Mathematics - 2018-05-26 (page 3 of 4)

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1:26 PM

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Hi. How do they make this yellow statement? does that follow from convergence of $a_nx_0^n$ ?

Alessandro Codenotti

Just convergence of $|a_nx_0^n|$, you know it goes to zero so it must be smaller than $1$ (for example) for all $k\geq n_0$ for some $n_0$. Pick $C$ equal to the max between $1$ and $|a_ix_0^i|$ with $i<k$

Got it, thanks!

2:07 PM

@LeakyNun hi @LeakyNun

Maneesh Narayanan

2:42 PM

@LeakyNun From Cauchy-Schwartz inequality, we get $t^2-2(a.b)t+(a.b)^2-||a||^2||b||^2\leq t^2-2(a.b)t$. How can we deduce that it has distinct roots from here?

@ManeeshNarayanan discriminant?

Maneesh Narayanan

okay. Thank you.

3:07 PM

user image

How does the convergence of $\sum |a_nx^n|$ imply that of $\sum |a_n|r^n$ if $|a_nx^n| \lt |a_n|r^n$ (shouldn't it be true in case $|a_n|r^n \lt |a_nx^n|$ )?

3:19 PM

ah okay, got it, there's nothing wrong actually

Abhas Kumar Sinha

What's the problem here? Everything seems fine....

Yeap

Maneesh Narayanan

3:48 PM

Let $A$ be a square matrix such that $A^3 = 0$, but $A^2\ne 0.$ Then which of the following statements is not necessarily true? (A) $A^ 2\ne A$ (B) Eigenvalues of $A^2$ are all zero (C)$ rank(A) > rank(A^2)$ (D) $rank(A) > trace(A)$

only (B) is true. right?

rest are all false if I take $A=O$. right?

MatheinBoulomenos

But if $A=0$, then $A^2=0$

Maneesh Narayanan

sorry

then $D,C,B$ are always true.

MatheinBoulomenos

all are true

Hi

MatheinBoulomenos

Hi @BalarkaSen

Maneesh Narayanan

3:53 PM

How $A$ is true? if $A^2=A$ the it is idempotent, then minimal polynomial is of the form $x(x-1)$ or $x$ or $x-1$. but here minimal polynomial is $x^3$. can I use this argument?@MatheinBoulomenos

If $A^2 = A$ then $A^3 = A^2 = A$.

MatheinBoulomenos

$(A^2)^2=A^4=A(A^3)=A\cdot 0= 0$, so $A^2$ squares to $0$, but $A^2\neq 0$

Also it follows from C

Is it true that a null-set (Lebesgue or Jordan) is bounded?

MatheinBoulomenos

consider $\Bbb Q$

Maneesh Narayanan

okay. Thank you @BalarkaSen @MatheinBoulomenos

what about my argument?

MatheinBoulomenos

3:55 PM

it also works

Maneesh Narayanan

okay.

Oh right, it's a Lebesgue null-set and unbounded. But are Jordan null-sets bounded?

Because there are finitely many open cuboids which cover it?

MatheinBoulomenos

aren't Jordan-measurable sets bounded by definition?

Yeah well I'm trying to prove the equivalence of a Jordan null-set to a Jordan measurable set with zero volume right now so I can't use that... :P

MatheinBoulomenos

what's your definition of Jordan-measurable?

The definition I know assumes bounded

3:58 PM

Our definition of the Jordan null set is like a Lebesgue null-set just with finitely many open cuboids covering it.

(I'm trying to argue that a Jordan null-set is bounded)

MatheinBoulomenos

the union of finitely many open cuboids is bounded

Ok thx

Abhas Kumar Sinha

4:28 PM

How can less intelligent students excel in mathematics?

Maneesh Narayanan

@AbhasKumarSinha through hard work. "There is no substitute for hard work"-Thomas A. Edison

4:43 PM

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Sanity check please, there's an error in the above right?

where does the second to last equality come from?

ah I see

@MikeMiller Some trigonemtric identity I think

$1 - \sin^2 \theta = \cos^2 \theta$, so now square the expression $\sin \theta \cos \theta = \frac 12 \sin(2\theta)$

The only thing that I don't know about (because I don't know what $r$ is) was the first equality :P

Everything else is fine

Thanks! @MikeMiller

If I know that $J$ is bounded why does it follow that the closure $\overline J$ is bounded too?

