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Jakobian
Jakobian
00:00
@XanderHenderson forget what I said about "they", that was untrue
Xander Henderson
Xander Henderson
@Jakobian In English, the second person pronoun is "you". We no longer have a distinction between singular and plural in the second person (historically, I believe that "thou" is the singular, and "you" the plural; in modern English, some dialects have an explicitly plural form, such as "y'all" or "youse", but these are pretty regional).
In Russian, the second person is either ты (singular or familiar) or вы (plural or formal / polite).
"You'uns", too.
And my favorite second person pronoun: "all y'all", which is distinct from "y'all".
Jakobian
Jakobian
that's the same as for Poland but like I said, the second word isn't used to refer to someone formally anymore, and that usage is archaic and replaced by "Pan/Pani"
Ted Shifrin
Ted Shifrin
@Semiclassical Well, I was thinking it should be in a plane parallel to the projection plane. But perhaps that was hasty.
Semiclassical
Semiclassical
i have to say, checking these is getting me mixed up
Jakobian
Jakobian
@Jakobian I believe, sometimes in movies a Slavic foreigner (probably Russian soldier) will call Poles pretentious because of this, because for them it sounds like we refer to each other as some kind of royalty
Ted Shifrin
Ted Shifrin
00:13
If you slice by a plane not parallel to the plane you're projecting to, it's not clear to me why you get the same bounding curves.
Semiclassical
Semiclassical
00:49
"Japanese is flat" /「日本語はフラットです」
►
Semiclassical
Semiclassical
01:35
@TedShifrin i fear i'm just repeating myself, but here's how i'd summarize what i now know. let $F(x,y,z)=2z^2-1-x y-\sqrt{1-x^2}\sqrt{1-y^2}$. then $F(x,y,z)=0$ for $z\geq 0$ defines a surface. this contains the space curve $g(t):=(4t^3-3t,t,t)$ for $1/2<t<1$, which obviously lies within the $y=z$ plane. the nontrivial claim is that $2\partial_y F=\partial_z F$ when evaluated at any point on the curve
so along the curve there's one linear relation between the coordinates, and another between the derivatives w/r/t to these coordinates
Ted Shifrin
Ted Shifrin
Just think about the kernel of the derivative of the projection map on the surface. Of course, projection is a linear map, so it is its own derivative,
user 85795
user 85795
Wow, the rendering in this room is really made for apple devices.
Android renders the above right into the comment's side bar.
Nice contrast, actually :-)
Semiclassical
Semiclassical
01:58
@TedShifrin sure, but i don't think you'd generally have a linear relation between the derivatives. (if only b/c i tried a different example and found no such relation)
 
4 hours later…
Dannyu NDos
Dannyu NDos
05:42
My professors say regarding only CW complexes makes algebraic topology too trivial.
That's why I'm writing about point-set topology despite having learnt algebraic topology.
Ted Shifrin
Ted Shifrin
05:57
@Semiclassical I don’t understand this at all.
@DannyuNDos This makes no sense. Point-set topology is of marginal interest and a subject of relatively little research in the modern day.
Semiclassical
Semiclassical
i mean, take a generic slice of a convex set, and look at the surface normals to the original set along the boundary of the sliced set. there's no reason that these normals have to lie in a common plane
e.g., slice a cone horizontally. the surface normals along the resulting circle will point down and radially out
Ted Shifrin
Ted Shifrin
So what? We were discussing a special circumstance.
Semiclassical
Semiclassical
sure. i was trying to say what was special about it
(and i just realized my cone example is a bad one, since all these normals have the same z-coordinate. maybe for a general conic section it'd be different, but a circle is a dumb choice)
one potato two potato
one potato two potato
06:20
I think CW complex alone can construct some higher concepts in AT. Caring only something in any field is rejected these days.
any it's not too trivial. any manifold admits a CW decomposition. So he's claiming that studying manifolds using AT is too trivial.
because understanding manifolds using only CW complexes in AT is too hard, people developed and created different methods.
