Well, assuming we're talking about individual pixels, you have a discrete number of points to work with. There isn't an automatic way to translate it into an infinitely differentiable (presumably continuous) surface
Arbitrary ways, on the other hand, might be more possible
@MikeMiller I meant $(p)$ is the unique maximal ideal of $k[x]/(p^n)$, period. There are no other ideals with quotient field $k[x]/(p)$ because such an ideal would have to be maximal, and hence in turn would be $(p)$
I mean, you can do it with reduction mod 4 I think
but it's cooler if you write $(y + \sqrt{-5})(y - \sqrt{-5}) = (x)^3$ and show that the ideals on the LHS are coprime, so they are both equal to an ideal cubed, but the fact that $\Bbb Z[\sqrt{-5}]$ has class number $2$ means that they must be the cube of a principal ideal, so you get $y + \sqrt{-5} = \pm \alpha^3$ (unit group is $\pm 1$)
and then
do some shit from there
one reduces mod 4 to show those ideals are coprime though so that's probably the connection
Interesting, let me try to understand. Why can't $(y + \sqrt{-5})$ be cube of a non-principal ideal? Because class number $2$ implies square of a non-principal ideal is always principal in $\Bbb Z[\sqrt{-5}]$, so cube has to be non-principal as well?
so if you have a principal ideal that is the cube of an ideal then the ideal of which it is the cube must itself be a principal ideal by class number 2
hahaha
Ahhh I see what you said
I misread I think
So then from that you get $y + \sqrt{-5} = \pm\alpha^3$ for some $\alpha \in \Bbb Z[\sqrt{-5}]$
and $y - \sqrt{-5} = \pm \beta^3$
Ah nice, if you set $\alpha = a + b\sqrt{-5}$ then I think you can just compare coefficients and get that $a = b = 0$ which is a contradiction
@ÍgjøgnumMeg Here's what I had in mind earlier: $y^2 = x^3 - 1$ mod $4$ implies $y$ is even and $x \equiv 1 \pmod{4}$. So write it as $y^2 + 4 = x^3 - 1 = (x - 1)(x^2 + x + 1)$. But $x^2 + x + 1$ is $3 \pmod{4}$ so has to have a prime factor which is $3 \pmod{4}$ as well, which hits $y^2 + 4$, so that $-4$ is a quadratic residue mod $p$ - so $-1$ is a quadratic residue mod $p$.
Your proof is way cooler. I wonder if there's a concrete link between these passing to various integral extensions and doing factorization of ideals magic there and quadratic reciprocity
I have seen this happen before, you can do most of these problems in two ways
a,b = 1,0
for n in range(20):
res = "$(1+\sqrt2)^{%d} = %d+%d\sqrt2$"%(n,a,b)
if b%2 == 0: res += "; $1+2+...+%d = %d^2$"%((a-1)//2, b//2)
print(res)
a,b = a+2*b,a+b
There is a dimension N beyond which a sphere of radius 1000km is smaller than a cube of sidelength 0.001cm. When this first happens, what is the length of the cube's diagonal?
(Hypersphere, hypercube, "smaller" means smaller hypervolume)
Most of the volume of the cube $[-1, 1]^n$ can be equated to volume of an extremely thin annulus of radius $\sqrt{n/3}$ in $\Bbb R^n$.
The thickness going to $0$ as $n$ grows huge
One way to understand this is to set up iid random variables $X_1, \cdots, X_n \sim \text{Unif}([-1, 1])$. Then by law of large numbers, $(X_1^2 + \cdots + X_n^2)/n$ converges to $\Bbb E X_1^2 = 1/3$.
So with high probability a random point from $[-1, 1]^n$ will have norm in $(\sqrt{n/3} - \varepsilon, \sqrt{n/3} + \varepsilon)$
Probability going to $1$ as $n$ going to $\infty$, for any fixed $\varepsilon > 0$.
I was thinking about some probability a while back. It's very hot topic nowadays to understand expected Betti numbers of random simplicial complexes. Me, being a pea-brain, wanted to do the calculation for a random graph.
So take an Erdos-Reyni random graph $G(n, p)$ with $n$ vertices and edge probability $p$. I want to compute the expected $b_1$, or the expected number of cycles.
The expected number of $k$-cycles is clearly $\binom{n}{k} p^k$, which falls out if you associate an indicator variable $T_{ij}$ for every edge $(i, j)$ in $K_n$, and basically count $T = \sum T_{i_1 i_2} \cdots T_{i_k i_1}$, the sum running over all edges $(i_1, i_2), \cdots, (i_k, i_1)$ in a $k$-cycle in $K_n$.
$\Bbb E T$ counts the expected number of $k$-cycles, which is $\binom{n}{k} p^k$ by additivity of expectation.
So the expected number of cycles is $\sum_{k = 3}^n \binom{n}{k} p^k$.
There are two questions you can ask now: (1) What is the scaling limit? Say you let $p = \lambda/n$ and $n \to \infty$. Then the expected number of cycles become something like $e^\lambda - 1 - \lambda - \lambda^2/2$.
(2) What is the rate of convergence? How fast does expected number of cycles concentrate around $\sum_{k = 3}^n \binom{n}{k} p^k$, where $p$ is fixed, and you're letting $n$ grow? If you just use plain vanilla estimations I think you get an $O(1/n^2)$ error term, i.e., polynomial concentration around the central tendency.
But I believe you can do some trick like Chernoff bounds and get exponential concentration as well.
It's just all a bit hard because the indicator random variables corresponding to each $k$-cycle are not independent; they are "weakly" independent, so you can somehow make the proof of law of large number still work because the covariance terms are so small.
There is a dimension N beyond which a sphere of radius 1000km is smaller than a cube of sidelength 0.001cm. When this first happens, what is the length of the cube's diagonal?
