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12:10 AM
hello if we have f(x,y)
then f'y (x,0)
what does it mean
 
12:38 AM
Would it be acceptable to finish a question asking me to solve a system of equations (which I do using a matrix) as a matrix/vector form result? i.e. $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{5} \begin{bmatrix} -3 \\ 5b - 8 \\ 5b - 1 \end{bmatrix}$
 
12:52 AM
Hilbert's paradox of the Grand Hotel is a veridical paradox (a valid argument with a seemingly absurd conclusion, as opposed to a falsidical paradox, which is a seemingly valid demonstration of an actual contradiction) about infinite sets meant to illustrate certain counterintuitive properties of infinite sets. The idea was introduced by David Hilbert in a lecture he gave in 1924 and was popularized through George Gamow's 1947 book One Two Three... Infinity. == The paradox == Consider a hypothetical hotel with a countably infinite number of rooms, all of which are occupied. One might be tempted...
cool
 
1:11 AM
@Paul, are you sure?
In Aluffi there were no functors there.
And I'm pretty sure a slice category is a special case of a comma category (or v.v.?)
 
1:29 AM
Any method to prove the following?A square was divided into 1000 triangles with equal areas. Then the same square
was divided into 1000 rectangles with equal areas. Prove that there exist 1000 points inside the square such that every triangle from the first division and every rectangle from the second division.
contains some of these points inside.
please advice
 
What do you mean by "first" and "second division"?
 
first division of triangles and second division for rectangles
first 1000 triangles from a square, and then 1000 rectangles from a square
 
1:46 AM
The "probabilistic method" might be useful here.
Morning, @Paul.
 
Slice is a special case of comma, yours is a special case of comma, and aluffi's is a special case of comma
Hey
 
Hmm.
So the one with functors I showed you, what's that called?
$\require{AMScd}
\begin{CD}
F(X) @>{F(f)}>> F(Y)\\
@VsVV @VtVV\\
A @= A
\end{CD}$
 
Pretty sure slice category, or you would just specify the type of comma category
Not everything needs a fancy name
Actually, maybe just comma, at least looking at the wiki def
Like I said, not everything needs a special name
 
Oh, yes!
This is a kind of comma category.
I should've checked wiki before :/
 
Categories for the Working Mathematician just calls this, $F$-over $A$
or $(F \downarrow A)$
 
1:58 AM
Hmm, thanks.
 
(and they point out that it is a special comma category)
 
2:14 AM
I am trying to prove the following, but I wonder if this relates to an identity matrix.
would you please help... All entries of an n × n matrix are non-negative. It is known that the sum of numbers
at any row and at any column of this matrix is exactly 1. Prove that you can choose n positive
entries such that all of them are from different rows and from different columns.
 
2:30 AM
Uh, the diagonal?
I think you meant n positive entries summing to 1.
 
He never said the entries are positive.
 
 
2 hours later…
4:24 AM
I wish there is some like summer school where you just pick a topic with a prof and solve numerous questions
in particular topic
 
 
1 hour later…
5:54 AM
@MikeMiller uh-oh :)
 
6:42 AM
@MikeMiller So my question was that if, given an $X$ with $Z(\pi_1) = 1$, any self-homotopy of $\text{id}_X$ is homotopic (modulo endpoints of the homotopy) to the identity homotopy. You suggested that the counterexample should be $X = S^2$, with the homotopy given by a full rotation of $S^2$ along the axis joining the poles.
How are you sure that this is not homotopic to the identity homotopy? At least, it's not obvious to me that it's not.
 
That represents an element of $\pi_1 hAut(S^2)$. The claim is that it's a nontrivial element; your claim is that it's trivial.
 
I don't know what $hAut(-)$ means.
 
6:57 AM
The group of self-homotopy equivalences. For a compact space this is topologized as a subset of the space of maps $X \to X$.
 
oh, ok. so you're giving it the compact-open topology?
 
Yes.
To prove that this represents a nontrivial element you should prove that $O(3) \hookrightarrow hAut(S^2)$ is a homotopy equivalence; and because this factors through $\text{Homeo}(S^2)$ it suffices to prove $\text{Homeo}(S^2) \hookrightarrow hAut(S^2)$ is a homotopy equivalence. This is surely true.
Lurie's proof for $g \geq 2$ is essentially standard and something different has to be done for $g \leq 1$. But there's other proofs of related things one might try to modify.
 
