>implying the reverse isn't true

Lol

You've asked me the question.

whatever it's abelian

@Karl thanksies

let's move on with out lives

It follows very simply, given that there is only one non-identity auto.

Anywho.

you never explained how "some conjugation map is the unique non-identity auto" implies the group is abelian

*sighs.

don't bother

It's easy to show.

I shan't clarify who that's written ro

It's not an obvious result, but it is, in fact, easy to show.

I would say: If ${\rm Aut}(G)=\{1,\alpha\}$ and ${\rm Inn}(G)\ne1$ then $|{\rm Inn}(G)|=2$ implies $[G:Z(G)]=2$ and then either invoke the fact or argue that this is impossible. Proof of impossibility: suppose $G=Z(G)\sqcup aZ(G)$. Conjugation-by-$a$ acts trivially on $a$ and on $Z(G)$ hence on $G=\langle a,Z(G)\rangle$, hence $a\in Z(G)$, a contradiction.

Okay?

Does R/Q have any non-trivial automorphisms?

Under addition.

like negation?

Other than inversion.

Besides that, I mean.

what a terrible group

multiplication by two?

why would you show me something like that

might find what you want in Q/Z

Yes.

I was thinking of R/Q or Q/Z.

I also need to assert whether or not there exists an uncountable such group.

@anon quotient G/Z cyclic implies trivial wiiiiii

With a unique non trivial auto

@pedro implies G/Z trivial?

Q/Z also has many nontrivial autos, not easy to see in that form though

cuz that ain't true

Like, for example?

in fact I think ${\rm Aut}(\Bbb Q/\Bbb Z)\cong\widehat{\Bbb Z}$, which is pretty big

Multiplication by a rational wouldn't work, because it would send something other than 0 to 0 also.

write $\displaystyle\frac{\Bbb Q}{\Bbb Z}\cong\bigoplus_{p\text{ prime}}\frac{\Bbb Z[1/p]}{\Bbb Z}$ to see where nontrivial autos may come from

(this is the so-called Prufer decomposition)

or even better, $\Bbb Q/\Bbb Z\cong \bigoplus \Bbb Q_p/\Bbb Z_p$

@PedroTamaroff Oh, but Fld has the property that the product of any pair of objects doesn't exist, while my example has products like $a\times a$. I think if we take the category with just one object and two morphisms, $1,f$ satisfying $f^2=f$, then we have another example. Namely, there is no arrow to map one the following two diagrams to the other. $$\require{AMScd}\begin{CD}\cdot @>f>>\cdot\\ @VV1V \\\cdot\end{CD}\text{ and }\begin{CD}\cdot @>1>>\cdot\\@VV1V\\ \cdot\end{CD}$$

Fld is the worst IMO

Ok... Cannot render mega tex in phone and must sleep and run in the morning

Bubyes

lol ok

(the idea here is that every rational number $x$ can be written as $\sum a_p/p^{e_p}$ for some exponents $e_p$ and $0\le a_p<p^{e_p}$ (only finitely many of which are nonzero): this is partial fraction expansion applied to rational numbers instead of rational functions; an example of a nontrivial auto of $\Bbb Q/\Bbb Z$ I think is where we take the $a_p/p^n$ part for some $p\ne 2$ and multiply just that part by $2$)

@Karl what's your favorite category

@Mike Don't have one. lol

Although, doing numerology but with categories (so categorology) would be cool.

what would that be?

it's a sign of poor health if the category of pointed spaces shows up in your life?

Ach.

A Categorologist sees categories in everything, like a numerologist sees numbers. Then you just have to make artificial links between these events/objects by comparing the categories.

pointed spaces and its opposite? your health will fluctuate wildly

Yes, you're getting the hang of it.

so I'll ask the obvious question

what does BanAnaMan signify

All your problems in life can be solved by eating more bananas.

perhaps wealth

there's always money in the bananaman

eh? eh? nobody?

