Mathematics - 2023-08-07

geocalc33

00:00

at least for self inverse distributions

Xander Henderson

What do you mean by "self inverse"?

Do you really mean $f \circ f = \operatorname{id}$?

There aren't a lot of probability distributions which satisfy this. I've certainly never heard of anyone studying them...

Jakobian

@XanderHenderson you're acting like discrete random variables don't exist

Xander Henderson

@Jakobian Nope. They fall into the same framework. That's a Lebesgue integral over a space $\Omega$ equipped with a suitable $\sigma$-algebra.

geocalc33

It's actually quite weird to define the mean for $f \circ f=\mathrm{id}$ distributions by using this symmetry

you have to "tilt the euclidean plane"

Jakobian

@XanderHenderson but not over the Lebesgue measure

Xander Henderson

00:14

@Jakobian No, it'll be equipped with some other measure.

Would you prefer that I write something like $\int_{\Omega} x f_X(x)\,\mathrm{d}P$?

Where $P$ is some probability measure?

or maybe $\int_{\Omega} x \,\mathrm{d}P$.

Jakobian

well, the latter if we're talking about probability measures

Xander Henderson

It's been a while since I've thought about what these integrals are supposed to "mean". But it is all the same theory---Lebesgue integration, more or less.

Jakobian

but I'd be satisfied with $\int_\Omega x f_X(x)\mathrm{d}\mu$ where $f_X$ is a probability mass function

Xander Henderson

@Jakobian Sure.

geocalc33

$\int_{\Omega} x f_X(x)\,\mathrm{d}P(x)$

Xander Henderson

00:16

I mean, do you want to regard this as a Lebesgue integral with respect to a probability measure, or a Lebesgue-Stiltjes integral with respect to a probability measure and some kind of mass / density function?

@geocalc33 Po-tay-to, po-tah-to.

The basic idea is the same, whatever notation you use.

Jakobian

I don't want $\mu$ to be a probability measure

Xander Henderson

@Jakobian Oh, sure. You could do that, too.

Oh, yeah. Okay, I wasn't reading carefully. Yes. I see what you have done.

Jakobian

I want $\mu$ to be a $\sigma$-finite measure, and $f_X = \frac{d P}{d \mu}$

Xander Henderson

@Jakobian Yeah. That's fine. I've no problem with that.

Still, it is the same basic idea.

Anywho... gotta go make dinner.

冥王 Hades

00:38

Koro must be having a blast right now

Akiva Weinberger

01:07

@PM2Ring Nice!! Can you post this as a comment under the question?

PM 2Ring

01:39

@AkivaWeinberger The URL is too big to fit in a comment. But I can post a comment with the chat link.

Akiva Weinberger

TinyURL?

Or chat link works too

@PM2Ring Actually honestly if you post it as an "answer" I wouldn't mind

PM 2Ring

@AkivaWeinberger I'm a bit hesitant to do that, since it doesn't even attempt to answer the question. ;)

I was hoping that a URL for the SVG output would fit in a comment, but it's a bit too long. Here's an example of a symmetrical solution for 6×6 with a single fault line.

sagecell.sagemath.org/…

Temporary SVG file:

That discussion about various ways to define the mean of a distribution reminded me of the various ways to define the mean radius of an elliptical orbit. From my answer astronomy.stackexchange.com/a/49267/16685

> It is often said that the semi-major axis is the "average" distance between the primary focus of the ellipse and the orbiting body. This is not quite accurate, because it depends on what the average is taken over.

冥王 Hades

02:05

When reading the question is harder than the question itself

Thomas Finley

03:04

Does anyone know how to prove that the inverse of the set of quaternions mod p, have an inverse?

My aim is to show that, if a (not equal to 0) be a quaternion mod p then, there must an $\tilde a$ such that $a\tilde a=1$

leslie townes

03:21

i don't know about your specific aim, but i note that the exercise doesn't seem to be asking you to do that.

more specific to that aim, what happens if you compute (i + j + k)^2 mod 3

Ted Shifrin

Leslie for the … win.

Thomas Finley

04:57

0

Q: Show that the possible ideals of a ring of quaternions over integers modulo an odd prime is either the $(0)$ or the set itself.

Using the ring of real quaternions as a model, we define the quaternions over the integers $\text{mod}$ $p,$ $p$ an odd prime number, in exactly the same way; however, now considering all symbols of the form $a_0 + a_1i + a_2j + a_3k,$ where $a_0,a_1,a_2,a_3$ are integers mod $p.$ (a) Prove that ...

