00:00 - 20:0020:00 - 00:00

12:00 AM
exhaust manifolds are more tiring, though, than intake manifolds.

(Just like numbers are interesting, to give an example.)

They're more exhausting?

Zach, that makes the humor totally unsubtle and much less clever.
@MATHASKER: I'm about to leave. I assume you'll have it all figured out.

@TedShifrin if iknow that the angle is 30 degrees how do i convert it to a bearing angle?

What angle is 30º?

12:02 AM
the third side was a 30 degree

That makes no sense.

EXACTLY.
Just sayin'

i found the third side to be 627.7 miles long and then to find the third angle as i had 3 sides i did law of cosines

You don't need law of cosines because you have a right triangle.
Just use the angles in the picture to figure out the angle between the third side (at the right vertex) and due north.
It's just like what you've been doing.
You know all three angles in the triangle.

but the 150 degrees its slanted, can i just turn it into like a straight line
I just know the middle one

12:04 AM
One is 90º-60º10'. One is 90º. So the last is ...
You know every angle in the picture if you just use the kind of reasoning we've already used.

Remember you have to use all the pieces of information!
OK, I'm outta here.

ok
thanks for helping :D
was it supposed to be 60 degrees 10 mins?

0

Here is the Cayley graph of an associated polynomial $f = xy + z$. As you can see the graph is based at any one of its elements and has generating set red arrow = $(2,3,5)(11,13,17)$, and blue arrows change and are shown in the image. See, I can't seem to fit the green arrows into the group be...

Prime relationships form a group is all I'm sayin

12:46 AM

1:10 AM
@PhysicsGuy .

Shared by the AoPS Facebook page.
The TL;DR is that a retired professor solved a classical conjecture in probability theory (something that some people spent decades of their lives on), using apparently not-too-complicated ideas.
Title: "A Retiree Discovers an Elusive Math Proof—And Nobody Notices"
(The TL;DR part of that is a combination of people being skeptical of proposed proofs of big conjectures, and him being bad at publishing the proof)
> Known as the Gaussian correlation inequality (GCI), the conjecture originated in the 1950s, was posed in its most elegant form in 1972 and has held mathematicians in its thrall ever since. “I know of people who worked on it for 40 years,” said Donald Richards, a statistician at Pennsylvania State University. “I myself worked on it for 30 years.”
The statement of the conjecture itself looks to be pretty simple, as well.

A Long-Sought Proof, Found and Almost Lost. @AkivaWeinberger same article. So that proof is with people or lost now?
2

> Original story reprinted with permission from Quanta Magazine, an editorially independent publication of the Simons Foundation whose mission is to enhance public understanding of science by covering research developments and trends in mathematics and the physical and life sciences. - Wired
@Fawad Almost lost, but with people

7 hours later…
8:29 AM
Suppose we have the following graph of a function:
We see that the function is concave, right?

9:13 AM
[Superrandom]
The northern cardinal (Cardinalis cardinalis) is a North American bird in the genus Cardinalis; it is also known colloquially as the redbird or common cardinal. It can be found in southern Canada, through the eastern United States from Maine to Texas and south through Mexico. It is found in woodlands, gardens, shrublands, and swamps. The northern cardinal is a mid-sized songbird with a body length of 21–23 cm (8.3–9.1 in). It has a distinctive crest on the head and a mask on the face which is black in the male and gray in the female. The male is a vibrant red, while the female is a dull reddish...

9:25 AM
@LeilaHatami Sorry for pinging you about completely unrelated issue, but I thought that it might be reasonable to let you know that there was a discussion on meta about a tag-synonym you have recently suggested: Do we really want (discrete-mathematics) $\to$ (combinatorics) synonym?
2
If there is some further discussion needed about this you can comment (or answer) on meta or we can discuss this in the tagging chat room.
Wow, I am a bit surprised how quickly the above message got starred.

Anonymous
10:07 AM
Is there a list of mathematical topic from easiest to hardest?

10:33 AM
How does (1-x)^2 > 0 follow from 2(1+x^2)>(1+x)^2?

@MartinSleziak Thank you dear Martin. I will read it.

Ok, I see. It is simple.

