Mathematics - 2018-09-25 (page 2 of 5)

1:03 PM

@CaptainAmerica16 Long story short, you can bookmark a script that turns it on constantly, or you can bookmark a script that renders all visible markup when you click it

CaptainAmerica16

$2$

@TobiasKildetoft Thanks for the help. I figured it out now! :)

Leaky Nun

1:31 PM

@Abcd yes

Abcd

1:44 PM

@LeakyNun Never mind I got my mistake.

@LeakyNun What is the statement of the Tower of Hanoi problem in Programming?

I mean I can find the algorithm everywhere

But cant find the problem statement

Like where to begin your program of Tower of Hanoi, what to do?

Leaky Nun

Tower of Hanoi

The Tower of Hanoi (also called the Tower of Brahma or Lucas' Tower and sometimes pluralized) is a mathematical game or puzzle. It consists of three rods and a number of disks of different sizes, which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape. The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules: Only one disk can be moved at a time. Each move consists of taking the upper disk from one of the stacks and placing it on top...

Abcd

@LeakyNun How to represent rods and discs in java man

user193319

Let $f : \Bbb{R} \to \Bbb{R}$ be some function. Is it true that $f$ is not differentiable at some point $x_0$ if and only if the derivative $f'$ is not continuous at $x_0$

Leaky Nun

output the steps

(1,2) for moving the top piece from the 1st tower to the 2nd tower

@user193319 no

Aubyn

"If the [Riemann] hypothesis is proven to be correct, mathematicians would be armed with a map to the location of all such prime numbers, a breakthrough with far-reaching repercussions in the field."

What does that mean?

user193319

1:47 PM

@LeakyNun Shoot...Do you have a counterexample in mind?

Leaky Nun

Sir Michael Atiyah's Proof of the Rie…

For all those who are interested in Sir Michael Atiyah's proof...

let's keep that discussion out of this room @Aubyn

3

@user193319 yes

Aubyn

@LeakyNun What does that have to do with what I've said?

Leaky Nun

ask the two people who starred my message

user193319

@LeakyNun Mind sharing?

Leaky Nun

@user193319 I don't

user193319

1:49 PM

@LeakyNun So, what does the function look like?

Aubyn

@LeakyNun No need to be cheeky. I didn't mention Atiyah's proof.

2

Leaky Nun

@user193319 desmos.com/calculator/lvaytm68sq

Secret

[Random]

$e^e = \frac{p}{q} \implies e = \ln (\frac{p}{q})$

$\ln (\frac{p}{q}) = \frac{r}{s}$

$\frac{p}{q} = e^{\frac{r}{s}}$ contradiction

$\therefore \ln (\frac{p}{q}) \in \Bbb{T}$

Holo

@Secret why is that true?

Secret

@Holo I forgot to put the words "assume"

user 2646

1:54 PM

@Aubyn he's not being "cheeky," just sensitive to 89 year-old mathematician

Holo

And T is irrationals?

Secret

nope, transcendentals

Next trial: Find $p,q \in \Bbb{Z}$ such that $e = \ln (\frac{p}{q})$

$\ln$ is monotonic under reals

Leaky Nun

@user2646 actually I'm sensitive to other people's sensitivity

2

user 2646

:-)

Aubyn

@user2646 You're guys are the ones mentioning this guy. I didn't mention him, the quote wasn't from him or his paper.

ÍgjøgnumMeg

1:57 PM

He has a point..

user 2646

nobody wants to talk about it

ÍgjøgnumMeg

so we just don't talk about RH anymore? lol

5

Semiclassical

yeah, let's not overreact: questions about why the Riemann hypothesis is considered important may be influenced by the present context, but they're still distinct from it

5

Leaky Nun

I'm glad to answer anything related to that topic, as long as it is not in this chatroom.

Semiclassical

(My own grounds for not wanting to answer that question is not that it's impertinent but that it's been done to death. If you want to learn about the Riemann hypothesis, there's plenty of sources on- and offline)

Leaky Nun

1:59 PM

Also @Aubyn we all know where that quote is from

3

Holo

@Secret monotonic property is not enough

ÍgjøgnumMeg

star spam

Secret

thus if there exists such $\frac{p}{q}$ it is unique

Holo

Oh, true

Assuming $q,p$ are coprimes

Secret

But, I am not sure if there are other things about ln that can be used. Most proof of transcendentals that does not make use of lindlemann Weistrass often involve crazy integrals defined and then somehow shown to violate some inequalities

Still yet to comprehend those

I wonder...