@PedroTamaroff yup :p

@TedShifrin it is in $\mathbb{R}^{1}$.

@alex what is

@Mike the dot product of two vectors is a vector in $\mathbb{R}^1$

@anon haha i just now noticed this. i'm surprised you didn't know yet. i do that all the time. i think i've done it to you before.

$\Bbb R\ne \Bbb R^1$

@AlexanderGruber The scalar product of two vectors is also a matrix ;) or an element of any (nontrivial) $\mathbb R$-algebra for that matter

@anon y'know here is a question

what is a meaningful difference there

i can't think of one.

is anything really different, maaaan?

@anon i mean set theoretically

booring

i can't think of a difference between $\mathbb{R}$ and $\mathbb{R}^1$ other than I write one with parenthesis

which is not really a very mathematical difference

How is $\dfrac{\ln 4-\ln 2}{\ln 4}=\dfrac 12$?

@Sush $\ln{x}/\ln{y}=\operatorname{log}_y(x)$

@alexander ones a vector space, ones a field

@Sush $\log(a^b) = b\log(a)$

@Mike but they're both both.

there aren't any properties of vector spaces that don't hold in $\mathbb{R}$

@alexander i don't buy it

@AlexanderGruber, @Mike, thank you so much.

@alex Ah, now you're talking about uniqueness of additional structure on a less structured $\mathbb R^1$, so you first have to admit $\mathbb R^1\ne \mathbb R$.

@KarlKronenfeld idk, am i? just because something's one algebraic structure doesn't mean it's not another.

if i talk about $\mathbb{Z}$ as a group that doesn't mean it still isn't a ring.

it's neither

lol

it's clearly a $0$-module

@AlexanderGruber We're moving very quickly onward to the philosophical question of what it means to be.

@KarlKronenfeld get Clinton in here!

@AlexanderGruber My perspective is this: it can be preferable to distinguish $(\mathbb Z,+)$ from $(\mathbb Z,+,\cdot)$. We want to prove the ring structure on the abelian group $\mathbb Z$ is canonical, so that we can really treat $\mathbb Z$ as a ring in contexts where we are using it as a group. Then we can naturally lift to ring homomorphisms if the other things turn out to be rings as well.

My recent sequence ans series answer got 7 upvotes.

I never thought it would be that hot

Hello?

@anon Hey man, you on?

Hullo, @N3bu

@N3buchadnezzar What's up for integrals today?

@BalarkaSen heya

@BalarkaSen hi

Hi @robjohn does this inequality make sense $$\mathrm{dist}(x_0, y) \le \left\|x_0+ \frac {y}{\alpha}\right\|$$ where $y\in Y$ a closed subset , $\alpha >0 \in \mathbb R $

@Sawarnik Whassup?

@N3buchadnezzar Liked that.

Yep, I recalled it from my sheer memory

@BalarkaSen Mind helping me find a question on main?

@N3buchadnezzar I'll try.

$$ \int \frac{1+x^5}{1+x^8}\mathrm{d}x $$

Yes?

You have to do what?

Evaluate it?

Yeh, I know it was posted on main

Partial fractions, I'd guess.

over $\Bbb Q(i)$

@BalarkaSen Again with the heavy machinery

There was a trick to solving it and I can not quite recall the finer details of it

Hmm, you want lesser Machines, do you?

Or laser Machines?

Hey, @Sawarnik whachha doin'?

@DanielFischer Bonjour

@BalarkaSen Watching you and some other writing some ununderstadable stuff.

I want Picasso and Bach, not a painting machine and soundwaves

@Complexanalysis Bonjour aussi.

I know it was asked on main, but it is impossible to search after latex >.<

I'd then split it up in $$\frac{1}{1 + x^8} + \frac{x^5}{1+x^8}$$

@BalarkaSen Do you retain some interest in lower questions?

@Sawarnik Maybe not.