11:03 AM
@LeilaHatami Thanks for the response. If needed, we can discuss this further in the Tagging chat room. In fact, I have already left there a few messages about direction of synonym. (Making two tags synonymous is not a symmetric operation.)

11:42 AM
@MartinSleziak Dear martin, it's not necessary, I have already convinced.

@LeilaHatami Ok. Still, in case you plan to suggest some other synonyms in the future, I guess both things are useful to know: 1) It is not symmetric, so at least in some cases the decision which of them will be master tag matters. 2) It might be useful to post on meta first before suggesting the synonym - at least in the cases of large tags where the creation of the synonym would influence many questions.

12:29 PM
@SimplyBeautifulArt hi, how are you

12:41 PM
If i have 98 miles per 1 hr = w * 10 in can i divide the 10 out and get 98 miles / 10 inc.?

Hello@MartinSleziak , you byusy now a days?

Yes.

ok,I see without you the rooms are empty
:)

lol
Without some proper context, that sounds quite funny. However, I am glad to give some other people an opportunity to talk in various chat rooms. (And I noticed that calculus chat room was quite active even without me.)

yes that was!
ok,all the best for your work.
If you will be available sometimes we can discuss ring theory ?

12:47 PM
I'd guess that there are probably quite a few romantic ballads about empty room without you.
@BAYMAX Why not simply posting your question here on in algebra chat room (or both) and see whether somebody will respond?
You know, I am not the only person on this site...

The question was As Zach speeds away from the police, who are in hot pursuit of him, he calculates his speed to be 98 mph. If he is
sporting 20-inch (diameter) wheels on his ride, find the speed of the wheel in rpm as he eludes the po-po.
So 98 mph is the v rigjt?

Yes Will post when I start reading , lol :) , ok.
@MartinSleziak

BTW nice background music @MartinSleziak
wait how do we do this link thing like "empty room with you " , instead of pasting the link ?

12:56 PM

@BAYMAX For any message in chat, you can view the source: chat.stackexchange.com/messages/36512419/history (Just go to transcript, click on the little arrow and then click on "history")

thanks @MartinSleziak @skillpatrol

np pal :-)

@skillpatrol how would i convert the inches to minutes i know i can multiply 1 rev/2pi but idk what to do after that

@MATHASKER Start by making a sketch.

1:08 PM
I did draw but I'm confused how would i turn the inches to minutes
Maybe convert the miles to inches, cancel the inches out

do you know what rpm means?

1:34 PM
Can someone suggest something similar to these stunning results?
4

I recently find some stunning visualizations... I prefer to share them all: $5)$ Mean inequalities [from Proof without words] $4)$ Streographic projection [by H.Segerman] $3)$ Farey-Ford Tessellation in non-euclidean geometry [by F.Bonahon] $2)$ Steiner Porism [by Wikipedia] $1)$ P...

1:46 PM
@Alucard as I always say, I am doing largely fine :-)

@MATHASKER have a look at this

2:17 PM
Hi guys
out of curiosity
if a positive sequence a_n -> 1 (strictly positive) does this happen if and only if log(a_n) -> 0?
I've tried to prove it by myself using the epsilon delta definitions
but I can't work out all the details
for each epsilon there an m such that -epsilon < a_n - 1 < epsilon for all n >= m
which is equivalent to
1 - espilon < a_n < 1 + epsilon iff log(1 - epsilon) < log(a_n) < log(1+epsilon)
assuming natural logs I have log(1 + epsilon) < epsilon, but I don't have log(1 - epsilon) > -epsilon

2:29 PM
@SimplyBeautifulArt that is very good to hear. I wish you that your luck never depletes :D

3:22 PM
Hello everyone, I am a highschool student who can handle basic algebra and trigonometry, geometry, calculus pretty well but find it hard to do discrete mathematics

4

discrete mathematics, mostly combinatorics, binomial theorem application
I eat a lot of chocolate already

@samjoe can you explain a little bit further where yur problem is? also, follow Mats advice

:)
The problem is that I find it hard to solve multinomial theorem problems, and binomial problems (not the basic ones), also I have hard time handling a complete new series

@Alucard I meant large, as in FAIL

3:27 PM
Interesting: something of negative mass will feel an upwards force due to gravity, but still accelerate downwards.
Taking $g\approx10\rm\frac ms$, the force $F$ will be $-mg\hat\imath$, which would be upwards. The acceleration will be $\frac Fm=-g\hat\imath$, which would be downwards.

