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12:00 AM
what is the other side of the paper becoming? it rolls into the circle, so it's the circumference of the circle!
 
but if we unroll the cylinder it turns just into a rectangle right like the circle dissapears
 
right, but the long part of the rectangle rolls to make the circumference of a circle, so that's $2\pi r$. Area of the cylinder: $(2\pi r)(h) = 2\pi rh$. Presto.
 
oh
so how do we get 2 pi r h
for surface area
we add it right
 
We just did. The length of the rectangle is $2\pi r$ and the height is $h$.
 
i mean 2pi * r^2
isn't that added to 2pi r h to get the surface area
 
12:06 AM
Oh, you add one or both circles if the can has a bottom or top and bottom.
You have to decide that from context.
 
ohh ones area the others circuference nvm i think i got it
thanks for helping me understand @TedShifrin
 
not circumference — $\pi r^2$ is area of a circle (think about units ... length$^2$)
The $2\pi r$ we used earlier has units of length — that's circumference.
 
All the Latin I know is Et tu, Brute? and Civis romanus sum, plus some phrases from the Catholic traditional liturgy
 
Et tu, Brute is fancy, though. That's the vocative case.
How's math going, Brody?
 
area of cylinder is 2 pi r * h because the circle the cylinder makes needs to be taken into account right so we can multiply it with the length
 
12:11 AM
Hah! cases
 
with the height, @MATHASKER, yeah.
 
@TedShifrin I'm still struggling a bit with definitions from analysis
 
really? like which?
 
Several, lol. And it's only basic stuff in $\mathbb{R}$
 
to get the area, so to find surface area we just find area of top and bottom circle and add it to the area of the cylinder, @TedShifrin
 
12:13 AM
@MATHASKER: If they ask you for a cylindrical can with top and bottom. But be prepared to think.
@Brody: You ain't tellin'?
 
@TedShifrin Compactness
 
oh ok thanks i'll try lol @TedShifrin
 
Didn't we talk about something with compactness a few weeks ago, Brody?
 
Not really. It was more about denseness and isolated points. We hadn't learned about compact sets yet
 
Back
 
12:15 AM
oh, ok, Brody. So is there something you wanna discuss?
wb Zach
 
Let $X\subset\mathbb{R}$. $X$ is compact if every open cover $C$ of $X$ has a finite subcover.
 
Sure.
 
That's the definition I know. Also, I think Heine-Borel gives the closed and bounded characterization.
 
Yeah.
So have you had to do some proofs with these?
 
cool
 
12:17 AM
Thanks Ted
Time to do some work.
 
Set $A=\{\frac{1}{n}:n=1,2,3,4,\ldots\}$.
Prove $A$ is not compact by the above definition.
 
OK.
 
Consider $A$ has open cover $C=\{(\frac{1}{n},1+\frac{1}{n}):n=2,3,4,\ldots\}$.
 
OK, it might be easier just to make the right-hand endpoint be 3/2 or something not varying.
 
Yeah, true.
 
12:20 AM
What's your argument that finitely many can't do?
 
Continuing anyway, the union of any finite subcollection of these intervals has infimum $\frac{1}{m}$ for some $m\in\mathbb{N}_{\ge 2}$, so basically many many points of $A$ are not covered here. QED?
 
Correct. You need to write a bit more explicitly, but sure.
 
Yeah, just offering the idea. We may consider $\frac{1}{m+1}\in A$
 
So if you name the sets $U_n$ you could take the maximum $n$ (which you seem to be calling $m$).
 
Ah, right. The index would be $S=\{1,\ldots,m\}$, so take $\max S$ for the infimum of the union.
mmm wait
 
12:26 AM
Well, no, $S=\{n_1,\dots,n_k\}$ for some $n_i$.
 
Yeah, that's how we need to formulate it :/
Let $n^*=\max S$ and consider $\frac{1}{n^*+1}\in A$, but $\frac{1}{n^*+1}\notin \bigcup_{j=1}^k U_{n_i}$
 
Right.
 
whoohoo
thanks
 
I didn't do much.
 
felt weird at first putting subscripts on subscripts, but it makes total sense
 
12:31 AM
You could write it in words, but otherwise, ... yeah.
 
