Mathematics - 2021-04-08 (page 1 of 8)

leslie townes

2:03 AM

today was better than yesterday. high hopes for tomorrow.

user863565

2:25 AM

@TedShifrin now I understand it properly in terms of probability

the thing is I didn't think it properly that's why I couldn't retain it

a single isotope has probability of decaying of $k\delta t$ in interval $\delta t$

implies for N(t) isotope has probability of $N(t)k\delta t$ in interval $\delta t$

so we can say N(t)-N(t+\delta t) is approximately $k\delta t N(t)$

now we want to know about change per unit time we take limit

so we get $\frac{dN(t)}{dt}=-kN(t)$

then differential equation chits happens

copper.hat

2:52 AM

the radioactivity killed the room

leslie townes

i tasted metal for a minute and now i'm slightly glowing. i assume it's fine.

copper.hat

quick potassium iodide.

Silent

3:17 AM

I am watching Complex analysis lectures by Prof Bernd Schroder, and I have doubt regarding this theorem:

But, this proof suggests that if power series converges at ANY point on the circle of convergence, then it is uniformly convergent on entire open disk of convergence!
i.e., if power series not uniformly convergent in entire disk, then it does not converge on any point on the circle of convergence!

Is it true?

Semiclassical

That doesn't sound unreasonable to me. It doesn't (and shouldn't!) guarantee convergence on the boundary of said open disk, but the interior seems fine

leslie townes

i don't see it dealing with the circle of convergence. the condition is an open condition, |z - z_0| < R, where R is some distance from z_0 at which the series is known to converge. but it's < and not $\leq$.

Semiclassical

@leslietownes well, it does matter for the premise

it just doesn't matter for the conclusion

leslie townes

i agree with $\leq$ it's not a good theorem anymore. but i do not see the issue with $<$. i am open to thinking more deeply about it.

Semiclassical

if you have convergence at one point on the circle, you have convergence inside the circle (but not necessarily on the rest of the circle)

leslie townes

3:23 AM

note that q is strictly less than 1. it's not treating the case where something is R away from z_0.

Semiclassical

right

to cite the obvious example: x+x^2/2+x^3/3+... converges at x=-1, and converges for any |x|<1, but does not converge for x=1

copper.hat

@leslietownes Ted answered something for me a day or so ago and at the time I thought Oh yes, but now Thomas has returned with his doubts.

It was a questions about some $z^n$ being real only for $n=0$. And Ted's response was z is a root of an irreducible quartic. So could it be a root of unity?

Silent

@Semiclassical So, is it uniformly convergent in |x|<1, where x is any complex number? (as is suggested by this theore, since it is convergent on one point on circle of convergence)

leslie townes

i remember that remark. was it true? when he says things i just go with them.

Semiclassical

Well, the theorem only seems to claim absolute convergence? (I forget the distinction between absolute and uniform)

copper.hat

3:26 AM

So, it may warrant further investigcation...

leslie townes

at least i don't remember the actual quartic.

Semiclassical

@copper.hat What was the full question?

leslie townes

here we go again.

Silent

@Semiclassical yes, but we can infer from the proof uniform convergence as well, as is done later in other context by prof himself. (I am posting link to entire pdf)

copper.hat

There was no quartic, but $z={1 + \sqrt{2}i \over \sqrt{3}}$, and the question was, can $z^n$ be real for $n$ non zero (positive integer).

Semiclassical

3:27 AM

fair

copper.hat

sry

leslie townes

the circle of convergence can be a very messed up set. note that the theorem does not exclude convergence outside of the stated region, or try to identify the full set of z for which the series converges. maybe that is helpful.

Silent

Link to pdf:

web.archive.org/web/20160407105937/http://www2.latech.edu/…

copper.hat

i am looking for Ted's answer, not more discussion really, I know the result is true and can prove it in the most awful way.

