Mathematics - 2017-08-17 (page 1 of 3)

12:00 AM

It's OK, we can still talk.

Well, take an 8x8 chessboard, which can easily be tiled by non-overlapping dominoes (small 2x1 rectangles). Remove two corner squares from opposite corners, leaving you with 62 squares. Can you still tile it with dominoes?

@JasperLoy

Jasper Loy

Oh, that's too hard for me to think now. =)

I am stupidata.

@AkivaWeinberger I read your profile, the last line made me LOL.

People here are mean as well, downvoting without comment and closevoting perfectly fine questions.

Maybe their high rep points make them think they are much smarter than they really are...

Simply Beautiful Art

12:36 AM

@AkivaWeinberger Giving people puzzles is not how they fall asleep usually. In fact, for me, it has the opposite affect.

Akiva Weinberger

@SimplyBeautifulArt Yeah but it makes the experience of being unable to fall asleep more pleasant

Simply Beautiful Art

xD

Akiva Weinberger

Like, if you're already gonna be awake, it's better than being awake and bored

Simply Beautiful Art

tru i guess

@JasperLoy :P At least I usually leave a lot of comments and stuff, so I hope I'm not horrible

Akiva Weinberger

12:53 AM

Desmos art

If you surround each dot with a square of just the right size, the boundary makes this large curve that travels through the whole square

Balarka Sen

3:49 AM

@Daminark m8

Typhon

4:25 AM

0

Q: How to build a literal collossus of disease?

Say you're a single celled being with the following characteristics. Able to control disease based cells around it Human-level intelligence Virtually immortal You've been around for millennia. You've "possessed" humans many times over, using their bodies to interact with the populous and what...

wassup?

Alex K Chen

6:48 AM

Anybody here ?

Leaky Nun

@AlexKChen sure

I wonder if $\displaystyle \sum_{r \in \Bbb Q} r$ is conditionally convergent. I mean, if you pair each element with its negative, then the sum would be zero, right.

Will Hunting

@LeakyNun Oh, you still haven't uploaded your pic yet.

Alex K Chen

Where I can do some mapping of the complex plane ? i.e I give it some function $f(z) $(say, $f(z) = z^3-z^2+1$), and it shows me how the complex plane is trasformed ?

Leaky Nun

@AlexKChen sounds like a good idea. let me write a desmos program about it :p

Will Hunting

@AlexKChen Do you mean a website where you key it in and it shows the output?

Alex K Chen

6:52 AM

@WillHunting Yeah, an website or a software

(Also, @LeakyNun, how to run the applets given here in Windows 8.1 with Java Version >7 ?)

I could only manage CirclePack, and the others are showing security errors :(

Leaky Nun

@AlexKChen java is mostly security errors

Will Hunting

I don't even install Java anymore.

Alex K Chen

So how to run the applets ? They look interesting.

Will Hunting

In fact, the Java plugin can't even be used on newer browsers.

Leaky Nun

@AlexKChen I have no idea

Alex K Chen

6:55 AM

Here's a tio program I wrote:

(Wait link too big, shortening)

goo.gl/skjqIw

Will Hunting

@AlexKChen By the way, there shouldn't be a space before the question mark.

Alex K Chen

:P

OK, I have found some here:

davidbau.com/archives/2013/02/10/conformal_map_viewer.html

math.ucla.edu/~tao/java/Complex.html

3d-xplormath.org/j/applets/en

Though either too slow/doesn't works because of sjssse

Tobias Kildetoft

@LeakyNun How would you even define a sum indexed by the rationals?

Leaky Nun

@TobiasKildetoft firstly biject $\Bbb N$ with $\Bbb Q$

let the bijection be $g$

then my requested sum is $\displaystyle \sum_{n \in \Bbb N} g(n)$

the definition being the limit of the partial sums

Tobias Kildetoft

@LeakyNun that will make it depend heavily on the choice of bijection

Leaky Nun

7:02 AM

@TobiasKildetoft which is what I mean when I said "conditionally convergent"

Tobias Kildetoft

@LeakyNun That is not usually what that term means

Leaky Nun

just like how $1-\dfrac12+\dfrac13-\dfrac14+\cdots$ can have any limit

@TobiasKildetoft oh, sorry

@TobiasKildetoft how would I express that concept then?

