@Akiva I have a solution to your contour! Break it at the intersection near the bottom left between the purple and black parts. Then, both contain only the green singularity, but are in reverse orientation, thus they cancel out
@AkivaWeinberger isn't it false on $S^2$ just by taking $2$ points as the $2$ sets? the complement of each is $\Bbb R^2$ while the complement of the union is a punctured $\Bbb R^2$
I think we mean different things when we say "sliding". I agree that if it bounds a cylinder you can do it fixing a basepoint by conjugating at each level.
(take a moment to compare with the original to make sure that the orientations make sense — clockwise around the region on the right and counterclockwise around the region on the left) @MeowMix
then we end up with two regions, neither of which contain either singularity.
Akiva: Actually, not all is lost, if $n>1$, $a(x) \in \mathcal{C}^{\infty}(\Bbb{C}or \Bbb{R})$ and $\lim_{R\to\infty}a^{R+1}(x) < \infty$ (These conditions kill off the $R+1$th term), then
I suspect it will work for all polynomials $P(x)$ since polynomials are nilpotent under differentiation, thus the sequence of derivatives will be eventually constant (=0 )
BTW, I think for R^3 your subset needs to be sufficient bad otherwise you'd probably disprove it by Alexander duality. If S^3 - X is simply connected, H^1(S^3 - X) = 0. By duality, H_1(X) = 0. Same for Y. H_1(X U Y) = 0 because it's a disjoint union. Again dualize and you get H^1(S^3-(X U Y)) = 0.
Well, that doesn't say $\pi_1$(S^3 - (X U Y)) = 0 but at least it's something.
It's easy to see that each set must be clopen in their union
which means they're open in their union, which means (by the argument above) that, if the union is the Antoine necklace, each one doesn't have a simply-connected complement
(the argument above being that if either one contains an open ball intersected with the necklace, then they contain a miniature version of the necklace due to the self-similarity)
@AkivaWeinberger Is it super-easy to see that complement of Fox-Artin arc has trivial homology, by hand (without Alexander duality)? Like, what's a surface which a thing in the middle bounds? Maybe you can do something with the Fox-Artin wild sphere
Or, maybe it bounds an infinite genus surface on one end.
so I was wondering... would I be able to get admission into MIT's PhD program if I had a GPA of 2.7 but by some miracle was able to solve one of the millenial conjectures?
@SoumyoB Just saying, "thinking you solved one of the millennium conjectures" is a tad more likely than "actually having solved one of the millenium conjectures"
A 10 litre container holds a mixture of water and sugar,the volume of
sugar being 15% of total volume.A few litres of the mixture is
released and an equal amount of water is added.Then the same amount of
the mixture as before is released and replaced with water for a second
time.As a r...
@BalarkaSen Hi! Do you know how to write equations of the superposition of two SHMs along axes which make a certain angle $\theta$ w.r.t to each other? I know it when the two SHMS are along perpendicular axes like x and y axes. I am bit confused about this situation though. Say $s_1=a\sin(\omega t)$ and $s_2=b\sin(\omega t)$ (which is at an angle $\theta$ w.r.t) each other.
I think I need to rotate the axis....
Ah! I guess this will do: $-x\sin(\theta) + y \cos(\theta) = b\sin(\omega t)$ and $x=a\sin(\omega t)$ :-)
[Integral symmetries] The aforementioned formula passed the product of sin, cos and exponential test,as shown in the sin example: $$\int \sin (bx)e^{-nx}dx=-e^{-nx}\left(\frac{\sin bx}{n}+\frac{b\cos bx}{n^2}+\frac{-b^2\sin bx}{n^3}+\frac{-b^3\cos bx}{n^4}+\frac{b^4\sin bx}{n^5}+\frac{b^5\cos bx}{n^6}+\frac{-b^6\sin bx}{n^7}+\frac{-b^7\cos bx}{n^8}+\cdots\right)=-\frac{e^{-nx}}{b}\left(\sin bx\left(\sum_{k=0}^{\infty}(-1)^k\left(\frac{b}{n}\right)^{2k+1}\right)+\cos bx \left(\left(\sum_{k=0}^{\infty}(-1)^k\left(\frac{b}{n}\right)^{2k+2}\right)\right)\right)=-\frac{e^{-nx}}{b}\left(\sin bx\l…
However:
The above requires the convergence of geometric series, which has a radius of convergence of 1. But the actual formula holds for arbitrary n and b (specifically, it holds for $b > n$ also). More investigation is needed on how it avoided divergence
Hey everybody. I've joined because I'm taking an MIT edX course that's heavy on statistics, probability, R, and economics. I imagine this room would be ideal if I run into any particularly sticky questions, or if I'm just not grasping something conceptually. I don't have anything to ask right now, but I hope that's the kind of thing that's welcome here.
If I recall, this room has the capacity to handle some probability theory and some type of statistics. however if you are into more hardcore economics and finance and statistics, then the following room might be a better choice:
Excellent, thanks Secret. This is actually the first time I've picked up a quantitative subject in along while, so how hardcore the course is, I have no idea. I've certainly found it challenging. I'll bear the suggestion in mind, though.
Hi. Could someone give an opinion as to whether this question: math.stackexchange.com/questions/2189710/… would be better suited for math overflow? Since posting it, I have noticed quite a few quadrature/cubature questions over there.
He guys, anyone familiar with the Chinese remainder theorem?
