oh, here's a (probably simple) complex analysis question i've had in my head since yesterday
if i consider the function $f(x)=e^{1/x}$, which is analytic everywhere except at zero, then by inspection one finds that $f(x)\to 0$ as $x\to 0^-$ and $f(x)\to +\infty$ as $x\to 0^+$.
now, in that example, at zero there's an essential singularity. what i'm wondering is whether that's a necessary condition given those limits
i.e. can i cook up an example with the same limiting behavior but with a non-essential singularity
the person doesn't quote the paper further, but the next few lines of it indicate that the analogue of the (linear) hypergeometric equation in that context is the (nonlinear) sixth Painleve equation
which would be neat, if i actually understood it :P
mostly what's confusing me is that in the hypergeometric case, one can take the periods of your elliptic curve as the 'objects' of the differential equation
it's not at all obvious to me what kind of interpretation one should take in the nonlinear case
i think the elliptic curve-hypergeometric equation thing comes from the fact that the inverse function of the $j$ invariant can be expressed as a rational function of hypergeometric functions.
i'm probably just asking the wrong question, of course
that's probably true, but i'm not familiar with $j$-invariant stuff
the way i know it is through picard-fuchs. namely, that there should be some second-order differential operator which annihilates periods of the riemann surface
nod. though the statement i know re: the cohomology of an elliptic curve is just
"hey, an elliptic curve is topologically a torus, so its homology is $\mathbb{Z}\times\mathbb{Z}$. since homology and cohomology are dual via de Rham's theorem, the cohomology of the torus must also be two-dimensional."
which includes a whole bunch of handwaving, i'm sure
but, if the cohomology is two-d, then any three one-forms on the riemann surface should be linearly dependent up to an exact form. so if i take a one-form and its two derivatives with respect to a parameter, there should be some linear combination which is equal to an exact form $df$.
then just integrate that around some closed loop to get a period of the riemann surface, and notice that any integral of an exact form over a closed cycle is zero
so, bam, the period must satisfy some second-order ODE
alas, i really don't have a clue how takes that kind of story and moves it to a nonabelian context
I don't know what Helly's theorem is, @Balarka. Do you mind stating the exercises you're referring to when you talk about them, by the way? It's not a big deal for me to go to the document, but just a little annoying.
@MikeMiller It says that if you have $n$ convex subsets of $\Bbb R^n$ for $n > d$ such that every $d+1$ of the sets has nonempty intersection, then the intersection of all of the $n$ sets is nonempty.
@BalarkaSen The results above come from that integral. Well, I need to talk to some mathematicians that study the prime numbers first, and see how important it is. I also investigate it, I'm just at the beginning with that. Then I'll post it for sure.
I'm currently working towards setting up a meeting with my academic adviser for next week regarding my independent math work, I'm hoping that they will be able to set up a meeting with a math professor to check my work
I won't keep my hopes too high, becoming a research assistant is the best-case scenario. The most likely scenario is getting the attention of the department head, and skipping a few courses that I don't need to take
@Chris'ssis: the thing i worry about is that, while you can write each integral as a sum of prime logs, it doesn't immediately follow that that's the 'right' way to understand it. you can rearrange the coefficients to get it into a sum of logs of all numbers up to $n+1$, not just the primes
i.e. you can write $\log(2)+\log(3)+\cdots +\log(6)$ as $4\log(2)+2\log 3+\log 5$. but that doesn't mean that the second expression is what's remarkable
@Semiclassical Yeah, I got your point, and I also agree with that, but after more terms, there is no skip of the primes. I mean one has a sum of 100 log of primes in their perfect order. This is strange to me.
From what I can gather, she's sending it to a mathematician to get a professional opinion first. Then maybe she'll share the integral, but it's not likely
eh, that's not terribly shocking to me. if I take some product/ratio of the first 100 numbers, then all the primes in that set will definitely show up in the factorization
@Semiclassical I mean you don't have $r_{i+1}\log(p_{i+1})+ r_{i+3} \log(p_{i+3})+r_{i+2} \log(p_{i+2})$ but always $r_{i+1}\log(p_{i+1})+ r_{i+2} \log(p_{i+2})+r_{i+3} \log(p_{i+3})$ for the first 100 primes. It cannot be a simple coincidence.
If $X$ is a set and $A, B \subset X$, is there a (compact) name for the complement $A - B$? 'Local complement' seems accurate but I've never heard it before.
Good afternoon folks, I have a question about a name for a concept. Suppose we have a topological space. We could endow this space with many different metrics. Is there a name for the collection of all the metrics which a space could be endowed with?
@Semiclassical I'm calculating now its closed form.
I just looked at it with the wrong eyes without assuming that sum can be taken over all values of a certain $x$ in a certain set of log. Well, it happens sometimes.
Find $f_m(x)$ such that $$\int_0^1 \frac{1}{\log(x)}+\frac{1}{\log^2(x)}+\cdots+\frac{1}{\log^m(x)}+f_m(x) \ dx$$ converges.
