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4:01 PM
@Chris'ssis Is he getting anywhere?
 
@robjohn He edited his answer but no progress. Splitting the integral and computing the three components makes the job pretty easy.
 
@Chris'ssis You have a way to compute $J_3$?
 
@robjohn Sure. It's obvious.
 
@Chris'ssis I guess I should look at it.
 
@robjohn Definitely. Just look at it. It's not hard at all.
 
4:09 PM
@Chris'ssis fyi, that integral i posted before (the one i said was in terms of elliptic integrals) is one of the parts of a bigger calculation, which is some rather difficult contour integral
what i'm trying to do now is figure out how to compute/sensibly approximate the last, really difficult part
 
@Semiclassical I noted that. It didn't seem easy to do.
 
well, it's not so bad if you can figure out how to write it in terms of complete elliptic integrals. (and that's really all one can hope for)
the integral i'm trying to do write now can presumably be written as an infinite sum of such elliptic integrals, but that's just too tedious for me
to be precise about it, the integral i'm considering now:
ugh, latex-typo hunting
$$\int_{\gamma_-}\frac{\sin k}{\sqrt{t+t^{-1}+2\cos k}}\log\left(1-e^{-2n i k} \frac{t+e^{-i k}}{t+e^{i k}}\right) \frac{dk}{2\pi}$$
ahah
 
When you put the differential at the beginning, it is hard to know where the integral ends...
 
yeah, physics habits
 
@Semiclassical Thanks :-)
@Semiclassical If you make a typo and run out of time to fix it, just ask me (or any mod), mods can fix things after the time limit.
 
4:20 PM
kk
right now i'm trying to figure out the right way to say what the integration cycle is
it's from 0 to 2pi in the real part, but there's an imaginary offset
if the square root out front were in the numerator, i'd say the imaginary part was negatie and chosen such that the square root vanishes when the real part is $\pi$. but in this case, i think i should say it's some sufficiently small $\epsilon>0$ above that
it needs to lie below (in imaginary part) all the poles of the function inside the logarithm, but still be above the branch point in the negative half-strip.
not exactly the nicest contour integral i've had to deal with :)
 
@robjohn I can post a solution to the whole problem without using contour integration at all, but it's much to write, nothing spectacular.
 
(i call that a contour integral since it corresponds to a closed cycle when I either take $z=e^{i k}$ or take the domain to be a cylinder rather than a strip)
 
@robjohn I mean $$\int_0^{\pi/2}\frac{x\log{\sin{(x)}}}{\sin(x)}\,dx.$$
 
@Chris'ssis Is this the one that is pages of text?
 
@robjohn Well that is just an answer by Mathematica, not done by my way.
@Semiclassical Interesting looking there.
 
4:27 PM
Hi guys. I've got a small reference request for some well-known result. I'm looking for a paper by Bonner from 1910 establishing the invariance of dimension under homeomorphisms.
 
@Chris'ssis Oh, Mathematica 8 failed to compute the integral, or second derivatives
 
@robjohn hello, can i ask you a question about compacity and continuity ?
 
compactness*
 
@robjohn In my opinion the best way to tackle to whole story is to begin by letting $-\log(\sin(x))=y$. After that some integration by parts combined with other substitutions (especially Weierstrass substitution) should work pretty fine. It's just much to write, that's all. It's not just "maybe it works" but it works that way.
 
yeah. what's really neat is that, if you introduce $\delta t= \frac{t-1}{t+1}$, then the result is approximately a function of $n\,\delta t$ alone
i know that numerically
(there's an overall prefactor of $n$, but beyond that it's just something like $f(n \delta t)$
 
4:31 PM
@robjohn Most of the time it's enough to make a few steps inside of an approach to see if the ends looks fine. That's the case here too.
@robjohn $$J_3=\frac{1}{2}\int_{1}^{\infty}\frac{\frac{1}{2}\log(q)(\log(2) -\frac{1}{2}\log(q)+\log(1+q)}{(1+q)\sqrt{q}}dq\\$$
2 of the components are of the type $$\int_1^{\infty} \frac{q^s}{(1+q)} \ dq=\frac{1}{2} \left(\psi ^{(0)}\left(\frac{1}{2}-\frac{s}{2}\right)-\psi ^{(0)}\left(-\frac{s}{2}\right)\right) $$
 