I'd have to look through Lurie's notes, but I don't think I'd understand all of this. Apparently, I don't understand why $\text{Homeo}(S^2) \hookrightarrow hAut(S^2)$ should be a homotopy equivalence at all.
 
7:17 AM
Generally such things are true when "the homotopy theory is the same as the topological theory". in dimensions at most 3, every closed topogical manifold has a unique smooth structure, and appropriately $\text{Diff}(M) \hookrightarrow \text{Homeo}(M)$ is a homotopy equivalence for closed $M$ in dimensions at most 3.
So, because every closed surface is homotopy equivalent to only itself, I suspect this should be true for closed surfaces. (It's true for genus at least 2, but for perhaps different reasons.)
 
interesting.
 
I think it's probably true that $Homeo_\partial(\Sigma) \hookrightarrow hAut_\partial(\Sigma)$ where the $\partial$ means we restrict to the things that are the identity on the boundary. This is more because I say so than any moral reason as above, though I guess you could justify it as "there's only one compact surface with $n$ boundary components in a given homotopy type"
I don't have time to check but you might try to mimic the argument given by Cerf in "Groupes d’automorphismes et groupes de difféomorphismes des variétés compactes de dimension 3" but in dimension 2 and with different groups.
 
Most likely I wouldn't be able to make anything of it, but thanks for the time and the reference!
 
7:32 AM
What is algeraic topology, briefly? @BalarkaSen
 
Study of topological spaces upto homeomorphisms using algebraic homeomorphism invariants.
 
Diving together multiple values in no given order, or a three part fraction, have any of thease been explored?
I've come up with two possible aproaches.
 
i don't know much about abstract algebra (all i know is Lagrange theorem), but isn't homeomorphism from topology? What is algebraic homeomorphism?
 
There's nothing called algebraic homeomorphism. I mean homeomorphism invariants of algebraic nature. If you don't know much abstract algebra/topology, probably you shouldn't bother about algebraic topology.
 
It's not that I don't have time right now, it's more that it's a day project, and I have more pressing things to spend a day on. I'd really like to know that it's true, though. (It predictably fails in dimension 3, but that's "because" you can have homotopy equivalent but not homeomorphic closed 3-folds.)
 
7:37 AM
bother myself? what do you mean @BalarkaSen? i'm just curious about how topology plays in algebra.
homeomorphism invariant of algebraic nature
 
@MikeMiller No, I understand. I'd probably ask prof, I think. (Also, how does having htpy equivalent but not homeomorphic 3-manifold imply that this fails in dim 3? I mean, I know examples of such manifolds (lens spaces!) but I don't see how having such a thing has anything to do with this problem)
@ShimKim I mean, I don't think you should think about algebraic topology without knowing both of algebra or topology to some extend. Also, topology doesn't "play" in algebra. Algebra is used to resolve standard topological problems.
 
Hello @MikeMiller I am back to SE...
 
hi @Jasper.
 
Hello @BalarkaSen. I could be an imposter, but I am not.
 
@BalarkaSen ah, now i see.
 
7:41 AM
@Jasper Where were you all this time?
 
btw it's Shin, not Shim
XD
 
@BalarkaSen In the same place. Just not on SE.
 
@Balarka: Recall what I said above. Morally you should expect that the automorphism groups are homotopy equivalent when the "theories" are the same, like Diff(M) and Homeo(M) for manifolds of dimension at most 3, where each topological manifold has a unique smooth structure.
Hi @Jasper.
 
@Jasper How're you, then?
 
@BalarkaSen Still the same, trying to get better. My colour is called "dodgerblue" by the way.
 
7:46 AM
@MikeMiller Oh, I see. So you're trying to say that "hAut(M) and Homeo(M) 'should' be the same when the homotopy theory of M is not-so-different from the topological theory of M"
But as these can be different for 3-manifolds, one shouldn't expect it to hold for 3-manifolds.
That makes sense.
 
Something like that.
 
@Jasper I like the colour.
 
