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12:48 AM
@Mats Granvik, I wanna counter that so bad, but I just can't.
2 hours later…
3:09 AM
It's pretty quiet in here tonight.
@robjohn good evening sir
U was watching this video
@Yuvraj hey
Is there any mathematical approach we can do it?
@Yuvraj I'd have to see the rules
Rules are simple, only thing we are given that mouse is in water
And cat can walk at speed 4times the maximum speed of mouse
Mouse have to escape
3:25 AM
what constitutes escape?
getting to a point on the edge before the cat?
Yes sir
is there any rule for what the cat does depending on where the mouse is?
How does the cat move, does it predict where the mouse will be, or just move in the direction of the mouse?
It moves on the circle towards the point where the cat is moving.
oh, the cat is constrained to the circle? I think ive seen the solution before but in the context of some predator and a person in a boat
3:29 AM
yup that was enough information to use some googlefu
3:44 AM
That is the circle of points to which a mouse can get to first if it is traveling $\frac14$ as fast.
This does not account for the circle restriction on the cat
2 hours later…
5:48 AM
@Thorgott sorry. I don't play higher topos theoretic Counter Strike.
3 hours later…
8:25 AM
How to search for a sequence? I have this sequence: 1,-1,-1,1,1,-1,-1,1.. starting from n=1. How do I find it's general term?
8:50 AM
Flatten[Position [(1 + {1, -1, -1, 1, 1, -1, -1, 1}), 2]]
{1, 4, 5, 8}
Flatten[Position [(1 + {1, -1, -1, 1, 1, -1, -1, 1}), 0]]
{2, 3, 6, 7}
9:09 AM
I don't understand that, but I actually found the sequence, on the same site
1 hour later…
10:14 AM
@feynhat Counter-Strike: Global Section
11:07 AM
Let $R$ be a noetherian ring and $I \subseteq R$ an ideal, and $S \subset (R\setminus \{ 0\}, 1, \dot)$ a submonoid with $S \cap I = \varnothing$. Then $0 \to I \hookrightarrow R \to R/I \to 0$ is a short exact sequence of f.g. $R$-modules, so $\operatorname{Hom}_{S^{-1}R}(S^{-1}(-), S^{-1}N)$ is exact and we get that $\operatorname{Hom}_{S^{-1}R}(S^{-1}I, S^{-1}N) \subseteq \operatorname{Hom}_{S^{-1}R}(S^{-1}R, S^{-1}N)$..
And then Baer it
So $S^{-1}N$ is injective in $_{S^{-1}R}\text{Mod}$ ?
@Balarka @Thorgott or no
1 hour later…
12:11 PM
quick question can one change the order in the following vector operation or not?

