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12:48 AM
Man, don't you love it when you give up on something only to revisit it later and find something better than Plan B:tm:?
So... I gave up a while ago on trying to efficiently compute $2^x\bmod y$. Well, I converted it to a recurrence relation, and wolfram spat out an explicit formula for it. It can be used to compute the modulus quite cheaply provided that I can find a solution for n: wolframalpha.com/input/…
Specifically, I need to substitute n with a function of x and y such that $a(n-1) \geq y$ and $a(n) < y$ .
Other than that, this is the home stretch. I compute this 64 times and I can compute a reciprocal, or generalize it to avoid the cost of a multiply and allow it to compute the quotient of two numbers.
 
 
2 hours later…
2:35 AM
$$\operatorname{sign}\left(x\right)\left(1-\frac{x}{2^{\operatorname{floor}\left(\log_{2}\left(\left|x\right|\right)\right)+1}}\right)^{\frac{\ln\left(\left|\left(1-x\operatorname{floor}\left(\frac{1}{x}\right)\right)\right|\right)}{\ln\left(\left|1-\frac{x}{2^{\operatorname{floor}\left(\log_{2}\left(\left|x\right|\right)\right)+1}}\right|\right)}} = 1\bmod x$$
Obviously, this is unsatisfactory to evaluate $n$ (the exponent here) in closed form directly unless you can compute floored reciprocal or $1\bmod x$ already, however, I've determined that the upper-bound for $n$ given an input of $2^{64}\bmod 3$ as a worst case is $n=32$. The worst I could do is the obvious multiply and check algorithm that is immediately apparent with a better variation using an implementation of $2^x$.
 
2:57 AM
@geocalc33 I think I remember learning something in diff. topology that the answer to this question is "yes" because one can think about the set of equations as a map and then the inverse image of 0 under this map must be a submanifold, almost at least. If 0 is a regular value the inverse image is a submanifold.
If 0 is a critical point, then since a single point is itself a submanifold of the codomain you can make an arbitrarily small perturbation to a map that is transverse to 0, and then do the pre-image is a submanifold business.
Couple question from me to the math chat tonight:
 
3:18 AM
1. I am a little confused about how one should think about the Lie Group U(1) as having a 'special' element called the identity that is its own inverse and ALSO think about said group as a manifold that is just $S^1$. These two facts seem somewhat... not contradictory but incongruous to me because when I think of $S^1$, all the points are 'equivalent' in some sense, but the group structure does not seem to respect that.

2. I noticed that both $\pi_1(S^1) = \mathbb{Z}$ and also that the irreps of $U(1)$ are indexed by $\mathbb{Z}$ and I was wondering if this is a coincidence or not. My inst
 
3:42 AM
There’s nothing wrong with a distinguished point. The multiplication law is not homogeneous.
Yes, the reps correspond to degree, which in this case is $\pi_1$.
 
Is there a way of defining the group multiplication as some kind of additional structure that goes along with the manifold, so that the distinguishing is explicit on the "topology" side and not just the "algebra" side?
 
the group multiplication is additional structure
it's not intrinsic
 
Right, but I don't know how to think about the group multiplication in a topological context
 
It isn’t.
 
Good evening, all.
 
3:55 AM
granted, the Lie group structure on $S^1$ is unique
 
It is not uncommon to find, in reviews, the phrase "this paper fills a much-needed gap in the literature"
 
(but unique means, of course, unique up to isomorphism, so it's not like I'm implying there's an intrinsic basepoint)
 
4:22 AM
@TedShifrin this is how Spectral theorem has been proven in Axler’s by considering that orthogonal complement to invariant space is invariant under T and that the restriction to the complement is also self adjoint. But I tried proving it differently using Schur’s theorem but as I said already my method was wrong.
 
I think the proof of Schur gives the same proof. That’s what I was saying.
 
its a Schur thing
i would guess all boil down to invariant subspaces in one way or another?
 
Yuppers.
 
4:49 AM
@AkivaWeinberger I never heard of voronoi before. I'll understand your comment more after I know voronoi. Thanks :).
 
in america we call them dirichlet tesselations. or freedom diagrams.
 
Akiva is also from America. I think he told so once.
 
I am
@leslietownes I… heh?
 
:)
 
i was just riffing on how sometimes america and the former USSR have slightly different names for stuff. not trying to impugn anyone's americanness or lack thereof.
cauchy-schwarz-bunyakovsky? not on this side of the pacific, thank you very much
the eagle-eyed people on this chat.
 