@BalarkaSen It is sad that over 2000 people died from the Indian heat wave.
 
(In fact, you can prove that hAut cannot always be homotopy equivalent to Homeo - it will fail for, say, at least one of L(7,1) and L(7,2).)
 
Indeed it is. It's good that it has been raining hard for the past few days though.
definitely interesting, @MikeMiller. I am interested to read about these, although I know I wouldn't understand much. I googled and found your question here.
Probably I shouldn't go down in a rat hole trying to understand this (although I want to :P).
Once again, thanks!
 
8:03 AM
Sure. I don't really understand how Cerf proves the theorem, since I don't understand that Lemma 0. I should sit down some weekend and try to really understand it.
 
8:15 AM
I see that Hatcher has deleted is old proof of $\pi_1(S^n) = 0$ and shifted the a bit of the van Kampen theorem (the one about the surjectivity of the map $\pi_1(A) * \pi_1(B) \to \pi_1(X)$) and placed it in here.
I have never really read his proof of $\pi_1(S^n)$ being $0$. As far as I remember, he worked with pathological curves on $S^n$ and proved that it can be locally perturbed to something nice, which in turn can be homotoped to the basepoint.
I am not sure why he deleted it. Is it incorrect (unlikely!) or is it just because it's too complicated (I agree that just doing baby van Kampen on the two hemispheres is easier)
I guess I will read up the old proof from my hard copy of the book.
 
He used the same argument later to show $\pi_m S^n = 0$ for $m < n$. The technique is known as simplicial approximation.
All one needs is to pass to a class of maps that can't possibly be surjective. Smooth, simplicial, cellular all work.
 
I only skimmed through 2.C. when I did homology. I guess I will do it this time.
That's actually a pretty neat proof.
 
8:37 AM
I am going to have some coffee...
 
9:16 AM
@Jasper long time I haven't seen you around. How are you doing? :-)
Perhaps it's a sign of overloading, I feel my arms and legs very heavy, and I feel I move myself in slow motion. :-)
@robjohn I started on that proof but I didn't finalize it, hope to do it today. Less energy here.
(it's just a matter of writing things nicely - since it can be calculated without pen and paper)
 
9:56 AM
@Chris'ssistheartist Hi. I am still trying to get better. How is your book progressing?
 
@Jasper Pretty well, thanks. In the last period of time I came across some extraordinary results that I'm going to add to my book. :-)
 
user147690
@Jasper JASPERRR!Hi!!
 
@AlexClark Hi!
 
user147690
You've come back!
 
I just answered some lhf so that I can chat, lol.
 
user147690
10:00 AM
You have more rep than I do :P
 
I see you have many accounts, lol. You should just use one.
 
user147690
I do just use one now :P
 
I am now using Fedora 22 Mate desktop.
 
user147690
Just this one until the semester ends and I work out what I am doing with the old main one
 
user147690
@Jasper I am still using ol' windows 7
 
10:01 AM
When I am rich, I will return to Windows and Office, but now it is Linux and LibreOffice.
Actually, Linux is really easy. More people should try it out.
 
user147690
What made you come back?
 
user147690
Missed us too much?
 
@AlexClark Nothing, just mood swings. Do you have a roof over your head?
 
user147690
Indeed I do, still live with the ex-gf
 
I see. If you truly love each other, you will be together again soon.
 
user147690
10:04 AM
We won't be, but that's alright, I'm over it now mostly
 
A girl I like very much got married yesterday, lol.
I don't expect too much now, just hope to get well and live a normal life...
 
user147690
Do you and her talk often?
 
No, I only met her online, lol. Still, I can "like her very much".
 
user147690
I am only really hoping these days that I get a TA job next semester
 
You should go to the US for grad school.
 
user147690
10:06 AM
I might, but the pay is much worse
 
user147690
We have a top 20 uni in australia, I might just go there
 
I never thought of going to Australia. If I get well, I will reconsider my options.
 
user147690
Pay is pretty much double here
 
I just wanna say, I will try to get well soon, but it might never happen, that's all...
In the meantime, I will try to answer some lhf on SE and get some rep for fun.
 
Hello @Jasper !! How are you?
 
10:09 AM
@MaryStar Hi, I see you are still a star.
 