@leslietownes If you need more context.

Check the above post where I explained my strategy.

I think it's a common one tho

leslie townes

05:10

oh, i don't need more context. i was just pointing out that the exercise above did not ask you to prove that nonzero elements of your thing were invertible (which is good, because not all of them are). this issue doesn't seem to be a live issue on your linked question. for that, i suggest implementing eric's hint (the first comment).

Thomas Finley

05:24

@leslietownes Mind checkin out the solution I have added in the OP? I think this sorts out the problem!

leslie townes

"as a is nonzero, we have a_0^2 + ... + a_3^2 is nonzero." this is not true in finite fields. see the example i suggested above.

Dannyu NDos

I was playing with the golden ratio base numeral system, and have sought for a signed modification of it.

I tried using 1 and -1 as digits, and found that 8 doesn't have unique representation.

11TT1.1TT1, 11TT1.T111, 1T111.1TT1, and 11TT1.T111 are all 8.

one potato two potato

06:30

Is it ok to ask textbook (graduate level) statement in MO? I'm told that MO is usually for research-level questions.

The textbook here is not like books that cover topics in quals for example.

leslie townes

if it is something particularly tricky, i would think it was OK. for me the line is not so much "in a textbook or not" or "is this 'research level' [whatever that means] or not," but "is this a routine kind of problem or not."

one potato two potato

I'll just ask first in MSE and if nobody answers then move on to MO

leslie townes

i don't think people on MO are too invested in policing the exact boundaries of "research level." they don't like it when people post routine homework problems there (even if they are in "graduate-level" textbooks) and might migrate that kind of thing to MSE.

but semi-obscure or particularly difficult problems that might happen to exist in some textbook, or otherwise be known consequences of known things... i think a lot of MO questions probably satisfy that description.

leslie townes

07:08

if it's that thing about geometric group theory, my guess is nobody answers on MSE who would not also have answered on MO. :)

one potato two potato

'this post will self destruct in 5 seconds'

Koro

07:33

Suppose that $f(x_1,x_2,...,x_m, y_1,y_2,...,y_n)= \sum x_i^2 -\sum y_i^2$. For c=0, f^{-1}(0) is connected (path connected actually) but not sure if f^{-1}(c) is connected or not if $c\ne 0$

leslie townes

07:47

hi koro :) i have no idea about that problem but haven't seen you in a while.

Koro

hi Leslie :).

one potato two potato

08:00

The graph of $x^2-y^2=1$ is not connected

one potato two potato

08:11

there'll be a special lecture in my college about conformal field theory. this kind of class needs some confidence and boldness to register.

Thomas Finley

10:21

Hallo, I am an undergraduate student pursuing a Bachelor's degree in Mathematics and am currently in my 2nd year of study. I have a long dream of pursuing a mathematical carrer abroad specifically in the renowned colleges of Ivy League. I have heard that admissions in masters program or a phd program is highly competitive

Unfortunately, I have no research based backgrounds and neither do I have done any internships. I have a friend, who claims to be an undergrad researcher and has already written/published 1 or 2 papers. I think I am falling behind. Since this place has a great group of mathematical enthusiast/ educators/professionals/researchers, I want some suggestions on what to do?

I mean how do I proceed to fulfill my dreams as if given the chance to study there, I still require scholarships, as I am a non-US resident coming from a lower-middle class family.

And as far as I have heard, the scholarships programs are even more competitive!

Dannyu NDos

What about taking double/dual degree?

I have double degree on Bachelor's degree, one being Applied Math, and the other being Electronic & Informational Engineering.

Sophomore's year is a perfect year to decide whether to do this.

PM 2Ring

@AkivaWeinberger There are 100 tilings with a single fault line. When the fault line is the centre line, all the solutions are minor variations on the one I posted earlier. Here's one with a more obvious colouring:

sagecell.sagemath.org/…

Dannyu NDos

10:51

I mean "double major"; sorry.

Thomas Finley

@DannyuNDos Double major doesn't interest me much as I am not so prepared to do two things at one time

Dannyu NDos

I'd say double major is a kind of insurance, especially for math students. Sadly enough, only a little portion of math students actually become professional mathematicians.

As such, a second major is often helpful. Prominent choices include Physics, Computer Science, and Economics.

Math skills can help learning other disciplines very well. The only exception is pure liberal arts.

