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12:01 AM
I see your point, but I just can't get excited about it. I don't know a single person who learned Lebesgue integration before Gauss' law, and I haven't encountered a single physics class that treats Gauss formally.
 
I know, it's pretty insane stuff, been ignoring it for ages, but I just thought of that point and it really makes sense, that Gauss law thing really upset me, if upending the entire theory of integration is the solution, so be it I guess :p
 
@EricStucky Turns out, I was looking to calculate the radius, not necessarily the diagonal line of a polygon. It appears much simpler. mathopenref.com/polygonradius.html
 
12:28 AM
I'm studying homeomorphisms between topological spaces and I'm wondering: are they essentially just relabelings of the elements the same way isomorphisms between groups are just relabelings of the group elements?
 
Yeah
In the early days you could think of topology as the group of bijections that preserved limit points, group elements called homeomorphisms..
 
12:47 AM
hmmmm interesting
 
 
2 hours later…
2:40 AM
Why is this statement false? The "line separation property" asserts that a line has two sides.
 
It asserts that a point splits a line into two rays, right? That if A, B, and C are colinear, with B between the other two, then the line through those points is the union of the rays BA and BC.
 
Yes
 
Not sure exactly how I would interpret the statement "a line has two sides"
My first instinct was that it meant that a line separates a plane into two pieces, or something like that
 
But the plane separation axiom does define sideness?
 
Does it? OK, I admit, I don't really know this stuff, sorry. Looking it up now
Ah, I see. So, then, it's the plane separation axiom that says a line has two sides, not the line separation property. And that's why the statement is false. I think.
@notorious (This statement)
 
2:50 AM
Oh, yes of course. That makes sense. Thank you!
 
I'm not sure, though
 
I was trying to wrap up my head around the calculus of variations and apparently one can define a gradient operator for vector spaces like so? $\left(\nabla_x f\left(x\right)\right) \cdot v =\left. \frac{\mathrm{d} f\left(x + \varepsilon v\right)}{\mathrm{d} \varepsilon} \right\rvert_{\varepsilon = 0}$ I'm not sure this makes sense to me.
 
 
1 hour later…
4:10 AM
hi
 
user147690
4:36 AM
What is 'the cocycle condition'?
 
user147690
Context: Line bundles on complex manifolds
 
4:47 AM
$g_{ab}g_{bc}=g_{ac}$
 
 
1 hour later…
6:07 AM
 
6:31 AM
@MikeMiller How does that actually relate to cocycles from cohomology?
 
no idea
 
What's logarithms used for?
 
user147690
@Danu I couldn't work out how that answered my question, but was hoping more reading would deduce what any of those letters mean
 
user147690
I think they are transition functions
 
Yeah
But I want to know why that's called the cocycle condition. I guess I'll google around a bit.
 
user147690
6:43 AM
@Danu Should be in Hatcher?
 
Wiki leads me here
 
user147690
I tried to read from Hatcher page 198 and made no progress
 
The subsection "cocycle" has the thing I was looking for.
You could have a look at the book on Riemann surfaces by Forster; the first part of the second chapter. It only deals with Riem. surf. (hence 1D, connected), but it's quite elementary in its wording and stuff.
Maybe you can get a feel for the basics from there.
 
user147690
That's actually where I am now haha
 
user147690
Finding my Complex analysis lacking a little
 
user147690
6:46 AM
(Read a-lot)
 
I'm taking Forster's course this semester ^^
 
user147690
Jealous
 
user147690
Envious*
 
I'm unable to attend any of the lectures though :( Schedule clashes...
So I'm just attending the exercise classes (I read most of the book over the break, though I wasn't very precise in my reading/skipped technical sections occasionally)
 
user147690
Now that I know that, you can give me an exposition of the necessary content :D
 
user147690
6:50 AM
So a Riemann surface is a 2 dimensional manifold that has complex charts for each open subset of the manifold
 
Holomorphically compatible, yes
 
user147690
7:10 AM
Do you know where Mike's $g_{ab}$ etc map from and to? @Danu
 
Check out pages 96-97 of Forster
Of course, it depends on which things you're looking at. Are you considering simply the complex structure on the manifold (i.e. transition functions between charts) or some sheaf?
(though the charts stuff may also yield a sheaf, for all I know)
But typically the transition functions map from overlaps of domains of definition/images thereof
 
7:48 AM
So is the support of any test function a ball with center a Lebesque point ad radius $\epsilon$ ?