Nobody happens to know about the topology reference? :\
 
@Chris'ssis They would all be derivatives of the Beta function or second derivatives, if the integral went to $0$
 
Sounds like a paper that's mostly of value to historians, @Danu, so I would be surprised to see someone know it
 
@MikeMiller Sometimes, the first paper becomes famous in its own right
 
@robjohn :-)
@robjohn Anyway, I never like the domain over $1$ to $\infty$. Changing it from $0$ to $1$ is far better.
 
4:41 PM
...I guess this is not one such case
 
yeah, but nowadays such a result can be given as an exercise in a first topilogy class - I doubt the paper gives any profound insight into compactness. More likely it's interesting because it demonstrates the development of the idea - a historian's pursuit :)
 
@MikeMiller Yeah, true. I'm mainly interested because my professor mentioned it and I'm editing the lecture notes on that course now.
 
@Chris'ssis one of the approaches I tried involved computing $$\int_0^{\pi/2}\frac{x}{\sin(x)}\,\mathrm{d}x$$ :-) That's when I thought that Catalan's constant might be in the answer
 
@robjohn Well, one cannot avoid here the use of a row of polylogarithms ... (unfortunately) At a certain point, one integral has to be computed in terms of polylogarithms.
:-(
 
@Chris'ssis unless you use hypergeometrics ;-)
 
4:45 PM
@robjohn :-) I remind you that you don't like hypergeometrics at all.
 
can anyone please suggest me a good book for linear algebra
 
@Chris'ssis Yes, but it sure simplifies that integral...
 
@robjohn Yes, but not as much as the polylogarithms.
 
@Chris'ssis really? The whole integral is the negative of a single hypergeometric function.
 
4:49 PM
@robjohn If talking of closed forms, it's clear that the hypergeomtric form can be brought down to more pieces where the most sophisticated one is the trilogarithm with the complex argument.
 
@Chris'ssis I don't think of hypergeometrics as anything but a way to rewrite a particular kind of series.
 
@robjohn Yes, but it that series can be split such that you get some known closed forms + something that is less sophisticated than the hypergeometric one you have?
 
@Chris'ssis Polylogarithms are the same. At certain arguments, they have nice values, but both Polylogarithms and Hypergeometrics are rewritten series.
 
@robjohn I just wanna say that you can get simpler forms from that point.
 
@Gato $\overline{\lim}_\infty\sqrt[n]{\sin(n)}$
 
4:53 PM
@Chris'ssis I like answers devoid of Polylogarithms and Hypergeometrics, but if that is all you can do, then that is fine. Someone may find a nice value for the particular argument you need.
 
@robjohn I like that too.
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ mais pour hadamard c'est lim sup de $\vert a_n\vert^{1/n}$ non ?
 
@Gato hadamard ? quel lien ?
 
Hey guys do uall use MIT OCW to learn mathmatics?
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ Hadamard c'est le rayon de convergence est égale à 1/lim sup
 
4:56 PM
Is that a good resource?
 
@Gato ah ok. c'est bien ça alors. et ? c'est bien ce qu'on a là non ?
 
comment tu fais pour en déduire la lim sup alors ? il faudrait $\vert sin(n)\vert$ non ?
 
@Gato Ah oui en effet ça doit être $\overline{\lim}_\infty\sqrt[n]{|\sin(n)|}$. Après tout, pas de $\sqrt{\sin n}$ si $\sin(n)<0$ hehe
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ haha oui mais ça chance quand même la valeur je pense.. En tout cas je trouve l'astuce génial!
 
Bah j'y ai juste pensé parceque j'ai eu l'exercice dans l'autre sens (RCV de $\sum\dfrac{z^n}{\sin(n\pi\sqrt{3})}$)
 
5:02 PM
@ɧɿρρԹʅȝՇԵՐՎԾՌ D'accord. Mais tout de même généralement les limp sup avec sin et compagnie c'est chaud et la ça trivialise presque le truc :D
 
Le RCV n'est pas si trivial que ça, si ?
 