4:56 AM
Never even heard of Буняковский
 
i did like spivak citing himself in cyrillic in his differential geometry book.
that's one time i found spivak entertaining.
 
I’m too lazy to switch keyboards.
I copied that in my multivariable book. :)
 
I also wish to learn Russian :)
 
Observation: Latin, Cyrillic, and Greek all have capital letters that look like H, but different corresponding lowercase letters: Hh, Нн, Ηη
(different pronunciations: h, n, Ancient ē / Modern i)
 
my daughter's learning a bit about hanukkah in her day care. she colored in a drawing of people playing with a dreidel. she yelled at me when i suggested that shin was not an M.
 
4:59 AM
Henceforth, you may call me Фёдор Шифрин.
 
sha isn't an M either.
 
Fyodor…?
Oh. I see it
Woulda thought Теодор
 
I am actually Russian (75%), so it’s not cheating.
 
@leslietownes Hah
 
Nope DogAteMy
 
5:02 AM
i was going to say, i don't think my name transliterates well into russian. over there i might be the equivalent of someone who moves here with a complicated name in their native language and then in the USA we call them Kevin.
i would have a name that i 'go by.'
 
ahhh, the Буняковский Bieber inequality
 
Good resource:
This page should evolve into a list of common Russian geographical and personal names, so that the standard policy on conventional naming could be evolved. It would eventually be incorporated into Transliteration of Russian into English or given a separate article if deemed appropriate. == Personal names == The origin of most modern given Russian names lies in Calendar of Saints that mentions various names from Biblical sources, translated/transcribed/transliterated and adopted from Greek, Latin and Hebrew, on a daily basis. These names were forced into use over Old Slavonic (mostly with ...
 
surely hellenisation would be much more appropriate?
 
No! We fought a war about this!
Diminutive: Fedya. Hm…
 
i was thinking more of the origins of cyrillic...
 
5:10 AM
And I was thinking Hanukkah
 
Chanukah?
 
חנוכה
Were I more brave I'd write Chanuka with no final h
and braver still, Chanuká
 
5:37 AM
the home there of 🇺🇸
 
6:00 AM
someone deleted a question as i was adding a hint.
 
Someone deleted an interesting (sorta) question on differential forms after I hinted in a comment.
 
look if you have a problem with the way i ask my questions, just say it to my face
it was a crummy comment anyway
no answer, no nothing, just who knows what about forms
phhbhbhbht
2
 
Go to your room.
 
leslie the closer.
 
the heat came on twice during the day today. this really is december.
 
6:14 AM
it was warm here today. small craft advisory until 9am, but otherwise very calm
 
my best work friend and i used to refer to ourselves as 'the closers' because we were routinely scheduled to take potential hires out to lunch or dinner as their last part of an interview day. the firm stopped doing this because nobody we had lunch or dinner with and gave an offer to ever ended up accepting.
i maintain that this had nothing to do with me.
 
my partner & myself were sort of good copy bad cop. i was very slow to hire, he was incredible at persuading both them & me. if i ever hinted that a potential was a possibility from my perspective he would close them.
 
you're a good judge of character
 
if i had to fit that dichotomy, i was the good cop
my pal was, more than once, the reason people didn't get offers, when i was like "really?"
maybe we were more like two aggressively weird cops
 
i do like the character 娱
2
 
6:29 AM
lol
the pleasure is all mine
🙏
 
 
1 hour later…
7:47 AM
Does anyone here know of a decent site that carries old-school calculators, those with an immediate-execution input input method (also called "algebraic entry system (AES))?
 
old calculators used to do trig immediately, so to get $\sin$ of $30^\circ$ you typed 30 followed by $\sin$ not some goofy $\sin($ followed by number...
 
 
1 hour later…
9:14 AM
@amWhy If you're still seeing the pumpkin pie, it is your cache that is stale. Refresh, and it will go away.
 
9:31 AM
The pie has been squared; welcome back @robjohn
 
@user726941 pie are round... cornbread are square.
 
TIL cornbread was created by native Americans.
 
@copper.hat amusing...
@user726941 I just learned that the American Meteorological Society says this about Indian Summer: "The use of this term is discouraged. It is considered a relic of the past and disrespectful of Native American people. The recommended term to describe this phenomenon is Second Summer."
 
Thanks for sharing.
 
@user726941: how are you doing these days?
 
