Mathematics - 2023-09-30

4:43 AM

Hello everyone, I am unable to access math.se website and I get a page which says the connection between SE servers and cloudflare is timed out. Is anyone facing this issue? I hope this is temporary.

leslie townes

paramanand: the main site loads for me right now. i haven't tried posting in weeks, but commenting seems to work as of a few minutes ago.

Paramanand Singh

Maybe the issue lies with the cloudflare servers in Chennai, India. At least that's what the error page says. I checked on stackstatus.net and it says all the stackexchange sites are working fine.

Ted Shifrin

Yeah, I’ve been on the site off and on all day.

Happy goose to Munchkin.

leslie townes

munchkin is turning 5 soon.

Ted Shifrin

Time doth fleet!

She’ll soon be kicking her parents out.

leslie townes

5:06 AM

yes.

Thomas Finley

0

Q: Let R be a ring with unit element, R not necessarily commutative, such that the only right-ideals of R are (0) and R. Prove that R is a division ring.

Let $R$ be a ring with unit element, $R$ not necessarily commutative, such that the only right-ideals of R are $(0)$ and $R.$ Prove that R is a division ring. This was a problem from the book, "Topics in Algebra" by I.N Herstein from Chapter- 3 (Ring Theory) , Page number- 139 (2nd Edition). I tr...

I need some help with this excercise from Herstein's book.

Paramanand Singh

I am able to access the site again. So it was more of a temporary and local problem.

Thomas Finley

6:14 AM

0

Q: A problem from the book, "Topics in Algebra" by I.N Herstein from Chapter- 3 (Ring Theory) , Section-3.5, Page number- 140 (2nd Edition)

Let $R$ be a ring such that the only right ideals of $R$ are $(0)$ and $R.$ Prove that either $R$ is a division ring or that $R$ is a ring with a prime number of elements in which $ab = 0$ for every $a,b\in R.$ This was a problem from the book, "Topics in Algebra" by I.N Herstein from Chapter- 3 ...

abstract-algebra ring-theory

I need some help with this excercise from Herstein's book.

ZaWarudo

6:46 AM

I would like to ask for (the slightest) hint to prove this: "Let $f_n$ be a sequence of continuous functions such that $f_n \to f$ uniformly on $(a,b)$. Show that $f_n \to f$ uniformly on $[a,b]$." I actually have an idea, but I don't know if it works. My idea is that $\lim_{n \to +\infty} \sup_{x \in [a,b]}|f_n(x)-f(x)|=\lim_{n \to +\infty} \sup \left(|f_n(a)-f(a)|,\sup_{x \in (a,b)}|f_n(x)-f(x)|,|f_n(b)-f(b)|\right)$.

Using that $f_n \to f$ uniformly on $(a,b)$ and that $f_n$ is continuous, I deduce that $f$ is continuous on $(a,b)$ and so $|f_n-f|$ is continuous on $(a,b)$, hence $\lim_{n\to+\infty} |f_n(a)-f(a)|=\lim_{n\to+\infty} \lim_{x \to a^+} |f_n(x)-f(x)|$.

Being $x \to a^+$, I can assume $a<x<b$ and so I can now interchange the limits because of the uniform convergence on $(a,b)$, so $\lim_{n\to+\infty} \lim_{x \to a^+} |f_n(x)-f(x)|=\lim_{x \to a^+} \lim_{n\to+\infty} |f_n(x)-f(x)| \le \lim_{x \to a^+} \lim_{n\to+\infty} \sup_{x \in (a,b)}|f_n(x)-f(x)|=0$ (because $f_n \to f$ uniformly on $(a,b)$).

Similarly, $\lim_{n \to +\infty} |f_n(b)-f(b)|=0$. Hence, $\lim_{n \to +\infty} \sup \left(|f_n(a)-f(a)|,\sup_{x \in (a,b)}|f_n(x)-f(x)|,|f_n(b)-f(b)|\right)=0$ because each one of the three sequences in the supremum tends to $0$. So, $f_n \to f$ uniformly on $[a,b]$.

Oh no, this doesn't work I'm afraid. By the continuity on $(a,b)$ I cannot deduce that $|f_n(a)-f(a)|=\lim_{x \to a^+} |f_n(x)-f(x)|$; I need the continuity at $x=a$ too.

leslie townes

putting aside any question about uniformity, can you show that the limit lim_n f_n(a) exists and is f(a) [and similarly for b]? what are the domains of f_n and where are they assumed continuous?

ZaWarudo

Sorry for not writing that, the domains are $[a,b]$ and they are continuous on $[a,b]$ for each $n\in\mathbb{N}$.