Is the elementary abelian group: E8 the same thing as $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$?
 
@ɧɿρρԹʅȝՇԵՐՎԾՌ Non d'où mon presque.
 
@ABeautifulMind Hi!
 
5:06 PM
Yes, @El'endiaStarman. An elementary abelian group is one of the form $\Bbb Z_p^n$, where $p$ is prime; they're uniquely determined by their order.
 
@MikeMiller Ah, awesome. Thanks! :)
 
Sure.
 
@user91500 Hi.
 
hello to all
 
salut @Gato, @Hippa ... Good night, @Mike, good night Jasper :)
 
5:10 PM
morning Ted
inputting grades...
 
@ABeautifulMind 4383-3=4380, Nice!
 
You can help me grade, @Mike. My exam's done in 10 minutes.
 
@TedShifrin Salut. How are you ?
 
Manifolds and inverse function theorem are so much more interesting :)
In pain, @Gato, but otherwise ok. Thanks. And you?
 
5:12 PM
Why is there no tag creation date in the tag info on MSE?
 
LOL @ "ça trivialise presque le truc." :P
 
or is it there and i missed it?
 
hi @Sayan
 
hey @sayan
 
5:13 PM
OK, going to collect my exams ...
 
@Ted Winter Storm Thor is going to miss your area completely...
@Ted Hello and goodbye
 
Hi @teadawg1337
 
Hey @Sayan
 
@TedShifrin Are you sick ? I hope you will be better then. :) Fine, working my course.
 
Did anyone notice the new 'Graphic' and 'Animated' tags?
 
5:14 PM
@TedShifrin Lol yes, belle astuce non ?
 
The storms name is Thor....the Norse god @teadawg1337
 
@Sayan Yup. This is the twentieth winter storm of the season in the US
 
Mean score 32.4, @Ted. Given the test I handed out, shoulda been 40.
 
@MikeMiller I believe you. Just thinking about it for the sake of thinking.
 
5:17 PM
Are people ignoring me or what?
 
I have nothing interesting to respond with to the things you're saying, @G-man, so I'm not responding.
 
@G-man I'm not, I just didn't answer because I haven't seen the new tags you're referring to
 
$$
\int_0^1 \int_0^1 \frac{2xy}{x^2 + y^2}\,\mathrm{d}x\,\mathrm{d}y
= 4 \int_0^1 \int_0^y \frac{2xy}{x^2 + y^2} \,\mathrm{d}x\,\mathrm{d}y
$$
any idea why the above holds?
 
@Mike is a Wittgensteiner.
 
oh. then its all right.
 
5:19 PM
Oh..u know @BalarkaSen I saw something
 
meh, it was wrong
 
I know the calculus which is required in 11 .....@BalarkaSen
 
@SayanChattopadhyay math.stackexchange.com/questions/1173794/… q_q he's from India too
 
well, you never gave me the proof of fundamental theorem of calculus, @Sayan.
 
@Balarka There are several fundamental theorems of calculus
 
5:24 PM
uh
in what sense
 
Hello @Chris'ssis. How is your book?
 
Well, not several. Just the single-variable version and its multi-variable counterparts
 
i am familiar with only one.
 
@ABeautifulMind Hi. Working on it. How are you?
 
well, if you're including generalizations of the fundamental theorem of calculus, there's a wide swath of things that could get that name
 
5:25 PM
that's stupid, @teadawg1337. there is even an FTC for forms.
:P
 
@Chris'ssis I am feeling scared now. Because I don't know if I will be able to do the things I need to do to solve my mental problems.
 
but there's one thing called the fundamental theorem of calculus
 
The generalizations all fit under the same roof of FTC, so my initial statement is incorrect
 
@ABeautifulMind ^^^ No disease can stop you from thinking like a winner.
 
@Ted!
 