Also why do we have to assume that $x$ is a Lbesque point?
 
8:03 AM
vector bundles are the same thing as elements of $H^1(X,GL_n)$, where this needs to be interpreted carefully as the latter is a sheaf of nonabelian groups; the point is instead of having an additive condition that a 1-cochain is a cocycle...
you have a multiplicative condition as I wrote above.
this just reflects the fact that the transition functions of your local trivialization satisfy that condition
 
8:14 AM
@Semiclassica Basics. One dimensional motion, vectors, etc.
@AkivaWeinberger The main idea behind why two free resolutions are chain homotopy equivalent is the special property of free modules that every free module is a projective module.
Once you start thinking about it like that, you can reconstruct the proof by yourself.
This is not a geometric fact, but free resolutions can be interpreted geometrically: I motivate them by group cohomology. I can tell you that if you want.
 
how is group cohomology geonetric
 
if you take a K(G, 1), take it's universal cover, then look at the free group on the n-cells of \tilde K(G, 1). G acts on this group by deck transformations on the n-cells for each n. The cellular chain complex then gives me a free ZG-resolution of Z, cohomology of which is precisely group cohomology.
This is of course the same as Ext_ZG(Z, Z).
I should clarify: this is an example of a free resolution coming from geometry.
 
Hi I have been reading Apostol calculus vol 1 and we assumed the existence of measurable sets saying these are sets in plane to which area can be assigned. What's a set in a plane?
 
A subset of $\Bbb R^2$
 
Ah so it's a ordered pair
Am i correct?
 
8:24 AM
A set is not an ordered pair, no.
A single point in $\Bbb R^2$ is an ordered pair
 
I just finished reading dostoyevsky's "the gambler".
 
No I meant by a set in a plane did the book meant a set of ordered pairs of reals?
 
Sure
 
so how does a set of ordered pair of reals is measured or is measurable? Can't all of them be joined by line segments which are axiomatically said to have zero area (in the book)
so all should have zero area
it just isn't making sense
 
Think about a disk.
 
8:33 AM
Alright but a disk cannot be represented in R^2 or R^3 by a finite set of ordered pairs or triplets can it?
 
Who said anything about a finite set?
 
^
 
Oh so all measurable sets consist infinite elements?
 
No... finite sets also have well defined measure. It's just that they have measure 0.
 
(all finite sets do)
 
8:36 AM
Should I read anything else before starting apostol to make it more clear?
 
9:18 AM
hI
 
hi @iwriteonbananas
 
The suspension of an infinite wedge sum is homotopy equivalent to the infinite wedge sum of the suspensions, is that true @MikeMiller?
Hi @BalarkaSen, what're you up to?
 
I have a question
I need to find the ortogontrixaal m
can anybody help me?
 
ortogontrixaal sounds like an alien species from outer space
@iwriteonbananas what am i upto? hmm. not anything particularly mischievous.
 
I have a problem for you: Find a delta complex structure on the presentation complex $X_G$ of a group $G$
and use that to prove that simplicial first homology is the abelianization of $G$
 
user116211
9:22 AM
So, can Riemennian geometry yield Euclidean postulates?
 
Presentaton complex being universal cover of the thing obtained from taking a wedge of circles labelled by elts of $G$ and attaching 2-cells corresponding to relators?
I confuse which one is presentation cplx and which one is cayley cplx
 
Not the universal cover of it. Just that space itself.
 
OK, great.
OK, so intuition says the cells can be chosen to be polygonal. Then I have to triangulate each polygon.
 
That's the right idea
 
That's what I am trying to ponder on. Give me a few minutes.
 