Semiclassical
Semiclassical
found an (admittedly sorta silly looking) way to reformulate my problem. Suppose $\cos(2\gamma)<\cos(\alpha-\beta)$ for $(\alpha,\beta,\gamma)\in[0,\pi]^3$. Then there exists $\delta\in[0,\pi]$ such that $\cos \delta = 1/3(2\cos\gamma+\cos\beta)$ (since the RHS is definitely in $[-1,1]$. Must $\cos(2\delta)<\cos(\alpha-\delta)$?
the angle $\delta$ definitely has to lie between $\gamma$ and $\beta$, so this is effectively asking whether the inequality remains true if we replace $\gamma,\beta$ by this "average" of them
as far as i can tell it is true, but i can find numerical counterexamples if i replace $\cos \delta$ with any other such averaging (e.g., weights 1/4 and 3/4)
 
1 hour later…
Dannyu NDos
Dannyu NDos
07:44
@TedShifrin It's just my master’s dissertation. Tbh, I can't find a single new result, so I'm just writing solutions to Chapter 7 of Munkres.
one potato two potato
one potato two potato
well, many of the master's dissertations are survey papers of their research interests unless he/she is very lucky or very talented to produce actually new theorems.
 
2 hours later…
Alessandro Codenotti
Alessandro Codenotti
09:48
@TedShifrin :(
Jam
Jam
10:32
WHy the change of variable in integrals the new region is not a region where the first region is transformed but a region which is transformed to the starting one? Like a pullback region why is this like going backwards?
Jam
Jam
10:45
its kinda counterintuitive. its more like a change of variable of (u,v) to (x,y) and not for (x,y) to (u,v)
or is it because the easy transformations of shapes for example rectangle to a disk and not a disk to a rectangle so instead of working how to make the disk to a rectangle we ask what region is transflrmed to a disk .
Jam
Jam
11:16
neverm ind i figured it out. when setting x=u+v, y=u+3v or whatever is a transformation of the u,v plane to the u+v,u+3v plane. It is different from setting x=2x+y which is a transformation of the x,y plane.
 
2 hours later…
Jam
Jam
12:47
Prove there exists m such that $ a<\frac{2+m}{2+n}<b $ for large enough n.
Sonozaki
Sonozaki
13:24
I was trying to build a simple example to show that even if we find a $\delta$ dependent on $x,x_0$ in the definition of continuity for a function $f$, this does not imply that $f$ is not uniformly continuous. I thought about this: let $f(x)=x$, this function is clearly uniformly continuous on $\mathbb{R}$ (it is enough to choose $\delta=\epsilon$).
But for arbitrary $\epsilon>0$, setting $\delta^*=\epsilon\left(1-\frac{1}{|x|+|x_0|+2}\right)$ if $|x-x_0|<\delta^*$ then $|x-x_0|<\epsilon\left(1-\frac{1}{|x_0|+2}\right) \le \epsilon/2 <\epsilon$; hence, showing a $\delta$ dependent on $x_0$ is not enough to conclude that a function is not uniformly continuous. Is this reasoning correct?
Some typos in the latter message, sorry: I meant "$|x-x_0|<\epsilon\left(1-\frac{1}{|x|+|x_0|+2}\right)<\epsilon$" and "showing a $\delta$ dependent on $x,x_0$".
Xander Henderson
Xander Henderson
14:06
In showing that the identity function is continuous $x_0$, you should not need to choose a $\delta$ which is dependent on $x_0$. You can choose such a $\delta$, but why?
For any $\varepsilon > 0$, choose $\delta < \varepsilon$. If $|x-x_0| < \delta$, then $|f(x) - f(x_0)| = |x - x_0| < \delta < \varepsilon$. Done.
Don't over complicate it.
I'm confused about what you are trying to show (and the repeated negations in your first sentence are giving me a headache). :/
user223626865
user223626865
14:21
@Sonozaki why even try to build your own example when textbook writers do that for you?
