Mathematics - 2021-03-03 (page 1 of 2)

12:00 AM

@CroCo The matrix representation is the same. The left side is different, but the right side is the same. They are either wasteful or its a misprint.

BigSocks

@user2103480 so messed up because this one lulls you into believing it could be something normal, but it has so much random crap tossed around...

user2103480

wym this is completely cryptic to me

Mike Miller

@user2103480 Is that Hairer

user2103480

Yes, some 2016 paper on regularity structures

arxiv.org/abs/1612.08138 page 64

BigSocks

basically just that the characters at least look similar to usual characters/notation. $Ker$ is in there. There is an integral and a product in there. some disjoint unions, stuff like that. the other thing uses so many nonstandard characters it's just sad to look at

but yeah, I won't pretend like it is any more reasonable

user2103480

12:09 AM

his notation really is the worst

Alessandro Codenotti

@user2103480 it's theoretical CS from a Cousot & Cousot paper

user2103480

thanks sounds way more reasonable

Alessandro Codenotti

Here di.ens.fr/~cousot/COUSOTpapers/publications.www/…

Filthy casuals posting memes without proper sources

It's on page 22 of the pdf

BigSocks

based Alessandro

CroCo

after some digging, it seems in extrinsic rotations (i.e. eq 2.63), the rotation matrices multiplication should be done starting from the most-left but in intrinsic rotations (i.e. eq 2.70) is the vice versa.
https://en.wikipedia.org/wiki/Euler_angles#Definition_by_intrinsic_rotations

Some thing the author ignored to mention it.

Mike Miller

12:24 AM

I'm not sure what an intrinsic or extrinsic rotation is supposed to mean

user2103480

@AlessandroCodenotti this

physics undergrad I think

Astyx

of course the authors are french

CroCo

@MikeMiller my understanding is that in extrinsic rotations, the rotating frame rotates relative to a fixed frame. In intrinsic rotations, the rotating frame rotates relative to a moving frame.

user2103480

@Astyx looks like a fine paper though

Astyx

If they are profs at ENS and Polytechnique, it means they're good

Mike Miller

12:28 AM

Sorry, I don't understand the language. It seems like you have reached an understanding now though.

CroCo

@MikeMiller
See this

This is extrinsic rotation if I understood it correctly

Notice ZA,YA,XA is fixed

^ this is intrinsic rotation.

Mike Miller

It's ok, you don't have to explain it to me. As far as I can tell there's no mathematical content to this distinction.

CroCo

There is actually in the matrices multiplications order. That is why I've asked.

Mike Miller

But that doesn't mean there's not content to it in computer graphics.

CroCo

I think there are different but yield same final rotation matrix.

they* are different

leslie townes

1:06 AM

if you get your hands dirty with atan2, is it possible to still call yourself a mathematician

i submit, no

Ryan Unger

3:21 AM

@leslietownes you're not a mathematician unless you're doing langlands

user 85795

Have you met him?

i submit, yes

BigSocks

@RyanUnger did you see the Scholze paper that came out recently

leslie townes

4:24 AM

i'm not doing langlands :(

Ryan Unger

@BigSocks of course

leslie townes

never even met the dude

i did meet VI Arnold once

BigSocks

@RyanUnger yeah I don't really know any of that but it looks impressive huh

Leonhard Euler

5:14 AM

finally I am able to compute the exact values of Bernoulli numbers in python

OneCold Ruben

congrats

Leonhard Euler

Now I will try to compute the exact value of zeta(s) at even and negative integers

making a computer algebra system is hard

I am still not able to define a "symbol" without using third party libraries

OneCold Ruben

I have used Python to create a collatz numbers to 1X10^20 but its all been done before... I also write simple programs.

leslie townes

there's an enormous amount of collatz stuff on math.se lately. people seem to have been bitten by the bug

Leonhard Euler

collatz bug

OneCold Ruben

5:19 AM

I have worked on it for 6 years to be convinced its true, looking at stopping time, Mod 2 Mod 3 Mod 4, I created a new notation, and my professors say I'm crazy. but you know... silly math problems.

leslie townes

i don't judge, i put some crap about collatz in my phd thesis. i think i have been inoculated against further investigations

OneCold Ruben

I even have a visual representation of it using silos and shoots, But I don't have the computer expertise to make it come to life in a 3d model.

leslie townes

graphviz is a very useful package, i would love to see more of it. i don't know that it goes into the third dimension.

OneCold Ruben

The cool thing about chasing a unsolved problem is how much it forces you to learn, I never knew Python, anything about number theory, or dynamics, chaos. It spurred the investigation process.

I also looked at binary strings, and the start of my mod investigation, I have filled 50 - 50MB excel sheets with calculations, they take about 5 mins to load, I all this stuff because I was curious about a problem.