9:26 AM
Sure
 
Yeah, I mean, given a relator $g_1g_2 \cdots g_n = 1$, I can take a polygon with edges $g_i$ (possibly with multiple labellings, and one has to take care of the direction on the edges), and attach that. This has the same effect as attaching a cell. Now if I triangulate this, what do I get? I think I get something like a triangle for each relation $g_i g_j = g_k$.
 
Hmm, no, that's what happens at the level of universal cover.
 
So we need to subdivide that polygon somehow to get a bunch of triangles
 
Right.
 
9:37 AM
How many triangles do we end up with?
 
I mean look at the simplest: $abab^{-1}$. That's a square. If I cut open along the diagonal, what does that tell me? The middle edge corresponds to the fact that $ab = ba$, no?
The upper triangle is the triple $a, b, ab$ and the lower triangle is the triple $b, a, ba$.
Not sure if that makes sense.
 
Don't we need to introduce a new edge if we cut along the diagonal?
 
@iwriteonbananas Yikes, I can't tell you immediately. How many triangles do I need to triangulate a n-gon?
@iwriteonbananas Yes. By "middle edge" I mean the "diagonal edge".
 
@BalarkaSen I actually don't know
 
It's n-1, right?
Pick a vertex, join it with the other n - 1vertices. You get n - 1 edges.
Sorry, n triangles. n - 1 more edges.
 
9:43 AM
Yeah, that sounds like a valid way to triangulate the thing
But I don't think we get n triangles
 
All I am saying is if you subdivide a polygon corresponding to relator $g_1 g_2 \cdots g_n$, then after subdividing that into triangles the "new edges" should tell me $g_1 \cdots g_i = g_{i+1}^{-1} \cdots g_n^{-1}$ for each $i$.
 
Why do the new edges tell you that?
 
@iwriteonbananas Yeah, whoops, stupid count. Anyway it's easy to count what it is.
@iwriteonbananas It's a "commutative diagram", if you think of the edges of the polygon labelled by $g_i$. After you draw a subdivisional edge, you get 2 polygons and the middle edge is the same if you go round one polygon or go round the other. That's precisely what that $g_1 \cdots g_i = g_{i+1}^{-1} \cdots g_n^{-1}$ means.
@BalarkaSen E.g., look at this example.
Oh, now I actually think that makes sense. Wasn't this how you construct BG or EG or whatever that is?
Not sure if remembering right.
 
@BalarkaSen Oh, I see what you mean now
 
Mhm. So it should be, at least on the universal cover, a triangle for each relation $g_i g_j = g_k$.
 
9:50 AM
@BalarkaSen Yeah we start like this but then we need to attach a bunch of higher-dimensional cells to kill higher homotopy groups
What about the first simplicial homology of this space, now that we have a $\Delta$-complex structure?
 
Hrm. Hrm hrm hrm.
@iwriteonbananas It's not immediately obvious to me how to compute the simplicial homology of this space from just the defn.
 
What's $\ker \partial_1$?
 
just cycles around the 1-skeleton of our complex.
cycles as in graph theory
 
How many vertices does our $\Delta$-complex structure have?
 
1, right? because we started out with a single 0-simplex, and attached stuff to it.
I am not sure how is this relevant though.
 
10:05 AM
Right, we started with a vertex and attached a bunch of $S^1$'s to it. Then for each relator we take a polygon, triangulate it, and glue it to the wedge of spheres
 
Mhm.
 
so all vertices of all polygons get identified with that one vertex in the wedge
So $\partial_1=0$ since there's only 1 vertex
 
Right, sure.
 
So $\ker \partial_1$ is free abelian on the set of edges. How many edges do we have?
And what's $\operatorname{im} \partial_2$?
 