For the other thing you asked, I think that the reasoning I used above works. Since $f_n$ is continuous on $[a,b]$ for each $n\in\mathbb{N}$, we have $f_n(a)=\lim_{x \to a^+} f_n(x)$. If $x \to a^+$, we have $x>a$ and so by uniform convergence of $f_n \to f$ in $(a,b)$ I can exchange the limits and get $\lim_{n \to +\infty} f_n (a)=\lim_{n \to +\infty} \lim_{x \to a^+} f_n(x)=\lim_{x \to a^+} \lim_{n \to +\infty} f_n(x)=\lim_{x \to a^+} f(x)=f(a)$.

No wait, again the same mistake. I need the continuity of $f$ in $x=a$ now to say that $\lim_{x \to a^+} f(x)=f(a)$.

Jakobian

11:02 AM

I see a lot of symbols in chat

Sine of the Time

Any idea to prove the quasi triangle inequality for this quasi-norm: $N(x,y,z)=(\sqrt {|x|}+\sqrt {|y|}+\sqrt {|z|})^2$

Jakobian

so the $\ell^{1/2}$ one

Sine of the Time

I'm not familiar with that, we use $\ell$ for spaces of sequences

Jakobian

its standard to use $\|x\|_p = \sum |x_i|^p$ for the quasi-norm in $\ell^p$ for $0 < p < 1$, just here its squared

here its squared but I don't think it'll be a problem

Sine of the Time

yes, that's true

Jakobian

11:08 AM

I guess this one isn't homogeneous though so that's why the square

Sine of the Time

it's homogeneous

but I can't prove the quasi triangle inequality

Jakobian

the one I wrote isn't because I didn't take power of 1/p

Sine of the Time

yes

Jakobian

you have some inequality of the form $(x+y)^p \leq 2^{p-1}(x^p+y^p)$ for $x, y\geq 0$

Sine of the Time

it's proven in a chapter after this exercise, but it does the job

Jakobian

11:14 AM

$f(x) = x^p+1 - (x+1)^p$ then $f'(x) = px^{p-1} - p(x+1)^{p-1}$ and $f'(x)\geq 0$ for $0 < p < 1$ so that $f$ is increasing. Moreover, $f(0) = 0$.

Sine of the Time

thank you

user123234

does someone know about affine varieties?

Jakobian

@SineoftheTime yeah I mixed it up because you use $d(x, y) = \sum |x_i-y_i|^p$ to have a metric so it satisfies triangle inequality

what I've wrote proves that $(x+y)^p \leq x^p + y^p$ for $0 < p < 1$ and $x, y\geq 0$

But you still have squares so you need something like... $(x+y)^2 \leq 2x^2+2y^2$

Sine of the Time

yes, I' m working on the squares

Jakobian

this should go from Jensen's inequality and convexity of $x^p$ for $p > 1$

@ZaWarudo If $f_n\to f$ uniformly on $A$ and on $B$, then $f_n\to f$ uniformly on $A\cup B$. And $f_n(x)\to f(x)$ iff $f_n\to f$ uniformly on $\{x\}$. Thus if we show that $f_n \to f$ pointwise on $[a, b]$, then $f_n\to f$ uniformly on $[a, b]$ as well

@ZaWarudo is $f$ assumed to be continuous as well? If so, then you can prove that $f_n$ converges pointwise to $f$ on $[a, b]$

I assume so since otherwise the question wouldn't make much sense

The essential thing to use here is the following inequality: $$|f(a)-f_n(a)| \leq |f(a)-f(x)| + |f(x)-f_n(x)| + |f_n(x)-f_n(a)|$$

where $x \in (a, b)$

you can pick $N$ large enough so that for $n \geq N$ the middle term is small, and the other two terms will be small from continuity, by picking $x\in (a, b)$ close enough to $a$

porridgemathematics

1:11 PM

is it true that if $u : \mathbb{C} \rightarrow \mathbb{R}$ is a harmonic function for which $0$ is not a critical value, then $u^{-1}{0} \neq \emptyset$ has finitely many components? I know that there should be at most countably many components.

nvm, that was a silly question

what I mean to ask is this, suppose $u$ is harmonic on some open subset of the Riemann sphere, $a < b$ are non-critical values of $u$. Consider the regular domain $\{ a \leq u \leq b \}$. Is it possible for this things boundary to have infinitely many components?

also we may assume its a compact regular domain.

