Mathematics

Huy

11:18 AM

what's the simplest way to achieve this in LaTeX?

Tobias Kildetoft

@Huy Hmm, possibly some sort of array

or possibly stackrel, though that is probably not the best option

user 1618033

Short announcement (for the record): very close to an amazing result in mathematics. These day should happen if I'm lucky enough.

Tobias Kildetoft

You always seem to be

user 1618033

Yeah, it's just natural to be so after working a huge lot.

Huy

@TobiasKildetoft: good idea, the Bmatrix environment seems to work pretty good, just gotta work on the spacing

AlpArslan

11:32 AM

Does anybody happen to know what the cofinite dual of an algebra is? I can't seem to find a definition anywhere.

Tobias Kildetoft

@AlpArslan In what context? I can think of what it might mean with some extra structure (like a grading or something like that)

AlpArslan

Well, I'm meant to compute the cofinite dual of the Tensor Algebra on a 2D vector space modulo some relations.

Tobias Kildetoft

Ok, so that is a graded algebra with finite-dimensional homogeneous components

so the natural thing to do is to take the dual of each graded component

user 1618033

@TobiasKildetoft However just to know that I'm at the very beginning of mathematical activity, self-educated, I have hundreds, thousands of new results in mathematics.

I think this is a promising activity.

I have to finish some stuff now.

BBL

Tobias Kildetoft

@AlpArslan What are you meant to do with this dual?

AlpArslan

11:45 AM

Taking the dual of each graded component just gives me the standard dual, doesn't it?

Well, all in all, Im trying to compute a universal measuring coalgebra.

That measures \mathbb{Q}(\zeta) to itself as \mathbb{Q}-vector spaces, where \zeta is a primitive 3rd root of unity.

Tobias Kildetoft

@AlpArslan No, taking sum of duals of each graded components preserves the dimension as being countable. If you just took the dual, you would get something of uncountable dimension

I think you mean mapping, not measuring

AlpArslan

Ach, no, you're right.

And, no...?

Hmm.

@Tob

@TobiasKildetoft * what exactly do you mean by mapping coalgebra?

Tobias Kildetoft

not sure, but it seemed like it made more sense that measure, especially given the following line where it should definitely be maps, not measures

Leaky Nun

@AlpArslan wrap your latex expressions with dollar signs ^^

Kari

12:07 PM

@Huy Maybe two align environments with \left\{ ... \right\} to encase the two?

Leaky Nun

$\displaystyle\left\{\begin{array}{c}\text{Free homotopy classes of}\\\text{(unbased) maps }S\to S\end{array}\right\}\longleftrightarrow\left\{\begin{array}{c}\text{Conjugacy classes of homo-}\\\text{morphisms }\pi_1(S)\to\pi_1(S)\end{array}\right\}$

@Huy

Huy

12:25 PM

thanks for your suggestions

Mats Granvik

2:30 PM

Let $\mu(n)$ be the Möbius function, then:

$$a(n) = \sum\limits_{d|n} d \cdot \mu(d)$$

$$T(n,k)=a(GCD(n,k))$$

$$T = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^z} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{s \rightarrow z} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n^c} = \frac{\zeta(z) \cdot \zeta(c)}{\zeta(c + z - 1)}$$

which is part of the limit:

$$\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\zeta (c)-\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}\right)$$

Integrate:

$$\int \frac{\zeta '(s)}{\zeta (s)} \, \mathrm{d}s =\log \zeta(s)$$

$$\exp (\log \zeta(s))$$
$$=\zeta(s)$$

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$Mathematics$ Mathematics