Mike Miller

@OneColdRuben That's great

I have never been sure why people find Collatz interesting

But if you are learning something from investigating / understanding things more, that's more than enough reason

Semiclassical

5:41 AM

@MikeMiller I get why it’s curious at a glance: seemingly intractable despite being so easy to pose. I find it harder to understand why people -stay- obsessed with it

I mean, one must have a hobby. But it seems like a hobby with better prospects would he more appealing

Leonhard Euler

Mathematics is puzzling

Mathematicians can proved super hard conjectures but the simple sounding collatz conjecture is still unproved

Mike Miller

Yes, I am fond of assembling things, since then you can look afterwards

Semiclassical

I like stuff I can compute

Makes it feel like progress :P

Mike Miller

Still, all that matters is that your hobbies are fulfilling and enriching

And it sounds like that's the case here

So I can't give a hard time for that

Semiclassical

Ya, fair

Leonhard Euler

5:51 AM

is there a nice algorithm for algebraic simplification

Semiclassical

that's pretty vague

Leonhard Euler

I don't need an algorithm for very complicated expressions rn

love_sodam

9:27 AM

Is My proof valid?: Let $R$ be PID and $(p),(q)\subset R$ be two distinct nonzero prime ideals. Then I want to show $(p)\not\subset (q)$ and $(q)\not\subset (p)$. But this follows from the fact that $(p)\subset(q)\iff q|p$ and R is UFD

I assume $R$ is not a field

Astyx

you could be more explicit, but this works

love_sodam

Since $R$ is UFD and $p$ is prime, $p = uq$ where $u$ is unit. Hence $(p)=(q)$ which is a contradiction if we assume $(p)\subset (q)$.

Leaky Nun

you could be less explicit, but that works

Leaky Nun

9:55 AM

Any prime ideal in $R$ is principal, so by Krull's principal ideal theorem they all have height $\le 1$, so $\dim R = 1$ where $\dim$ is Krull dimension, so every non-zero prime ideal is maximal, so $(p)$ and $(q)$ are distinct maximal ideals, so they cannot contain each other.

probably don't need the step $\dim R = 1$

if $(p) < (q)$ then we can form a chain of prime ideals $(0) < (p) < (q)$, contradicting the fact that $(q)$ has height $\le 1$

love_sodam

Actually I wanted to show any PID which is not a field has Krull dimension 1

Leaky Nun

lol

accidentally proving your theorem

love_sodam

Any example of ring homomorphism from $\Bbb Q\to \Bbb Z$?

Want to see such homomorphism is injective

But it seems finding such map is not that easy

Astyx

let p/q be in the kernel

love_sodam

I first wanted to see that map is injective but now I want to find a ring homomorphism from $Q\to Z$

Astyx

10:07 AM

look at the image of 1/2

love_sodam

what do you mean

Astyx

Assume you have such a homomorphism f. What is f(1/2) ?

or rather, what relation does f(1/2) satisfy?

love_sodam

f(1/2)= 1/f(2)

f(2) = 1 then

or -1..

Astyx

I was thinking about 2f(1/2) = 1

love_sodam

Oh such ring homom does not exist

Astyx

10:13 AM

right. stronger than that: there's not group hom

(other than the zero map ofc)

love_sodam

Then what would be an example of ring homom. such that $F\to R$

$R$ is commutative ring with unity

Astyx

What's F ?

love_sodam

Field. sorry

Astyx

That's too vague a question. You can always take F = R

love_sodam

I will let $R$ be a commutative ring with unity that is not a field

I want to see an example of $F$ and $R$ such that ring homom $F\to R$ exsits

Astyx

10:20 AM

$F\to F[X]$

love_sodam

Oh

Leaky Nun

10:38 AM

any ring hom $f: \Bbb Q \to \Bbb Z$ must induce a group hom $\Bbb Q^\times = \Bbb Q \setminus \{0\} \to \Bbb Z^\times = \{-1,1\}$, which means $f(x) = \pm1$ for all $x \ne 0 \in \Bbb Q$, which means the image of $f$ is contained in $\{-1, 0, 1\} \subseteq \Bbb Z$, but no subset of that is a subring, contradiction.

Leonhard Euler

11:35 AM

The moving sofa problem is the most important thing in mathematics

it can help you to shift your furniture

robjohn

12:11 PM

@LeakyNun I was thinking of groups, and when I realized, I deleted.

anakhro

12:27 PM

@robjohn probably a quick question, but is uniform convergence of an infinite product on K identical to saying the "partial products" are uniformly convergent on K?

I can't think up a better definition.

User

12:45 PM

Let $X,Y,Z$ be connected metric spaces and $f:X\to Y$ be an embedding and $g:Y\to Z$ be continuous such that the restriction map $g:\text{Im}(f)\to Z$ is a proper map. Is $g\circ f:X\to Z$ a proper map?

