Mathematics - 2016-02-10 (page 3 of 3)

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Akiva Weinberger

8:01 PM

@BalarkaSen Quick! What's the capital of Sindh?

Um. Jamaica?

Akiva Weinberger

buzzer

(It's Karachi, if you're curious.) Sorry to hear that the geography section didn't go well.

sindh is a country ?

Akiva Weinberger

No

I know Karachi is in Pakistan. I didn't know it was the capital of Sindh.

Akiva Weinberger

8:03 PM

But its capital is Karachi.

@BalarkaSen Me neither. I just looked it up :P

Fair enough!

How are you doing?

Akiva Weinberger

Good! Can't talk now

@PVAL: It seems to. I'm very upset about this. Why is there an embedding of the punctured 3-fold?

I'm less upset because I see no reason to believe there's an embedding of the punctured thing.

1

Q: If $(z_{n}) \in \overline{ \mathbb{C}}$, $z_{n} \to \infty$ as $n \to \infty$, what happens to $|z_{n}|$, $Re(z_{n})$, $Im(z_{n})$, $Arg(z_{n})$?

Suppose the sequence $(z_{n}) \in \overline{\mathbb{C}}$ (where $\overline{\mathbb{C}}$ is the extended complex plane) converges to infinity as $n \to \infty$. I need to determine what this implies about $|z_{n}|$, $Re(z_{n})$, $Im(z_{n})$, $Arg(z_{n})$. I know that a sequence $(z_{n})$ converg...

sequences-and-series complex-analysis limits complex-numbers

I'm really, really not seeing anything about what happens to Arg z_{n}, and the guy who gave me an answer that I really don't think is complete or elaborate enough is now ignoring me.

Nice.

So if anybody here could help me out, I would be so grateful!!!!!

Akiva Weinberger

The $(2i)^n$ thing also works for the argument.

For that, the argument doesn't converge to anything.

Another example is $z_n=nc$, for some $c\in\Bbb C$.

The argument is ${\rm Arg}(c)$ for all $n$.

@JessyCat

8:14 PM

Thank you!!

@AkivaWeinberger, dude actually just added an edit. But, thank you!

Akiva Weinberger

@BalarkaSen I'm doing well. In a week I'm taking the AMC 12 (a math contest).

Ah. Good luck!

Akiva Weinberger

Thanks!

8:32 PM

:)

Akiva Weinberger

$$\frac12+\frac25+\frac3{31} +\frac4{1241}+\frac5{1923551} +\frac6{4440055831261} \approx\\0.99999999999 999999999999969565$$

I don't see any pattern in the denominator. Elaborate?

Akiva Weinberger

@BalarkaSen I'm actually trying to figure that out now.

They're each one more than a multiple of the previous…

And one more than a multiple of the numerator.

Hi, DogAteMy ...

Akiva Weinberger

Oh, I see it.

$a_n=a_1a_2\dotsb a_{n-1}n+1$.

@BalarkaSen That's the reccurence.

Don't know how to prove that it converges to $1$, though.

Rather, that $\sum\frac n{a_n}$ converges to $1$.

8:45 PM

@TedShifrin Bonsoir ! Aurais-tu une idée pour ça ?

Bonsoir, M le Méchant.

Your beginning sentence is wrong. It's only continuous from the left at $R$.

@TedShifrin Oops I indeed meant to write $R^-$

Akiva Weinberger

Alternatively, $a_n=\frac n{n-1}a_{n-1}^2-\frac n{n-1}a_{n-1}+1$…

@TedShifrin Wait, were you referring to me?

Sorry, didn't notice!

In the complex case, Abel's Theorem generalizes to approaching within a wedge from inside the disk.

Akiva Weinberger

("DogAteMy")

8:48 PM

Yes, @Akiva, that was you :)

@AkivaWeinberger Yeah, I don't know how to prove that either.

Akiva Weinberger

Hi!

@TedShifrin Hi, can I ask you a question related to Algebra?

@TedShifrin A wedge isn't enough for my case :(

@Hippa: The other questions are similar to things I have assigned for homework when I've taught graduate complex analysis.

8:50 PM

@TedShifrin My teacher was unable (well, in 5 mins) to find (counter)-examples to my question(s), and I didn't find any myself

You have to distinguish the questions of whether the series converges and whether the function represented by the series can be analytically continued.