Now, use the definition to prove $A\cup\{0\}$ is compact.
 
So now you must work with an abstract, arbitrary open covering.
 
Yeah :(
 
So why's it gonna work?
 
$B=A\cup\{0\}$ is closed (contains the accumulation point 0) and bounded.
 
12:34 AM
But open covering ...
 
Dunno...
hmm
 
What's different about having $0$ there?
 
Well, I don't really grasp the definition intuitively
 
But think about my question!
 
that's the problem. I don't see how this relates to open (sub)covers
 
12:39 AM
What does it mean to have an open cover, though?
 
I only know the definition
 
Go on.
What does "cover" mean?
 
the compact set is a subset of the cover?
 
Literally, each point has to be in some open set.
 
Okay, right
 
12:41 AM
Now revisit my question.
 
It just occurred to me that atop next year will likely conflict with combo :/
 
Since 0 belongs to at least one member of the open cover, it is interior to the cover.
 
There's some open set in the cover containing $0$. Why does that help you?
Well, you might ask your adviser to let the person who schedules classes know that you'd like to take both, or tell that person yourself, Demonark.
 
Does it guarantee all the points by the accumulation will be taken care of?
 
Huh?
 
12:46 AM
I mean all the $\frac{1}{n}$ near zero can be covered finitely. But this is not guaranteed
 
What does it mean for $0\in U$ and $U$ open?
 
I will try, though I think class times have basically all been decided for years
 
well, then prioritize, Demonark
 
Yeah, this will be a bit of a debate for sure. Though I'll say it leans at least 65% to combinatorics
 
0 is interior to $U$, so some neighborhood around zero is a subset of $U$.
 
12:48 AM
Right, @Brody. Write down what that tells you and then you're almost done.
 
I said that above :(
well, sort of
 
The whole point at this stage is saying things carefully and correctly.
Well, a lot of the whole point.
 
True. It's interior to some member of the open cover
 
I need to eat dinner and go play bridge, so I'm gonna let you finish. But if you write down that $(-\epsilon,\epsilon)\subset U$ and think about it, you'll be two sentences away from done.
Well, maybe three.
Keep me posted, @Brody.
 
See you @Ted!
 
12:50 AM
See ya, Demonark.
 
I'm about to dinner as well, @Ted. Thanks for the help so far. Will do
 
See ya.
 
1:08 AM
Is the AMS website mantienence extended?
 
1:27 AM
I've finished dinner and written up a proof @Ted. Define $B=\{0\}\cup\{\frac{1}{n}:n=1,2,3,\ldots\}$. $B$ is compact. Proof: Suppose $C=\{U_s\}_{s\in S}$ is an open cover of $B$. Then, $\exists r\in S:0\in U_r$. Since $U_r$ is open, $\exists\epsilon >0:N_\epsilon(0)\subset U_r$. Thus, $B'=\{0;\frac{1}{n},\frac{1}{n+1},\frac{1}{n+2},\ldots\}\subset U_r$ for $n=\lceil \frac{1}{\epsilon}\rceil$. The remainder $B\setminus B'$ is finite and trivially has a finite open cover.
Call that cover $C'$. And $B\subset C'\cup U_r$.
 