:-D
 
@Jasper I often think I should finish a uni and become a professor, or a researcher that I like much more.
 
@Chris'ssistheartist Yes, you should, like I said before.
 
user147690
@BalarkaSen the last few questions on the past exam make me think it was actually fairly hard, just on time constaints
 
Yeah, I think of that.
 
user147690
10:22 AM
The first half is pretty damn chill
 
user147690
@BalarkaSen imgur.com/iFgIahH
 
user147690
Q6 is about a page
 
user147690
I suppose Q7 isn't too bad, but I had to think for awhile to get it
 
user147690
and its 2hours with no perusal
 
Q6 is not that big.
 
user147690
10:24 AM
Not a page?
 
nah
 
user147690
It's an if and only if proof though?
 
it's one-liner actually
 
user147690
What? An actual proof isn't?
 
hint : a field has no nontrivial ideal.
I mean, writing it should take a few lines. But it's not that huge.
 
user147690
10:26 AM
Write it up and you'll see it's longer than you think :P
 
user147690
I didn't do it the proof wiki way, but there way is really long for example proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field
 
$I$ is maximal implies there is no intermediate $J$ between $I$ and $R$ implies there is no $R/J$ contained in $R/I$ implies $R/I$ is a field (every ideal of a comm. ring is a quotient, iirc)
The other direction is equally easy
That didn't take a page.
@AlexClark yeesh
 
user147690
Starting with maximal is actually the shorter direction admittedly, but I don't understand your proof so give me a min
 
user147690
Seems like you used a theorem since I don't see anything about multiplicative inverses, which seems like the logical direction
 
Hi ha Howdy @Jasper
 
10:31 AM
the other direction : $R/I$ is field implies there is no nontrivial ideal $R/J$ in $R/I$ implies there is no intermediate $J$ between $I$ and $R$ implies $I$ is maximal
done and done
@AlexClark what kind of multiplicative inverse?
oh, you want to write down an explicit proof. nah. look at my hint above.
 
One uses multiplicative inverses to show that a field has no nontrivial ideals. The entire content of the theorem you're talking about is the fact that the ideals of $R/I$ are of the form $J/I$, where $J \supset I$ is an ideal of $R$. This is a computation.
 
$R$, a commutative ring, is a field iff it has no nontrivial ideal.
 
user147690
I've already written an explicit proof, it's just long is all
 
user147690
@BalarkaSen That's true
 
There are shorter proofs, that's the point of my proof.
 
r9m
10:33 AM
@robjohn ROFL :P !!! ..
3
 
I am just trying to justify that the 2hrs is a fair amount of time for doing the exam :P
 
user147690
@BalarkaSen Well he would expect us to do the long explicit ones always admittedly
 
well, why would they?
 
user147690
and 7b doesn't mean write it, it means write and prove it for example
 
user147690
@BalarkaSen Expect that?
 
10:35 AM
yes?
@AlexClark Yes, I get it, but it won't be huge either.
 
user147690
@BalarkaSen Why not? It's easier to remember a bunch of theorems than to know what's happening
 
user147690
I'll probably be doing a bit of both :P
 
I disagree that explicit proofs say more about what's happening.
 
user147690
Uni is about jumping through hoops in the hopes that we learn something along the way
 
the correspondence theorem for ideals is useful and a powerful tool
 
user147690
10:36 AM
@BalarkaSen Really?
 
This is a weird debate, because the two proofs you're talking about are literally the same thing, just written differently.
 
user147690
@MikeMiller I know, but explicitly writing the steps seems more insightful than using theorems.
 
@AlexClark Yes. The real idea will get hidden behind a bunch of computations.
 
