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12:38 AM
@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…
2:09 AM
disaster has been averted. we have maybe 2 months worth of wet food. all of olivia's favorite varieties.
 
2:23 AM
In a commutative ring if x=ay and y=ax does it imply a is a unit?
 
consider a = -1
unless i'm misunderstanding what a unit is
 
unit is an element a such that ab=1 for some b
 
I believe you are, -1 is a unit
 
my memory of this is rusty, rip
would 0=0x0, 0=0x0 satisfy your problem here or are there conditions on x and y
 
2:26 AM
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?
 
y=a^2 y does not imply a^2=1
And yeah f, g are not 0…
 
What's the counter-example?
 
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
 
We ask for the existence of a unit a
(1,0)*(1,1)=(1,0)
Oh ok u r answering his question nvm
 
I think that was also supposed to be a counterexample to the claim
 
2:36 AM
yes. a is not a unit in my example
 
I’ve tested on all subrings of C
And all subrings of Set(X, \Bbb R)
X topological space \Bbb R the reals
 
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
 
And obviously looking at symmetric difference and intersection from a power set doesn’t work since A=BC and C=BA implies A=C
 
or in any ring without zero divisors
 
yea, I gotta admit I'm a little salty I forgot something as silly as 2*2=0mod4
 
2:48 AM
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…
3:52 AM
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?
 
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'
 
yes, you can use fewer
just watched it and I don't understand the proof at all
 
4:08 AM
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
 
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
 
4:24 AM
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.
 
@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.
 
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
 
That is definitely how I read that.
 
so if f' changes sign infinitely often in any such neighborhood it would not be said to 'change sign' at c
 
$x\sin(1/x)$
 
4:32 AM
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
 
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?
 
@Koro Then $f'$ is not continuous on $[a,b]$
 
Yes! That's what I thought. f' does not have to be continuous.
 
@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.
 
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.
 
4:44 AM
Ok, I figured my real problem out. I don't understand the probabilistic method.
 
@Koro A derivative needs to obey the IVT, however.
If the derivative exists at all points.
 
@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.
 
@Koro you might look at Darboux's theorem
 
I have already stated that in my last comment. That gives you IVT of f'.
 
5:02 AM
extremum or local extremum?
 
5:13 AM
I was thinking local extremum.
 
5:36 AM
@Koro did you look at the proof math.stackexchange.com/a/407698/27978
 
@leslietownes Disaster indeed averted. Mazltov!
 
they even had the big bag of her favorite brand of kitty litter.
 
@copper.hat yes, I have. But I want to understand GH Hardy's proof. That seems so intuitive. :)
 
@leslietownes in a field, most elements are units....
 
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?

hmm. (1) posted to math.GM (2) on April 1st... it's not looking good.
 
6:08 AM
I have posted my question here math.stackexchange.com/questions/4322642/….
Please let me know what I'm missing. Thanks.
 
6:34 AM
@Koro i added an answer. no doubt with some mistakes.
 
7:23 AM
Thanks a lot @copper. :)
 
 
1 hour later…
8:24 AM
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. :)
 
8:39 AM
@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…
1:17 PM
21 hours ago, by Wolgwang
Can a square be divided into 5 squares(may not be congruent) ?
 
 
1 hour later…
2:33 PM
@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.
 
3:15 PM
@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
 
@PM2Ring Does this answer say "Yes"?
 
3:55 PM
@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
 
@PM2Ring Hmm I think by 'glue' in that MO question, the OP mean that pieces can be glued together to form squares (?)
 
Note that Bill Cook's solution is closely related to Pythagorean tilings. en.wikipedia.org/wiki/Pythagorean_tiling
 
> Is it impossible to divide a square into (may not be congruent) n squares, if n = 5?
This was the exact question.
 
@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
$u(x)=\frac{1}{4\pi} \int_{\mathbb{R^3} \frac{e^{-|x-y|}}{|x-y|}f(y)dy$
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$
 
4:02 PM
@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.
 
Ohk Thanks :)
 
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?
 
int_{ delete that {
or close it
 
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?
 
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,
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.
 
4:27 PM
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
 
4:43 PM
@Jam try substituting $y\mapsto x-y$ first.
 
Jam
good idea
i need to calculate $ \Delta u$
can you help me?
 
@PM2Ring That hurts my eyes.
 
@PM2Ring Fascinating!
Hello! Prof. X
 
5:01 PM
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.
 
@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
 
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.
:(
 
@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. :)
 
they should have dared to have bigger dreams.
my daughter would love that
 
@XanderHenderson Feel free to use your mod powers to edit it so it doesn't one-box.
 
5:10 PM
@PM2Ring It is mostly gone now.
Two or three more messages, and will be all gone.
Bye bye spinny squares of doom!
 