5:30 PM
hi @teadawg: Hope you're still not frozen in a snow-drift :P
hi Balarka
 
i'm trying to remember what form the FTC takes on in the context of differential forms. my instinct would be "all differential forms are locally exact" but i suspect i'm remembering wrong
 
@TedShifrin Not anymore, temps have been above freezing over the weekend
 
No, no, very wrong @semiclassical
it's called Stokes's Theorem
 
<--- knows nothing about forms, but hopes to study it.
 
and what you're trying to remember is that locally any closed form is exact
 
5:31 PM
yeah, i was about to say
knew i was forgetting an important qualification
 
Forms are fun, @Balarka, because you mix calculus and (exterior) algebra
 
Although there's still a little bit of snow left on my back porch....
 
if you were in Boston, @teadawg, you'd still be saying that in May, I think
 
well, there are two bits to the FTC, though
 
Sharpe's book (the one you pointed me at, @Ted) has something it amusingly calls a nonabelian FTC
 
5:32 PM
@TedShifrin yes, and i dunno anything about exterior algebra...
 
one is that the integral of a derivative is just the difference of the values at the endpoints. and that's just stokes theorem, yes
 
sort of like the Ambrose-Singer Theorem, @Mike
 
Hell if I remember what the theorem was anymore, @Ted
I just remember being amused
 
Holonomy = integral of curvature form, @Mike
 
@Ted Allow me to give you a brief run-down of the weather headed my way: 1-2 inches of rain today through tomorrow morning, temps dropping all day on Wednesday, 1-3 inches of snow/ice accumulating Wed PM/Thurs AM
 
5:33 PM
ah, yeah
 
= berry phase, sayeth the physicist
 
Below freezing all day Thursday, and all of the rain falling today and tomorrow will freeze over
 
well, @teadawg, you can come visit me in CA :)
 
Hey guys do u all use MIT OCW to study mathmatics?
 
I used to love snow when I lived in New England, but I hate ice.
 
5:34 PM
Is that a good learning tool?
 
Ice is the WORST
 
When I went to MIT there was no OCW
 
@yswong never heard of it
 
U know what im refeering to?
Its the online website
That contains maths resources
 
IMO, OpenCourseWare is useless without the proper textbook to accompany the material
 
5:36 PM
well, @teadawg, that's probably usually true, but I assume you could follow my lectures ok without any book.
 
@BalarkaSen I gave you the proof
 
oh, no, @Ramanewbie is here ... time to leave
 
Then where do u all learn math then?
 
haha ... I found the method to repulse you @ted !
 
Well, @Mike, it is kinda obvious that $\Omega_0(pt)$ is $\Bbb Z/2$, as $[M] + [M] = [\partial (M \times [0,1])] = [0]$.
 
5:37 PM
bien sûr, @Ramanewbie
most people still use books, @yswong
 
I decided to restart medication. I just made an appointment to see the doctor.
 
I'm too old to learn.
Do you feel happy about your decision, Jasper?
 
Ted: im sure there are textbooks available online
 
@Ted I can watch your lectures without issues
 
@BalarkaSen Uh, you just write two unrelated clauses.
 
5:38 PM
@TedShifrin Well, I thought I would try it again, since I have been feeling so bad.
 
@MikeMiller Well pick up a bunch of points.
 
You might have issues when you get to the hard stuff, @teadawg :P
 
That's a 0-manifold.
 
@TedShifrin: i guess what i probably have in mind is just the statement that, if I integrate a function, then i can differentiate to get that function back. which i suppose should generalize to something like: you can take an $n$-form, integrate with respect to some variable to get an $(n-1)$-form, and then observe that $d$ maps this back to the $n$-form.
 
Ted: How do i find them ?
 
5:38 PM
ok, Jasper .. I wish you and your doctor the best of fortune!
 
@BalarkaSen when I was solving some questions in spivak I just thought of something....
 
No you didn't give me a proof, @Sayan
 
We'll see about that @Ted
 
@ABeautifulMind good luck
 
hi @sayan
 
5:39 PM
Ted: Where did u learn mathmatics from?
 
That's not necessarily right, @Semiclassical. You need to start with a closed form.
 
You just proved that $\int_a^b F(x) dx = f(b) - f(a)$, assuming FTC
 
Books and courses, @yswong. I'm old.
 
hmm. so i guess that bit of the FTC doesn't generalize as directly?
 