Why is the theorem true, @AlexClark?
 
user147690
@MikeMiller Well if I have read and understand the proof it is fine for sure to just use the theorem
 
user147690
10:39 AM
But often I can't remember the proof, so writing it explicitly should mean I remember the proof
 
Haven't you learned the correspondence theorem for ideals?
 
user147690
No
 
user147690
Reading it now
 
it's a super-useful tool. so useful that one should adapt is as a generic axiom while doing commutative algebra, but I guess it's just a matter of taste and practice.
 
user147690
Haha sure
 
10:44 AM
How did you do 7(b)?
 
user147690
I so happened to know the answer :3
 
user147690
That's a past exam, but it was also an assignment question remember
 
True. We talked about it here.
 
user147690
My assignment was showing $(p)$ and $(f)$ were prime ideals, + showing $(p,f)$ is prime sometimes
 
right. but $(f)$ might not always be maximal.
 
user147690
10:46 AM
Yep since Z[x] isnt a pid
 
nods.
 
user147690
I suppose I have learned a thing or two
 
user147690
Hopefully I'll learn a third thing next sem :P
 
you have :)
I'd say your algebra is pretty strong up 'till now.
 
user147690
Thanks, that actually means alot for some reason
 
user147690
10:50 AM
I'll get atleast a distinction for the class
 
user147690
Then I will do both algebra classes, functional and differential geo next sem
 
user147690
Or try to, and drop either alg physics or differ
 
Hi @Balarka .. Thanks for the explanation yesterday
 
no problem, but as Mike said, it's the Nash-Kuiper embedding theorem, not Nash embedding theorem. The latter adds an extra condition on smoothness.
@AlexClark Definitely, try both, and see what you like as the class moves along :)
 
Is there any quick way of finding a cubic's roots when one is known such as $2\lambda^3 + \lambda^2 - 13\lambda + 6 = 0$ other than going through the process of polynomial division then factorising?
I know $(\lambda - 2)$ is a root and one of them must be $(2\lambda \pm y)$ and the other $(\lambda \pm z)$ but besides that I'm stumped
 
10:56 AM
@Hippalectryon tu connais le théorème des fonctions implicites ?
 
I could form a few questions such as $-2zy = 6$ but I figure I'd probably end up doing a contrived method
 
r9m
@robjohn @Chris'ssistheartist (or anyone else interested) .. could anyone suggest a badass way of solving: $$x^3+1 = 2\sqrt[3]{2x-1}$$ for real $x$.
@Rememberme you might like this one ^^
 
@r9m A badass? :-) OK, let me think ...
 
Is a badass a bad person's ass?
 
r9m
@Jasper :P lol ... a bad person's ass would be badass if it was intimidating and impressive at the same time I guess :P
 
11:05 AM
Hello@Jasper
Remember me?@Jasper
 
@Rememberme Yes.
 
How are you?@Jasper
 
Still trying to get better.
 
Hmm.. I still have few months before I am totally free of anemia@Jasper
 
Can one be free of it?
 
11:09 AM
Kinda yes and kinda no
 
A good answer.
 
@r9m btw, did you have time to look on my question I showed you some days ago?
 
r9m
@Chris'ssistheartist posted a message in the other chat room just now
 
@r9m Yeah, right. :-)
 
The SE sites are pretty boring on the weekends.
 
11:13 AM
@Balarka But you said Nash embedding theorem is really not about embedding but according to what you last said you said that it says
That there is an isometric embedding of $S^1\times S^1$ in $\Bbb R^3$.
roughly
 
Yes.
The essence of the "surprise" in Nash/Nash-Kuiper is isometrical embedding not just topological embedding.
Nash is surprising still as you add smoothness.
 
oh.. though I didnt get what you mean by smoothness
 
You don't have to care about it right now.
 
Okay bubye
 
Also, note that it not only says that $S^1 \times S^1$ can be isometrically embedded in R^3, but also that any Riemannian manifold can be done so in an appropriately large euclidean space.
 
11:19 AM
Are coefficients of matrices commutative? - $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \frac{1}{2} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \frac{1}{2} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix}$?
 
crl
of course, A*(kB) = k.A*B
 
@r9m did you see that from the first time? Just trying to see how difficult the problem is.
 
@BalarkaSen: I'm going to fight about this one, too. You should be careful how you state this, as these are two very different theorems. The Nash theorem says that given any Riemannian manifold $M$, there is a $C^k$ isometric embedding into some arbitrarily large $\mathbb R^N$, where the $N$ the theorem gives is quadratic in the dimension of $M$. This is for some level of smoothness - maybe $4 \leq k \leq \infty$? I don't remember what the lower bound is other than that it's bigger than 2.
You can even extend this to be an analytic isometric embedding.
 