We have the convergence

Even your question doesn't make sense. You say $f(z_0)$ converges. What do you mean? Do you mean the series converges at $z_0$? Or do you mean the analytic function represented by the series on the interior of the disk makes sense at that point?

Hi @Ted.

Goodnight @MikeM

8:51 PM

I mean that $\sum a_n z_0^n$ exists (i.e. converges)

Sorry, @Paradox — in the middle of something at the moment.

Ok sorry

@Hippa: So $f(z)=\sum z^n/n^2$ ... the series converges on the circle, but the function it represents does not make sense at $z=1$.

@paradox101: what's the question?

@TedShifrin Sure, but that's the opposite of what I need. That one converges everywhere except on some points

8:52 PM

The other standard example (which you're alluding to, I think) is $\sum z^{n!}$, which cannot be analytically continued at all.

Akiva Weinberger

@BalarkaSen Oh, I think I see how to prove it. I think that the sum of the first $n$ terms is $1-\frac{n+1}{a_{n+1}-1}$. If I can prove that, which I guess would be done via induction, it would imply that the series tends to $1$.

So, Hippa, in one easy sentence, what exactly am I looking for? Your post is way too long.

@TedShifrin Indeed my teacher led me to think about for instance $\sum z^{n!}/n^a$, but it didn't lead anywhere really.

@TedShifrin Basically, can I find a power series such that it diverges everywhere except in a finite (positive) number of points on the circle ?

But you require that it converge at some point?

Yes

8:55 PM

This is a nontrivial question. It's related to convergence questions for Fourier series.

@Huy Hi. Prove $G$ is abelian $(ab)^n= a^n b^n$ such that $a, b \in G$ and $n=i, i+1, i+2$ where $i$ is an integer. In this question can we assume that $(ab)^i= a^i b^i$ is true and implies that commutativity holds and use this fact to prove for the other consecutive powers? Or is it essential to first prove it for $i$ and then proceed to $i+1$ and $i+2$?

I've read this question in chat before

at least a week ago

was that you too?

No, I just started this question now

@TedShifrin Surely it has been studied already though hasn't it ?

Yes, @Hippa, it has. I just don't remember the answer. But you should find it on MO. Maybe look under Hardy spaces.

8:59 PM

@TedShifrin Doing random problems from 2+3+4. 3.3.12 says if $f :U \to \Bbb R^m$ is differentiable, $U$ convex and open, $Df(x) = 0$, then $f$ is constant. This is true in general if $U$ is path connected (i.e., every pair of pts can be connected through a path). Just choose $\gamma : [-1, 1] \to U$ between $x \neq y$, $f \circ \gamma : \Bbb [-1, 1] \to \Bbb R$ is differentiable and has derivative everywhere $0$ via chain rule. So by MVT, it's constant and in particular $f(x) = f(y)$.

Also, don't remember if I ever did the ladybug problem so here goes: $X$ be where the ladybug is, $A, B$ arbitrary with angle $AXB$ right-angle. We're looking at $f(X) = AX + BX$. $D_vf(X) = \nabla f(X) v = (AX/\|AX\| + BX/\|BX\|) \cdot v = v/\|v\| \cdot v = \|v\|^2/\|v\| = \|v\| = 5$.

By the way, the Marx brother problem on the exam is just done by noting that we're looking for the directional derivative of the sum of Groucho's distance from Harpo and Chico at the direction of $[3, 1]$, no? Then this can be computed just from the dot product of the gradient with $[3, 1]$.

@Paradox101: can you check the statement you're trying to prove? something's off, maybe some words missing

Morning @Ted.

Akiva Weinberger

Where do you live that it's morning?

?

Mike says "morning" even at 11 PM.

@Balarka: Of course you need piecewise-$C^1$ path connected.

Akiva Weinberger

9:01 PM

Ah.

@Hippalectryon I got amazing results these days.

Your answer to the ladybug is wrong, I believe, @Balarka.

