Q: Algorithm to count irreducibles in a non-UFD?
Consider a non-UFD that only has 2 units ( $-1,1$ ) and the min difference between 2 elements is $1$. Also there are only a finite amount of elements for any given fixed norm. ( Maybe that follows from the other 2 conditions ? ) I wonder about counting the irreducible elements bounded by a lower...
Any ideas ??
@mick what does the min difference between elements mean?
What does the notation $\times_{G}$ mean?
@quallenjäger Depends on context. Here it likely means the orbit space under $G$-action.
If a and b are coprime, does that mean that b and a are prime?
@MikeMiller Haha sure, let me know if you're ever not feeling lazy :p.
Yep I will prod him about it shortly. Have to make coffee or obtain an energy drink.
@MikeMiller What is $P \times_G F$? I am building the cross product over a principal bundle?
Aka, the quotient/orbit space of the product.
and hence $4\sin s\sin (s-a)\sin(s-b)\sin(s-c)=\sin A\sin b\sin c$. cyclic permutation gives two other such formulas.
@Secret agreed.
is that really heron's formula analogue in spherical trig?
I don't know.
A cycloid is the locus of a point on the circumference of a rolling circle, and it's arc length for one revolution of the circle is 8R, does this result change if the circle is not moving with constant velocity?
@TedShifrin I have the solution.
Hey guys, for a function that is $1$ for positive values of $x$ in $\mathbb{R}$ and $0$ for non-positive $x$, is $f((x_n))_{n \in \mathbb{N}}$ with $x_n = (-1)^n\frac{1}{n}$ divergent or convegent? It alternates between $1$ and $0$, so for every $\epsilon$ there is an $n$ such that for all $N>n$ there is an $N_1$ such that $f(x_{N_1}) - f(x) \ge \epsilon$, so yes it deverges. But $\lim_{n \to \infty}f(x_n) = 0$, so the limit exists, so it conveges. What is right?
Hello is anyone on?
@PrittBalagopal Is it along the lines of, what is the meaning of life?
Or, is it more directly related to math?
Ah, now your question is bothering me, simply because it is shrouded in mystery.
OhhKayy, so basically, how do you find minimum distance of a point from a surface defined by 3d cartesian equation?
with the distance squared being the objective function and the surface being the constraint.
Darn it, @Semiclassical beat me to it.
Oh, man, I got distracted. I was looking something up in a mse question, and then I found my way into the chat!
@KarlKronenfeld Ah, fair enough.
Okay, I currently don't know what Lagrangian Multipliers are, trying to read a Wikipedia article and see if I can understand anything. Btw, I am just a college first year student, so I know very little math.
If a series converges uniformly on $[a+\varepsilon,b) $ for all $\varepsilon>0$ and pointwise on $[a,b)$, is then the convergence uniform on $[a,b) $?
@PrittBalagopal the geometric idea goes like this. suppose you've got a surface and a point not on that surface.
Yes, and we have to find the radius right?
Oh I see.
Is it like a slope analogy for 3D?
Question: suppose you had a curve represented by a pair of parametric equations. This curve goes to the same point as t goes to either positive or negative infinity. How would you go about finding a point that is equidistant from all points on that curve in the xy plane?
Q: Doubt on the equation 8.11.9 ( Green's function's construction ). How do they take limit of the integral?
Reference: Linear Partial Differential Equations for Scientists and Engineers(Authors:-Tyn Myint-U, Lokenath Debnath) I understood till the red box. I have doubt on the equation in the red box. How do they take limits of the integral? I don't understand. Please help me. What is the motivatio...
you end up with the following equations: $\frac{\partial F}{\partial x} = \lambda \frac{\partial G}{\partial x}$, $\frac{\partial F}{\partial y} = \lambda \frac{\partial G}{\partial y}$, $\frac{\partial F}{\partial z} = \lambda \frac{\partial G}{\partial z}$, and the constraint equation written as $G(x,y,z)=0$.
the surface constraint is that $G(x,y,z)=z=0$ and the objective function is $F(x,y,z)=(x-0)^2+(y-0)^2+(z-h)^2=x^2+y^2+(z-h)^2$ (i.e. the distance squared between $(0,0,h)$ and a particular point on the surface).