Semiclassical

intuitively it doesn't look like a root of unity

leslie townes

3:29 AM

oh i think i'm recalling the polynomial. give me a minute.

copper.hat

My proof stopped with a similar conclusion but since it was for a high schooler I coudl not leave it there.

no worries, just if it clicked. at the time it seemed like a chorus of 'its so obvious'.

including myself.

leslie townes

it was never obvious, it was deceptively simple.

it just floated by and we all nodded at it.

Silent

@leslietownes yes yes! But if $z_1$ is considered to be sitting on circle of convergence =, which is indeed a possibility, then maybe we can infer something strong, I think. Hence my original doubt

copper.hat

@Silent what exactly was your doubt? if the series converges at some $z_1$ then you know that is is absolute for $|z-z_0| < |z_1-z_0|$.

Silent

No, not exactly.

leslie townes

3:32 AM

it's possible for the series to converge at one or more complex numbers exactly R away from z_0. it's just not generally guaranteed. and should the series diverge somewhere on that circle the behavior at the boundary will be potentially annoying to analyze.

copper.hat

Sry, getting ahead of myself.

leslie townes

it's conceivable for example that the series converges for all z. the hypotheses don't rule that out. they just don't assume it.

Semiclassical

only thingi can say is that it's a zero of (z^2+1/3)^2+8/9=0

copper.hat

Yeah, but I presume Ted meant something with the quartic and irreducible. And have been digging since :-).

leslie townes

it lets us write z^4 and all higher powers of z in terms of small powers of z. but i'm not seeing the lightbulb on top of my head yet.

Silent

3:36 AM

@leslietownes I know that part!! My question is not regarding convergence of power series on circle of convergence! nor regarding absolute convergence inside circle! Its regrading uniform convergence on disk.

Anyway thanks for giving it a shot.

bye

copper.hat

no worries. that partly addresses my concern :-)

@Silent where did you get uniform convergence from?

Semiclassical

the moderately nicer form is 3z^4+2z^2+3=0

user863565

@Silent where is link for video? I want to review complex analysis.

copper.hat

there are a few people on a few continents scratching their heads on this at the moment. was not my intent.

Semiclassical

lol

copper.hat

3:39 AM

there are probably real problems to be solved :-).

leslie townes

you have opened pandora's box.

Semiclassical

z^4 = (-2/3)z^2-1

copper.hat

ahhh, pandora...

user863565

I don't remember topology $hits although topology is intuitive topic

copper.hat

reminds me of a joke i cannot tell here.

leslie townes

3:40 AM

i like the bernd schroeder guy. the beamer templates he is using resemble the ones i used. impeccable taste.

copper.hat

:-)

Semiclassical

so z^6 = (-2/3)z^4-z^2 = (-2/3)((-2/3)z^2-1)-z^2=(2/3)^2 z^2-z^2+2/3 = (-5/9)z^2+2/3

copper.hat

like m brother's rollex he got from hong kong last century

Semiclassical

i guess this does lead to something like $z^{2k}=A(k)z^2+B(k)$

and you'd need $A(k)=0$

at least to get it via an even power

leslie townes

david letterman did a variant about that bit. he was out somewhere dealing with shady characters in queens. buying things. "paul, does rollex have two l's?"

narrator: rolex does not have two l's.

Semiclassical

3:42 AM

oh, and thus $z^{2k+1}=A(k)z^3+B(k)z$

so you'd need $A(k)=B(k)=0$ in that case, which is not plausible

copper.hat

:-) i think someone resembling my brother acquired all of microsoft's software on a few dvds in whatever the street was called back then.

Semiclassical

so that makes it seem like even powers are the only possibility

with that in mind, $w=z^2=\frac13(-1+i\sqrt{8})$

copper.hat

@Semiclassical the original question was math.stackexchange.com/questions/4090491/show-that-for-any-n%e2%88%88-bbb-n-numbers-x-n-and-y-n-are-nonzero/4093455#4093455

leslie townes

one time i had a friend visit from sweden and the only thing he wanted was a fake rolex. we went to chinatown in SF and could not find the right stuff. he was so disappointed.

we found a multitude of cheap watches but not ones that purported to be what they were not.

copper.hat

how awful. did the ip irony of his accomplice's occupation escape him?

leslie townes

3:46 AM

another guy i knew is of chinese ancestry and worked for the office of the US trade representative. they were in beijing for some kind of meeting, and he was shopping, and he said "where can i get some dvds? i want iron man 2 [or whatever]" and the owner said "those are gone because the americans are in town this week for some kind of trade conference."

lesson learned.

copper.hat

bitorrent

leslie townes

this reflected, to me, a surprising degree of sophistication for a shop owner.