My guess is that $g\circ f$ is proper map.

anakhro

@User sounds unlikely to me.

Can't you just do something like $B\hookrightarrow\mathbb R^n\to\mathbb R^n$ where $B$ is the open ball in $\mathbb R^n$, the first arrow the inclusion, and the second arrow the identity?

Mike Miller

Obviously not, take X = Z = R and Y = R^2, with f(x) = (x,0) and g(x,y) = x.

A month or two into talking about properness I would have hoped you'd have some intuition about what that means? At the very least you need to develop the skill to think of (counter)examples whenever you have a question. Playing with examples builds more intuition than pure formality.

anakhro

Admittedly, it's kind of a hard definition to wrap your head around, sometimes.

Mike Miller

For sure, but if it's a major topic you're working with you have to get there eventually!

(The intuition here was: proper means "sequences going to infinity get sent to sequences going to infinity." Since Im(f) doesn't know most sequences going to infinity, I had the map be well-behaved on Im(f) but crush everything else down to Im(f) to destroy properness.)

anakhro

I am sitting here struggling with uniform convergence stuff, so "eventually" seems to sometimes be down the line a little (only maybe if you are as bad as I am, I guess :P).

ery cerious matherfunker

12:59 PM

think of delta epsilon as fixed box

I don't remember if delta or epsilon or both should be fixed

I am half brain dead after car accident

User

Probably I am the only person who talks here proper related things. Thanks, @Mike miller for your suggestion. I should play with counter example while writing my master thesis.

Mike Miller

Most people here who have studied manifolds in a lot of detail have seen properness, since it is the appropriate condition to generalize many standard results beyond the assumption of compact domain. (Isotopy extension, well-definedness of intersection numbers, etc.) That's not to say that it's instantly understandable but one ought to work with examples for the sake of understanding it

I just tend to think that if I can produce a counterexample within 10 seconds you can produce a counterexample in some number of minutes if you attempt to!

Astyx

I don't get your counterexample. Isn't $g\circ f$ the identity? Is the identity not proper?

Or was he asking about $g$ before editing?

Mike Miller

Lol I'm being a dick and didn't even read the question correctly

Astyx

$g\circ f$ is proper because an embedding gives a diffeo on the image (so the restriction is proper) and the composition of two proper maps is proper

Mike Miller

1:12 PM

But then the answer is yes by pushing definitions? If g|_{Im(f)} is proper, then for K subset Z compact, g^-1(K) cap Im(f) is compact. Hence f^-1(g^-1(K)) = f^-1(g^-1(K) cap Im(f)) is a preimage of a compact set, and f is proper, so this is again compact.

Astyx

Right

Mike Miller

OK, so examples/counterexamples don't help here and my advice on that doesn't help so much, but it's still a definition push

User

that is ok. I asked this type of question before, as I am getting contradictory statements before while using proposition 10.1.17 and do not know whether it is right or not. google.co.in/books/edition/Topological_Methods_in_Group_Theory/…

geocalc33

1:42 PM

what do you call a biholomorphism s.t. $f(z)=f^{-1}(z)$

is it a "holomorphic automorphism?"

Mike Miller

Read the proof. If you can understand all of the details then it is right. If you don't understand a detail pursue a counterexample. If that pursuit doesn't work try proving the detail yourself.

Leaky Nun

@geocalc33 involution

geocalc33

okay so it's the same

ty

so $f(z)=1/z$ is an automorphism of the punctured complex plane $\Bbb C^*$

robjohn

2:26 PM

@anakhro That's what it means. Same for infinite sums.

anakhro

2:37 PM

@robjohn thanks. That's what I thought (that if it is that way for infinite sums, then it should be that way for products).

robjohn

It's about the convergence of the partial sums/products.

Edward Evans

math.umd.edu/~laskow/Pubs/713/…

Theorem 1 of this lmao

Astyx

It's a pretty impressive theorem no ?

Edward Evans

Yeah definitely

Astyx

Although outright horrible

anakhro

2:45 PM

@EdwardEvans is something like this something you can improve on? Like it's deg 25 in 26 variables, so can you possibly make 25 or 26 smaller?

Edward Evans

I think thereÄ's one on 20 variables

anakhro

Does the degree go up as a result?

Edward Evans

Yeah, apparently some dude showed it's possible to do it on 10 variables but the degree hits like 10^45

The record for the lowest degree of such a polynomial is 5 (with 42 variables [JSWW76]), and the record for fewest variables is 10 (with degree about 1.6 x 10^45 [Matijasevic77]).

anakhro

Wild.

Are any of these minimal, or still possibly something smaller?

Edward Evans

I dunno, someone just posted this theorem on facebook and I thought it was wild lmao

Thorgott

2:50 PM

nUmBeR tHeOrY

Edward Evans

literally

anakhro

Q: what does alternating caps like that mean in general?