@Semiclassical Is there any way to render the LaTeX commands? I'm not really able to translate all those equations you wrote.
the above equations then become: $2x=\lambda\cdot 0,$ $2y=\lambda\cdot 0$, $2(z-h)=\lambda\cdot 1$, $z=0$.
so $x=y=z=0$ in which case $F(0,0,0)=h^2$. So the minimum distance from a point at height $h$ to the horizontal plane $z=0$ is just $h$ (as it has to be).
This example is trivial, but the same idea can work for $F,G$ which are decidedly nontrivial.
Any book on multivariable calculus should discuss Lagrangian multipliers, so I'd look there for further examples.
Wow thanks @Semiclassical ! You told me alot of stuff, I'll take quite a while to fully understand. Thanks for your help, really appreciate it!
Please help me.
I don't understand.
@KevinDriscoll did you see a new kind of science in the night sky lately?
@BalarkaSen uuuummmmm now that I know of?
then $e_2^T(I+S)e_1=S_{12}$.
Is there an easy way to argue from that that there exist unit vectors $v_1,v_2$ such that $v_2^Tv_1 = S_{12}$?
Is the automorphism group of a topological group always topological?
okay, got it: since $I+S$ is positive semidefinite with $S$ symmetric (and real---forgot to say that) there's a cholesky factorization $I+S=V^T V$. Hence $S_{12}=e_2^T (I+S)e_1=e_2^T V^T V e_1=(Ve_2)^T (Ve_1)=v_2^T v_1$.
I felt like there's a whole world of matrix factorisation I did not knew exists until now...
hmm...
The post office metric space, $P$ has the distance function defined as follows: $$ d_P (\mathbf{x},\mathbf{y}) := \begin{cases} 0 & \mathbf{x} = \mathbf{y}\\ \Vert \mathbf{x}\Vert_2+\Vert \mathbf{y}\Vert_2 & \mathbf{x}\neq \mathbf{y} \end{cases} $$ where $\Vert \mathbf{x}\Vert_2 = \sqrt{x_...
hmm, very strange...
@MatheinBoulomenos Ok, I see. I guess I thought there was some kind of induced topology.
i need to show that the decomposition of harmonic and analytic function is harmonic, someone can help?
so why would it be harmonic?
Here is theorem 1.14 from Rudin's RCA: If $f_n : X \to [-\infty, \infty]$ is measurable for $n=1,2,3,...$, and $g = \sup_{n \ge 1} f_n$ and $h = \lim_{n \to \infty} \sup f_n$, then $g$ and $h$ are measurable. Immediately following this is the corollary (a): The limit of every pointwis...
@MatheinBoulomenos so you say $im(g) \circ f = im(g\circ f ) $ ?
Here's one for you, Eric.
@MatheinBoulomenos sounds right, i'll try to prove it, thanks!
Eric: I think my suggestion of Riemann-Hurwitz is the fastest way using just alg top.
Do you guys get something similar?
@TedShifrin It's just lifting the orientation from the base using the local biholomorphism, yes?
I'm lost, Balarka. You were suggesting classification of surfaces. Because of the branch points, you fail to have a local biholomorphism everywhere.
What's the proof that orientability still follows even if you've pulled out finitely many points?
Don't start telling me $w_1$ stuff.
Can't you use the generalization of Liouville's theorem from Krantz?
I don't have my resources any more to look up, Mathei.
@Maneesh: That's a long post. Can you ask a specific question? It looks like it follows immediately by substitution, as the authors say.
@TedShifrin I was going to tell you that non-orientability implies there is an embedded circle in your manifold which has nontrivial normal bundle.
Yeah, so you need to push that circle off the missing points.
I think there's a direct way to get it from the Ahlfors Lemma, as the OP suggested.
@TedShifrin I guess you can use compactly supported cohomology too.
That's getting fancier, @Balarka.
Easier to construct a nowhere-0 2-form in the first place, namely, the restriction of the Kähler form on $\Bbb P^2$.
Yippee.
and define green function like that?
Yes, that is suspicious, @Maneesh. They seem to have the wrong sign on the derivative of $u_1$, since $x<b$.
But it works out right when they put the negative sign in the second line. Then you get two negatives when you use the fundamental theorem of calculus.