@AkivaWeinberger but how do you explain negative mass? couldn't this be just another force in the opposite direction of gravity?

The force would be upwards, yes. But the acceleration would be downwards.
It would also have negative kinetic energy.
Pretty sure that such an object is very impossible.

If one is the object, probably not very good for the stomach...

4:04 PM
Hello!!
We have the following relations:

If we double i then s will also be doubled.
If d gets halved, then n be multiplied by 4.
If s gets multiplied by 4, then d will be doubled.

Can we get from these relations a formula?

You can get a system from those descriptions

What system? @user8469759
Could you give me a hint?

I'm thinking
You have
s = f(i) s.t. 2s = f(2i)
n = g(d) s.t. 4n = g(d/4)
d = h(s) s.t. 2d = h(4s)
sorry
4n = g(d/2)
you probably have to use those relationshion to extrapolates the formula

And how could we solve that system without knowing the functions f, g, h? @user8469759

4:17 PM
well as a guess it seems to me that s is a linear function of i, n is somekind of reciprocal
because
s = f(i) and 2s = f(2i) implies s = f(2i)/2 = f(i) then 2f(i) = f(2i)
so I would just say s = i...
sorry s = constant*i
while for n = g(d) and 4n = g(d/4) implies n = g(d/4)/4 = g(d) then 4g(d) = g(d/4)
let me do it again
4n = g(d/2) => n = g(d/2)/4 = g(d) => 4g(d) = g(d/2)
you can solve these functional equations
do you have any conditions on i,d,n,s ?
are they integers?

No. But I want to check also if the following are correct:

- d*i^2= s^2*n
-d^2*i=s*n
-d^2*i^2=s^2/n
- d^2*i^2=s*n
- d^2*i=s/n
How could we check these conditions?

as I said you have f(i) = f(2i)/2
you can iterate and you have
f(i) = f(2i)/2 = f(2*2*i)/2*2 = f(2*2*2*i)/2*2*2 = ... = f(2^k*i)/2^k
but anyway if you have those five
you'd probably better just veryfing the condition
or maybe not...
let me think

Holy sh.. :D

I could have sworn I saw that paper months ago
ah, it was submitted a year ago. it just got revised.

4:49 PM
Hi guys. Say we have $o(\vert t\vert)$ as $t\to0$. Now my teacher says that \begin{aligned}\lim_{t\to0}\frac{o(\vert t\vert)}{t}=0 \end{aligned}. But first of all, we know that \begin{aligned}\lim_{t\to0}\frac{o(\vert t\vert)}{\vert t\vert}=0 \end{aligned}. How do we know for sure that we can get rid of the absolute values? What if the function indeed goes to zero in the right-hand limit, but doesn't go to zero in the left-hand limit?
this is where my question comes from^
never mind, I asked on the main site!

@ShaVuklia do you know what $o(|t|)$ means?
means it represents an $f$ such that $f/|t|\to0$
surely if $f/|t|\to0$ then $f/t\to0$

oh right, let me see

$f/|t|\to0$ implies $|f/|t||\to0$ i.e. $|f/t|\to0$ implies $f/t\to0$

oh of course!!
but that means that $o(\vert t\vert)=o(t)$ right?

yes

5:01 PM
cool, thanks!

Hello friends
I just posted about an article
I quoted relevant information
But I'm not sure if I should lnik the article or not
0

I'm looking at an algorithm and its implementation. In the article: $$\Delta_t(a) = \frac{\partial V(\pi)}{\partial \pi(a)} . \eta$$ where $V$ is a standard value function using a function $Q$ to represent the expected value of each action $a$, and a probability $\pi$ for playing each action ...

@BlueMoon93 Why you have doubt that this might not be good to link the article?