David Raveh

11:10

@ThomasFinley I found (as a current senior who has published several papers) that the best way to get involved in research is to explain your position to a professor you have a good relationship with; they may offer you a small project to work on, or they may have a good suggestion of whom to approach. To really get involved required reading a lot to catch up. To be fair, my research is in (mathematical) physics, but I imagine it is similar.

"Double major doesn't interest me much as I am not so prepared to do two things at one time" with physics, these things go together. I solved more PDE's in quantum then I did in PDE's. Often the most intuitive understanding of mathematical theorems came from my physics professors. And a single problem in physics can require many tools from separate fields in math to solve.

Jakobian

11:32

@onepotatotwopotato ew. Physics

Jai Sri Krishna

11:42

Hello Guys, My doubts may seem silly but pls help me out, To prove root is irrational I assume root 2 to be rational and is equal to p/q, and P AND Q are coprime which leads to a contradiction, My doubt is why should p and q be coprime, Even 2/4 is rational....Why can't root 2 be something of the form 2/4?

Jakobian

you can assume they are co-prime

if $d = \gcd(p, q)$ then $p/q = p'/q'$ where $p' = p/d, q' = q/d$ and $q', p'$ are co-prime

123

Hello Everyone..

one potato two potato

@Jakobian good point

Jai Sri Krishna

How Can I convert latex to normal readable form?

user 85795

Like unrender?

Jakobian

11:53

@JaiSriKrishna in chat?

Jai Sri Krishna

Yes :-)

in chat

Jakobian

in the description, you have "LaTeX in chat: link"

user 85795

Refresh

Jakobian

if you're on browser there should be no problem to render it

Jai Sri Krishna

Oh

s.harp

12:08

Anybody got a simplification for $\sum_{k=0}^m \binom{a}{2k+1} \binom{b}{2(m-k)}$ ? trying to find it via google is driving me nuts

Jakobian

looks a bit like en.wikipedia.org/wiki/Vandermonde%27s_identity

s.harp

yeah, the factors of 2 ruin it though

Jakobian

do you need it like this?

s.harp

more or less, honestly i just want the rank of $A_{a,m} =$ above expression with $b=N-a$ to be at least $\lfloor (N-1)/2\rfloor$ (and $a\leq N, m\leq \lfloor (N-1)/2\rfloor$)

easiest way to get there should be to simplify the sum, but there are so many formulas for sums of binomial coefficients that I can't find this one on the net

Jakobian

I wouldn't be surprised if it wasn't on the net

s.harp

12:18

maybe, I also wouldnt be surprised if it was, but I think finding it might be harder than finding the simplification by hand

Jakobian

maybe try $(1+ix)^a(1+ix)^b$ ?

kind of like you prove Chu-Vandermonde but with imaginary number there to assure that we have some even/odd values

after taking real/imaginary part

s.harp

yeah, I was now thinking along the lines of $((1+x)^a-(1-x)^a)(1+x)^b = (1+x)^{a+b}-(1-x^2)^{a}(1+x)^{b-a}$

Jakobian

ah right, negatives not imaginary numbers
imaginary numbers work for multiples of $4$, my bad

I'm not sure if this will work since we're only adding till $m$

the closed form might not exist

s.harp

it probably doesn't, but the sum $k=0$ to $m$ gives $2$ times the $x^m$ term on the left-hand side

user 85795

12:34

Jam

13:05

i understand that open sets in a subspace topology are intersections of an open with the subspace and it is clear.But intuitvely such open sets do not have enough space to be open. For example an open set o [0,1] is say (1/2,1] but what is an epsilon ball withc centre 1 that is a subest of (1/2,1] ? I know i have to use the metric induced for that ball but still dont see it.

Jakobian

13:26

$(1/2, 1]$ is such ball

sunny

Is $$s(x)=\sum_{k=0}^\infty 2^{k}{\sin(3^{-k}x)}$$ uniformly convergent on $\mathbb{R}$? It is pointwise convergent since it is absolutely convergent (using $\lvert\sin(x)\rvert\leq |x|$ we obtain a geometric series). Since the terms are also continuously differentiable and the term-wise differentiated series converges uniformly, we have that $s(x)$ is differentiable. My suspicion is it isn't uniformly convergent, but I can't prove it.

Akiva Weinberger

@PM2Ring Wait, exactly 100?

