@robjohn Do you know wether the $\zeta(4)$ accelerated series mencioned on section 7 of Alf van der poorten's paper A proof that Euler missed ..., transcribed at ega-math.narod.ru/Apery1.htm, can be obtain by similar techniques of those used in the same paper to derive the corresponding accelerated series for $\zeta (3),\zeta(2)$?
... There's a 1981 paper by Henri Cohen availble at numdam, where he used the Euler-Maclaurin summation formula to prove the accelerated formula for $\zeta(4)$.
teach my class, teach two of a colleagues' classes because he'll be in texas, but also need to do a bit more prep before my meeting tomorrow with my adviser
Using the Marsden-Weinstien Mayer reduction we can get a symplectic form on CP^n from the diagonal action by $U(1)$ on $C^{n+1}$. The question is to find the function $f(\lambda) = Vol(M_\lambda)/Vol(M_1)$ Where $Vol(M_x)$ is the symplectic volume of the reduction at value $x$ (I can't get the latex to work here, so im sorry if it doesn't render properly)
@Akiva I see. it's not $a^n,b^m$ where there is two different orders for the order of $(a,b)$ it must be a common order and by definition it must be the least common
@Obliv Ya. For example, in $\Bbb Z_2\times\Bbb Z_3$ (switching to additive notation), the order of $(1,1)$ is $6$ because $6\cdot(1,1)=(1,1)+(1,1)+\dotsb+(1,1)=(0,0)$
@MikeMiller The latex works, thanks. I meant to say this from the Marden-Weinstien Mayer theorem we get a symplectic form $\omega_{red}$ on $M_x$. The volume of $M_x$ is then the integral of the volume form $\frac{\omega_{red}^n}{n!}$ (though the $n!$ doesn't matter here since we are interested in the fraction of volumes)
Im not sure how to get $\omega_{red}$, all i know is $i^*\omega = \pi^* \omega_{red}$. Where $i$ is the inclusion $\mu^-1(x) \to \mathbb{C}^{n+1}$ and $\pi$ the quotient map.
Hey folks – in what way is the inverse differentiation rule (df/dx = 1/(dx/df)) applicable to multivariable calculus? When looking at spherical coordinates, I get ∂r/∂x (r,θ,φ) = ∂x/∂r (= sinθcosφ), which really confuses me.
Any proof of the reduction theorem should ezplicitly construct the form. But the way you probably want to do it is to get a volume form $\omega^n/n! \wedge dt$ on the fibers $S^{2n+1}$, where $dt$ is a 1-form that eats the tangent vector of the action and spits out 1. This should be moderately computable. The integral of this...
...will be the same as the volume of the quotient. Then you just need to compare the volumes of these. I expect your answer is going to be "they differ by a factor of $\lambda$" or something like that.
If your function is analytic, why do you need any theorem to say that it has a power series expansion? It's definitional that there is. (Mike beat me to it).
Taylor's theorem identified what the power series is and gets some control over its partial expansions for smooth functions, which is weaker than analytic.
@MikeMiller Thanks, i will try to do that. Can you point to a theorem that would say why the integral of $\omega^n /n! \wedge dt$ would be the same as the volume of the quotient?
I have to prove that any finite group $G$ of even order contains an element of order $2$. Doesn't that require there to be an element that is its own inverse? If $G$ is of even order how am I supposed to deduce that such an element exists?
@semiclassical So if that always held then there would always be a subgroup of order 1 since any order of any group is always divisible by 1? Then I can deduce that if it's of order 1 the element must be its own inverse and so its order would be 2.
Saying let $t(G)$ be the set $\{g \in G~|~g \ne g^{-1}\}$ and then it's easy to show any nonidentity remaining elements of $G - t(G)$ must have order $2$. But, we're assuming $G$ to be even so $G - t(G)$ could technically result in an empty set.
hi. let $f(\alpha)=\sum_{\ell=1}^{\sqrt{K}} \ell^{\alpha-1}(\sqrt{K}-\ell)^2$. since $f(0)$ and $f(1)$ kind of have closed forms (the former involves the digamma function), I know that they are of (asymptotic) order $K \log K$ and $K^{3/2}$, respectively. what I would like to know is the exact (I can see bounds) order of $f$ for $0<\alpha<1$. are there some techniques for that?
