Mathematics - 2021-12-02 (page 1 of 2)

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.

AMDG

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 $2^x$.

Kevin Driscoll

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:

Kevin Driscoll

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…

Ted Shifrin

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 $\pi_1$.

Kevin Driscoll

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?

Thorgott

the group multiplication is additional structure

it's not intrinsic

Kevin Driscoll

Right, but I don't know how to think about the group multiplication in a topological context

Ted Shifrin

It isn’t.

UnderMathUate

Good evening, all.

Thorgott

3:55 AM

granted, the Lie group structure on $S^1$ is unique

Akiva Weinberger

It is not uncommon to find, in reviews, the phrase "this paper fills a much-needed gap in the literature"

Thorgott

(but unique means, of course, unique up to isomorphism, so it's not like I'm implying there's an intrinsic basepoint)

Koro

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.

Ted Shifrin

I think the proof of Schur gives the same proof. That’s what I was saying.

copper.hat

its a Schur thing

i would guess all boil down to invariant subspaces in one way or another?

Ted Shifrin

Yuppers.

Koro

4:49 AM

@AkivaWeinberger I never heard of voronoi before. I'll understand your comment more after I know voronoi. Thanks :).

leslie townes

in america we call them dirichlet tesselations. or freedom diagrams.

Koro

Akiva is also from America. I think he told so once.

Akiva Weinberger

I am

@leslietownes I… heh?

Koro

:)

leslie townes

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.

Akiva Weinberger

4:56 AM

Never even heard of Буняковский

leslie townes

i did like spivak citing himself in cyrillic in his differential geometry book.

that's one time i found spivak entertaining.

Ted Shifrin

I’m too lazy to switch keyboards.

I copied that in my multivariable book. :)

Koro

I also wish to learn Russian :)

Akiva Weinberger

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)

leslie townes

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.

Ted Shifrin

4:59 AM

Henceforth, you may call me Фёдор Шифрин.

leslie townes

sha isn't an M either.

Akiva Weinberger

Fyodor…?

Oh. I see it

Woulda thought Теодор

Ted Shifrin

I am actually Russian (75%), so it’s not cheating.

Akiva Weinberger

@leslietownes Hah

Ted Shifrin

Nope DogAteMy

leslie townes

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.'

copper.hat

ahhh, the Буняковский Bieber inequality

Akiva Weinberger

Good resource:

Wikipedia:Romanization of Russian/Harmonization

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 ...

copper.hat

surely hellenisation would be much more appropriate?

Akiva Weinberger

No! We fought a war about this!

Diminutive: Fedya. Hm…

copper.hat

i was thinking more of the origins of cyrillic...

Akiva Weinberger

5:10 AM

And I was thinking Hanukkah

Ted Shifrin

Chanukah?

Akiva Weinberger

חנוכה

Were I more brave I'd write Chanuka with no final h

and braver still, Chanuká

user 726941

5:37 AM

the home there of 🇺🇸

copper.hat

6:00 AM

someone deleted a question as i was adding a hint.

Ted Shifrin

Someone deleted an interesting (sorta) question on differential forms after I hinted in a comment.

leslie townes

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

Ted Shifrin

Go to your room.

copper.hat

leslie the closer.

leslie townes

the heat came on twice during the day today. this really is december.

copper.hat

6:14 AM

it was warm here today. small craft advisory until 9am, but otherwise very calm

leslie townes

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.

copper.hat

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.

user 726941

you're a good judge of character

leslie townes

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

copper.hat

i do like the character 娱

2

user 726941

6:29 AM

lol

the pleasure is all mine

🙏

ryang

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))?

copper.hat

only one i could find scientific601.altervista.org/sr50/sr50.html

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...

robjohn

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.

user 726941

9:31 AM

The pie has been squared; welcome back @robjohn

robjohn

@user726941 pie are round... cornbread are square.

user 726941

TIL cornbread was created by native Americans.

robjohn

@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."

user 726941

Thanks for sharing.

robjohn

@user726941: how are you doing these days?

user 726941

9:48 AM

Fine thanks, how are you? @robjohn

robjohn

Doing pretty well. Still observing the progress of omicron and waiting to see how mild/bad it will be.

user 726941

Indeed, "these" days can be stressful.

How are your pets? @robjohn

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.

user 726941

I'm sorry to hear that.

Bunnies are my favorite.

🐰

robjohn

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.

user 726941

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

robjohn

10:21 AM

@user726941 Just read it. It seems to contradict a negative connotation about Native Americans.

Thanks.

user 726941

Yup, I would use "late" summer.

Have you heard from Chris's sis @robjohn

Since the pandemic started.

