I think the only thing of note personally for myself is that I figured out a square function set with its own sine, cosine, etc.
The function doesn't do angles, or at least I don't think it does without some temporal scalar function no doubt involving circular functions. It does, however, allow relating a unique location on a "unit square" of "radius" 1.
Does someone have a more constructive proof that a closed subgroup of a profinite group is again profinite? Something more enlightening than "A group is profinite iff Hausdorff, compact, totally disconnected, and a closed subgroup of a compact group is compact, so again profinite"
Let $G = \varprojlim G_i$ be a profinite group, $H$ a closed subgroup of $G$, $\pi_i : G \to G_i$ the canonical projections. Suppose $x \in H$. Then $\pi_i x \in \pi_i H$ obviously, so $x \in \prod_i \pi_i H$ and the profiniteness of $G$ means the $\pi_i x$ satisfy the relevant inverse system so $x \in \varprojlim \pi_i H$..
Now let $x \in \varprojlim \pi_i H$ and let $x_i \in H$ be such that $\pi_i x = \pi_i x_i$.
$\frac{\pi_i x}{\pi_i x_i} = 1$, cancel the $\pi_i$ so that $x = x_i$ qed
I listened to 10 hours of Skyrim - Music & Ambience - Day today
I have an idea
Let's say $y$ instead of $x_i$ because _ is too hard to type
Okay $\pi_i x = \pi_i y$ so $xy^{-1} \in \ker \pi_i$
does that even make sense
I showed that $H \subseteq \varprojlim \pi_i H$ so I can actually multiply those elements can't I ? Let's pretend I can
now $\pi_i : H \to \pi_i H$ is an open, surjective map, so $\varprojlim \pi_i H / \ker \pi_i \cong \pi_i H$, and $\pi_i H \leq \pi_i G$, and $\pi_i G$ is finite ($G$ is profinite so $\pi_i G$ is finite by definition)
err
right so $\pi_i H$ is finite, i.e. $\ker\pi_i$ has finite index in $\varprojlim \pi_i H$, so it is an open subgroup
I thought I had an idea but I've lost track of it
well $\ker\pi_i$ is an open neighbourhood of $1$, and $xy^{-1} \in \ker\pi_i$, and by the characterisation of the closure of $H$ that I recently found on google, since $\ker\pi_i \cap \overline{H}$ is non-empty, $xy^{-1} \in \overline{H} = H$ (since $H$ is closed) so $x \in Hy = H$ (since $y \in H$)
if I can't multiply $x$ and $y^{-1}$ then I can just say that $x \in y\ker\pi_i$ and $y\ker\pi_i \cap H$ is non-empty anyway so
Hence $H \supseteq \varprojlim \pi_i H$, so $H = \varprojlim \pi_i H$ is profinite
the end
idk if that works because I need that every open neighbourhood of $x$ meets $H$ for $x$ to be in $H$
(I can't read apparently)
$\pi_i y = \pi_i x$ means $y \in x\ker\pi_i$ and these are a basis of open neighbourhoods of $x$, so that gives me that $x \in H$
I have a very basic query regarding dimension of null and range spaces with regard to $Ax = y$ where $x$ is vector $(n\times 1)$ and $y$ is also vector $(m \times 1)$. So is it true that dimesion of range space is always $\le \text{min}(m, n)$ and also is it true dimension of null space = $\text{max}(m, n) - \text{dimension of range space}$ ??
Let $\boldsymbol{X}$ be a random vector in $\mathbb{R}^{m}$ such that no elements of $\boldsymbol{X}$ is
linear combination of others, and $\mathbf{Y} \in \mathbb{R}^{n}$ another random vector. Then $\exists$ an $n \times m$ matrix $M$ such that (Covariance) $\operatorname{Cov}(\boldsymbol{X}, \b...
So, I've thought myself into a knot: A binary operation on a set $S$ is a function from $S\times S\to S$, making it a relation on $S\times S\times S$. I've recently read somewhere that relation composition is associative, but I know that there are binary operations that are not associative, such as division, cross product, and set difference. What precisely am I missing? Is what I read about relation composition being associative incorrect?