Okay No.

$$\int \frac{1+x^2}{1+x^4}\,\mathrm{d}x = \int \frac{1+1/x^2}{(x - 1/x)^2 - 2}\,\mathrm{d}x $$

@DanielFischer does this inequality make sense $$dist(x_0, y) \le ||x_0+ \frac {y}{\alpha}||$$ where $y\ in Y$ a closed subset , $\alpha >0 \in \mathbb R $

@N3buchadnezzar Now, sub $t = 1 + x^8$ for the second integral

@Complexanalysis Not sure.

@N3buchadnezzar I'd avoid those kind of simplification tricks. It makes my head spin.

@BalarkaSen They are lovely

$$\int \frac{dx}{1 + x^8}$$

Evaluate that.

$$\operatorname{dist}(x_0,Y) \leqslant \left\lVert x_0 + \frac{y}{\alpha}\right\rVert$$ would make sense if $Y$ is a linear subspace.

Integrals need a certain je ne sais quoi to be attractive.

@BalarkaSen Over $R$ i could do it quite easilly

@DanielFischer Yes i meant that . But why would that hold, i don't see it ?

Integrals are not my thing.

Give me series.

@N3buchadnezzar $R$?

(-\infty,\infty)

You mean $\Bbb R$?

yeah

Same thing, you could view R as the contour.

Yeah, sure.

@Complexanalysis If $Y$ is a linear subspace, $-\alpha^{-1}\cdot y \in Y$ if $y\in Y$, and the distance is the infimum of all distances of $x_0$ to a point of $Y$.

=P

=P

Well the only integrals that interest me are those whose solution is clever, unexpected, part genious.

All other integrals are mostly boring =P

@N3buchadnezzar True.

Tedious.

@N3buchadnezzar You remember saying me that I&S always picks the heavy guns?

Mmm?

Well, S has a heavy gun in the avatar.

=P

@DanielFischer So it doesn't make sense to talk about dist$(x, y)$ is it ? or is it just a conventional issue ?

:p

...and he just changed it. Odd.

@Complexanalysis For points, it's just the distance with respect to the metric, $\operatorname{dist}(x,y) = d(x,y)$.

$$\int_0^\infty \frac{a-\cos (x)}{a^2-2a \cos x +1}\cdot \frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}\exp \left( \frac{1}{\sqrt{2}}\right)\frac{a \exp \left(\frac{1}{\sqrt{2}} \right)+\sin \left( \frac{1}{\sqrt{2}}\right)-\cos \left( \frac{1}{\sqrt{2}}\right)}{1-2a \exp \left(\frac{1}{\sqrt{2}}\right)\cos \left( \frac{1}{\sqrt{2}}\right) +a^2 \exp \left(\sqrt{2}\right)}$$

(Bam, ohhhhh, my head! Boom, aaaahha!)

@DanielFischer Ok .

This is what I like. Clean and clear.

I think you might need it to be bigger than $b>a>1$ but whatevs

@BalarkaSen Not that bad if you remove the 1/(1+x^4) part

What's the most bizzare identity you ever saw?

Möbius function?

?

What about it?

Darn.

Is sigma the mobious functions?

Repost : $$\frac{\sigma(nz)}{\sigma^{n^2}(z)} = \frac{(-1)^{n-1}}{\left(1!2!3!\cdots(n-1)!\right)^2}\begin{vmatrix} \wp'(z) & \wp''(z) & \cdots & \wp^{(n-1)}(z)\\ \wp''(z) & \wp'''(z) & \cdots & \wp^{(n)}(z)\\ . & . & \ldots & .\\ . & . & \ldots & .\\ \wp^{(n-1)}(z) & \wp^{(n)}(z) & \cdots & \wp^{(2n-3)}(z) \end{vmatrix}$$

Accidentally clicked on delete instead of edit.

@N3buchadnezzar No. Google [Weierstrass sigma]

@N3buchadnezzar I though mobious function is denoted by mu?

$\mu$?

Perhaps you wanted to refer to the sum of powers of prime factor function?

