Mathematics - 2023-05-29

12:00 AM

@TedShifrin Who, me?

I have a graph. You assign a probability distribution to the vertices - each vertex gets a probability between 0 and 1 inclusive, and they add up to one.

I will choose two vertices independently at random according to your probability distribution.

You win if those vertices are distinct and joined by an edge. I win if the vertices are either equal, or are distinct and not joined by an edge.

(a) What is the formula for your probability of winning, given the probability distribution? (b) How do you maximize your chance of winning?

geocalc33

12:32 AM

@AkivaWeinberger I would like to use the K-distribution here

$f_X(x; b, v)= \frac{2b}{\Gamma(v)} \left( \sqrt{bx} \right)^{v-1} K_{v-1} (2 \sqrt{bx} )$

Akiva Weinberger

1:53 AM

@geocalc33 No idea what that is but I don't think it's right

Koro

5:57 AM

I want to show that given a manifold with the maximal atlas $\{(U_a, \phi_a)\}$, and a point $p\in M$ with a neighborhood $U\ni p$, there exists a $U_b$ such that $p\in U_b\subset U$.
I think the following works: Since the atlas is maximal, there is some a such that $p\in U_a\implies p\in U_a\cap U.$ Define $g=\phi_a|_{U_a\cap U}$. For any $b,$ if $U_b\cap U_a\cap U$ is empty, then the charts $(U_a\cap U, g)$ and $(U_b,\phi_b)$ are compatible . Suppose the intersection to be non empty. Then $g\circ \phi_b^{-1}$ is $C^\infty$ at any $x$ in the intersection (because $g$ is restriction of $\ph…

any objections to this attempt of the proof?

porridgemathematics

@Koro its fine - i would give full marks to someone who said it follows because inclusion maps from open subsets of manifolds are smooth

or in other words, differentiability at a point is a local property at that point

Koro

6:43 AM

thanks :-)

one potato two potato

8:34 AM

Instead of taking algebra 2 course, I decided to take number theory course. Probably I regret my decision a lot, but whatever.

wait, I decided not to take any number theory class because of the shock of hensel's lemma in elementary number theory class. Interesting.

Jakobian

8:56 AM

@MatsGranvik isn't it more about the limits

Koro

9:13 AM

How to construct an example of a continuous function $f:[0,\infty)\to R$ such that $\lim_{t\to \infty} \int_{[0, t]} f d\lambda$ exists but the integral $\int_R f d\lambda, \lambda$ is Lebesgue measure does not.?

Sourav Ghosh

one potato two potato

Why $||f_n||_{L^2}\leq 1$ implies ${f_n(x)\over n}\to 0$ as $n\to\infty$ for a.e. $x$?

Sourav Ghosh

Clearly $(0, 0,0) \in D$

Koro

@onepotatotwopotato $|f_n|(x) \le 1$ for a.e. x.

Sourav Ghosh

I have to find a path from $(0, 0,0) $ to $(a, b, c) \in D$

one potato two potato

9:24 AM

@Koro hmm... why?

porridgemathematics

@Koro think about the alternating series test and why it works.

@Koro this should give you ample intuition to construct such a function

then what you want is for $\lim_{t \rightarrow \infty} \int_{0}^t f d\lambda$ to converge because of the alternating series test, but you want the lebesgue integral to not exist, because both the positive and negative series diverge.

so for probably the easiest concrete case of this happening, use the harmonic series, so that $1 - \frac{1}{2} + \frac{1}{3} .... $ converges, the positive series is $1 + \frac{1}{3} + \frac{1}{5} + ... $ and the negative series $-\frac{1}{2} - \frac{1}{4} - ...$, both of which by comparison to the harmonic series, diverge

Koro

The other day I was trying to construct a sequence of functions $f_n$ such that $f_k(x)\to 0$ as $k\to \infty$, and $\lim_{k\to \infty}\int f_k=\infty$.
I was thinking of triangles with increasing height and it doesn't work...

porridgemathematics

you can actually turn this into a concrete example of a function $f$ satisfying your requirements (i leave the details to you)

Koro

One of my friends looked at it and said: take $f_k(x)= 1/k$...

it worked. I don't know why it didn't occur to me then 😅

porridgemathematics

well, the harmonic series is basically the go to for these kinds of things, because its the boundary point at which $\frac{1}{n^s}$ series converge/diverge

Koro

9:30 AM

@porridgemathematics hmm thinking...

