can you solve at once whole problem and upload it?_(as I do)_
 
@Ramanujan i'll just solve it here
 
Because it's taking time to get from you in this way
 
when n is odd, n+1 is even, so the (n+1)-th term is $2(n+1)^2$
From the given theorem, the first $n+1$ terms sum to $\dfrac{(n+1)(n+2)^2}2$
The sum of the first $n$ terms, plus the $(n+1)$-th term, is the sum of the first $n+1$ terms
Sum of first $n$ terms $+ 2(n+1)^2 = \dfrac{(n+1)(n+2)^2}2$
@Ramanujan ok?
 
What? Sum of first n+1 terms is $\dfrac{(n+1)(n+2)^2}2$
 
@Ramanujan from the given theorem.
$n$ is odd, so $n+1$ is even.
you can just let $N=n+1$ if you are not satisfied
$N$ is even
 
2:38 PM
So you mean that it is sum of even number?
 
@Ramanujan sum of the first $N$ terms where $N$ is even
 
I didn't came to that before
I mean that formula of summation of even number
 
it is given
 
0
Q: n x n symmetric matrix inversion

Saesun KimI am trying to solve the inverse of 16x16 matrix, but Mathemtica cannot handle it since it is not numerical variable, and it is very large. I know that my matrix is symmetric with lots of zero, does it have any good method or approximation to solve this faster? I don't really need exact solutio...

Is it opening?
Sum of natural numbers is n(n+1)/2
I don't know sum of even number is (n+1)(n+2)^2/2
 
no
not the sum of even numbers
do you understand what you question means
 
2:44 PM
Wait,pls
 
2:54 PM
Ok
So we have to substitute n+1 wherever n is right?
 
yes
 
Then? I didn't get that
 
so the sum of the first n+1 terms is (n+1)(n+2)^2/2 if n is odd
 
n+1 th term is even
 
yes
 
2:58 PM
In question it is asked for when n is odd
 
yes
so n+1 is even
 
Then why you take the sum of n+1 term?and then said if n is odd
OK,sorry,
I got it now
 
good
 
@DHMO Are you interested in Möbius transformations?
 
@Lozansky seems interesting
do you have a question?
 
3:02 PM
Just ask , don't ask to ask
 
nice music!
 
@Lozansky DHMO is all in one , just ask it
 
@DHMO Of course I rather discuss back-and-forth than get a straight answer
 
3:04 PM
Let me think
 
@Ramanujan Lol yeah, he's pretty great
 
> If I have seen further, it is by standing on the shoulders of giants. -- Isaac Newton
 
And if I ever want to see further , i stand on the shoulders of DHMO
 
@Lozansky any luck letting $z-1=r(\cos\theta+i\sin\theta)$ where $0<r<1$?
$f(z)=1-\dfrac{2i}{z+i}$ by the way
 
Right
Haven't tried that
I think I have seen a similar case
And they somehow worked out the intersection between the circles
 
3:11 PM
I see
 
But I can't remember exactly how that was done
Anyway
Is my reasoning so far correct?
 
@PVAL Yesterday my new idea fell apart but now I think it works again.
 
@Lozansky sorry but I don't understand your reasoning
 
@DHMO Anything specific that's not clear?
 
@Lozansky that you mapped some points and somehow extended it to your result
 
3:19 PM
@DHMO That's because I expected a halfplane, so first I check which axis is dividing the halfplane and then which side it's on
There are 4 halfplanes
 
why did you expect a half-plane?
@Lozansky my calculations give me the plane x<1
 
@DHMO I used a result from the book where they show that if you maps the interior of a circle you get a halfplane
@DHMO That's weird
 
@Lozansky my caulculations might be wrong
 
@DHMO Did you get that for the intersection between the images of $C_1$ and $C_2$?
 
@Lozansky oh, I just meant the image of $C_1$
 
3:33 PM
Ok
You can try and find the inverse to $f$
And put in some value say $z=0.5$ and see if it gets mapped back to the unit disk
 
$y=\dfrac{x-i}{x+i} \iff i\dfrac{y+1}{y-1}=x$
 
The inverse is directly given by $\frac{dw - bx}{-cw+a}$
With $w=0.5$ I get $3i$ which is not on the unit disk
 
strange
 
I wish I knew how to plot the mapping graphically
 
4:07 PM
okay i have figured out somethings about my question now after talking to my professor
i am searching for the chromatic polynomial of $G \square K_{2}$. xD
 
4:33 PM
what is the permanent of the empty matrix?
 