Semiclassical

which is a root of $(w+1/3)^2+8/9=w^2+(2/3)w+1=0$

or $3w^2=-2w-1$

copper.hat

i visited a friend who was a ceo in the eda industry and he showed me a plot of illigal licenses used vs country.

basically flat until a big $\delta$ at the end. guess where.

leslie townes

hahaha

Semiclassical

3:48 AM

If $F=Aw+B$, then $(3w)F=A(3w^2)+3Bw=A(-2w-1)+3Bw = (3B-2A)w-A$

leslie townes

one time i got a phone call from someone who somehow knew that i had a bunch of CAD files on my computer for a diagnostic device and insisted that i email them over. i was like, "who are you, again?" about ten times, and he never said anything that made sense so i hung up. i don't know how he suspected i had those files. that keeps me up at night.

Semiclassical

so the map takes $(A,B)\to (3B-2A,-A)$

copper.hat

we used to joke that we should hire the folks who figured out how to use our software without any training.

@Semiclassical sry, i really did not mean to use your time on this :-)

Semiclassical

which I guess would be implemented as the matrix $$\begin{pmatrix} -2 & 3 \\ -1 & 0\end{pmatrix}$$

lol

copper.hat

look at the link above and you will see a few echos of that

Semiclassical

3:50 AM

ah nice

I have a sneaking suspicion i'm about to loop back on myself anyways

because if I want to iterate that map then i'd want to diagonalize it

copper.hat

:-). i thought it was entirely trivial, but it needs a little thought.

Silent

:57581759 From this proof, can't we infer that? Since $M\sum_{n=N+1}^{\infty}q_n\to0$, as $n\to\infty$, we see that for any $\epsilon>0$, there is $N_0$ such that for any $n>N_0$ we have $\sum_{n=N+1}^{\infty}|a_n(z-z_0)^n|\le|M\sum_{n=N+1}^{\infty}q_n|<\epsilon$ for any $|z-z_0|<R$, implying uniform convergence.

Semiclassical

and i'm guessing that characteristic polynomial gives me back my original one

Silent

@user863565 web.archive.org/web/20160407171007/http://www2.latech.edu/…

leslie townes

there's a potential thing about uniform convergence on an open disc vs. uniform convergence on any closed sub-disc of that disc.

i'm mildly thinking about this. i've had a long day and a weird week.

Semiclassical

3:53 AM

@copper.hat ah yeah, i see that matrix showing up

Silent

I was afraid of CA, but this is an illuminating course. Its teeming with awesome commentary by professor @user863565

copper.hat

@Silent the quantity $q$ depends on $z$.

leslie townes

something can converge uniformly for all |z| <= r < R for all r < R without converging uniformly on the open disc |z| < R. i think. unless part of my brain fell out.

copper.hat

It is true that the convergence is uniform on compact subsets of $B(z_0,|z_1-z_0|)$.

leslie townes

complex analysis is probably my favorite subject, as a class to take in a university. i really loved it.

copper.hat

3:54 AM

mine was real analysis.

Silent

@copper.hat Wow! Thank you very much. That clears my doubt. :)

leslie townes

i had complex analysis from marina ratner, who was stern but amazing.

copper.hat

esp fixed pt theorems and implicit fn theorem.

leslie townes

my real analysis class was a clueless postdoc who had never thought about analysis and turned it into a point set topology course, which i enjoyed somewhat but wasn't analysis.