Wait then I will write it again@BalarkaSen
 
5:40 PM
path independence is definitely an issue, @Semiclassical. Think about even $1$-forms (vector fields).
 
@TedShifrin If I finish my PhD at 45 and want to work from 45 to 75 in a math department, is that possible?
 
Ted: Specifically which books and what type of courses? Online ones or university ones?
 
@Mike Now if you have an even number of points, it appears as a boundary of a 1-manifold, obtained from crossing your points with [0, 1]
 
well, yes. i should be inserting the word 'locally'
though even there, hmm
 
So the only bordism classes are [0] and [pt].
 
5:41 PM
All things are possible, Jasper, although in this country more and more jobs are becoming adjuncts rather than tenure-track.
 
You haven't proven those are distinct.
 
@semiclassical: No, even locally, you only have path-independence if the $1$-form is closed (and there are analogues for higher forms).
 
Hey guys lets say i want to learn mathematics independently online, then where should i start?
 
@TedShifrin Thank you. Of course, I will need to stay physically healthy to do that, and that should not be a problem. I just need to solve my mental problems and then I can start living again.
 
I am not a fan of on-line courses, @yswong. But there are very informative lectures out there.
Jasper, you know I'm rooting for you.
6
 
5:43 PM
the perils of knowing differential forms only from an intuitive/physics perspective
 
Ted: Like which ones?
 
given an application, i'd probably come to the right conclusions. but in the abstract it becomes a lot easier to say nonsense
 
@Mike That's trivial.
 
well, @Semiclassical, even you lowly physicists know that some force fields are not conservative :P
 
I don't have the energy for you this morning.
 
5:43 PM
@TedShifrin Thank you. You know, all these years, maybe I made many many mistakes in trying to solve my mental problems, but I just want to say that I think I have tried my best, and I will continue trying.
 
LOL ... poor enervated @Mike.
 
what did i do? looks around
tries to pull an innocent face
 
I suspect @Balarka drives his parents mad.
2
I am not an expert, @yswong. All I know is that one of my courses is now on the web, and lots of professors are doing such things, even without OCW.
 
i'm not gonna comment on that.
 
@DanielFischer Hi just a quick question. Do you know why is it that for parametric curves, say $r(t)$, it follows that the derivative $r'(t)$ is orthogonal to the curve but the second derivative $r''(t)$ is not necessarily orthogonal to the derivative $r'(t)$?
 
5:45 PM
well, yes. but when i think about that in the context of EM, it's usually b/c i end up integrating around something like an electric current.
 
@JohnJack: Your first statement is wrong.
 
Ted:Do u have the links to those webpages?
 
$f(t)$ is orthogonal to $f'(t)$ if and only if $\|f(t)\|$ is constant.
 
I've learned that online courses don't fit my learning style at all
 
and there i'd still say that, if I take a point next to a line current, a sufficiently small circle will still fail to enclose that current
 
5:46 PM
I am opposed to on-line courses, @teadawg :P
 
and so the circulation around that small circle vanishes
 
Look on my profile page, @yswong, for mine.
 
Ted: That can teach me from Calculus/ Linear Algebra all the way up to more advanced levels.
 
right, @semiclassical, you need curl = 0 to get path independence.
 
@TedShifrin Why only for $f|(t)|$ constant?
 
5:48 PM
facepalm dur. and outside the electric current that's obviously true
 
@ted I have to do national service until 40. =( They can call me up for two weeks a year until then.
 
For starters, @yswong, you need to do lots of exercises and get them critiqued. So this is why on-line learning is limited.
Surely you get a medical excuse, Jasper?
@JohnJack: Differentiate $f(t)\cdot f(t) = \text{constant}$.
 
@TedShifrin I do less, but I still need to do. Mostly in the office.
 
Ah, ok, Jasper.
 
I tried to skip it altogether, but they gave me the lowest grade just above not doing anything at all.
 
5:49 PM
so $\mathbf{B}\cdot d\mathbf{l}$ outside the current is obviously closed, etc. etc.
 
but... interesting... i don't know how to prove that a point and the nullset aren't bordant!
 
if i have a continuous function $f:\Bbb{R}\rightarrow \Bbb{R}$ such that $f^2=1$. Does it implies that $f$ is constant ?
 
sweats
as impossible as it seems
 
It's called mod-2 arithmetic, @Balarka.
 