5:17 PM
I dont hang out often in MathExchange

5:33 PM
Just found out next semester the topology seminar is about fibre bundles, looking forward to it :)

Why is it true that $o(\Vert t\vec u\Vert)=o(\vert t\vert)\cdot\Vert\vec u\Vert$? As far as I can rewrite, we have:
$$\lim_{t\to0}\frac{o(\Vert t\vec u\Vert)}{\Vert t\vec u\Vert}=\lim_{t\to0}\frac{1}{\Vert\vec u\Vert}\cdot\frac{o(\Vert t\vec u\Vert)}{\vert t\vert}=0,$$
so why don't we say \begin{align}o(\Vert t\vec u\Vert)=\frac{1}{\Vert\vec u\Vert} o(\vert t\vert)\end{align}?
I know it doesn't matter if we multiply by a constant $\Vert\vec u\Vert$, but I'm just confused why they multiply and don't divide by the constant. In the end it doesn't matter of course, because the constant disa

6:12 PM
Hi, how can i find the angle between s1 and s2? s1: $xy^2z+3x/7+z^4-9=0 s2: 2ln(xy-z^2)+1-sqrt(x^2-3)=0$

Hi chat

Hi @Astyx

Hi @Astyx , do you know the answer to my question?

Salut @Astyx. Hi @Alessandro.

Hi @Ted

6:21 PM
@user379685: What does your question mean? Can you tell me in words?

I'm given two equations and need to find the angle between them.

What does that mean?

Hi @Ted

heya @PVAL

So conjugation $\Bbb C\to\Bbb C$ can't be written as a polynomial because of holomorphicity (is that a word?) reasons, what about conjugation $\Bbb H\to\Bbb H$?

6:28 PM
Same argument should apply.
Restrict to various copies of $\Bbb C\subset\Bbb H$?

Is every function $\Bbb H \to \Bbb H$ which is differentiable (in terms of $\Bbb H$) expressible as a power series (with powers taken in $\Bbb H$)?

I agree it can't be done when restricted to a copy of $\Bbb C$ (because conjugation in $\Bbb H$ agrees with the complex one there) but I'm not convinced we don't lose anything when restricting

Non-commutative power series? WTF?

Well the powers are commutative.

Yeah, but when you put in coefficients, ugh ...

6:31 PM
they don't commute with the coefficents though

Yeah but I'm pretty sure right mult. by some element
is equivalent to left mult. by some other element.

It means that for a point where the functions intersect we take a crosssection and measure the angle between two curves?
@TedShifrin?

@user379685: The functions do not intersect. The surfaces intersect. And you take the angle between their normal vectors.
But did they give you a particular point? I have no idea how you'd find if those surfaces intersect and/or where.

Hi @Ted, @PVAL, @Alessandro

Hi @Balarka

6:33 PM
Hi @Balarka

Okay even that question I don't know about.
Is every linear map $\Bbb H \to \Bbb H$ coming from a right multiplacation

equivalent to one coming from a left multiplacation?
certainly its true if your scalar is invertible.

Well, can we solve that for multiplication by, say, $i$ (or $j$ or $k$)?
This is a skew-field. Aren't all nonzero elements invertible?

right

6:35 PM
they are

@TedShifrin and what should i do if i had a point

But I don't see how it works for $i$. $f(x)=xi$. Plug in obvious things for $x$.
@user379685: Find the gradient vectors of the functions at that point.
Find the angle between them.

hm hm hm what should i do now

@TedShifrin ok thanks

@PVAL what do you mean with "differentiable in $\Bbb H$"? We saw a few different and incompatible ways to define that

6:36 PM
Left mult i sends i to -1 which forces the induced right mult to be i
@Ted
so it can't work

That's what I'm saying, @PVAL :P

foliation or differential geometry or ...

or un-sleep, @Balarka?

higher category theory @Balarka

@Alessandro I mean some difference quotient goes to 0 with the limit taken in H
I don' t know if it depends on the difference quotient you use.

6:37 PM
I don't even know how to define a difference quotient.
Left derivative versus right derivative?

or like f(z)-f(z_0)/z-z_0
thats all I mean

How do you divide, silly :) ?

According to my professor, which stated this as a fact without proof, this is way too strong and gives only affine transformation as $\Bbb H$-differentiable (and there are differences I forgot by doing it on the left or on the right)

(f(z)-f(z_0))*(z-z_0)^-1
or the other one
whichever one works.