Tobias Kildetoft

@LeakyNun Not sure

Will Hunting

There is something called an unordered sum, where you can sum over an uncountable set.

But this concept appears only in very few textbooks on analysis.

Leaky Nun

desmos.com/calculator/vywxwldgp5

@AlexKChen $z_0$ is the coefficient of $z^0$, etc.

The points are from $(-n,-n)$ to $(n,n)$, where $n$ is a value you can change

Tobias Kildetoft

7:17 AM

@WillHunting Usually those are not really "sums" any longer (the usual series are close enough to not be sums themselves)

Alex K Chen

@LeakyNun Nice, thanks !

Leaky Nun

@AlexKChen I tried to make it more user-friendly, but I can't

just tell me what features you would like to have

Alex K Chen

(I should learn Desmos - can I simulate what happens here ? If I can, then from where what things should I learn to be able to code that ?)

@Le Hey that's cheating !!!

Leaky Nun

@AlexKChen Desmos is really a joke. I just made it for fun.

Alex K Chen

You're just doing stuff in Cartesian plane.

(But thanks for your efforts)

Leaky Nun

7:25 AM

desmos.com/calculator/ds65rlozi6

@AlexKChen here, more user-friendly, still only for cubic polynomials.

challenge: find out how I extracted the coefficients from the polynomial

Alex K Chen

Hehe.

Use Viete

They rheyme

Leaky Nun

no, I didn't use Vieta

Alex K Chen

Okay, but you can do. Eg f(0) gives the f[0]

Leaky Nun

@AlexKChen no, the "e" in "Viète" is supposed to be silent

@AlexKChen that isn't Vieta

Tobias Kildetoft

Seems like the latest attempt at a P vs NP proof is at least being taken seriously and is not obviously wrong

2

Leaky Nun

7:28 AM

@TobiasKildetoft is it = or $\ne$?

$\neq$

interesting

Corollary 1.

If it had been $=$ I don't expect it to have been taken nearly as seriously

Leaky Nun

@AlexKChen oops, my program is completely wrong. brb fixing it

Martin Sleziak

7:31 AM

On a nearby site: Is Norbert Blum's 2017 proof that $P \ne NP$ correct?

Leaky Nun

desmos.com/calculator/twtlmkqtfv

@AlexKChen here, see how $i$ is a fixed point of $z^2+z+1$

(it will still work for cubic polynomials)

one method to verify that it is correct for $f(z)=z^3-z^2+1$ is to verify that $1$ is a fixed point

$f(z)-z = z^3-z^2-z+1 = (z-1)(z^2-1) = (z-1)(z+1)^2$

So $-1$ and $1$ are the only fixed points

@AlexKChen look at how $1+i$ and $1-i$ all go to $-1$... so beautiful

a verification of the fundamental theorem of algebra

Tobias Kildetoft

7:53 AM

It seems like the answer cstheory.stackexchange.com/a/38812 and the comments are pointing out something that might be a fatal error in the paper. But I don't know anything about the topic

Will Hunting

8:05 AM

I don't trust SE answers to point out errors in such a paper.

Tobias Kildetoft

@WillHunting Why not?

Will Hunting

@TobiasKildetoft I think I have met enough high rep users on this site who know shit about math.

Tobias Kildetoft

How are those two statement at all related?

Will Hunting

Just talking about SE in general. I make random vague associations all the time.

Leaky Nun

@WillHunting ad hominem.

Will Hunting

8:09 AM

@LeakyNun Sorry, my Latin is limited to i.e. and e.g. LOL

Leaky Nun

Ad hominem

Ad hominem (Latin for "to the man" or "to the person"), short for argumentum ad hominem, is in which an argument is rebutted by attacking the character, motive, or other attribute of the person making the argument, or persons associated with the argument, rather than attacking the substance of the argument itself. However, its original meaning was an argument "calculated to appeal to the person addressed more than to impartial reason". Fallacious ad hominem reasoning is normally categorized as an informal fallacy, more precisely as a genetic fallacy, a subcategory of fallacies of irrelevance. However...