$ a,b,c$ vectors $a.(b\times c)$
1 hour later…
1:41 PM
$S^1 \subset \Bbb C$ acts on $S^3 \subset \Bbb C^2$ as: $w \cdot (z_1, z_2) = (wz_1, w^2z_2)$. What is $S^3/S^1$?
@Edward I'm not super familiar with injective modules, but that sounds reasonable
since every ideal in $S^{-1}R$ is of the form $S^{-1}I$
1:53 PM
Right, and Baer only requires me to Check that property for one ideal right?
@feynhat Hopf reacc only
I don't really know Bear, but Wikipedia says all primes
No for an arbitrary ideal lol
Yeah for noetherian rings it suffices to Check primes
2:55 PM
@EdwardEvans What are das alphabets? Are English alphabets called das alphabets?
3:06 PM
@EdwardEvans das alphabet
The English alphabet doesn't have a special name
It is just referred to as "The Alphabet"
Okay, but please see this google.com/…
the English alphabets A, B , C .... are named as das alphabet
In Germany, "the" is "das"
Oh okay!
Thank you so much!
What does this mean "All letters of the German alphabet have the same article: das."
letters don't have articles before them, words have.
3:18 PM
I don't speak German but I presume it means, the letters are called "das A, das B, das C" like how we call "the letter A, the letter B, the letter C"
3:34 PM
I'm trying to arrange tiles with a total area of 125 units into a 2d shape. Since 125 is not a square or rectangular number, they would not fit there. Any suggestions?
@LeakyNun I see. As a topological space its just $\mathbb{CP}^1$.
What else would you be considering it as?
3:50 PM
I tried pixel circles but they only work with 124 and 126. Triangular numbers also produced areas of a slightly incorrect size
The action is free though.
Since 125 is a cube number, it works with cube nets but I'm looking for an actual shape.
@Knight yes, letters take the article "das"
Is it?
(0, 1) is a fixed point.
4:51 PM
@Thorgott PlayerUnknown's Classifying Spaces
5:08 PM
5:24 PM
Gaussian distribution, Students T distribution, Binomial distribution etc. are so same looking. Are they actually same?
their graph i mean. looks so same
(I dont have a mathematical background btw. i m trying to understand biometry in conceptual way)
Also the Witch of Agnesi looks so same as Gaussian
no, they're different distributions (one even is discrete whereas the other two are continuous)
what you are observing is the central limit theorem
if you take a number of independent random variables with the same distribution, add them up and properly normalize this, you will get something that's Gaussian in the limit
What factors determine a probability distribution will follow which distinct distribution pattern?
such as when it will not be a normal distribution
Is it the "bias" that shift a distribution from gaussian (normal) to something else?
5:40 PM
@RichieBendall or der or die
Unlike in English, there is a thing called gender, and in German and Latin there are three options, not just two. das is neuter, der is masculine, die is feminine.
Das Kapital. Der Mann. Die Frau.
the classic examples, of course
And, just because it makes so much sense, das Fräulein, das Mädchen.
That's because of the diminuitive suffix in those cases, right?
heya @Fargle
5:54 PM
Wow, I remember two things from high school German. And heya Ted.
(The other, incidentally, is Rindfleischettiketierungsueberwachungsaufgabenuebertragungsgesetz.)
hi chat
that's a great one to remember
Salut, @Astyx
5:57 PM
@Thorgott My ability for recalling useless information is astounding. It's also why I love pure math
Thankfully, in my five semesters of university German, I never was subjected to memorizing page-long words :P
It was just a curiosity for extra credit. If I really wanted to be a stickler I'd have put the umlauts too.
Yes, I discounted your contribution for the lack of umlauts.
But I properly Romanized it!
Yes, now why no ß turned into ss?
5:59 PM
Partial credit for knowing that it's "ueber", at least, come on.
Any of you all ever read Apostol's Cal 1?
No double s's. Blame the Deutsch for that one.
I read both volumes, @Naganite, but no longer own the books.
Right so I'm peeking at the first few pages, and it's talking about solving for the area under the function $f(x) = x^2$.
I think I've thumbed through a bit of 1, but a long time ago.
6:01 PM
you messed the t's up though
it's Etikett, not Ettiket
Basically the guy goes through this whole process to solve for the true area.
This is what Archimedes in fact did a long long time ago, @Naganite.
(Inner and outer sums).

So he ends up with something like this: $s_n < A < S_n$
@Thorgott I'll take a 61/63.
Yeah, so I've heard, @ted
6:02 PM
they shall be granted to you
But my question is, how does he come up with the notion of the area being $\frac{b^3}{3}$?
I've got my copy around somewhere. Let me look.
Because $S_n$ and $s_n$ both approach that number as $n$ gets bigger.
I mean, it makes sense once he assumes that it is the area, because then he disproves that it is the area for all %n%, no less and no greater.
Whoops, don't know how to italicize.
I don't understand your sentence, however.
6:04 PM
But IDK, it feels pulled out of thin air.
No, no. $b^3/3$ is the UNIQUE number $A$ satisfying $s_n<A<S_n$ for all $n$.
Since $S_n-s_n$ can be made arbitrarily small by making $n$ big enough, there can only be one single number $A$ that satisfies that inequality for all $n$.
Right right, the only area that works for any and all $n \ge 1$.
I get that part, and it's really elegant.
It's just the act of *choosing* $\frac{b^3}{3}$ that throws me off.

It makes sense a priori but not at first glance.
What's the form of $s_n$ as he gives it?
Right. You look at the formulas for $s_n$ and $S_n$ and the $b^3/3$ falls out.
Fargle asked the key question.
Does anyone know a $K(\Bbb Z/2 \ast \Bbb Z/2, 1)$?
6:08 PM
$$s_n = \frac{b^3}{3}(1^2+2^2+ \cdots + (n - 1)^2)$$
$$S_n = \frac{b^3}{3}(1^2+2^2+ \cdots + n^2)$$
Can I take wedge of two $\mathbb{RP}^\infty$'s?
You can.
Those are the two sums.
Surely those have got to be divided by something in terms of n.
Now you have to show that if $\pi_k(X)=0$, then $\pi_k(X\vee X) = 0$.
@Naganite, no, not quite right.
6:09 PM
(Still can't find my copy. Might be at my folks' place...)
Oh dang, did I make a mistake?
I mean that's just van Kampen.
@feynhat: VanKampen is for $\pi_1$, no?
I need to show it has a contractible universal cover.
I tried editing my message but I messed it up further. :(
6:11 PM
You should have $S_n = \frac{b^3}{n^3}(1^2+2^2+\dots+n^2)$.
Oh, yes. $\frac{b^3}{n^3}$, not $\frac{b^3}{3}$, my bad.
And then you happen to "remember" a formula for $1^2+2^2+\dots+n^2$.
Right. He does derive the formula for the sum of the squares of the first $n$ natural numbers.
So, what does $\dfrac{1^2+2^2+\dots+n^2}{n^3}$ look like as $n$ gets bigger and bigger?
The inequality he makes with the true area and the approximate ones is clean.