$1 = \ln (\ln (\frac{p}{q}))$

Semiclassical

2:04 PM

also, the above holds so long as one is asking for the meaning of the quotation. the moment it crosses over to questions about the writer of said quotation, then it's no longer about the Riemann hypothesis as such

and at that point I definitely agree that there's no reason for it to be here

Abcd

@LeakyNun Could you please explain how to execute the algorithm of tower of H in Java

Leaky Nun

I thought you said algorithms are everywhere

Abcd

@LeakyNun Yes I understand how to solve it but am unable to put it in form of code

Leaky Nun

@Abcd sorry, please ask somebody else

Secret

we knew if $q$ is rational, then $\ln (q)$ is transcendental

Abcd

2:07 PM

@LeakyNun OK ...

Semiclassical

@Secret $q=1$ is rational...

(I think that's neverthless true in general, but there is that one exception)

Secret

ah great... how does that escaped the contradiction...

$\frac{p}{q} = e^{\frac{r}{s}}$

rschwieb

@LeakyNun Hey, I have a naive question about the proof-software that we talked about the other day

Secret

ah, $e^q$ is transcendental except for $q=0$

Leaky Nun

@rschwieb go ahead

Semiclassical

2:09 PM

@Secret for rational q, you mean

rschwieb

@LeakyNun In a conversation with a developer of a different-although-related technology, they said there were some complications with programming in what a finite set is.

Leaky Nun

with constructive maths maybe

rschwieb

@LeakyNun Do you know if there's any difficulty with that in the thing you were working on?

Leaky Nun

I know of at least 4 inequivalent definitions of a finite set

Secret

@Semiclassical yeah, I was reusing labels without declaring them, in the p/q line, p, q are both integers, while in the e^q line, q is any rational

Semiclassical

2:10 PM

right

Leaky Nun

@rschwieb well sometimes

rschwieb

Ones which can safely live in that proof-software context?

Leaky Nun

each of them can do that

Secret

In a series perspective, it is easy to see why $e^q$ when $q=0$ is rational, because it kills off the infinite sum, thus there is no chance for it to be irrational or even transcendental

rschwieb

@LeakyNun I guess I have an even more naive question. What's the interaction with the software like? Do you just type up an encoded version of the human proof, and the software looks at it and gives a thumbs up or dowN?

Leaky Nun

2:11 PM

@rschwieb you can see the "context" of the proof

i.e. the list of assumptions and your current goal

Secret

So for $p\neq q$, $\ln (\frac{p}{q})$ is transcendental

Leaky Nun

@rschwieb here's what you see with Coq

I don't use Coq though

(so why did I put a screenshot of Coq?)

rschwieb

Whew, pretty dense

Secret

Let $\ln (\frac{p}{q}) = t$ where $t \in \Bbb{T}$, what can we say about $\frac{p}{q} = e^t$

$\frac{tp}{q} = te^t \implies W(\frac{tp}{q}) = e$

Leaky Nun

@rschwieb LHS is the code, RHS is the context

rschwieb

2:15 PM

@LeakyNun So is it verifying proofs or is it actually constructing them?

Secret

We also knew that $W(e^{1+e}) = e$

Thus $\frac{tp}{q} = ee^e$

Leaky Nun

@rschwieb both...

well we sort of write programs to construct proofs

or we can construct the proofs ourselves

or both

Secret

$\ln (\frac{tp}{q}) = \ln e + e \ln e = 1+e$

ok this is not useful

Holo

Secret you are moving in circles

user 2646

don't disturb his circles

rschwieb

2:21 PM

@LeakyNun Have you ever seen anyone prove a finite Desarguesian geometry is Pappian in such software?

Leaky Nun

@rschwieb maybe the curry howard correspondence will help you understand more

I don't know much about projective geometry

is the proof hard?

rschwieb

@LeakyNun No, it's definitely something that can be proven in an introductory book

But the thing is that where I saw it proven, they were forced to go through Wedderburn's little theorem, and they commented that it would be interesting to see a direct proof.