4:53 PM

what would it mean for it to be unbounded?

I guess that the boundary is unbounded since $\overline J = J \cup \partial J$?

Alessandro Codenotti

$X\subseteq Y\implies \overline{X}\subseteq\overline{Y}$. What does bounded mean?

That there is a $j \in J$ such that there exists $R \gt 0$ with $J \subseteq B_R(j)$...

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user image

mmh I guess I can take the closed set $\overline B_R(j)$ which contains $B_R(j)$ and therefore $J$ and since $\overline J$ is the smallest closed set containing $J$ it must be inside $\overline B_R(j)$

Can I argue that $\overline J$ is also inside $B_R(j)$?

MatheinBoulomenos

5:05 PM

it may not be inside $B_R(j)$, but it's inside $\overline{B_R(j)} \subset B_{R+1}(j)$

if $J = B_R(j)$ then ...

According to Weierstrass M-test, how could we find a $M_m$ to show that the convergence of $\sum \frac{(-1)^mt^{2m}}{(2m+1)!}$ is uniform ?

@MatheinBoulomenos Mmm why would I know that $R+1$ is enough?

MatheinBoulomenos

Every element in $\overline{B_R(j)}$ can be approximated by a sequence $x_n$ of elements with $d(x_n,j) < R$

by continuity of the function $x \mapsto d(x,j)$ (which follows from reverse triangle inequality), this implies that every element $x \in \overline{B_R(j)}$ satisfies $d(x,j) \leq R$

So actually $\overline{B_R(j)} \subset \{x \in X \mid d(x,j) \leq R\} \subset B_{R+1}(j)$

the $R+1$ is just one choice, any $R+\varepsilon$ with $\varepsilon > 0$ works

Ok thx! :) I feel like I forgot a lot of topology already.

MatheinBoulomenos

5:15 PM

Important note: $\overline{B_R(j)}$ may not be the same as $\{x \in X \mid d(x,j) \leq R\}$

Ok thx

MatheinBoulomenos

A counterexample is the discrete metric $d(x,y) = 1$ if $x\neq y$ and $d(x,y)= 0$ if $x=y$. Then let $x$ be any element and consider $R=1$. We have $B_1(x)=\{x\}$, so $\overline{B_1(x)} = \{x\}$, but $\{y \in X \mid d(x,y) \leq 1 \} = X$

Anderson Felipe Viveiros

5:40 PM

Hello, guys. How do I show the following: "Let $M$ be a three-dimensional compact orientable (riemannian) manifold and $\Sigma\subset M$ a connected embedded surface. Then $M\backslash \Sigma$ consists of one or two connected components."?

MatheinBoulomenos

If $M$ is a 3-sphere, this follows from Alexander duality. Not sure about the general case, though

Anderson Felipe Viveiros

Indeed, I think maybe the text I'm reading has changed the terms...

I don't know

Because I think I proved the result when $M$ is connected (of course we need this) and $\Sigma$ is orientable (and connected)

I mean, the symbols $M$ and $\Sigma$ may be one in the place of the other, in the statement...

MatheinBoulomenos

do non-orientable surfaces even embed into 3-dimensional orientable manifolds? (At least they don't embed into $\Bbb R^3$)

Anderson Felipe Viveiros

I don't know...

RP^2 embeds in RP^3

MatheinBoulomenos

5:51 PM

is $\Bbb RP^3$ orientable?

Yes.

MatheinBoulomenos

oh

Anderson Felipe Viveiros

That's right.

MatheinBoulomenos

oh yeah it's a Lie group, right

$SO(3)$

Anderson Felipe Viveiros

Well, I don't know if it is worthy, but I've posted the question here...
https://math.stackexchange.com/questions/2797110/does-a-surface-separates-the-ambient-space-in-at-most-two-components

5:52 PM

alternatively you can look at its top homology w/ \Z coefficients

@AndersonFelipeViveiros I think you should be able to prove your thing by doing a Mayer-Vietoris argument on $M - \Sigma$ and $N(\Sigma)$ (a tubular neighborhood of $\Sigma$).

If $M - \Sigma$ has more than two connected components, that's the same as claiming $H_n(M - \Sigma)$ has rank > 2.