Semiclassical

gross

leslie townes

3:56 AM

there were two kinds of undergrads at berkeley when i went there, people who tried to enroll in classes taught by postdocs for everything, because they were perceived to be easier, and people who tried to avoid them and get classes from real professors. i fell into the latter group.

copper.hat

when i took it at berkeley (which was really my 3rd exposure) the lecturer was visiting from brasil and was very good, did not let me off the hook.

leslie townes

it wasn't a status thing they just tended to be better at teaching.

copper.hat

i was using the material at the time in other classes & research, so it was all good.

i took functional analysis from kobayashi. excellent teacher.

leslie townes

i wish i could have been there. i learned most of what i know from paul chernoff.

RIP.

copper.hat

and measure theorey from chernoff. i was top of the class :-)

yeah, sad.

he could be pretty derisive.

leslie townes

3:59 AM

he'd sometimes yell at us. he was somewhat unstable. he was also gone a lot. but he was a very kind person.

copper.hat

but mild compared to my own advisor (and vel kahan).

i did not really know him personally.

leslie townes

he had a lot of heart trouble while i was learning from him. one time i was in the campus health center for an eye exam and he'd had a heart attack, or something, i saw him on a stretcher. i forgave him for a lot of his angry moments after that.

copper.hat

his ta hated to see me because my homework was a bit of a challenge, but he was very generous and helpful nonethless.

i escaped his derision thankfully, but was a little surprised by him sometimes (and, i might add, that is saying something).

leslie townes

he seemed to be a very lonely person. he lashed out at me once or twice. he also presented a homework exercise i had done to the entire class as an example of how to do homework.

copper.hat

his notes were awesome.

leslie townes

4:01 AM

yes i still have them.

i regret not being able to go to a number of funerals or memorial services. i got the emails for all of them, i just didn't have the money or the time.

copper.hat

yep. one colleague in the ucb me dept passed away but i found out after the arrangements.

leslie townes

that's the worst, when you find out kind of too late to even have tried to make arrangements.

copper.hat

unfortunately two of my research group colleagues passed away while i was there.

i mean students

obv i attended their services.

one was self inflicted. very very sad.

leslie townes

there's way, way, way too much of that in graduate school. this touches a nerve, my step brother attempted suicide this week by stepping in front of a train.

people just need to reach out and ask for help.

copper.hat

i must say, my stay as a student in berkeley had a lot of unexpected stuff happen.

so sad to hear.

leslie townes

4:06 AM

there's something about the culture of graduate school about not asking for help. it's horrific. i hope my daughter doesn't go. or she goes to a good school and not some kind of pissing contest school.

copper.hat

its hard.

Ted Shifrin

Chernoff was very overweight!

copper.hat

no sh*t

leslie townes

yes a lot of it was fluid. particularly near the end. with fluid stuff you can just balloon up.

Ted Shifrin

Even in the 70s when I was there.

copper.hat

4:08 AM

that was his ta's name, or something similar.

balon or something i think. jeez, its been so long.

leslie townes

my mother had a thyroid condition and that's actually when i realized she needed to go to the doctor, she visually resembled chernoff.

copper.hat

my mom & br. are drs and both have said that your family can deteriorate in front of their eyes and they will never notice.

i'm getting my gates chip in the morning supposedly.

Ted Shifrin

Oh, exciting!

copper.hat

then i'm going to infiltrate the evangelists.

leslie townes

but he was very clever, if sometimes nasty, he seemed to be wrestling with demons none of us knew about.

my wife has the bill gates microchip. we have really good 5G in the house now.

copper.hat

4:11 AM

he was super quick.

i am afraid i will suffer the bsod.

Ted Shifrin

Bsod?

leslie townes

my wife's second shot was non consequential but my father, mother, and step mother all were completely out of commission after the second shot, for about 48 hours.

copper.hat

sry, blue screen of death. too long in the tech world

for the astrazeneca, 38% reported side effects, for the placebo, 28% did.

Ted Shifrin

I had headache and fever for a few days but, being a retired bum, I was not out of commission.

copper.hat

sry to hear it.