That book was written perhaps in the 50's or 60's bya . famous mathematician

Leaky Nun

@rschwieb maybe you can tell me the definitions and the proof and I can tell you how long it would take me to prove it in the software :P

rschwieb

I've been thinking of asking if anyone took up his challenge and solved it between now an dthen.

Secret

If $e^t$ is algebraic, then there exists some polynomial $P$ such that $P(e^t) = \sum_{n < \omega} a_ne^{nt} = 0$

rschwieb

2:26 PM

@LeakyNun Uhh, the ideas for a proof that I know in a nutshell are "desarguesian planes" can be coordinatized by a division ring, and "pappian planes" can be coordinatized by a field, so if your plane is finite Wedderburn's little theorem says your division ring is commutative. As for the actual definitions of desarguesian and pappian, I'd have to look those up again myself :)

Leaky Nun

lol

rschwieb

I remember the desargues theorem in terms of Euclidean geometry, but of course it has a projective version

Secret

Multiply $P$ by the least common multiple $\ell$ (which there always exist one since we only have finite many $a_n$ to get rid of denominators, we have:

$\sum_{n <\omega} \ell a_n e^{nt} = 0$

Relabel $\ell a_n = b_n \in \Bbb{Z}$

Also typo earlier, let $e^t$ be rational (as I don't know how to handle algebraic $a_n$s)

so... $\sum_{n < \omega} b_n e^{nt} = 0$

$b_0+b_1e^t+b_2e^{2t}+\cdots = 0$

that looks familar...

Holo

Completely unrelated problem: can $\Bbb R^+$ be separated to $\aleph_0$ disjoint sets such that all of them closed under addition?

I successfully proved it if you change $\aleph_0$ to any natural number but I am stuck there

What I did is induction, and I showed that if you can separate it into partition $P$ of size $n$ I can divide at least one of the elements in $P$ into 2 and so I get $n+1$ elements

Problem is that to prove it to $\aleph_0$ using similar method means I have to use transfinite induction

And I have no idea what the limit step should be

(I am assuming the axiom of choice)

Secret

2:57 PM

Algebraic function

In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power. Examples of such functions are: f ( x ) = 1 / x {\displaystyle f(x)=1/x} f ( x ) = ...

Sharath Zotis

Let $G$ be a group. Suppose that $G$ has a unique element of order 2, denoted by $a$. Prove that $ag = ga$ for all $g \in G$

How do I proceed here, I am not sure what exactly having a unique element of order 2 gives me here to show $ag = ga$?

Secret

If $f$ is an algebraic function, and $t$ is transcendental, what condition is $f(t)$ algebraic

Sharath Zotis

3:12 PM

anyone have any ideas?

Paul Plummer

What affect does conjugating an element have on its order? @SharathZotis

Sharath Zotis

What do you mean by conjugating?

Semiclassical

$b$ and $a$ are conjugate means that $b=g a g^{-1}$ for some $g$

Sharath Zotis

I see

Now what relation does this have to order?

Semiclassical

Well, let $b=gag^{-1}$. You want to show that $b=a$

Secret

3:22 PM

Semiclassical

But, what can you say about the order of $b$?

Secret

Actually, in positive terms, what does the contradiction mean. Does it mean that any difference between $e$ and its $b$th partial sum is smaller than all conceivable rationals?

Sharath Zotis

Im not seeing it, what can we say about the order of b?

Semiclassical

Well, what's the definition of the order of b?

Sharath Zotis

oh its also 2?

right

definition of order of b

Semiclassical

3:25 PM

Yep. Why?

The basic point is that $b=gag^{-1}\implies b^n =g a^n g^{-1}$

Sharath Zotis

the definition of order of b is that b^ord(b) = e

Semiclassical

Not quite---it's the smallest such positive integer for which that's true

Sharath Zotis

oh yeah smallest right

Semiclassical

But anyways. It's easy to see that if $a^n =e$ then the same is true for $b^n$ and vice versa

which in slogan terms means that conjugation preserves order

So a and b will have the same order (2)

Sharath Zotis

Right

Semiclassical

3:28 PM

But what do we know about elements of order 2 here?