There are $|G|$ many edges, I think.
@iwriteonbananas Are we doing an order argument here?
I mean the 1st simplicial homology has to have order smaller or equal to the abelianization, if equality is attained then they are the same (because 1st simplicial homology is abelian, so obviously has the commutators as relators, but it can hav some more things in it in which case the order gets smaller)
 
10:12 AM
I'm not sure :P But there aren't $|G|$ many edges in general. Look at the torus, we only have finitely many edges
@BalarkaSen That's smart, didn't even consider that
 
@iwriteonbananas Yelp, I meant in the universal cover. Down below things are identified because of more relators. Sigh.
Note that this can be easily proved by using the same argument as we do in $H_1(X) \cong \pi_1(X)^{ab}$ in this context. I am just wondering whether there is a distinct, more transparent than the already transparent proof in this context.
I have to go now, gotta get some non-math things done. If I get the time, I'll think about it.
 
@BalarkaSen Right. It should be possible to do this completely simplicially.
Sure, talk to you later
 
The $H_1(X) \cong \pi_1(X)^{ab}$ thing can be done simplicially too, if $X$ is a delta complex.
One has to use the simplicial approximation theorem to get a map $\pi_1(X) \to H_1(X)$.
Alright, byes.
 
Oh, I see
 
10:50 AM
 
 
1 hour later…
12:00 PM
What would you say if I told you "The more intuitive notion of limit, namely the injective or direct limit, arises when all arrows are reversed (i.e. when we have a sequence of embeddings rather than projections), and is the algebraic analogue of taking derivatives"?
"Projective or inverse limits are the way algebra ‘integrates’ an infinite tower of structures into a single structure"
 
Doesn't really make much sense.
 
"integration" means "glue", not integration as in calculus :P
projective limit of things glues stuff togather. on the other hand direct limit "zooms in".
 
"The very rough slogan is: direct limits is what happens when all you care about is the local behavior near a point. Inverse limits is what happens when you patch together local information to get a global object.

*ach*

It honestly takes years to develop this intuition..."

http://qr.ae/8zfbaX

Not buying your criticism ;)
 
12:15 PM
No, your analogy is far-fetched and not of much use. Those quotes do not really mean you should identify one with "integration" and the other with "differentiation". But I am not going to argue with you anymore.
 
It's not my analogy, it's his amazon.com/Moonshine-beyond-Monster-Connecting-Mathematical/dp/… give it a few years
 
OK, let me correct one of my points. Direct limit has more to do with differentiation than inverse limit with integration (the latter is the not useful and far-fetched idea, which is what you originally asked).
 
12:32 PM
Apparently it's better to think of inverse limits as Cauchy sequences, or Cauchy completions math.stackexchange.com/questions/38517/… math.stackexchange.com/q/839272/82615 math.stackexchange.com/q/887613/82615 which is a bit strange, and leaves you wondering what a direct limit is, and wondering why you'd construct such a thin in the first place, yet there's also that diff/int analogue which has something to it
 
12:45 PM
Quick question but hard to google: ||v|| was the square of the length of v right?
 
If $v = (2,3)$ then $||\vec{v}|| = \sqrt{2^2 + 3^2}$
 
So it's the geometric length thanks (2³+3²=13) sqrt(13)=3.6....
 
So not only have inverse limits been described by gluing, integration, completion and cauchy sequences, we now have a power series interpretation too:
"The ring of formal power series can be described as an inverse limit"
http://planetmath.org/formalpowerseriesasinverselimits
What a versatile concept :p
 
It is power series.
It's almost definitional that it is.
 
So if an inverse limit is a formal way to define power series, a direct limit is? Evaluating the polynomial? Truncating?
 
1:00 PM
I think you typoed. You meant direct limit there, not?
 
Thanks
 
Direct limit doesn't have a straightforward power series interpretation.
Think of it as collection of points which "eventually become the same" as you push it forward by the morphisms in the directed set. "Zooming in" is the most appropriate interpretation.
 
In Bourbaki's set theory, 3rd chapter they define orders, ordered sets, well-ordered sets, cardinals, then finite sets, finite integers, euclidean division, basic combinatorics, infinite sets, then inverse limits & direct limits, as though they are all one unified whole in some fashion, so I'm guessing they thought of inverse limits as a set-theoretic way to define formal power series for integers, without having even defined algebra yet
I guess this is risky, but 'double inverse limit' would be composition of power series :p
 
I don't know the history.
 