Will Hunting

@TobiasKildetoft But anyway, thanks for telling us. It's interesting to know. I hope someone proves the Riemann hypothesis soon. =)

Tobias Kildetoft

I am just hoping that Geordie Williamson gets a Fields medal next year.

2

Will Hunting

@LeakyNun By the way, I didn't say the answer or comments are wrong. I just said I don't trust them in general.

Tobias Kildetoft

And someone proving the Riemann hypothesis might get in the way of that if there are enough other good candidates

Will Hunting

8:11 AM

I know the idea of the Fields medal, but I think awards that are ageless make more sense.

@TobiasKildetoft So far, I have only encountered unordered sums in 3 books on analysis. =)

skullpatrol

Why are you using a movie character's name if you've decided to stop watching movies?

Will Hunting

@skullpatrol I haven't decided to stop watching movies. I just happened to stop watching them for a while.

skullpatrol

OK

Will Hunting

@LeakyNun At least your avatar doesn't blow up right now.

Leaky Nun

@WillHunting I've used my identicon

Will Hunting

8:18 AM

@LeakyNun The previous one looked terrible because it didn't work properly.

Interesting fact: Someone once wrote Qiaochu's name as Quicho. LOL.

Martin Sleziak

@WillHunting Of course, there are a few posts on this site about them, such as use of $\sum $ for uncountable indexing set or The sum of an uncountable number of positive numbers or Does uncountable summation, with a finite sum, ever occur in mathematics?

2

And many others, if you look at linked questions in the sidebar.

Will Hunting

I see, good.

Tobias Kildetoft

8:32 AM

@WillHunting I can never recall how to spell Qiaochu's name properly, and I always have to look it up if I need to

Will Hunting

@TobiasKildetoft It's actually very simple if you know Chinese. =)

Tobias Kildetoft

Yeah, I don't know any Chinese apart from a few words and symbols

Leaky Nun

袁翘楚

Tobias Kildetoft

And the few words I know I keep mixing with the Japanese words for the same

Leaky Nun

@TobiasKildetoft benefit of being Chinese? :P

Will Hunting

8:34 AM

Of course, that name is written in hanyupinyin, which is the most common form of transliteration.

skullpatrol

What? A "q" with no "u" :P

Leaky Nun

@skullpatrol it isn't a "k" sound in Chinese; it's somewhat like a "ch" sound

@WillHunting you're Chinese also?

Will Hunting

But some people have pointed out to me that hanyupinyin is not totally consistent, which I marvelled at when I realised.

Leaky Nun

@WillHunting something like "iu" for "iou"?

Will Hunting

@LeakyNun Yes, by race, not by nationality though.

Leaky Nun

8:35 AM

@WillHunting I see

Will Hunting

@LeakyNun Like they write qu without the two dots on top when they really mean the two dots on top.

Leaky Nun

@WillHunting oh, I see

those only affect j- q- x-

skullpatrol

Arabic has the "q" no "u" combo

Leaky Nun

right, the "qu" convention is from Latin

and then English started spelling /kw/ as "qu", even for native English words

Old English "cwic" > English "quick"

Old English "cwen" > English "queen"

Will Hunting

@skullpatrol Arabic has a letter ghain, where the gh is pronounced like the French and German guttural R.

Leaky Nun

8:42 AM

@skullpatrol Arabic letter "quf" and Latin letter "q" come from the same letter

skullpatrol

neat

TIL

Will Hunting

@LeakyNun Yeah, actually to be consistent it should just be zh, ch, and sh followed by the u with two dots.

Leaky Nun

@WillHunting this letter ghain and the Latin letter "O" come from the same letter

@WillHunting jqx can only be followed by -i- and -ü- while zhchshrzcs can only be followed by -nothing- and -u-

in a sense this is a complementary distribution

Will Hunting

@LeakyNun You used a linguistic term there, awesome.