Astyx

@EdwardEvans ferocious even

Edward Evans

weyyy

@anakhro en.wikipedia.org/wiki/Alternating_caps

geocalc33

when you HaVe LUncH

anakhro

3:00 PM

> It is typically used online to convey a mocking tone.

geocalc33

NuMber TheorY

if I capitalize less letters that means I'm mocking less

Edward Evans

fEwEr

geocalc33

dIfFerEnce gEomEtRy

anakhro

Is there an easy way of seeing that $e^{z/n}\to 1$ uniformly on $|z|\leq r$?

Of course, since it is on a compact disk, it feels like it has to be uniform.

Something like "for sufficiently large $n$, $1 - e^{z/n} \geq 0$ and then use some branch of the logarithm?

Thorgott

it's bounded between $e^{\pm R/n}$, no?

anakhro

3:11 PM

That would seem to make sense.

Oh yeah, of course.

user2103480

plug in taylor?

Astyx

That's just continuity of exp at 0 no?

anakhro

The amount of question marks make me think these might not be as "obvious".

user2103480

$\|1-e^{z/n}\| \leq |z|/n \sum_{k \geq 0} \frac{(|z/n|)^k}{(k+1)!} \leq |z|/n e^{|z|/n}$

Thorgott

what's wrong with what I said?

Mike Miller

3:15 PM

Thorgott just gave a complete argument, I'm not sure why people are continuing to go

anakhro

@Thorgott Nothing in particular, I am joking because with question marks, everything seems to be voiced in a doubtful tone.

Thorgott

oh, I see

anakhro

Do you see, Thorgott, or are you a blind mathematician.

Joining the ranks of Giroux and Pontryagin.

user2103480

and the late euler

Mike Miller

It converges uniformly to 1 on the box $|\text{Re}(z)| \leq r, |\text{Im}(z)| \leq r$, since $$e^{z/n} = e^{\text{Re}(z)/n} (\cos(\text{Im}(z)/n) + i \sin(\text{Im}(z)/n);$$ the first term is a real number bounded between $e^{-r/n}$ and $e^{r/n}$ and the second term is a unit complex number with argument in $[-r/n, r/n]$.

These both give you concrete estimates on how far each term is from 1, multiplying them you again get a uniform estimate (converging to 0 as n goes to infinity) about how far these values from 1.

This is Thorgott's argument adjusted to correct for the imaginary terms he forgot about (his argument only bounds magnitude)

anakhro

3:26 PM

So possible concrete estimates are respectively $\leq |e^{-r/n} - 1|$ and $\leq 2$, right?

And then of course, the product converges to zero, and depends only on n so it's a successful uniform estimate.

Mike Miller

$|zw - 1| = |zw - z - w + 1 + [z-1] + [w-1]| \leq |z-1||w-1| + |z-1| + |w-1|$ which explains why you need to show both of these terms go to 1

Then $|e^{\text{Re}(z)/n} - 1| \leq \text{max}(1 - e^{-r/n}, e^{r/n} - 1)$, I dunno which is bigger

But they both go to zero

And $|e^{i \text{Im}(z)/n} - 1| \leq |1-\cos(\text{Im}(z)/n)| + |\sin(\text{Im}(z)/n)| \leq 2/n$ or something.

Make it 4/n if you're paranoid.

Your whole expression is bounded above by something that looks like C/n.

The punchline is that you need to get your hands dirty but none of the individual estimates are very hard. If this looks like your intro real analysis class that's because it basically is.

anakhro

Thanks Mike. I will make sure to get my hands dirty.

Mike Miller

That should say 2r/n since Im(z) can be as big as r.

And this is implicitly for large n.

The bounds I'm getting there can be argued from the Taylor expansion but you really only need a first derivative.

anakhro

3:42 PM

In all my analysis courses, arguments about simple things like this always got swept under the rug as "it's easy to see".

And of course, I fell for the fool's view that "of course, if it is easy, I shouldn't need to bother to check myself".

<--- bad person

Astyx

If f is continuous at 0, $x\mapsto f(x/n)$ converges uniformly to f(0) on any compact set

Of course you need to know that $\exp$ is continuous at 0, which Thorgott/Mike's method shows

anakhro

@Astyx wouldn't showing that implication be just as much work as Thorgott/Mike's argument (modulo the continuous at zero part).

Or is there a really easy way of seeing that implication?

