hello guys

Hey @Ted :<

Whoa, is that @Alessandro past midnight?

hi Zach

Past 1 actually

Hi @Ted

i should get more sleep

Well, past 1 is also past midnight :)

All you ever do is nap, Zach.

but at night i don't sleep

Oh, you're becoming another Balarka?

So it's balanced I guess...

It's truly unhealthy.

@Adeek can't you write a (piecewise) linear homeomorphism even (just to make sure the inverse is nice too)? By turning the sides 90 degrees to get a closed segment on the x-axis

yeah

tht is what I did @AlessandroCodenotti

@TedShifrin can't argue with that

awesome

Zach, you'll be amused to know that one of the 4th graders asked as he was leaving if he could call me Mr. Teddy Bear ... :D (Of course, I'll never be back at that school again, but I might see him ...)

Ahahaha

Do your students address you as just Ted or "Mr. Shifrin"?

My college students used to call me either Ted or Dr. Shifrin ... a few said Dr. Ted.

All the kids I tutor here call me Ted.

Dr. Ted Shifrin, PhD.

That's redundant.

I know

It's like calling me Mr. Zach "Meow Mix" Hauk

…the first

??

No, Zach, that's not redundant. It just gives you more gravitas than you deserve.

@AkivaWeinberger The guy I was teaching isn't online

Hi, DogAteMy!

Hi!

So, shall we talk about the QQ questions?

Hey @Ted

If you're interested

Heya, Demonark.

I might get called back up soon, so I don't know how much time I'll have

called back up where?

Oh. I guess tomorrow then

Today I ate a yogurt, half of a bagel, and a slice of pizza

I'm in a super long choir rehearsal, since we have a concert on Sunday

and like half of one of those string cheese sticks

ohhh ... you have way more patience than I, DogAteMy

Right now the adult choir ("Kol Ram") is practicing a thing, but when they're done, me and the rest of Chamber Choir need to rehearse a little dance thingy

Zach, you'll never turn into the food connoisseur and snob that I am at that rate!

DogAteMy, so I take it this at the temple and not at school.

No, it's at school

Is it a Jewish school?

Oh, so you're not an adult yet?

Yeah @MeowMix

Does anyone recognize the decimal 0.7966 as anything noteworthy?

I didn't bother applying to some prestigious high school

@TedShifrin There's the lower and middle school choirs, the high school choir (which I'm in), the chamber choir (a subset of the HS choir which I'm also in), and Kol Ram (the adult choir of people not necessarily connected to the school)

and we're all performing together on Sunday

Because most of the people I know are going to the public high school

Ohhhh, what is a school doing with an adult choir of non-student adults?

Also I probably wouldn't get in

Doesn't immediately ring a bell

Zach, I was perfectly happy going to a public high school. ...

And I will be too.

@TedShifrin I guess the choir director (a very nice guy by the name of Mr. Henkin) wanted to give people in the community a choir to sing in

@Sophie: Mathematica tells me the answer to your question is 0.7966, but I have no idea what that is.

Hi

G'night, @MikeM.

My mom's in it

It's a new set of numbers. The set of fake numbers

Well, now the government is going to defund Americorps, the humanities, the arts, and scientific/medical research. But we will be so great.

Oh, not to mention school lunches for the kids who have no food and Meals on Wheels for the adults too poor and weak to feed themselves.

There's a website where you write your decimal and it tries to find a close expression for it in terms of known constants, I forgot the name though

I was thinking it was just for sequences, @Alessandro.

@AlessandroCodenotti Would it happen to be "The website where you write your decimal and it tries to find a close expression for it in terms of known constants?"

puts Zach on ignore

snaps nice one

Oh lol

@Daminark Encouraging my behavior will probably result in a Ted-nore also

I like making portmanteaus of words with "Ted" in front.

@Alessandro: This might be it.

Wow, a huge list of outputs. Amazing.

Now you have to find which ones are correct :P

Hey @TedShifrin

Well, I need to get Sophie's question with more decimal places.

Hi @Ali

@TedShifrin It doesn't load on mobile but it rings a bell, I think that's the one I meant

So I'm skipping Britain this trip, @Ali, but maybe next time.

@TedShifrin It's ok, the plane might miss anyway as we will be leaving the EU

Well, I'm actually flying home via Heathrow, but that doesn't count. ... LOL, the world is such a mess :(

I'm off to bed, good night (or day) to everyone!

Where are you,

@TedShifrin They have good shopping so I'm sure you'll be fine!

Night, @Alessandro

@MikeM: I'm talking about returning from Europe in June.

I just found out that the Wreath product of two groups contains a subgroup, for every group extension of the two groups, isomorphic to the extension.

Ah. You sure you want to?

Well, I'm not sure of much these days, Mike.

