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Kevin Driscoll
Kevin Driscoll
00:38
@TedShifrin It is sad that mathed majors as a group seem so reluctant to to y'know like explore math. Kinda reveals why its difficult for so many teachers to communicate the spirit of mathematical learning
 
2 hours later…
leslie townes
leslie townes
02:09
disaster has been averted. we have maybe 2 months worth of wet food. all of olivia's favorite varieties.
William Sun
William Sun
02:23
In a commutative ring if x=ay and y=ax does it imply a is a unit?
hyper-neutrino
hyper-neutrino
consider a = -1
unless i'm misunderstanding what a unit is
William Sun
William Sun
unit is an element a such that ab=1 for some b
hyper-neutrino
hyper-neutrino
oh
Kevin Driscoll
Kevin Driscoll
I believe you are, -1 is a unit
hyper-neutrino
hyper-neutrino
my memory of this is rusty, rip
would 0=0x0, 0=0x0 satisfy your problem here or are there conditions on x and y
Kevin Driscoll
Kevin Driscoll
02:26
I mean maybe this is a dumb observation, but $x=ay$, $y=ax$ implies $y = a^2 y$ so $a^2 = 1$, I guess, and so $a$ is its own inverse?
William Sun
William Sun
y=a^2 y does not imply a^2=1
And yeah f, g are not 0…
Kevin Driscoll
Kevin Driscoll
What's the counter-example?
leslie townes
leslie townes
commutative rings can have nonzero things that multiply to be zero
y = a^2 y tells you that y and 1 - a^2, are at least such things, even if they are nonzero
silly example, consider ordered pairs of elements of your favorite unital commutative ring, multiplied and added componentwise, if a = y = x = (1,0) then x = ay and y = ax but a is not a unit
in this example the image of a is a unit in the quotient of the ring by the ideal generated by 1 - x, maybe there is something like that you can do in general with your a, y, x
William Sun
William Sun
We ask for the existence of a unit a
(1,0)*(1,1)=(1,0)
Oh ok u r answering his question nvm
Kevin Driscoll
Kevin Driscoll
I think that was also supposed to be a counterexample to the claim
leslie townes
leslie townes
02:36
yes. a is not a unit in my example
William Sun
William Sun
I’ve tested on all subrings of C
And all subrings of Set(X, \Bbb R)
X topological space \Bbb R the reals
leslie townes
leslie townes
in a field, kevin's argument works and deduces from y = a^2 y that either y = 0 (which you seem to be excluding?) or that a = +/- 1 is a unit
William Sun
William Sun
And obviously looking at symmetric difference and intersection from a power set doesn’t work since A=BC and C=BA implies A=C
leslie townes
leslie townes
or in any ring without zero divisors
Kevin Driscoll
Kevin Driscoll
yea, I gotta admit I'm a little salty I forgot something as silly as 2*2=0mod4
leslie townes
leslie townes
02:48
Z/4 is an interesting example because there aren't elements a for which 1 - a^2 is both nonzero and a zero divisor. it's either 0 (a = 1 or 3) or 1 (a = 0 or 2)
 
1 hour later…
SAJW
SAJW
03:52
I don't understand why 10 points can always be covered by 10 non-overlapping unit-circles
Why does the 45 configuration-technique not apply to ten?
ibb.co/WkNsLLm the 3 upper points can't all be covered by nonoverlapping circles
if the red line is the radius of the unit circle
then just place the rest of the points somewhere, such that the down point is the middle of a slightly bigger circle
Where am I wrong?
leslie townes
leslie townes
SAJW, where does this claim originate?
does every circle have to have a point in it?
this isn't inherent in the usual meaning of 'covered,' or maybe it is
you could cover those points in your example with fewer circles, which wouldn't break a statement that allowed for some circles to enclose no points
google led me to math.stackexchange.com/questions/6737/… which suggests that 'at most 10 non-overlapping unit circles' is the intended meaning, not 'exactly 10 non-overlapping unit circles, each with one point in it'
SAJW
SAJW
yes, you can use fewer
Can you always cover 10 points with 10 equally sized (non-overlapping) coins?
►
just watched it and I don't understand the proof at all
leslie townes
leslie townes
04:08
i haven't clicked through, but it wouldn't surprise me if a post on se were better than a video for something like this
SAJW
SAJW
Probably!
But that's what youtube decided to throw in my suggestions :D
...Still sceptical, but I guess it's right haha
Didn't even know you can proof something with probability. Lucky day
Koro
Koro
04:24
is there an example of a function f defined on [a,b] such that $f'$ changes sign in passing through c in (a,b) but still c is not a point of extremum of f?
For example: $f(x\ne 0):=x^2(2+\sin 1/x) \land f(0):=0$ doesn't retain sign in any left or right nbd. of zero so there are some a<0 and b>0 such that $f'(a)>0$ and $f'(b)<0$ and if we consider c to be 0 then c is a point of extremum.
robjohn
robjohn
@Koro That would mean the function is increasing on one side of $c$ and decreasing on the other side of $c$. That seems like a local extremum.
It may not be a global extremum.
leslie townes
leslie townes
koro, a lot of books would not say that "f' changes sign at c" unless both f'(c) = 0 and f' is nonzero in some punctured neighborhood of c
robjohn
robjohn
That is definitely how I read that.
leslie townes
leslie townes
so if f' changes sign infinitely often in any such neighborhood it would not be said to 'change sign' at c
robjohn
robjohn
$x\sin(1/x)$
leslie townes
leslie townes
04:32
and once you're in the land of f' being nonzero a little to the left of c, and nonzero of opposite sign to the right of c, you're in the land where robjohn's comment applies. just MVT or whatever to deduce the increasing/decreasing behavior and hence a local max or min
Koro
Koro
Is there a function f on [a,b] such that f'(a)>0 and f'(b)<0 but there is no c in (a,b) such that f'(c)=0 and " f' retains sign in some left and some right nbd. of c (that is f' is -ve in some left nbd of c and +ve on some right nbd of c) " and c is not a point of extremum?