1:52 AM
Hey, I'm an amateur mathematician, but I have quite a bit of formal math education too. including a math minor. I just don't have a PhD or Masters degree.
so I posted a question on math stack exchange, but now I'm second guessing, thinking maybe it should have gone on math overflow
is there anyone who has a good sense of this who could help me figure it out?
 
well, 6 hours isn't much time for something like clifford algebras (which is specialized/advanced for MSE). I'll take a look.
quaternions is denoted by H, not Q
"isomorphism $f:X\mapsto A$" doesn't make sense; you mean isomorphism $f:C\ell(p,q)\to A$
 
ahh, good looking
(thanks for the feedback)
 
the isomorphism is an isomorphism of k-algebras (k being reals or complex numbers). the only grading that Cl has is the "super", Z_2 grading, and it only respects that grading if you put the appropriate grading on the matrix algebra for that to happen
besides being an isomorphism of algebras, I'm not sure what you mean by its "character"
it seems like in you want explicit formulas for where the isomorphism sends standard basis matrices and standard basis elements of the clifford algebra
since the isomorphism is built recursively, it will probably not be very pretty, and will also be "unnatural" (dependent on many choices built into constructing the isomorphism)
 
exactly. i meant "character" in a qualitative sense, since i'm also curious about the surrounding theory.
interesting, so it's built recursively, huh? see that's an answer about the "character" right there :)
I'll put "character" in quotes
 
indeed, that also gets to (3); the fact that various clifford algebras are isomorphic is a manifestation of the recursive relations they satisfy
the physicists would have more experience handling coordinates for everything
 
2:03 AM
so to explicitly construct the isomorphisms for higher and higher degrees, you need to build on the isomorphisms of lower degrees, according to the relations from the bott periodicity?
that's fascinating
it reminds me of a fractal
(that's partly because I study fractals, so everything reminds me of a fractal in some way or another)
 
there's more recursion than just bott periodicity
I have them written out somewhere, I'll look
 
I'm so excited that I came in here
haha
 
C(2,0)⊗C(r,s)=C(s+2,r)
C(1,1)⊗C(r,s)=C(r+1,s+1)
C(0,2)⊗C(r,s)=C(s,r+2)
with C(2,0)=H, C(1,1)=R(2), C(0,2)=R(2). (where R(2) means 2x2 matrices over R)
note also C⊗H=H⊗C=C(2) and H⊗H=R(4)
I can write an answer in the next few days
 
wow this is so exciting! I'd love to read your answer!
 
(-:
 
2:09 AM
and you're saying these relationships are not solely related to the bott periodicity?
(of course my understanding of the bott periodicity is pretty meager)
 
well of course they're related; you can derive bott preiodicity (meaning the mod 8 periodicity of the morita equivalence class of clifford algebras) immediately from them
if by bott periodicity you mean the mod 8 periodicity of the homotopy groups of O(inf), then that's another story
(still related, just not easily)
 
gotcha, that makes sense. related deeply to bott periodicity, but not as much in a way relevant to the specific manifestations of these isomorphisms
i think this website has the potential to really kick my math education to the next level
 
if you want references, John Baez has quite a number of online posts on them, Lawson & Michelson's Spin Geometry has introductory chapters on clifford algebras, as do the online notes of the same name by Figueroa
 
oh yes, I should really check his posts. I saw a few actually and they were very helpful, but I'll read them in more detail
hey, thank you so much. I gotta bounce out for a while, but I'll be back around!
 
also, if you use the "super" tensor product instead of the normal one (which gives different results), you have the dumb recursion C(r,s)⊗C(p,q)=C(r+p,s+q). :)
see ya
 
2:42 AM
@arctictern see now that's fascinating
 
3:09 AM
hello, (ladies) and gentlemen!
Will be happy to get a response on some questions! 1) What is a tail sum? 2) What is a telescoping identity? 3) How it should be read here: $\sum_{j=k}^{\infty}\left( \frac{1}{b_j}-\frac{1}{b_{j+1}}\right)=\frac{1}{b_k}$? If $j=k$, why just not to exchange all $j$-s to $k$-s? But then how $b_k^{-1}-b_{k+1}^{-1}$ be $b_k^{-1}$ again? Thanks.
 
 
5 hours later…
7:59 AM
@PVAL @MikeM I went to sleep yesterday. Thanks, that's very interesting.
 