Jakobian

Weierstrass function

In mathematics, the Weierstrass function is an example of a real-valued function that is continuous everywhere but differentiable nowhere. It is an example of a fractal curve. It is named after its discoverer Karl Weierstrass. The Weierstrass function has historically served the role of a pathological function, being the first published example (1872) specifically concocted to challenge the notion that every continuous function is differentiable except on a set of isolated points. Weierstrass's demonstration that continuity did not imply almost-everywhere differentiability upended mathematics,...

this is related

Jakobian

13:46

If you input $x_m = 3^{m+1}\cdot \pi/2$, then $s_m(x_m) = 2^{m+1}-1$ and $s(x_m)-s_m(x_m) = 2^{m+1} + y_m$ where $|y_m|\leq \sum_{k=m+2}^\infty \left(\frac{2}{3}\right)^k x_m = 3(2/3)^{m+2} $

sunny

I was just about to say...since the function is differentiable it is continuous and since the terms are continuous too, it has to be uniformly convergent.

Jakobian

@Jakobian this is wrong, the last equality is not true

okay no, $y_m$ is positive

$s(x_m)-s_m(x_m) > 2^{m+1}$

this shows that $\|s-s_m\| > 2^{m+1}\to \infty$

well not exactly it's not true that $s_m(x_m) = 2^{m+1}-1$ because the sum will oscillate

should be $s_m(x_m) = \pm (2^{m-1}-2)$

sunny

The theorem I'm using states:

> Theorem. Let $\left(f_n\right)_1^{\infty}$ be a sequence of continuously differentiable functions in the interval $E$. Assume that the sequence is pointwise convergent to the function $f$ and that the differentiated sequence $\left(f_n^{\prime}\right)_1^{\infty}$ is uniformly convergent to the function $g$ in $E$. Then $f$ is continuously differentiable in $E$ and $f^{\prime}=g$.

Jakobian

well, what matters is that this is the type of argument you want to prove it's not uniformly convergent

sunny

ok, but we agree on that the above series is uniformly convergent, or?

Jakobian

13:59

I don't agree

sunny

But if $s(x)$ is differentiable it is continuous and since the terms are continuous, it has to be uniformly convergent, no? This is the uniform limit theorem.

Jakobian

ah alright

you're using that pointwise convergence of $s_m$ and uniform convergence of $s_m'$ (from Weierstrass M-test) implies uniform convergence of $s_m$

okay I agree now

@sunny no, it's not

if $s_m$ is uniformly convergent to $s$, and $s_m$ are continuous, then $s$ is continuous

this is the theorem you cite

but $s_m$ can all be continuous, converge to continuous function, but not uniformly

sunny

ok, I see

Jakobian

okay I misunderstood you, but maybe that's for the better since we see it's uniformly convergent now

sunny

wait...I don't, can we reiterate why it is uniformly convergent?

Jakobian

14:08

Uniform convergence

In the mathematical field of analysis, uniform convergence is a mode of convergence of functions stronger than pointwise convergence. A sequence of functions ( f n ) {\displaystyle (f_{n})} converges uniformly to a limiting function f {\displaystyle f} on a set E {\displaystyle E} as the function domain if, given any arbitrarily small positive number ϵ {\d...

if a sequence $f_n$ converges pointwise, and $f_n'$ converge uniformly, then $f_n$ converges uniformly

this is part of theorem about uniform convergence of derivatives

Jam

@Jakobian whats the radius? around 1? 0?

sunny

@Jakobian ok, thanks a lot

Jakobian

@Jam 1/2

@sunny you said you're using Spivak, did he get to that theorem yet?

Jam

1+1/2 then is not a subset of the interval

Jakobian

@Jam what do you mean

Jam

14:10

i mean intuitevly there is no room around 1 in the subset

Jakobian

your intuition is not mine intuition and honestly I don't get it

Jam

we say that an open set in a metric space need to have for every point ||x-xo||<r as a subset

d(x,xo)<r

Jakobian

element

not subset

sunny

@Jakobian hmm, I don't think it's in his book, at least I haven't found it in the way you stated it under the Wikipedia link (maybe he has hidden it in some exercises). There is a similar theorem to the one I stated above in bold.