 
2 hours later…
6:16 PM
Hullo?
 
heya
 
Find all Möbiustransformations that map the right halfplane to the unit disk $|z|<1$
Anyone know where to even start??
 
Hi @Danu.
 
Heyo
Disappointing news---not going to learn about Donaldson's theorem after all
We won't cover the topology side (Freedman's work) :\
 
ask me about it
 
6:21 PM
After the course, I guess :)
Once I'm ready to be enlightened
Other than that, great news: Kotschick told me he might start doing a (bi?)weekly meeting for his students
Where he either tells us something or has one of us present (there are apparently 3 PhD's and 4 MSc's)
Super happy about that possibility
Other good news: I got further options for topics
and I like the papers he gave me now a lot
They're mostly about interplay between topological/geometrical conditions and characteristic classes
 
@Danu Want to tell me?
 
Err, sure---I don't really know anything as usual.
I read one of the papers
 
Well, what are the papers?
 
The papers are all very short (5 pages-ish)
The one I read was by Ping Li, about "minimal" conditions to be imposed on a manifold so that it must be biholomorphic to $\Bbb CP^n$ here
 
That sounds interesting
 
6:30 PM
Then one by my supervisor here, relating curvature conditions and Pontryagin numbers
 
Seems like they're sort of classifying homotopy CP^n's.
 
Ping Li also wrote a paper on something related to that, but using Ricci curvature instead of sectional curvature, here
Then finally Kotschick gave me a really nice & coolsounding thingy which is an unfinished manuscript so I probably shouldn't talk too much about it. It's again something about characteristic classes and how they relate to certain geometric/topological conditions
I like the way all of them look a lot.
 
Where are all the conformal mapping experts?
 
good evening everybody
 
@MikeMiller Any comments?
 
6:40 PM
Hello @Axoren
 
Hello @Evinda, long time no see.
 
@Danu Sorry, I'm chatting with a colleague right now. Give me a bit.
 
Yes, that's true... @Axoren
@Axoren I have a question... Have you had computability theory? Because I will have a presentation at this subject and wanted to ask you if you have some topics in consideration.
 
I think I'm going to read ahead and learn connections.
 
@Evinda I've done a bit of computability theory, but I'm rather busy today. I was in a hiking accident last Friday and I've been trying to recover and catch up with work that hasn't gotten done as a result of me being in forced-bed rest.
If you've got minor questions, you can ask me a couple
 
6:44 PM
@Danu Those all look really really cool.
 
Question one:
 
I like results that show something is automatically CP^n.
 
@BalarkaSen That is not really the case, and that's already done through other methods. No homotopy CP^n can have a Kahler structure IIRC>
 
@Axoren Ok...Are you for example familiar with inductive inference?
 
6:45 PM
@MikeMiller Except CP^n, you mean? Interesting.
 
You're violently changing the smooth structure.
 
@Axoren It's a theory of Solomonoff
 
This is at least certainly true on CP^2.
 
yeah, that's certainly believable (in all dimensions, I mean, not just 2)
 
@Evinda This might be above my knowledge. That's a predictive theory, isn't it?
 
6:47 PM
By the way @MikeMiller
 
I'm not familiar with computability in a predictive context.
 
What is the signature of an arbitrary closed, oriented manifold
(I only heard about it in the case dimension is 0 mod 4)
 
signature doesn't make sense for odd dimensional manifolds.
 
I don't know. I haven't studied the topic by now... If you think of some topic, you can tell me... @Axoren
 
so for one, you need to restrict to 0 mod 2 dimensions.
 
6:50 PM
They don't do that in these papers.
 
@Evinda I recommend that you focus on topics that your existing study more easily covers, like the concept of a Quine.
 
defined to be 0. because to define the quadratic form in homology of some dimension, you need the middle dimension.
also I think for 2*odd dimension, the middle dimension has trivial self-intersection (being odd dimensional), making it not interesting
 
Yeah, I know
it's symplectic (in the 2*odd case)
 
@BalarkaSen That's nonsense. The intersection form is nondegenerate, it's just skew-symmetric instead of symmetric.
 
He's right taht it's not of special interest though
 
6:55 PM
@MikeMiller Which bit is nonsense? I just said it has trivial self-intersection, aka $\alpha \cup \alpha = 0$.
that's by skew-symmetry, yes
 
Oh, fine. I thought you meant the intersection form was identically zero.
Oops.
 
