@user10478 I don't know if it will be helpful to you, but this answer shows how the Poisson probability is related to a limit of Binomial distributions.
@amWhy Thanks. I have candy that won't be given out tonight because we won't be having any ToT tonight.
What is the Laurent series expansion of an entire function? Is it just the power series? I.e., are no there no negative powers of $x$ in the expansion?
Okay. Here's a problem I am working on: Show each branch of $f(z) = (z^2 - 1)^{5/2}$ is meromorphic at $\infty$, and obtain the series expansion for each branch at $\infty$...So I gather I need to find the order of pole $w=0$ of $g(w) = f(1/w) = \frac{(1-w^2)^{5/2}}{w^5}$, right? How do I deal with the square root in the numerator?
For diffeomorphism between regular sufaces $f:S->\tilde{S}$, an orientation for S naturally induces a well-defined orientation for \tilde{S} w.r.t. which $f$ is orientation preserving.
If $N$ is the given orientation then $N\circ f^{-1}$ seems to be the natural orientation for $\tilde{S}$
But <(N\circ f^{-1})(f(p)),v> = 0? for $v\in T_{f(p)}\tilde{S}
Only know $<N(p),v>=0$ for $v\in T_pS$
My definition of orientation is unit normal field on regular surface
And normal field N is a vector field such that <N(p),v> = 0 for all p\in S(regular surface) and $v\in T_pS$
@love_sodam No, there's no reason to believe that the pullback of a normal field is a normal field. However, the pullback of a normal field is never parallel to the surface, so always picks out one particular side of the orthocomplement. You can use that to determine what the induced orientation is.
Consider for S the top quarter of the sphere, S-tilde a disc in the xy-plane, and f the projection, to see what's going on.
How can I show that $2k$ twisted Mobius band are all diffeomorphic and $2k+1$ twisted Mobius band are all diffeomorphic? where $k\in\mathbb{N}$
Edit: Parametrization of $n$-twisted Mobius band: $f(u,v) = (cos(u)(2+vsin(n\frac{u}{2})),sin(u)(2+vsin(n\frac{u}{2})),vcos(n\frac{u}{2}))$
In the comment, someone said that even twisted Mobius strips are all diffeomorphic and so is odd twisted strips. How can I find the diffeomorphism between thoese things?
@TedShifrin would it be taboo to ask a potential supervisor who I have not yet been in contact with if he had a topic for a Ph.D. in a particular subfield? The professor in mind wrote a paper on something I actually have deeper experience with in 2010, but since then has seemingly not worked on it according to their list of recent publications. I do not know if it is worth mentioning or if I should just be prepared to do something they are currently working on. Any advice?
Is the reason $C^\infty(M)$ over some manifold is a ring and not a field because defining "pointwise division" of two functions would be ill defined for any functions that pass through zero?
The reason it is a ring is because you have not yet defined the multiplicative inverses. The problem with defining multiplicative inverses is what you stated.
This is not about how algebraists see things. This is a fundamental difference between something being additional structure vs satisfying additional conditions
The property of being a unit is intrinsic to the operation, not to the element. Why bother defining a "division" operation when all of the information you need is in the multiplication you already have?
@Thorgott there is a tangible difference in constructive mathematics. But I'd definitely agree that there is not necessarily a philosophically tangible difference
That is, it takes a special intangible something to insist it ought to be the constructive way.
Let $V$ be the vector space of functions that are holomorphic on the extended complex plane except possibly at the points $0$ and $i$, where they have poles of order at most two. What is the dimension of $V$ ? Give an explicit basis for this vector space.
I don't really know where to start w...
It's part of the Laurent series expansion at $z_0$. But how do an expansion when I have two singularities.
Another excerpt from my book: "Suppose that $f(z)$ has an isolated singularity at $z_0$. Then $f(z)$ has a Laurent series expansion $$f(z) = \sum_{k= -\infty}^{\infty} a_k(z-z_0)^k$$"
I guess the moral reason this should hold true is that if you paste two fundamental squares of the Möbius strip together you end up with the fundamental square of the cylinder
Because any three dimensional embedding of the two looks different, but adding an additional dimension allows you to untwist the strip. But if you wee only considering this in three dimensions, it looks like you are cutting the strip.
like, if you cut the doubly twisted strip up at one point, you can just un-twist it to get a rectangle, but this is compatible with the orientation on the ends you cut up (unlike in the case where you have only one twist), so you can reglue them
@anakhro ah, so they're even ambiently isotopic in 4 dimensions?
There is a question on this site about the distinctions between the full-twisted Mobius band and the cylinder, but I would like to ask something different, so I start a new question.