Astyx

It's analytic at 0

Mike Miller

3:59 PM

Ultimately you have to do something somewhere, I used a definition of exp which doesn't use power series

If you use the definition based on power series then you have to work with that instead

Either works

Astyx has the mmore sound phrasing

@anakhro I always say "CHECK: " when I say something like this. "Easy to see" is really an instruction.

geocalc33

Hope everyone's having a good day

polynomials are holomorphic mappings from C to (shining) C

anakhro

@MikeMiller one really hopes in hindsight that it is easy to see that "easy to see" is really an instruction.

geocalc33

4:19 PM

can someone help me set up an integration problem correctly

anakhro

Depends on the integration problem. :P

geocalc33

actually I think I can set it up correctly just had to think for a sec

I need to evaluate it via contour integration!

this is going to be the difficult part

linear_combinatori_probabi

Hi, does anyone who know how to prove an algorithm called IDA*(iterative deepening A*), which is a graph searching algorithm in computer science. I really like to know a formal proof of it. If anyone could share it to me, I'm willing to formulate this into a question and provide bounty. Thanks!

anakhro

4:36 PM

@linear_combinatori_probabi cse.sc.edu/~mgv/csce580f09/gradPres/korf_IDAStar_1985.pdf

No bounty required.

famesyasd

5:25 PM

I have a function $F(z_1,z_2) := z_1^2\cos^2(z_2)+\frac{3}{4}z_2^2$.

I need to show that $\forall \alpha > 0\, \exists \gamma > 0\, \forall z\, \|z\| \geq \alpha \implies F(z_1,z_2) \geq \gamma $.

Aside from using Lagrange multipliers for this function $F$ with restriction $z_1^2 + z_2^2 \geq \alpha^2$ that give a basically unsolvable (?) equation over $z_2$:
$-(\alpha^2 - z_2^2)\sin(2z_2) + \frac{3}{2}z_2 -z_2\cos^2(z_2) = 0$ do you guys have any other ideas I could use here?

I'm asking this because this is an example taken from a book so I feel like it should not be that hard, maybe so…

Sayan Chattopadhyay

In the universal coefficient theorem for Homology, to show that $H_n(C;G) \cong H_n(C) \otimes G \oplus \operatorname{Tor}(H_{n-1}(C) , G)$ this splitting is not natural, I was asked to examine the map given by identifying the boundary of the disc in, $H_2(\Bbb{R}P^2, \Bbb{Z}_2) \to H_2(S^2, \Bbb{Z}_2)$. I have reduced it to showing that $H_2(\Bbb{R}P^2, \Bbb{Z}_2) \to H_2(S^2, \Bbb{Z}_2)$ is not the trivial map. How does one do that?

Balarka Sen

Compute kernel :)

The map is $\Bbb{RP}^2 \to \Bbb{RP}^2/\Bbb{RP}^1$

Ted Shifrin

Greetings, a @Balarka, @Sayan, demonic A.

jeea

Does someone know quick intuition for Cayleys formula for number of spanning trees

Cayley's formula

In mathematics, Cayley's formula is a result in graph theory named after Arthur Cayley. It states that for every positive integer n {\displaystyle n} , the number of trees on n {\displaystyle n} labeled vertices is n n − 2 {\displaystyle n^{n-2}} . The formula equivalently counts the number of spanning trees of a complete graph with labeled vertices (sequence A000272 in the OEIS). == Proof... ==

Jakobian

proof should give intuition

Balarka Sen

5:38 PM

A typical way to product a spanning tree in $K_n$ would be to pick a vertex, flail your arm in $n$ directions, then stand at each of the other vertices, flail your arm in $n$ directions, so on. At some point you need to stop else you're going to create cycles, which spanning trees do not have. You have to stop at the $(n-2)$-th stage exactly, whence $n^{n-2}$

Roughly, this is the intution

leslie townes

famesyasd, are you able to show that $(0,0)$ is the only root of $F$? (is that true?) if so it would seem by continuity of $F$ enough to establish that $\inf F$ is bounded away from zero on any set of the form $\{z: \|z\| \geq \alpha\}$

Ted Shifrin

@famesyasd Just to be sure. Your $z_1,z_2$ are real, right?

Alessandro Codenotti

Isn't there a neat double counting proof we talked about once? @Balarka

Jakobian

@famesyasd if $\cos(z_2)\geq 3/4$, then $F$ is bounded by $\alpha$. If not, you have lower bounds on $z_2^2$

Balarka Sen

The usual proof is some coding thing

I never remember, but I am also not trying hard

Ted Shifrin

5:42 PM

@Jakobian You mean $|\cos z_2|\ge\sqrt3/2$?

Jakobian

sorry, $\cos(z_2)^2 \geq 3/4$, not that it matters much

Ted Shifrin

Well, it matters to someone who isn't knowing how to do it.

To be clear, I'm clarifying for famesyasd, not for you.

leslie townes

when i see "yasd" all i can think of is yet another sudden death. a phenomenon common to some of the computer games i like to play

Ted Shifrin

Good thing I lead a sheltered life.

leslie townes

the games are uncompromising. like math, sort of. i do think that identifying the set of roots of $F$ is close to solving the problem. i would not want to explicitly construct the best $\gamma$ or anything like that. but i could be missing something

Ted Shifrin

5:48 PM

It's equivalent just to bound below when $z_2\approx 0$ and then have an obvious lower bound when $|z_2|$ is bounded below. Again, I'm assuming all real variables.