Jakobian

@Jam getting back to this, $1+1/2$ is not in the open ball

If $d(x, y) = |x-y|$ is the Euclidean metric then on $[0, 1]$ we consider the restriction $\rho = d\restriction_{[0, 1]}$ instead, that is $\rho(x, y) = d(x, y)$ for $x, y\in [0, 1]$

Thus $B_\rho(x, r) = \{y\in [0, 1] : \rho(x, y) < r\}$

Jam

14:15

ok now it is clear

thanks

Jakobian

it wouldn't make sense to ask if $1+1/2$ is in $B_\rho(x, r)$ because you can't plug in $1+1/2$ into $\rho$

Jam

exactly thanks

so balls in subspaces are also half balls etc

Jakobian

@sunny could you provide the theorem?

sunny

@Jakobian yup, give me a moment

Jakobian

@Jam for any non-empty open set $U$ in a metric space $(X, d)$ there is an equivalent metric $\rho$ on $X$ such that $U = B_\rho(x, 1)$ for some $x\in U$

in some sense any open set can be a ball

Jam

14:17

yes is just that intuitevly ud want a open set to have some room around every of its elements

but if you whole space doesnt have room to begin with u dont mind it

Jakobian

this is the place where your intuition fails I suppose

Jam

:(

Jakobian

I don't think of it as a property that a metric space should posses

quite the opposite, it's what makes it interesting to study

in fact the property of being an endpoint of closed interval is a topological property

sunny

@Jakobian Ok, this is exactly how it's stated in theorem 3, chapter 24 in his book:

> Theorem 3 Suppose that $\{f_n\}$ is a sequence of functions which are differentiable on $[a,b]$, with integrable derivatives $f_n'$, and that $\{f_n\}$ converges uniformly on $[a,b]$ to some continuous function $g$. Then $f$ is differentiable and $$f'(x)=\lim_{n\to\infty} f_n'(x).$$

Contrast this to:

13 mins ago, by Jakobian

if a sequence $f_n$ converges pointwise, and $f_n'$ converge uniformly, then $f_n$ converges uniformly

I think he's leaving something out.

Jakobian

it's a different kind of theorem

it has stronger assumption on convergence of $f_n$, something we don't want

it's still interesting because it gives a condition on exchanging derivative with a limit without demanding uniform convergence of derivatives

sunny

14:26

The Wikipedia article you linked seems to cite from Rudin's. It's indeed a different theorem.

Jakobian

it should be in Spivak

it's a very standard theorem

@sunny I've checked in Spivak and this is a different theorem

he assumes $f_n$ converges pointwise to $f$ and $f_n'$ converge uniformly to $g$

it's theorem 3 in Spivak but you wrote it wrong

not sure why he assumes $f_n'$ are integrable though

sunny

oh wow, I did indeed write it completely wrong, let me rewrite it:

> Theorem 3 Suppose that $\{f_n\}$ is a sequence of functions which are differentiable on $[a,b]$, with integrable derivatives $f_n'$, and that $\{f_n\}$ converges (pointwise) to $f$. Suppose, moreover, $\{f_n'\}$ converges uniformly on $[a,b]$ to some continuous function $g$. Then $f$ is differentiable and $$f'(x)=\lim_{n\to\infty} f_n'(x).$$

Jakobian

it's basically the theorem on wikipedia but with extra (unneeded) assumptions

sunny

Well, he leaves out the most important part I think. He doesn't state that $f_n$ in fact converges uniformly as a consequence.

@Jakobian ah, I found it, problem 25b) in chapter 24:

> In Theorem 3 we assumed only that $\{f_n\}$ converges pointwise to $f$. Show that the remaining hypotheses ensure that $\{f_n\}$ actually converges uniformly to $f$.

Jakobian

14:42

Try something like $|f_n(x)-f(x)|\leq |f_n(x)-f_n(a) - (f(x)-f(a))| + |f_n(a)-f(a)| \leq \int_a^x |f'-f_n'| + |f_n(a)-f(a)| \leq |x-a| \|f_n'-f'\| + |f_n(a)-f(a)|$

ah, sure

Martin Sleziak

16:04

@s.harp Have you tried searching with Approach0? (Or SearchOnMath - but that one isn't free.)

This seems a bit similar, but not the same: Combinatorics That Looks Similar to Vandermonde's Identity and Combinatorics simplification.

one potato two potato

what is the motivation of automatic groups? Give me one surprising theorem

Jakobian

@Thorgott math.stackexchange.com/questions/4749032/…

do you know perhaps if I can find $\Phi$ such that $\Phi(a_i) = b_i$ for all $i$?