I'm just making my question more precise. I understand the confusion caused. Leslie and Robjohn.
Have I expressed my question well now?
robjohn
robjohn
@Koro Then $f'$ is not continuous on $[a,b]$
Koro
Koro
Yes! That's what I thought. f' does not have to be continuous.
robjohn
robjohn
@Koro $|x|$
but it does have a local extremum
Changing sign would mean that left of $c$ $f(x)$ is one sign and right of $c$ is has the other sign.
That is what it means to me.
Koro
Koro
Background of the question: I'm trying to understand GH Hardy's proof of $\lim_{x\to \infty} (f(x)+f'(x))=L\implies \lim f(x)=L \land \lim f'(x)=0$. I believe that the argument for the case when f' changes signs for large x is somewhat shaky. Here is professor GH Hardy's proof of the result math.stackexchange.com/questions/840957/… in the accepted answer.
SAJW
SAJW
04:44
Ok, I figured my real problem out. I don't understand the probabilistic method.
robjohn
robjohn
@Koro A derivative needs to obey the IVT, however.
If the derivative exists at all points.
Koro
Koro
@robjohn In my comment just before your comment "Then f' is not continuous ...", I have tried to fix "changing sign" and tried to state what I meant by that.
Yes, f' must follow IVT (Darboux's rule) but doesn't have to be continuous. I hope my confusion/question regarding the last part of GH Hardy's proof is clear now.
copper.hat
copper.hat
@Koro you might look at Darboux's theorem
Koro
Koro
I have already stated that in my last comment. That gives you IVT of f'.
copper.hat
copper.hat
05:02
extremum or local extremum?
Koro
Koro
05:13
I was thinking local extremum.
copper.hat
copper.hat
05:36
@Koro did you look at the proof math.stackexchange.com/a/407698/27978
Ted Shifrin
Ted Shifrin
@leslietownes Disaster indeed averted. Mazltov!
leslie townes
leslie townes
they even had the big bag of her favorite brand of kitty litter.
Koro
Koro
@copper.hat yes, I have. But I want to understand GH Hardy's proof. That seems so intuitive. :)
robjohn
robjohn
@leslietownes in a field, most elements are units....
leslie townes
leslie townes
i improved that comment in the next line
1
Q: Elementary Proof of No Odd Perfect Numbers

Raj RainaI came across this proof on the Arxiv that there are no odd perfect numbers. It is elementary and easy to follow and looks correct to me? Of course there must be a mistake there somewhere but I am not catching it - could someone else take a look?

number-theory perfect-numbers
hmm. (1) posted to math.GM (2) on April 1st... it's not looking good.
Koro
Koro
06:08
I have posted my question here math.stackexchange.com/questions/4322642/….
Please let me know what I'm missing. Thanks.
copper.hat
copper.hat
06:34
@Koro i added an answer. no doubt with some mistakes.
Koro
Koro
07:23
Thanks a lot @copper. :)
 
1 hour later…
PM 2Ring
PM 2Ring
08:24
Thanks @robjohn for your answer to that lunar precession question astronomy.stackexchange.com/a/47674 It was nice to get the bounty on my answer there, I put a fair bit of work into creating the 3D anim code. Sadly, it's hard to score much rep on mathematical answers on Astronomy. In contrast, this answer on lunar libration astronomy.stackexchange.com/a/47584/16685 took a lot less work, but scored a lot more rep, thanks to the HNQ. :)
love_sodam
love_sodam
08:39
@copper.hat This is really the first time I've seen to use L'Hopital's rule to prove something rather than calculating limits.
 
5 hours later…
Wolgwang
Wolgwang
13:17
21 hours ago, by Wolgwang
Can a square be divided into 5 squares(may not be congruent) ?
 
1 hour later…
robjohn
robjohn
14:33
@PM2Ring I had worked quite a bit on that answer, to get things right and the animations the way I wanted. I think I started while the bounty was still active, but I did not finish until yesterday. Your answer deserved the bounty; after all, the bounty was started to be given to your answer.
PM 2Ring
PM 2Ring
15:15
@robjohn I originally intended to post a GIF anim, too. The Sage 3D plotter can easily save single frames as PNG, but its "save as anim" functionality is buggy. It produces very low quality, badly dithered frames, and the scale isn't even consistent. :( I guess I could've got it to save single frames, so I could download them & produce a good anim on my own computer, but I wanted to do everything on the Sage server, using my phone. :)
I must admit it was very gratifying to get that bounty. I haven't won many bounties, and that was the 1st time a bounty was specifically created to reward my answer.
@Wolgwang I don't think so. See mathworld.wolfram.com/MrsPerkinssQuilt.html
Wolgwang
Wolgwang
@PM2Ring Does this answer say "Yes"?
PM 2Ring
PM 2Ring
15:55
@Wolgwang Sure, but that's a slightly different problem to what I thought you meant. ;) There are various diagrams here: math.stackexchange.com/q/96776/207316
Wolgwang
Wolgwang
@PM2Ring Hmm I think by 'glue' in that MO question, the OP mean that pieces can be glued together to form squares (?)
PM 2Ring
PM 2Ring
Note that Bill Cook's solution is closely related to Pythagorean tilings. en.wikipedia.org/wiki/Pythagorean_tiling
Wolgwang
Wolgwang
> Is it impossible to divide a square into (may not be congruent) n squares, if n = 5?
This was the exact question.
PM 2Ring
PM 2Ring
@Wolgwang I don't quite understand that MO answer. But once you have those 3 pieces, you need to make further cuts to create the pieces for the 5 squares.