MatheinBoulomenos

so basically a bit like the proof of Alexander duality?

I'd need pen and paper to do it though. I'm rusty on computations.

@MatheinBoulomenos Is that how you prove Alexander duality? God I have forgotten.

Maybe that plus Poincare duality, yeah

MatheinBoulomenos

yeah, you do an argument like that and then apply Poincare duality

Cool cool

Anderson Felipe Viveiros

6:00 PM

@BalarkaSen I'm not very familiar with homology :'(

But ok, thanks!

I'll try the computation out after I'm done with this apple juice.

Anderson Felipe Viveiros

I need to study it a little bit ^^

6:11 PM

Ok, I'm back. I just realized $M - \Sigma$ is noncompact so what I said is not quite correct. The rank of $H_0(M - \Sigma)$ is not the same as $H_n(M - \Sigma)$.

But maybe I can just try to compute $H_0$ using the unreduced Mayer-Vietoris?

$H_1(M) \to H_0(\partial N(\Sigma)) \to H_0(M \setminus \Sigma) \oplus H_0(N(\Sigma)) \to H_0(M) \to 0$

If $\Sigma$ is orientable, $\partial N(\Sigma)$ is $\Sigma \times \{0, 1\}$ (two components) and if $\Sigma$ is nonorientable it's just the orientation double cover of $\Sigma$ (so a single component).

Does anyone know of a good way of systematically generating polyhedral (i.e. planar, 3-connected) graphs?

So, $H_1(M) \to \Bbb Z^{1, 2} \to H_0(M \setminus \Sigma) \oplus \Bbb Z \to \Bbb Z \to 0$.

if f(f(x))=x is f(x) necessarily the identity function?

@tatan, No - consider $f(x) = 1/x$

Let's call $G$ to be the image of the first map. Then $0 \to \Bbb Z^{1, 2}/G \to \Bbb Z^{1+n} \to \Bbb Z \to 0$ is a short exact sequence. What can I say about $n$?

@MatheinBoulomenos This seems to be the algebraic problem.

MatheinBoulomenos

6:18 PM

rank of abelian groups is additive in short exact sequences

so $n \leq 2$

Ah, bingo.

Fantastic.

MatheinBoulomenos

one slick way to see this is to tensor with $\Bbb Q$ (which is flat over $\Bbb Z$, so this preserves exactness) and then apply rank-nullity. The rank of an abelian group $A$ is the same as $\operatorname{dim}_{\Bbb Q}(\Bbb Q \otimes_{\Bbb Z} A)$

Ahh yeah

I remember this

Very cool

MatheinBoulomenos

@AndersonFelipeViveiros Balarka solved your problem, in case you didn't notice

Anderson Felipe Viveiros

Hahaha I've noticed it. Thank you guys

I'll take a look

6:25 PM

@MatheinBoulomenos Also, the end group $\Bbb Z$ is free in this case, so I think the sequence also splits.

MatheinBoulomenos

sure

So you can so more, actually. Eg, that $G$ has to be one of those "trivial" subgroups quotienting by which doesn't introduce torsion.

if f(x+f(y))=f(x)+y for all x,y in R; then f(100)=? . I tried the following:Putting x,y=0 we get f(0)=0. Now substituting x=0 we get, f(f(y))=y . What to do next?

MatheinBoulomenos

true

Hi @Tobias

Tobias Kildetoft

@MatheinBoulomenos Hi

6:27 PM

This chat is too helpful. I have probably recovered more of my forgotten math in the last few hours than I could have in a week on my own.

Anyone willing to help me with my question ;-)?

Anderson Felipe Viveiros

@BalarkaSen Maybe it is that apple juice of yours haha

lmao

@tatan: You've answered it yourself with what you have there !

hi @Tobias, a @Balarka, @Mathein, @Anderson

10/10 would drink apple juice instead of coffee from now on

MatheinBoulomenos

6:30 PM

Hi @Ted

Hi @TedShifrin!

Hey everyone!

Tobias Kildetoft

@tatan How did you get $f(0) = 0$? Setting both equal to $0$ gives $f(f(0)) = f(0)$

Oh, and now there's Demonark.