@crl Thanks! I was pretty certain, but exam tomorrow and don't want to throw away quarter of the paper from doing something basic wrong :P
 
But the $N$ has to be big. There is no smooth embedding of the flat torus into $\mathbb R^3$.
 
11:23 AM
@AshleyDavies Just don't tear the paper.
 
r9m
@Chris'ssistheartist well to be honest .. even though the trick is old to me .. it's still an interesting and highly entertaining problem :) .. and also after I got to the square analogue .. I knew I was done simply because I had attended that in the past .. even that should be an interesting challenge to students who are trying it for the first time .. (aka a novel question)
 
Hahaha @Jasper I'll make sure not to do that :') I'll probably do what I did last year and go in, forget everything I know about calculus, and fail it horrendously, but there we go
 
The other one - first Nash, then Nash-Kuiper - says, roughly, that $C^1$ isometric immersions are dense in the space of $C^1$ immersions, provided that the domain is codimension at least 1 in the codomain. In particular this implies that if you have a $C^1$ embedding $M \hookrightarrow N$, $\text{dim} M < \text{dim} N$, then you can perturb this an arbitrarily small amount to make this into a $C^1$ isometric embedding. This is where we get the flat torus in $\mathbb R^3$ business.
 
ok, ok, I get you. maybe I should desist from talking about anything without knowing the right statement :P
 
11:26 AM
So this says for $C^1$ stuff, isometric immersions and isometric embeddings are not much different than immersions and embeddings with no statement about the metric.
 
@BalarkaSen Yes, that is why I say nothing, so people don't know how ignorant I am.
 
@r9m Yeah, indeed. It's an entertaining problem. :-)
 
@Chris'ssistheartist As entertaining as your pole dancer.
 
But this is, I think, wildly false for $C^k$, $k \geq $ whatever. I don't know if Nash's quadratic bound on $N$ is tight; probably not. But the way you prove this for $C^1$ immersions fails for higher $k$.
 
@MikeMiller So Kuiper theorem is actually a general statement which has nothing to do with metric spaces?
Weird.
 
11:27 AM
@Jasper :-)))))))))))))
 
I don't understand the question.
 
@Chris'ssistheartist I think I should watch La Vita E Bella again and see Monica Belluci, lol.
 
I mean, it doesn't assume anything about $M$, $N$ (about them being geodesice metric spaces/Riemannian manifolds)
 
Oh, I see. I think I phrased it poorly. I meant that the Nash-Kuiper theorem says that you can promote any $C^1$ immersion or embedding to an isometric one by an arbitrarily small perturbation.
 
@r9m in my book I also wanna add the alternating version with $(-1)^{m+n+p}$. :-)))
 
11:28 AM
Isometric doesn't make sense unless they're Riemannian manifolds.
 
@Jasper hehehe, a very good movie! And don't miss Malena! :-)
 
haha, fair.
I'm being stoopid.
 
r9m
@Chris'ssistheartist I see .. that's interesting :-)
 
That's extra structure on a manifold. It doesn't make any assumption about the underlying smooth manifold, though. Every smooth manifold can be given a metric.
 
@Chris'ssistheartist My favourite German movie is Sommersturm, or Summer Storm.
 
11:29 AM
@r9m It is! I actually I added it in my folder with the problems.
@r9m By the way, would you consider that version I showed you a good problem to add in a book? Of course, I plan to add it to my book.
 
r9m
@Chris'ssistheartist you have a folder with your problems?! :D awesome ! I'd like to see that someday (if you are willing to show me that is)
 
@Jasper I didn't watch them.
 
r9m
@Chris'ssistheartist indeed!
 
@r9m Yes, over 300 ready for my book. Still undecided about some of them. I might replace some on the way.
 
r9m
@Chris'ssistheartist oh!! Then y u no publish them as soon as you have finalized which ones you wanna put them in book ?!
 
11:31 AM
@r9m I said that I plan to increase the number of problems from 300 up to 500. 300 is too less.
 
Wait, I don't believe you. Any $C^1$ embedding of $\Bbb RP^2$ in $\Bbb R^4$ can be perturbed to an isometric one?
 
r9m
@Chris'ssistheartist OHO!!! Thats fantastic !!!! :D
 
I intend to keep this account for life, LOL.
 