porridgemathematics

btw @onepotatotwopotato just because the L^2 norm is less than or equal to one, does not give you that inequality there..

as in |f_n(x)|<=1 a.e. is not true just because |f_n|_{L^2|<=1

one potato two potato

It's stated in Stein's RA so I wondered why

porridgemathematics

you mean your original question

because the function constantly equal to 10 on [0,1/100] and zero elsewhere has L^2 norm equal to 1

but is certainly not less than 1 a.e. in absolute value.

one potato two potato

Oh you mean $|f_n(x)|\leq 1$. I didn't think it's true.

porridgemathematics

yeah just wanted to point out that is definitely not true

your original question im still thinking about (its probably true by virtue of sum 1/n^2 converging or something..)

what are the other hypothesis, what measure space are you on etc

or is it completely general measure space

one potato two potato

9:36 AM

Anyway my original statement is in Stein's book saying 'we saw that ...'

I didn't see that before.

Nothing. It's just stated like that: 'We saw that if $||f_n||_{L^2}\leq 1$, then ${f_n(x)\over n}\to 0$ as $n\to\infty$ for a.e. $x$.'

porridgemathematics

ohhh, maybe try looking at f_n^2/n^2

sum _{n=1}^infinity |f_n|^2/n^2 is in L^2

uh i mean L^1

by comparison with sum_{n=1}^infinite 1/n^2

and the triangle inequality

therefore \sum_{n=1}^infinity |f_n|^2/n^2 is finite for a.e. x

i.e. is convergent for a.e. x

hence lim_n |f_n|/n = 0 for a.e. x

i think that works @onepotatotwopotato

(something like that should work at least)

one potato two potato

It seems correct to me too. Thanks

porridgemathematics

np

Koro

@porridgemathematics ohh

@onepotatotwopotato nvm

porridgemathematics

@Koro yeah its a nice thing to remember for these types of problems

replacing integral with sum can be helpful

(it turned out both of your problems were examples where thinking about sums helped)

porridgemathematics

10:16 AM

@Koro by the way, for your most recent question, there is a simple way to directly see why that union is open. What happens when $||x||$ is very small? You should be able to see that you automatically get that when this is the case, those points lie in the second set besides $\{0 \}$, so that your second set actually contains a small punctured disk at $0$

so you can certainly proceed directly, using your argument

Koro

@porridgemathematics I don't understand how to construct the example yet.

here is an answer math.stackexchange.com/a/4339003/266435

porridgemathematics

use a riemann sum

Koro

this uses complex integral.

@porridgemathematics for what function?

porridgemathematics

on $[0,1]$ set your function to $1$, on $[1,2]$ to $-\frac{1}{2}$, and so on

Koro

ohh!!

porridgemathematics

10:19 AM

:)

its really just the sum

you barely have to work any harder

you can convince yourself that the limit equals 1 - 1/2 + blah

but the lebesgue integral doesnt exist

because the positive and negative series are also just sums

which diverge

i think this is a better example than those answers

but yeah the accepted one is a common example too

its the dirichlet integral

which is useful in fourier analysis

of course there will be jumps in my case

but you can smoothen my function out at the jumps

and get it as smooth as you like

Koro

I understand that the limit exists.

but I don't yet understand why Lebesgue integral does not exist.

Ahh

I got it.

$f= f^+- f^-$

$\int_0^\infty f^+=\infty=\int_0^\infty f^-=\infty$

so the Lebesgue integral is not defined.

porridgemathematics

right

this is cooked up precisely for this question

Koro

@porridgemathematics sum and integrals are so closely related.

porridgemathematics

but you can do it with any conditionally convergent series

try and prove it

that is, you can get infinitely many counterexamples, by choosing the values of the function to be equal to the sequential values of conditionally convergent series

whose positive series and negative series need to both diverge

(otherwise they wouldnt be conditionally convergent in the sense im talking about)

Koro

yes, for conditionally cgt. series, either the series of positive or negative terms will diverge.

porridgemathematics

10:26 AM

both will for me

otherwise technically you can rearrange the terms

and the thing will still blow up

so im allowing +infinity

and -infinity as limits

as long as they are well defined limits

meaning no matter how you order the summation

you converge to the same thing in $[-\infty,+\infty]$

in that case, a series is only conditionally convergent when both its positive and negative series converge

*diverge

sorry!