Sharath Zotis

they are the same

a = b

Semiclassical

there's only one such element, so yeah---b must just be a

Sharath Zotis

since $G$ has a Unique element of order 2

exactly

Secret

Transcendental number

In mathematics, a transcendental number is a real or complex number that is not algebraic—that is, it is not a root of a nonzero polynomial equation with integer (or, equivalently, rational) coefficients. The best-known transcendental numbers are π and e. Though only a few classes of transcendental numbers are known (in part because it can be extremely difficult to show that a given number is transcendental), transcendental numbers are not rare. Indeed, almost all real and complex numbers are transcendental, since the algebraic numbers are countable while the sets of real and complex numbers are...

Semiclassical

so $b=ga g^{-1}=a$ and you're done

Secret

3:29 PM

and where does that integral operator came from othter than it works

Rithaniel

A quick, random question: Are there any sequences which you would describe as "interesting" which remain in mod1000?

Semiclassical

What?

Sharath Zotis

wait how does this imply $ag =ga$?

Semiclassical

@SharathZotis Rearrange it

You're always free to multiply on the left or right by $g$ or $g^{-1}$

Sharath Zotis

ok

so $bg = ga = ag$

Semiclassical

3:31 PM

ya

In more formal terms: The center of a group is defined as the set of elements which commute with everything else in the group

So you've shown that, if there's a unique element of order 2, then it must lie in the center

Sharath Zotis

$\mathbb{U}_5$

this group is cyclic correct? and generated by $<3>$

Holo

@Secret no, it says that the difference between $e$ and $S_b$ must be rational number with denominator both smaller than $b!$(so it will be able to divides $b!$ and be an integer) and greater then the numerator times $b!$

Sharath Zotis

How do i listing all of its sub groups

Semiclassical

I don't know what that group is

Sharath Zotis

$\mathbb{U}_n$ multiplicative group of integers relatively prime to $n$ modulo $n$

Semiclassical

3:38 PM

Ah. I usually see that written as $\mathbb{Z}_n^\times$

Sharath Zotis

so $\mathbb{U}_5 = {1,2,3,4}$

Semiclassical

3,3^2=9=4,3^3=27=2,3^4=81=1

checks out

Sharath Zotis

yup

Semiclassical

There's another such generator tho

Sharath Zotis

how do i list its subgroups tho

is the other generator 4?

Semiclassical

3:41 PM

Well, what do you know about the order of a subgroup in relation to the order of a group?

@SharathZotis what are the powers of 4 mod 5?

Sharath Zotis

are they the same

Semiclassical

write them out

Sharath Zotis

order of subgroup in relation to order of a group

Semiclassical

@SharathZotis they needn't be identical, no

Sharath Zotis

4^0, 4^1 = 4, 4^2 =1, 4^3 = 4, 4^4, hmm i see

4 doesn't generate

Semiclassical

3:43 PM

correct. nevertheless, <4> is still a subgroup

Sharath Zotis

2 does generate i believe

Semiclassical

since 4^n 4^m = 4^(n+m)

right

Sharath Zotis

so does each generator generate a subgroup?

Semiclassical

one easy way to see that is that 2=-3 mod 5

@SharathZotis well, you should be able to test that

If you pick two powers of some $a\in U_5$, then you have $a^m a^n =a^{m+n}$

So the set of powers is closed under the group operation

You should check for yourself that any <a> generates a subgroup

What I"m not remembering is if that generates all the subgroups (up to isomorphism)

i.e. if A is a subgroup of a group G, then is there an element $g\in G$ such that $A=\langle a \rangle$

The answer is probably "of course not" but I forget the specifics

Paul Plummer

Well not all groups are cyclic @Semiclassical

Semiclassical

3:52 PM

ya, true

I mean, the group U5 is small enough that there's not really a lot of options

Paul Plummer

Every subgroup of a cyclic group is cyclic (that is a good exercise)

Semiclassical

oh, nice

So it's fine in this context. (On the other hand, counting the subgroups of something like S6 presumably would require more finesse)

user525966

can anyone help me better understand how to prove completeness in a logical system? I've already got soundness down I think

(not a student / just a bored person at work self-learning)

Paul Plummer

Definitely, I suspect that could be done by hand without to much work, but it would be more work

Semiclassical

for reference: oeis.org/A005432

in particular, for S6 one gets 1455 subgroups (yikes)

it only goes up to S18, at which point they basically just say "nope"