9:48 AM
Fine thanks, how are you? @robjohn
 
Doing pretty well. Still observing the progress of omicron and waiting to see how mild/bad it will be.
 
Indeed, "these" days can be stressful.
How are your pets? @robjohn
 
@user726941 we lost a dog we've had for over 8 years at the end of October. I take these things pretty hard since I bond closely with our pets.
we still have one dog, three cats, and a bunny.
 
I'm sorry to hear that.
Bunnies are my favorite.
🐰
 
10:05 AM
@user726941 we get a shipment of veggies and stuff from a local farm today. The bunny always knows and comes to be cute and get carrot tops, etc.
 
coolio
I found this long discussion about the term indian summer
> JUST WHAT IS INDIAN SUMMER AND DID INDIANS REALLY HAVE ANYTHING TO DO WITH IT?
sorry for the all caps
 
10:21 AM
@user726941 Just read it. It seems to contradict a negative connotation about Native Americans.
Thanks.
 
Yup, I would use "late" summer.
Have you heard from Chris's sis @robjohn
Since the pandemic started.
Wow, coming up on 2 years ...
 
 
2 hours later…
12:15 PM
 
 
2 hours later…
2:13 PM
Ignore the above post.
 
If $x_1,...,x_n \in \Bbb{R}$ satisfy $x_1 + .... + x_n =0$, why is $\sum x_i x_j \ge 0$?
 
2:31 PM
Problem : Show there is a map $\Bbb RP^\infty\to\Bbb CP^\infty$ which induces the trivial map on $\tilde{H}_*(-,\Bbb Z)$ and nontrivial map on $\tilde{H}^*(-;\Bbb Z)$.
 
I can show it's true for $n=2$ and $n=3$.
Sorry, I am trying to prove that $\sum x_i x_j \le 0$
If $x_1 + x_2 =0$, then $x_1x_2 + x_2 x_1 = x_1 (-x_1) + (-x_1)x_1 = - 2x_1^2 \le 0$.
 
My sol: $H_n(\Bbb RP^\infty) = 0$ if $n$ : even and $H_n(\Bbb CP^\infty) =0$ if $n$ : odd. Hence, whatever $f$ is, the induced map $f_*$ in homology is trivial. Now, the cohomology ring $H^*(\Bbb RP^\infty;\Bbb Z) =\Bbb Z[x^2]/(2x^2)$. Hence, $H^{2n}(\Bbb RP^\infty;\Bbb Z) =\{0,x^{2n}\}$. Since $H^{2n}(\Bbb RP^\infty;\Bbb Z)\simeq \langle \Bbb RP^\infty,K(\Bbb Z,2n)\rangle$, there is a map $f:\Bbb RP^\infty\to K(\Bbb Z,2n)$ such that $f^*(\alpha) = x^{2n}\neq 0$
where $\alpha$ is a fundamental class
In particular, there is a map $f:\Bbb RP^\infty\to K(\Bbb Z,2)=\Bbb CP^\infty$ such that $f^*(\alpha) = x^{2}\neq 0$ so such $f$ is the desired map.
What do you people think of this argument?
 
If $x_1 + x_2 + x_3 = 0$, then

\begin{align}
x_1 x_2 + x_2 x_1 + x_1 x_3 + x_3 x_1 + x_2 x_3 + x_3 x_2 &= 2 (x_1 x_2 + x_1 x_3 + x_2 x_3) \\
&= 2 (x_1 x_2 + x_1(-x_1 - x_2) + x_2(-x_1-x_2) \\
&= -2(x_1^2 + x_1 x_2 + x_2^2) \\
&= -2 ((x_1 + \frac{x_2}{2})^2 + \frac{3x_2^2}{4}) \\
&\le 0
\end{align}
But I don't see how to generalize this...
I wonder if there is a way to solve this using inner products and matrices.
If $v = (x_1,...,x_n)$ and $w = (1,...,1)$, then the first condition is equivalent to $\langle v,w \rangle = 0$, where $\langle , \rangle$ is the standard inner product on $\Bbb{R}^n$.
So, $v$ and $w$ are orthogonal.
 
2:47 PM
@user193319 How about this? $0=(x_1+\cdots+x_n)^2 = \sum x_i^2+2\sum x_ix_j$.
 
D'oh!
That looks much better!
@love_sodam Thanks!
 