What is it about ? Is it a difficult topic? @Axoren
 
@Evinda It's a program whose sole purpose is to produce itself as output in some encoding. In the standard setting: as text.
 
Is there a lot to present about this topic?
 
7:01 PM
How much of a presentation do you need to fill?
 
For one hour. @Axoren
 
@Danu So I prefer to think of the signature as the index of the operator $d+d^*$ from $\oplus_{odd} \Lambda^n T*M \to \oplus_{even} \Lambda^n T^*M$.
 
@MikeMiller Index...
I keep on encountering this concept :P
The analysis is catching up with me
 
@Evinda You could do a program-description presentation, where lots of wacky things happen when you use a program's description as input in odd ways. The Fixed-point theorem, the recursion theorem, quines, and multi-quines.
You'd only have to talk about each for 15 minutes and build up to each of them in order
 
@MikeMiller Can I think in terms of Hodge decomposition?
I don't see why the kernel is finite-dimensional---I thought the "finite part" was in the harmonic forms
 
7:13 PM
That's not Fredholm when $n$ is odd. When $n$ is $4n+2$-dimensional, it's self-adjoint.
@Danu Sure, think in terms of that.
 
I've got to go.
Peace all
 
Ok, thank you... See you @Axoren
 
@MikeMiller So why is the kernel finite-dimensional?
 
Wait, I'm an idiot. This is the Euler charavteristic.
And is always defined.
 
What is the Euler characteristic??
 
7:17 PM
@Danu Write down the Hodge decomp!
the index of this operator
 
So $dA^{n-1}\oplus H\oplus d^*A^{n+1}$
 
And what'a the kernel of d+d*?
 
@MikeMiller I am trying to prove if we have a banach space which is seperable then it has a subset A which is of 2nd category and has non-empty interior.
I was wondering do you have any hint for this problem ?
 
Durr
So
$d$ is zero on first and second, and $d^*$ on second and third
 
what is the permanent of the empty matrix. Is it 1?
 
7:21 PM
Hi all
1
Q: Solving for best fit value $C$ in $\sqrt {Exp_a^{[1/2]} (x) \cdot Exp_b^{[1/2]} (x )}$ ~~ $ Exp_C^{[1/2]} (x).$

mickLet $Exp_t^{[y]} (x) $ denote the $y$ th iteration of the exponential function with base $t$ : $t^x.$ For example $Exp_t^{[1]} (x) = t^x. $ Let ~~ denote best fit. Now as $x$ Goes to positive infinity and a pair $(a,b)$ with $e<a<b$ Is given , I wonder how to find the best fit base value $C$ ...

 
But that doesn't tell me what the kernel is @Danu
@Karim Ask me in an hour
 
@MikeMiller I guess you get things that contain things from different cohomology groups
I mean... definitely the harmonic forms should be in the kernel I guess.
But perhaps you can get some kind of cancellation too
You probably can, in fact
 
@Danu Try to work out precisely what the kernel is. It's not too hard.
It's a generally useful idea.
 
@MikeMiller So in every degree the harmonic forms are in the kernel at least, I guess that much is right. Then for other things to be in it I need something like $d^*d\alpha=-d d^*\beta$
For $\alpha,\beta \in A^n$
Then $d\alpha+d^*\beta$ would be in the kernel
I don't know how to investigate if this can happen or not
Actually, I should be able to say something
Using Hodge decomposition again on $A^n$, maybe
 
7:46 PM
Bye
 
@MikeMiller How's stuff? I vaguely recall you writing something above about your idea not working but then working again.
 
No, I don't think I can get anything except the harmonic forms @MikeMiller.
 
@Danu Do you have a proof?
@AndrewThompson I think it works! I want to check all the details tonight so I can show it proudly to my advisor tomorrow.
 
@MikeMiller I wrote out cases
Ya know $\omega\in H^n$ (I'm using $H$ for harmonic)
 
@MikeMiller Wee, best of luck!
 
7:55 PM
Then I tried $\omega\in dA^{n-1}\oplus d^* A^{n+1}$, got that the components vanish
Then I tried $\omega\in d^*A^n\oplus d A^n$, which is the only other case where I could see something happening, but that doesn't work either.
So I'd say that exhausts the cases, no?
 
@Danu Probably yuou could afford to be a little more careful, but roughly, yes.
The kernel is the even-dimensional harmonic forms. The cokernel will be the odd-dimensional harmonic forms.
@AndrewThompson Thanks.
 
@MikeMiller So what's cokernel?
 
I believe in you.
 

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