Rithaniel

4:01 PM

Bah, 125889331236297288 isn't all that many. :P

Semiclassical

in the Holt paper linked in that OEIS entry: "The complete calculation of the 5 808 293 fixed-point-free subgroups of S18 using these techniques took approximately four cpu-days."

user193319

4:16 PM

Problem: For any continuous function $f : [-1,1] \to \Bbb{R}$, show that $\int_{-1}^{1} f(x^2)dx = 0$...Question: Is continuity necessary? All that's needed is Riemann integrability, right? To prove it, I just noted that $xf(x^2$ is an odd function, so it suffices to prove that the integral of an odd integrable function over an interval symmetric about $0$ is $0$.

Wait! The composition $f(x^2)$ isn't necessarily integrable if $f$ is just assumed to be integrable, right?

So continuity is necessary?

Ted Shifrin

@user193319, I don't follow your "to prove it, I just noted that ..." ... Also, you should be able to prove that if $f$ is integrable and $g$ is continuous, then $f\circ g$ is integrable (on $[a,b]$). As a simple case, take $g$ monotone.

user193319

@TedShifrin I thought it was the other way around: $g \circ f$ is guaranteed to be integrable on $[a,b]$.

Ted Shifrin

Yeah, what I said is wrong. But it is correct when $g$ is monotone.

Neither $f\circ g$ nor $g\circ f$ is right, actually, in general.

I still don't understand your argument, regardless.

Leaky Nun

hi @Ted

Secret

[Random]

Tobias Kildetoft

4:29 PM

@user193319 Was the integral meant to be of $xf(x^2)$ instead?

Ted Shifrin

By the way, @user193319, the problem you're trying to prove is just flat wrong.

Ohhhh ....

That would explain it ... @Tobias

user193319

Well, since $xf(x^2)$ is odd, then if I just prove that whenever $h : [-a,a] \to \Bbb{R}$ is an odd integrable function, $\int_{-a}^{a} h(x)=0$.

Ted Shifrin

Yeah, you typed the problem wrong.

user193319

Oh, I see: I dropped a parenthesis.

Ted Shifrin

No, you dropped an $x$.

Tobias Kildetoft

4:30 PM

No, you dropped an $x$

LOL

Oh, whoops...

oO

Damn. Sniped by a more grammatically correct version.

user193319

I did drop a parenthesis in my OP, though; so I was right about that!

Ted Shifrin

4:31 PM

Yes, but even I am not that picky if I understand what you are saying.

So, anyhow, I do believe that if $f$ is integrable, then $f\circ s$ is integrable for $s$ monotone, and then even for $s(x)=x^2$ (why?).

Secret

"Polynomial flow" iterated function:
Motivation: Consider a sequence of iteration of a function $f$ e.g. $f \circ f^2 \circ f^3 \circ f^2 \circ f$
Ok fail, because the fact that $f^n \circ f^m = f^{n+m}$ holds always means that even if I do something as crazy as $f^{\int_a^b g(x)dx}$ I can always find some real number $r = \int_a^b g(x)dx$

and that is already captured in the study of flows

Ted Shifrin

I don't know what that means for most functions $f$.

Leaky Nun

@Secret I have a function $f : A \to A$ such that for any $x$, there is $n$ such that $f^{n!}(x) = x$, so I can write $\cdots \circ f^{4!} \circ f^{3!} \circ f^{2!} \circ f^{1!} \circ f^{0!}$

this function shows up in field theory

oh and of course, $f^{n!} \ne id$ for any $n$

(@TedShifrin am I doing the right thing?)

Tobias Kildetoft

@LeakyNun Well, you can construct such a function without needing the factorials as well quite easily (though possibly less naturally)

Leaky Nun

sure

we don't know if $\sum n!$ is an integer right? @TobiasKildetoft

Secret

4:39 PM

I am guessing there isn't any point to introduce an uncountable analogue of an algebraic function (the countable analogue of an algebraic function is a transcendental function) because of the fact that any real number can always be approximated by a sequence under the open interval topology

Ted Shifrin

I have no idea what you're doing, @Leaky, nor why the infinite composition makes sense.

Tobias Kildetoft

@LeakyNun Sum over what?

