Mathematics

12:38 AM

I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$

$G=G^0$ and I want to prove that $G^n =\{1\}$ at some $n$

I have played around, and the inverse is long and doesn't seem to have an obvious simplification(from being an ugly set of sums and products produced by the augmented identity method)

I have done some examples on 4x4 matrices with mathematica and I can see it works

We introduce a $0$ superdiagonal when we obtain $G^1$

Perhaps a proof could restrict attention to removing the superdiagonal and then the diagonal above that, while treating the remaining upper triangular elements as irrelevant

Stan Shunpike

1:29 AM

So @TedShifrin why is Rudin going to be challenging for me? I mean it is. I am not understanding the implicit function theorem. But I can read John Lee's books on manifolds so I don't get why I am having more trouble with Rudin. Rudin has some weird notations tho, or at least ones I haven't seen

Kevin

1:44 AM

Hello! What is the sum of $1^k + 2^k + \ldots + n^k$, assuming $k$ is a positive integer?

Mike Miller

@Kevin en.wikipedia.org/wiki/Faulhaber%27s_formula

Kevin

1:58 AM

Thanks!

r9m

2:25 AM

@Chris'ssis which problem was that? (I have totally forgotten ... sorry :( ... )

Mike Miller

3:32 AM

@robjohn I saw you on campus this morning :)

Incurrence

@MikeMiller Have you talked to Terry Tao?

robjohn

@MikeMiller Did you? Where was I at the time?

Mike Miller

@Incurrence Sure, he's human like the rest of us :)

@robjohn You were talking with a colleague, on the sidewalk between Powell and the physics building

I would have said hello if you were alone

Not the physics building - the building that's after IPAM, coming from the math dept; I was going the opposite way

Incurrence

@robjohn Oh I just wanted to ask how well he socialises out of curiosity

robjohn

@MikeMiller The new "upside down basket" building?

Mike Miller

3:42 AM

I don't remember...

Incurrence

Oh that was meant to be directed at Mike whoops

@robjohn Do you think you could give me a hint on my problem above?

3 hours ago, by Incurrence

11 hours ago, by Incurrence

I have a group $G$ defined inductively $G^0$ is the group of upper triangular matrices, with $1$'s on the diagonal, and $G^1 = [G^0,G^0], G^{i+1}=[G^0,G^i]$

$G=G^0$

I can see that when I do each iteration of $G^i$ I lose the next diagonal up

robjohn

@MikeMiller I was just having some chai, then went back to proctor an exam.

Incurrence

But that is using mathematica for computing the steps

@MikeMiller I chose tensor products of finite groups lol

@MikeMiller I'll have 10days ish from Monday to learn about them enough for a 5 minute presentation

Kaj Hansen

6:19 AM

@robjohn, thanks again for the hint last night. It proved to be just what I needed to finish the problem I was working on.

Chris's sis

7:25 AM

Greetings

@r9m it's about this one $$\int_0^{\infty} \frac{\cos(x)}{x} \left(\int_0^x \frac{\sin(t)}{t} \ dt \right)^2 \ dx$$ and the cubic sister. :-)

Remember me

Its given that the $span(a_1,a_2.......,a_n)=V$ where V is a vector space so ${a_1,a_2,a_3........,a_n}$ is a basis for V , so since the basis is finite,V is finite dimensional right?

Chris's sis

@r9m There were months when I didn't know how to do it elementarily. Finally my research opened amazing simple doors to crack them all. :-) Indeed, without having at hand a proper tool, that is proving to be horrible. I initially did that on 6 pages or so, and using CAS.

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$Mathematics$ Mathematics