 
4 hours later…
11:32 AM
Well, MSE has hit a new low I guess. An OP literally stating "I have no intention of doing this exercise."
 
lol, nice
You could answer with "I have no intention of doing this exercise for you"
I would give it a +1
 
Uhhh
I may have done just that
Like, before you mentioned it
-2
Q: Prove the solution set of a system is a two dimensional linear space

Smiley Denote $\zeta_0 = \dfrac{c}{2\sqrt{km}}, \omega = \sqrt{\dfrac km}, x_1= x, x_2=x'$. Transform the equation $x'' + \dfrac cm x' + \dfrac km x =0$ to a linear system of the state vector $\xi = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ $$ \dot \xi = \begin{bmatrix} 0 & 1 \\ -\omega ^2 & -2\zet...

 
Make it into an answer and I'll vote you up :P
 
maybe their English is really bad, and they meant something else, like "I have no idea how to do this exercise"
maybe they mixed up "intuition" with "intention"? :P
 
11:55 AM
Heh... posting that as an answer would probably get the answer flagged/removed, though. I'll stick with a comment
And, yeah, I wish that's what they meant...
 
12:18 PM
8
Q: Writing challenge - test run! April 20, 2017 - May 11

MithrandirOver on this post, it was discussed whether or not we should have writing challenges on Meta. The result: we give it a try, and see how it works. That's the point of this post! Here's how this will work: The plan was for there to be a (optional, I think) prompt, to help inspire you to write. ...

For anyone looking to write something fun.
 
i have something saved in my drafts but it's terrible :P
 
12:47 PM
@BalarkaSen I believe in you!
 
(Be prepared that some time later after I get the chem stuff working there will once again be a long post on my attempt on $C_0(\alpha)$. Meanwhile, if my workings in my previous post is correct, it seems epsilon mapping is not growing at the same rate as pentation which is contrary to what I expected)
 
1:03 PM
@Secret Just be careful on the $\alpha$ and stuff, weird things happen if you aren't careful
 
I think I might be seeing the sign of one weird thing happening: Is the following correct?

$$\epsilon_{\epsilon_0}=\omega^{\epsilon_0}=\epsilon_0$$?
 
@Secret That is wrong
$$\varepsilon_{\varepsilon_0}=\sup\{\varepsilon_\omega,\varepsilon_{ \omega^\omega},\varepsilon_{\omega^{\omega^\omega}},\dots\}$$
I also have another definition for an ordinal collapsing function:
$C(\alpha)_0=\{0,1\}$
$C(\alpha)_{n+1}=C(\alpha)_n\cup\{\gamma+\delta,\gamma\delta,\gamma^\delta ,\omega_\gamma,\sup(C(\eta)),\psi_\gamma(\eta)|\gamma,\delta,\eta\in C(\alpha)_n, \eta<\alpha\}$
$C(\alpha)=\bigcup\limits_{n<\omega}C(\alpha)_n$
$\psi_\beta(\alpha)=\sup\{\gamma|\gamma\in C(\alpha),\gamma<\omega_{\beta+1}\}$
As always, it hasn't effected much of your calculations yet, so if you want to use this instead of whatever you've been using, you should be fine and still getting the same results.
 
ok sorry, I missread the definition of epsilon numbers. They are $\alpha=\omega^{\alpha}$, not $\epsilon_{\alpha}=\omega^{\alpha}$
 
no wonder I get the wrong calculation
I will try the new def later, cause the new one looks even more confusing due to there are more terms depending on the inequalities, and I really suck at recursively defined sets involving inequalities
 
1:12 PM
lol, okay
It's more of a simultaneous thing
so the inequalities shouldn't be that bad.
 