@Huy The exact statement goes that there exists an $i \in \mathbb Z$ such that for all $a,b \in G$ (where $G$ is a group) $(ab)^n= a^n b^n$ for $n=i, i+1, i+2$. If $G$ staisfies this then it's abelian

ok

I don't really understand your confusion then

you wrote down pretty precisely what to prove

@Huy my confusion is just whether I also need to show the above for $n=i$ or whether I can assume that to be true and merely prove for the other two.

9:07 PM

@TedShifrin By path-connected, I did mean C^1 path connected. Thanks for pointing that out.

@TedShifrin I am a bit confused.

Is path-connected equivalent to piecewise-$C^1$ path connected? What hypothesis do we have on our domain?

@Paradox101: what? you need to show that $G$ is abelian

Oh, of course that ladybug thing was wrong. Whoops-a-daisy.

That was silly.

@TedShifrin What I meant was if $U$ is $C^1$-path connected and open, then if $f$ has everywhere 0 derivative and is diffable, then is everywhere constant.

C^1-path connectedness is stronger than piecewise-C^1 path connectedness.

Is it? And is a connected open subset of $\Bbb R^n$ $C^1$-path connected?

@TedShifrin Ugh, so far I've only found (on MO) a function that converges almost everywhere on the circle and diverges in a dense set... which does answer the last part of my MSE question, but not the first one :(

9:14 PM

@TedShifrin By the way, the correct answer to the ladybug problem is $5\sqrt{2}$.

@Huy yes I do. I just got what I was doing wrong. Thanks

Closer, @Balarka.

ok

I guess I missed a $-$ sign, because this ladybug fellow is decreasing.

indeed

9:17 PM

The first mistake was shameful. Apparently I forgot there was a person called Pythagoras.

Evidently so.

@TedShifrin Uh, if any two points can be joined by a C^1 path, they can also be joined by a piecewise C^1 path, right? Any C^1 path is automatically piecewise-C^1?

sure... but I'm asking the other way

@Ted: Slmeone has checked out out copy of Besse. :(

Oh. No, I doubt the other way is true. I also strongly doubt that any path connected open set in R^n is C^1 path connected.

If you give me a few minutes, I can come up with ctrexamples.

9:20 PM

@BalarkaSen That I'd love to see.

@Hippa: I presume you found this.

@MikeM: It's a shame when libraries aren't there only for your personal use.

@TedShifrin I was talking about this one actually, but I did see the above link too

Ultimately, @Hippa, you've stumbled onto a nontrivial question.

Aw :(

Haven't seen le petit Méchant in ages ... he still alive?

I wonder :P - I'm not sure why he doesn't come anymore.

9:26 PM

@Ted: I agree!

I'm mildly curious who has it. I don't think Petersen has students.

Could be another faculty member ...

Fair

@Hippalectryon do you think I might be one day as great as Ted? It's not an easy thing though.

@I'manartist I'm definitely not qualified to estimate Ted's level haha

@TedShifrin Actually, I think it may be true that piecewise C^1 path connected open implies C^1 path connected open. Let $U$ be a piecewise C^1 path connected open set. $x, y \in U$. Join by a piecewise C^1 path $\gamma : [0, 1] \to U$. So break up $[0, 1] = \bigcup I_k$, $\gamma$ C^1 on each $I_k$. So non-C^1 points are precisely the endpoints of $I_k$.

Let's take some endpoint $x_k$ of $I_k$ which is also startpoint of $I_{k+1}$. $B$ be an nbhd of $x_k$ contained in $U$. I need to modify $\gamma$ around $x_k$ inside $B$ so that $\gamma|I_k$ and $\gamma|I_{k+1}$ has equal 1st derivatives at $x_k$.

This doesn't sound so impossible.

9:30 PM

Just do it for two pieces.

@Hippalectryon did you try my integral above?

1

Q: Another way of doing integration

What's your option for calculating this integral? No full solution is necessary, it's optional as usual. Calculate $$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...

calculus real-analysis integration definite-integrals polylogarithm

@Hippalectryon ^^^

@I'manartist I saved it :D but as you know until the end of the year I have other priorities :-)

@Hippalectryon aaaaaaaaaaaaaaaaaaa, way disappointed to hear that :-(((

@I'manartist Sorry :P I'm just busy with other problems

Can't do everything at once >:o

@Hippalectryon OK ;)

Just kidding with Ted. He knows well that my aim is Ramanujan.

Anyway.