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
 
@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.
 
@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...
 
Oh.. indeed. David found a number of closely related questions.
Arguably, the question should have been closed as a duplicate of one of those.
 
5:26 PM
@XanderHenderson Your eyes or your brain?
 
@TedShifrin Initially, the area just behind my left eye. And then I started getting vertigo.
 
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
 
@Govind75 When you encountered the question, did you give it a biscuit, pat it on the head, and say "Nice doggy"?
 
I got that $c = \frac{(v,v)_1}{(v,v)_2}$, but surely this constant is dependent on the vectors I am choosing
 
(You really need to do that, or they can get mean.)
 
5:32 PM
@XanderHenderson Oops I forgot to do that, I'll keep it in mind for next time
 
koro did this exercise the other day. i think there are copies of it on math.se
 
yeah I found some solutions
But surely my constant is dependent on the vectors I am taking the inner product of?
 
@Govind75 Don't call me Shirley. Also, what makes you so sure?
 
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?

 
Isn't the interesting thing about the exercise the fact that the constant doesn't depend on the vectors?
 
5:34 PM
@XanderHenderson Airplane movie reference!
 
Ohhhh I see
Cheers <3
 
@AdilMohammed Seems to to date the man! Actually, another Xanderism. A reference to Xanderisms is in progress! ;D
 
@XanderHenderson On a related note, this is currently on the HNQ: How is "Dogs dogs fight fight" grammatical?
 
@amWhy Thanks, I made a mental note of movies that needs a reboot/reshoot in 2021 and Airplanes was on top of it
 
@Govind75 That wouldn't be much of a constant, it seems to me.
Besides, it looks as if your fraction is upside-down
 
5:41 PM
@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.
 
@robjohn Oh yeah
 
@AdilMohammed What's our vector, Victor?
 
@PM2Ring Ah, Scraps is a boy dog...
 
@robjohn What are your pets' names? I know you have a kitty.
 
@amWhy Yes, it's from EL&U. And the comments are currently locked. ;)
 
5:43 PM
@robjohn Maybe it is a variable constant?
A vari-stant, if you will?
A constable?
 
@PM2Ring Hah! That's an excellent site.
@XanderHenderson "variable-constant means constable" Xanderism # 549! :D
 
@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.
 
@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".
 
On a related note:
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. ;)
 
5:59 PM
@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.
 
6:17 PM
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.
 
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
 
@leslietownes What if the only ones on the internet are people who are wrong? =P
 
nah, couldn't happen.
 
@leslietownes For all you know, it might be descriptive of the internet.
But if so, then we have a paradox. ^
 
i only like descriptivists who don't like themselves.
 
6:25 PM
@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! ;)
 
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
 
@user21820 They need to mingle more!! ;D
 
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.
 
@leslietownes Lol. I would have used "cahoots". =D
 
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.
 
6:32 PM
@leslietownes Agreed, @leslietownes
 
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.
 
@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.
 
and show the source.
 
that is disturbing
 
that got dark pretty quickly.
 
6:37 PM
i'm in a delicate mood
 
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.
 
my folk theory is divided responsibility
 
@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.
 
i told my kids if they need help in a crowded situation to ask a specific person not just say help
 
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.
 
6:38 PM
irregardless
 
@copper.hat I'm just saying we need to know what reality, sometimes, before prescribing remedies.
 
its like literally true
@amWhy yup
 
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.
 
i like wilde's "i am not young enough to know everything"
 
PM: i literally died when you wrote that
 
6:40 PM
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.
 
i yearn for the days when i could be prescriptive
 
@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.
 
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".
 
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.
 
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.
 
6:44 PM
you mean conor mcgregor, of course
 
@PM2Ring Agreed!
@copper.hat Were you asking me?
 
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
 
@copper.hat Thanks. This is where the prescriptivist in me comes out: Notorious \neq notable!!!
 
unfortunately (for me, at least) language evolves.
 
@PM2Ring We need to teach our youth.... !!!
 
6:48 PM
even in more subtle ways. i just moved phone and am trying to decipher a plethora of minute status symbols.
 
@copper.hat I agree; I was just saying how quickly describing and prescribing can flip.
 
y
i think the rate of change of such things is beyond where cognitive decline has left me
 
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!
 
it is a losing battle, i think :-)
 
That "evil" thing has already happened with "bad" and "sick".
 
6:54 PM
@PM2Ring True.
 
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.
 
well, things will change, but not to the ideal model they have.
their presumption of correctness is what needs to change :-)
 
@copper.hat I'm good with that. (your first sentence). Evolution of ideas, often takes longer than one's lifetime.
 
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).
 
7:03 PM
divided loyalties: the copper.hat story
 
@copper.hat I think I understand.
 
of course, if i waited long enough it would have been moot.
 