Wow, coming up on 2 years ...

love_sodam

12:15 PM

For interested parties math.stackexchange.com/q/4321711/668308

love_sodam

2:13 PM

Ignore the above post.

user193319

If $x_1,...,x_n \in \Bbb{R}$ satisfy $x_1 + .... + x_n =0$, why is $\sum x_i x_j \ge 0$?

love_sodam

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 $\tilde{H}^*(-;\Bbb Z)$.

user193319

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 $x_1x_2 + x_2 x_1 = x_1 (-x_1) + (-x_1)x_1 = - 2x_1^2 \le 0$.

love_sodam

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?

user193319

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.

love_sodam

2:47 PM

@user193319 How about this? $0=(x_1+\cdots+x_n)^2 = \sum x_i^2+2\sum x_ix_j$.

user193319

D'oh!

That looks much better!

@love_sodam Thanks!

Typo

3:27 PM

Can I get a hint on how to solve this? Thank you
$e^x=11\ln{x}$

user193319

If you search "smacks" in the chat, it's basically 13 pages of Ted smacking people.

4

Typo

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

robjohn

@Typo How do you solve it without the $11$?

Typo

@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)

robjohn

@Typo and do you know what $\operatorname{W}(-1)$ is?

Typo

@robjohn I don't know any of its properties, just that it's the inverse of xe^x

robjohn

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

Typo

@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?

robjohn

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

Typo

ah that is unfortunate :(

I got the question from someone else. Now I think they were just trolling me

robjohn

It can be solved numerically as far as one wishes.

Wolgwang

Can a square be divided into 5 squares(may not be congruent) ?

Typo

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 :)

robjohn

1.5076614562177928862

@Typo Since the minimum of $xe^x$ is $-\frac1e$, $\operatorname{W}(-1)$ is not real.

Typo

4:25 PM

Ooh, I see

Well thank you again, have a nice day

robjohn

you, too

user193319

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?

Thorgott

what does $d$ mean to you

user193319

@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.

Thorgott

5:02 PM

this is true, but explaining why depends on how you define that symbol

ryang

@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?

Thorgott

in reality this is an equality of complex-valued 1-forms, but I'm not sure if that's the formalism you want

user193319

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.

Thorgott

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

user193319

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.

C.F.G

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)

leslie townes

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").

user193319

So, it is possible he isn't being very rigorous with the differential notation?

leslie townes

i wouldn't regard it as a use of 'differential notation'

user193319

Hmm...I see...

leslie townes

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

Thorgott

5:16 PM

I've not read all of these books, but the notion that they're comparable in content strikes me as weird

leslie townes

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

user193319

How would I combine $\int_{a}^{b} f d(Re \gamma) + i \int_{a}^{b} f d(Im ~ \gamma)$ to get $\int_{a}^{b} f d(Re ~ \gamma + i Im ~ \gamma ) = \int_{a}^{b} f d \gamma$?

leslie townes

i've read conway. i don't know what goes on in random books but that's what's going on in conway

Thorgott

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

leslie townes

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.

Thorgott

5:28 PM

also, I agree with what leslie is saying about the integration thing, just listen to him

leslie townes

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.

geocalc33

I just scored a 64 on my test

C.F.G

@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.

leslie townes

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

love_sodam

5:58 PM

@user193319 I know that book. It's conway functions of one complex variable

user193319

Yup!

leslie townes

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

Koro

jstor.org/stable/2689813

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 $x$.

Thorgott

6:21 PM

whats g

Koro

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.

Ted Shifrin

@Koro Have you ever proved L'Hôpital's rule in the $\infty$ cases?

Koro

yes, I have.

I even have a post on mse on the same.

Ted Shifrin

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.

Koro

$\lim_{x\to \infty}f(x)=\lim_{x\to \infty}\frac{e^x f(x)}{e^x}=\lim_{x\to \infty}(f(x)+f'(x))$.

leslie townes

6:33 PM

the l'hopital bit is a clever argument given on the first page of the link.

Ted Shifrin

How do we know $e^x f(x)\to\infty$?

leslie townes

yeah, it's far more fun to just go prove various cases in l'hopitals theorem.

Koro

Why do you want that professor Ted?

leslie townes

l'hopitals theorem has hypotheses.

Koro

L'Hospital requires that only denominator goes to infty.

Ted Shifrin

6:34 PM

WRONG.

leslie townes

they do vary depending upon who is proving it. i forget what rudin does.

Koro

$0/0$ form or (anything/infty) form.

Ted Shifrin

I've never seen anything/infinity.

Does anything have to have any limit at all?

Koro

0

Q: L'Hospital's rule proof review

Let $f,g$ 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.

Ted Shifrin

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 $\infty$.

Koro

6:41 PM

Of course Rudin proves for a= infty.

Ted Shifrin

Where?

Koro

$-\infty\le a\lt b\le +\infty$ (theorem 5.13)

Ted Shifrin

He claims it holds. He does not prove it.