Now even I know 0,1 arithmetic, lol.
 
5:51 PM
What do you think, @Gato?
Why am I getting so many stars? :(
 
hmm. for the FTC, though, we essentially have $F=\int_0^x \omega \implies dF=\omega$ where $\omega=f(x)\,dx$. though in that case one is just integrating along the real line.
 
@TedShifrin i don't get it.
 
@TedShifrin Do your lectures go into vector calc? I need a better understanding of the concepts than I could squeeze out of Stewart Calc
 
@TedShifrin graphically it seems it is always orthogonal since $r(t)$'s point draws out the curve and $r'(t)$ is a tangent to this curve.
 
@Semiclassical: Try integrating $-y dx$ from the origin to $(X,Y)$ along two different paths.
 
5:52 PM
@TedShifrin I think it's correct.
 
Nonsense, @JohnJack: Only if your curve is a circle centered at the origin.
 
It is showing the. Message is too long
 
actually, i guess it's the difference between having a one-form in exactly one variable versus more than one
 
Yes, @teadawg. You mean line integrals, surface integrals, etc.?
 
@TedShifrin Nuff said.
 
5:54 PM
Path-independence is automatic on $\Bbb R$, @semiclassical.
 
@Ted Yes, I'm well-versed with double and triple integration
 
yeah, i was about to say
 
Yes, @teadawg ... That's what I do all the differential forms stuff for. But there's also a lot more theoretical material than you've seen.
 
LetF be an antiderivative of f,as in the statement of the theorem. Now define a new function gas follows: g(x)= x a f(t)dt 3 ByFTC PartI,g is continuous on [a,b]and differentiable on(a,b)and g(x)=f(x) for every x in(a,b). now define another new function as follows: h(x)=g(x)−F(x) Then h is continuous on[a,b]and differentiable on (a,b) as a difference of two functions with those two.Moreover,ifx∈(a,b),h(x)=g(x)−F(x),but g(x)=f(x)byFTCPartI,andF(x)=f(x)bydefinitionofantiderivative.
 
unreadable, @Sayan. go learn latex.
 
5:55 PM
so would the correct statement be "locally path-independent$\implies$ closed"? which does seem right
 
Any place to
 
and you didn't even state the theorem.
 
the perils of being a lazy physicist
 
@semiclassical: Yes, closed is equivalent to locally path-independent (for $1$-forms).
@Balarka: You're a lot harder on your young compatriots than we were on you !
 
@TedShifrin Theoretical material? Isn't much of higher mathematics theoretical anyway?
 
5:58 PM
hmm. let me see if I can come up with an example, then, where that fails for 2-forms
 
Ted: So where do u think is the best place i should start learning maths?
 
@SayanChattopadhyay codecogs might be a good place. anyway, here's an exercise for you to do, if you think you know enough calculus :
 
Well @BalarkaSen how do I differentiate something like this
$$f(x)={\sin{x}}^{i}$$ as x tends to infinity
 
sure, @teadawg. I'm just saying there's lots in my course that does not appear in a standard multivariable course at all.
 
f : [0, 1] \to [0, 1] be a continuous function. Prove that there is a point x in [0, 1] such that f(x) = x.
 
5:59 PM
@yswong: I have no idea how old you are or what you know or what you have access to. I'm not capable of answering such a general question.
 
Don't google.
 
Bring on the challenge, @Ted :D
 
Ted: Well im a 20 year old university student
 
@SayanChattopadhyay You differentiate functions. $\lim_{x \to \infty} \sin(x^i)$ is not even a function.
 
Ted: Basically i need to brush up on my maths
 
6:00 PM
@TedShifrin He ignored that, it seems.
 
Ignored what?
 
Office hours. The hordes are descending.
 
Ah, god, not looking forward to Thursday.
 
I smell conspiracies going around to set me up on ignore again. OK, I apologize to spelling the word "trivial" :P
 
Ted: My math knowledge is at the calculus/ linear algebra level and i want to extend that knowledge
 
6:03 PM
@Mike $\Omega_1(pt)$ would be tremendous. There are a lot of $1$-manifolds out there.
 