Leaky Nun

@WillHunting I do know some linguistics

Will Hunting

8:50 AM

Sometimes, I don't want to sleep and I drink some coffee, but after drinking coffee, I feel tired and want to sleep.

Leaky Nun

wait, drinking coffee makes you tired?

Will Hunting

It doesn't. My sleep is just random.

@TobiasKildetoft I think it's better to hope that you yourself win it and not hope for somebody else. =)

Leaky Nun

Tobias Kildetoft

@WillHunting The difference is that I think Geordie has good shot at it. Also, I have until the one after that to get mine anyway :)

Will Hunting

9:20 AM

Just did another song today you can hear if you are bored:

Da hai

►

Leaky Nun

@WillHunting wait, you are Jasper Loy?

Will Hunting

@LeakyNun Of course I am, LOL. I thought everyone knew.

Leaky Nun

you know, "hai" means vagina in Cantonese, and then "da" looks like English "the", so

Will Hunting

@LeakyNun It means big sea in Mandarin, LOL.

Leaky Nun

@WillHunting this song was used for my form 1 class song competition

form 1 = grade 7

Will Hunting

9:24 AM

@LeakyNun Wonderful!

Martin Sleziak

9:52 AM

C.r.u.d.e chat room is intended to help with various "janitorial" task such as closing, reopening, (un)deleting, editing and improving post. Some users suggested to revive this room a bit. If you want to join these efforts, you're more than welcome there.

Leaky Nun

3

Q: Is the proof that $f(x)$ has no root correct?

Let $f: I \to \Bbb R$ be differentiable, where $I$ is an open interval in $\Bbb R$. Let $a \in I$. Let $g: \Bbb R \to \Bbb R$ be continuous. If $f(a)=1$ and $f'(x) = g(f(x)+x) f(x)$, then prove that $f(x)=0$ does not have any solutions in $I$. Is this correct? Let's say we have a $t$ that belon...

calculus proof-verification

This problem looks interesting

skullpatrol

Thanks @MartinSleziak :-)

Perhaps, we can get a mod or room owner to pin a link to the room.

Martin Sleziak

I'm not sure whether it is that important. (Let's leave the decision to mods and room owners.)

skullpatrol

Ok.

Akiva Weinberger

11:48 AM

@LeakyNun You look blue

s.harp

12:09 PM

@Akiva you look purple

Does anybody have an argument why the group action of translations on $L^1(\Bbb R^n)$ is continuous?

ie $(f,x)\mapsto f\circ \tau_x$ where $\tau_x: y\mapsto x+y$

or in general on $L^1(G)$ where $G$ is our locally compact Hausdorff abelian friend

Balarka Sen

@AkivaWeinberger plays Eiffel 65 in the background

s.harp

(I don't know if the statement is even true, I just firmly believe it is)

Totally unrelated: Has anybody looked at this thing called "Global Calculus", which to me right now appears to be using sheaves to talk about differential manifolds?

Secret

12:28 PM

So.. today in the maths society in uni, there's this weird and dangerous discussion about the infamous $-\frac{1}{12}$. One student claimed that it is possible the correct way to think about the value of $\zeta (-1)$ is that it is actually a hyperreal number of the form $-\frac{1}{12}+\infty$. He however is not sure whether there exists a representation of $\zeta (-1)$ that can produce both the transfinite and real part without the $\infty$ drowning out the real part.
This question then makes me curious about that, but then I am not very good at infinite series, especially when it is divergent

s.harp

I thought $\zeta(-1)$ is just a regular number $-1/12$

The $\zeta$ function has a unique analytic continuation to the point $-1$

its just the expression $\sum_n 1/n^{-s}$ which a priori doesnt make sense when evaluated at $s=-1$

Secret

12:48 PM

Sometimes, I don't get how analytic continuation and exponential regularisation make sense. It is known that $\zeta(1)$ is a pole, thus even if the sums $\sum_{s=1}^{\infty}se^{-sx}$ and $\sum_{s=1}^{\infty}\frac{1}{n^s}$ make sense when $s$ is in the neighbourhood of $1$ but diverge at $1$, why we can go with that and just assume $s=1$ will behave like the finite values for some neighboring $s \neq 1$.