Koro

@porridgemathematics yes

thanks :)

porridgemathematics

np

Koro

This example is not continuous though but I think that's not a problem as we can smoothen it.

porridgemathematics

its really not an issue yeah

you can smoothen it by drawing a picture and making sure your smoothenings dont contribute more than finitely to the integral (i.e. the lebesgue integral of the smoothened portions is finite)

so for continuity, its enough to just draw a bunch of straight lines in between the positive and negative ends

with gradient getting steeper and steeper from left to right of the graph

so that your area contributions there are not important

(an easy way is to make their area contributions sum like a geometric series)

porridgemathematics

10:55 AM

@Koro i tried to answer your question about why the 'direct' method wasnt working (it also works)

i mean your definition method

Koro

11:11 AM

@porridgemathematics I suspected that: the union should absorb 0 somewhere to give an open set. Thanks.

porridgemathematics

np, and yeah thats what has to happen to get openness

Koro

0

Q: In $(X,S,\mu), \int f^r d\mu<\infty\implies \int f^q d\mu<\infty$, where $\mu(X)<\infty$.

Suppose that $(X,S,\mu)$ is a measure space with $\mu(X)<\infty$. Suppose that $f:X\to [0,\infty)$ is $S-$ measurable. Suppose that $q>r$ are given positive numbers. Then, $\int f^rd\mu<\infty\implies \int f^q d\mu<\infty.$ I tried to prove it the following way: For brevity, let $\{f>k\}$ stand f...

I think this is correct. But I posted it to get it vetted.

Sine of the Time

11:23 AM

If I've a finite field and I want to find the number of primitive elements of the multiplicative group, I have to use the totient function right?

Koro

12:17 PM

@SineoftheTime primitive elements?

porridgemathematics

@SineoftheTime thats right, because you are computing the number of generators of a cyclic group

Koro

$f(x)= \sin x/x $ is not Lebesgue integrable on (0,\infty).

Robjohn style answer: $f=f^+-f^-$.

$\begin{align} \int_{[0,\infty)} f^+=&\sum_{k=0}^\infty \int_{2k\pi}^{(2k+1)\pi} \sin x/x\tag 1\\ \ge & \sum_{k=0}^\infty \frac 1{(2k+1)\pi}\int_0^\pi \sin x dx\tag 2 \\=&2\sum_{k=0}^\infty \frac 1{(2k+1)\pi}\gt \infty\tag 3\end{align}$

porridgemathematics

its interesting that the harmonic series appears again

Koro

:)

0

Q: Textbooks to read after Armstrong Topology

I just finished reading through Armstrong - Topology and I'm looking for a logical next step to keep going in learning more topology. I've taken courses in Real + Complex Analysis, Algebra, Geometry and others and am a third-year undergrad for reference. I'm considering Hatcher - Algebraic Topo...

general-topology geometry algebraic-topology

Sourav Ghosh

1:05 PM

Mats Granvik

@Jakobian Yes the Euler Gamma constant is a limit, the von Mangoldt function is a limit, the natural logarithm is a limit, the logarthmic derivative of the Riemann zeta function is a limit, the Riemann zeta zeros are a limit, the Riemann zeta function is a limit.

Sourav Ghosh

b) is definitely true ( by using Vieta's formulas)

Now I want to show that $D$ is path connected.

Is $D$ a linear subspace? Is $D$ is a cone? Is $D$ a convex set?

$(0, 0,0) \in D$

Let $(a, b, c) \in D$

That's why my algebra professor didn't like me.

Sine of the Time

1:44 PM

@porridgemathematics ty

@Koro yes

porridgemathematics

@SouravGhosh you dont need anything special to see $D$ is connected

just use that $D$ is the image of a continuous function from $\mathbb{R}^3$

one potato two potato

Any general remarks or advice on graduate life you wanna tell? e.g, exercise regularly, sleep enough, etc, or something in academics. @TedShifrin @leslietownes

porridgemathematics

@SouravGhosh the coefficients are continuous functions of the roots

being connected is a more basic property than being a cone or being a linear subspace or being a convex set. so asking those questions is asking something more specific to get information to assist you with something a lot more basic about $D$

Sourav Ghosh

2:06 PM

@porridgemathematics Thanks. Can you write the explicit map?

porridgemathematics

its just +/- times elementary symmetric functions

i.e. from vietes formulas

ill write it out

(x_1,x_2,x_3) -> (-x_1-x_2-x_3,x_1x_2 + x_1x_3 + x_2x_3 , -x_1x_2x_3)