Basically in the process of picking a point under the graph of $f$ from the points under the graph of $cg$ the distribution following $f$ is already being normalized
The numerical instability in solving that volume seemed to stabilise at the nonsensical n=-193 only if I pre divided the n by at least $10^80$, but clearly that is not enough
Is there any version of the structure theorem for modules which are not finitely generated?
It would have been nice to expect something like that for operators. Like I could look at the C star algebra associated with the operator and do something like the structure theorem
Fun fact: wolfram alpha give up very quickly with unable to parse, if if ask to solve for n expressions of the form $s \Gamma (n+1)^\frac{1}{n} < 1$ where s has a string length >2
It feels so dissatisfying we can only get bounds for that question when it is structurally speaking, not a complicated problem
Hey, does anyone know if the implicit function theorem (applied on a function F(x,y) that satisfies the assumptions) gives us the formula for f'(x), or just its value at (x_0,y_0)? Thanks in advance
So given picture I have to check whether divergence and curl is positive, negative or zero.
So I check vector field at some point ; for figure b I can see that vector field goes outward so divergence is positive but for figure a I am not able to predict. Also how to conclude about curl?
It will tell you the existence of such an $f$ (the implicit solution). From that the derivative at the point of interest is clear from composition rule.
That's the sort of thing you're trying to train yourself to be able to tell --- not something you're trying to learn how to ask someone else to tell you :)
my argument is. $F(z)=log(z-z_0)$ is the primitive of $f$ therefore integral around closed curves is zero therefore around $z_0$ is zero . contradiction
So given picture I have to check whether divergence and curl is positive, negative or zero.
So I check vector field at some point ; for figure b I can see that vector field goes outward so divergence is positive but for figure a I am not able to predict. Also how to conclude about curl?
well so far it's looking a little more clear, no thoughts? it's hard to stay on topic when on topic gets null response $$\mathcal H={\{h_k}\}_{k=1..N} \subset \mathbb N \land N \gt 1$$
Let $G$ be a topological group, and let $H$ be the closure of the set $\{1\}$. I am trying to prove that if $f : G \to Y$ is a continuous map, where $Y$ is a $T_1$-space (i.e., all singletons are closed), then $f(gh) = f(g)$ for every $g \in G$ and $h \in H$.
Here's what the authors state but I don't follow their reasoning: "For $y \in Y$, the singleton $\{y\}$ is closed, so $f^{-1}(\{y\})$, hence of the form $AH$ for some set $A \subset G$. This implies that $f(gh) = f(g)$ for every $g \in G$ and $h \in H$."
@LeakyNun There's an exercise in Dummit-Foote which is asking me to prove that if $K/F$ is a finite extension, then if $K$ is a splitting field of some polynomial over $F$, then for every irreducible $f \in F[x]$, $f$ has a root in $K$ implies it splits completely in $K$.
I don't see a super-easy way to do this, but this line is coherent, right? Say $K$ is splitting field of the polynomial $g \in F[x]$. Then if the roots of $g$ are $\theta_1, \cdots, \theta_n$ in $\overline{F}$, $K = F(\theta_1, \cdots, \theta_n)$. Suppose $f \in F[x]$ is irreducible and $\alpha$ is a root of $f$ which belongs to $K$; then $\alpha$ is some polynomial on $\theta_1, \cdots, \theta_n$ with coefficients in $F$.
For any element $\sigma \in \text{Aut}(\overline{F}/F)$, $\sigma$ must permute $\{\theta_1, \cdots, \theta_n\}$, therefore $\sigma(\alpha) \in K$ as well, being a polynomial on $\theta_1, \cdots, \theta_n$ with coefficients in $F$.
I claim $\text{Aut}(\overline{F}/F)$ acts transitively on the roots of $f$. This, along with the previous conclusion, will prove the result.
This is because if $\alpha$ and $\beta$ are two roots of $f$ over $F$, then there is an isomorphism $F(\alpha) \to F(\beta)$. Compose with the inclusion $F(\beta) \to \overline{F}$ to get an embedding $F(\alpha) \to \overline{F}$, and extend that to an embedding $\overline{F} \to \overline{F}$ which will be an isomorphism.
This is an automorphism of $\overline{F}$ which fixes $F$ and sends $\alpha$ to $\beta$.
Therefore all the roots of $f$ are contained in $K$, hence $f$ splits completely in $K$.
For the sake of completeness: If $E$ is an algebraic extension of $F$, any embedding $i : E \to \overline{F}$ can be extended to an isomorphism $\overline{F} \to \overline{F}$ as follows. Take the set of tuples $(L, j)$ consisting of an extension $L/F$ and an embedding $j : L \to \overline{F}$ extending $i$.
Then this is nonempty, every chain has an upper bound, so Zorn's lemma applies and you get a maximal element which has to be $(\overline{F}, \phi)$ otherwise we can just add another algebraic. The final embedding $\phi : \overline{F} \to \overline{F}$ is surjective because $\overline{F}$ is an algebraic extension of $\phi(\overline{F})$, which being isomorphic to $\overline{F}$ is already algebraically closed, so cannot be proper.
I feel like I cheated and used Galois theory but I also don't see an easier way to solve this problem. I don't know what Dummit-Foote has in mind.
I suppose the key insight out of this mess is that if $K$ is splitting field of some polynomial (I suppose these are called normal extensions?) over $F$ then $\text{Gal}(K/F)$ acts transitively on the geometric points of $\text{Spec}\, K$, that is, every $F$-embedding $K \to \overline{F}$ must preserve $K$.
So given picture I have to check whether divergence and curl is positive, negative or zero.
So I check vector field at some point ; for figure b I can see that vector field goes outward so divergence is positive but for figure a I am not able to predict. Also how to conclude about curl?