Yeah, the divisor function. I always mess up the names

Divisor function is probably not what you want.

It's $\sigma_0$ or $d$ or $\tau$

We usually denote $\sigma_1$ as $\sigma$

@N3buchadnezzar This is not hard to derive.

I find this releation somewhat strange

@BalarkaSen Well I have never seen anyone done it :p

It is not. It comes from the complementary properties of the integral

@DanielFischer I hope to see you in a while .

@N3buchadnezzar You want me to do it?

If you want, I can show right away.

Sure

I have done it before, let me pull out my notes.

I know they are double integrals but the finer workings are a tad wonky

Hmm, I can't find them. Looks like I have to rederive it from nothing then.

@N3buchadnezzar I don't think I did it with double integrals, but I will consider your suggestion.

Like i said this identity stumpled me

if you want I could ask it as a question ?

Wait, let me derive it.

Then I will ask you to ask it as a question =D

I think one can get that by expressing the elliptic blahs into Jacobi thetas.

Oh, hey @robjohn

@Complexanalysis that allows $\mathrm{dist}(x_0,y)=0$ even though $x_0\ne y$

Nevermind. Ignore them.

=D

I am going senile these days.

@Karl How does one define the dimension of a [perhaps only local?] field? What is meant by "higher-dimensional fields"?

Hey Balarka I've got an integral for you to compute :P

@GabrielR. OMG, I am just a kid!

See my latest post

Nevertheless, throw it. I am already stumped on N3bu's problem

On Math se

Which problem is it?

I think it would suffice to use the $\Gamma - \zeta$ formula

Which is? Never heard of it...

Wait, this is the gamma zeta formula I have to prove,

Just use series expansion.

Series expansion of whom?

1/(1+e^t) can not be expanded since t>0

Take a look at Bernoulli numbers. (Oh, and @-ping me always, you can't find me otherwise)

@N3buchadnezzar Darling's product might help

Through Clausen, perhaps, if nessesary.

Yep, Darling definitely helps.

Have you tried it, @N3buchadnezzar?

@BalarkaSen Can you help me with this: math.stackexchange.com/questions/666279/…

Or anyone else :)

See Pinilla's comment.

Stewart took the approach by mimicking Barrow.

That's why I do not consider it a very nice book.

It confuses people a lot.

Personally, I like to keep things in order.

I saw a book ones that assumed a whole lot of theorem as axioms and then proved the axioms as theorems in geometry.

@BalarkaSen Nope, do you think one can apply this technique to integrals of the form

Not sure.

$$ \int_0^\infty \frac{1+x^{n-2}}{1+x^n}\,\mathrm{d}x $$ ?

@N3buchadnezzar I think I have found out a way. Differentiate w.r.t $k$ to prove this.

@BalarkaSen Then you would need to know both sides of the equation at forehand

Okay, you want other way not verification. I see.

I can't recall even bits of what I did before, darn.

I am sure this comes from hypergeometric what nots though, and the differential equations.

Well it is not trivial atleast :p

darn I can not seem to make the substitution on the general integral

darn

@N3buchadnezzar I am not sure about that.

It could be you know

I find hypergeometric linear dependence pretty amusing.

Don't you?

meh Hypergeometric functions are just constructions to make life easier, they were made that way.

I didn't know that.

Still, that construction surpasses our controlling devices now.

A simple example can be seen in one of my highly upvoted answer.

@Sawarnik You are getting addicted of chat!! =p

@BalarkaSen Yes possibly, its interesting. But how do you manage your time, since you are online all the time. When do you study your mind boggling theories, do your research, answer and chat here, and manage schoolwork among other things.

I am really getting addicted of this site, though.

@Sawarnik I do several things at once, that's true. If I can get a screen shot of my screen at this time, you'll see at least 50 tabs open in Firefox, almost in a crashing position.

And... I think I just solved your problem @N3buchadnezzar

It involves tedious double integrals, as you suggested.

How do you manage all that in one head?