Tobias Kildetoft

Hi @TedShifrin

6:30 PM

o. .o

RAWR

MatheinBoulomenos

@TobiasKildetoft I've noticed that the definition of basic algebra given in Assem-Simson-Skowronski seems to be different from the one you gave me

Oh no don't tell me Daminark's here :(

MatheinBoulomenos

Hey @Daminark

@Daminark He didn't

Tobias Kildetoft

@MatheinBoulomenos Then whatever I claimed was wrong :)

6:31 PM

He told you Demonark's here

Oic

Congratulations, you played yourself

Tobias Kildetoft

@MatheinBoulomenos What did I claim?

add MLG noises here

@TobiasKildetoft Doesn't taking f^-1 both sides imply it?

MatheinBoulomenos

6:32 PM

@TobiasKildetoft right after the definition, there's a lemma that says that a finite-dimensional $K$-algebra $A$ is basic iff $A/\operatorname{rad}(A)$ is some direct product of copies of $K$

Tobias Kildetoft

@tatan Is the function assumed to be bijective?

I doubt it's bijective.

if or is? @TobiasKildetoft

MatheinBoulomenos

@TobiasKildetoft I remember you said that there might be non-basic three-dimensional algebras with $A/\operatorname{rad}(A) \cong \Bbb C \times \Bbb C$

Tobias Kildetoft

@tatan Woops, fixed

MatheinBoulomenos

6:33 PM

doesn't this imply that all three-dimensional algebras over an algebraically closed field are basic?

Tobias Kildetoft

@MatheinBoulomenos Hmm, in that case I misremembered which of the properties is basic.

@TobiasKildetoft Its not mentioned

Tobias Kildetoft

(for some reason I keep hitting f when I mean to hit the s)

We also get $f(0) = -f(f(0))$, so what does that tell us?

@TedShifrin How?

6:34 PM

Combined with what @Tobias said, that tells us that $f(0)=0$. :P

Set $y=0$ and $x=-f(0)$.

Tobias Kildetoft

@MatheinBoulomenos Yes, since there are no simple ones of dimension less than $4$

(no other simple ones)

MatheinBoulomenos

yeah

Tobias Kildetoft

How do they define basic?

@Balarka: You mean you're remembering things? Not all is lost?

@TedShifrin Okay... we have f(0)=0. Now? From here all we have is f(f(y))=y

6:36 PM

That answers your question, though.

MatheinBoulomenos

If $e_1, \dots, e_n$ is a complete set of primitive orthogonal idempotents in $A$, then $A$ is called basic if $e_iA \not \cong e_jA$ for $i \neq j$

Write down a few computations.

Tobias Kildetoft

@MatheinBoulomenos Ok, so I remembered the definition correctly, but forgot what it really meant for the quotient by the radical.

Alessandro Codenotti

@Mathei an interesting question from the logic chat. If you have a system of polynomial equations $P(x)$ with coefficents in an algebraically closed field $K$ then it has a solution in a field $L\supseteq K$ iff it has a solution in $K$. What if I have infinitely many polynomials?

boo, demonic @Alessandro

Alessandro Codenotti

6:37 PM

Boo?

Hi @Ted

@AlessandroCodenotti i think hilbert basis theorem sorts out that issue?

I'm confused. If it's algebraically closed, who needs logic?

MatheinBoulomenos

polynomial equation in how many variables?

if it's finitely many variables, then any infinite system has a solution iff an associated finite system has a solution, by the Hilbert basis theorem, as loch mentioned

If the solution is in $K$, I'm guessing one variable, @Mathein?

This sounds very sloppy altogether.

Alessandro Codenotti

@MatheinBoulomenos ah, right

MatheinBoulomenos

6:39 PM

the question is kinda boring for one variable

I know. But it's ill-posed with more.

resigns from logic/algebra land

Alessandro Codenotti

Finitely many variables, solutions in $K^n$

Aha.

MatheinBoulomenos

so Hilbert basis theorem, then

is there a way to do this with compactness?

Alessandro Codenotti

Right

6:39 PM

@tatan: You see it?

Alessandro Codenotti

@MatheinBoulomenos maybe but I don't see it

@TedShifrin Nope

@MatheinBoulomenos What if infinitely many variables?

What's $f(f(100))$?