Jam
Jam
$u(x)=\frac{1}{4\pi} \int_{\mathbb{R^3} \frac{e^{-|x-y|}}{|x-y|}f(y)dy$
i want to differentiatiate this function 2 times i know it is a convolution and the derivative goes to $f(y)$ but what would $\frac{\partial f}{\partial x_i}$ it be
$ u(x)=\frac{1}{4 \pi} \int_{\mathbb{R^3} \frac{e^{-|x-y|}}{|x-y|}f(y)dy$
PM 2Ring
PM 2Ring
16:02
@Wolgwang Yes, and that question doesn't say that it's permitted to rearrange the pieces. ;) Which is why I linked to the article about Mrs Perkins's Quilts.
Wolgwang
Wolgwang
Ohk Thanks :)
Jam
Jam
$u(x)=\frac{1}{4 \pi} \int_{\mathbb{R^3} \frac{e^{-|x-y|}}{|x-y|}f(y)dy$
why latex does not compile?
Odestheory12
Odestheory12
int_{ delete that {
or close it
Jam
Jam
$u(x)=\frac{1}{4\pi} \int_\mathbb{R^3} \frac{e^{-|x-y|}}{|x-y|}f(y)dy $
it is a convolution what is the partial derivative. I know it goes to the partial of $f(y)$ but what is $\partial f(y) $ with respect to x
thanks
it will be parti y_i right?
PM 2Ring
PM 2Ring
Coincidentally, I've been playing with Pythagorean tilings in SVG, for the last day or two, trying to find an efficient way to make them. Eg,
sagecell.sagemath.org/…
Ultimately, I'd like to animate it in pure SVG, but that might be tricky, until trig functions in SVG get browser support. Here's a GIF one I did several years ago. IIRC, I used Cairo in Python.
user image
bus busman
bus busman
16:27
I have to take the logarithm of a sum of exponentials : $\log (\sum_{i=1}^M \exp(-C_i)$ the problem is I am doing this on matlab and $C_i\in(0, E+06)$ i.e they can be very very large. This causes underflow where matlab reads $exp(-C_i)$ as zero. I would like to first apply the logarithm and then the exponential but since its a sum of exponentials I cant see how to do this
robjohn
robjohn
16:43
@Jam try substituting $y\mapsto x-y$ first.
Jam
Jam
good idea
i need to calculate $ \Delta u$
can you help me?
Xander Henderson
Xander Henderson
@PM2Ring That hurts my eyes.
Wolgwang
Wolgwang
@PM2Ring Fascinating!
Hello! Prof. X
leslie townes
leslie townes
17:01
you could sell children's clothing, particularly pajamas, with those patterns. there's a lot of money to be made in loud patterns on children's clothing.
PM 2Ring
PM 2Ring
@XanderHenderson Sorry about that. I must admit that it is a bit intense, especially with those colours. Here's another version, with some explanation, newtonexcelbach.com/2012/02/13/animated-pythagoras and here are some static tilings (including 3D versions on the Riemann sphere) with more sedate colour schemes. newtonexcelbach.com/2008/06/30/pythagoras-penrose-and-pov-ray
Xander Henderson
Xander Henderson
I am just going to post a few more messages to get that off the top of my screen.
It is legitimately making me feel some vertigo.
:(
PM 2Ring
PM 2Ring
@leslietownes I guess that pattern would make good anti-camouflage. ;) FWIW, in the late 1990s, my mum & a couple of my sisters had a small company that manufactured kid's clothing. But they never made anything quite that loud. :)
leslie townes
leslie townes
they should have dared to have bigger dreams.
my daughter would love that
PM 2Ring
PM 2Ring
@XanderHenderson Feel free to use your mod powers to edit it so it doesn't one-box.
Xander Henderson
Xander Henderson
17:10
@PM2Ring It is mostly gone now.
Two or three more messages, and will be all gone.
Bye bye spinny squares of doom!
aaksaksyk
aaksaksyk
math.stackexchange.com/q/4322004/1000002 my question maths SE with tThe reason provided for cosing the question is “ This question is not about mathematics”. Even though it’s clearly about maths. It’s about proportionality. But even if it’s not I was not explained properly as to how it’s not mathematical. So even if it’s not mathematical someone could provide an answer and explain how it’s not mathematical. Can Someone please help me re open my question
Sorry for the typos
my question was closed
and I am trying to re open it
Xander Henderson
Xander Henderson
@aaksaksyk The question is, basically, if $a \propto b$ and $a \propto c$, then $a \propto bc$, yes?
I think that the physics-y stuff at the start distracted some people. On the other hand, I would not be surprised to discover that there is already a question on the site which addresses this.
PM 2Ring
PM 2Ring
@leslietownes They did ok for a while, but it was a lot of work relative to the amount of profit. They did a lot of the sewing themselves, with a few employees, and they sold the garments via a party plan scheme. So they closed down the business, but mum kept the company name registered, just in case...
Xander Henderson
Xander Henderson
Oh.. indeed. David found a number of closely related questions.
Relevant and/or related: How does one combine proportionality? AND Proportionality AND If $A$ is proportional to $B$ when $C$ is constant and $A$ is proportional to $C$ when $B$ is constant then how is $A$ is proportional to $BC$? — Dave L. Renfro yesterday
Arguably, the question should have been closed as a duplicate of one of those.
Ted Shifrin
Ted Shifrin
17:26
@XanderHenderson Your eyes or your brain?
Xander Henderson
Xander Henderson
@TedShifrin Initially, the area just behind my left eye. And then I started getting vertigo.
Govind75
Govind75
I encountered the following question today: Let $V$ be a vector space on $\mathbb{C}$ with two inner products $(,)_1$ and $(,)_2$. Given $(v,w)_1 = 0 \iff (v,w)_2 = 0$ for all $v,w \in V$. Show $(v,w)_2 = c(v,w)_1$, where $c$ is a positive real
Xander Henderson
Xander Henderson
@Govind75 When you encountered the question, did you give it a biscuit, pat it on the head, and say "Nice doggy"?
