Picky @robjohn, and patient @TedShifrin, I feel I might be X-Y'ing huh. The ultimate goal us to get an expression for the time derivative of the expectation of momemtum. Ie $d/dt \int{\Psi^* [-i\hbar\frac{\partial}{\partial x}]\Psi}dx$ (the solution to this has been given to me, my task is to try get the method)
@AndrewMicallef As Ted said, you have to use that $\Psi$ satisfies the Schrodinger equation. Of course, the answer is what you expect (quite literally; expectation recovers classical equations)
i'm sometimes interested in formal analysis of coefficients of power series when it's integer coefficients. rational coefficients is ramanujan territory.
Let $T$ be bounded self-adjoint on $H$. There are two different formulations for the spectral theorem, one says there is a projection-valued measure $\mu$ such that $T = \int_{\sigma(T)} \lambda d\mu(\lambda)$, the other says there is a real-valued measure $\nu$ such that $H = \int^{\oplus}_{\sigma(T)} H_\lambda d\nu(\lambda)$, and $T$ acts on this direct integral as multiplication by $\lambda$.
I would like it if for Borel subsets $B \subset \sigma(T)$ if $\int_B^{\oplus} H_\lambda d\nu(\lambda)$ was equal to the spectral resolution $E(\mathcal{B}) = \text{im} \mu(B)$.
Is that true?
I am guessing no but I am also having a hard time pinning down the precise relation between the two formulations above
Maybe this is uniqueness of the spectral resolution? We can define from the second formulation a new spectral resolution for $T$ which associates to each borel subset $B \subset \sigma(T)$, the projection to the subspace (maybe closure of this) $\int_B^{\oplus} H_\lambda d\nu(\lambda)$ of $H$.
That seems to be a new spectral measure, but spectral measure is unique
in (very, very weak) defense of this question, that integral seems way harder to do with Stokes than it does to do the line integral directly @TedShifrin
if $(h,k)=1,k=0$, prove that there is a constant $A$ depending on $h$ and $k$ such that if $x\ge 2$ $$\sum_{p\le x, p\,\equiv\, h(\text{mod}\, k)}\frac{1}{q}=\frac{1}{\phi(k)}\log\log x+A+O\left(\frac{1}{\log x}\right)$$
The Kempner series is a modification of the harmonic series, formed by omitting all terms whose denominator expressed in base 10 contains the digit 9. That is, it is the sum
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∑
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n...
I feel like "$\sum_{n=1}^\infty' 1/n$ where the prime indicates that n takes only values whose decimal expansion has no nines" is about the most useless application of summation notation i've seen
Sample a Poisson random variable (with some rate constant) and use that to pick your first integer. Then generate another sample and add that to your first integer to get your second. Repeat ad infinitum. What can be said about convergence of the sum of reciprocals?
more mathematical statement of it: Consider a sequence $X_1,X_2,\cdots$ of iid Poisson random variables, and let $Y_k$ be the $k$th partial sum. What is the probability that $\sum_k 1/Y_k$ converges?
@robjohn I took me a while, and it wasn't intuitively obvious, but I see how that works. Not sure why I would use that strategy though, but hey that was fun
Can anyone please help me understand why $\mathbb{R} ^2$ minus $n$ points deformation retracts onto a wedge sum of circles? This has been bothering me for 2 days
i think im missing something really obvious, if $f_n \rightarrow f$ uniformly on $\mathbb{R}$, and $f_n$ converges to something in $L^2(\mathbb{R})$, then it must be the case that this function is $f$, right?
is there a way to prove this besides using that there is a subsequence of $f_n$ converging a.e. to its limit in $L^2$, and therefore its limit in $L^2$ is equal a.e. to $f$?
@GeorgeRevingston it doesn't matter where the $n$ points are, up to homeomorphism, so you can fan them out around the origin, now draw a wedge of $n$-petals , each petal around one of the points
now there is an easy choice of deformation retraction
inside each petal, you project radially outwards from the point onto the petal, and outside the petals you split the plane into $n$ regions, and within each region you move along the line between a point and the point in the petal corresponding to that region
of course this isn't the formal answer in terms of some actual function, but you can probably turn it into one
and on the lines that separate the region, you move towards the wedge sum along that line, so you'll eventually reach the wedge point from there
you make the best decision given the info available. if you are having a hard time deciding you either need to get more info or there is no clear answer and either decision is good at that time.
if you've studied something several times before in math, its usually pretty easy to remind yourself of the material right? Like say you go to a uni where students do eight courses a year and exams are at the very end of the year, if you learned the first four courses (taught in the first half of the year) well at the time, it should be doable to review that material within a few days right?