Leaky Nun

@TedShifrin because it's well-defined at every point

@TobiasKildetoft $\sum_n n! \in \overline{\Bbb Z}$

Ted Shifrin

I don't see why things stabilize when you compose functions.

Secret

but $\sum n!$ should blow up to infinity, no?

Leaky Nun

4:40 PM

@TedShifrin because of the property of my function

Ted Shifrin

If $f^6(x)=x$, what is $f^{4!}(x)$?

Leaky Nun

x, of course

Secret

@TedShifrin en.wikipedia.org/wiki/Iterated_function iterated function compositions always obey index rule

Ted Shifrin

I have no idea what you're doing.

I withdraw from this discussion.

Leaky Nun

$f^{24}(x) = f^6(f^6(f^6(f^6(x)))) = x$

Secret

4:41 PM

so as long the countable operation converges to some number $r$, then $f^r$ is well defined

Ted Shifrin

Oh, right.

I still withdraw.

Leaky Nun

@TedShifrin that isn't what I mean

I mean, am I doing the right thing by encouraging him, lol

Secret

Hypertranscendental number

A complex number is said to be hypertranscendental if it is not the value at an algebraic point of a function which is the solution of an algebraic differential equation with coefficients in Z[r] and with algebraic initial conditions. The term was introduced by D. D. Morduhai-Boltovskoi in "Hypertranscendental numbers and hypertranscendental functions" (1949). The term is related to transcendental numbers, which are numbers which are not a solution of a non-zero polynomial equation with rational coefficients. The number e is transcendental but not hypertranscendental, as it can be generated from...

ok I did not expected that...

but it is not an uncountable thing

rabbit hole way too deep, better stick with transcendentals for now

2

Q: Product of Uncountably Infinite Number of 1s

Just like the title says: What is the product of an uncountable number of 1s? Intuitively the answer is 1, but how does one go about defining such a product in general?

infinite-product

technically, uncountable products are well defined with a net, but it is not really necessary

This is because all real numbers are either algebraic (which can be mapped to zero under finite number of algebraic operations) or transcendental (which can be mapped to zero under countably many of these)

The reals simply does not contain any intrinsically uncountable structure in it, other than its cardinality

I think we need to get to the surreals or banach spaces in order to have an intrinsically uncountable structure, but we will deal with this later...

Secret

(Unrelated)

Let $f$ be an algebraic function. That is given polynomials $a_n$, $\sum_{n < \omega}a_nf^n =0$. Suppose $t$ is transcendental:

Let $f(t)=r$. Then we have: $\sum_{n < \omega}a_n(t)r^n=0$

Suppose $r$ is algebraic, then $r^n$ is also algebraic by field closure of algebraic numbers

Expanding the sum, we thus have a polynomial of algebraic coefficients $Q(t) = \sum_{n,m < \omega} r^nb_{nm} t^m =0$

But this is a contradiction since this will mean there is a polynomial $Q$ which $t$ is a root

Semiclassical

5:08 PM

The following is discombobulating: I thought that I was out of a medication, to the extent that I was having to skip doses to keep it up longer. But the pharmacy says that what I picked up at the end of August should've been enough to keep me through this month

wtf?

Secret

It depends on how much medication you have left and what is the original instruction

Mike Miller

Any chance they gave it in multiple bottles and you misplaced some?

@Secret that feels tautological lol

Semiclassical

@MikeMiller that's what I have to wonder

Lozansky

Maybe you got a ghost problem

Semiclassical

I think I could see how it would happen

Secret

5:10 PM

@MikeMiller ah right, your's is a more likely possibility since some pharmacy sometimes give you slightly more than necessary

Semiclassical

But it throws you for a loop when the mismatch between the reality implied by the medical records is so off from what you think is going on

Mike Miller

@Semiclassical You can surely call your doctor for an emergency refill, though. That's certainly better than shipping meds

N..

so, what's the answer to @user193319 original question? Is continuity necessary to prove that $\int_{-1}^{1} \! x f(x^2) dx \equiv 0$? Naively I would have said that an integrable $f$ has antiderivative $F$, the integrand is $(F \circ \varphi)'(x)$ and used the fundamental theorem

bphi

"Find kernel of f: G -> G, f(x) = x^2 for G abelian" I can't be more precise than {x in G | x^2 = e} without knowing more about the group right?