1:24 PM
morning
my definition of formal power series: "Those things that let me use generating functions without worrying about silly things like convergence" :p
 
1:45 PM
need confirmation and constructive ideas by clicking: math.stackexchange.com/questions/1806831/…
 
2:04 PM
Guys is there a way to see history of this chat room, in text? Or to see my post history here? I don't want to search through each day seperately
 
@Jake1234 In the room info there is link to transcript and search field. And there is also search field in your profile.
To see your recent messages, you can try this: chat.stackexchange.com/users/125616/jake1234?tab=recent chat.stackexchange.com/users/125616/… etc. (Just change the page number.)
 
Thanks.
 
And Krijn's suggestion is very reasonable too. You can search for your messages posted specifically in this room: chat.stackexchange.com/…
Of course, you can do the same with any user/any room.
 
2:26 PM
How do you guys get yourself started?
2
I have been procrastinating for more than 4 hours by now
2
 
3:01 PM
Well, I was going to respond, but hopefully you've figured something out in the last 35 minutes :)
 
@Krijn There is productivity.SE site. Maybe you can find some advice there. (But more likely this is just a different possibility how to procrastinate.)
 
@Eric Watching Narcos now instead of studying Étale Cohomology, soooooo
@Martin I'll check it out!
 
Are you studying from a book Krijn?
For me, I find that reading from a computer is too distracting, because of Netflix &c.
So even if I only have an electronic copy, I tend to print out pages, so that I can actually get reading done.
Anyway, I'm going to go eat breakfast and then start in on my work today :P
 
Well, it's Milne's online notes, but I don't have a printer around :(
 
3:37 PM
Even some e-book readers are relatively good for math books. And you can read this way away from computer. (So far I only bought a reader twice. Always one of the things I was looking for was whether it can handle djvu format.)
 
For me, writing is where my procrastination has become a real obsctacle
partly because there's usually not an alternative to being on a computer for that
and partly because i just find writing really really really frustrating
 
Guys, quick sanity check: $\sqrt{n+1}-\sqrt n\to0$, right?
 
Jup
 
As $n\to\infty$?
 
3:51 PM
Then yeah.
more precisely, $\sqrt{n+1}-\sqrt{n}=\sqrt{n}\left[\sqrt{1+\frac1n}-1\right] \sim \sqrt{n}\frac{1}{2n} = \frac{1}{2\sqrt{n}}$
 
4:04 PM
@Semiclassical using its conjugate might seem more elegant to answer the question :D
 
eh, i like having an error estimate as well :p
 
Let $(x_n)_{n\ge1}$ be a sequence of real numbers with $x_1=1$ and $x_{n+1}=1+x_1x_2\cdots x_n$. Calculate $$\sum_{n=1}^{\infty} \frac{1}{x_n}$$
@Semiclassical this one is cute.
 
i like that one, especially if i can solve it :p
 
4:21 PM
the first thing i see is that, for $n>1$, the recursion can be rewritten as $x_{n+1}=x_n^2-x_n+1$
which is convenient to work with if nothing else
 
@Semiclassical Indeed.
 
@user1618033 btw, that sequence shows up on OEIS as A129871
 
@Semiclassical nice, I didn't know it has a name.
 
which links to A000058 i.e. Sylvester's sequence
and that one has a ton of stuff
 
@Semiclassical @user1618033 Yeah, I did it with the conjugates, but there was someone who was pretty sure it diverged so I was afraid I made a mistake somewhere
 
4:31 PM
@user1618033 see especially the third sentence of the entry on Sylvester's sequence :)
I didn't know the 'greedy Egyptian representation' was a thing
 
@AkivaWeinberger limits are sometimes deceiving
 
@user1618033 At a guess, 2?
 