6:39 PM
Hence my saying left- and right- derivatives?
What a god-awful mess.

why are we trying to do calculus on H

Hi @Alessandro @Ted @user37

@BalarkaSen We had a couple of (non mandatory) lectures on $\Bbb H$, held by a professor who's doing research in those topics

ah ok

I think I still believe $\Bbb H$ is a subset of $\Bbb C$.
thats at least what I was taught.

6:42 PM
If it's the upper half-plane, I guess it is. Problem solved.

Are you getting better @Ted ?

Not tremendously; I postponed the oral surgery. How're you doing, @Astyx?

Working mostly

Well, that's not a bad thing ...

No it isn't
It's just a bit depressing considering the amazing weather we have

6:48 PM
Well, take an hour every day for a good walk!

Well I go walk to the university and back home, so that's settled :p

I meant a recreational walk, but never mind :)
I love walking in Paris :)

I need to get up and walk for 5-10 minutes every hour and a half or so.
or I will be in pain.

Inside Paris ? I find it stressing more than anything ..

Ah. Well, @PVAL, stop chatting and go walk!

6:52 PM
It's dark now

@Ted I still don't get what Thurston is saying though.

Well, @Astyx, I have favorite places I like to meander ... but you're trying to get somewhere fast, and I'm rarely doing that there.
Huh? @PVAL ... am I supposed to know what this refers to?

I've never done that.

What are these places ?

6:54 PM
5th and 6th, parts of the 3rd, I guess ... and even out near the Bois de Bologne ...

Interesting, Balarka. Lots of famous allusions in there. I'll look carefully later.

I don't really know Paris that much anyway

Yeah I didn't get it because of the allusions which are unfamiliar to me.

6:56 PM
I like the style a lot

Are you from Paris or just studying there?

I live in a town just next south of Paris
Not far from the IHES

I commuted to the École Polytechnique for two weeks back in summer 1980 :P
Nice hike up that hill every day.

Well I live approx 1 km away from there :p
And I hope to go there next year

Cool :)
OK, lunch time for me. See you all a bit later.

7:00 PM
Bon appétit ! :)

See ya

buon appetito
@Astyx what's the IHES?

Hi

c'est coté non

7:04 PM
Un peu

tu es élèves là-bas ?

Il y a une bonne dizaine de médailles Fields qui y sont :p

un peu lol
énormément alors

Non je ne suis pas du tout à ce niveau là

Grothendieck y était non ?

7:05 PM
Je suis en prépa pas loin

ah ok :)

Oui il y était

@BalarkaSen Do you know the infinite sphere being contractible?

Oui oui, francais :)

@JeSuis I know that's true, yes.

7:15 PM
@BalarkaSen Ok
@Alucard en prépa aussi ?

Did you have a question? :)

i saw in the internet that when integrating over a ball $B(0,R)$ we have $\int_0^R \int_{-\sqrt(R \ ^ 2 -x ^ 2}^{\sqrt(R \ ^ 2 -x ^ 2}1 dy dx$
my question is : why $0$ to $R$ and not $-R$ to $R$ ?
it seems to me like we getting half the area, and im wrong :P but i dont see why

Yeah, I have $S^{\infty}\setminus \{b_1\}$ is homeomorphic to $H^{\infty}$, $S^{\infty}$ is a weak retract of $\Bbb{R}^{\infty}$ and there is a homotopy from $S^{\infty}$ vers $S^{\infty}\cap H^{\infty}$ where $H^{\infty}$ is the hyperplane of equation $x_1=0$. I was wondering if I can deduce from that the contractibility of the sphere $S^{\infty}$.

Let $f\colon\mathbb R^n\setminus\{0\}\to\mathbb R$ be continuous and homogeneous of degree $\alpha>1$. Show that $f$ is differentiable at $\vec 0$ with derivative $\vec 0$.
I know that $\lim_{\vec x\to0}f(\vec x)=0$, so it follows that $f(\vec 0)=0$ (due to the continuity of $f$). So I basically need to show that:
$$f(\vec h)=o(\Vert\vec h\Vert).$$
So in other words, we need:
$$\lim_{\vec h\to0}\frac{f(\vec h)}{\Vert\vec h\Vert}=0.$$
I don't know how I can proceed from here? I feel like I've used all the properties of $f$ already. Any hints?