BTW, Zach, in French, the plural of a word ending in -eau is -eaux. :D

I tried playing a French word game

I lasted two minutes...

The only french english pun you need to know is $\Bbb F_1$

When I was in Morocco I learned some French but it was so confusing

I used to be as fluent in French as in English, but that was in college.

I'm still pretty good when my tongue stays untied.

@Ali: Tu devras me l'expliquer ...

It's been some time now so I don't know it too well anymore. I'm not really that good at any foreign language at this point

Though it'll prob be a good idea to actually learn French if I go to grad school, right?

Isn't the french currency called "Francs" or is that something else?

@TedShifrin Fun

@Meow It used to be

To be franc, it's not any more

Ba dum tss

Now they're using the Euro but I think until 2001 it was the franc. Switzerland still uses it if I'm not mistaken

It's interesting because the field with one element 'should' exist as many constructions in geometry and algerbra become nicer with it

That's a nice one

Zach, it's all Euros now.

Like compactification of Spec Z is a kind of curve over F_1

smacks Ali

All non-abelian finite simple groups that aren't sporadic are groups of Lie type

and so on

Oh and something about the Riemann hypothesis

You're Lie-ing to yourself, @ALie Caglayan

The second one was better than the first :P

Zach, don't you have actual math to do?

Uhhh yes

I agree with Ted this isn't actual math

kindof what you say to people when waiting in an elevator

Would you say the math is totally irrational?

or before a seminar starts

Okay I'm going to do some LA

That way I don't get @Ted too mad at me

If linear algebra is about arrows how is it any different than category theory?

waits for impending smack

@Ali An arrow is a complete metric space under the subspace topology, right?

I'm going to get banned from the site for all the smacks.

Because if so, that would make it second category in itself

I heard smacks were contravariant functors from lalaland to reality

contravariant, huh?

Contrabass saxaphone?

yeah I guess its the twisting force

remands Zach to math calisthenics for a week

Calisthenics? I'd rather do Jazzercise-math

@TedShifrin do they have jazz bars in Ca?

They have like 3 in london

I honestly thought there would be more

I have no idea, @Ali. I'm sure. I went to one years ago in Atlanta.

@TedShifrin I put that number on an inverse symbolic calculator and nothing convincing came out

@Sophie: Yes, that's what I've been doing.

What number is this?

The last number I saw was a page number

Sophie had a weird limit of integrals.

$\displaystyle\lim_{x\to\infty}\int_{1-1/x}^1 \frac{1-t^x}{1-t}dt$

$\lim\limits_{x\to 0^+} \int_{1-x}^1 \dfrac{1-t^{1/x}}{1-t}dt$

I rewrote to be clever :P

I haven't even proven it exists yet

Only analysts care if it exists lets work it out

Well, keep me informed, @Sophie :)

Today I did one of those cool things where you slide down the rail of a stair case

like, a long staircase

That's p cool (make sure you don't converge to the ground too quickly)

$\lim_{\text{face}\to\text{ground}} \text{Zach}$

if a teacher gives you time to work in groups and review for a test, do you think it's rude to read stuff on your phone instead?

@Sophie I'd generally say it's best not to

@Sophie Just ask the teacher

At the end of the day they probably don't care

and if it betters you for the test then whats the point

the test was about introductory calculus topics and I was reading about the russians being genocidal in the Caucasus so I don't think it helped

> An interesting perspective that I heard recently was that people choose not to be self-aware because they are scared of the can of worms it would open up.

- someone on Reddit

lol

@Sophie Just sit tight and think about calculus then when class is over you are free to learn about genocides

Only in the Caucasus?

also I am starting to think your limit doesn't exist

What happened to Chris'sis he usually did stuff like this

$\lim\limits_{\rm-bo~time!}$

@AkivaWeinberger a sudden fascination about the Caucasus hit me. One needs to start somewhere

hums anthem of Ingushetia

I should have called him a statist and tell him anecdote about tanzanian farmers and invisible hands and chinese famines

sorry typo

$y''=y'\textrm{ln}x$

$y'(x)=(1+x'())y(x-1)$

Now consider $(1+x'())$

$(1+x'())'=(x'())'=('()+x''())$

$(1+x'())''=('()+x''())'=(''()+(''()+x'''())$

so yeah @MikeMiller once you get that $I \times \{0\}$ is homeomorphic to $I \times \{0,1\} \cup \{0\}\times I$ then we are done

Sorry typo again:

Am I allowed to brag about someone thinking that I'm humble

or does that defeat the purpose

@Adeek I agree.

@MikeMiller I was wondering if we have two maps which satisfies the HLP then are they vertically homotopic ?

@Adeek Yes, I think that's probably what the point of this proposition is.

In fact I think that's literally what you're proving?