Sourav Ghosh

$(\alpha, \beta, \gamma) \to (-a, b, -c) $

@porridgemathematics Thanks :)

porridgemathematics

@onepotatotwopotato unsolicited, but spend less time on MSE and general 'procrastaworking' (i would tell myself when I started, lol)

basically ignore free dopamine from answering random peoples questions and answer your own (usually harder dopamine)

Mats Granvik

4:18 PM

How would one classify/categorize the symmetries in the GCD(n,k) matrix?
Greatest Common Divisor=GCD

geocalc33

$\nabla$

$\Delta$

$$ \Delta \varphi = - \frac{xn}{\sum_{i=1}^n t_i} \varphi_x $$

with $\varphi(x,t_1,t_2,\cdot\cdot\cdot,t_n)$

Thomas Finley

4:38 PM

Hi everyone! Can anyone suggsest me some Mathematical Mahazine/Periodicals, where undergrads are allowed to publish any article explaining a particular topic or get problems published in it,

or solutions to problems in the magazine/journal maybe something like "Crux Mathematicorum." I examined about the "American Mathematical Monthly " but it seems one has to subscribe to that in the website of MAA, to read it. This means, one cannot contribute if one doesn't subscribe, for then one wouldn't be aware what's written there i.e unaware of the contents of it. So, any other popular mathematics based journals/periodicals/magazine if suggested, would be really helpful!

冥王 Hades

5:45 PM

@ThomasFinley Romanian Mathematical Magazine is one. Has some of the most insane looking integrals I've ever seen.

Ted Shifrin

6:23 PM

American Math Monthly is probably higher level, anyhow. But the AMS/MAA do have a few journals aimed more at undergraduates, like Mathematics Magazine and College Mathematics Journal.

Jaakko Seppälä

7:43 PM

How can I show this: Let $\varphi_0$ be a homomorphism from an integral domain $R$ to algebraically closed field $M$ and $F$ be a field containing $R$. Then $\varphi_0$ extends either to an embedding $\varphi$ of $F$ into $M$ or to a place $\varphi$ of $F$ into $M\cup \{\infty\}$?

Ted Shifrin

7:53 PM

@JaakkoSeppälä I don’t understand your English. Do you have examples.?

Jaakko Seppälä

I try to write it on the other way: Consider an integral domain $R$ and an algebraically closed field $M$. Let $\varphi_0$ be a homomorphism from $R$ to $M$, and let $F$ be a field that contains $R$. In this context, we can explore two possibilities for extending $\varphi_0$: either as an embedding $\varphi$ of $F$ into $M$, or as a place $\varphi$ of $F$ into $M\cup {\infty}$."

Is it clearer?

Well, I don't have examples, I'm just reading a book.

Ted Shifrin

Place $\varphi$ ?

Into suggests injection, which is wrong, too.

Jaakko Seppälä

A place of a field F is a map φ of F into a set M ∪ {∞}, where M is a
field, with these properties:
(a) φ(a + b) = φ(a) + φ(b).
(b) φ(ab) = φ(a)φ(b).
(c) There exist a, b ∈ F with φ(a) = ∞ and φ(b) not equal to 0, ∞.

Ted Shifrin

I suggest you always try to understand new statements with examples.

Strange word. So it’s a homomorphism but we have to define addition/mult of infinity.

So try to give an example where you get an embedding and an example where you get a “place” — maybe those examples will give you understanding.

Jaakko Seppälä

Okay. I try to do that. Thanks!

Ted Shifrin

8:09 PM

@leslie Is today duck pond day?

leslie townes

we went on saturday and saw ten baby ducks.

Ted Shifrin

Wow ... no joke about Make Way for Ducklings.

leslie townes

yeah, it was cute. they even came over to us, and got pretty close, once munchkin realized that they might not flee into the water if she just sat still.

Ted Shifrin

She will learn from ducks, but not from humans.

leslie townes

yes, ducks are a big influence

Ted Shifrin

8:22 PM

Next, it'll be cats.

copper.hat

9:43 PM

your need an ai duck

robjohn

@冥王Hades I just looked back at that problem. Did you get $x=48$?

Oh, I see it in the image.