100

6:41 PM

@BalarkaSen hmm how do i see this? i can convince myself the former by just imagining orientable surfaces in $\mathbb{R}^3$ - but im not sure how to see the latter claim

@TedShifrin Apparently so

Oh, I was misremembering the question.

Maybe

Sorry, I thought we wanted to iterate $f$ $100$ times.

nothing to be sorry actually

;-)

6:43 PM

I'll ponder more in a bit.

Sure

@loch The normal bundle of $\Sigma^2$ inside $M^3$ is a real line bundle over $\Sigma$. Give it a fiberwise Riemannian metric and take the unit $S^0$-bundle that's sitting inside it. This is a double cover of $\Sigma$, which can either be trivial (in which case it has two components) or be nontrivial (in which case it has one component)

If you call the line bundle $E$, $E - \Sigma$, where $\Sigma$ is the zero section of $E$, precisely deformation retracts to this $S^0$-bundle.

Maybe I was a little blase about the orientation thing. I meant normally oriented (that determines if the $S^0$-bundle is trivial or not), not orientability of $\Sigma$.

MatheinBoulomenos

@AlessandroCodenotti @user21820 If we allow infinitely variables, then this is no longer true. Let $X$ be any non-empty set and consider the collection $Y$ of all non-zero polynomials with variables taken from $X$. Then consider the collection of polynomials with coefficients taken from $X \times Y$, where we write down each $f \in Y$ the equation $f(x_1, \dots, x_n)y_f=1$, where $y_f$ is the variable that corresponds to $f$.
If we have a solution to that system of equations in some field $L$ containing $K$, then the elements of $L$ which correspond to the elements in $X$ will be algebraica…

But $M$ is oriented, so if $\Sigma$ is normally orientable, it is orientable.

May as well add to the chaos and give some commentary as I read through some ANT

6:47 PM

Is there a simple argument to show that 6^x + 6^|x| is many-one?

@Daminark c a c o p h o n y

MatheinBoulomenos

so the transcendence degree over $K$ of any field $L$ that contains a solution to that system of equations will be at least the cardinality of $X$

So let's say $f(t) \le ct^{\alpha}$ for some $\alpha \ge 0$. Define $I(s) = \int_1^{\infty} \frac{f(t)}{t^{s+1}}dt$ and $I_x(s) = \int_1^x \frac{f(t)}{t^{s+1}}dt$

MatheinBoulomenos

These extra variables who correspond to polynomials are just a trick to turn the inequality $f(x_1, \dots, x_n) \neq 0$ into a polynomial equality

(similarly to how you show that $K^\times$ is an affine variety)

@MatheinBoulomenos Interesting. I thought of the idea of a transcendence base bigger than K, but didn't realize we could use another variable to make those inequalities.

Thanks!

MatheinBoulomenos

6:52 PM

@TobiasKildetoft what does the confusion over basic algebras say about our problem with positive bases? you said that you can show the existence for basic algebras, is this still true with the other definition?

So these notes have typos, >:(, but by the looks of it, we're letting $\sigma = \text{Re}(s)$ and then saying that the above series converges when $\sigma > \alpha$, $|I_x(s)| \le \frac{c}{\sigma - \alpha}$, and $|I(s)-I_x(s)| \le \frac{c}{\sigma - \alpha}\frac{1}{x^{\sigma - \alpha}}$

So $|I_x(s)| = |\int_1^x \frac{f(t)}{t^{s+1}}dt| \le c\int_1^x t^{\alpha - \sigma - 1} dt$

@BalarkaSen hmmmmm how do i see this

Hm, decompose $TM|_\Sigma = T\Sigma \oplus E$ where $E$ is the normal line bundle to $\Sigma$.

Ah no it turns out $|I(s)| \le \frac{c}{\sigma - \alpha}$, but the bound on $|I(s)-I_x(s)|$ is true

If direct product of two bundles is orientable, and one of the factors is orientable, the other factor is orientable too.

7:03 PM

@BalarkaSen Is this the same Balarka Sen here-mymathforum.com/new-users/37415-rather-late-introduction.html

@tatan A newer version, yes

That was me from uh

52 years back

How did you study those things in 8th grade? @BalarkaSen

@loch Explicitly, pick a fiberwise orientation of $T\Sigma \oplus E$. That's constant on $E$ ("always outward pointing"). So that in turn gives an orientation on $T\Sigma$.