It feels like as if it is analogous to you have a single variable function with a single point jump discontniuity, that while the limit exists, and so is the function value, but it is disco…

anon

> and so is the function value

well, zeta(1) is not defined a prior, humans define functions

to understand analytic continuation, treat the geometric series 1/(1-x)=1+x+x^2+... first

the power series clearly only converges when |x|<1 but 1/(1-x) is defined everywhere except x=1

s.harp

@Secret The $\zeta$ function did not fall from the sky, it was defined in a certain way. In the region with real part of $s$ greater than $1$ it is supposed to be the sum over $1/n^s$. It turns out that there exists exactly one analytic function on all of $\Bbb C$ minus some points that is equal to this on on that region. Since analyticity is super super nice this must be the nicest function that agrees with that above definition when the domains coiincide, so we identify the two ideas

Secret

But won't there will be some risks of trading away points that should blow up to infinity with a nice analytic function even if it is the unique one, to me it still feels like they are technically different?

Leaky Nun

@Secret we already know what points will blow up to infinity

Secret

same argument about the series representation of the geometric series, I cannot see any way we can justify to force a diverging series to equal to its analytic continuation as that will be effectly equating the concept of infinity to many different real values at once?

s.harp

12:57 PM

If a context appears where you have a function that agrees with $\zeta$ on real part $>1$ but differs from the analytic continuation, then you will use this function and call it something like $\hat\zeta$. It has a relation to the sum that is probably fantastic and interesting, but right now the word $\zeta$ function refers to the analytic continuation of that sum

Leaky Nun

@Secret of course a diverging series is not equal to its analytic continuation.

after all, it isn't us who claimed that $\zeta(-1) = -\frac1{12}+\infty$

anon

we do not "force the diverging series to equal its analytic continuation." the geometric series already equals 1/(1-x) on its domain of convergence, outside of which the series diverges (hence has no values) but 1/(1-x) exists

hence the word continuation

Secret

Right, now I can see how it is treated carefully

hmm, re s.harp's point, I wonder if an existence proof on $\hat \zeta$ can be found, to be checked...

Secret

Another interesting question came across in the club is the notion of symmetric group with infinitely many elements with different cardinalities.

While I know $S_{\infty}$ exists, it was the first time that I heard about trying to do something like $S_{\aleph_{\alpha}}$

anon

"infinite sets" or "sets with infinitely many elements," not "infinite elements"

2

Leaky Nun

1:04 PM

@Secret symmetric group is simply the set of permutations

and a permutation is simply a bijective function

Secret

Ah right, so it wil mean the set under consideration will be a subset of $\aleph_{\alpha}^{\aleph_{\alpha}}$

s.harp

> “Divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever... Yet for the most part, the results [from using them] are valid... I am looking for the reason, a most interesting problem.” Niels Henrik Abel 1828

Abel then died in 1829

Leaky Nun

@Secret subset of, not in.

s.harp

@Secret there exist many such, what you need for a definition to be sensible is to have a problem or scenario that warrants the use of such a function

Secret

Well afaik, they are doign a purely abstract algebraic treatment on it, so I am pretty sure they have not thoguht of specific conditions on the underlying set except for its cardinality yet

It still surprised me why $S_{\infty}$ is often discussed without reference to its cardinality, perhaps countable and uncountable versions of it are not very different group theorically

s.harp

1:11 PM

You can define $S_\infty :=\varinjlim_n S_n$ ($"=" \bigcup_n S_n$), or $S_{|K|}:= Bij(K)$.

the first definition make sense without any reference to an underlying set

and if you unpack the definition it becomes the group of bijections on $\Bbb N$ such that only finitely many elements are not kept fixed

Secret

Interesting

s.harp

as such it is a countable group :)

btw I just found this: math.ucr.edu/home/baez/history.pdf (history of n-categories)

like everything Baez writes it seems to be amazing

Secret

1:32 PM

[Joke]

Infinity likes to eat but not all who likes to eat are infinities

Martin Sleziak

2:24 PM

From the calculus and analysis chat room - in case somebody would like to take this:

in Calculus and analysis, 14 mins ago, by user8469759

can anyone explain to me why the banach steinhaus theorem is important?