Instead, trying doing the examples on your own before reading the solutions that are given following the examples.
geocalc33
geocalc33
15:15
$$\zeta(s)=\frac{1}{\Gamma}\mathcal{ M} \bigg [\sum_{\alpha \in \Bbb N} \Phi_\alpha(t)\bigg] (s) $$
$\Phi_{\alpha}(t)$ is a 1-parameter family of Riemannian metrics. Sum over the parameter restricted to the spectrum of a certain ring, take the Mellin transform and you obtain a zeta function
actually the family of metrics is carefully crafted so that pre-multiplying by a Gamma factor $1/\Gamma$ gives $\zeta(s)$
Q: Let $\Phi_{\alpha}(t)$ satisfy a geometric flow. I wonder if that allows you to say something more about $\zeta(s)$?
geocalc33
geocalc33
15:39
for example exploit some symmetry and get a functional equation
 
2 hours later…
Ted Shifrin
Ted Shifrin
17:41
@Sonozaki Just to emphasize. To show that a function is not uniformly continuous you must show that it is impossible to find a $\delta$ that is independent of $x_0$.
Shaun
Shaun
17:55
Hi :) How do you get the normal subgroup generated by a set, in GAP?
Assume the group is finite for my purposes.
I'm writing a programme in GAP for my research, mainly so that I can gather data to inform conjectures.
Surely someone must've asked this before . . .
Ted Shifrin
Ted Shifrin
Many of us have no idea what GAP is. Algebra software is pretty recondite.
Shaun
Shaun
18:20
Fair enough, @TedShifrin. I've asked on GAP Forum. I'm surprised it's not an FAQ.
Alessandro Codenotti
Alessandro Codenotti
@TedShifrin It's even less popular than general topology! :P
Koro
Koro
Suppose that $f(x_1,x_2,...,x_m, y_1,y_2,...,y_n)= \sum x_i^2 -\sum y_i^2$. Then 1) when is $f^{-1}(c)$ an n+m-1 surface? 2) when is $f^{-1}(c) $connected?
Definition: For an open set U of R^{n+1}, given a smooth map f: U-->R, for a c in image f, the set f^{-1}(c):=M is said to be regular if for every p in M, gradient f at p is non zero. In this case, M is said to be an n space.
So for 1): Fix a c in R and denote f^{-1}(c) by M. For M to be regular, for every p in M, gradient of f at p must be non zero. It is observed that it is possible iff norm of p is non zero. In other words, M should not contain 0. That is, f(0)= 0 \ne c. It follows that for all non zero c, M is regular hence an m+n-1 space.
Shaun
Shaun
18:36
It's "NormalClosure(G, S);" - I forgot that synonym.
Ted Shifrin
Ted Shifrin
@AlessandroCodenotti That might be, but I wouldn’t know. :)
Alessandro Codenotti
Alessandro Codenotti
I'm just joking, I do agree general topology is not very popular nowadays
Ted Shifrin
Ted Shifrin
@Koro Start with small values of $m,n$.
Alessandro Codenotti
Alessandro Codenotti
I wouldn't really call myself a topologist either, even though I spent a lot of time recently going through some hard topology from the 50s/60s (Bing & Co.)
Ted Shifrin
Ted Shifrin
People forget that topology has moved quite a bit away from both general and algebraic. Lots of Seiberg-Witten invariants and symplectic tooology.
Alessandro Codenotti
Alessandro Codenotti
18:46
All the people who do algebraic topology here are doing $\infty$-categories and other scary things 24/7
Ted Shifrin
Ted Shifrin
Of course, basic point-set and a year or more of alg top are tools in all sorts of math.
Koro
Koro
I did that. So for m=n=1, we get connectedness if c=0 else we are looking at a hyperbola x_1^2-y_1^2=c which is not connected. For m=2, n=1: if c=0, then we get $x_1^2+x_2^2=y_1^2$, which is connected, if c\ne 0 then $x_1^2+x_2^2=y_1^2+c$, not sure what to conclude from this without looking at graph.
For higher dimensions, even a graph won't help.
Ted Shifrin
Ted Shifrin
$c=0$ gives a cone, so always connected. You can do your example without a graph, Think basic inequalities to see disconnectedness in some cases.
Koro
Koro
you mean in higher dimensions too?
the set of all (bold(x), bold (y)) in R^{m+n} such f(bold(x), bold(y))>c is open and so is for f(bold(x), bold(y))<c and these are disjoint. But I don't see how this helps if that's what you meant by inequalities.