assuming you're not relearning stuff, just reviewing material
maybe ive phrased my question poorly, its normal to not be able to recall little details of maths you've learnt before but haven't kept fresh in your memory right? does that count as 'forgetting' that piece of mathematics? but how could it if you can refresh your memory much faster than it would take to learn it the first time
oh yeah for sure, this will depend on the speed at which someone can parse mathematics, and lots of other personal differences
lets say you've been able to review material at the rate of three or four days before, and its material of a similar difficulty for you (or slightly harder)
well, i studied in ireland back on the early 80's so this may not be relevant. but all of our (engineering) exams were right before summer (no semester/quarter thingy) so it was a different sort of problem. i would made detailed notes of the issues that i needed refreshing on and basically look at them frequently and just before the exam.
thats pretty relevant, its the exact same system im currently in actually
I see, yeah I happen to have made those notes, and I find myself doing a similar thing
in retrospect I wish I looked at notes for my first four courses more in the second half of my year, but at the same time i literally couldnt find the time because learning the latter four courses were hard enough
fortunately they give us a month to prepare for exams, but thats basically four days for each course, which is why I asked - im basically just able to finish the course content in those four days so it seems doable
i really get stuck on using the same word over and over again , i wonder what that says about the way I think
easy to say, but focus on what you need to do and do not let the unhelpful feelings get in the way. i am definitely not a ahhhoooommmm sort of person, but i found some simple mind settling sort of focus/meditations to help.
of course, L^infty convergence need not imply L^p convergence for any other p globally, but if that convergence is given, the limits have to agree a.e. by this argument
actually, now I'm unsure
what's a sequence of L^2 functions converging to something non L^2 in L^infty
I've been trying to approximate a constant function but that didn't work due to the L^2 functions vanishing. A nonintegrable, vanishing function sounds like the way to go
maybe im misinterpreting something, I thought all @Thorgott was saying is if you have convergence in L^infinity and L^p for some finite p, the limits coincide a.e.
anyone have objections to defining identical vector spaces as: if the subspaces of a vector space have as possible dimensions the same number of dimensions as the subspaces of another vector space, then the two vector spaces are the same?
@TedShifrin that's the definition, I'm trying to see if it works out intuitively: two vector spaces are the same if their subspaces have the same number of possible dimensions
@porridgemathematics that's definitely true, the additional question I just raised is whether converging in L^p is implied by converging in L^infty or whether that is a genuine hypothesis
@porridgemathematics so if $f_n$ converges in $L_\infty$ and some other $L_p$ then the limits must be the same ae. The $L_\infty$ convergence dominates the $L_p$ convergence on finite measure sets.
@TedShifrin what do you think of this one: two vector spaces with dim=n are the same if they can be generated from the same subset of vectors from a vector space of dim>n
I know several people at CUNY, but most of the ones I know are surely retired. And saying geometry/topology is awfully broad. Unless you truly are interested in anything/everything in those fields.
@shintuku: You shouldn't make it a question of living inside someone bigger. Why are you doing that? Just say that you can choose identical bases. Period. And yes, that's the same. Isomorphic, of course, but also same.
Some do, some don't know fields at all. If you were committing to a particular adviser at that Canadian university, that is damn specific.
I basically knew I wanted to do complex geometry, and I did, although I was slightly tempted by one person in dynamical systems from whom I took a course first year.
@porridgemathematics in general, i think you need something like if $A$ is measureable non null set then there is some non null $B \subset A$ of finite measure.
@feynhat: I suspect either one would be fine for you. Big difference between living in NYC and Bloomington. Expense, culture, ambiance. Ignoring all things Covid.
This semester I will compute the Euler class of a certain bundle over a certain grassmannian which will tell me the number of lines in a quintic threefold.
Ah, yes, that's beautiful stuff. Enumerative geometry is far more a part of algebraic geometry than of algebraic topology. My papers are full of Chern class computations.
@TedShifrin I listened to a seminar talk on Chern-Weil theory and its applications to Kähler manifolds earlier. I understood very little, but it was quite interesting.
so, how about this: Proof. Notice that the dimension of any subspace of $\mathbb{R}^1$ is either 0 or 1. Let $U$ be a nontrivial subspace of $\mathbb{R}^1$. Since it is nontrivial, its dimension is 1. Therefore, since $dim(U) = dim(\mathbb{R}^1)$ and $U \subset \mathbb{R}^1$, they are identical.
those are valid ways to visualize ways these concepts, but sometimes (and this is one of those times), there are distinct ways of visualizing the same mathematical object
@shintuku what made me change?...one word: Ted :p...............but being more serious about it, looking at things in higher dimensions. you can't describe a position with just one value in a higher dimension
was such a convoluted term needed to describe a complex differentiable function at a point @copper.hat?
eventually you get to a place where for the same set of things, there might be one or more choices of what your 'scalars' are going to be. that's all. it doesn't have to be today.
@dc3rd my comment on meromorphic functions reminded me that there is a nice, relatively informal discussion of the pitfalls of modern control system design called "Respect the unstable" by Gunter Stein. not entirely apropos, but a good reminder that algebraic methods applied without understanding the underlying dynamics/physics can have fairly disastrous consequences.
anyone know of complex functions that cause a rotation away from the imaginary axis? so first quadrant imaginary numbers get closer to positive real axis, second quadrant imaginary numbers get closer to negative real axis