Josh Wonser

2:35 PM

Hey, quick question- what is the matrix notation |A| called?
It's not absolute value, and I believe it notes |A| = (A*A)^1/2

s.harp

I call that the absolute value @JoshWonser

Huy

@JoshWonser: it probably denotes a matrix norm but what you wrote in the second line is not one. need more context to find out which

s.harp

or modulus

Bratteli and Robinson refer to it as absolute value or modulus

Josh Wonser

Thanks guys, the context is this is to calculate a Dice Coefficient, and assumed it meant matrices, but it meant sets. So that notation is the count of items in a set.
"where |X| and |Y| are the numbers of elements in the two samples"

https://en.wikipedia.org/wiki/S%C3%B8rensen%E2%80%93Dice_coefficient

Also, matrix norms are two bars according to my googling ||x||

user193319

3:02 PM

Okay. An element in a cyclic group is a generator of that group if and only if its order and the groups order are relatively prime. How could this be so? Wouldn't this contradict Lagrange's theorem since the order of an element must divide the order of the group?

Leaky Nun

@user193319 where did you see the first statement?

An element in a cyclic group is a generator of that group if and only if its order and the group's order are the same

user193319

I read it here: proofwiki.org/wiki/…

Oh! Wait!

$k$ isn't the order of $a$ in the link, right?

Huy

right

user193319

Ah! I see now. Thanks!

Leaky Nun

@user193319 simple proof that $\Bbb R$ isn't cyclic?

user193319

3:09 PM

@LeakyNun I am not sure I understand what you are asking. $\Bbb{R}$ is not cyclic.

Leaky Nun

@user193319 I'm asking you to prove it

user193319

@LeakyNun I would probably prove it by noting that if $\Bbb{R}$ were cyclic, then any subgroup would be cyclic, which would include $\Bbb{Q}$. But $\Bbb{Q}$ is not cyclic.

Leaky Nun

@user193319 there's a much simpler proof :P

user193319

@LeakyNun Hmm...Are $1$ and $\infty$ relatively prime?

Leaky Nun

@user193319 $\infty$ isn't a number

user193319

3:15 PM

Hmm...Well, if it were, then $1$ would have to be a generator of $\Bbb{R}$, which is the contradiction I was hoping for.

SteamyRoot

wut o.O

Leaky Nun

@user193319 no, the link you just cited only works for finite cyclic groups

user193319

Ah! You are right. That would have been a nice proof though...

I am not sure how to prove that at the moment. I'll have to think about it.

Leaky Nun

@user193319 and that isn't even what the theorem says

even if it worked for infinite cyclic groups, this still isn't what it's saying

Leaky Nun

3:38 PM

Ten Illegal Things To Do In London

►

Kirill

3:49 PM

@TobiasKildetoft, I remember the question correct now, it was a statement "is that true that for $A \in\mathbb{C}^{n \times n}$: if $(A-I_n)^n=0$, then 1 is the only eigenvalue." So, without $\lambda$ before the identity matrix.

Huy

@Kirill: what did you answer?

Kirill

@Huy it's correct, hi @Huy

Huy

hi. why? @Kirill

Kirill

@Huy I couldn't find any counter-example. But to prove as well.

for $n=1$ it is true. I think about how to show this for $n+1$. (or to prove that it is not true for $n+1$)

Ted Shifrin

It's correct, @Kirill. What do you know about the minimal polynomial of $A$?

Kirill

4:00 PM

@TedShifrin three things, no, four

Ted Shifrin

For this case :)

Kirill

@TedShifrin 1) it is the smallest polynomial $\chi$ such that $\chi(A)=0$

2) it is a divisor of $\chi_A$.

3) one can define it with the length of Jordan blocks, - as I did for another exercise.

4) one can do that with Eigenspaces.

Ted Shifrin

So what do you know about this particular minimal poly?

Kirill

@TedShifrin I do not even know how it looks like (now). This difference can be a characteristic matrix for the eigenvalue $1$, but it can be also just a term, without any connection to eigenvalues. Hint?