Super weird that a Turing machine $T$ can halt in one mode of PA and go on forever in another
 
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
 
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
 
7:06 PM
no, it's 24. 24 is the largest number
 
@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!
 
i think it harks back to something you said earlier
data needs to be gathered first
 
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)
 
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.
 
You can also replace PA with ZFC or whatever
 
7:14 PM
@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.
 
i think the conflict is inevitable, unfortunately.
 
@copper.hat But the conflict undermines the goals of both.
 
it appears to, but i presume there is some greater evolutionary truth that i cannot see...
 
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 $pq(p-q) + qr(q-r) + rp(r-p)$
 
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.
 
7:24 PM
@amWhy as i mentioned above, i'm in a delicate mood :-)
 
@LearningCHelpMeV2 I don't know specifically, but (and these are super broad) ring theory and algebraic geometry may be up your alley
or
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...
 
@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!
 
This seems semirelevant also
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...
 
@amWhy no need to be. i enjoy discussion.
 
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
 
7:39 PM
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
 
It also depends on what underlying scalars we allow. But generically things will be irreducible.
 
uh oh, ted's here
 
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.
 
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
 
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 $-1$.
 
7:52 PM
could be, but i'll have you know that i've never made a sign error in my life.
 
So: Write down the lines passing through pairs of $[\omega,\omega^2,1]$, $[1,\omega,\omega^2]$, $[\omega,1,\omega^2]$ in $\Bbb CP^2$.
 
never drove through a stop sign?
 
Ugh, I got a $100 or so ticket in 1978 for a rolling stop a block from my apartment on Hillegass.
 
my daughter told on my wife for running a stop sign. she said it the minute they got home.
 
those sign errors
 
7:55 PM
$100 is insane for 1978.
no wonder you remember.
 
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.
 
my last ticket was for driving at felony speeds.
 
even in the mid 2000s, i think i usually had no more than about that left over at the end of the month
 
i managed to get loans from the international house administration.
 
is that what they called the IRA in those days
 
7:58 PM
Oh, my three points are in error.
 
well, they knew where i lived...
 
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.
 
i thought we were going to stop encouraging people to do lines. not even once, kids.
 
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.
 
@leslietownes What is a symmetric polynomial? Does symmetric mean cyclic? Whereby if you change the variables in a cyclic order the polynomial remains unchanged?
 
8:04 PM
Oh, hmm, Mathematica says I'm wrong. The only solution seems to be the point $x=y=z=1$.
 
and what are the prerequisites for algebraic geometry? can it be learned with high school math background
 
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.
 
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
 
the coefficients are likely to get messy anyway. you are computing an orthonormal basis with respect to a goofy inner product.
 
;( I guess I'll just continue to grind through it
 
8:08 PM
surely Gram Schmidt would help?
 
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.
 
@copper.hat Yeah absolutely
 
@LearningCHelpMeV2: So with $\omega$ the primitive cube root of unity ($\omega = -1/2 + i\sqrt3/2$), write out $$(x+y+z)(x+\omega y + \omega^2 z)(x+\omega^2 y+\omega z).$$
 
But it gets really messy because you run it using horrible eigenvectors
 
anything that involves orthogonalization by hand, in general, is going to involve messy coefficents. unless the operator is 'choice'
 
8:10 PM
Better not to use the numerical values. Just use basic algebra ...
 
@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?
 
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.
 
Enjoy
🍽️
 
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.
 
8:27 PM
Ted! ;v
or @Thorgott
 
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.
 
I need to share what I showed today x'D
 
Hi @Sha. I'm making lunch, only here briefly.
 
Perfect!
 
dtn
 
8:30 PM
This is an impossibly difficult question, @dtn
 
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)
 
This sounds like it's going to be more for @Thor than for me, @Sha.
 
Lol oops
 
I am better at differential topology and more concrete alg top, not foundational picky stuff.
 
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?
 
8:33 PM
Fair
 
No earthly idea, @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?
 
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.
 
@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
@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.
 
8:40 PM
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.
 
Ah I see
 
Discriminant, not determinant. No, degree 8 is beyond control.
 
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...

 
This was the resource I was following btw: web.math.ucsb.edu/~padraic/ucsb_2013_14/math108b_w2014/…
 
dtn
@TedShifrin yes, of course discriminant, sealed up
 
8:46 PM
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
@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 ...
 
As I said at the outset, this is an impossibly difficult question.
 
@ShaVuklia A conceptual reason is that, the pushout of a map along with a cofibration is a homotopy pushout.
 
dtn
@TedShifrin thank you for the advice. I'll try to simplify the task, suddenly something works out.
 
you may find a learning opportunity; but, don't expect too much :-)
 
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