Govind75
Govind75
I got that $c = \frac{(v,v)_1}{(v,v)_2}$, but surely this constant is dependent on the vectors I am choosing
Xander Henderson
Xander Henderson
(You really need to do that, or they can get mean.)
Govind75
Govind75
17:32
@XanderHenderson Oops I forgot to do that, I'll keep it in mind for next time
leslie townes
leslie townes
koro did this exercise the other day. i think there are copies of it on math.se
Govind75
Govind75
yeah I found some solutions
But surely my constant is dependent on the vectors I am taking the inner product of?
Xander Henderson
Xander Henderson
@Govind75 Don't call me Shirley. Also, what makes you so sure?
leslie townes
leslie townes
govind: it turns out not to be, but from one perspective, that's essentially the entire exercise
2
Q: Orthogonality in different inner products

holaGiven a vector space V where two vectors v and u are orthogonal with respect to a given inner product. Will the same vectors be orthogonal with respect to all the other inner products?

linear-algebra vector-spaces vectors inner-products orthogonality
Xander Henderson
Xander Henderson
Isn't the interesting thing about the exercise the fact that the constant doesn't depend on the vectors?
Adil Mohammed
Adil Mohammed
17:34
@XanderHenderson Airplane movie reference!
Govind75
Govind75
Ohhhh I see
Cheers <3
amWhy
amWhy
@AdilMohammed Seems to to date the man! Actually, another Xanderism. A reference to Xanderisms is in progress! ;D
PM 2Ring
PM 2Ring
@XanderHenderson On a related note, this is currently on the HNQ: How is "Dogs dogs fight fight" grammatical?
Adil Mohammed
Adil Mohammed
@amWhy Thanks, I made a mental note of movies that needs a reboot/reshoot in 2021 and Airplanes was on top of it
robjohn
robjohn
@Govind75 That wouldn't be much of a constant, it seems to me.
Besides, it looks as if your fraction is upside-down
amWhy
amWhy
17:41
@PM2Ring That's terrible. (OPINIONATED). Is that a post from EL&U? It seems a tad troll-ish. But that's up to the users there.
Govind75
Govind75
@robjohn Oh yeah
PM 2Ring
PM 2Ring
@AdilMohammed What's our vector, Victor?
robjohn
robjohn
@PM2Ring Ah, Scraps is a boy dog...
amWhy
amWhy
@robjohn What are your pets' names? I know you have a kitty.
PM 2Ring
PM 2Ring
@amWhy Yes, it's from EL&U. And the comments are currently locked. ;)
Xander Henderson
Xander Henderson
17:43
@robjohn Maybe it is a variable constant?
A vari-stant, if you will?
A constable?
amWhy
amWhy
@PM2Ring Hah! That's an excellent site.
@XanderHenderson "variable-constant means constable" Xanderism # 549! :D
PM 2Ring
PM 2Ring
@amWhy I must admit I mostly only visit it from the HNQ. One of the prominent regulars is Peter Shor (inventor of Shor's algorithm for factorising numbers on a quantum computer). He posts a lot more on EL&U than he does on Physics.SE or Math.SE.
amWhy
amWhy
@PM2Ring Not surprising. quite a handful of mathematicians/scientists have very could mastery of the English language, written.
Typos plaguing my keyboard (and me) recently. ^^^ meant "good mastery of written English".
PM 2Ring
PM 2Ring
On a related note:
Muphry's law
Muphry's law is an adage that states: "If you write anything criticizing editing or proofreading, there will be a fault of some kind in what you have written." The name is a deliberate misspelling of "Murphy's law". Names for variations on the principle have also been coined, usually in the context of online communication, including: Umhoefer's or Umhöfer's rule: "Articles on writing are themselves badly written." Named after editor Joseph A. Umhoefer.: 357  Skitt's law: "Any post correcting an error in another post will contain at least one error itself." Named after Skitt, a contributor to...
Aka the law of recursive pedantry. ;)
amWhy
amWhy
17:59
@PM2Ring Love it!!
I'm guessing the biggest battle in EL&U would be between the prescriptive and the descriptive. I'd toss a bet in, but I do not follow the site. It creeps up in many fields: "This is what should be" vs. "This is the way it is". Oversimplification, no doubt.
PM 2Ring
PM 2Ring
18:17
There were some good discussions on prescriptivism vs descriptivism on the old xkcd forums. Sadly, the site went offline after a data breach and never returned, although it's archived on the Wayback Machine.
leslie townes
leslie townes
you can't say DECIMATE unless you mean exactly one in ten things were destroyed!!!! THAT'S WHAT DECIMATE MEANS!!! points wildly at chalkboard full of latin ahhhhhhh!
amwhy: you mean between the descriptive and people who are wrong on the internet? :D
user21820
user21820
@leslietownes What if the only ones on the internet are people who are wrong? =P
leslie townes
leslie townes
nah, couldn't happen.
user21820
user21820
@leslietownes For all you know, it might be descriptive of the internet.
But if so, then we have a paradox. ^
leslie townes
leslie townes
i only like descriptivists who don't like themselves.
amWhy
amWhy
18:25
@leslietownes I think it varies by discipline. I mean social science would go no where, if not for descriptivists. I think the best is achieved when precriptivists dance with descriptivists, work together, biases aside. Maybe even procreate! ;)
user21820
user21820
Most people are probably either taught to be prescriptivist (my teacher said ...) or taught to be descriptivist (the corpus says ...). And then there are the few who are descriptivists on weekdays and prescriptivists on weekends. =P
amWhy
amWhy
@user21820 They need to mingle more!! ;D
leslie townes
leslie townes
weirdly the last person who found fault with something i said was a social scientist. i referred to a person's "cohorts," meaning slightly suspicious affiliates, partners in crime. i was informed that "cohort" was only a collective noun for a group of people, like an entering class in a graduate school. there was some detour through latin and what it meant in the roman army, or some such.
all i gathered from that was that this social scientist had never heard the word until grad school, in a social science context.
