Ah. Well, I've written all sorts of papers with McCrory and others on singularities. It's just such a technical area. I suppose one could try to read Arnol'd, too. shrug
Of course, in the context of Gauss maps, there's some built-in symmetry a general map won't have.
I have Arnol'd et al's two volume tome that I look at from time to time. Gay has some other perspective from what the usual singularity theorists thought about for decades though.
monoids $S_1,\dots,S_n$ in a unital commutative ring $A$ are comaximal when for any choice of $s_i\in S_i$, there exist $a_i\in A$ with $\sum a_is_i=1$. an $A$-module $M$ is locally cyclic if after localization at comaximal monoids $S_1,\dots,S_n$, $M$ is generated by a single element. is this definition dependent on the choice of monoids? i'm guessing no but i'm not sure how to prove it
The ethos of his work is Morse functions on 3-manifolds give a bisection of the manifold (Heegaard decomposition), and Morse 2-functions, valued in $\Bbb R^2$, should likewise give trisections on 4-manifolds.
I didn't understand the idea before I encountered the following example. Take $f : \Bbb{CP}^2 \to \Bbb C$, $f([z_0 : z_1 : z_2]) = (|z_1|^2, |z_2|^2)/|\mathbf{z}|^2$.
This takes the whole of $\Bbb{CP}^2$ to the right-triangle with vertices $(0,0)$, $(1, 0)$ and $(0, 1)$
Take the barycenter of the right triangle and drop perpendicular bisectors to each edge. You subdivide the triangle in three regions, let's call them $H_1, H_2, H_3$ containing $(0, 0), (1, 0)$ and $(0, 1)$ resp.
Then $\Bbb{CP}^2 = f^{-1} H_1 \cup f^{-1} H_2 \cup f^{-1} H_3$ is a trisection. In fact, a very familiar one, as it's just the usual handle decomposition $B^4 \cup B^2 \times B^2 \cup B^4$.
Incidentally, $f$ is the moment map of the $T^2$-action on $\Bbb{CP}^2$ by $(\theta, \phi) [z_0 : z_1 : z_2] = [z_0 : e^{i\theta} z_1 : e^{i\phi} z_2]$.
@LukasHeger: Suppose $X$ is loc. compact Hausdorff, and has a filtration $X_0 \subset X_1 \subset \cdots \subset X_{n-1} \subset X_n = X$. Do you think the collection of all the open sets in $X_i$ forms a site?
@copper.hat. I have a bunch of names, but I realized my issue was a lot of names has the same last letter, so all of them got hash to the same index :(
Is the power series expansion of a polynomial uniquely that polynomial? I haven't done The Method of Frobenius in a few months and I'm struggling to get solutions even to trivial examples that match the regular solution. For example, to $(x - \pi)y' = 0$, which should yield $y = C$, I'm getting $y = \sum_{n = 0}^\infty \frac{Cx^n}{n\pi^n}$. This can't be right, can it?
Soooo, I just came by to say that I (finally) understand Swan's theorem, and now a lot of results in differential geometry make sense conceptually to me, which I'm very happy about! Also, I hope you're doing well, Ted!
@TedShifrin Without dividing by $(x - \pi)$ to retain at least some non-triviality, I plugged in the guess $y = x^s \sum_{n = 0}^\infty a_n x^n$. After the calc/algebra/shifting indices, I got the indicial equation $-\pi s a_0 x^{s - 1} = 0$, yielding $s = 0$. Plugging this in I got the recurrence $a_{n + 1} = \frac{n}{\pi (n + 1)} a_n$, with solution $a_n = \frac{C}{n\pi^n}$, which I plugged into the guess.
Oh, I'm sorry, @user10478. I read the $\pi$ in the denominator as an $x$. My apologies. But your recurrence will tell you that $a_1=0$ and hence $a_2=\dots = 0$.
Demonstrate that quadratic probing will guarantee a successful addition, if the hash table is at most half full and its size is a prime number. So, for the following quadratic probing,
$$(k+j^2) \bmod{m}, m \in \{primes\}, k=2, j\ge0$$
"A closed form of $\pi(n)$ is as follows... y̵̮̘͖̹̖͇̺̩͔̬̰̲̤̲͙̎͠ơ̸̧̛̲̪̞̖̺̞̮̻͒̓̋̑͗͋̌̅͐͋̈̅͊͜ͅu̵̦͎̮̓̾̈́̕͠͝ẘ̵̧̰̦͉͇͙̣̳͉̞̼̹̖͖͛̐̾̏̀̉̂̈́̀ȉ̷̡͕͇͕̺̖̜͔̰͉̠̀́́̀͌͌̀͐̋̏͒͝s̴̢̺̱̐̄͛́̈̑̑̈͘͘h̵̛͕̝̺̲̳̱̓͆͋̈́̑͊͒̿͝ͅͅ"
Imagine you're a govt research firm and you spend millions hoping to uncover a secret to further progress humanity and after finally figuring out what was written, all you learn is that it says "loljk" in an ancient language that no one knows or understands.
@monoidaltransform do you have any conditions you implicitly want to impose beyond 'isomorphic to'? that is a finitely generated torsion free abelian group. it's determined by its rank.
More elements added to hash table will produce longer probing sequences as above equation shows. Probably, this is why it's recommended to keep hash table half full.
did you know Robert Langlands was one of the first ones to conjecture conformal invariance of 2D percolation, one of the cornerstone conjectures in probability still to date
> -vingt quand il est suivi d'un autre nombre: quatre-vingt-cinq ou quand il est empl. comme adj. numéral ordinal: page quatre-vingt; mais quatre-vingts pages.
@TedShifrin Take a topological quadrilateral in the plane and subdivide it into a very fine grid; let water run from one side to the other on this grid while randomly opening or closing any edge with probability $1/2$; there'll be some probability that water indeed percolates from one side to the other. There are two pairs of sides in a quadrilateral, so consider the minimum of the two corresponding probabilities; this is called the crossing probability.
The conjecture is that this has a limit as mesh size goes to 0 in the grid, and this is conformally invariant
i borrowed some results from CFT in some calculations i did, but
that's about it
mostly stuff like "according to CFT people, the leading form of this Taylor expansion should look like X and hey, our numerical calculations reproduce this"
informally, if $f$ is a scalar function, and $A = U D U^T$ where $U$ is orthogonal and $D$ diagonal, we define $f(A) = U f(D) U^T$, where $f$ is applied to the diagonal elements of $D$.
you can check that in the case of psd A that the above operation will give a matrix $S$ that satisfies $S^2 = A$.
the above is very informal, but in the case you are looking at it is easy to check.
If we have a hash table that is half full, we want to reshape it so that we increase the size to ensure that it can have more elements and still be half full. In this regards, how we can copy the elements from the previous hash table to the new one? I saw a suggedtion saying that by using the new hash function to compute the entries indices in the prev column :/
The prev hash table that is to be resized already has its elements in the table, so how we can rehash them again to the new hash table which should have same hash function but with larget size?
I solved this by proposing a for loop to iterate over all elements, get the keys and hash again, but this does not seem to be the same solution suggested above though