Semiclassical

@MikeMiller the tricky thing is that I'm in the middle of a transition between clinics right now

and haven't been able to get set up with the new one yet

Secret

5:13 PM

(unrelated cont.) and therefore $f(t)$ must be transcendental.

Semiclassical

(the old one was the university clinic, which I don't have access to now that it's fall semester and I've graduated)

so it's a bit delicate

Secret

Combined with the fact that $f(k)$ is algebraic for all algebraic number $k$, it means the image of any algebraic functions will preserve the transcendence of its arguments

ok, that could be useful...

Ultradark

Secret is churning out knowledge here

Lozansky

Can't you just do the substitution $u = x^2$? :> Since the lower limit = upper limit and $f(u)$ is continuous you would have $0$

Secret

Hmm...

Leaky Nun

5:17 PM

@bphi yes

Lozansky

I know $0$ analysis so it's almost certainly wrong but :P

bphi

Thank you @LeakyNun

N..

@Lozansky for continuous $f$ you can split the integral and do that I suppose, can you do that when $f$ is only integrable?

Ultradark

Can someone explain the entropy formula

Shannon entropy

Secret

Let $g$ be a transcendental function. Suppose $t$ is transcendental. Let $g(t)$ be algebraic, that is there exists some polynomial $P$ such that $P(g(t)) = \sum_{n < \omega} a_n[g(t)]^n=0$

N..

5:20 PM

@Lozansky I thought the initial question "does it work if $f$ is integrable" more interesting than the actual proof for continuous $f$

Soham Chowdhury

where i come from, the state of chat rn would be described by most high school teachers as a "fish market"

Lozansky

@N.. Not sure, but I can't come up with a counter-example

Mike Miller

One pound fish!

Leaky Nun

where i come from, it would be "wet market"

but even that term is not common English, so this is the definition:

> In Hong Kong English and Singapore English, a wet market is a market selling fresh meat and produce, distinguished from dry markets which sell durable goods such as cloth and electronics.

Perturbative

5:36 PM

Hey everyone

So there's this theorem that says "If $H \leq G$ such that $[G : H]= 2$, then $H$ is normal in $G$. The proof starts as follows "Since $[G : H] = 2$, there are only two left cosets of $H$ in $G$, namely $H$ and $G \setminus H$"

Now I fail to see how $G \setminus H$ is a left coset of $H$ in $G$.

Tobias Kildetoft

@Perturbative Do you see that there are two cosets?

Perturbative

@TobiasKildetoft Yeah, that part I'm fine with

Tobias Kildetoft

And that $H$ is one of them?

Leaky Nun

@Perturbative maybe work with an example

work with $S_3 = D_{3,6}$

Perturbative

@TobiasKildetoft Actually let me just think again why $H$ is one of them, I thought it was trivial initially

Leaky Nun

5:45 PM

it is.

Perturbative

@TobiasKildetoft Okay yeah so initially $G/H = \{aH, bH\}$ for some $a, b \in G$. Then $aH = H \iff a \in H$, but since $1_G \in H$ (because $H \leq G$) it follows that $a \in H$, so $aH = H$. So $G/H = \{H, bH\}$ for some $b \in G$

Oskar Tegby

I'm still a bit stuck with this one.

Secret

Let $g$ be a transcendental function. Let $q$ be algebraic. Suppose $g(q)=a$ is algebraic. Then there exists polynomial $P$ such that $P(g(q))=0$. Inverting this equation using the preimage, we get $g(q)=P^{\leftarrow}(0)$. Inverting $g$ using the preimage, we have $q = g^{\leftarrow}(P^{\leftarrow}(0))$

Since $P$ is an algebraic function. It follows $P^{\leftarrow}$ is also an algebraic relation. Since $0$ is algebraic, $P^{\leftarrow}(0)$ must also be algebraic

Oskar Tegby

@TedShifrin: Do you have any hints?

Soham Chowdhury

@LeakyNun that's the image i was going for, but TIL "wet market"

Leaky Nun

5:57 PM

@SohamChowdhury you're from singapura?