Let us call $C$ the standard cylinder embedded in $\mathbb{R}^3$, and call $F$ the full-twisted Mobius band (aka....
@love_sodam it's not an exercise to be proven rigorously, that's for sure. What Thorgott said about the gluing with the correct orientation is probably what the book intended.
Also Ted is here and he might have a great simple explanation.
I like to think about orientation in terms of local homology. With this I can clearly see why the double twisty Mobius band should be orientable. You can continuously choose generators of $H_2(X, X-x)$, just take a small oriented circle about $x$, when you go around the double twister, the orientation reverses twice and you get back to where you started from with your circles having the same orientation.
@anakhro In response to your question, I think you could contact the prof and say that you're really interested in his 2010 paper — even better if you have specific comments or even a question. You could ask to "meet" to talk with him about it. His response should tell you something.
@anakhro You should ask Mike your isotopy question, but I would be the answer is $4$. This question should probably be settled by old work of people like Hirsch and Smale, but it's not stuff I know.
The fancy way of looking at this is to look at real line bundles on the circle. They are classified by $\Bbb Z_2$, and the double-twist is the tensor product of the generator with itself.
Oh, well, no. But you can draw a sequence of pictures if you want to.
@Thorgott A diffeomorphism is clear. Consider $S^1 \times D^2 \to S^1 \times D^2$, $(\theta, z) \mapsto (\theta, e^{i\theta} z)$. Image of $S^1 \times I$ where $I \subset D^2$ is a diameter under this diffeomorphism is double-twisted Mobius strip.
I think
Ya
Imagine both $S^1 \times D^2 \subset \Bbb R^3$ embedded as the standard solid torus.
@anakhro Another way to see it would be to note that if $H \subset \Bbb R^3$ is the Hopf link and $U \subset \Bbb R^3$ is the unlink consisting of two circles, then embedding $\Bbb R^3 = \Bbb R^3 \times 0 \subset \Bbb R^4$, you have an ambient isotopy of $\Bbb R^4$ taking $H$ to $U$. This is because $H$ and $U$ are isotopic as links inside $\Bbb R^4$ (this is easy). $H$ bounds the double-twisted Mobius strip as Seifert surface in $\Bbb R^3$, whereas $U$ bounds the annulus.
The ambient isotopy must take one Siefert surface to another, so you're done.
Was anybody asking for this ambient isotopy though? I thought someone was looking for an explanation of orientability, which doesn't need any of this. I also wrote down a diffeomorphism.
@Thorgott Okay, I'll try verifying that later. However, I have another concern. I followed what Kavi suggested to find a basis for $V$. Given what he showed, I can write $$f(z) = \frac{a_4 z^4 + a_3 z^3 + a_2 z^2 + a_1 z + a_0}{z^2(z-i)^2} = a_4 \frac{z^2}{(z-i)^2} + a_3 \frac{z}{(z-i)^2} + a_2 \frac{1}{(z-i)^2} + a_1 \frac{1}{z(z-i)^2} + a_0 \frac{1}{z^2(z-i)^2}$$
It's clear that those five functions span $V$, but how do I show they are linearly independent? It doesn't seem like an easy task.
I may have asked this already, but my brain is running on fumes at this point: If $f$ is analytic at $\infty$, does this mean that there exists $r > 0$ and $C > 0$ such that $|f(z)| \le C$ for $|z| \ge r$?
the importance of lim sup and lim inf while studying sequences of real numbers is because they gave us the power to see "how far" we can make a subsequence go with a given "parent" sequence?
@EduardoC. The point is that you want something to replace LIMIT when a sequence does not converge. So knowing liminf and limsup allows you to do estimates and, also, to prove convergence when they are equal.
This is an "order-2 branched cover". "Order 2" means a path that wraps around a piece of the knot twice gets you back where you started. For the unknot, that gives you two worlds; for the trefoil knot, you get six.
How can I show that $2k$ twisted Mobius band are all diffeomorphic and $2k+1$ twisted Mobius band are all diffeomorphic? where $k\in\mathbb{N}$
Edit: Parametrization of $n$-twisted Mobius band: $f(u,v) = (cos(u)(2+vsin(n\frac{u}{2})),sin(u)(2+vsin(n\frac{u}{2})),vcos(n\frac{u}{2}))$
Let $V$ be the vector space of functions that are holomorphic on the extended complex plane except possibly at the points $0$ and $i$, where they have poles of order at most two. What is the dimension of $V$ ? Give an explicit basis for this vector space.
I don't really know where to start w...