Ted Shifrin

You said $A$ satisfies $(t-1)^n=0$.

Kirill

4:12 PM

@TedShifrin sorry, what is $t$ and what is this term? (I remember from the videos that you use it for $\chi_A$)

Ted Shifrin

$p(t) = (t-1)^n$

Kirill

what is $p(t)$? (besides the computation of the polynomial $p$ in the point $t$)?

Leaky Nun

@Kirill characteristic polynomial

Ted Shifrin

$p(A)=0$

Kirill

hi @LeakyNun, $p(t)$ is a computation for $t$.

Leaky Nun

4:21 PM

@Kirill $p$ is the characteristic polynomial of the matrix

$p:t \mapsto (t-1)^n$ if you want to be pedantic

Will Hunting

Pedantic is a word I seldom hear.

Huy

That's because you don't listen to me.

Will Hunting

I listened to you last night, nice playing. =)

Kirill

so, $p(A)=0$ is that what is given, if we use $p$ for this. But it is not a char. pol., @TedShifrin

Leaky Nun

@Kirill the fact that it is is Cayley-Hamilton theorem

(as long as $p$ has degree $n$)

Kirill

4:27 PM

@LeakyNun Cayley just says: if $\chi_A$, then $\chi_A(A)=0$.

Leaky Nun

@Kirill hmm.. interesting

I refer you to @Ted :)

Kirill

@LeakyNun sorry for pinging you yesterday, I thought you played the piano, all the time.

Leaky Nun

@Kirill ??

Kirill

@LeakyNun Huy sent me a link where he plays the piano yesterday. I thought you had sent that and spoke to you about that.

Leaky Nun

@Kirill whatever

Akiva Weinberger

4:35 PM

It's kinda cool that the center of mass of a triangle is the center of mass of its vertices

(and of a tetrahedron)

Not true for arbitrary polygons, I think

Kirill

@LeakyNun check the Cayley-Hamilton theorem in German and English at Wikipedia- these are two "different"Cayles and Hamiltons. They state the same, but somehow totally differ.

Ted Shifrin

Not characteristic poly a priori!

Semiclassical

This is pretty good: theatlantic.com/science/archive/2017/08/…

Secret

[Chemistry] I currently experiencing an annoying problem where my molecule (pics later) rotate in some weird back forth back forth fashion while it should be rotating 3 deg for each step

Akiva Weinberger

What quadrilaterals have their center of mass equal the center of mass of their vertices?

Clearly parallelograms. Are there any others?

(Parallelograms because they have that 180 degree rotational symmetry about their centers of mass)

Semiclassical

4:53 PM

@AkivaWeinberger By 'center of mass of a quadrilateral' I presume you're taking the quadrilateral to have a uniform mass density.

Akiva Weinberger

Yes.

Semiclassical

kk

Akiva Weinberger

It's also called the barycenter IIRC, and perhaps some others I can't remember

Secret

Ted Shifrin

Hi DogAteMy and Semiclassic

Secret

5:02 PM

Any thoughts on why it is doing this weird flip flopping behaviour which result it to rotate by 3 deg only every even steps and flip nearly 180 deg every odd -> even or even -> odd step?

(matrix shown is what is coded, and the unit vector is pointing out of paper)

Ted Shifrin

You said cm of vertices.

Oh, DogAteMy, think affine geometry.

Akiva Weinberger

@TedShifrin That shows that parallelograms work, it doesn't show that nothing else does. At least, not easily

Secret

5:27 PM

Neverminded, solved, it is because python default is radians and all those numbers there are radians

20

Q: Python: converting radians to degrees

In the python.org math library, I could only find math.cos(x), with cos/sin/tan/acos/asin/atan. This returns the answer in radians. How can I get the answer in degrees? Here's my code: import math x = math.cos(1) y = x * 180 / math.pi print y 30.9570417874 My calculator, on deg, gives me: c...

python math

Secret

desmos.com/calculator/osp9poxfuu

The flip flopping behaviour of cos(3x) for integer x is summarised here. Not sure if it can be expressed as a commutator of something as it seemed to be a constant precession

BeginningMath

can someone help me with a very simple boolean algebra question? how can you more simplify b'cd'+bc'd' ?

i've been studying for hours for the exam and my mind is not working well lol

Leaky Nun

@BeginningMath b'cd' + bc'd' = d'(b'c+bc') = d'(b xor c)

BeginningMath

is it possible without xor?

