April 19

I ran across this statement: If ${\displaystyle N(a)}$ is a prime number, then ${\displaystyle a}$ is an irreducible element in ${\displaystyle Z[i]}$. What function does ${\displaystyle N}$ denote? 76.14.188.230 (talk) 06:14, 19 April 2011 (UTC)[reply]

Nevermind, I believe I've found the answer. 76.14.188.230 (talk) 06:23, 19 April 2011 (UTC)[reply]

Hi all, was hoping you could help with the following:

We say a sequence ${\displaystyle X_{n))$ of random variables converges completely to X if

${\displaystyle \forall \epsilon >0,\,\sum _{n\in \mathbb {N} }\mathbb {P} (A_{n}(\epsilon ))<\infty }$, where ${\displaystyle A_{n}(\epsilon )=\{|X_{n}-X|>\epsilon \))$.

Two parts of this problem I'm stuck on are:

I) Show that complete convergence implies almost sure convergence (defined in the usual sense).

II)Find a sequence of (dependent) random variables converging almost surely but not completely.

These are smaller parts of a bigger question, including things like proving the two Borel-Cantelli lemmas, which I believe are going to be highly relevant. For I), I can certainly fix any epsilon, and then show for that epsilon that ${\displaystyle \mathbb {P} (A_{n}(\epsilon ){\text{ occurs infinitely often)))=0}$ - this follows immediately from Borel-Cantelli 1. However, my trouble is translating that to the fact that ${\displaystyle \mathbb {P} (X_{n}\to X)=1}$: that surely requires going from a statement like ${\displaystyle (\forall \epsilon >0)\,(\mathbb {P} ({\text{something happens)))={\text{something)))}$ ('something' being for example that ${\displaystyle \exists N_{\epsilon }:n>N_{\epsilon }\Rightarrow |X_{n}-X|<\epsilon }$ with probability 1) to ${\displaystyle \mathbb {P} ((\forall \epsilon >0{\text{ something happens)))={\text{something)))}$. Isn't that a reversal of two limiting processes in some sense? I thought of using Borel-Cantelli again taking epsilon = 1/n, but I didn't get anywhere with it.

For II), I have just shown that in the case of independent variables, almost sure convergence implies complete convergence too, so certainly we will need to take dependent variables. However, I haven't a clue where to start, I'm not sure I have enough of a clear picture in my head of the differences between these two modes of convergence to sensibly construct such a sequence. Could anyone help me please? I've tried to show my working, or at least my thinking, as best as I could. Thank you very much, Staytime101 (talk) 12:02, 19 April 2011 (UTC)[reply]

For (II), I would just use the fact that convergence of the terms of a series to zero need not imply convergence of the series itself to zero. (For instance ${\displaystyle X_{n}=\mathbb {I} _{[0,1/n]}(U)}$ where U is a uniform random variable on [0,1]. So ${\displaystyle X_{n}\to 0}$ a.s., but not completely.) Sławomir Biały (talk) 20:56, 19 April 2011 (UTC)[reply]

For (I), the following works. Note that ${\displaystyle X_{n}(\omega )\not \to X(\omega )}$ if and only if there exists an ${\displaystyle \epsilon >0}$ such that ${\displaystyle |X_{n}(\omega )-X(\omega )|>\epsilon }$ for infinitely many n. In fact, we can choose ${\displaystyle \epsilon =1/k}$ for some integer here, since we can certainly make ${\displaystyle \epsilon }$ smaller if necessary. Hence the event ${\displaystyle E=\{X_{n}\not \to X\))$ is the union of the events ${\displaystyle E_{k}=\{|X-X_{n}|>1/k\,{\text{for infinitely many ))n\))$ as k varies. Now you can use Borel-Cantelli to conclude that each ${\displaystyle E_{k))$ is null, and therefore their countable union ${\displaystyle E}$ is also null. Sławomir Biały (talk) 21:21, 19 April 2011 (UTC)[reply]

I forgot to say thankyou for this at the time: thank you very much for the help! :) Staytime101 (talk) 16:38, 21 April 2011 (UTC)[reply]

Dear Wikipedians:

I am looking at the multivariate function ${\displaystyle f(x,y)=x^{2}+y-e^{y))$, applying the second derivative test I found that

${\displaystyle f_{x}(x,y)=2x=0\Rightarrow x=0}$

${\displaystyle f_{y}(x,y)=1-e^{y}=0\Rightarrow y=0}$

${\displaystyle f_{xx}(x,y)=2}$

${\displaystyle f_{yy}(x,y)=-e^{y))$

${\displaystyle f_{xy}(x,y)=0}$

Thus the only critical point that I can find for the above function is the point (0, 0). But when the point (0, 1) is under consideration, could I simply say that (0, 1) is not a critical point or can I only conservatively conclude that the behaviour of the point (0, 1) under the function ${\displaystyle f(x,y)}$ is not detected by the second derivative test?