I will write the full proof for $X^1_1$ in your notation (just to not get cray with indices). So we have that locally:
$$ X_1^1(p)=\sum_{i,j} C^i_j(p) dx^j_p\otimes \partial x_i^p$$
Assume the $C^i_j$ are smooth functions. Pick $X,\alpha$ smooth sections of $TM,T^*M$ respectively. Which locally ...
Hello, Does anyone here know how $\partial x_{i_r}^p(X_p^r)$ can be defied on my question above?
it was just to show that since the order is even and there existed an identity element who is self invertible, there had to exist another element that was also self invertible
Let $x$ be an element of finite order $n$ in $G$. Prove that if $n = 2k$ and $1\leq i < n$, then $x^{i} = x^{-i}$ if and only if $i = k$ <-- is the exact thing i'm trying to prove
it's basically saying if the order is even, $x^i$ must be self invertible
For the converse suppose x^{i} = x^{-i}. You will find x^{2i}=1 and the next step is the tricky bit. Hopefully you have seen a result that tells you if x^{a}=1 then order (x) divides a. You then can use that and the bound on i in the question to deduce i=1. Hope this helps.
Hey guys, I just have a small question about birational equivalences, and it would be much appreciated if somebody could point me in the right direction.
Let k = \mathbb{C}, be the ground field. So, if I have the set of nilpotent $2 \times 2$ matrices of order 2, then I can show that this has the structure of a variety $X = Z(x_1^2 - x_2x_3) \subset k^3$. I need to try and find a birational equivalence between this variety and k^2.
Does anybody have any idea what I should be thinking about?
This is off topic: How would you go about proving that an affine space $\Bbb{A}^m$ in $\Bbb{R}^n$ can be contained in a 'hypersphere' $\Bbb{S}^k, k\geq 2$, with $n-m\geq 2$?
Is that even true?
I think that the limiting case ($m=n-2$) is actually a 'hypercylinder', but these are all abstractions from the simple (which can be visualized) cases.
@Semiclassical I reached the same point as Ramanujan as soon as is shown that $$2\int_0^{\sqrt{2}-1} \frac{\arctan(x)}{x} \, dx=\int_0^1 \frac{\arctan(x)}{x \sqrt{x^2+1}} \, dx$$
the route i was trying to take goes like this: Suppose $A(s)=\sum_{n=1}^\infty a_n n^{-s}$ for some sequence $\{a_n\}$. (not worrying about convergence)
@Semiclassical yeah, but the point is that the closed form is related to a particular case of the inverse tangent integral which is not known to have a nice form in known constants.
These values are known for many years, and unknown in terms of known constants.
@AlpArslan You want to prove $xy = z^2$ is birational to $\Bbb A^2$, right?
I don't know of a very rigorous proof off the top of my head, but consider this: $xy = z^2$ is the cone inside $\Bbb A^3$. Take the $xy$-plane inside $\Bbb A^3$, let $p$ be the cone point which also lies on that plane (we have chosen an implicit coordinate system while we do this, but whatever).
Consider a line $\ell$ tangent to $p$, which lies on the cone. Now you can start "projecting off" from the line $\ell$ by choosing a "pencil of circles" on the cone hitting $\ell$ at a unique point each, and then throwing lines from each point on $\ell$ which stereographically projects each of those circles off.
This should definitely be a rational function, with the singular locus $\ell$, projective the cone to the $xy$ plane (which is the copy of $\Bbb A^2$). I think it should not be hard to see what the inverse map is also.