3:27 PM
Can I get a hint on how to solve this? Thank you
$e^x=11\ln{x}$
 
If you search "smacks" in the chat, it's basically 13 pages of Ted smacking people.
4
 
3:40 PM
I could have solved it if that 11 was not there using lambert W, but the 11 is making anything I try hideous
 
@Typo How do you solve it without the $11$?
 
@robjohn e^x=lnx leads to (infinite power tower of e) = x. from there to e^x=x, then xe^(-x)=1, so x=-W(-1)
 
@Typo and do you know what $\operatorname{W}(-1)$ is?
 
@robjohn I don't know any of its properties, just that it's the inverse of xe^x
 
3:57 PM
@Typo You know that the inverse of $x^2$ is $\sqrt{x}$, and are asked to find the solution of $x^2=-1$.
can you find the minimum of $xe^x$?
@user726941 I have from time to time, here.
Last time was at the end of July
 
@robjohn -1/e?
@robjohn Are you talking about the fact that there are complex solutions? the equation with the 11 has real solutions. there must be some way to express it?
 
4:18 PM
@Typo the inverse to $xe^x$ requires a special function. I bet $\log(x)e^{-x}$ also requires a special function.
whether it has been given a name yet is uncertain
 
ah that is unfortunate :(
I got the question from someone else. Now I think they were just trolling me
 
It can be solved numerically as far as one wishes.
 
Can a square be divided into 5 squares(may not be congruent) ?
 
Alright, thank you for that
@robjohn May I ask why you brought this up though? If you were leading up to some lesson, I still want to learn :)
 
1.5076614562177928862
@Typo Since the minimum of $xe^x$ is $-\frac1e$, $\operatorname{W}(-1)$ is not real.
 
4:25 PM
Ooh, I see
Well thank you again, have a nice day
 
you, too
 
4:42 PM
If $\gamma : [a,b] \to \Bbb{C}$ is a smooth path, is it true that $d \gamma = d(Re~\gamma) + id(Im ~ \gamma)$? Does such an equation even make sense? How does one prove this?
 
what does $d$ mean to you
 
@Thorgott Differential I guess. I am trying to make sense of a theorem I'm working through
The theorem is the following: If $\gamma$ is piecewise smooth and $f : [a,b] \to \Bbb{C}$ is continuous, then $$\int_{a}{b} f d \gamma = \int_{a}^{b} f(t) \gamma ' (t) dt.$$
In the proof, they say it suffices to assume $\gamma ([a,b]) \subseteq \Bbb{R}$ by looking at the real and imaginary parts of $\gamma$.
And that $\gamma$ is smooth simpliciter.
If $\gamma$ is smooth, then $\gamma '$ exists and is continuous. But $\gamma '$ exists if and only if $(Re ~ \gamma )'$ and $(Im ~ \gamma)'$ exist and $\gamma ' (t) = (Re ~ \gamma)'(t) + i (Im ~ \gamma)'(t)$. Moreover, $\gamma'$ is continuous if and only if the real and imaginary parts are continuous, hence $Re ~ \gamma$ and $Im ~ \gamma$ are smooth paths.
So, $$\int_{a}^{b} f d(Re ~ \gamma) = \int_{a}^{b} f(t) (Re ~ \gamma)'(t) dt.$$ Same equation holds for the imaginary part. I want to multiply the equation involving the imaginary part by $i$ and then add the two equations together. But to show that this reduction is true, it seems that I would need something like $d \gamma = d(Re~\gamma) + id(Im ~ \gamma)$ to be true.
 
5:02 PM
this is true, but explaining why depends on how you define that symbol
 
@copper.hat Thanks for that calculator, that's cute! Yes that input method is called immediate-execution. I am rather looking to buy actual physical old-school calculators though. Has there been any online discussion/protest re: why such calculators don't seem to be manufactured AT ALL anymore?
 
in reality this is an equality of complex-valued 1-forms, but I'm not sure if that's the formalism you want
 
Hmm...I'm working through Conway's book on Complex Analysis, but I didn't see any rigorous treatment of the symbol.
Yeah, that formalism is fine, I think.
 
ok, then it's just a placeholder symbol as part of the integral expression, in which case you need to spell out how you define those integrals
 
I've seen a little bit of it before, but I'm kind of rusty on it.
I think this theorem contains the definition, but I'm not entirely sure. Let me keep looking through the book.
 