BBL (to my research)

@Hippalectryon Ah, almost forgot

9:35 PM

@TedShifrin I'd be hard-pushed to write down a formula right now, but: take $\gamma : [0, 1] \to \Bbb R^n$, $\gamma|[0, 1/2]$ C^1 and $\gamma|[1/2, 1]$ C^1. Pick a nbhd $B$ of $\gamma(1/2)$. Delete the middle stuff between two points $\gamma(x_0), \gamma(x_1)$ in the image of $\gamma$ in $B$, $x_0 < 1/2$, $x_1 > 1/2$. The goal is to plug in another curve $\sigma$ between $\gamma(x_0)$, $\gamma(x_1)$ so that derivative of $\sigma$ matches up with 1st derivative of $\gamma$ at $\gamma(x_i)$'s.

Intuitively, this is quite possible, contrary to impossible, but I don't think I can write down a reasonable formula especially on 3 AM :)

I can try tomorrow.

@Hippalectryon Did you find a nice series representation of $$\frac{\log(1-x)}{1+x}$$?

This is referred to as "rounding the corners," and, in fact, you can do it $C^\infty$.

Ah.

Were you thinking there was a connected open subset of $\Bbb R^n$ that was not smoothly path-connected?

Are you going to tell me the procedure, or do you think I can come up with it myself?

@TedShifrin Yes, I still do. I don't believe that's true.

9:38 PM

Better think about that one first, then. ...

Do you know about bump functions and partitions of unity?

Nope.

OK, you'll get to that in Chapter 8.

@I'manartist Ah, I don't think I had something 'nice'... but I did forget to try one way, give me a few minutes

;( looking forward to it.

@Hippalectryon If you have that, don't post it here.

9:39 PM

ok

a crossed sad face looks even sadder.

Moral of the story: Don't make faces.

Random Variable

@DanielFischer I know by direct calculation that $$\text{Res}\left[e^{iz} \sin \left(\frac{1}{z} \right),0 \right] = \frac{J_{1}(2)+I_{1}(2)}{2} \tag{1}.$$ And I know indirectly that $$\lim_{r \to 0} \int_{C_{r}} e^{iz} \sin \left(\frac{1}{z} \right) \, dz = \pi J_{1}(2), \tag{2}$$ where $C_{r}$ is the upper half of the circle $|z|=r$, traversed counterclockwise. But does $(2)$ somehow follow from $(1)$?

@TedShifrin Oh, I guess it is true that if there is a path then there is a piecewise-linear one.

It's a standard connectedness argument.

9:48 PM

You can open cover your path by balls and go through the balls by tiny piecewise linear paths, I believe.

And once you have a piecewise linear path, you have smooth one modulo believing this rounding the corner business which I don't know yet.

There are fancier ways of smoothing continuous functions, too. Sometime you'll learn more real analysis and learn about convolution.

Interesting. Where exactly does one use smoothing methods? Pardon my ignorance, but I don't know much about $C^\infty$.

Thus the question.

It shows up a lot in topology and analysis (differential equations) and geometry. You can certainly imagine wanting to prove smooth objects are dense (in an appropriate topology) in the space of all continuous objects.

Aha.

Thanks, that's exactly what I was looking for. Very fun.

@I'manartist You don't want any $\Phi$ right ?

9:57 PM

@Hippalectryon Not sure the meaning of that symbol for you. I usually use it as a golden ratio.

@I'manartist Lerch (I don't know the exact latex letter)

@Hippalectryon $\Phi$

@Hippalectryon No, I don't. :-)

Then I've failed so far :P I'll keep searching when I can

@Hippalectryon OK :D

Also it's been ages since I've asked... how's the book going ? :-DDD @I'manartist

9:59 PM

@RandomVariable Not sure. Could be that you can exploit some symmetry without explicit calculations, but I don't see it.

Guten Abend, @DanielF :)

Evening @Ted.

@Hippalectryon In a great shape! Thanks! :-) It takes a while to be published. :-)

Yay \o/

@Hippalectryon just to have an idea

I never saw an elegant, marvellous solution to that series. Then, I thought it's time to have one in my book.

10:07 PM

Then I'll wait eagerly to see it :-)

@Hippalectryon Presently I have 3 solutions to that series.