Soham Chowdhury

no, lol

i meant the "fresh meat and produce" thing

Leaky Nun

I see

Soham Chowdhury

Kolkata, India

"fish market" is a word-for-word translation of the Bengali machher bajar (where bajar is cognate with Hindi bazaar, as you might expect)

Leaky Nun

indeed, that is exactly what I might expect :P

Soham Chowdhury

"Ever-so-slightly-broken Indian high school teacher English is a joy forever"

Secret

6:01 PM

$e^e=q \implies e = \ln q \implies (\ln q)^{\ln q} = q$

Perturbative

Okay I got it, so to prove $G \setminus H = bH$ observe first that since $bH \neq H$, it follows that $b \not\in H$, so $b \in G\setminus H$. I'll first show that $bH \subseteq G \setminus H$. Suppose that for some $x \in bH$ we have $x \not\in G \setminus H \iff x \in H$. Then $x = b\cdot h$ for some $h \in H$. But then $xh^{-1} = b$ and hence $b \in H$ (since $x \in H$ and $h^{-1}$ and $H \leq G$). Thus $\forall x \in bH$ we have $x \in G\setminus H$. So $bH \subseteq G\setminus H$.

Now conversely to show $G\setminus H \subseteq bH$ choose $y \in G \setminus H$ and suppose that $y \not\in bH$. Since the cosets partition $G$, we must then have $y \in H$ a contradiction. Hence $y \in bH$ so $G \setminus H \subseteq bH$. Hence $G \setminus H = bH$

Secret

$e^e=q \implies q = \ln q \implies (\ln q)^{\ln q} = q \implies$

$\ln q^{\ln q} = e^{\ln q} \implies \ln q^{\ln q +1} = (\ln q)e^{\ln q} \implies W(\ln q^{\ln q +1}) = \ln q \implies$

bumpbumpbump

Holo

6:22 PM

@Secret $e^e=q \implies q = \ln q???$

Secret

oops typo, should be $e = \ln q$

nevertheless it leads to nowhere

Holo

@Secret Proving that number is transcendental is usually extremely hard :)

Secret

Lindemann–Weierstrass theorem

In transcendental number theory, the Lindemann–Weierstrass theorem is a result that is very useful in establishing the transcendence of numbers. It states the following. In other words the extension field ℚ(eα1, ..., eαn) has transcendence degree n over ℚ. An equivalent formulation (Baker 1990, Chapter 1, Theorem 1.4), is the following. This equivalence transforms a linear relation over the algebraic numbers into an algebraic relation over ℚ: by using the fact that a symmetric polynomial whose arguments are all conjugates of one another gives a rational number. The theorem is named...

Meanwhile still trying to comprehend this proof

Why so many integrals out of nowhere

Ted Shifrin

6:51 PM

@Oskar: How far did you get?

Ultradark

How do you factor x^2+x-1

Ted Shifrin

Use the quadratic formula to find the roots.

Ultradark

Completely forgotten how to complete square

Leaky Nun

@Ultradark $x^2+x-1 = (x-t)(x+t+1)$ in $(\Bbb Q[t]/(t^2+t-1))[X]$

(it surely has no roots in $\Bbb Q$, so let's factorizing it in a splitting field)

ÍgjøgnumMeg

@Leaky do ya remember the tip I gave you?

Leaky Nun

6:55 PM

(why is $\Bbb Q(\sqrt5)$ the canonical splitting field? what makes my choice not valid?)

@ÍgjøgnumMeg completely forgotten :P

Ted Shifrin

Leaky is practicing being uber-obnoxious.

5

ÍgjøgnumMeg

cough pedagogy cough

Alessandro Codenotti

Hi everyone

Ted Shifrin

hi demonic @Alessandro!

ÍgjøgnumMeg

Hey @Alessandro

Perturbative

6:56 PM

Hey @AlessandroCodenotti @TedShifrin

Ted Shifrin

hi Perturb

Leaky Nun

@ÍgjøgnumMeg omg is your throat alright

Ultradark

Lol

Perturbative

@Ted My algebra lecturer said something interesting today, apparently nobody knows how to classify finite groups of order greater than $2000$ (according to him), and the main tools to classify them at order near $2000$ rely on cohomology (I assume he was talking about group cohomology)

Mike Miller

what does it mean to classify groups of fixed order?

or to classify groups?

Sir Michael Atiyah's Proof of the Rie…

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$Mathematics$ Mathematics