@tatan Ehhhh I doubt I did. I liked a little math, but mostly I was bluffing about the stuff I learnt from a cursory reading of wikipedia lmao

Balarka is this chat's leading expert in spectral zeta functions

I wanna die.

Let's never talk about that forum again, okay?

7:08 PM

No

Not okay

@BalarkaSen If you don't mind can I ask something? Where are you studying now?

We must discuss this forum daily

MatheinBoulomenos

These are some pretty advanced topics for an 8-th grader. When I was in 8th grade, I still did calculus

I think any lessons to be learned have long been past though

When I was in 8th grade I did linear algebra

7:09 PM

I get the desire to move on, I made enough jokes in my day

In the sense of y = mx + b

@Daminark Atlast I have found someone who did things like me ;-)

Let's put it like this. When I was in 8th grade I did my best to be a little shit

MatheinBoulomenos

lol

But you had a better taste back then. "Math Focus: Number Theory"

:P

@BalarkaSen Its great that you tried to do it ,atleast

7:11 PM

@BalarkaSen yeah - and you're saying that if we take the $S^0$-bundle of the normal bundle of $\Sigma$ and if it's connected, then it is non-orientable as a bundle over $\Sigma$?

MatheinBoulomenos

@Daminark Gaussian elimination on a 1x1 matrix seems pretty pointless

@loch Yep!

:thinking:

@tatan I won't deny that I liked math.

MatheinBoulomenos

*thonking

7:12 PM

That's probably the only unchanging factor throughout time

Whoops :/

@BalarkaSen Where are you studying math right now?

graduated high school a few weeks back. Applied for a few uni's in my country

I thought the lack of response the first time was intentional to defuse the conversation

As long as it pushes away the link above I'm happy

7:14 PM

lets not discuss the link if it upsets you

Knew you'd be the one flagged and banned for that

@BalarkaSen Sorry if i disturbed you

PSA: If someone asks you to switch topics or not mention something embarrassing anymore, the correct response is to drop the topic. Period.

^

okay... lets not discuss that anymore

7:18 PM

I doubt Balarka was seriously offended but just overstating things

@BalarkaSen ah ok i think i see it now

I know he feels strongly that he doesn't want to be associated with himself from then though

@loch Can I give you an exercise?

sure

Calculate the unit tangent bundle of RP^2 explicitly

Whether or not the unit bundle of a vector bundle is orientable as a manifold governs whether the total space is orientable or not

You can use that T^1 S^2 is identified with SO(3) (proof: the latter acts on the former freely and transitively), which is identified with SU(2)/(Z/2) = S^3/(+-1) = RP^3

so im thinking of $\mathbb{R} \mathbb{P}^2$ as the quotient of $S^2$ under the antipodal map $x\mapsto -x$ - so i think the tangent bundle of $\mathbb{R}\mathbb{P}^2$ is obtained by quotienting out the action of $\mathbb{Z}/2$ on $TS^2$ - which acts by $(x,v) \mapsto (-x,-v)$

so i just have to see what this action translates to when I identify $TS^2$ with $SO(3)$

7:28 PM

hi, im trying to show that $<Fr> \le Gal(\overline{F_p}/F_p)$ is infinite, someone can help?

@Liad what are the elements fixed by a power of the frobeinus?

MatheinBoulomenos

@Liad hint: a polynomial in one variable of degree $n$ has at most $n$ solutions. What does that tell you about the order of the Frobenius on an infinite extension?

$F_p$ @loch

MatheinBoulomenos

Do you know that $\overline{F_p}$ has infinitely many elements?

if f(a+b)=f(ab) and f(-1/2)=-1/2 , is f(x) a constant function?

7:31 PM

@MatheinBoulomenos i dont :/

MatheinBoulomenos

@Liad let $K$ be a finite field, then what can you say about the polynomial $1+\prod_{a \in K}(x-a)$?

@tatan maybe

I think so

@MatheinBoulomenos it doesnt have roots

MatheinBoulomenos

so can $K$ be algebraically closed?

nope

MatheinBoulomenos

7:33 PM

right

so any closed field is infinite

@LeakyNun is there an easy way to show 6^x+6^|x| is many-one?

MatheinBoulomenos

@Liad now assume that the Frobenius on some infinite field has finite order. Can you get a contradiction?