Waiting

@Secret Related to the previous discussion, I also developed new series that bear my name and that are prepared for publication. The process of publishing anything is terribly slow.

Secret

5:33 PM

nice

BeginningMath

i need it tofind some don't cares in a function

Secret

Typical, just need to be patient

Waiting

@Secret For my last article I waited for almost a year which is a record (to me) for waiting to publish.

Secret

We chemist typical need 1-3 years, and that does not include the referees sending back to us for corrections

Tobias Kildetoft

@Waiting Was that for the peer review part or in total from submission to publication?

Waiting

5:36 PM

@TobiasKildetoft From submission to publication.

BeginningMath

i got $f(w,x,y,z)=\sum(0,1,2,4,5,10,12)$ with the boolean simplification of a'c'+c'd'+b'd' and i need to verify if (i)15 might be a don't care (ii)8 must be a don't care

Tobias Kildetoft

@Waiting Ok, so not terrible by math standards at least

Waiting

@Secret God, that is terribly long.

Tobias Kildetoft

I have a paper that was put on arXiv in May last year and which is still under review

BeginningMath

so i've drawn my KMAP and trying to obtain the given boolean expression to check those

Waiting

5:37 PM

@TobiasKildetoft Sometimes it happened faster like 5, 6 months.

SteamyRoot

1-2 years is standard for my subfield of math, apparently.

Secret

The Cos precession phenomenon every multiple of 6 radians will be investigated soon, I expect there should be a comutator that describe the precession

Tobias Kildetoft

@Waiting that is pretty fast for math

Waiting

@TobiasKildetoft It depends on the journal policy and (pretty much) of course on the content of the article.

Will Hunting

@Waiting Hi there, I changed my username =D

Waiting

5:46 PM

@WillHunting That's a nice one.

Tobias Kildetoft

@Waiting I don't think I have seen any journal whose policies would help with the long review times

Will Hunting

@Waiting Now our usernames both start with W- and end with -ting, LOL.

Ted Shifrin

DogAteMy — plenty of other shapes work.

Waiting

@TobiasKildetoft I saw somewhere a paper by Mathematical Association of America with timing at mathematical journals around the world. The timing for the main publishing steps might be pretty different (in terms of a difference of many months).

Will Hunting

@TedShifrin I thought you went travelling overseas.

Perturbative

5:50 PM

Hey everyone!

Will Hunting

@Waiting So far I only read one paper by MAA, which proves the Jordan curve theorem using graph theoretic methods.

Waiting

@TobiasKildetoft If I'm not wrong I saw it in a post on academia.stackexchange.com. Not able to find now the post.

Tobias Kildetoft

@Waiting Hmm, I only think I have seen such a list from the AMS, but there the times certain did vary a lot

Waiting

@WillHunting Are you still affiliated to any university by chance? Lots of nice journals which are not open access unfortunately.

user193319

I read somewhere that a function is Lipschitz continuous if and only if the first derivative is bounded. Does the domain of the function matter?

Will Hunting

5:56 PM

@Waiting No, of course not. I left my university after finishing my degree long ago. I don't even have access to any math library in real life.

@Waiting My knowledge is limited to simple arithmetic mostly. I don't need to read journals. =D

Waiting

@WillHunting I wanted to read some paper published in Ramanujan journal like 1, 2 years ago. Couldn't do it.

Will Hunting

@Waiting I see. Well, maybe you can find it on some Russian servers? I dunno, LOL.

Waiting

@WillHunting Eventually I managed to get that paper.

Will Hunting

@Waiting You do now about Genesis Library right? I use it mostly for books though, not papers.

Akiva Weinberger

@TedShifrin Plenty of other quadrilaterals work?

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$Mathematics$ Mathematics