@AkivaWeinberger Yeap.
 
in fact, the Wiki page on Sylvester's sequence includes a proof of the sum
 
@Semiclassical interesting
 
4:35 PM
u16, do you know if that sum has any combinatorial meaning? I wish it were, like, a probability. Maybe if you cut off the first term?
 
there's a meaning in terms of Egyptian fractions, at least.
so if that has a combinatorial meaning then so does the original one
 
@EricStucky It might have, I didn't think of it enough.
 
4:58 PM
I thought that looked familiar
but it turns out I was thinking about a different reccurence:
$x_n=1+x_1x_2\dotsb x_{n-1}n$
 
morning
 
Evenin'
 
Feb 10 at 20:34, by Akiva Weinberger
$$\frac12+\frac25+\frac3{31} +\frac4{1241}+\frac5{1923551} +\frac6{4440055831261} \approx\\0.99999999999 999999999999969565$$
afternoon
@Semiclassical @user1618033 Can you prove that adds to one?
 
not if it's only an approximation :)
 
$x_n=1+x_1x_2\dotsb x_{n-1}n$
 
5:01 PM
but i presume you mean the summation to infinity not just 6
 
If you sum the infinite series it goes to 1, but I didn't know it when I posted that
(Feb 10)
$\sum\dfrac n{x_n}$
 
and no, i don't.
 
@AkivaWeinberger I don't think it's hard. Let me try it first without pen and paper.
 
I don't remember how to do it but I think I remember it being easy
I just posted it because it's similar to the Sylvester thing
 
It does still grow more slowly than Sylvester, though
 
5:10 PM
Done.
 
Yeah, it wasn't particularly hard
 
Anyone know the answer to this?
118
Q: Super Mario Galaxy problem

JeffεSuppose Mario is walking on the surface of a planet. If he starts walking from a known location, in a fixed direction, for a predetermined distance, how quickly can we determine where he will stop? More formally, suppose we are given a convex polytope $P$ in 3-space, a starting point $s$ on t...

I put a 100 rep bounty on it and it's about to be wasted...
 
5:28 PM
god help me, but while looking up stuff on Egyptian fractions i stumbled upon this
and now my head hurts
@EricStucky pretty sure what i just linked does imply a combinatorial interpretation in terms of 'groupoid cardinality'
but that makes my head hurt
...kind've tempted to do it as an MO question :P
 
@AkivaWeinberger before writing down the key relation, well, it's not the kind of question anyone can do.
Does any want to write down the answer before my post?
(I wouldn't like to distroy your fun)
 
I think @Mike might be trying to solve it, but I'm not sure
 
Just let me know when to write it down ...
OK, let's not wait anymore
$$\frac{1}{(n+1) (x_{n+1}-1)}=\frac{1}{n (x_{n}-1)}-\frac{1}{n x_n}$$
@AkivaWeinberger ^^^ Q.E.D.
 
Yeah, QED
 
I'll send it now by email to some students I help once in a while.
(although I doubt anyone will finish it)
 
5:44 PM
You made a typo, I think. It should be $\dfrac1{nx_n-1}$, without parentheses
We also have $\dfrac n{x_n-1}-\dfrac n{x_n}=\dfrac{n+1}{x_{n+1}-1}\\$
 
For $f \in k[X_1, \ldots, X_n], f_*$ is the homogeneous part of $f$ of the lowest degree
What does that mean?
 
@user1618033
 
@AkivaWeinberger let me put it on paper and check it
 
what was the ping about?
 
@Krijn Thanks!
 
5:52 PM
@AkivaWeinberger True, there was a typo at the fraction $n/(n+1)$. It was turned upside down.
$$\frac{n+1}{x_{n+1}-1}=\frac{n}{x_n-1}-\frac{n}{x_n}$$
like yours.
@Semiclassical you should try that, it's pretty cool.
 
6:27 PM
And note that the fraction on the left can always be simplified; $(x_{n+1}-1)/(n+1)$ is an integer.
 
@AkivaWeinberger How long did it take to you to get a solution for the first time?
 