Ted Shifrin
Ted Shifrin
Use the fact that (sums of) squares are nonnegative.
Koro
Koro
18:55
ohh so effectively I'm looking at $(x_1,...,x_m)$ such that $\sum x_i^2\ge c$?
(because $\sum x_i^2 =\sum y_i^2 +c \ge c$ since $\sum y_i^2\ge 0$.)
$f^{-1}(c)= \{(\bar x, \bar y): f(\bar x, \bar y)=c\}=\bigcup_{r\ge 0}\{(\bar x, \bar y): \|x\|^2=c+r\}= \{(\bar x, \bar y): \|x\|^2\ge c\}$
Ted Shifrin
Ted Shifrin
More to the point, suppose $c<0$; what does this tell you about $y$?
Koro
Koro
yeah, I'm thinking how to go from here.
Thanks
Ted Shifrin
Ted Shifrin
Anyhow, do some concrete stuff in small numbers before you feel like you should do the general argument.
Koro
Koro
For m=n=1, we get disconnectedness.
@TedShifrin sure
Koro
Koro
19:29
one problem though: according to the above set of equalities, I'm getting $f^{-1}(0)=\{(\bar x, \bar y): \|x\|^2\ge 0\}= R^{m+n}$, which can not be true as this would imply that $f\equiv 0$.
why this paradox?
sunny
sunny
I want to verify the uniform convergence of $$\sum_{k=1}^\infty (1-x)x^{k^2}, \quad 0\leq x\leq 1.$$ In a previous exercise, I computed $$\sup_{0\leq x\leq 1} (1-x)x^c,\quad c\geq 1$$ which I found was equal to $\frac{c^c}{(c+1)^{c+1}}$. Now I'm a little unsure how to apply this to the series above. I assume we want to use the Weierstrass M-test and so bound the absolute value of the terms.
However, when I let $c=k^2$, I get a very weird-looking fraction that I cannot determine the convergence of to use the test. Appreciate any help or hints.
Jakobian
Jakobian
$$\sum_{k=1}^\infty \frac{(k^2)^{k^2}}{(k^2+1)^{k^2+1}} \leq \sum_{k=1}^\infty \frac{1}{k^2+1} < \infty $$
@sunny here you go
sunny
sunny
@Jakobian I need to process this, but thank you!
Jakobian
Jakobian
I hope this isn't bad that I'm giving you the answer straight-out
sunny
sunny
no, not at all, I appreciate your help a lot!
sunny
sunny
19:57
ok, I see how we have $(1-x)x^{k^2}\leq \frac{(k^2)^{k^2}}{(k^2+1)^{k^2+1}}$, however, why is $\frac{(k^2)^{k^2}}{(k^2+1)^{k^2+1}}\leq \frac{1}{k^2+1}$?
Jakobian
Jakobian
take one factor of $k^1+1$ from denominator
sunny
sunny
got it!
algbr
algbr
20:41
Hi.
Jakobian
Jakobian
21:02
hello
sunny
sunny
Just want to double check this. I want to show that $$s(x)=\sum_{k=1}^\infty \frac{\arctan kx}{1+k^2x^2},$$ is continuous for $x>0$. I have shown previously that it is uniformly convergent in the interval $x\geq c$ where $c>0$. Does the result now simply follow from the uniform limit theorem?
Jakobian
Jakobian
21:17
@sunny yes
sunny
sunny
easy peasy then, thanks for confirming
D.C. the III
D.C. the III
21:47
@TedShifrin More or less than these new words you expose us to, to enlarge our vernacular?...................😊
geocalc33
geocalc33
Is it reasonable to expect that $\theta_3 \circ f$ satisfies a functional equation, given that $\theta_3$ is the Jacobi theta function?
$\theta(x)=\frac{1}{\sqrt{x}}\theta(1/x)$
no there probably not reasonable
algbr
algbr
22:04
I wonder why so few people chat here. There can't be so few people who are interested in mathematics. Perhaps the stackexchange chat is too unkown. :)
Gwyn
Gwyn
22:20
When proving limits, we can assume $\epsilon$ "small" without losing generality; that is, there is a logical equivalence between two definitions of limit, the first with "for each $\epsilon>0$" and the second with "for each $0<\epsilon \le \epsilon_0$". Is this assumption still possibile for the characterization of the supremum?