It's just the whole "so yes let's assume the true area is $\frac{b^3}{3}$ that throws me out of wack.
6:13 PM
No, no, there's no "assume."
He asserts, as I said, that that is the unique number satisfying the inequality for all $n$.
He then proves it afterwards.
Is it true that if $\widetilde X$ is a universal cover of $X$, then $\widetilde X \vee \widetilde X$ is a universal cover for $X \vee X$?
Guess not.
So then would it be solvable if you assert an incorrect area?
What do you mean "it would be solvable"?
How do connected sums work? Say we have two manifolds $M,M^{\prime}$ and charts $\varphi\colon B^n\rightarrow M$, $\varphi^{\prime}\colon B^n\rightarrow M^{\prime}$. Then we glue together $M\setminus\varphi(0)$ and $M^{\prime}\setminus\varphi^{\prime}(0)$ by identifying $\varphi(x)\sim\varphi^{\prime}((1-||x||)x)$.
On its own, I think this is just identifying one ball minus origin with the other ball minus origin turned inside out, but what does this actually do when these balls are embedded into manifolds? And why do we need to do it this way?
6:16 PM
That's not the way I usually think of it, @Thorgott. I think of gluing the actual balls together.
Like, supposing he said $\frac{b^3}{4}$ for the area, could we still somehow arrive at $\frac{b^3}{3}$ being the one true area (given that it is applicable for all $n \ge 1$)?
You're messing up the logic, @Naganite. He's arriving at the $b^3/3$ from the $s_n$ and $S_n$ formulas in the first place.
Or is it a necessary step in the process to first bring up $\frac{b^3}{3}$?
No, you discover the $b^3/3$ from these formulas and then you check, as I said, that there's a unique number that works because $S_n-s_n$ gets arbitrarily small.
But why do we turn one of the balls inside out, so to say? As in, what does this $1-||x||$ term accomplish?
6:18 PM
As $n$ increases.
Because the ball is "inside" its boundary for one manifold and "outside" for the other. Think about just gluing together two spheres.
Yes, @Naganite.
But looking at the formulas, if I recall, you'd get this:

$$s_n = \frac{b^3}{3} - \frac{b^3}{2n} + \frac{b^3}{6n^2}$$
$$S_n = \frac{b^3}{3} + \frac{b^3}{2n} + \frac{b^3}{6n^2}$$
I can't use the word "limit" because Apostol saves that for later.
Yes, and as $n$ gets bigger and bigger, each of those gets closer and closer to $b^3/3$.
(BTW, for your future work, keep the $b^3$ factored out once and for all.)
And I mean, yeah, $\frac{b^3}{3}$ is in there,
but what's to say you don't guess $\frac{b^3}{3} + \frac{b^3}{6n^2}$, you get me?
The answer can't have $n$ in it.
6:21 PM
@TedShifrin Yeah my bad I'll factor it out next time.
It's a habit that will save you lots of sloppy errors as you proceed.
Ah, I see now.
The "true" area can't be dependent on $n$?
@TedShifrin Any hints for this? I only know one way for any pi_k computation: homotopy LES of a fiber bundle.
I haven't thought about this in a while. I'm sure Hatcher addresses this.
You need a contractible universal covering, of course.
@Naganite There is one and only one number that is below all upper sums and above all lower sums (in the event that you have something where it makes sense to integrate). If it depended on n, there'd be infinitely many such numbers, and therefore there wouldn't be a well-defined area.
6:24 PM
So, if $Y$ and $Z$ are contractible, is $Y\vee Z$ contractible?
@Fargle: That doesn't logically make sense. We need a number $A$ to start with, not a function of $n$.
I suppose, yes.
Oops, typo.
Makes sense.
Can't you write a proof, @feynhat?
Having the integral depend on $n$ doesn't make sense in the first place, so I reserve my right to babble.
6:26 PM
And I mean maybe this is opening a can of worms on its own, but then doesn't that beg the question as to whether or not we can assume there is only one such area?
He's not assuming it. He's proving it.
He asserts it. He doesn't assume it.
You need to learn to be careful with the logic.
Yeah, still working on that, hahaha.
I'm not seeing the issue
An instructive example might be the function
$$f(x) = \begin{cases}0 & \text{if }x \in \Bbb Q \\ 1 & \text{if }x \notin \Bbb Q\end{cases}$$
Instructive because, in this case, there is no number which is less than all lower sums and greater than all upper sums. (Why?)
6:31 PM
@Thor: Think about the orientation on the boundary circles.
@Fargle: Specify an interval, say $[0,1]$.
Can't edit anymore, but yeah, pretend I did that.
It's annoying that we get only milliseconds to edit.
Ouchies, having a tough time understanding that.

