@anon No, I was looking at the review of group actions at the beginning of a book and realized that you can make a groupoid similarly to how you construct the fundamental groupoid of a topological space. I am wondering if there is any representation theory of groups via groupoids.
you may be interested in groupoid cardinality (as it relates to combinatorics - generating functions, species, stuff types etc.) or as a more category-theoretic approach to quantum theory through groupoidification
let $D$ be the derivative operator. then $e^{hD}f(x)=f(x+h)$ by Taylor's formula, so $e^{hD}-1$ is the discrete forward-difference operator $\Delta_hf(x):=f(x+h)-f(x)$. in particular, we have the "discretized derivative" $\frac{e^{hD}-1}{h}f(x)=\frac{f(x+h)-f(x)}{h}$. Thus $x/(e^x-1)$ at "$x=hD$" is essentially taking the derivative after first inverting the process of a discretized derivative
me? Finding math interesting I guess. Additionally, There are often questions that I think I can learn from other users about general strategy. That's all I think... By the way: this room isn't about math?
@AntonioVargas I don't know the theory of infinite-dimensional vector space. For example, we see that $D$ is a linear operator on $V=\mathcal C^\infty$. Is $e^D$ well-defined on it?
@FrankScience I haven't thought about that kind of thing for a while, but I would guess that you'd need to restrict your space to analytic functions for $e^D$ to operate on it.
$\displaystyle\left(e^{hD}f\right)(x)=\sum_{n=0}^\infty h^n\frac{f^{(n)}(x)}{n!}$ is the Taylor expansion of $f(x+h)$ around $h=0$ when $f$ is analytic at $x$
the partial sums of the formal series expansion converge pointwise to the shift operator on the space of analytic functions, I am not sure if they converge any more "tightly" than that
We can assume that $R$ is a UFD if we cannot answer the general answer immediately.
@anon I conceive an ugly proof. Suppose $\phi(t)=a_0+a_1t+\dotsb+a_mt^m$, $\psi(t)=b_0+b_1t+\dotsb+b_nt^n$, where $a_j,b_k$ are indeterminate, then $f(t)$ should be a polynomial of variables $a_j,b_k,t$ over $\mathbb Z$. Since when $a_j,b_k\in\mathbb C$, it's zero, therefore it's zero polynomial. Right?
Well, step-by-step. First, let $\phi(t)=a_0+a_1t+\dotsb+a_mt^m$ and $\psi(t)=b_0+b_1t+\dotsb+b_nt^n$, where $\phi,\psi\in R'[t]$ where $R'=\mathbb Z[\{a_j\},\{b_k\}]$.
I think this question was mistakenly closed. math.stackexchange.com/q/453808/8581. The OP asked his/her question properly, however others didn't think that. Thanks.
@BabakS. I had a feeling that was the case. I couldn't understand the question but my sense was that fewer people would understand the nature of the question than there were people voting on it.
If you provide an explanation of the question in the comments for others I'd cast my reopen vote.
ultimately yes, functions can take any sort of value. f() could take values that are numbers, that are fruits, that are sequences or functions or teletubbies or sets or whatever
if a random person you've never met before tells you they're thinking of a number 1-1000, there's no way you can just completely guess it with no information
however, if you know something extra about this function, for example that it's on an IQ test or a human person invented it, then you can guess that it follows a pattern
@anon He puts an ordering on the tableau and then say suppose $T$ has strictly increasing columns but is not a tableau. Then we may suppose that the $k$-th entry of the $j$-th column is strictly larger than the $k$ - entry of the $j+1$ th column (I get this). But then he says we have a relation $e_T = \sum e_S$, the sum over all $S$ obtained from $T$ such that...
The thing is I don't think we can use the relations obtained from $Q$ to do this. For example consider the young diagram $(3,3)$ with tableau (from left to right, top to bottom) $1,2,3,5,4,6$
@Nick why wouldn't it? the notation means "write the numbers from 1 to x down in a tuple"; just because you see 1, 2, and 3 in the notation as supposedly distinct from x doesn't mean x doesn't necessarily take the values of 1, 2, or 3
Suppose $f,g\in\mathbb C[x,y]$ have no common factor. How can we show that $R=\mathbb C[x,y]/(f,g)$ is a finite-dimensional vector space over $\mathbb C$? It seems that the problem is related to the finitude of the intersections of $f$ and $g$.
You can actually avoid such terms for most of the proof, but we need the result that $V(I)$ is finite if and only if $k[X_1,\dots,X_n]/I$ is a finite dimensional $k$-vector space.
@FrankScience In the exercise, only one direction is needed, $|V(I)|=n<\infty$ implies $\dim k[X_1,\dots,X_n]/I$ is finite. Do you know the Nullstellensatz?
Does Hilbert's nullstellensatz helps? It says that the maximal ideals $\mathbb C[x,y]$ corresponds to the points in $\mathbb C^2$, the rule is: $M_a\mapsto a$, where $M_a$ is the kernel of the substitution.
Every coset of $(x^2+y^2-1)$ has a representative of the form $A+Bx$, with $A$ and $B$ polynomials in $\mathbb R[y]$. I think you can use this to show that $x$ is irreducible.
Hey, quick question. Given any non-empty finite set $S$ and partial order relation $R$ on $S$, must $S$ always contain a maximal element? Is it the case that the only way for for $S$ to not have a maximal element is if $S$ is infinite (for example, when $S=\Bbb{Z}$ and $R=~>$)?
@PeterTamaroff I guess it doesn't have to be unique. Although I had been thinking about the problem of constructing a poset that contains a unique maximal element but no greatest element.