Thanks,

70.31.154.4 (talk) 15:02, 19 April 2011 (UTC)[reply]

Your function ƒ : R² → R is given by ƒ(x,y) = x² + y − e^y. As you say, the only critical point is (x,y) = (0,0). All of the other points in R², including (x,y) = (0,1), are regular points. The image of a critical point is called a critical value. Since ƒ(0,0) = −1, it follows that −1 is the only critical value of ƒ. All other points in R are regular values. It follows that the level set ƒ⁻¹(c) is a smooth, parametrisable manifold in a neighbourhood of each of its points for all c ≠ −1. Since (x,y) = (0,1) is a regular point we know that the curve ƒ(x,y) = 1 − e¹ will pass through (x,y) = (0,1) and will be a smooth curve in a neighbourhood of (x,y) = (0,1). The only reason for you to use the Hessian matrix is to determine if the critical point (x,y) = (0,0) is a degenerate critical point or not. The determinant of the Hessian matrix is −2e^y, which is non-zero at (x,y) = (0,0), meaning that you have a non-degenerate critical point. Because you have a non-degenerate critical point you know that ƒ⁻¹(c) undergoes a Morse transition (named after Marston Morse) as c varies between −1−ε and −1+ε (for some very small, positive number ε). Because the determinant of the Hessian matrix is negative at the critical point (x,y) = (0,0) we have a hyperbolic Morse transition. The canonical example of a hyperbolic Morse transition is given by the one-parameter family of curves x² − y² = k, where k is small and passes through zero. I've added a picture to show how ƒ⁻¹(c) changes as c passes through −1. The moment of Morse transition is shown by the green curve, when c = −1. — Fly by Night (talk) 15:49, 19 April 2011 (UTC)[reply]

Whilst it is not a critical point the point is special in a particular kind of way. One way to observe this is that the level set through the point ${\displaystyle f^{-1}(c)}$ where ${\displaystyle c=f(0,1)}$ has a vertex. --Salix (talk): 21:21, 19 April 2011 (UTC)[reply]

Further, the point lies on a "height ridge", this is roughly equivalent to the ridges you get on hills. There are various definitions of these, see [1]. One definition is to find the eigen vectors of the Hessian matrix, u, v. You will get a ridge point if ${\displaystyle \nabla f\cdot \mathbf {u} =0}$. This fits with the point (0,1) as the Hessian is ${\displaystyle {\begin{pmatrix}2&0\\0&-e^{y}\end{pmatrix))}$ so (1,0) and (0,1) are the eigen vectors, and ${\displaystyle \nabla f\cdot (0,1)=0}$ at your point.--Salix (talk): 22:15, 19 April 2011 (UTC)[reply]

I did some sums, and the branch of ƒ⁻¹(1–e) that passes through (x,y) = (0,1) can locally be parametrised by x(t) = t and

${\displaystyle y(t)=1+{\frac {1}{e-1))\,t^{2}-{\frac {e}{2(e-1)^{3))}\,t^{4}+{\frac {2e^{2}+e}{6(e-1)^{5))}\,t^{6}+O(t^{7})\,.}$

After some calculation, I also notice that (x,y) = (0,1) is an ordinary affine vertex; i.e. the affine curvature is non-zero while the first derivative is zero and the second derivative is non-zero. In fact the affine curvature is negative at (0,1) so that means that there exists a hyperbola having exactly five point contact with ƒ⁻¹(1–e) at (x,y) = (0,1). The geometric interpretation of a ridge point (of the surface z = ƒ(x,y)) is that one of the principal curvatures has a maximum in the other principal direction. As nice as this all is, vertices, affine vertices and ridge points involve higher order terms than just the 2-jet, which is what the OP was asking about. Hopefully we haven't confused him/her. — Fly by Night (talk) 22:45, 19 April 2011 (UTC)[reply]

Wow, thanks so much. I am clear now. L33th4x0r (talk) 01:43, 20 April 2011 (UTC)[reply]

@FBN "height ridges" are somewhat different to ridges. Height-ridges are are specific to height functions and are invariant under rescaling the height, rather than rotation of the surface in 3D. It seems the definition of these does just involve second derivative terms. There is also a "watershed ridge" which are more connected with the way water might flow down a hillside. These are characterised as boundaries between water flowing into one basin or another, like the continental divide. Alas these don't have a local characterisation.--Salix (talk): 07:27, 20 April 2011 (UTC)[reply]

Oh, okay. I've never heard of a "height ridge" in this setting. I thought you meant Ridge points; but on second thoughts they're characterised by the family of distance squared functions, and not the height functions. Do we have an article on them? Would you like to start one? — Fly by Night (talk) 12:36, 20 April 2011 (UTC)[reply]

Ridge detection covers some aspect of the subject, but tends to focus on the image processing aspects and does not cover the watershed ridge.--Salix (talk): 13:45, 20 April 2011 (UTC)[reply]

Resolved

${\displaystyle {\sqrt {\frac {1-\sin \theta }{1+\sin \theta ))}={\frac {1-\tan(\theta /2)}{1+\tan(\theta /2))).}$

What are the occasions for the use of this particular version of the tangent half-angle formula? Where does it appear in the literature? (Eccentricity of ellipses? Other things?) Michael Hardy (talk) 16:04, 19 April 2011 (UTC)[reply]