5:06 PM
Hi friends. I am in a dilemma that which one is preferable. Morita geometry of diff forms or bott-tu Differential Forms in Algebraic Topology GTM81? Both cover almost same material according to their TOC. Also Milnor's Morse theory or Matsomotu An intro to Morse theory? (FOR students of Limited budget)
 
user: that looks like a definition to me. he's definitely not defining "dgamma" or "dgamma(t)" outside of that expression, anyway.
presumably later he shows that reparametrization doesn't affect the result when you're integrating f(gamma(t)), in which case some people might drop the a^b and any reference to a parametrization (e.g. no "t").
 
So, it is possible he isn't being very rigorous with the differential notation?
 
i wouldn't regard it as a use of 'differential notation'
 
Hmm...I see...
 
one way of thinking of it is that in a few pages, he's going to define something like int_gamma f where f is a function defined on gamma. it doesn't really depend on inputs other than gamma and f
as it happens you can write this in terms of normal riemann integrals. the riemann integral has a "d" in it which in a calculus class often is not given independent meaning. you could write int_a^b f, but instead, people write int_a^b f(x) dx
 
5:16 PM
I've not read all of these books, but the notion that they're comparable in content strikes me as weird
 
you don't have to go down a rabbit hole of defining what dx or dgamma or d is, to make sense of it in terms of the usual notation for riemann integrals
 
How would I combine $\int_{a}^{b} f d(Re \gamma) + i \int_{a}^{b} f d(Im ~ \gamma)$ to get $\int_{a}^{b} f d(Re ~ \gamma + i Im ~ \gamma ) = \int_{a}^{b} f d \gamma$?
 
i've read conway. i don't know what goes on in random books but that's what's going on in conway
 
after the basics of Morse theory, Milnor spends the rest of the book discussing geodesics, studying them via Morse theory and applying this to prove Bott periodicity and the like. whereas Matsumoto, after the basics, goes into an in-depth discussion on handlebodies and applications to low-dimensional manifolds
Morita's book seems much more differential-geometric in nature, whereas Bott-Tu is much more algebro-topological. Morita has chapters on Hodge theory, Connections and Curvature and his treatment of characteristic classes seems to be entirely Chern-Weil theory. The only mention of any singular theory seems to be in the chapter on the de Rham theorem (something which Bott-Tu don't discuss in that amount of detail).
On the other hand, Bott-Tu contains extensive discussion of singular homology/cohomology theories, sheaf cohomology spectral sequences, homotopy theory and their approach to charac
this is just from a brief-ish glance and not claiming any experience or exhaustiveness, but they strike me as different books
 
user: if you write gamma(t) = c(t) + i d(t), then integral f(t) gamma'(t) = integral f(t) c'(t) dt + i integral f(t) d'(t) dt. if you know the real result and he's defined int f dgamma in terms of partitions, given e, you can choose partitions of your interval where the integrals of f c' and f d' are within e/2 or whatever of sums of f(t_n) (c(t_n) - c(t_{n-1}) and f(t_n) (d(t_n) - d(t_{n-1}) respectively.
if you pass to a common refinement you ought to get within e of a sum of f(t_n) (c(t_n) - c(t_{n-1} + i (d(t_n) - d(t_{n-1})) = f(t_n) (gamma(t_n) - gamma(t_{n-1}), which is then a sum over a partition in the sense of the partition-based definition of int f dgamma.
 
5:28 PM
also, I agree with what leslie is saying about the integration thing, just listen to him
 
and then some crap about mesh sizes or common refinements. he's burying an annoying fact about riemann integrals, elementary reasoning with them tends to require a lot of sum notation. good for him.
 
I just scored a 64 on my test
 
@Thorgott tnx a lot. All about Minor book say that it is a Bible. This tempt me to pay for it but I believe in Japanese teaching way.
 
rather than saying he isn't being rigorous, i might say, he's not trying to introduce more than he absolutely needs to to make sense of the notation. if all you care about is line integrals of complex-valued functions on curves, he's giving enough rigor to give that meaning. that the notation is suggestive and appears to point in other directions can safely be ignored if he doesn't intend to explore them.
i forget if conway ever integrates a function over a region in the plane that isn't a curve in that book
 
5:58 PM
@user193319 I know that book. It's conway functions of one complex variable
 
Yup!
 
conway gang in the house
who needs green's theorem? not conway
hah, i just followed the starred suggestion and searched 'smacks' in the chat
a good number of those are me :o
 