Btw, is this previous ramble of mine correct?
18 hours ago, by Secret
[Epsilon Towers]
Let ${}^n\omega=\underbrace{\omega^{\omega^{\omega^{\dots}}}}_{\textrm{n times}}$
Let $\epsilon_0=\sup(\omega, {}^2\omega, {}^3\omega, ...)={}^{\omega}\omega$
Then:
$\epsilon_{\alpha+1}=\sup(\{{}^j\epsilon_{\alpha}|j\in \Bbb{N},\alpha < \epsilon_0\})=\underbrace{{}^{^{^{^{\omega}}}}({}^{^{^{\dots}}}({}^{^\omega} ({}^{\omega}\omega)))}_{\textrm{$\alpha$ times}}$

However: To be checked:
$\underbrace{{}^{^{^{^{\omega}}}}({}^{^{^{\dots}}}({}^{^\omega} ({}^{\omega}\omega)))}_{\textrm{$\alpha$ times}}\stackrel{?}{=}
I am trying to "visualise" epsilon number back there
 
Repost to render the latex
$$\underbrace{{}^{^{^{^{\omega}}}}({}^{^{^{\dots}}}({}^{^\omega} ({}^{\omega}\omega)))}_{\textrm{$\alpha$ times}}\stackrel{?}{=} \underbrace{({}^{({}^{({}^{\omega}\omega)}\omega)\dots}\omega)}_{\textrm{$\alpha$ times}}=\omega \uparrow^3 \alpha$$
 
@Secret Knuth arrows do not work like that
the ? step is indeed wrong
One can easily check inductively that for any $\alpha>1$ and $\beta>2$, we have:
$$\omega\uparrow^\beta\alpha=\varepsilon_0$$
Since $\varepsilon_0=\omega^{\varepsilon_0}$
Nope, it doesn't grow at all lol
Of course, you could define up arrows to be recursive in that manner, rather than in the normal manner so that you could write things like that
At least define it like that for transfinite ordinals
 
Hmm, so that means after tetration of $\omega$ it got stuck because it hit $\epsilon_0$ which is a fixed point under exponentiating $\omega$
 
1:19 PM
mhm
But using up-arrows is quitely... troublesome IMO
 
So why doesn't higher epsilon in terms of lowers ones get stuck since you need a countable exponential tower of $\epsilon_0$ (itself is a countable tetration of $\omega$) to get to $\epsilon_1$?
 
It is not, since exponentiation is not communitive like that
$$\varepsilon_0^{\varepsilon_0}=(\omega^{\varepsilon_0})^{\varepsilon_0}= \omega^{\varepsilon_0^2}$$
You could also use a different definition of $\varepsilon_1$, such as
$$\varepsilon_1=\sup\{\varepsilon_0+1, \omega^{\varepsilon_0+1}, \omega^{\omega^{\varepsilon_0+1}},\dots\}$$
@Secret Perhaps it would be most constructive to look at all the ordinals between $\varepsilon_0$ and $\varepsilon_1$?
 
thought so, I always get lost when I end up at the epsilon numbers because of my knowledge on tetration and exponentiatin get muddled up
 
mhm?
Well, what's the smallest ordinal larger than $\varepsilon_0$?
...or rather, what's the smallest ordinal larger than $\omega$?
 
1:40 PM
$\epsilon_0+1 > \epsilon_0$ and $\omega+1 > \omega$. The rule breaks for $\zeta_0$ as you shown very earlier, however (to be worried later)
 
No no, $\zeta_0+1>\zeta_0$, but $\psi(\alpha+1)$ needn't be greater than $\psi(\alpha)$.
You may conclude that for $\beta>\alpha$, then $\psi(\beta)\ge\psi(\alpha)$, but not $\psi(\beta)\color{red}>\psi(\alpha)$
 
ah ok
 
Certainly $\omega_1>\zeta_0$, but $\psi(\omega_1)=\psi(\zeta_0)$.
 
Re: that link. I am currently at $\Psi_0(\epsilon_0)$ before I got tripped due to misremembering epsilon number definitions. I do get the same results on that link for everything before, though my $\Psi_0(\omega)$ is wrote in terms of tetration (which I most likely got it wrong, will show that later)
 
$\psi_0(\omega)$ is usually given by $\sup\{\psi_0(1),\psi_0(2),\dots\}$
 
1:46 PM
is that $\epsilon_{\omega}$ since all $\Psi_0(k)=\epsilon_{k},k<\omega$ (that was based on what I have worked so far using $C_0(\alpha)$?
 