@leslie Can you make heads or tails of this? ... I'm not going to tell the OP, but I'm guessing what someone told him/her to do is to look at the ODE that $\int_{-\infty}^\infty e^{-\pi x^2}e^{2\pi ix\xi}\,dx$ satisfies.

puts Balarka, Sayan, and demonic Alessandro on the naughty list

leslie townes

i'm confused by the problem statement, as you are. i wonder if they are hiding the ball a bit here. "someone told me to do X" would be much more preferable to "here's my hat, and now a rabbit is hopping out of it."

Ted Shifrin

Yes, I tried to make my accusation less accusatory :P

leslie townes

i also personally cringe on 'thanks in advance.' i'm working with my therapist on that, too.

Ted Shifrin

It's the people who want us to do their homeworks/exams who tend to say that ... or very polite people.

I truly was confused by the $\xi\in\Bbb C$ at the beginning and the $\xi\in\Bbb R$ at the end.

leslie townes

also the superfluous inclusion of a theorem number. it's fishy. i'm not saying it stinks, but it's fishy

Ted Shifrin

5:56 PM

Well, they are citing Stein & Stakarchi.

Jakobian

didn't know 'thanks in advance' was so controversial

Ted Shifrin

To me, it feels presumptuous, @Jakobian.

Especially given the current "do my exam question for me, and hurry" all over MSE.

leslie townes

for some people, 'thanks in advance' sounds more or less like an advance guarantee that no acknowledgment of anything will be coming later.

which yeah, 'do my homework,' i don't know. i do see it in my professional life where it is less cringeworthy.

Ted Shifrin

Anyhow, do you agree that they're probably angling for (or being told to do so) the ODE approach?

I think Stein/Stakarchi do prove differentiation under the integral sign for integrals $\int_a^b$.

I no longer have that book.

leslie townes

yes that does appear to be the setup. it's so weird when people link solutions to the problems they're asking, and just say 'instead of'

the USPS lost about half of my math books during a cross-country move. i have approximately 50% of my previous mathematical knowledge. the curse was i managed to get most of the books i didn't even want anymore.

Ted Shifrin

6:00 PM

They actually mistyped the theorem. They have $[0,1]$ and then $[a,b]$.

I got rid of literally thousands of my beloved books when I retired. Sigh.

leslie townes

there wasn't time. the answer is needed NOW. a,b,0,1, just answer.

i did find one of my lost books in a bookstore. that was a trip. i wanted to tell the guy at the counter that i was simply repossessing stolen property.

Ted Shifrin

LOL ... so you packed all the good books in the boxes they lost, huh?

I nostalgically remember many hours browsing at Cody Books on Telegraph. Back when we had books and bookstores.

leslie townes

cody's was good. and moe's. and shakespeare & co (not great for math, but great)

i saw kurt vonnegut in moe's once. he was browsing history books

Ted Shifrin

Moe's must be after my time.

leslie townes

same block as cody's i think. and now all of it is gone

i literally lost all of the expensive books i bought and retained books that i had been given for free

i have something henry helson gave me on dirichlet series. it looks like it's a book. i don't know if it was really published. i lost all of the $100 tomes they made me study in graduate school.

Ted Shifrin

6:09 PM

Oh, Moe's webpage says they're still there. The webpage says they've been there since 1959. I honestly do not remember it at all.

leslie townes

they did not have a mathematics section. which cody's definitely did (halfway in the back on the first floor)

Ted Shifrin

I was a French lit major in college, so I know how to read a non-math book :)

robjohn

@TedShifrin That is one book I don't have. That was after I was there.

Ted Shifrin

I taught graduate complex out of it one time, @robjohn. Halfway through the course, I regretted my choice pretty seriously.

robjohn

I may get it at some point because it does present a lot of the things he taught in class that are not in his previous books.

Ted Shifrin

6:11 PM

I knew that the geometry side would be weak, but I didn't realize how much they underplay the multivaluedness of log, which is something I want to play up from week 2.

There were a few serious mathematical errors, but I guess the strength (which didn't matter to me) is the inclusion of the number theory material.

leslie townes

when i was in graduate school i embarked upon a fairly aggressive campaign of copyright infringement, in partial revenge for losing my books. they are on a hard drive under my desk now. my favorite complex analysis book is conway.

Ted Shifrin

I never liked Conway, either.

But since he's a functional analyst, I imagine he'd appeal to you :)

leslie townes

bak and newman is a really good undergraduate treatment.

and yes, that's how he hooked me

robjohn

@TedShifrin I attended one of his seminars (about FRACTRAN) when he visited Princeton, and I wrote a Life implementation, but other than that, I don't know much of what he did.