This is a famous problem based on derivatives: The statement to be proven is $\lim_{x\to \infty} (f(x)+f'(x))=L\implies \lim_{x\to \infty}f(x)=L \land \lim_{x\to \infty} f'(x)=0$
I know that it can be proven using L'Hospital's rule.
But I wanted to discuss/clear one confusion I have in Hardy's proof of the above theorem.
Having shown that, $\lim_{x\to \infty} g(x)=L$ and given that $\lim g'(x)$
exists then $g'(x)\to 0$, Hardy's proof goes somewhat like this: WLOG, let L=0. If $f'$ is positive for all large $x$ then $f$ is increasing for all large x. Hence, $f$ either converges to a limit L(as x tends to $\infty$) or diverges to $+\infty$ (as x tends to $\infty$). If f diverges to $\infty$ then $f'$ must diverge to $-\infty$, which is not possible as $f'$ is positive for all large $x$.
If $f$ converges to a limit $L$ then $f'$ must converge to $-L$ but this is not possible until L=0.
The case when $f'$ is negative for all large $x$ can be handled similarly.
Now comes the case when $f'$ oscillates between positive and negative values for all large $x$.
 
6:21 PM
whats g
 
Thor: I have taken $g$ to state a theorem that: if limit of a function exists as $x\to \infty$ and it is given that limit of derivative of the function exists as $x\to \infty$ then the limit of the derivative must be zero.
For the last case: f' must attain $0$ also for all infinitely many $x$, then these are the maxima and minima of f(x). "If x has a large value corresponding to a maximum or a minimum of f, then f(x)+f'(x) is small and f'(x)=0 so that f(x) is small. A fortiori the other values of f(x) are small when x is large."
That's what I don't understand. If f'(x)=0 then why is x a point of maxima or minima? Clearly for f(x)=x^3 on (-2,2), f'(0)=0 but 0 is neither a point of maxima nor minima.
 
@Koro Have you ever proved L'Hôpital's rule in the $\infty$ cases?
 
yes, I have.
I even have a post on mse on the same.
 
OK, most people have not. I don't see how this problem has anything to do with L'Hôpital, though.
I think I should smack @user193319 for searching my smacks.
 
$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{e^x f(x)}{e^x}=\lim_{x\to \infty}(f(x)+f'(x))$.
 
6:33 PM
the l'hopital bit is a clever argument given on the first page of the link.
 
How do we know $e^x f(x)\to\infty$?
 
yeah, it's far more fun to just go prove various cases in l'hopitals theorem.
 
Why do you want that professor Ted?
 
l'hopitals theorem has hypotheses.
 
L'Hospital requires that only denominator goes to infty.
 
6:34 PM
WRONG.
 
they do vary depending upon who is proving it. i forget what rudin does.
 
$0/0$ form or (anything/infty) form.
 
I've never seen anything/infinity.
Does anything have to have any limit at all?
 
0
Q: L'Hospital's rule proof review

KoroLet $f,g$ be real valued functions defined on $(a,b)$ and $f,g $ be differentiable on $(a,b)$ such that $g'\ne 0$ on $(a,b)$. If $\frac{f'(t)}{g'(t)}\to A$, as $t\to a$. If $1) f(t)\to0, g(t)\to 0$ as $t\to a$. Or 2) $g(t)\to \infty$ as $t\to a$ Then $\lim_{t\to a}\frac {f(t)}{g(t)}=A$ Proof: L...

Professor Ted: see this proof. I posted the proof in this post.
Rudin does that, Leslie.
 
I see Rudin states it for $x\to a$, but does he prove it for $a=\infty$?
Interesting, though, that I never noticed that generality in Rudin.
Your post does not address $\infty$.
 
6:41 PM
Of course Rudin proves for a= infty.
 
Where?
 
$-\infty\le a\lt b\le +\infty$ (theorem 5.13)
 
He claims it holds. He does not prove it.
 
Hiya Ted
 
heya @Sha
The proof I've taught for $a=\infty$ is rather quite a different flavor.
 
6:44 PM
I think the same proof works for a= infty also. Rudin is considering extended reals.
16 mins ago, by Koro
That's what I don't understand. If f'(x)=0 then why is x a point of maxima or minima? Clearly for f(x)=x^3 on (-2,2), f'(0)=0 but 0 is neither a point of maxima nor minima.
Anything on this though?
 
I think you'd better check that carefully before making the glib statement.
Your English is faulty: It is a maximum point or a minimum point. Maxima and minima are plural. So what is the precise question you're asking? Of course it's correct to say that $f'(c)=0$ does not imply $c$ is a local max/min.
I cannot decipher all that stuff.
 