Yeah
$\varepsilon_\omega$ is the supremum of all those, so yeah...
 
So... I wrote my $\Psi_0(\omega)$ in the following bizzare fashion...

$$\Psi_0(\omega) =\sup_{j<\omega}(\{{}^j\epsilon_k|k\in\Bbb{N}\})=\epsilon_{\omega}$$

Bad thing for getting too used to indices...
 
mhm.......
Why not just this?
$$\psi_0(\omega)=\sup\{\varepsilon_k|k\in\mathbb N\}$$
 
because...
\begin{align}
C_0(1)_1 & =\{j,\omega[i]j,...,{}^2\omega,...,\epsilon_0[\omega_1]|j\in\Bbb{N},i=\{1,2,3\}\},\eta=0\\
C_0(1)_2 & =\{j,\omega[i]j,...,{}^3\omega,...,{}^2\epsilon_0,[\omega_1]|j\in\Bbb{N},i=\{1,2,3\}\},\eta=0\\
C_0(1) & =\{j,\omega[i]j,...,{}^j\omega,...,{}^j\epsilon_0,[\omega_1]|j\in\Bbb{N},i=\{1,2,3\}\},\eta=0\\
\Psi_0(1) & =\sup_{j<\omega}(\{{}^j\epsilon_0\})=\epsilon_1
\end{align}
and the same pattern continues all the way to where I am at so far
(I literally try to build my step up using expoenential towers...)
But yeah, I could have just put my j in the bracket so it becomes more like the one above
 
@Secret but notice that if you use my definition with supremum, then the towers of epsilons are not needed
Since higher epsilons are larger than towers
:P
 
1:57 PM
Btw, I think I need to spend more time on understanding this one

$$\varepsilon_1=\sup\{\varepsilon_0+1, \omega^{\varepsilon_0+1}, \omega^{\omega^{\varepsilon_0+1}},\dots\}$$

since $\alpha=\varepsilon_0+1$ is not a fixed point of $\omega^{\alpha}$
 
Mhm
That's the point of it
I personally like the other definition more
 
yeah, I also like the tower of epsilons because the rules are simpler and there is technically only one operation is done: exponentiation
 
Poem/song thing I wrote:
Sad hummingbird,
Born in the rain, all alone.
Your nest was poisoned years ago.
Everyone you touch is dead.
But I'll stay by you,
Sad hummingbird.

You've lost so much
And you've got no home
But you've got me, as you can see
And snacks to eat together.

Don't mind the blood. Don't mind the death.
Everything'll be all right
As long as you're with me...
Friends forever.

Goodnight dear sad hummingbird.
 
That's a really beautiful poem @Simply
 
Thanks @ShaVuklia
This might interest you @ShaVuklia writers.meta.stackexchange.com/questions/1283/…
 
2:05 PM
is the hummingbird and the poet one and the same?
 
Square matrix $A$ is a diagonally dominant matrix iff the spectral radius of $A $ is less than one ?
 
@ShaVuklia it's tied to an anime/manga called "Deadman Wonderland"
And no, I'm not the hummingbird =P
 
@Simply oh, too bad I'm not familiar with anime/manga :P I personally often write to myself, so the one who is being addressed is also the addresser, hence my question :P but we're all different, luckily:)
 
Then you should contribute to writing.SE
:D
 
you know what, I might actually do that. I like it that it's not a "contest", but just free and spontaneous
 
2:14 PM
Yup
 
No dream too big no dreamer too small!
Turbo
 
Hello!!!

Suppose that we have $a \equiv 1 \pmod{p}$ and $a \equiv x \pmod{q}$, where $p<q$ and $x>1$. Will it always hold that $a \equiv x \pmod{pq}$ ?
 
@Semiclassical Finally learning some physics again.
 
What kind of physics?
We started doing stuff in complex analysis that feels like there is some homology going on behind the scenes
 
2:32 PM
Electric fields, etc.
Right now learning electric flux and Gauss's theorem.
@Alessandro Yeah complex analysis does have something like homology going on. What are you doing?
 