Sorry typo:

$y^{k-1}(x)=\sum_{l=0}^k\sum_{i=1}^s \begin{pmatrix}k\\ l\end{pmatrix}(i+1)y^{(l+i)}(x-1)+\begin{pmatrix}k\\ l\end{pmatrix} xy^{(l+i+1)}(x-1)$

$\int^{(n)}y^{(k-1)}(x)d^nx=\sum_{l=0}^k\sum_{i=1}^s \begin{pmatrix}k\\ l\end{pmatrix}(i+1)\int^{(n)}y^{(l+i)}(u)d^nu+\begin{pmatrix}k\\ l\end{pmatrix} \int^{(n)} uy^{(l+i+1)}(u)d^nu+\begin{pmatrix}k\\ l\end{pmatrix}\int^{(n)} y^{(l+i+1)}(u)d^nu$

why would you want to do all that?

that kind of math upsets me... I need a safe space

I am trying to try out a generalisation of the method of snake oil for the continous case, which involve integrals of the form $\int a(n)x^n dn$

The idea is that if I can get a reduction formula out of this mess, then integration by parts will quickly kill this integral

and hopefully, a sequences of derivatives of $a(n)$

NB note that here $y=x^x$

and then... I think I need to grab some lunch...

What is the right notation for a set of the integers that are divisible by a or (divisible by a) + 1? I came up with the following notation but I think its incorrect:

$\left\{ a\ell,a\ell+1\mid\ell\in\mathbb{Z}\right\} $

that's fine

can also write $a\Bbb Z\cup (1+a\Bbb Z)$

Yeah @MikeMiller I think it follows from this

we can see that the derivatives of y(x-1) increases for each iteration, and the integrals of the differential operator (1+x'()) also increases. Both cases have well defined recursion formulae (derivatives are easy to determine, the integral is a combination of lower order integrals, thus this suggest integrate by $m$ times of $x^x$ is always elementary, if not tedious to compute

I'm pretty sure I'm half screwed on my assignment.

Because I'm half done.

ok wait I think now I can see why letting $x=e^{iu}$ is much easier. I will work out the Laplace transform version of this later

but as you can envision, the -(k+1) term becomes extremely easy to integrate since $e^{ax}$ is an eigenfunction of the functional $\int dx$

therefore at the end of the day you end up with an infinite exponential series where all coefficients are derivatives

and I suspect, it will work for all functions

@Kaumudi.H The condition is in fact $a \ge 0$. Visualise $e^x+a$ as shifting the graph of $e^x$ by a. It should be clear that for $a \ge 0$ alone does this graph stay above the x axis $\forall x \in \mathbb R$

$-a \le 0 \implies a \ge 0$ not $a \le 0$

$e^x \ge -a$ does not imply $-a \le 0$. But $-a \le e^x \in [0, \infty )$ does because you want $a$ to be less than the entire range of $e^x$ which has a lower bound of 0.

Because if $a$ is some negative value, then $-a$ is some positive value and then $\forall x< \mathrm{ln}(-a) , \space e^x < -a \implies e^x+a < 0$

This stuff is a lot easier if you try to visualise it first. :)

Any sort of a diagram

Its the qualitative features you're after

Hey guys, can I get some help with proving this

I've been doing math all day for my assignment, I am exhausted. I'm not even sure how to start this.

Any help is appreciated.

@Dragneel Whats $\mathbb P$?

The set of prime numbers

Are you done with the last problem you posted. Because you can use the result here

You referring to me?

yeah

the one with $n=kl, k<l \implies k \le \sqrt{n}$

Wow that was this morning. How did you even remember that?

I'm curious

Because I read it about half an hour ago when I woke up :P. Different time zones.

Haha nice

Yea so I just manipulated the equality $l=k$ and proved that it equals $sqrt(n)=k$

I'm not sure how I can use this fact to prove the expression I posted above

I didn't think it was related

Can you give me a hint please :)

n is composite so there exist integers l and k ( between 1 and n ) so n=lk. Now because of the symmetry between l and k (because of commutativity). You can assume something that reduces this problem to the result of the previous one.

@TimTheEnchanter If $1 \le k \le n$ and $1 \le l \le n$, then $1 \le k \le l \le n$, which is the same scenario as the first problem. Did I get it right?

@Dragneel Yes, now that you've assumed $k \le l$ you can apply the earlier result to say $k \le \sqrt{n}$. Now you have to show there is a prime number p that can fill the shoes of k, can you use the same arguments you've already made to show such a p exists?

To show that $k$ is prime, do I use the fact that the square root of primes is irrational?

And since $k$ can equal the square root of $n$, then the square root of $n$ is irrational because we established that $k$ is an integer

You cant do that because k does not need to be prime, for example 100 = 4 * 25 has 4 < 10 but 4 isn’t prime. However you can say something about the magnitude of the prime factors of k.