Ted Shifrin

4:13 AM

It was no big deal.

leslie townes

my dad almost couldn't move. he was texting me from his bedroom. "this kicked my ass," he said. but a day later, back at work.

copper.hat

i have zero billable today. very undisciplined.

Ted Shifrin

Bad copper.

copper.hat

i am.

Ted Shifrin

Kobayashi taught functional analysis ... out of area but great!

copper.hat

4:16 AM

i remember once he looked up, beamed that great smile of his, and said "sets are not like doors, they can be open and closed".

i believe his daughter was a student at the time, she took a class with of hald's with me.

leslie townes

i'd love to see those notes. i'm very tempted to scan or somehow reproduce my notes of chernoff's unbounded operators. he'd say the textbook had a proof of x, and then do it in a paragraph, where the textbook took something like three pages.

copper.hat

mine were basic functional analsisi & measure theory.

fourier stuff, etc.

Ted Shifrin

Shy but super friendly. I loved his wife Grace. Super spunky. She and Mrs Chern pretended to fight over who could see more bridges from their respective living rooms.

copper.hat

i liked him. he knew i was remedial but it didn't bother him :-).

i was willing to do the work.

leslie townes

that's most of it in the end

Ted Shifrin

4:19 AM

He was not on my committee but he knew me well .

copper.hat

he didn't get lost in detail, which was perfect for my style. i will do the detail later.

i want to do grad school 2.0.

this time i will do it right.

leslie townes

law school! they pay you to bother people. it's amazing.

copper.hat

i could not do it for long.

i think i would do aeronautical engineering or fluids.

leslie townes

it was socially isolating to be older than everyone in my class by several years, although i sometimes got outdated jokes that the instructors made. i don't actually recommend it.

there's a lot of cool stuff happening in microfluidics right now.

copper.hat

yeah, plus the computational side is so much better

Semiclassical

4:23 AM

trying to figure out what sort of math to bracket this under. one example i have from my very old experimental physics book is to start with the equation $\omega^2 =\frac{gh}{h^2+k^2}$

leslie townes

the bioinformatics is better and semiconductor folks are able to cheaply fabricate detectors in 2021 that they could not do in 2001.

copper.hat

odd units for $\omega^2$?

leslie townes

in the old days you'd need a couple of lenses and a very expensive camera to detect what a biochemical reaction was doing. now you just put it on something resembling a digital camera chip of pixels, and you throw it away after the picture is taken, because that's cheap now.

Semiclassical

not really. g has units of m/s^2 and h has units of m, so omega has units of 1/s

leslie townes

i do wonder about units in that equation.

Ted Shifrin

4:26 AM

What is $k$

Semiclassical

radius of gyration, so another length

copper.hat

the $h^2$ and $k^2$ are side by side with diff units???

Semiclassical

um, no? $h$ and $k$ are both lengths

copper.hat

sry, can't read.

Ted Shifrin

so what is your query?

Semiclassical

4:26 AM

was going to get to that :P

the experimental variables are $\omega$ and $h$, so it's seemingly not a linear relationship. however, we can write $h^2 = g(h/\omega^2)-k^2$

so there is a linear relationship between the variables $Y:=h^2$ and $X:=h/\omega^2$

leslie townes

where is it going? what is the desired end result?

in an ideal world, what would the quantities be that you know about?

Ted Shifrin

so you're aiming for least squares plot of data?

Semiclassical

you'd measure $h$ and $\omega$, and by plotting $X,Y$ you'd be able to read off $g,k$

that's the idea of the example, yes. what i'm wondering is what you'd call all of this.

as in

copper.hat

and $h,\omega$ are distributed how?

are $g,k$ constants?

Semiclassical

yes

suppose I have some polynomial condition $H(x,y)=0$. Is there a name for the scenario: there exist rational functions $f(x,y)$ and $g(x,y)$ such that $H(x,y)=0$ is equivalent to a linear relationship between $f(x,y)$ and $g(x,y)$

basically, is there a name for this kind of construction?