Given an absolute CW complex $X$ with skeleta $X_n$, I want to prove that I have the following homeomorphism: If $X_n$ arrises from $X_{n-1}$ by attaching $n$-cells with index set $J_n$, and with characteristic map $\chi_i$ for each $i\in J_n$, then I want to show that the map induced by the characteristic maps
$$
(J_n\times D^n)/(J_n\times\partial D^n)\to X_n/X_{n-1}
$$
is a homeomorphism. I've shown that this map is bijective, and continuity is easy too. However, when $J_n$ is infinite, I don't know how to argue that closed sets are mapped to closed sets (using that the induced characteri
 
Ugh. I don't think about stuff like that. Aren't CW complexes finitely generated in each dimension?
 
You mean that $J_n$ can be taken finite?
 
paging @Thor
 
6:51 PM
what does the adjective absolute mean here?
 
That it's not a relative CW complex
Relative would mean that we have a pair $(X,A)$
And then $A=X_{-1}\subset X_0\subset X_1\subset\dots$
So in our case $A=\emptyset$, and that means in particular that $X$ is Hausdorff
 
Oh @Koro. If the derivative oscillates between positive and negative, then the zeroes of the derivative MUST be local extrema.
 
So we don't have an underlying space that we attach cells to, but we just start with cells
 
I'm confused, are you saying you have an argument when $J_n$ is finite?
because the cardinality of $J_n$ actually doesn't make a difference, the domain is just a disjoint union of copies of $D^n/\partial D^n$ and the argument for each one is the same
 
Well, each characteristic map is closed, so I would then that we would have a finite union of closed sets then
 
6:56 PM
ah, well, ok
argue they're open instead
 
Hi.Anyone help with this problem?
0
Q: Continuous map on metric spaces.

unit 1991Theorem. Let $(X, d_X)$ and $(Y, d_Y )$ be metric spaces. A function $f : X → Y$ is continuous at $a ∈ X$ if for every $\epsilon$ > 0 there exists $δ > 0$ such that $d_X(x, a) < δ$ implies that $d_Y (f(x), f(a)) < \epsilon$. Book does not prove this theorem but says that proof follows from defini...

 
there's two cases: you either look at an open subset in the interior of $D^n$ or an open neighborhood of $\partial D^n$ inside $D^n$
 
@unit1991 Ask a very specific question. What particular statement can you not prove?
 
@Thorgott They're not I think
Oh
maybe they are
I never thought about it because they're not bijective anyways
 
I don't mean the characteristic maps, I mean the induced maps
 
6:59 PM
Yea so open subset of interior of $D^n$ is fine
My issue is when the open subset contains $\partial D^n$
 
@TedShifrin Ah, I meant to say a=$-\infty$ there.
 
Hm, ah
maybe it makes sense now
Because
as soon as the (saturated) open intersects $\partial D^n$, then it has to intersect $\partial D^n$ for each $i\in J_n$
so that boils down to
 
@TedShifrin About that: Let's say on (a,b), we have f'(c)=0 for some c in (a,b) then if f' retains same sign on some left nbd of c and opposite sign on some right nbd of c then yes c is a point of an extremum. But in my question what confirms the existence of such neighborhood?
 
The words you typed said that $f'$ changed sign from positive to negative infinitely often.
 
ok I see it, but I can't phrase it succintly
 
7:03 PM
Start with that, not with your statement.
 
thanks @Thorgott
@Thorgott wait, I actually don't see it after all
how to argue that if $U$ is an open neighbourhood of $\partial D^n$, then $\chi_i(U)$ (where $\chi_i$ is induced) is open?
 
you don't, what I said earlier is wrong, sorry about that
you need to argue directly that $\coprod_{J_n}D^n/\coprod_{J_n}\partial D^n\rightarrow X_n/X_{n-1}$ is open
 
right, any idea how? X'd
 
7:19 PM
True or false: For any function $f$, there exists a Turing machine $T$ such that the statement "$T$ computes $f$" can't be proven or disproven from PA
 
@ryang i don't know, i suspect most people don't use calculators after they leave school so no one really notices the changes. i am sorry i did not keep my old calculators, i had a sinclair cambridge a long time ago. (i still have a slide rule which predates this stuff.) look at vintagecalculators.com for some history.
 