That is, in "$\sup A$ is the supremum of $A$ (bounded above and nonempty) if and only if for each $\epsilon>0$ there exists $a_\epsilon \in A$ such that $a_\epsilon>\sup A-\epsilon$" is the latter proposition equivalent to "for each $0<\epsilon \le \epsilon_0$ there exists $a_\epsilon \in A$ such that $a_\epsilon>\sup A-\epsilon$"?
I think it is true: the right implication is obvious (since it holds for each $\epsilon>0$ with $a_\epsilon \in A$, it holds for $0<\epsilon \le \epsilon_0$ with the same $a_\epsilon \in A$).
We must prove that the left implication holds only for $\epsilon>\epsilon_0$, since for $0<\epsilon \le \epsilon_0$ is true by hypothesis. Let $\epsilon>0$ be arbitrary. By hypothesis, there exists $a_{\epsilon_0} \in A$ such that $a_{\epsilon_0}>\sup A-\epsilon_0$; being $\epsilon>\epsilon_0$, we have $-\epsilon_0>-\epsilon$ and so $\sup A-\epsilon_0>\sup A-\epsilon$. That is, there exists $a_{\epsilon_0}\in A$ such that $a_{\epsilon_0}>\sup A-\epsilon$. Is this correct?
leslie townes
leslie townes
i didn't read the proof, but yes, the same kind of thing can be done for the supremum
and if that's what you're doing there then you got it
user 858770
user 858770
Requires mathjax.
Gwyn
Gwyn
@leslietownes Thanks for this confirmation leslie)
Jakobian
Jakobian
@algbr That's probably a part of it
another thing is that people might not want to chat about mathematics
user 858770
user 858770
Like Professor Wiles did for 7 years :P
Truly old school.
mick
mick
22:46
some topics are poison

when you talk about prime twins or collatz people will downvote you. anything less then a formal proof is a downvote. So a downvote.

such a shame
user 858770
user 858770
Stay in the mainstream or be prepared to work alone.
sunny
sunny
23:05
@robjohn I have a related question to one of your answers. Is the function $$f(x)=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)$$ continuous on $[0,\infty)$? How do we verify this? My suspicion is it isn't, since I can't get an upper bound on the terms and apply the Weierstrass M-test.
On the other hand, neither can I get a lower bound for $\lVert f-f_n\rVert=\sup_{x\in [0,\infty)} \left|\sum_{k=n+1}^\infty \left(\frac1k-\frac1{k+x}\right)\right|$ to show it isn't uniformly convergent...
Jakobian
Jakobian
the only troublesome point seems to be $0$
@sunny $f(x) = x\cdot\sum_{k=1}^\infty \frac{1}{k(k+x)}$ where $\frac{1}{k(k+x)}\leq \frac{1}{k^2}$ so that the series $\sum \frac{1}{k(k+x)}$converges uniformly
conclusion: f is continuous
algbr
algbr
@mick I try to disprove periodicity of the Jacobi-Perron algorithm. Is it a downvote? ;)
sunny
sunny
23:23
@Jakobian I kind of buy it, however, can you move the $x$ outside of the summation before knowing it is uniformly convergent? After all, $\sum c a_n=c\sum a_n$ holds only if $\sum a_n$ converges (we don't know yet if it does, or?).
Jakobian
Jakobian
@sunny everything here obviously converges
pointwise
and the reason can be traced down to $\frac{1}{k(k+x)}\leq \frac{1}{k^2}$
robjohn
robjohn
@sunny both converge or both diverge
@sunny this function is meromorphic, with poles at the negative integers.
sunny
sunny
@robjohn alright, so $\sum c a_n=c\sum a_n$ holds even if $\sum a_n $ diverges. You learn something new every day, thanks :)
robjohn
robjohn
Just look at the sequence of partial sums.
sunny
sunny
yeah
fun! :)

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