nobody learned anything.
user21820
user21820
@leslietownes Lol. I would have used "cahoots". =D
leslie townes
leslie townes
i think it's possible to do prescriptivism right, but it shouldn't involve detouring into what a word may have meant at one time in another language. weakens the argument a little and suggests that a string of characters has to be the same thing no matter where or when it is found. this is not necessary for prescriptivism to work.
amWhy
amWhy
18:32
@leslietownes Agreed, @leslietownes
leslie townes
leslie townes
i think it was augustus demorgan, or one of his contemporaries. i was reading something they'd written of their university education.
apparently a popular thing to do would be to find some odd turn of phrase from one of the classic scholars in greek or latin, and use it in coursework, and then wait for the instructor to pounce on it.
then you'd say, if this is inappropriate, well, so is virgil, or whoever.
amWhy
amWhy
@leslietownes Hah! But much of social sciences research is designed to help us learn how people actually respond, say to a woman being publicly raped, in the middle of the day, with many many people present.
leslie townes
leslie townes
and show the source.
copper.hat
copper.hat
that is disturbing
leslie townes
leslie townes
that got dark pretty quickly.
copper.hat
copper.hat
18:37
i'm in a delicate mood
amWhy
amWhy
Or more recently, how people on a subway pretended nothing was wrong, failed to assist, etc., when one passenger was sexually assaulting another passenger. I'm just saying that we need to know, how in fact, people behave, vote, etc., before any prescriptive folk preach.
copper.hat
copper.hat
my folk theory is divided responsibility
PM 2Ring
PM 2Ring
@leslietownes That's called the etymological fallacy. Meanings drift and adapt, over space & time. But I must admit that I do get a little annoyed when a word or phrase with a fairly specific meaning becomes ambiguous, unless there are adequate replacement terms that can be used when you need to be precise & unambiguous. Eg, how "literally" is now often used to mean "figuratively". We had some heated discussions about that one on xkcd.
copper.hat
copper.hat
i told my kids if they need help in a crowded situation to ask a specific person not just say help
leslie townes
leslie townes
well, the person who instructed me on 'cohorts' did not do this kind of work. her main work seemed to be bothering people at parties. i do know someone in a psychology department who studies that kind of stuff.
copper.hat
copper.hat
18:38
irregardless
amWhy
amWhy
@copper.hat I'm just saying we need to know what reality, sometimes, before prescribing remedies.
copper.hat
copper.hat
its like literally true
@amWhy yup
leslie townes
leslie townes
some people go into academia because they like correcting people. math profs get a bad rap for this but i think it's actually worse in other fields.
not that math students don't do this, but the worst of them tend not to become professors.
copper.hat
copper.hat
i like wilde's "i am not young enough to know everything"
leslie townes
leslie townes
PM: i literally died when you wrote that
PM 2Ring
PM 2Ring
18:40
If someone's a shady character, then there's a good chance that they habitually associate with other shady characters, so both meanings of "cohort" apply.
copper.hat
copper.hat
i yearn for the days when i could be prescriptive
amWhy
amWhy
@copper.hat That's smart. The problem is the phenomenon descriptive social scientists termed "diffusion of responsibility". People are more likely to respond when they know it is up to them to help. Else, everyone is waiting for someone else to take action.
@copper.hat I do too.
copper.hat
copper.hat
yes, i recognise that bahaviour in myself, unfortuately
i used to engage much more in such situations, in the back of my mind runs the thought "maybe someone else can step up today".
PM 2Ring
PM 2Ring
On a related note, I've noticed in recent years that "notorious" is being used as a synonym of "notable". When I was a kid, "notorious" was only used to refer to disreputable people, but kids these days will use it to refer to celebrities that they admire, without intending to imply any negative connotations.
amWhy
amWhy
I think descriptivism in social science is not studied to blame people, but to understand what might explain it better, so we are better able to interrupt such patterns.
copper.hat
copper.hat
18:44
you mean conor mcgregor, of course
amWhy
amWhy
@PM2Ring Agreed!
@copper.hat Were you asking me?
copper.hat
copper.hat
well, i think it is a perfectly rational behaviour, if against the principles my mother raised me with
@amWhy i'm just spouting...
oh,no that was areponse to PM
mcgregor called himself the notorious
amWhy
amWhy
@copper.hat Thanks. This is where the prescriptivist in me comes out: Notorious \neq notable!!!
copper.hat
copper.hat
unfortunately (for me, at least) language evolves.
amWhy
amWhy
@PM2Ring We need to teach our youth.... !!!
copper.hat
copper.hat
18:48
even in more subtle ways. i just moved phone and am trying to decipher a plethora of minute status symbols.
amWhy
amWhy
@copper.hat I agree; I was just saying how quickly describing and prescribing can flip.
copper.hat
copper.hat
y
i think the rate of change of such things is beyond where cognitive decline has left me
amWhy
amWhy
Okay, we need another word to replace notorious. Maybe EVIL will do the trick? (half joking). But we must take care not to put "evil" at risk of evolving to "good". Wow, this is tiring work!
copper.hat
copper.hat
it is a losing battle, i think :-)
PM 2Ring
PM 2Ring
That "evil" thing has already happened with "bad" and "sick".
amWhy
amWhy
18:54
@PM2Ring True.
PM 2Ring
PM 2Ring
eartha kitt, i want to be evil
amWhy
amWhy
I merely propose the question: *Is it inevitable that even the most idealistic youth, ultimately, invariably, conclude, at some point in their life, "nothing will change; it's a losing battle." ? I refuse to believe it's "inevitable, or invariably true". But descriptively, that's the case.
copper.hat
copper.hat
well, things will change, but not to the ideal model they have.
their presumption of correctness is what needs to change :-)
amWhy
amWhy
@copper.hat I'm good with that. (your first sentence). Evolution of ideas, often takes longer than one's lifetime.
copper.hat
copper.hat
i think having the ability to simultaneously have conflicting models in one's head is important if not quite the consistent ideal we would like.
i have difficulty with the latter
for example, i am a us & and irish citizen simultaneously.
at some point that would have been inconceivable for me.
then i had a daughter and the possibility of being separated overrode.