Sha Vuklia

Hiya Ted

Ted Shifrin

heya @Sha

The proof I've taught for $a=\infty$ is rather quite a different flavor.

Koro

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?

Ted Shifrin

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.

Sha Vuklia

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…

Ted Shifrin

Ugh. I don't think about stuff like that. Aren't CW complexes finitely generated in each dimension?

Sha Vuklia

You mean that $J_n$ can be taken finite?

Ted Shifrin

paging @Thor

Thorgott

6:51 PM

what does the adjective absolute mean here?

Sha Vuklia

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

Ted Shifrin

Oh @Koro. If the derivative oscillates between positive and negative, then the zeroes of the derivative MUST be local extrema.

Sha Vuklia

So we don't have an underlying space that we attach cells to, but we just start with cells

Thorgott

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

Sha Vuklia

Well, each characteristic map is closed, so I would then that we would have a finite union of closed sets then

Thorgott

6:56 PM

ah, well, ok

argue they're open instead

unit 1991

Hi.Anyone help with this problem?

0

Q: Continuous map on metric spaces.

Theorem. 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...

real-analysis general-topology

Thorgott

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 $D^n$

Ted Shifrin

@unit1991 Ask a very specific question. What particular statement can you not prove?

Sha Vuklia

@Thorgott They're not I think

Oh

maybe they are

I never thought about it because they're not bijective anyways

Thorgott

I don't mean the characteristic maps, I mean the induced maps

Sha Vuklia

6:59 PM

Yea so open subset of interior of $D^n$ is fine

My issue is when the open subset contains $\partial D^n$

Koro

@TedShifrin Ah, I meant to say a=$-\infty$ there.

Sha Vuklia

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

Koro

@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?

Ted Shifrin

The words you typed said that $f'$ changed sign from positive to negative infinitely often.

Sha Vuklia

ok I see it, but I can't phrase it succintly

Ted Shifrin

7:03 PM

Start with that, not with your statement.

Sha Vuklia

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?

Thorgott

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

Sha Vuklia

right, any idea how? X'd

Akiva Weinberger

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

copper.hat

@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.

Akiva Weinberger

Even if $f$ is uncomputable

Thorgott

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 $X_{n-1}\sqcup\coprod_{J_n}D^n\rightarrow X_n\rightarrow X_n/X_{n-1}$?

Akiva Weinberger

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

Sha Vuklia

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

Thorgott

7:25 PM

ok, but what's the preimage I ask about?

cause if that preimage is open, we're good

(by definition)

Sha Vuklia

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?

Thorgott

well, how do you define a CW complex?

Sha Vuklia

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!

Ted Shifrin

Thanks to Thor for his service :)

He deserves an unsmack.

Sha Vuklia

x''D

copper.hat

7:38 PM

something amusing this mornin.

Ted Shifrin

Are you amused?

copper.hat

i got a text saying that a fedex package will arrive next week

no idea from who or what

leslie townes

christmas is coming early

Ted Shifrin

Oh, it could be a gift. Or it could be a mistake. Both have happened to me.

copper.hat

eventually ended up calling fedex and wading through the support tree

Ted Shifrin

7:39 PM

If you look at the tracking number, you can figure out from whence it comes.

copper.hat

the guy had a hard time pronouncing the shipper, steck hexcenge

leslie townes

haha

copper.hat

@TedShifrin were life so simple

Ted Shifrin

LOL ... oh, you're getting your gift from SE?

copper.hat

i'm getting a second round of swag

Ted Shifrin

7:40 PM

What did you do to earn the swag?

copper.hat

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

Ted Shifrin

No one here loves me, so I'm swag-free.

leslie townes

it's your jumpsuit

copper.hat

@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...

Ted Shifrin

Well, maybe that's incentive for me to stay around to 100k and then disappear.

copper.hat

7:43 PM

:-)

i'm getting an orange jumpsuit, and will then be collected and delivered to a chinese remainder prison.

Thorgott

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

copper.hat

where gcds are never one

and even hermitian matrices are not diagonalisable

Ted Shifrin

Well, diagonalizability is overrated, anyhow.

Sha Vuklia

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'

Thorgott

open/closed doesn't make a difference, either works

Sha Vuklia

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

monoidaltransform

what does it mean for a vector field to be covariantly constant?

Ted Shifrin

Covariant derivative is zero.

We would need more specific wording to know in what sense to interpret that.

monoidaltransform

it's for the GR class, so I don't think I should elaborate

Ted Shifrin

The question is whether it's on an open set or on a particular curve.

monoidaltransform

curve

copper.hat

7:58 PM

curves are dangerous

Ted Shifrin

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."

Slereah

Hello math people

monoidaltransform

Okay, this makes sense. Thanks

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