@Evinda $5 \equiv 1 \pmod 2$, $5 \equiv 2 \pmod 3$. Indeed, $2 < 3$ and $2 > 1$. But $5 \not\equiv 2 \pmod 6$
 
Cauchy's integral formula
But we also saw that the integral of a function along homologous (I hope that's a word in english) chains is the same
 
homotopic
 
No, it holds for homologous chains too
 
Perhaps, but you won't see that pop up in an introductory course on complex analysis
At least, in my experience
 
2:40 PM
@SteamyRoot Oh yes, right... One can also find a counterexample when both p,q are odd, right?
 
We didn't really talk about homology though, we just defined chains as formal linear combinations of piecewise $C^1$ curves and defined a chain as homologous to 0 if it has winding number 0 for every point
 
@BalarkaSen What was interesting that you pinged me and PVAL about?
 
Oh, your answer to the homology sphere admitting taut foliation thing.
 
@SteamyRoot that's a weird course though, the first semester was general topology and the second semester is divided between intro to algebraic topology and intro to complex analysis
 
I see
 
2:42 PM
Morning, by the way. I have a silly question; say you look at the wrong foliation by circles on the Klein bottle (which looks like the foliation by circles on the Moebius strip, tangent to the boundary circle). Is this analytic?
I think it is.
 
It just surprises me - at least in my university and the other Belgian ones, complex analysis is taught way earlier than anything related to homology
We do see the same result, I guess, but we just work with homotopies and show that you can always replace a path by a piecewise linear one...
 
@Alessandro You're secretly learning de Rham cohomology, in the complex variables.
 
@SteamyRoot We didn't talk about homology in the algebraic topology part of the course, just homotopy and fundamental groups. In fact we'll prove that $\pi_1(S^1)=\Bbb Z$ with complex analysis next week or in an upcoming lecture.
 
Pochhammer contour is my favorite example of a contour which is not nullhomotopic but is nullhomologous.
 
Ah I think that it always hold when $x<q$, do you agree? @SteamyRoot
 
2:46 PM
@AlessandroCodenotti That's... an interesting order of teaching those subjects. Not the one I'd want, but of course I'm heavily biased towards the order I was taught and now teach :P
 
@SteamyRoot fair enough :P
 
I like this order, personally.
 
Makes sense, given how much you're into homology and homotopy and so
@Evinda Uhhh... dunno, let me think
 
@BalarkaSen Yeah, the holonomy is linear ;P
 
@Evinda $7 \equiv 1 \pmod 3$, $7 \equiv 2 \pmod 5$, but $7 \not\equiv 2 \pmod{15}$
It's really easy to make counterexamples if $p*q > a$
But even if it's not...
 
2:58 PM
I see.. But if we would have $a \equiv 1 \pmod{p}$ and $a \equiv x \pmod{q}$ where $p<q$ and $x>q$ wouldn't we get that $a \equiv x \pmod{pq}$ ? @SteamyRoot
 
@MikeMiller Right. Thanks.
 
Oh, $x > q$ now?
wait, what
$x$ is the remainder when you divide by $q$. You can't have $x > q$.
 
Oh yes, right... Thank you :) @SteamyRoot
 
3:19 PM
guys, I know it's Dutch, but maybe someone could help me; what do they mean by $D_1, \Pi_{12},E_{12}$?
so the matrices have entries from $\mathbb F_2$
that's really all you need to know
 
Uhhh... apart from that notation being awful imo, I'm guessing $D$ stands for diagonal matrix and $E$ for elementary
 
But then what is the difference between $D_1$ and $D_2$?
 
the sub-index on $E$ seems to stand for the coordinates in the matrix where you have a nonzero entry
 
ah right, that makes sense
 
Those are the elementary matrices, those corresponding to the operations you do in Gaussian elimination
 
3:22 PM
I have no idea what the superindices are, or what those $D$'s are, though
 
but... we have 3 non-zero entries
 
Yeah, but the diagonal isn't important here
 
$\Pi$ swaps the rows numbered by thr subscripts
 
Because that's the matrix you multiply with, if you want to add a row to another row
it always has diagonal being ones
 
so $\Pi^2$ is twice swapping of rows?
 
