Proving that it is in the dual of $K^{n\times n}$ for each $A$, that if $f_A(X)=0$ for each $X$ then $A=0$ and finally that $A\mapsto f_A$ is an iso $K^{n\times n}\to (K^{n\times n})^*$. Thought you might have something interesting to say about it.
every map $K^{n\times n}\to K$ will be of the form $[x_{ij}]\mapsto \sum_{ij}c_{ij}x_{ij}$. the map $f_A$ is of the form $[x_{ij}]\mapsto\sum_{ij}a_{ij}x_{ji}$...
The first point shows it is a linear transformation. The second shows it is injective. But whenever we have a transformation in equal dimensions $$\text{ injective }\iff\text{ surjective }\iff \text{ bijective }$$
@PeterTamaroff Okay, assume $\ker f^n=\ker f^{n+1}$ and let $x\in\ker f\cap f^n(V)$ (yes I generalized). Write $x=f^n(z)$. Then $0=f(x)=f^{n+1}(z)$, so $x=f^n(z)=0$.
The assertion $\ker g=\ker fg$ implies $\ker f\cap g(V)=0$ holds using the same proof. And if we also have $fg=gf$, then $f^2g(x)=0$ implies $fg(x)\in\ker f\cap g(V)=0$.
This shows that $\ker g=\ker fg$ implies $\ker fg=\ker f^2g$ when ever $f$ and $g$ commute like what I tried at the beginning.
Why is nobodoy just showing that the integral is strictly positive and decreasing for x>1 and by simple computation the answer is 4? math.stackexchange.com/questions/459828/…
the indefinite integral is strictly positive and increasing as the interval integrated over increases. the integrand is strictly positive and decreasing, which is obvious and says nothing about convergence.
robjohn mentioned that integration by parts yields the actual value
someone should perhaps point out the relevance of the gamma function in all this
one factor of 2 comes from the substitution, another factor of 2 comes from the interval being the whole line instead of just the positive half of the real line
This semester I am barely taking ny mathematics, It will be somewhat strange..
Only statistics, and some quantum mechanics. Although the latter is more or less pure mathematics. Combination of function analysis and algebra..
Any idea how to show $(1)$ from $(2)-(4)$ + Abels ? http://math.stackexchange.com/questions/459735/dilogarithm-inversionformula-textli-2z-textli-21-z-zeta2 I have already showed it with using the integral definition and derivation, but wanted to see if there were a purely algebraic approach =)
@anon Fine =) I was looking here arxiv.org/pdf/hep--th/9408113.pdf and it seems they have proven it on page 7. The problem is they use the L(x) = dilog(x) + log x log (1 - x) function instead. Following the same steps gives me a bit of mess thugh
And I am not able to come up with the expression between $(1.7)$ and $(1.8)$
one might also argue that C[C4] is iso to C^4 via factoring z^4-1 into four linear factors, then observe C[C2 x C2] has four LI idempotents (1, x, y, xy).
Hmm, it seems strange because the usual example given of isomorphic group algebras with non-isomorphic groups are for the non-abelian ones of order $8$
and it seems strange that the two rings should be isomorphic as the groups have very different characters
Does having isomorphic group rings not imply having the same character tables?
Or is that the other way, and we need the group algebras to be isomorphic as Hopf algebras to get identical character tables?
Hmm, no, since the representations only depend on the group ring, not the comultiplication, we should get the same characters if the rings are isomorphic. But those two groups have very different character tables, since one has only real entries and the other does not
@anon yea but now I want to understand why the Young symmetrizers form a complete set of idempotents without saying that the number of partitions is the number of isomorphism classes of irreps, etc
@TobiasKildetoft I'm not familiar with the minutia of information-recovery between algebras, tables and groups. might it be the scalars used?
actually, I'm not familiar with the overarching facts, let alone the details
Since ${\bf C}[A]\cong {\bf C}^{\oplus n}\cong{\bf C}[B]$ for any abelian groups $A$ and $B$ of order $n$, finite abelian group algebras over complex scalars shouldn't be expected to provide much information
@TobiasKildetoft I am not sure what you mean by "go back to getting"; C[G] is C-algebra isomorphic to a direct sum of matrix algebras End_C(V) over irreps V
ahh, and the whole question of characters requires a comultiplication, since we need conjugation to make sense (conjugation by the group-like elements that is)
so the example with the non-abelian groups of order $8$ is just usually chosen because it is the smallest example of non-abelian groups with isomorphic group rings
@PeterTamaroff I'm still trying to understand what you were saying about my last point. Put aside the uselessness of what I am trying to do. I understand that you cannot integrate using elementary integration methods is that what you were trying to say?
@PeterTamaroff hmm... I see. So if I were to carry on integrating the function It wouldn't be considered a combination of elementary functions because of the fact that it would contain infinite sums?
@Alizter I wouldn't tackle it. I don't know Differential Galois Theory. =)
user61230
So here's a question that came up recently in RPG.SE chat:
user61230
In the D&D 4E world, figures move in squares, including along diagonals. This is a change from D&D 3.5e, which measured diagonal distances as longer than linear distances.
user61230
7:41 PM
The theory goes that if a unit walks around a 5x5 square, then they're actually walking in a circle.
user61230
However, this results in a weird problem: one can have a 45-45-90 right triangle whose sides are the same length. So, we've begun theorizing about the nature of these conditions. We came across the Chebyshev distance, which is a close model. But, to truly understand what happens in this world, we need a formula for something fundamental: The area of a triangle.
There is a metric on the real plane which calculates the distance between $(x_1,y_1)$ and $(x_2,y_2)$ as $\max(|x_1-x_2|,|y_1-y_2|)$. This is sort of the limit of your distance metric with smaller and smaller squares. Odd metrics like this abound a lot. There really isn't a way to treat this intuitively as a geometry because there can be multiple shortest paths between two points, for example. So it violates a basic property of geometry.
user61230
Under the premise that it's an entirely different geometry than the one we know, I don't see a reason why that would make a difference.
Guys I wanted to let you know that I found an easy way to type in Latex. If you go to Desmos Graphing Calculator desmos.com/calculator you can type in the equation much more easily, and when you copy and paste it it shows it in Latex
hello. I just wanted to know if I proved this correctly. I have to prove if p|lcm(a,b) then p|a or p|b. So we assume p|lcm(a,b). But since ab is a multiple of a and b we have p|ab. Hence by euclids lemma we have p|a or p|b