@JeSuis Interesting, you are working with the unit sphere in $\Bbb R^\infty$, not the CW complex.
You have that $S^\infty$ is a retract of $\Bbb R^\infty$? Doesn't that already say it's contractible (because R^infty is contractible)?

7:22 PM
I didn't know the CW complex "case". Oopssss $\Bbb{R}^{\infty}\setminus {0}$

Ok.

@Sha You have that $f(tu) = t^\alpha f(u)$ for any unitary vector $u$
(Any norm you want)

yes, but I have $\vec h$, and not $t\vec h$ or something?

@JeSuis Let me try to translate to the "CW complex case" I have in mind (maybe @MikeM would want to verify if it's right) because that's what I know.

I don't get what you mean @ShaVuklia

7:24 PM
I need to show that limit, right?
because then I'm done

Pick a sequence of subspaces of R^\infty: $V_1 \subset V_2 \subset V_3 \subset \cdots$ - where $V_k$ is a $k$-dimensional subspace of $\Bbb R^\infty$ - which exhausts $\Bbb R^\infty$.

but the limit uses $f(\vec h)$ in the numerator
I don't know how I can rewrite that using the homogeneity of $f$

Oh, rewrite $f(h)$ as $f(||h||{h\over ||h||})$

oh right!
of course:')
thanks! I hope I will be able to solve it from there on

You need to use a little topology at some point

7:26 PM
$V_k \cap S^\infty \subset S^\infty$ is a standard $k$-sphere in $S^\infty$ (aka a linear subsphere). Let's call this $S^k$.

probably the fact that continuous functions are bounded on bounded sets?

That's not true

@ShaVuklia that's false, you want compact sets

@BalarkaSen but I think it's same as CW complex ? isn't it? $\mathbb{R}^\infty$ mean the vector space consisting of real tuples $(v_1, v_2, v_3, \ldots)$ such that all but finitely many $v_i$ are zero equipped with the coherent topology (so that $U \subseteq \mathbb{R}^\infty$ is open iff $U \cap \mathbb{R}^n \subseteq \mathbb{R}^n$ is open in the standard Euclidean topology on $\mathbb{R}^n)$

oh right. well $S^1$ is bounded and closed, so it's compact

7:27 PM
For instance $x\mapsto {1\over x}$ on $]0,1]$

$f:(0,1)\to\Bbb R$, sending $x\mapsto\frac1x$ surely is continuous

Because we're in finite dimension @ShaVuklia

lol we have a lot of fantasy with those examples

Then you have an exhaustion of $S^\infty$ by linear subspheres $S^1 \subset S^2 \subset S^3 \subset \cdots$ from the series of subspaces we chose before.

hahaha :P
let me see
oh alright, I see
handy example to keep in mind!

7:29 PM
yes @BalarkaSen

I think it's more or less clear than $S^{k+1} - S^k$ is a union of two disks (the upper and lower hemispheres respectively). So this should give a CW-structure on $S^\infty$ given by two 2-cells in each dimension (two 1-disks to S^0 give S^1, two 2-disks to S^1 gives S^2 etc etc).
Do you agree with this?

Ok I see your point, agree don't know now, but I am following :p

@JeSuis It's not a CW complex per se. Justifying that a space is a CW complex requires that it admits a decomposition into cells.

@BalarkaSen do you see link beetwen what I want to prove and the property that if $x$ and $y$ are in the same path component of $R^{\infty}$ then there are also on same path component of $R^{n}$ for $n$ big enough ?

In any case, if what I did right, you have this CW structure on $S^\infty$ (do you follow that CW structure?). Can you use that to show it's contractible?

7:34 PM
( I am asking that because we proved this in the problem which is strange that i didn't use it yet)

@JeSuis I don't get it. $\Bbb R^\infty$ is path-connected.
It was only one path component.
So any two points are on the same path component. That hypothesis is void.

arf, yeah the question was for a hausdorff topological space with the coherent topology. But here it's a vector space, lol.

ah ok.
Anyway you should try and understand the CW-complex structure on $S^\infty$. That's the key to solution.

Ok! Thanks!

I am confident about what I did but this makes me wonder how to do this for unit sphere of any infinite dimensional uh normed? space
It works here because you have the coherent topology, but...

7:48 PM
don't know, when I googled I found something call Lipschitz contractible

Thundercat likes money, so she gets money

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