I'm not sure if finding such $F$-homomorphism is possible

Xander Henderson

@onepotatotwopotato Theorem: Xander has no idea what an automatic group is, and he does not care to know.

Proof: Meh.

Frankly, I'm pretty shocked by that theorem!

冥王 Hades

16:24

Theorem: Touching boiling hot water causes pain

Proof: My finger covered in bandages

one potato two potato

Oh this is actually a good chance to learn something Xander doesn't know.

user 85795

Where Meh is by definition.

Or is it a given fact @XanderHenderson

Proof by Meh.

What justifies the use of meh, certainly not curiousity.

Xander Henderson

16:43

@user85795 I mean, I can't really be arsed. Hence... meh.

geocalc33

16:58

I just came up with a question but it's pretty meh

I don't think anyone can answer it in teh world

one potato two potato

well even if you don't know what that is, you can pretend you know. Just say the magic words: oh, it's ... a notion in a category theory. You're 100% correct.

geocalc33

oh it's ....category theory

Ted Shifrin

@冥王Hades Bandages and pain may be correlated, but I don’t think there is a valid implication.

Thorgott

@Jakobian what is $L$

Jakobian

a field

more precisely it's a field obtained by trying to extend the map $F\to R'$ to an order preserving map, and using Zorn's lemma

The purpose of taking $a\in R\setminus L$ is to obtain a contradiction

I want to prove $L = R$

for this I need $\Phi$ to be order-preserving

I've tried looking in Jacobson's third book, but he's doing some kind of magic there

Artin-Schreier theory is in the last chapter of that

I've also been looking in Squares by Rajwade, but they don't give the proof of uniqueness and redirect to Jacobson

@Jakobian the extension of $\Phi$ to $L(a)$ to be more precise

Thorgott

18:19

@Jakobian why are these roots all roots?

冥王 Hades

@TedShifrin you’re right. Should I try it with my right hand too?

Ted Shifrin

Why not? Maybe try flames on that?

冥王 Hades

18:36

For some reason touching flames isn’t as painful as touching boiling hot water

I probably learned the reason in thermodynamics classes but I forgor

Ted Shifrin

Yeah, true.

geocalc33

19:11

$$ \frac{1}{(n+1)!}E\big(|\det(Q)|\big). $$

I'm having a tough time with this expression. Does it necessarily go to zero as $n \to \infty$? I think it does but can't be sure yet. $Q$ is a random matrix, which I can give more details on.

copper.hat

depends in what random means.

geocalc33

where $Q$ is a $(n+1)\times(n+1)$ matrix whose $i$th row is $(1, \mathbf{ \bar X}^{(i)}),$ $i=1,2,\cdot\cdot\cdot , n+1.$

Here I'll use a normalized version of $\mathbf X$ denoted by $\mathbf{\bar X}$ such that $\bar X_i= X_i/E(X_i),$ $i=1,2,\cdot\cdot\cdot, n.$ Consider $n+1$ $\mathrm{iid}$ $n$-dim. random vectors $\mathbf{ \bar X}^{(1)},\cdot\cdot\cdot, \mathbf{ \bar X}^{(n+1)} $ each with the same distribution as $\mathbf{\bar X}.$

I'm just going to choose $\mathbf X$ to be Gaussian.

$\mathbf X$ is a random vector that's Gaussian distributed

Xander Henderson

What are the mean and variance of $\mathbf{X}$?

Oh, wait. I see... you are normalizing them to mean 1?

copper.hat

why make it so complicated, just start of saying $X$ is Gaussian with mean 1.

geocalc33

$\mathbf X$ is Gaussian with mean $1$

variance I want to be something simple like $1$

That's if all the random varaibles all have the exact same mean and variance.

Jakobian

19:58

@Thorgott wdym

Thorgott

you claim that all roots of $f$ are in $R$ in the post

Jakobian

I meant that they are all roots over the respective fields, I'm pretty sure

so that both $f$ and $g$ have $n$ roots in $R$ and $R'$ respectively

@Thorgott so when I'm saying that, I mean that I take all of the roots of $f$ which are in $R$

and similarly all of the roots of $g$ which are in $R'$

what I meant to point out here is that they have the same amount of roots in total

s.harp

20:15

@MartinSleziak that tool is surprisingly strong!

from the answers you link it seems there is no good closed form. mathematica gives me some nightmare expression with hypergeometric functions

Xander Henderson

20:48

Oh, I hear thunder!