so the boundary consists of a bunch of analytic jordan curves

PNDas

2:09 PM

Is $f:\Bbb R\to\Bbb R$ defined by $$f(x)=\begin{cases}\sin(x)&x\in\Bbb R\setminus\Bbb Q\\\frac1{|x|+1}&x\in\Bbb Q\end{cases}$$ measurable?

porridgemathematics

oh, again the answer is certainly yes.. because the open set $u$ is defined on to begin with could be disconnected..

im guessing even if it was a domain, the answer could still be yes, but an obvious example showing this may be doesnt come to mind

@PNDas yes it is

PNDas

@porridgemathematics How to show that?

porridgemathematics

use the definition of measurability

in general things like this will always be measurable, its only when you start constructing things using constructions that require uncountably many intersections do you need to be cautious

and usually for things like that there are ways to approximate your object by sequences of measurable functions

PNDas

$f^{-1}(O)$ is measurable for all open set $O$ in $\Bbb R$. If $O\subset [-1,1]^c$ then $f^{-1}(O)$ is countable or empty so its measurable but otherwise I'm not able to see it.

@porridgemathematics I agree.

porridgemathematics

continuous functions are measurable, your function has two branches, and its measurable on both of them

now use the definition of measurability

PNDas

2:18 PM

Ah yes.

They also asked is it true that $f(E)$ is measurable for every non empty bounded open subet $E$?

My guess is false but how do I show that?

I guess I have to consider $[0,2\pi]$ then it's image is of measure 1. so there exists a non measurable set in the image

and take its preimage?

porridgemathematics

i think its true

PNDas

@porridgemathematics Why?

porridgemathematics

because both of these functions are basically invertible

i.e. they both essentially have continuous inverses (so measurable inverses)

PNDas

Sorry I wrote it wrong. They also asked is it true that f(E) is measurable for every non empty bounded subset E ?

porridgemathematics

ah, this is likely to be false

if E doesnt even need to be measurable

PNDas

2:31 PM

Yes

porridgemathematics

so I think you could prove this is false as follows: consider [0,pi/2], take a non-measurable irrational subset of it, and take f of that subset, f restricted to [0,pi/2] - Q is bi-measurable, so the image wont be

PNDas

@porridgemathematics What I said is also correct right? chat.stackexchange.com/transcript/message/64504683#64504683

porridgemathematics

it can work, once you establish that f(G) is measurable whenever G is

its simpler to just establish that for one of the branches where its invertible imo

g2g

EE18

2:49 PM

Morning all! I'm just learning some formal theory about polynomials for the first time and my book alludes to but does not out and out say if the following is true, so I am guessing:

That is the lemma as I have guessed it. Is it correct?

Oops, should be $\leq$ in both

Lukas Heger

@EE18 with that correct it is correct

well the "if and only if" is a bit vague

EE18

I see, thank you. The part I am in particular worried about is the "if and only if". The book only states the "if"

Lukas Heger

it's true that $R$ has no zero divisors if and only if for all polynomials p,q the inequality holds

EE18

I guess you're saying for the "only if" I should have a "for all p,q"

Lukas Heger

@EE18 yes, exactly

if you put that it's correct with an if and only if

EE18

2:52 PM

OK, I will just rewrite and put back here for verification if that's OK

Lukas Heger

sure

EE18

Probably there are nicer ways to write it, which gives me a lot more respect for textbook authors who do so so easily!

Lukas Heger

there is another variation of this: in (b) you have equality as soon as the leading coefficients are non zero divisors, even if $R$ has some other elements which are zero divisors. But what you wrote is correct

EE18

Noted, I will put that in my notes right now. Thank you!

When you say leading coefficients are nonzero divisors, can we further weaken that to say "even if they are zero divisors, as long as the other is not an element which is such that $ab = 0$"?

That is, for any two specific $p,q$ we have equality if and only if the leading coefficients are not such that $p_{\deg{p}} q_{\deg{q}} = 0$.

Ted Shifrin

3:18 PM

Um, check the second one.

EE18

Oof, yes I missed a plus sign. That's what you mean right?

Ted Shifrin

Rather important, yes.

EE18

I missed it twice!

Thanks for catching that Prof. Shifrin

Ted Shifrin

You could do composition ….

EE18

Sounds painful! I wonder if they exclude that just because composition of polynomials doesn't come up so much in analysis proper? But I defer to you, I have no clue if it does or doesn't

EE18

3:33 PM

Also, am I right to be suspicious about informality as in the sums without upper bounds in (8.14) and (8.15) here:

That is, they should have bounds in terms of the degree of the given polynomial, no?