3:23 PM
Just once
I'm not sure what the super-indices are, but I don't think they're powers
 
$E_{nm}(k)$ adds the $m$-th row multiplied by $k$ to the $n$-th
I think the superscript just tells you the dimension of the matrices
 
Hmmm
 
oh yea, that sounds plausible!
 
Yeah, probably that. Or it refers to the field you're working over
 
about the superscript
 
3:25 PM
either way, they don't seem to convey any information if you already know you're working in $GL_2(\mathbb{F}_2)$
 
While $D_n(m)$ multiplies the $n$-th row by $m$
 
ahh
 
Oooh, yeah, that would explain it
 
haha, great notation!
 
and $\Pi_{ij}$ is swapping the $i$-th and $j$-th row
 
3:25 PM
thanks, guys
 
so in this case, you only have $\Pi_{12} = \Pi_{21}$
It surprises me this notation wouldn't be explained somewhere before, though...
 
well, it's from a syllabus
that happens
 
Like if you have a matrix $A\in M_n(\Bbb R)$ then $E_{34}^n(6)A$ is $A$ after adding $6$ times the 4th row to the 3rd and so on
(Unless you're using the backward convention and then you have to multiply on the right)
 
right, got it!
 
$D_n^m(k)$ is the same as the $m\times m$ identity, with the 1 on the $n$-th row replaced by $k$ since it's not written explicitely in your notes
 
3:30 PM
yes, I've added my own notes to the syllabus! thank you very much, Alessandro
oh, I see now where they've explained it. it's in a paragraph we skipped temporarily :P
 
It seemed weird to use never defined notations
 
3:56 PM
@BalarkaSen hi
 
4:32 PM
Hi all, i need some help on real analysis question
how do i derive $f''(d) < 0$ in case 1 and $f''(d)>0$ in case 2?
 
My syllabus says that $\hom_{\mathbb F_2}(\mathbb F_2^2,\mathbb F_2^2)$ contains 16 elements. I’m guessing this is because $\mathbb F_2^{2\times 2}$ contains 16 elements, but I don’t see why this should be the reason, because we also have three bases for $\mathbb F_2^2$, so shouldn’t there be $< 16$ linear maps? Because the same linear map can be represented by three different bases, so for some maps, there will be several matrix representations.
 
I need to brush up on my linear algebra.
I got an answer to this question wrong.
 
i've posted the question on the main site, btw
 
I thought that two matrices have the same Jordan normal form iff they have the same multiplicities. Now, there are three types of multiplicity, right? And if I remember correctly, they help determine the Jordan normal form, yes? Where can I get some revision done?
Am I right?
There's algebraic multiplicity, geometry multiplicity, and a third one I've forgotten.
It's something to do with the size and number of Jordan blocks for a given eigenvalue.
 
4:52 PM
Hey everyone!
 
0
Q: Show the power series converge uniformly on $\mathbb{R}$

Little RookieShow that the power series $\sum_{n=1}^{\infty}\frac{x}{n(1+nx^2)}$ converge uniformly on $\mathbb{R}$. How should i approach this question? Weirestrass M-test does not work for $|x|>1$ here.

 
@Daminark Hi :)
 
How's it going @Shaun?
 
hii @Daminark
 
bad. I'm working my way through a textbook in Semigroup Theory these days and messing up on questions in linear algebra.
*Not
 
4:58 PM
presses F to pay respects
 
How about you, @Daminark?
 
How's it going @Sha?
And lol manifolds midterm is tomorrow which is at least slightly disconcerting, today in analysis we finished Fubini's theorem on $\mathbb{R}^n$ and talked a bit about product measures more generally
 

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