@Hippalectryon :D

@Hippalectryon I would like to have some people working for me. It's very tiredsome to work on a huge list of results. Each result can be exploit a lot, then ... (a team would be better).

@Hippa: Tell M le petit Méchant that I say hi.

@TedShifrin Sure :-)

I have to decide when I am going to head to bed soon.

hi

I'm trying to determine if these statements are true. Are there any local experts online? math.stackexchange.com/questions/1610068/…

btw, hi hippalectron

10:13 PM

OK, 7 AM seems like the perfect hour.

@TheGreatDuck o/

stay up all night? damn that seems tough.

@TheGreatDuck What's new since last time ?

Night? You call 7 AM night?

more theorems. some removed due to inaccuracy. some noted as definitions now.

10:15 PM

Well, at least clearly the first theorem is true (I don't remember what was there and what wasn't there)

yeah, i think i just added a couple to the bottom

I changed some notations due to people being picky here math.stackexchange.com/questions/1636835/…

@TheGreatDuck I don't get what your 'Integration Continuity Theorem' is about though. What's the link with step functions ?

nothing

just saying integral needs to be continuous

the link is that floor is /discontinuous/

The integral of discontinuous functions is continuous anyway

i know

Random Variable

10:18 PM

@DanielFischer And numerically it appears that if $C_{r^{-}}$ is the bottom half of the circle, $$\lim_{r\to 0} \int_{C_{r^{-}}} e^{iz} \sin \left(\frac{1}{z} \right) \, dz = i \pi I_{1}(2). $$

@TheGreatDuck Btw I was thinking, i said it was clearly true but... for the Jump Series Resolution Theorem, by constant you mean a function right ? Not an actual constant real number ? (the 'constant' simply being minus the integral of the function that links the parts together)

you mean "floor treated as a constant"

i mean that you integrate it as if it werre a constant

So basically you're just splitting the integral. Ok.

yeah i guess. :p

wait no

i mean that the integral of floor is x*floor

like how the integral of c is x*c

Steven Stewart-Gallus

What symbol do people usually use to denote the axis of rotation?

10:21 PM

you draw a curved arrow around that aXIS AT THE ARROW

Steven Stewart-Gallus

@TheGreatDuck but what letter would you use to denote the unit vector?

you mean like integrations of rotated areas correct?

@StevenStewart-Gallus $\omega$ or $\Omega$

Steven Stewart-Gallus

@TheGreatDuck or for finding the moment of inertia.

umm..youre not referring to integrated calculus and analytical geometry are you?

10:23 PM

@StevenStewart-Gallus The moment of inertia would be $J$

Steven Stewart-Gallus

@Hippalectryon like with quarternions and I and J and K vectors?

wonders if TheGreatDuck is à l'orange ...

@StevenStewart-Gallus hmm.. maybe ? I'm not sure what the link is ... You don't need 4 dimensions

wat...

Ugh, I can't stay awake. Need tea.

10:25 PM

COFFEE

Allergic to coffee.

then YER DOOMED

Better just to go to sleep, Balarka.

@TedShifrin Everyone's celebrating the end of their exams with new racing cycles and video games (and girls...). But, innovative and creative as I am, I am celebrating mine by promptly fixing my biological clock back to normal.

rolls 6 of 8 eyes

10:28 PM

My natural habitat is nocturnal environment.

@TheGreatDuck That can't be generalized though. The integral of $e^{E(x)}$ isn't $xe^{E(x)}$

Duh, I might just go to sleep. I'll try fixing the clock step-by-step.

i know

it only applies to the implied integral of floor

@TheGreatDuck But you do have $\int_0^x f(t) E(g(t))dt=\sum_{n=E(g(0))+1}^{E(g(x))}n\int_{f(g^{-1}(n))}^{f(g^{-1}(n+1))} f(t)dt$

Hey @BalarkaSen

@TedShifrin hi

10:33 PM

hi Karim

finally understood some stuff I needed about Delta complexes

user image

user image

I don't understand why 3 gives that as consequences

but this is just

by the following

$X$ a $\Delta$-complex with simplices $(\sigma_\alpha\colon \Delta^{n_\alpha} \to X)_{\alpha}$, then
\begin{equation*}
\left(\coprod_\alpha \Delta^{n_\alpha}\right)\big/{\sim} \cong X
\end{equation*}
where $\Delta^{n_\alpha} \ni x \sim y \in \Delta^{n_\beta}$ iff $\sigma_\alpha(x)=\sigma_\beta(y)$.