It's zero.
 
jawdrop
 
And there aren't very many 1-manifolds out there.
 
Well, bunch of lines, bunch of circles.
 
The way you show [pt] \neq [0] is by classifying 1-manifolds. There are four connected ones: S^1, [0,1], [0,1), (0,1).
So the only closed 1-manifolds are disjoint unions of S^1s.
 
6:04 PM
Ah, so it wasn't quite trivial.
 
Which you can explicily write down null-bordisms of.
No, of course not, whence the frustration with your love of that word.
 
B..but that'd mean each time I compute $\Omega_n$, I'd have to classify $(n+1)$-manifolds! That's not computationally very efficient.
 
Or you could develop a bordism theory, which is highly nontrivial.
With some more algebraic topology you personally can calculate $\Omega^2(pt)$. $\Omega^3$ will still be out of your reach without developing tools built for such a purpiose.
 
Actually bordism groups of points are the only complicated bit, I suppose.
As it's only the dimension axiom that goes haywire.
 
The way one normally builds it up for CW complexes is by using Meyer-Vietoris, and now that one no longer has simple values on the point, I highly doubt values on CW complexes will be uniquely determined by values on a point.
 
6:11 PM
I can edit posts, now. Shall I go power-mad? (I realize such a spree would more than likely be short-lived)
 
No idea why one would want such a complicated theory.
 
@TedShifrin: i guess the relevant example for 2-forms is again a simple line current, but now considering the magnetic flux rather than the circulation
bah, i'm feeling silly for not being able to see it
 
6:32 PM
Hello!!

To write a natural number N in the decimal system we need $\lfloor \log_{10} N \rfloor +1$ decimal digits.

Can someone explain to me why this stands??
 
because a "decimal place" means a power of ten
 
And how do we get the logarithm??
 
For example, a three-digit number is between $10^2$ and $10^4$
log 10 is an increasing function, see?
So if $10^2 \leq x < 10^3$, what happens if you take logs?
 
If we take logs we get $2 \leq \log_{10} x \leq 3$
 
Now the floor function is non-decreasing, so the "strict" inequality holds
Adding the 1 takes case of the "end cases"
That is: $2 < \log_{10}x < 3 \implies 2 \leq \lfloor \log_{10}x \rfloor + 1 \leq 3$
It doesn't change until we get to 1000, in which case we "go up one digit"
Our "decimal number system" is "exponent based"-each "digit place" corresponds to a POWER of ten.
Which power? Take the log and find out.
 
6:48 PM
So, we want to write N in the form $a_0 10^0 +a_1 10^1 + \dots + a_n 10^n$, right?
 
It's a fancy way of saying what we already know
Yes, but the number of digits there is controlled by $n$, so only the "left most" matters
 
so we are looking for n...
 
Right-but remember $a_n \leq 9$, so $\log_{10}(a_n) < 1$
The smallest $n+1$ digit number is when $a_n = 1$, and $a_k = 0$ for $k < n$.
So, for example, in 3 digit numbers, the smallest one is 100.
$\log_{10}(100) = 2$, and the floor of 2 is 2. We have to add 1 "to correct".
$\log_{10}(x)$ is going to be more than 2, and less than 3, as long as $x$ is more than 100, and less than 1000.
In all these cases, the floor function will return 2, we add 1, and we find the number of digits is 3.
 
I see... Can we conclude that the number of digits is $\lfloor \log_{10} N \rfloor +1$ from the formula $N=a_0 10^0 +a_1 10^1 + \dots + a_n 10^n, a_n \leq 9$ ??
 
For more digits, the same pattern holds, we're "rounding down" to the nearest power of 10.
@MaryStar That's not so useful, the log of sums isn't easy to deal with.
 
6:58 PM
Ok... How can we prove then the formula??
 
Look, if a 3-digit number is $abc$, it doesn't make ANY difference what digits $b$ or $c$ are.
436 and 467 are both 3-digit numbers.
It's more useful to observe log is an INCREASING function.
 

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