Ted Shifrin

10:04 PM

Howdy @copper and @robjohn.

robjohn

@TedShifrin good afternoon

shintuku

10:16 PM

$\phi:R\to S$ surjective ring homomorphism, $\mathfrak b$ ideal of $R$. how come $\phi^{-1}(\phi(\mathfrak b)) = \mathfrak b + \ker \phi$?

i know we should have $\mathfrak b \subseteq \mathfrak b^{ec}$, but where is that equality coming from?

robjohn

Gads, the day has gotten away from me. It's just over an hour until I take my dog to the park.

Ted Shifrin

Almost the pooch’s dinner time!

robjohn

As soon as we get back from the park

Then it will be our dinner time, too

Ted Shifrin

I have no idea what your notation is, but just prove it. @shin

leslie townes

shin: anything in b + ker(phi) gets mapped to something in phi(b), right? that's one inclusion. conversely, if z is in phi^(-1)(phi(b)), there's w in b with phi(z) = phi(w), and then z = w + (z - w) exhibits z as a sum of an element (w) of b and an element (z - w) of ker(phi).

robjohn

10:23 PM

tough love vs spoon feeding

leslie townes

shin: think about where, if at all, phi being surjective might be used in this argument, or outside of this argument, why it would matter generally

shintuku

ultimately the point is to prove that surjective homomorphisms take prime ideals to primes, maximals to maximals, etc.; there's just this step i'm stuck at. currently reading what you wrote

leslie townes

OK, so there are additional areas where some or all of the hypotheses on phi might come up

Ted Shifrin

Surjectivity is crucial just to get out the door.

@robjohn Are you suggesting my spoon has jagged edges?

robjohn

A more universal tool: cutting and scooping

leslie townes

10:29 PM

shin: probably worth thinking generally about which pieces of your arguments use/need phi being surjective and b being an ideal, as opposed to just phi being a homomorphism or b being some more general subset of R

robjohn

dig in and scoop out the heart of the matter

shintuku

at leslie: on the first direction, we do get that b + ker(phi) gets mapped to something in phi(b), but i'm not seeing how the inverse map restores b. consider Z -> Z_8

with, e.g., the ideal <7>

Ted Shifrin

The whole ring?

shintuku

$\mathfrak b$ is the ideal of $R$ we're extending and then contracting

maybe the poster did not mean the entire ideal is restored, only its contraction

math.stackexchange.com/a/292857/755010 if you want to give it a go, but i might just go with a simpler proof, not sure where he's getting that equality from and i've been thinking a while

copper.hat

@TedShifrin Happy Memorial Day (albeit war is never a cause for celebration...)

Ted Shifrin

10:42 PM

Never.

shintuku

10:56 PM

oh i understood

i was making a tremendous derp

i woke up at 5 am today and it's late, for some reason i was thinking of inverse maps as mapping an element to an element instead of mapping an element to an ideal

Ted Shifrin

That is wrong.

Did you not learn about preimage of a set in your intro to higher math coursr?

shintuku

i know about preimages, they're not functions and map a single element to multiple others in case of surjective functions

for some reason brain was not computing

Ted Shifrin

That’s garbled, but this is that in the setting of a homomorphism.

shintuku

yeah in the case of Z -> Z_8 the preimages of the canonical homomorphism are ideals

leslie townes

11:11 PM

when f is a function from X to Y, some books will actually introduce distinct notation for the "preimage under f" function from the power set of Y to the power set of X, instead of writing f^{-1} for it. in my experience, this turns out to be more annoying than not doing it, and doesn't fix the issue of this being a significant stumbling block for a lot of students.

i had to teach out of a book that did this once. the trauma has erased the book's title from my memory.

Ted Shifrin

That sentence is no good. The preimage of a single element is only an ideal in one case.

shintuku

at ted: what about $\phi:\mathbb Z \to \mathbb Z_8: z \mapsto z + 8\mathbb Z$, and we take $\phi^{-1}(5+8\mathbb Z)$? don't we get the set containing the solutions to $x \mod 8 = 5$?

leslie townes

at shin: yes. and that's not an ideal (for example, because it doesn't contain 0, which is a problem you might encounter with preimages of many other single elements)

shintuku

ohhhhhhhh

that's why you add $\ker \phi$

cool cool, thanks a lot

noballpointpen

11:30 PM

shintuku, I asked you sometime ago but you missed my question. Are you in school or self-studying? I am questioning since Ted once mentioned that you have gaps in your math knowledge.

shintuku

self-studying, currently econ major

noballpointpen

Good.

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$Mathematics$ Mathematics