The bijection is given by $\Delta^{n_\alpha} \ni [x] \mapsto \sigma_\alpha(x)$ and (iii) tells you this is a homeomorphism.

you lost me there

im just saying that the implied integral treats floor as a constant

Sorry, Karim, I don't have time to do this now. Ask Balarka tomorrow.

@TheGreatDuck It's just a direct consequence of my last email (well it might be $(n-1)$ instead of $n$ but whatever)

oke

10:36 PM

@TheGreatDuck But have you proved that ? It does't seem clear at all for functions that aren't nice

well...

the only areas that dont have a derivative of 0 are the jumping points

hence the only thing to adjust for by implied integration would be to remove the jumpsz

and it's also a question. So, there is the whole possibility that someone might prove me wrong. :p

but im confident

@TheGreatDuck Ok, before going any further, remind me what your problem is exactly. You want to find a useful way to compute the integral of floor-based functions ?

that was the original thing

Your question (the MSE one) isn't really clear at all (hence the downvotes), you need to formalize it mathematically.

nowe im just trying to determine jump series rules and things

i dont care about the downvotes

10:39 PM

@TheGreatDuck Ok, so right now what's your actual question ?

If I use the implied integral on a function like f(x+ [x]) or just a composed function of [x], then how will the continuity change pre-integration to post-integration

"how will the continuity change pre-integration to post-integration" I don't have a clue what that's supposed to even mean

I probably am just asking something far too broad.

More like, far too unclear :(

I'll likely delete the question.

10:43 PM

The question is fine, if you write it so as to get understood :-)

"how will the continuity change pre-integration to post-integration" compare the continuity of the original function to the continuity of the integrated function

But the integrated function is always continuous .... ? How do you "compare" continuities except by saying "it is/it isn't contiuous" ?

that's why i am saying the implied integral

I realize though that "composite function" is so gigantic and vague that you cant answer it that simply

And I don't know what you mean by "defined integral". Remember, you might have thought about it for hours, but I haven't ;_; I don't know what your vocabulary refers to unless you've defined it for me

i understand

that theorem page has everything

:p

like i said, I think it's just too broad. I'll likely either delete it or just leave it.

10:47 PM

But the theorem page itself isn't clear :(

oh

ill look into it

It's not necessarily too broad though. And even if it was, you can always limit yourself to polynomials, and then use series to extend it to (nearly) any continuous function)

i was thinking of writing a much bigger document. A lot of the stuff /is/ pretty incendiary/confusing. Thanks for letting me know, though.

@TheGreatDuck As I suggested last time, you could write an actual article-like document in which you explain everything; it wouldn't be suitable for MSE questions anymore, but I could help :-)

For instance "The integral of a floor-based function is floor treated as a constant minus the appropriate jump series of the integral" is extremely unclear, yet everything you have worked on relies on this.

you can proofread. No, it's more the implied integral one that everything relies upon. It's more about removing jumps now then just integrating.

10:51 PM

Removing jumps have been solved already though. That's easy when $g$ is monotonous (I'm still considering $h(x)=f(x) E(g(x))$, since its already a pretty general case)

well sure. but im thinking more in the broader sense with general rules.

like how integrals have different rules and formulas

but that is one thing

there's a lot i can see with composed functions

but i gotta go

cya

But, as you can see from what you just said, you clearly have an idea in mind, but you need to put it on paper and make it clear for everyone :P

Anyway, you have my email, so contact me anytime

If a subgroup is non-abelian its group will also be non-abelian?

11:12 PM

any subgroup of abelian group must be abelian

if the subgroup is non abelian however, will the group then also be non abelian?

yes

what is the definition of non-abelian

is that there exists $a,b \in G$ such that $a + b \neq b + a$

commutation doesn't hold

what quantifier you use ?

you use for all quantifier

for all elements contained in the group

11:20 PM

yeah

Oh ok got it. thank you

so since you know this particular subgroup has elements which don't commute so its by definition your group is non-abelian

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