Even if $f$ is uncomputable
 
in the non-trivial case, you have an open set in the domain, which lifts to an open subset of $\coprod_{J_n}D^n$ that is a neighborhood of the boundary in each summand. take the image in $X_n/X_{n-1}$. what is its preimage under the canonical quotient map $X_{n-1}\sqcup\coprod_{J_n}D^n\rightarrow X_n\rightarrow X_n/X_{n-1}$?
 
Inequivalent restatement: for any function $f$, there exists a Turing machine $T$ such that no finite conjunction of statements of the form "$T$ on input $i$ halts and outputs $f(i)$" is disprovable from PA
 
Well it will send the open $U$ to opens in $X_n\setminus X_{n-1}$ along with $X_{n-1}$
So if I write $U_i=U\cap {i}\times D^n$, then $U$ is mapped to $\bigcup_i \chi_i(U_i)\cup X_{n-1}$
Those $\chi_i(U_j)$ are open, but $X_{n-1}$ is closed
 
7:25 PM
ok, but what's the preimage I ask about?
cause if that preimage is open, we're good
(by definition)
 
Ah, eh, right, then it's $X_{n-1}\sqcup\coprod_{J_n}U_i$, which is open, sure
But now you're mixing $X_{n-1}$ with $D^n$ - why?
The map $X_{n-1}\sqcup\coprod_{J_n}D^n\to X_n$ isn't open, right
But it seems like we need that for the argument to hold?
 
well, how do you define a CW complex?
 
Oh right
That is the underlying space of $X_n$ (without the relation)
Nice, thanks! I'll go through the argument once more, but I think I'm convinced!
 
Thanks to Thor for his service :)
He deserves an unsmack.
 
x''D
 
7:38 PM
something amusing this mornin.
 
Are you amused?
 
i got a text saying that a fedex package will arrive next week
no idea from who or what
 
christmas is coming early
 
Oh, it could be a gift. Or it could be a mistake. Both have happened to me.
 
eventually ended up calling fedex and wading through the support tree
 
7:39 PM
If you look at the tracking number, you can figure out from whence it comes.
 
the guy had a hard time pronouncing the shipper, steck hexcenge
 
@TedShifrin were life so simple
 
LOL ... oh, you're getting your gift from SE?
 
i'm getting a second round of swag
 
7:40 PM
What did you do to earn the swag?
 
no idea why
i pointed that out to them (stack exchange, i mean)
who knows, maybe it will be ticking when it arrives
i know how to deal with that, of course
 
No one here loves me, so I'm swag-free.
 
it's your jumpsuit
 
@TedShifrin 100k was the first round for me. why the second is anyone's guess.
the t shirt ripped in a few days. the mug is nice but not mse specific
it may be that my kids ripped the shirt...
 
Well, maybe that's incentive for me to stay around to 100k and then disappear.
 
7:43 PM
:-)
i'm getting an orange jumpsuit, and will then be collected and delivered to a chinese remainder prison.
 
most elementary statements about CW complexes are not difficult to prove, but there's a certain degree of technicality involved that one simply has to get used to via practice
 
where gcds are never one
and even hermitian matrices are not diagonalisable
 
Well, diagonalizability is overrated, anyhow.
 
Oh btw, @Thorgott I think we need to modify your argument to work with closed sets, but otherwise it can be left unchanged
Also, I think the reason it's a bit annoying at first to work with CW complexes, is because we have identifications going on when choosing a push-out that is homeomorphic to the 'original one'
 
open/closed doesn't make a difference, either works
 
7:51 PM
But closed seems easier to me, since each $\chi_i$ is a closed map. For open $U_i=U\cap\{i\}\times D^n$, we don't know that $\chi_i(U_i)$ is open
We would only know that $U_i'=U\cap \{i\}\times \operatorname{interior}D^n$ is mapped to an open, since $\chi_i$ is a homeo on $\operatorname{interior}D^n$
But I'm happy enough with understanding why closed works
 
what does it mean for a vector field to be covariantly constant?
 
Covariant derivative is zero.
We would need more specific wording to know in what sense to interpret that.
 
it's for the GR class, so I don't think I should elaborate
 
The question is whether it's on an open set or on a particular curve.
 
7:58 PM
curves are dangerous
 
Oh, so it's just parallel translation of the vector along the curve. The resulting vector field on the curve has covariant derivative zero, and is called "covariant constant."
 
Hello math people
 
Okay, this makes sense. Thanks
 
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