Yes. ($C^1$ isometric.)
 
11:35 AM
Weird!
 
I also stated the theorem incorrectly. Apologies. The domain needs to be $\mathbb R^q$, at least in every statement I checked. I don't know enough about the proof technique to say whether or not it should be extendable.
 
I can believe that there is C^1 isometric embedding, but that one can whip one up from every C^1 embedding is totally wacky.
 
@r9m Mathematics is the easiest part. The aim is to make the book looks like the book of Nahin. I'm not interested in the cold language of mathematics as regards my book, I need a friendly-looking book.
The reader should NOT tramble with the book in hands, but have fun! The effect of a roller coaster amongst integrals, series and limits.
 
@MikeMiller What d'you mean the domain need to be $\Bbb R^q$?
 
r9m
@Chris'ssistheartist okay! well I prefer challenging problems over friendly encounters .. :P
 
11:37 AM
Codomain. Sorry again.
 
phew!
 
Too many sorries today.
 
@r9m Well, yeah, but my point is to present the challanging problems in a friendly way. The book MUST NOT be a cold book.
 
r9m
@Chris'ssistheartist :D .. that's heck of a goal! :D
 
@r9m I like Nahin's book very much, one also has some fun there. And all is also very clear.
 
r9m
11:39 AM
@Chris'ssistheartist :-)
 
@r9m So, the hardest part is not mathematics, not at all, but the way you present the content to the reader. That makes the whole difference in my humble opinion. :-)
Solving problems is a piece of cake, presenting them in a very nice way to the reader is the incredibly hard part.
In many parts of the solutions you might talk to yourself and say "that's clear, obvious"
 
OK, yes, it fails wildly if you let even $C^k = C^2$. Given a $C^2$ isometric immersion of $S^2$ into $\mathbb R^3$, apparently composition with some affine orthogonal transformation (i.e. $x \mapsto Ax+b$, $A$ orthogonal) gives you the standard inclusion map $S^2 \hookrightarrow \mathbb R^3$.
 
but the question is : "Is it true for the reader too? Is it obvious to him/her? Maybe no."
 
@BalarkaSen A matrix is rank $0$ is it's the null matrix ?
 
In particular, the Nash-Kuiper embedding theorem gives you a $C^1$ isometric embedding that's contained inside some arbitrarily small ball in $\mathbb R^3$, but here you can't get any smaller than a ball of radius 1.
 
11:44 AM
@r9m First I put all my stuff in a terse way, and then I develop all in the reader's language if I can say that. :-)
 
@Gato Yes
 
@BalarkaSen and rank 1 if all the columns are collinear ?
 
I'll believe you, @Mike. I have heard that there's no $C^2$ isometric embedding of the flat torus in $\Bbb R^3$, iirc.
 
Sure, that's a curvature argument. You can only do that because curvature is a $C^2$ construct.
 
I am going to eat dinner now.
 
11:49 AM
Can someone give me a hint of how to integrate $cosh^2x$?
I've tried substitution but ended up with a mess of $\frac{u^2}{sinhx}$
 
@Gato The matrix has rank 1 if the columns generate a 1-dimensional vector space over R. That means each of the colums are R-multiple of some vector. That means they're all collinear, right?
 
@AshleyDavies using exponential ?
 
'Cause they're all sitting in a 1-dimensional subspace of R^n.
 
Aw crap I keep forgetting! Thanks!
I hate calculus :c
 
@BalarkaSen right, Ted was right, no need to use determinant. :)
thanks
 
11:52 AM
Nope, just use the definitions.
 
@AshleyDavies don't say this here :p
 
@Gatot Hahaha probably shouldn't :') I struggle to keep track of all the identities and what forms are differentiable/integrate-able - I remember all the identities and equivalences but never consider them in problems like that :P
Life was good until integration by parts and substitution invaded :'(
 
@AshleyDavies don't worry, in few weeks (with 'hard' work) you will say that the life is beautiful with integration ;).
 
Yeah hopefully :) @Gato I intend to practise loads over summer so I'm ready for university but that might just end up like another new year's resolution :P
 
@AshleyDavies haha :D, it's even harder during summer..
 

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