So if $x$ is rational, $f(x) = 0$, otherwise $f(x)=1$?
Not getting the relationship between that function and lower and upper sums.
6:33 PM
It's not a pretty function. And good luck drawing it.
Yeah, hahaha. The real number line is weird like that.
Ah, they're opposite
So is this just an orientability issue?
Yes. That's why the funny stuff happens with the gluing.
In algebraic geometry, this shows up explicitly when you talk about blow-up of a point in a complex $n$-manifold and this turns out to be connect sum with a $\Bbb CP^n$ with reverse orientation.
@Naganite I see that Apostol has not yet defined upper and lower sum rigorously. No matter---divide the interval [0,1] into pieces. The upper sum for that division is where you make the highest rectangles possible on those pieces, and the lower sum is the one where you make the lowest rectangles possible on those pieces.
(Found my copy!)
He has defined "circumscribed" and "inscribed," no? That's what you're doing.
6:36 PM
@Fargle Oh yeah, he does that post-Order Axioms, right?
Hey guys, would love to get some direction on the following problem I have to solve. The concept is that a Monster is positioned at (x1, y1) and the player is positioned around him (x2, y2). Once the Monster attacks the player, the player must get pushed backward by 3 points.

So given a slope - (y1 - y2) / (x1 - x2), starting point = (x2, y2) and center point (x1, y1), and d is the distance that the new point should be at.

Are there enough parameters to get this solution?
Possibly? Been a while.
I thought the function $f(x)$ was strictly defined to be either $0$ or $1$.
So will doing the gluing "naively" as in my picture give something diffeomorphic to the thing we get from doing the gluing properly, but simply not carry a natural orientation consistent with the orientation of the summands?
in this specific case, it's clearly still a sphere (as it should be)
@Naganite That's not an issue.
6:38 PM
Or is it the whole interval $\left[0,1\right]$?
$x$ is in $[0,1]$. $f(x)$ is always either zero or one, according to whether it's rational or not.
@Thor: Something non-orientable isn't diffeomorphic to something orientable.
Oh, that's where I was confused. I thought $x \in \mathbb{R}$ no strings attached.
it's still orientable, just not in a way that extends the orientation of the summands
@Naganite To be more precise: if you've split the interval into pieces, the upper sum is where each rectangle's height is the supremum of the function in each piece, and the lower sum is where each rectangle's height is the infimum et cetera.
Yeah, sorry, it was too late to edit the example. Ted pointed this out too.
6:42 PM
if you do the gluing in my above picture, you still get a sphere, which is orientable, but either orientation on the sphere will not extend the orientation of one of the summands in that picture
Yeah yeah, saw that but I didn't understand what you two were referring to.
Let $f: X \to \{x_0\}$ and $g : Y \to \{y_0\}$ be homotopy equivalences. Define $F : X\times Y \times I \to \{x_0\} \times \{y_0\} \times \{0\}$, $F(x, y, s) = (f(x), g(y), 0)$. This descends to the quotient $X\ast Y$ and has an obvious homotopy inverse.
So, if you've split the interval, say, in half, what do your upper rectangles and lower rectangles look like?
And then the true area is just the process of squeezing both sides (AKA $n$ is increasing).
By contractible, I mean "has homotopy type of a point" and not "deformation retracts to a point".
6:43 PM
No, like, what are their heights?
For this specific example of this specific function.
They'd be closer to eachother height-wise, right?
How can they be?
I never said deformation retract, @feynhat, did I? Still, the standard definition is that the identity map is homotopic to a constant map. Isn't that what "homotopy type of a point" means?
6:44 PM
The only values the function takes are zero and one. Those are the only possible rectangle heights.
But, anyhow, you're done, aren't you?
@Thorg: I'm not following. It's like saying a Möbius strip and a cylinder are the "same"? How can your thing be orientable if you glue wrong?
@TedShifrin Some texts write contractible for deformation retraction to a point (I guess Bredon). I was just clarifying what I meant.
Whoa. I've never heard of such a thing.
Do you agree that if you do the gluing as in my picture, you get a sphere (up to diffeomorphism)?
Well then, I must be wrong. Let me check.
6:48 PM
Oh, maybe in this dimension it will work, actually. You are flipping the "outside" of the disk in the second sphere into the inside of the first ball.
oh okay. This is the same thing as the weak deformation retraction.
How so? There's nothing really happening to the outside of the disk.
Anyway, how does this helps me conclude anything about wedge sums?
Is something interesting happening here?
@feynhat: Talk to @MikeM.
@Thorg: So how am I interpreting your picture?
6:59 PM
You have about 6 minutes, at which point I'll take a nap.
51 mins ago, by feynhat
Does anyone know a $K(\Bbb Z/2 \ast \Bbb Z/2, 1)$?
00:00 - 19:0019:00 - 00:00

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