Maybe we'll finally get some monsoonal rain!

geocalc33

20:59

I guess the limit is 0 under those conditions

It's expected to rain here too

leslie townes

monsooner or later

Xander Henderson

@leslietownes Booooo!

Jakobian

what I imagine is going to happen is either someone like Eric Wofsey is going to answer me, or I'm going to go through the proof in Jacobson and answer myself

I think I need primitive element theorem for this?

Jakobian

21:25

what Jacobson does, I think, is he takes a polynomial which has $a_i$ and $\sqrt{a_{i+1}-a_i}$ as roots, and then takes primitive element

then uses the theorem about extending isomorphism to simple extensions

sunny

I want to show $\log x \leq x-1$ holds for all $x>0$. It is simple for $x\geq 1$, since then $1/x\leq 1$ and so $$\log(x)=\int_{1}^{x}{\frac{1}{t}\ dt} \le \int_{1}^{x}1 \ dt =x-1. \tag1$$ However, when $0<x<1$ then I just get $$\log(x)=\int_{1}^{x}{\frac{1}{t} \ dt}=- \int_{x}^{1}{\frac{1}{t} \ dt}\le 0 < x, \tag2$$ which is not really what I want. Any ideas on how to resolve this?

Jakobian

@Thorgott see the update to my answer, I think this is solution I'm seeking

if you could just double check that what I wrote makes sense, I'd appreciate that

Xander Henderson

21:44

@sunny Here's a dumb idea: $x - 1 = \int_{1}^{x} \,\mathrm{d}x$, right? So $(x-1) - \log(x) = \int_{1}^{x} 1 - \frac{1}{t} \,\mathrm{d}t. $ The integrand is positive when $x > 1$ (so the integral is positive there), and the integrand is negative when $x < 1$ (so the integral is again positive).

Thorgott

@Jakobian why is that?

Jakobian

@Thorgott because of Sturm theorem

field isomorphisms commute with derivatives and gcd is mapped to gcd, so the Sturm sequence is mapped to Sturm sequence of the same length

coefficients of those polynomials don't change sign either since the isomorphism is order preserving

so from Sturm theorem you can conclude they have the same amount of roots

from what I've seen this is actually pretty crucial in proving the uniqueness of real-closure of a field

that's part of the reason why some people don't include the proof of uniqueness, like Rajwade in his Squares

they'd have to develop the theorems of basic calculus for polynomials over real-closed fields, and then prove Sturm theorem

well, you can just trust me on that one

after I wrote the thing with roots, I'm actually concerned because I don't really see how it proves that $\Phi$ is order-preserving

well, I'm sure this boils down to some kind of maximality of positive cone argument

sunny

22:01

@XanderHenderson I like it :) makes sense, thank you

Jakobian

I still don't get it, I don't know

CroCo

Hi everyone, I came across this question today
https://math.stackexchange.com/questions/4748751/when-is-a-a-top-psd-even-though-a-is-not-psd

The OP is asking "When is ($𝐴+𝐴^T$) positive semi definite (PSD) even though A is not PSD?"
I think this is impossible and this is my proof

CroCo

Let $x^T [A+A^T]x \geq 0$ and $x^TAx < 0$ (i.e. not positive semi-definite). We know that any matrix can be decomposed into symmetric and antisymmetric matrices that is $A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$ and $x^T A^T x=x^TAx$ (i.e. scalars are symmetric). We have
$$
x^T A x < 0 \implies \frac{1}{2} x^T(A+A^T)x < 0
$$
which contradicts our assumption; therefore, $A$ must be positive semi definite.

Is my proof correct before I post it?

Sorry I'm terrible at proofs.

Thorgott

22:23

@Jakobian ah ok, that's neat

@Jakobian what you wrote makes sense

but I don't see how it implies that $\Phi$ is order-preserving either

Jakobian

22:51

I think what Jacobson is getting at, is not that $\Phi$ is order-preserving, but we can use those type of extensions to define an isomorphism from $R$ into $R'$

For $a\in R$ you take minimal polynomial $f$ of $a$, and then send $a$ to root $b$ of $f$ in $R'$, but so that the order of roots is preserved like here

and then you want to prove this is actually an isomorphism

this way we don't need to prove it's order-preserving because ordering of a real-closed field depends only on squares in that field

so it's automatically order-preserving

this works I guess but I think it's a slightly different argument than what authors of that book about real algebraic geometry were getting at

still, I don't know what they were getting at

sunny

23:10

I'm trying to show $$\sum_{k=1}^\infty \log\left(1+\frac{1}{k^2x}\right)$$ converges uniformly on $(0,\infty)$. Is it then enough to only consider intervals of the form $[d,\infty)$ where $d>0$?

leslie townes

no, it isn't enough. i mean, maybe something specific that you are doing with that function on those intervals will work for the whole thing, but that won't work as a general strategy.