Ted Shifrin

Actually, Lang's book is surprisingly good. I was going to teach out of Gamelin my last time teaching the graduate course, but then I got cancer and almost didn't teach at all that semester, so I gave the course to someone else.

Different Conway, @robjohn.

robjohn

6:13 PM

Oh

leslie townes

there are too many conways

i like lang's algebra a lot. he was very prickly in person, i don't know that i would recommend him as a person. but the algebra book is good

Ted Shifrin

The one you're thinking of was the topologist/algebraist, who just died. The one leslie's talking about is (was?) a functional analyst.

I think the Algebra book is only good if you already know the material.

leslie townes

this is probably true

Ted Shifrin

Yes, he was prickly. I sorta liked him, though. Although the seminar talk he gave at Berkeley where he ("teasingly") put down Chern was a bit shocking to me. Chern just smiled in his usual way.

leslie townes

he was kind of creepy around certain people. i'll leave it there. he also, i think, prioritized saying unpopular or controversial things over the kinds of compromises that most of us make when communicating with an audience. he found some joy in that

he was very funny

Ted Shifrin

6:17 PM

I should point out to @robjohn that it's worse than he realizes. Both of these are John Conways. The algebra/topology/Life one is middle initial H, and the analyst is middle initial B.

Thorgott

Bak and Newman contains a really cool proof (by Newman) of the fact that the Fourier transform is injective using complex analysis in the last chapter

Ted Shifrin

Creepy? Just because he was closetedly gay?

leslie townes

bak and newman is the coolest book ever. i'm surprised more people aren't onto it.

Ted Shifrin

I actually never even saw that book, although I've heard of it, of course. I think it appeared after I taught undergraduate complex many, many years ago.

Astyx

@Thorgott Could you sketch it?

Ted Shifrin

6:18 PM

Salut @Astyx. Guten Abend @Thor.

Astyx

Salut Ted

leslie townes

absolutely not. nothing creepy about being closeted, especially for a person of his vintage. he was sometimes creepy on top of that.

but not too creepy. i'm not canceling him

serge lang struck me as the kind of person who never met a bull without wanting to wave red banners in front of it

famesyasd

@TedShifrin yes, $z_1,z_2$ are real
@jakobian didn't quite get your message

leslie townes

is yasd a reference to another sudden death?

my therapist and i can process this together if the answer is no. i'm ready for either answer

famesyasd

damn these messages

Jakobian

6:26 PM

@famesyasd Ted Shifrin elaborated on what I was saying, maybe that'll be more understandable for you

what kind of atlas do we give to a coordinate slice? The one map from non-constant coordinates of coordinate system?

the book doesn't really explain that

geocalc33

If we know that ${\displaystyle P(s)=\sum _{n>0}\mu (n){\frac {\log \zeta (ns)}{n}}},$ then what is $P(sz)?$

Thorgott

So $f\colon\mathbb{R}\rightarrow\mathbb{C}$ is in $L^1$ and we assume $\hat{f}(x)=\int_{-\infty}^{\infty}f(t)e^{2\pi ixt}dt=0$ for all $x$. Fix some auxiliary real $a$ and now define a function of a complex argument $z$ by $\int_{-\infty}^af(t)e^{2\pi iz(t-a)}dt$ for $\operatorname{Im}(z)\le0$ and $\int_a^{\infty}f(t)e^{2\pi iz(t-a)}dt$ for $\operatorname{Im}(z)\ge0$. For real $z$, these agree by the hypothesis.
Dominated convergence implies continuity of this function on $\mathbb{C}$ and we see it is bounded by $\int_{-\infty}^{\infty}|f(t)|dt$. In fact, it is holomorphic on the upper and …

leslie townes

The emoji with the hearts in the eyes

Ted Shifrin

@Jakobian Yes. If you're looking at the $(n-k)$-dimensional slice given by $x^{k+1}=\text{const}$, ..., $x^n = \text{const}$, then of course $(x^1,\dots,x^k)$ give coordinates on the slice.

Thorgott

I think the idea behind this proof is motivated by Paley-Wiener theory, but I never learned that proper

it's a very neat trick in any case

Jakobian

6:33 PM

ok, thanks

Astyx

"For real $z$ these agree by the hypothesis" you mean they're opposite of each other right?

Thorgott

oh, I forgot a minus sign on the second integral

leslie townes

My advisor once showed me a book from Wiener that he had had signed by Wiener. I do not have many mathematical heroes but it did feel cool to touch the page.

famesyasd

@leslietownes just to be clear we pick a sequence $z_n$ by contradiction that tends to zero and then apply continuity of $F$?

Astyx

Very neat trick indeed!

geocalc33

6:39 PM

If I ask question can 10 people agree to upvote it instantaneously?