I actually did it backwards — I first found the sequence, and then I had to figure out what the recurrence relation to my own sequence was!
I'll explain:
 
@Semiclassical: that looks like a particularly readable introduction to groupoids, actually. I'm going to read that on the airplane tomorrow.
 
This is what you get when you do the greedy algorithm. First term has to be $\dfrac12\\$, or else you're already at $1$. Second term has to be $\dfrac12+\dfrac25$, because $\dfrac12+\dfrac24$ is already $1$. Etc
The next term is $\dfrac12+\dfrac25+\dfrac3{31}$ because $\dfrac12+\dfrac25+\dfrac3{30}$ is $1$.
 
Time to do Guillemin-Pollack again.
 
6:39 PM
@AkivaWeinberger I see. I have a neat, very simple solution to it.
 
Of course, the numerators can be anything, not just the integers.
$$1=\frac{\sqrt2}2+\frac{\sqrt3}6+ \frac{\sqrt5}{531}+\frac{\sqrt7 }{376169}+\dotsb$$
 
I go out jogging. Let me know if you want me to post the solution.
 
I wonder: If you make the numerators $\sqrt n$ and do the greedy sequence, do you get a nice recurrence relation for the denominators? Doubt it.
Didn't you already do that? @user1618033
 
@AkivaWeinberger Some might not see the steps I used, how I started.
 
Ah.
Sure, why not
 
6:44 PM
$$x_n=1+x_1x_2\dotsb x_{n-1}n$$
$$x_{n+1}=1+x_1x_2\dotsb x_{n}(n+1)$$
$$=1+(1+x_1x_2\dotsb x_{n-1}(n+1)-1) x_n$$
$$=1+(\underbrace{1+x_1x_2\dotsb x_{n-1}n}_{x_n}+x_1x_2\dotsb x_{n-1}-1) x_n$$
 
Hey, could someone help me understand something with Algebra 1 because I simply not getting it? It involves "Write exponential functions: tables & graphs" on Kahn Academy?
 
$$=1+x_n^2+\underbrace{x_1 x_2\cdots x_n}_{(x_{n+1}-1)/(n+1)}-x_n$$
 
@AkivaWeinberger I am not sure if you saw what I pinged you a couple hours ago. If you did, it's ok.
 
@AkivaWeinberger that's the story.
(the rest is arranging things to the form I used)
 
6:50 PM
I guess that No one can help me?
 
@BalarkaSen This?:
11 hours ago, by Balarka Sen
@AkivaWeinberger The main idea behind why two free resolutions are chain homotopy equivalent is the special property of free modules that every free module is a projective module.
 
This, and the next messages.
 
Out for some jogging.
 
Did you expect me to remember/know what a projective module is?
 
No, I expect you to google the definition and see for yourself why a free module is always a projective module.
 
6:52 PM
Ah. I Googled it this morning.
 
And why it is relevant in the proof of ext being independent of choice of free resolution
 
Honestly, right now, I want to focus on other stuff. Mostly Hopf.
And Rudin's analysis book, though I don't need to think about that 'til Monday (school).
 
OK, that is fine. What's Hopf?
 
Hopf algebras. Sorry.
 
Analysis is good.
@AkivaWeinberger Fair enough.
 
6:54 PM
Also I picked up a group theory book on Saturday because I didn't know the Sylow theorems and I felt I should.
 
Good that you're filling up the basics.
 
I mean, I still don't. I spent that day reviewing the chapters before that.
But that was the intention.
 
I was reading up on this Donaldson/Yau/Sun vs. Tian controversy... Has anything been cleared up since 2013?
 
Artin does basic Sylow theory very lucidly and geometrically, in case you'd want to read it.
At the very least, so I thought when I read it.
 
Also, I really should be studying for chemistry, since I have the SAT2 test on it on Sunday
 
6:57 PM
yuck
but good luck
 
Acids are face-meltingly beautiful.
 
(that rhymed!)
 
@MathCubes: I don't have a KA account and so it won't let me see the question.
 
Oh that is new
Okay
 

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