Axoren

For any such $d$, if you can always show it converges, then you have an inductive proof. But do you have that?

Jakobian

@sunny it'd be enough if you were trying to show continuity but not uniform convergence

leslie townes

an example worth keeping in mind is 1 + x + x^2 + ... which converges uniformly (to 1/(1-x)) on [-a,a] for any 0 < a < 1, but not does not converge uniformly on (-1,1) (in fact, sup_{x in (-1,1)} |f_n(x) - f(x)| = +oo for every n, where f_n is the nth partial sum and f is the pointwise limit)

sunny

hmm ok

I was thinking about using $\log(x)\leq x-1$ and then, for $x\in [d,\infty)$, $$\left|\ln\left(1+\frac1{k^2x}\right)\right|\leq \left|\frac1{k^2x}\right|\leq \frac1{k^2d}.$$

Axoren

23:18

I missed the uniformly convergent. They may be a spike to upwards to infinity somewhere on $(0, d)$ for any such chosen $d$. Your argument won't show that that spike does not exist.

At least showing that that spike doesn't exist on $[d, \infty)$ doesn't show that it doesn't exist on $(0, \infty)$.

sunny

alright, so there's a problem

@Jakobian why is that?

Jakobian

@sunny because they're simply not equivalent

so you can't do that... because you can't

a better question would be, why would you be able to do that in the first place?

leslie townes

continuity is a purely local thing, so if it holds at each point of interest, it holds everywhere

uniformity of convergence isn't a purely local thing

Axoren

Very specifically, it's a global thing explicitly in the definition. It must be shown true at every point in the domain at the same time*

.

sunny

Ok, suppose I'd like to show $$s(x)=\sum_{k=1}^\infty \frac{\arctan kx}{1+k^2x^2}$$ is continuous on $x>0$. Then I have to show $s(x)$ converges uniformly on $(0,\infty)$, but then we're running into the same problem as above, don't we?

Axoren

23:27

Same problem when trying to show it's continuous? Or uniformly convergent?

leslie townes

maybe some wires got crossed above. to show that f is continuous on (0, oo) it's enough to show it's continuous on (d,oo) for all d > 0. to show that f_n converges uniformly to f on (0, oo) it's not enough to show that f_n converges uniformly to f on (d, 00) for all d > 0.

i think that's the content of jakobian's remark.

a common way of proving that a pointwise limit of a sequence of continuous functions is continuous is to prove that the convergence is uniform

and in situations where you can't prove that, you might not at first glance have any other way of knowing that the pointwise limit is continuous

so in that setting, proving continuity of the limit function might not be appreciably 'easier than' proving uniformity of the convergence

but that isn't quite what was going on up above

sunny

@leslietownes Ok, so how would you show $s(x)$ (the above series) is continuous on $x>0$? Also @Axoren

leslie townes

if you have shown the uniform convergence on (d,oo) for every d > 0 then you've done that. have you done that?

sunny

yes

more or less

I have shown it converges uniformly for $x\geq d $ where $d>0$.

Jakobian

@Thorgott math.stackexchange.com/a/4749242/476484 could you check this proof? Thanks

leslie townes

23:36

i mean, that's probably how i'd do it. then i'd look about what (if anything) i was actually using about positivity of d to get uniformity, and seeing if that means there is some obstacle to there being uniform converngece on (0, oo)

Axoren

For uniform convergence, you would need a singular $\varepsilon$ such that $|f_n(x) - f(x)| < \varepsilon$ for the entire domain $(0, \infty)$. Generally, with a spike at infinity at 0, a function could have an $\varepsilon_5$ = 5 at $d = 5$, but then may require a $\varepsilon_4 = 10$ for $d = 4$. The further towards $d \to 0$, the larger the $\varepsilon$ we must use to satisfy the partial domain. As you can see, your selection of $\varepsilon$ actually diverges as you continue this process!

As for whether or not uniform convergence of the elements of the cover is sufficient, I'd rather defer to leslie on that one.

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