(would have been the same for a son, of course).
leslie townes
leslie townes
19:03
divided loyalties: the copper.hat story
amWhy
amWhy
@copper.hat I think I understand.
copper.hat
copper.hat
of course, if i waited long enough it would have been moot.
Akiva Weinberger
Akiva Weinberger
Super weird that a Turing machine $T$ can halt in one mode of PA and go on forever in another
leslie townes
leslie townes
it becomes less weird if you realize that there's no such thing as a turing machine
for more on ultrafinitism, consult my podcast, leslie speaks
Akiva Weinberger
Akiva Weinberger
This is literally halting problem plus compactness, but it's odd in the same way that "It's consistent with PA that PA is inconsistent" is odd
@leslietownes There is no number greater than 6
Think you have ten fingers? No
leslie townes
leslie townes
19:06
no, it's 24. 24 is the largest number
amWhy
amWhy
@copper.hat I worry about the trajectory of Greta Thunberg, and the massive energy of the Parkland highschool seniors. It would be wonderful to couple the energy and fuel of youth with mentors with more experience.
Oops, I've likely taken up too much space here, but it's been lovely, the exchange!
copper.hat
copper.hat
i think it harks back to something you said earlier
data needs to be gathered first
Akiva Weinberger
Akiva Weinberger
Fun fact: there exists a Turing machine $T$ such that, for any function $f:\Bbb N\to\Bbb N$, there's a model $M$ of PA in which the machine $T$ computes the function $f$
(It never halts in the standard model, but by the above property always halts in these models)
copper.hat
copper.hat
for example, we are heavily dependent on oil, etc, and a move to improve certainly appropriate. however it needs to be staged, not just drop & run. that perspective only comes with experience.
Akiva Weinberger
Akiva Weinberger
You can also replace PA with ZFC or whatever
amWhy
amWhy
19:14
@copper.hat Oh, I agree. But I think we could do a better job of harnessing the energy of youth, to help them and us, find ways to accomplish that. So much energy is spent by both young idealists (and only a small portion of youth are idealists), and their "experienced counterparts" dismissing each other. I'd like us to recognized the value of both, so the young can mature into experience, if they are valued by the experienced.
copper.hat
copper.hat
i think the conflict is inevitable, unfortunately.
amWhy
amWhy
@copper.hat But the conflict undermines the goals of both.
copper.hat
copper.hat
it appears to, but i presume there is some greater evolutionary truth that i cannot see...
LearningCHelpMeV2
LearningCHelpMeV2
Does anyone know where I can go to learn about factorising polynomials? I don't mean simple polynomials in one variable. But things such as $a^3+b^3+c^3-3abc$ or $pq(p-q) + qr(q-r) + rp(r-p)$
amWhy
amWhy
And "inevitable" should be laid to rest as a word. The only things that are inevitable are the earth, and live as we know it, will come to an end, and each of us will die before then. The only relevant concerns are "and how do we proceed, knowing that our actions will speed up, or deter, the end of life on our planet.
For Most else, use of the word "inevitable" signals the speakers resignation.
@copper.hat Claiming "x" is inevitable, invariably gives license to do "x": racism, sexism, religious wars, etc.,; ignoring our youth, and dismissing them. Ignoring poverty, because "the poor will always be with us." All of such ideas, allow only some to live with a clear conscience, only in their own minds.
copper.hat
copper.hat
19:24
@amWhy as i mentioned above, i'm in a delicate mood :-)
Akiva Weinberger
Akiva Weinberger
@LearningCHelpMeV2 I don't know specifically, but (and these are super broad) ring theory and algebraic geometry may be up your alley
or
Factorization of polynomials
In mathematics and computer algebra, factorization of polynomials or polynomial factorization expresses a polynomial with coefficients in a given field or in the integers as the product of irreducible factors with coefficients in the same domain. Polynomial factorization is one of the fundamental components of computer algebra systems. The first polynomial factorization algorithm was published by Theodor von Schubert in 1793. Leopold Kronecker rediscovered Schubert's algorithm in 1882 and extended it to multivariate polynomials and coefficients in an algebraic extension. But most of the knowledge...
amWhy
amWhy
@copper.hat I'm sorry. I'll end now. Maybe some other day, down the road? Thanks, though, You've made a lot of excellent points!
Akiva Weinberger
Akiva Weinberger
This seems semirelevant also
Cantor–Zassenhaus algorithm
In computational algebra, the Cantor–Zassenhaus algorithm is a method for factoring polynomials over finite fields (also called Galois fields). The algorithm consists mainly of exponentiation and polynomial GCD computations. It was invented by David G. Cantor and Hans Zassenhaus in 1981. It is arguably the dominant algorithm for solving the problem, having replaced the earlier Berlekamp's algorithm of 1967. It is currently implemented in many computer algebra systems. == Overview == === Background === The Cantor–Zassenhaus algorithm takes as input a squarefree polynomial f...
copper.hat
copper.hat
@amWhy no need to be. i enjoy discussion.
Akiva Weinberger
Akiva Weinberger
Semirelevant because it seems that it's for one-variable polynomials over finite fields
but maybe you could use it by choosing a random large prime, taking your polynomial mod it, and plugging in other random numbers for all the variables except one
or something, I dunno
leslie townes
leslie townes
19:39
a lot of 'contest math' seems to involve symmetric polynomials and more general symmetric functions of relatively low 'complexity.' these are sometimes amenable to techniques that do not have general application.
in generality, i don't think factorization of multivariable polynomials is anything but hard
AG would definitely be one place to look for general stuff
Ted Shifrin
Ted Shifrin
It also depends on what underlying scalars we allow. But generically things will be irreducible.
leslie townes
leslie townes
uh oh, ted's here
Ted Shifrin
Ted Shifrin
AG is generally not going to help with a concrete question like this. In this case, maybe it can. You have a cubic curve in $\Bbb P^2$ and you want to know if it's irreducible. If we can use calculus to show it's a smooth curve (the three partial derivatives do not have any simultaneous zeroes away from $(0,0,0)$), then the curve must be irreducible.