Ted Shifrin

4:31 AM

Change of variables :)

Semiclassical

lol

fair

Ted Shifrin

Reparametrization

Semiclassical

yeah

the variables generate a parametrization of some linear relation

something like that

Ted Shifrin

interesting question ... must be a rational curve and linear in some embedding

leslie townes

the linearity seems tough to me. i'm not an algebraist or a geometer.

copper.hat

4:33 AM

can't you just regress on $k^2,g$?

Ted Shifrin

you're full of excuses, leslie

Semiclassical

@copper.hat g and k are fixed constants here

leslie townes

i realize it's a little off-brand to disclaim knowledge of linearity.

copper.hat

and you are trying to estimate them?

Semiclassical

in this example, yeah

the example which brought this to mind is one which i -don't- think i can place in linear form

Ted Shifrin

4:35 AM

The algebraic geometry is what I said above.

copper.hat

so you have data $\omega_k^2 h_k^2 + \omega_k^2 k^2 -g h_k$ and you want to minimise the sum of squares.

Semiclassical

the problem that brought this old example to mind is $\omega^2 = (ML+mx)/(MR^2+mx^2)$ with $\omega,x$

which is similar but worse

copper.hat

what have you tried?

Semiclassical

to be clear, this is not an experiment i've actually done

Ted Shifrin

Obviously the square on $\omega$ can be removed.

Semiclassical

4:37 AM

just trying to decide whether one could do an experiment based on a model like that and do linear least squares

yeah, that's trivial

Ted Shifrin

$y(a+bx^2)=c+bx$

Semiclassical

in principle I could write it as $MR^2 (\omega^2)+m(\omega^2 x^2) =mx+ML$

copper.hat

why not? don't ask, do.

Semiclassical

um

copper.hat

i've done worse for a friend doing low temp physics at berkeley.

leslie townes

4:39 AM

don't do, ask.

Ted Shifrin

so probably elliptic curve, so not rational

Semiclassical

@TedShifrin yeah

i can sorta stretch the definition and regard the linear relationship as amongst $\omega^2, \omega^2 x^2, x$

leslie townes

copper.hat you should tell them to "chill out" and then go for a high five that will not come.

Semiclassical

but that seems goofy since there's only two experimental variables there

leslie townes

i promise you, this will be a comedic bit that works.

or not. terms and conditions apply

Semiclassical

4:41 AM

that said, if i regard $m$ as another variable (which i really could) then that becomes more reasonable

copper.hat

:-) well, they did have an explosion in the lab once.

nothing to do with me i might add.

Semiclassical

so the best one can seem to do is seek a three-variable linear relationship rather than two

copper.hat

what are you hoping to get?

Ted Shifrin

What if we change $y$ to $1/y$? Then it becomes a conic.

Semiclassical

A set of variables suitable for linear least squares. But it's as much a pure math question at this point

Ted Shifrin

4:42 AM

So rational.

copper.hat

experimental data or a theoretical model?

Semiclassical

tbh, just the model. i'm not really that interested in actually taking data so much as deciding whether one -could- do it

Ted Shifrin

$a+bx^2=(c+bx)z$

copper.hat

sry, i'm not much use there. my hands are dirty.

Semiclassical

@TedShifrin yeah, that seems like it runs into the same problem

Ted Shifrin

4:44 AM

Subtract, complete the square and do the usual rational parametrization

Semiclassical

hmmmmmmm

i see what you mean, i think

Ted Shifrin

Conics are equivalent to lines by stereographic projection

Semiclassical

so $b(x-z/2)^2 = cz-a+bz^2/4=(b/4)(z^2+4c/b*z)-a = (b/4)(z+2c/b)^2 -b/4(4c^2/b^2)-a$

Ted Shifrin

But you don't care. A sum of squares is linear in the right variables.

copper.hat

haven't hd fesenjan for a while.

Semiclassical

4:47 AM

@TedShifrin exactly

so the "correct variables" here would seem to be $(x-z/2)^2$ and $(z+2c/b)^2$

Some Guy

Hello, does the Fibonacci sequence start 1,1 or 0,1?

Ted Shifrin

So I think we’ve solved this once we invert $y$.