@user1618033 wellll, following the direction I was taking gives the value of said series as $$\frac{1}{64}\left[\zeta(2,\frac18)+\zeta(2,\frac38)-\zeta(2,\frac58)-\zeta(2,\frac78)\right]$$
where $\zeta(s,m)$ is the Hurwitz zeta function. which is probably valid but doesn't actually seem useful :/
to clarify what I am saying: replace $xy = z^2$ by the isomorphic variety $C : x^2 + y^2 = z^2$ just because things are easier to write down. then consider the line $\ell : (x, y, z) = (t, t, t)$. The map eats a point $P$ on the cone $C$, spits out what you get when you project $P$ stereographically off from the point in $\ell$ which hits the circle $x^2 + y^2 = a^2$ which $P$ belongs to. @AlpArslan
if $x$ is an element of finite order $n$ in $G$, use the division algorithm to show that any integral power of $x$ is equal to one of the elements in the set $\{1,x,x^2,...,x^{n-1}\}$
if you're doing calculus, it'll mean integration. if you're doing something involving discrete math instead, it may well mean integers. if you're not doing math, it probably just means 'important.'
I wrote a code that calculates Euler's Totient ($\phi(n)$) and I was wondering why is it that $\phi(958324857389475983274569823476558973245)$ calculates very fast, but $\phi(43649128734981236498122387)$ very slow and $\phi(985763284578345834)$ very fast again ?
Yes, those were generated by me smashing my keyboard
it's probably short enough for a comment, namely that said 2-by-2 determinant is equivalent to a 3-by-3 one (and then with the interpretation in terms of linear dependence)
@Semiclassical Thanks for the suggestion. Commented now, but without additional interpretation, so as not to change the spirit of Martin-Blas's answer.
Well, one can still dualize the prism operator to get it working in singular cohomology, so I don't see how that makes a difference. But most of the time I cannot correlate what happens with forms with what happens with cochains, so I suppose it's fine.
That's why I say it work, work, work ..., because if you held a class about calculus and a student asked you to prove the one below you wouldn't know ...
Honestly speaking, what is the use of calculus hours if not able to prove such an identity? Just to go through the math stuff and learn new concepts? Then, in front of the practice the issues comes in front.
I talk to myself only. Pretend you don't see me. I'm not sure if I'm right (100%), maybe my thinking here is flawed (to some extent).
What would I do though if I student asked me to solve that? I have to give him/her a cool answer, right? Such little questions make math beautiful and attractive for the students.
That's why I say it is important to know to answer them.
Tomorrow I'll give this question to all students I tutor (curious to see how many will give me the solution).
Hello all, I am trying to understand this section of the chapter on Polynomial Rings in (Fine, Gaglione, Rosenburger, *Introduction to Abstract Algebra*, (2014), pp.377-9) http://docdro.id/jy8IJWN <- pdf scan of page
I don't understand the use of $r$ 2/3rds of the way down
I'm having a problem with https://qchu.wordpress.com/2015/11/04/the-categorical-exponential-formula/. In particular, that $|X^n/S^n|=|X|^n/|S_n|$. Is it correct that -$X^n$ is the category with objects $x_1 \otimes \cdots \otimes x_n$ and morphisms $g_1 \otimes \cdots \otimes g_n$, where $g_i$ are morphisms in $X$ from $x_i$, and that -$X^n/S^n$ has objects and morphisms the objects of $X^n$ modulo action of $S^n$ and the morphisms of $X^n$ modulo $S^n$, respectively?
I am just doing an project when I stopped to look Facebook where I saw a stupid gif. In the gif appear 2x2 which the result is the same as 2+2, the question is if there are more numbers that the multiplication result is the same as the sum. Sorry for the stupid question
Hmm, I wonder if there isn't a bigger lesson about AG to be learned from this question, since any answer is in the zero set of a 2-variable polynomial, but there are so few integral solutions...
I have seen on a similar site where you can filter by your favorite tags. It's one thing to have questions with your favorite tags highlighted, and it's another, and much better, to only see questions with your favorite tags. As of now, you can filter to see one tag by clicking the name in your...
@Semiclassical But I tell you what. I would like to see classes where the teachers come in place in expose themselves to the worst problems and then say: I don't know! I have no idea! It is very important for a student to understand that we all can fail without extremely much work.
@Semiclassical Yeah, but we all are vulnerable in front of the problems. Student shouldn't think he is more stupid than his teacher that KNOWS EVERYTHING TO SOLVE (at least how it might seem in a class).
I have seen on a similar site where you can filter by your favorite tags. It's one thing to have questions with your favorite tags highlighted, and it's another, and much better, to only see questions with your favorite tags. As of now, you can filter to see one tag by clicking the name in your...