Remember that you don't need to find an expression for p, only a way to show that it exists.

I multiply both sides of $k=sqrt(n)$ by $n$, resulting with $k*n=n$. Which means $k|n$.

@TimTheEnchanter I feel like the answer is staring me in the face. I give up. It's 3:47 am here, and I've been working on this assignment since this morning haha

@Dragneel We know k divides n from the expression $n = kl$. We have done all we want to with k, that is, prove that there exists some factor of n such that $k \le \sqrt{n}$. Its time to move on. Now we have to prove that 1) There is a prime factor p of n (true by fundamental theorem of arithmetic) and 2) That same prime factor p also satisfies the condition $p \le \sqrt{n}$. If k was prime, then k is the desired prime factor. But if k is composite, what can you say about the prime factors of k?

with respect to the two properties I've listed.

Hi, I'm new to this SE, so may I ask, is this answer of mine okay?

@Kaumudi Look at $x^3$ near the origin

The derivative is $0$ there, but positive before and after

Yet the function is strictly monotonic

No problem lmao

Hi chat!

Hey @Baymax

ha @Daminark

@Daminark Correct me if I'm wrong, but if a function is monotonous in an interval(open) it should be monotonous at each point in the interval right?

@BAYMAX Hey

yo @TimTheEnchanter

Monotonicity isn't specifically a property of a function at a point, it has to do with comparisons between points

yes

Given a point, you can talk about the function being locally monotonic, meaning on some open neighborhood of it

You mean an epsilon neighbourhood?

Yeah

it can also be tim neighborhood

:)

Ok. The name confused me a bit. Thanks

But if you're in an open interval where the function is monotonic, this is automatic for each point

notation change!

No problem, and yeah that's fair, I have just grown accustomed to saying "in an open set"

No, I meant that I thought the phrase "monotonous at a point" meant that the right and left hand derivatives had the same sign. I was just used to seeing monotonicity over an interval.

Ah, now I get it

I am terrible at terminology. It annoys me so much that closed and open sets are'nt just opposites.

Oh yeah that threw me for a loop to start with

I was like "Wai u do dis?!"

Topology in general threw me for a loop to start with

complement of open sets implies closed sets and vice versa

bt

not open doenot imply closed

also

not closed doesnot imply open

Thinking of some examples helps clear it all up

yes

Hi guys

morning

out of curiosity

do you know any online reference for unum?

(arith22.gforge.inria.fr/slides/06-gustafson.pdf

any kind of lecture notes or free book maybe

I assume since it's a relatively new subject

probably the book is quite hard to find

There are more red flags in that document than the red army used to carry around

what?

Like the error bound of the rectangle's method being often infinite, if you work with $C^2$ functions over a closed interval as you're supposed to that bound is always finite

Also the error is estimated using exact arithmetic, it's not an error deriving from floating point errors

The whole thing smells of "too good to be true"

Not to mention, if this were feasible, this would've been patented already, rather than made publicly available.

@user8469759 has it actually been implemented?

Old news: Eh what?

I don't know, but I'd like to find that out

I'd like to implement it by myself in software

but I have no clue how arithmetic works there

Morning guys

More seriously, is a circle uniformly curved?

oh god wait

:p

ok fixed

Could someone tell me why $\overline n=\overline0\in H$?

zero and n are within the same equivalence class $\overline{0}$

i think I get it

it's the neutral element

so it must be in $H$

Indeed, since the identity $1_H$ of a subgroup $H \leq G$ is the identity $1_G$ of the group $G$

is uniform curvature of a curve at any point implies constant curvature at any point of a manifold?

I'm not sure what you mean with "uniform curvature"

And you may also want to specify which curvatures exactly you're talking about

(and which manifolds)

If $G$ is $k$-line colourable and $k$ divides the number of lines, then every colour occurs equally often.

ok we can start with 1D curves, so the curvature if I recall is the radius of a circular arc tangent to that point?

Could someone help me out with this one? I was thinking of using contradiction; so say colour A occurs more often than some colour B.

But I can't come up with more than that. I haven't had a lot of practice in colouring graphs yet:l

A curve is always $1$D, otherwise it's not a curve.

It seems you're talking about a planar curve

Then, the curvature is the reciprocal of the radius of the circle tangent to the curve.

yes, that. So I am guessing being uniform curvature at all points will imply constant curvature for a planar curve?

I can show that a colour differs at most 1 from the other colours in terms of how often it is used, without using the fact that $d\mid n$

However, how can I use this divisibility property to restrict myself to the case that the colours don't differ in how often they are used?

Test: $a\_b$

$a\text_b$

$1_{H_i}$, $~1_{H\text_i}$, $~1\text_(H_i)$