With $z=1/y$

Semiclassical

yeah. it has the unfortunate problem that the 'linearizing variables' contain the constants you'd be wanting to determine

copper.hat

@SomeGuy wikipedia has it.

Semiclassical

so it solves the problem mathematically but in a way that wouldn't be great for experimental design

which, oh well

Ted Shifrin

4:49 AM

I can’t fix everything, Semiclassic

Some Guy

thanks

copper.hat

why not?

Semiclassical

yeah

there was no guarantee that there was a nice answer

if I was to pick an answer, it'd be to go to a 3-variable linear relationship and be satisfied with taht

copper.hat

pick a line in the plane. pick two distinct points not on the line. find a point on the line such that the sum of the distances from the distinct point on the line is minimised.

problem is straightforward.

but the first order conditions are a mess, yet there is a simple ('reflection') solution.

just curious if there is some detangling of the first order conditions that can give an explicit solution.

the angle condition is straighforward.

Semiclassical

oh, it should've been $MR^2\omega^2+mx^2 \omega^2= MgL+mgx$

copper.hat

4:57 AM

some rotational thing going on?

Semiclassical

yeah. basically: i've got a metal ruler with a hole in it, and I can tape a mass to it

so i can measure the period of oscillation

$L$ is the distance from the hole to the ruler's center of mass, and $R$ is its radius of gyration

so can one move the mass around on the rule, measure the period, and from that deduce the properties of the ruler

copper.hat

how are you measuring the period?

Semiclassical

timing

copper.hat

electronically?

Semiclassical

same as you'd measure how fast your heart beats

count how many times it beats in a minute

robjohn

5:00 AM

@copper.hat If the points are on the opposite side of the given line, the minimizing point is the intersection of the given line and the line between the points. If they are on the same side of the given line, move one of the points orthogonally to the opposite side, solve and move back.

Semiclassical

(or count how long for 10 beats, same idea)

copper.hat

@robjohn yeah, but i am perplexed why that does not appear clear in the first order conditions.

it only (as far as i can see) shows up as the equal angle condition.

@Semiclassical how old school :-).

Andrew Micallef

6:07 AM

what area of maths deals with operators on functions?

From the footnote in the book I am reading:
> An “operator” is an instruction to do something to the function that follows; it takes in one function, and spits out some other function.

WHere should I go to learn more about that?

user2103480

8:28 AM

@AndrewMicallef broadly, functional analysis

sonicboom

9:10 AM

Let $X$ be random variable supported in $[1,2]$. And Let $\varepsilon$ be the standard Gaussian distribution. Is the product $X\varepsilon$ a positive random variable (like $X$), or is it an unbounded random variable like $\varepsilon$?

user2103480

Imagine X is constant. How large is the support of X*eps?

And judging from that, can you bound the absolute value of X*eps reasonably and usefully below?

sonicboom

9:26 AM

Well if its constant the support is R. If I take X to be uniform on [1,2], then it seems that X*eps also has infinite support

@user2103480 How are you bounding X*eps below?

Monty

10:08 AM

Let $ f: E \to M$, where $E$ is some topological space and $(M,d)$ is a metric space.

also let the sequence $ f_n : E \to M$

assume for each $x\in E$ $d ( f_n(x),f(x) ) \to 0$ as $n\to \infty$, is it true that $\sup_{x\in \Omega} d ( f_n(x),f(x) ) \to 0 $

if $\Omega$ is compact

is this just convergence on compacts = uniform convergence

Srijan M.T

How did we get z+4 here ?

user3733558

11:07 AM

Polynomial long division

In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalized version of the familiar arithmetic technique called long division. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones. Sometimes using a shorthand version called synthetic division is faster, with less writing and fewer calculations. Another abbreviated method is polynomial short division (Blomqvist's method). Polynomial long division is an algorithm that implements the Euclidean division of polynomials...

user2103480

11:23 AM

@sonicboom $\Bbb P( |\varepsilon X| \geq M) \geq \Bbb P( |X| \geq M)$

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$Mathematics$ Mathematics