So: here's the assignment for DogAteMy and leslie. Does the system $$x^2-yz=0, y^2-xz=0, z^2-xy=0$$ have any nontrivial solutions?
Hint: The field we're working over is really going to matter here.
leslie townes
leslie townes
you can take x=y=z=t for any nonzero t. or if w^3 = -1 and w isn't -1, x = t, y = -wt, and z = w^2 t, t nonzero.
i guess that second bit depends on the field
Ted Shifrin
Ted Shifrin
So, projectively, we have a singular point at $[1,1,1]$, or, working over $\Bbb C$, three critical points at $[\omega,\omega^2,1]$, etc. So this will tell us that the cubic curve is singular in any event. In the real case, further examination is required. Working over $\Bbb C$, the three singular points tells us that this will be the union of three lines .... And that gives the factorization.
@leslie I think your equation should have been $\omega^3 = 1$, not $-1$.
leslie townes
leslie townes
19:52
could be, but i'll have you know that i've never made a sign error in my life.
Ted Shifrin
Ted Shifrin
So: Write down the lines passing through pairs of $[\omega,\omega^2,1]$, $[1,\omega,\omega^2]$, $[\omega,1,\omega^2]$ in $\Bbb CP^2$.
copper.hat
copper.hat
never drove through a stop sign?
Ted Shifrin
Ted Shifrin
Ugh, I got a $100 or so ticket in 1978 for a rolling stop a block from my apartment on Hillegass.
leslie townes
leslie townes
my daughter told on my wife for running a stop sign. she said it the minute they got home.
copper.hat
copper.hat
those sign errors
leslie townes
leslie townes
19:55
$100 is insane for 1978.
no wonder you remember.
Ted Shifrin
Ted Shifrin
Oh, speaking of errors, all three points lie on the single line $z_0+z_1+z_2=0$.
It may have been $75, but for a grad student, it was robbery.
copper.hat
copper.hat
my last ticket was for driving at felony speeds.
leslie townes
leslie townes
even in the mid 2000s, i think i usually had no more than about that left over at the end of the month
copper.hat
copper.hat
i managed to get loans from the international house administration.
leslie townes
leslie townes
is that what they called the IRA in those days
Ted Shifrin
Ted Shifrin
19:58
Oh, my three points are in error.
copper.hat
copper.hat
well, they knew where i lived...
Ted Shifrin
Ted Shifrin
They're projectively all the same point
The correct three points are $[1,1,1]$, $[1,\omega,\omega^2]$, $[1,\omega^2,\omega]$. Now do the three lines.
leslie townes
leslie townes
i thought we were going to stop encouraging people to do lines. not even once, kids.
Ted Shifrin
Ted Shifrin
At any rate, this will give the factorization of the polynomial over $\Bbb C$. Over $\Bbb R$, there's only one singular point, and I'm pretty sure it's a singular but irreducible cubic curve.
LearningCHelpMeV2
LearningCHelpMeV2
@leslietownes What is a symmetric polynomial? Does symmetric mean cyclic? Whereby if you change the variables in a cyclic order the polynomial remains unchanged?
Ted Shifrin
Ted Shifrin
20:04
Oh, hmm, Mathematica says I'm wrong. The only solution seems to be the point $x=y=z=1$.
LearningCHelpMeV2
LearningCHelpMeV2
and what are the prerequisites for algebraic geometry? can it be learned with high school math background
Ted Shifrin
Ted Shifrin
So, yes, the symmetry should suggest the factorization, but you need to use complex cube roots of unity.
You don't need to learn algebraic geometry. That is a serious advanced graduate course. But you do need to think about symmetry and roots of unity.
Govind75
Govind75
Anyone got an idea for this q? Let $V$ be an inner product space on polynomials over $\mathbb{R}$ of degree up to 2. With inner product $(f,g) = \frac{1}{2}\int_{-1}^{1}f(x)g(x).$ With linear transformation $T:V \rightarrow V$ such that $$T(1) = 1+x-2x^2$$ $$T(x) = 2+x-3x^2$$ $$T(x^2) = 1+x-2x^2$$. I need to find an orthonormal basis $B$ such that $[T]_B$ is upper triangular.
I tried using schurs theorem and following his method but my coefficients get very messy
especially when finding the norm using the inner product, was wondering if there's any quick trick that you guys could see
leslie townes
leslie townes
the coefficients are likely to get messy anyway. you are computing an orthonormal basis with respect to a goofy inner product.
Govind75
Govind75
;( I guess I'll just continue to grind through it
copper.hat
copper.hat
20:08
surely Gram Schmidt would help?
leslie townes
leslie townes
yes. bear in mind that if the numbers in front of 1, x, and x^2 in the definition of T were replaced with other numbers, this is essentially the question in its full generality, for any operator on any inner product space.
which is going to be calculations.
Govind75
Govind75
@copper.hat Yeah absolutely
Ted Shifrin
Ted Shifrin
@LearningCHelpMeV2: So with $\omega$ the primitive cube root of unity ($\omega = -1/2 + i\sqrt3/2$), write out $$(x+y+z)(x+\omega y + \omega^2 z)(x+\omega^2 y+\omega z).$$
Govind75
Govind75
But it gets really messy because you run it using horrible eigenvectors
leslie townes
leslie townes
anything that involves orthogonalization by hand, in general, is going to involve messy coefficents. unless the operator is 'choice'
Ted Shifrin
Ted Shifrin
20:10
Better not to use the numerical values. Just use basic algebra ...