Because I'm having some problems with an algorithm
Let $a_1,\dots,a_t\in\mathbb Z$ and $n_1,\dots,n_t\in\mathbb N$. We are looking for $a\in\mathbb Z$: \begin{align} a&\equiv a_1\mod n_1\\ &\dots\\ a&\equiv a_t\mod n_t. \end{align} Assume that for each $i\in\{1,\dots,t\}$ we've found $b_i\in\mathbb Z$ such that: \begin{align} b_i&\equiv 1\mod n_i\\ b_i&\equiv 0\mod n_j&(j\neq i). \end{align}
My book says that $a=a_1b_1+\dots+a_tb_t$ satisfies the condition for $a$ we've set above. I don't see this though.
or maybe...
maybe I do see it, I'm going to consider $a\equiv a_1\mod n_1$ first
ohh...
I see it
We know that $b_i=0\mod n_1$ for $i\in\{2,\dots,t\}$. Therefore $a_2b_2+\dots+a_tb_t$ is a multiple of $n_1$. And given that $b_1\equiv 1 \mod n_1$, we know that $a_1b_1\equiv a_1\mod n_1$
Fun fact: Rotation in 2D clearly doesn't fix any nonzero vector. All rotations in 3D fix some nonzero vector. There are rotations in 4D that don't fix any nonzero vector. I think rotations in 5D must fix some nonzero vector again.
@AkivaWeinberger Linking doesn't have anything to do with self-intersection. The null-homotopy has to intersect the Fox-Artin arc, not the simple loof.
No, the skipped Fox–Artin loop has a simply connected complement, so that that loop (when allowed to self-intersect) can be shrunk to a point without intersecting the arc. @MikeMiller
I just put a 100 point bounty out on this question if anybody's interested. It's a question somebody else asked three years ago, but I'm working on the same problem right now.
I'm asked to show that $\mathbb{Z}_m$ (the integers mod $m$) is a principal ideal ring for every $m > 0$
I see that it is the same discussion used in verifying that $\mathbb{Z}$ (the set of all integers) is.
Can anybody help me in writing a correct solution?
Are $\Bbb R^2\setminus\Bbb Q^2$ and the same set with $(0,0)$ homeomorphic?
@Zach there is a town with $32$ inhabitants that wish to form clubs, to prevent the formation of too many clubs the mayor decides that they must follow those $2$ rules: 1) every club must have an odd number of members 2) each pair of club must share an even number of members How many clubs can be formed?
it's a cool problem that I wouldn't have been able to solve the first time I saw it, but maybe you will
Probably really dumb, but I'm even just having trouble wrapping my head around how to show that if $J$ is an ideal of $\mathbb{Z}_{m}$, then if $J$ contains an element $k$ then $J$ must contain all elements of the form $nk$, $n \in \mathbb{Z}_{m}$. Although I'm probably overthinking it as usual.
@AlessandroCodenotti I feel like nah. I bet $\Bbb R^2 \setminus \Bbb Q^2$ has very few local homeomorphisms - that it's quite rigid. You could then try to argue that doing the thing-without-0 into -thing-with-zero would be impossible to add a single point like that, since the map is decided by what it does in some small neighborhood. Or something like that.
Hi @MikeMiller remember that thing about Pontryagin classes? I think maybe that thing @TedShifrin and you got confused about is that for the tangent bundle of a dimension $<4$ manifold, the Pontryagin class must vanish for dimension reasons.
No, I have a fibration of the "twistor space" (which is complex, and I think indeed even Kaehler in this case) over a quaternionic Kaehler (but not even almost complex) manifold, $G_2/SO(4)$. The fiber is $\Bbb CP^1$ and the fibers are complex submanifolds, I think.
It is not the projectivization of a complex rank 2 bundle, I'm pretty sure. But its definition is as the sphere bundle of an oriented, rank 3 bundle
Well, a twistor space in my setting is by definition a $\Bbb CP^1$ bundle over a quaternionic Kaehler manifold
I don't know any other possibility :P
I can probably get the Pontryagin class of the base, so all that's left is the vertical part. Since each fiber is of dimension two, the restriction of $p(T\pi)$ ($\pi$ the projection) to each fiber is trivial
I was thinking of $\Bbb{CP}^1 \times Q$, where $\Bbb{CP}^1$ parameterizes the complex structures on a hyperkahler manifold $Q$, and the complex structure on $Q_\theta$ is $J_\theta$.
I don't know the more general story for quaternionic Kahler things.
Okay. Well Now $Q$ is not even almost complex, but you can still use almost complex structures of the individual tangent spaces to get a complex structure on the total space. It's not a product anymore.
We know that the cohomology of a sphere bundle satisfies the Gysin sequence, and I think that fact about $p(T\pi)$ implies that the pushforward ("integration over the fiber") is zero, so $p(T\pi)$ is a pullback of a class on $Q$. Not sure what to say from here.
Once I figure out $p$ of the base I can check if $p$ of the vertical part should be zero, too, since I already know what the Pontryagin class of the total space has to be.
A 10 litre container holds a mixture of water and sugar,the volume of
sugar being 15% of total volume.A few litres of the mixture is
released and an equal amount of water is added.Then the same amount of
the mixture as before is released and replaced with water for a second
time.As a r...
I'm asked to show that $\mathbb{Z}_m$ (the integers mod $m$) is a principal ideal ring for every $m > 0$
I see that it is the same discussion used in verifying that $\mathbb{Z}$ (the set of all integers) is.
Can anybody help me in writing a correct solution?