@tatan no idea

i just want to make sure we are talking about the same thing , by "the Frobenius" you mean the subgroub generated by the transformation $x\to x\ ^ p$ ? @MatheinBoulomenos

MatheinBoulomenos

7:35 PM

I mean the Frobenius as just one element in the Galois group

one can talk about the order of an element in a subgroup

it's the smallest integer $n$ such that $\sigma^n=\operatorname{id}$

yea ok but the Frobenius is just the mapping $x \to x \ ^ p$ ?

@LeakyNun How do i find the range of x^2+x+1/x^4+1?

MatheinBoulomenos

yeah

ok

@MatheinBoulomenos if the order was finite

then there was an element $x \in \overline{F_p} $

that is not a root of any element of $<Fr>$

MatheinBoulomenos

what do you mean root of an element of $\langle Fr \rangle$?

try to write the statement $Fr^n=id$ as an equation that holds for all elements in $\overline{F_p}$

7:41 PM

Hmm, this is interesting. So if $a$ and $b$ are arithmetic functions and $l(n) = \ln(n)$, then $l\cdot (a*b) = (l\cdot a)*b + a*(l\cdot b)$

@Daminark dirichlet convolution?

Yeah

completely don't see why that is true...

@MatheinBoulomenos this cant happen

because the degree of $Fr \ ^ n$ is finite

(np)

MatheinBoulomenos

I don't know what you mean exactly @Liad

7:43 PM

doesnt it mean that each $x\in \overline{F_p} $ is a root of $Fr \ ^ n $?

MatheinBoulomenos

I have to go now, so I'll spoil it for you. If $Fr^n=id$, then $x^{p^n}-x=0$ for all $x \in \overline{F_p}$. But that polynomial can have at most $p^n$ roots

if by "root" you mean fix point, then yeah

ahhh

@Daminark what is $\cdot$?

yes ! @MatheinBoulomenos my bad ^^

$\ln(n)\sum_{kl = n} a(k)b(l) = \sum_{kl = n} \ln(n)a(k)b(l) = \sum_{kl = n} (\ln(k) + \ln(l))a(k)b(l)$

7:43 PM

@MatheinBoulomenos thanks!

$\cdot$ is just multiplication

oh, I thought you meant composition lol

I see it then

because you phrased it like it's a differential operator

so I thought you mean composition

MatheinBoulomenos

@Liad no problem

@Liad the important parts are really that $\overline{F_p}$ has infinitely many elements and that the Frobenius is given by a polynomial

@MatheinBoulomenos yea i can see that it would have worked with any other polynomial also

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7:46 PM

Yo

hi @BalarkaSen

hi

I'm back my dudes

@tatan Lol I was being comically uncomfortable with the link. No need to apologize.

welcome back, pal

So it's kinda related to differentiation, like if you have some Dirichlet series $f(s) = \sum_{n=1}^{\infty} \frac{a(n)}{n^s}$, its derivative is $f'(s) = -\sum_{n=1}^{\infty} \frac{\ln(n) a(n)}{n^s}$, but yeah the log isn't quite "operating" on it in the same way

7:47 PM

@ACuriousMind Sigh, I hope you can find a way to turn down the condescension knob in your tone.

7

@BalarkaSen I am happy

@tatan Are you a member of the forum that I should recognize?

I don't regularly go there anymore

Nope...

Mmk.

So if $X$ has a measure function $\mu$, then a measurable function $f : X \to \Bbb R$ is one where $\{ x \in X \mid f(x) > r \}$ is measurable for each real $r$

equivalently, for each rational $r$

if $f$ and $g$ are measurable, then so is $f+g$

Anderson Felipe Viveiros

7:53 PM

@BalarkaSen That's why we need this wikiwand.com/en/Irony_punctuation

So we consider $\{ x \in X \mid f(x) + g(x) > r \}$

Apparently, $\{ x \in X \mid f(x) + g(x) > r \} = \displaystyle \bigcup_{q \in \Bbb Q} \left( \{ x \in X \mid f(x) > q \} \cap \{ x \in X \mid g(x) + q > r \} \right)$

but can someone explain to me what this does graphically?

@AndersonFelipeViveiros Wow this is actually something we need

did you see that?

I did :)

:)

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