I wrote something like that instead for comparison to (8.14)

Soumik Mukherjee

why deg $p+q$?

It should be max$\{\textbf{deg} p, \textbf{deg} q\}$

EE18

3:49 PM

That would not be correct, because it could be strictly less than that maximum right?

Or I guess it would be with leading coefficient equals zero

True true, I see what you mean

Ted Shifrin

4:04 PM

You can cancel out the highest degree stuff, so not equal.

EE18

Per Soumik's suggestion, I went with the above. Is that what you mean Ted?

Ted Shifrin

Personally, I don’t write the formulas out like this. I look for the highest degree term. You still haven’t discussed why the degree may be <.

EE18

The strict inequality would obtain iff $p,q$ have the same degree and their leading coefficients are additive inverses of each other, right?

I agree that to look for highest degree term is good, but I guess I'm trying to be general and to write something down which holds both for the equality and inequality case, since $p,q$ are arbitrary here

Hopefully that makes sense?

Can anyone give a hint for how you would argue for the equality (the starred * equality) of the following expressions in some commutative ring $R$. That is, how to "prove" the Cauchy product is the correct expression. I'm thinking induction on the degree of one of the two polynomials?

$$\left(\sum_{j=0}^{\deg (p)}p_jx^j\right)\left(\sum_{i=0}^{\deg (q)}q_ix^i\right) = \sum_{j=0}^{\deg (p)}\sum_{i=0}^{\deg (q)}p_jq_ix^{j+i} \stackrel{*}{=} \sum_{n=0}^{\deg (pq)}\left(\sum_{j=0}^n p_jq_{n-j}\right)x^n.$$

Wikipedia and MathSE answers always talk about infinite series and how essentially this is the definition of products in such cases, but I am looking to prove it simply from the ring structure in the "finite sum" case.

sunny

6:03 PM

Is anyone familiar with Teschl's book Ordinary Differential Equations and Dynamical Systems? I have a question about his definition of a sub solution. It's quite short. Let me know if you are interested :)

Sha Vuklia

6:59 PM

I'm doing the exercise above. I managed to show (a), and (c) follows from (b). I've also noticed that (b) <=> (d). I could prove (d) by proving that $\mathbb Z[(1+\sqrt{-3})/2]$ is a PID (by showing it's Euclidean), and then I'm done, however I don't think this is how the exercise is intended. Does anyone see how (b) follows from (a), assuming that the exercise is set up logically?

I have an idea

I should focus on the second part of (a) I believe for the solution

(which I initially got wrong)

leslie townes

7:34 PM

i have no idea but yeah my first question would be does the case of equality in (a) have anything to do with the cases identified in (b)

Sha Vuklia

still working on that x'D

F.White

9:15 PM

@ShaVuklia Do let us know if you work it out.

Sha Vuklia

@F.White sure will do (so far no luck, but I'm going to discuss this exercise with two fellow students tomorrow, so hopefully that will yield something)

lone student

10:19 PM

@ParamanandSingh Hello, I needed some guidance here. If possible, can you leave a comment on my answer to confirm whether my answer to this question is correct conceptually rather than algebraically? How consistent is my solution? Does the solution contain a trivial conceptual error? Thank you .

shintuku

didn't someone here say something about the fact that the faculty to admin ratio shrank in the past couple of decades?

as a result, do faculty have fewer administrative tasks they had before?

leslie townes

11:08 PM

shin: "admin" might be shifting meanings there. a lot of for lack of a better term 'secretarial' work is now done by faculty themselves, or by software. "admin" can also refer to the bureaucratic apparatus that supports a university (e.g. presidents, vice presidents, assistants to those people, etc.). that has generally exploded over the last few decades, to the benefit of nobody.

if you look at the highest paid employees of most universities (outside of coaches if they have them, and profs at medical schools) it will be people with titles like executive VP to the vice dean and nobody will know what that person does or why they are paid so much.

if you get a job in management consulting maybe you can tell us. that will be your job.

shin: capitolweekly.net/tuition-uc-administrators-tripled-csu-data is an article with some examples of the size of the expansion, compared across two state systems.

Sidharth Ghoshal

What prevents math.stackexchange or mathoverflow from being overrrun with spam?

Is it just that nobody has bothered trying to attack the site?

leslie townes

11:24 PM

i think that they have some automated systems that work fairly well for the spam activity they do get, and the voting/flagging system does the rest. but that is probably a big part of it - for whatever reason it just isn't perceived of as a high value target for spam.

but enough about that, why don't you have a look at some products and services that might interest you. i have a whole website full of them.

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