So the first line is typical physicist nonsense, but they get the right answer.
@TedShifrin sir, i don't get the point.
I understood, Why there is negative sign.
So you're asking why they use $\int_a^x$ for one and $\int_x^b$ for the other? Because that actually gives the Green's function formula they want. The Green's function should depend on the entire interval.
ok sir. Now I understood. they have used the fundamental theorem of calculus to denote $u_1(x)$ and $u_2(x)$ right?
Yes.
Sure thing.
im trying to show that $v(z) = 1/(2\pi) \int_0 ^{2\pi} Re(\dfrac{e \ ^ {it}+z }{e \ ^ {it }-z}) u(e \ ^ {it})dt$ equals the real part of some analytic function. $u$ is harmonic.
Someone can help?
Hey there folks!
Looks a lot like Poisson's formula, @Liad.
So write the line integral without the real part as a line integral over the circle, @Liad. What do you have?
This is the same sort of thing you've been having repeatedly ... where we took the derivative under the integral sign months ago.
Ah, Kevin is psychic.
@TedShifrin you mean write this: $1/(2\pi) \int u(e \ ^ {it} ) dt$ as a line integral?
No, you need the other stuff in there with $z$ and $\zeta = e^{it}$.
Like what?
@TedShifrin what to do with $u$ ?
A-hah! @TedShifrin is here.
$u$ stays ... it's a continuous function.
heya @Antonios. I saw your ping a while ago.
@Mr.Xcoder: That's totally fishy.
?
@TedShifrin You know what they say. One man's fish is another man's poisson.
So you get the plane cubic, too, Mathei.
my fault ... not mathfrak.
=@Liad: why not?
You get the embedding of $\Bbb C/\Lambda$ as the projective plane cubic.
@EricSilva I can do decent $\mathfrak{p}$'s and $\mathfrak{q}$'s.
@Antonios: I believe JCF, but what is the right statement to prove? Remember that I suggested certain eigenspaces got combined for $p(A)$.
hm , we got something like $1/(2\pi) \int \dfrac{e \ ^ {it} +z }{e \ ^ {it} -z }u(e \ ^ {it}) dt$ , dont we?
yes, Balarka, biholomorphic.
Rewrite $\zeta = e^{it}$ and write the line integral.
You don't have to do that, @Balarka.
It follows from facts about the doubly-periodic function $\wp$.
Well, true.
@TedShifrin: Basically, after some thought, the claim is : If the characteristic polynomial of $A$ is $p(x)=\prod_{j=1}^k (x-\lambda_j)^{d_j}$, then the characteristic polynomial of $q(A)$ ($q$ some polynomial) is given by $\prod_{j=1}^k (x-q(\lambda_j))^{d_j}$.
I'm not sure what you mean by Weierstrass form. Why is that "pictorial"?
$A$ is over $\mathbf{C}$, so that using JNF $J$ for $A$, and taking $p(J)$ the result is immediate.
OK, Antonios. I believe that.
But that doesn't tell me the eigenspaces.
(replace $\lambda_j$ with $q(\lambda_j)$.) Sorry if that wasn't clear.
Ugh, @Liad. That doesn't look right.
Spare me the partial fractions. What is the formula?
I already showed that the eigenspaces are the same @TedShifrin. But, that also follows from the JNF.
But I claim they're NOT the same, @Antonios.
Oh, huh.
Why doesn't the JNF show that?
You may have to combine eigenspaces of $A$ to get eigenspaces of $q(A)$. That was my whole point.
This is what you get when you rewrite, removing the real part? Um, no.
I'm not asking for Cauchy's integral formula.
Oh, right, @Balarka: But we have to use the facts about zeroes/poles of $\wp$ to get such a thing.
I don't see that that is pictorial.
I think it's kind of pictorial because you can choose a slice of $\Bbb C^2$ by a 1-dimensional subspace so that it visually has two circle components (with one point on one of the circles at infinity, so that it's parabolic).
@TedShifrin Hm, really?
We're discussing totally different things.
$1/(2\pi) \int_0 ^{2\pi} \dfrac{e \ ^ {it}+z }{e \ ^ {it }-z} u(e \ ^ {it})dt$ @TedShifrin this is what i get from removing the real part, what did you want me to do from here?
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