The second point would have been like that
Evaluate in closed form such an integral after fixing $f_m(x)$ that doesn't include the terms already present in the integrand multiplied with any real number.
How such a closed form would look like, right?
Well ...
Here is a solution ...
I'll also add that in my book I think.
Let's take, say, $m=1$, and we get a closed form for
Please be gentle as I do not have any degree in maths.
By using a compass/straighedge method to construct Metatron's cube, a regular dodecahedron can be inferred from intersecting points. I'm looking for the ratio between the lengths of the edges (blue) of the dodecahedron and the radius of the ...
This is one of those solutions $$\int_0^1 \frac{1}{\log(x)}+\frac{1}{\log^2(x)}-\frac{x}{\log^2(x)} \ dx$$ One thing is to find them, but then one might like to also come up with closed forms. The structure of the integrand also allows solutions by differentiation under the integral sign. A more inspired one is based upon special functions.
How do we make it convergent? I wonder how many solutions we have. $$\int_0^1 \frac{1}{\log(x)}+\frac{1}{\log^2(x)}+\frac{1}{\log^3(x)}+\large{?} \ dx$$
@Committingtoachallenge Ok, so here is what I found out. We get the induced isomorphism by taking the derivative of the homomorphism at the identity, just like when we get the algebra from the group.
@Committingtoachallenge Like if a group homomorphism $\lambda:G\rightarrow G'$ acts as $\lambda(G)G'=GG'G^\dagger$ then the induced homomorphism $\lambda_*:\mathfrak{g}\rightarrow \mathfrak{g'}$ acts as $\lambda_*(g)g'=gg'+g'g^\dagger$.
I'm only able to perform it visually (i'm only doing recreational math), and i find interesting points where polyhedra meet under scaling/rotation (of interesting numbers like 55 degrees, -0.078 degrees and 1.9485 as a scale for instance)
this is a star tetrahedron rotated 1.9496 degrees over the x axis (red) and a star tetrahedron scaled at $/frac{1}{/sqrt2}/cdot2$
I think if you come up with two particular metrics, it's interesting to ask if they are comparable ... But really once one has one metric, one's happy.
@Ted Ah I see. I am asking because in political science it is common to represent individuals as some point or distribution in a high dimensional Euclidian space, where each dimension represents some policy position, say spending on the Iraq war, etc.
@Ted and then the model is that candidates are also a point or distribution in that space and voters will vote for the candidate "closest" to where they are
I agree. So, usually political scientists makes some VERY strong assumptions about the metric, basically that everyone's is the same and that preferences are equally important and totally seperable
So, for example, suppose we have a set of voters $v_i$ and a set of candidates $c_i$, I guess its possible to make any of the $c_i$ closest to any particular voter, $v_j$
Hmmm, I thought it went the other direction. We start with the same topology and then put different notions of 'nearness' onto it, but I guess I have it backward
Right, any time you don't care about candidate $j$, there's a vacuous notion of nearness for that coordinate. So it's the indiscrete topology in that coordinate.
Now that I've confused @Kevin, I'm going to go cook dinner.
So is it in principle possible to have the same topology, but 2 different metrics. My understanding was that when you specify the open sets, you're giving some notion of which points are 'near' others. But then the metric comes along and says precisely how far everyone is form everyone else
Let $p \in M$ be a a point in a manifold and let $\varphi^X_t$ and $\varphi^Y_t$ be the local flows of the vector fields $X$ and $Y$ respectfully. Define the commutator of flows: $\alpha(t)= \varphi^Y_{-t} \varphi^X_{-t}\varphi^Y_t\varphi^X_t$. I'm trying to prove:
$$\left .\frac{d}{dt} \right|_...
Question, if you simplify an expression why is it not considered simplification if you divide by a common number ? such as: -68r + 32 becomes -34r + 16
@Ropstah Hi :) I always feel like recreational math. Except I'm about to have some pizza :D
@ThomasAndrews Probably better than "How the $f_{uc}(k)$ should I know?" Probably best just not to comment I would imagine. "Rude" is obviously a very subjective term.
Nah, that's too obvious. I would never use the full world, but my initial inclination was the say "How the eff would I know," then realized where I was. @induktio
Given it was related to Enigma, I might have written $\phi_{u}(c_k)$, since $\phi_e$ is often the notation for the function defined by the $e$th Turing machine.
Chat is fine for discussion, but very few user use chat, so you would have to find someone who is interested in what you are interested in at the moment and somehow get the to chat
@Ropstah If it is computational, like computer graphics geometry, gaming and others might have some knowledge. Otherwise, this is the best place, I think.
I'm not really sure where I'm going with this, but i'm combining all sorts of polyhedra with stellations, the points where you find a sort of 'equilibrium' and other intersection points are all definitions for mathematical proofs
My final goal is to perform a two dimensional penrose tiling over the faces of a dodecahedron (five fold regular polygon), i think this relates to the prime number sequence
I can 'plot' (render) it, but it involves a lot of calculations and steps and I don't want to make mistakes