Govind75
Govind75
@leslietownes Just to be sure, the method is to find the evecs and evals, then use gram-schmidt to find an orthonormal basis for the matrix. Then on the remaning sub block calculate its evals and evecs, then run gram schmidt again on a new basis consisting of all the eigenvectors?
Ted Shifrin
Ted Shifrin
Good exercise for you, @LearningCHelpMeV2: Can you show that the only real solution of $x^3+y^3+1-3xy=0$ is $(1,1)$?
So consider $x=1+u$, $y=1+v$ and consider $3(u^2-uv+v^2)+u^3+v^3=0$. Any non-zero solutions?
Maybe @leslie will like that one, too.
Lunchtime for Ted.
user 726941
user 726941
Enjoy
🍽️
leslie townes
leslie townes
govind i would look at what the eigenstuff gives you before orthonormalizing anything. one eigenvector can certainly be the first element of the basis but don't orthonormalize yet.
if it's three distinct eigenvalues, then yes, orthonormalize the vectors and you're done. the phrasing of the question suggests you might have a repeated eigenvalue. i would solve matrix equations in that case to identify vectors that will do the job once orthorogonalized, before doing gram schmidt.
GS is a last resort.
you do end up doing it, but there's no advantage in orthonormalizing inputs to an algorithm that's oging to spit out different vectors that you will again have to GS.
Sha Vuklia
Sha Vuklia
20:27
Ted! ;v
or @Thorgott
dtn
dtn
Let me integrate into the conversation, I need a specialist in polynomials, their solution, etc. I asked this question in Mathematica Stack Exchenge, but did not know how to proceed with the problem, so I decided to start with the fundamental things. The problem eventually transformed into the following: is there a simple condition under which a polynomial of the 8th degree has only real roots (both positive and negative). I ask for help on what to look for.
Sha Vuklia
Sha Vuklia
I need to share what I showed today x'D
Ted Shifrin
Ted Shifrin
Hi @Sha. I'm making lunch, only here briefly.
Sha Vuklia
Sha Vuklia
Perfect!
dtn
dtn
user image
Ted Shifrin
Ted Shifrin
20:30
This is an impossibly difficult question, @dtn
Sha Vuklia
Sha Vuklia
Ehh, yea this bugged me for a while, but I showed today that if we attach $n$-cells via attaching maps $f$ or $g$ that are homotopic, then the resulting push-out spaces are homotopic (which I find reallyy satisfying)
Ted Shifrin
Ted Shifrin
This sounds like it's going to be more for @Thor than for me, @Sha.
Sha Vuklia
Sha Vuklia
Lol oops
Ted Shifrin
Ted Shifrin
I am better at differential topology and more concrete alg top, not foundational picky stuff.
dtn
dtn
@TedShifrin jstor.org/stable/2325063 in this article simple criteria are given, but for the case when the coefficients of the polynomial are positive, but what about negative ones?
Sha Vuklia
Sha Vuklia
20:33
Fair
Ted Shifrin
Ted Shifrin
No earthly idea, @dtn.
dtn
dtn
:5978244 and Is it possible to adapt Sturm's theorem to the solution of the problem by forcibly choosing such coefficients so that there are 8 real roots on the indicated interval?
Ted Shifrin
Ted Shifrin
This is classic theory of equations stuff that people did in the early- and mid-20th century, but I don't know if anyone thinks about it any more.
Govind75
Govind75
@leslietownes It's one repeated eigenvalue with one eigenvector, I extended it to a basis of the vector space. But do I not need to make this basis orthonormal as schurs theorem suggests?
dtn
dtn
@TedShifrin the positiveness of the determinant can speak of the real roots. in mathematica we can get this expression, but it is so monstrously huge that I just don't know how to work with it. trying to find some more advanced methods.
user image
leslie townes
leslie townes
20:40
govind: the question seems to be asking you for an orthonormal basis, so yes, you eventually need to produce an orthonormal basis. i don't know what proof of schur's theorem you are looking at. ideally, you would find linearly independent vectors u, v, w such that u is an eigenvector, such that Tv is a multiple of u, and such that Tw is a linear combination of u and v, before orthonormalizing anything.
i'm not saying you won't have to orthonormalize. i'm more focused on when you do it. ideally, you do it at the very end.
Govind75
Govind75
Ah I see
Ted Shifrin
Ted Shifrin
Discriminant, not determinant. No, degree 8 is beyond control.
dtn
dtn
31
Q: Criteria to determine whether a real-coefficient polynomial has real root?

WileyGiven a polynomial equation $x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0$, where $n$ is even and all the coefficients $a_i$ are real, what is the best way to determine whether it has a real root or not? I know Sturm's theorem, but I am wondering if it's possible to determine this by the sign of some fo...

polynomials ca.classical-analysis-and-odes
Govind75
Govind75
This was the resource I was following btw: web.math.ucsb.edu/~padraic/ucsb_2013_14/math108b_w2014/…
dtn
dtn
@TedShifrin yes, of course discriminant, sealed up
Ted Shifrin
Ted Shifrin
20:46
The only thing I would suggest is to think about curve sketching techniques you learn in calculus ... It will suffice for local minima to be below the $x$-axis and local maxima to be above (although that's not necessary). But the algebra for that is equally unwieldy.
dtn
dtn
@TedShifrin Is it possible to reformulate the problem in terms of multiobjective optimization? if yes, then you need criteria, oh shit, I'm back where I started the question ...
Ted Shifrin
Ted Shifrin
As I said at the outset, this is an impossibly difficult question.
Yai0Phah
Yai0Phah
@ShaVuklia A conceptual reason is that, the pushout of a map along with a cofibration is a homotopy pushout.
dtn
dtn
@TedShifrin thank you for the advice. I'll try to simplify the task, suddenly something works out.
user 726941
user 726941
you may find a learning opportunity; but, don't expect too much :-)
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