Astyx

no

Ted Shifrin

@famesyasd Too hard. Just do a lower bound when $z_2\approx 0$ (so Jakobian's suggestion is $|\cos z_2|\ge \sqrt3/2$) and then give a lower bound for the rest of $|z_2|$.

leslie townes

@famesyasd see Ted's comment. it is probably the simplest solution.

A key in that problem seems to be that $F$ is a continuous sum of nonnegative things. This does put a lot of pressure on when those things are both going to be zero.

geocalc33

I can't believe I got 10 upvotes on a question on the Mathoverflow

I must have asked an easy to state but hard to answer question

it's definitely not research level lol

Do you think if someone solved like the Novikov conjecture or the Hodge conjecture in a math S.E. exchange post it would garner many upvotes?

I think it would get like no upvotes cause nobody would understand lol

leslie townes

7:04 PM

you may be right about that. if i look at my upvotes it is mostly for the dumb stuff that i say and not the smart stuff that i say. which isn't to say that i say smart stuff. people do upvote dumb.

and if it seems technical and confusing, forget it

famesyasd

@leslietownes okay I think I get, thanks guys. btw, yes, "yasd" means "yet another stupid death", an acronym coming from roguelike games, I used to play them a lot when I have created this nickname. My favourite game is Rogue, I personally think it's the best game of all time. I have added yasd to my nickname when the system considered "Fames" being too short or taken. I don't really think it is a great addition but whatever.

BTW I was really edgy when I was creating my nickname so fames means "hunger" not fame and it's spelled diffrently not how you think

geocalc33

if the solution was posted it on Mathoverflow it would probably be recognized as legit and people would upvote it a lot

leslie townes

i love that yasd meant what i thought it meant. i am a nethack fan.

mathoverflow is a strange place. i forgot the password to my account there

geocalc33

geocalc33 is derived from geo-metry calc-ulus and the number 33

famesyasd

I don't like nethack don't beat me :)

geocalc33

7:09 PM

because calculus (with analytic geometry) was my favorite subject in college

leslie townes

Leslie Townes is the first and middle name of the actor known as "Bob Hope"

geocalc33

thought that name sounded familiar

leslie townes

i don't like the new nethack. 3.4.3 was OK

Sayan Chattopadhyay

7:37 PM

@BalarkaSen Okay I think I get it now. Tell me if this is correct. When I consider $\Bbb{R}P^2/\Bbb{R}P^1$, the 1-cell is being collapsed, so the CW structure is just $e^0,e^2$. Since the quotient map $q$ is taking the original 2-cell to the 2-cell in $S^2$, I am guessing it should induce an isomorphism between $H_2( e^1 \cup e^2 / e^1)$ (for $\Bbb{R}P^2$) and $H_2(e^2/e^0)$ (for $S^2$). Now when I take $\Bbb{Z}_2$ coefficients the degree maps are all 0 for $\Bbb{R}P^2$ and same for $S^2$

Which gives me that both $H_2(\Bbb{R}P^2,\Bbb{Z}_2)$ and $H_2(S^2,\Bbb{Z}_2)$ are isomorphic as there are no higher cells in the CW complex.

Mike Miller

Yes

If $A \subset X$ is a subcomplex then $C_*^{CW}(X/A) = C_*^{CW}(X)/C_*^{CW}(A)$

All chains reduced

So you're ust doing an algebra computation

(The natural map $C_*^{CW}(X) \to C_*^{CW}(X/A)$ is identified with the canonical quotient map)

Jakobian

8:07 PM

I've got 12 upvotes for a smart thing once

but I've put a lot of bounties on that question because no one seems to be able to answer it

only easy questions with hard answers attract audience

and it makes sense

not really a mystery that the further you go, the less and less people know what you're talking about

Edward Evans

I had a series of questions about modular forms that only one guy was answering

Thorgott

it's official: I passed my physics exam in the absence of understanding

Edward Evans

Glückwunsch

geocalc33

is anyone solid at evaluating contour integrals

I guess I have to draw a contour and sum the residues

Astyx

8:34 PM

@Thorgott what physics?

Thorgott

newtonian mechanics

geocalc33

langrangian wish I learned

instead of newtonian

1

Q: On evaluating $\rho(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)P(sz)\,dz.$

How do you evaluate the following integral? $$\rho(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(z)P(sz)\,dz.$$ $P(\cdot)$ is the prime zeta function. I got the integral from taking $\rho(s)=\sum_{\rho\in\Bbb P}\exp\big(-\rho^s\big)$ and using the Mellin transform to extend the domain o...

okay I just asked my contour integration exercise

I think I almost got it but I think I messed up somewhere

Ted Shifrin

8:58 PM

@Thorgott I take this statement with several grains of salt.

Transcript for

$Mathematics$ Mathematics