You might be thinking about (a-b)-c vs a-(b-c) But if subtraction is a composition here then, you actually have a+(-b-c) vs (a-b)-c which are the same
and for cross product axbxc you have a(xb(xc)) vs (a(xb))xc which are the same also. In particular, the domain of axbx and bx should agree with that of c
So, take a binary operation $\cdot$. My brain is telling me that I should be able to write the expression $a\cdot(b\cdot c)=(a\cdot b)\cdot c$ in terms of composition of relations.
Right?
See, this is where the "thinking myself into a knot" comes into play. I keep confusing myself whenever I think about this
I'm trying to piece together what the composition of relations would look like in this case, but I'm not sure if it makes sense
Both $a\cdot (b\cdot c)$ and $(a\cdot b)\cdot c$ would be relations on $S\times S\times S\times S\times S$
I guess these would end up being two different relations on that set, but how would they necessarily differ?
C composed by B to give BC and then composed by A to give A(BC)
A composed by B to give (AB) and then composed by C to give (AB)C
However, B's domain already has C's image, so there is no difference in ordering (since whatever goes into A include C as well as the image of C is a subset of B)
hence the relation composition do not lead to nonassociativity in the final image
I am currently working on a problem from my operator theory course and the problem reduces to calculating the following integral over the unit disk $\Bbb{D}$: $$\int_{\Bbb{D}} \frac{(1-|w|^2)}{\overline{w} (1-z \overline{w})^{2+ \alpha}} dA(w),$$ where $dA$ is the normalized area measure on $\Bbb{D}$.
The problem doesn't state what the integral should evaluate to, but apparently it has a nice form...but I don't see anything at the moment.
By the way, $z \in \Bbb{D}$ is fixed.
I tried polar coordinates, but that didn't seem helpful.
So, my thoughts are that the fact that I can get square sine and cosine only strengthen my conviction that I should be able to find sine and cosine for a circle. Pretty fun stuff: desmos.com/calculator/djkxiyi0tm
Ancient Indian mathematicians used to write math in the form of poetry. Major results were usually condensed into a shloka, which is a verse consisting of two bars of 16 syllables each.
I'm just not used to the topological arguments but that'll get better over the course of the seminar
Whilst making love a necklace broke. A row of pearls mislaid. One sixth fell to the floor. One fifth upon the bed. The young woman saved one third of them. One tenth were caught by her lover. If six pearls remained upon the string How many pearls were there altogether?
Also, not quite sure why, but in the case of this square sine and cosine, 2 on the opposite side is a limit, and if I add more than two decimal places of 9s, desmos freaks out.
Built a nonassociative division by zero algebra. The bad news: Most of the undesirable features from the associative cases that is worked on back in 2015 are still there mostly, in particular you can only have x+x=x for any x The good news: Nonassociativity allows a lot of nice things to be retained, and can narrowly avoid having any null semigroups
Langlands said in his interview after receiving the Abel prize that he didn't really know why people describe mathematics as beautiful and then proceeded to avoid the question
sometimes I wonder too what exactly I find beautiful because its actually hard to come uo with examples which dont seem totally dry even for other mathematicians
I think you realize something in Mathematics is beautiful not when you're in the process of doing it, but when you sit back and look where you started at and where you ended up with.
What I mean to say that (personal opinion) every step that you do should seem like the most obvious conclusion, but those most obvious conclusions take you some place different.
math is at its most beautiful when I wave my hands around like a maniac while trying to forcibly convince an innocent bystander that category theory is the answer to everything
I look back and see that, in primary school, I used arithmetic synonymously with addition and subtraction, and that I now use it synonymously with classifying extensions of number fields
could I have a hint for the following problem? Show that if T is locally compact hausdorff, and H \subset T is locally compact, then H = A \ B for some A and B closed in T
There's a very tiny range of values for which I can cycle through that makes the error function zero at specific points at a non-linear rate. Any ideas?
@EdwardEvans there's a papers such as "Generators of quantum stochastic flows and cyclic cohomology" and thats just so many buzzwords and I don't have a clue what that is
@AMDG If you would give us an idea of what you are doing, rather than asking for ways to compute functions with almost no error and not using series or recursion, then someone might be able to help.
The green plot is an error function, and you can see that I've found an approximate maximum value of a_1 (see a_max) that between some minimum and maximum value of a_1, the value is zero.
Uh, not really, or at least I don't think so... also exponentials can be trivially computed (with integers) using simple bitshifts.
given ln(x) = log_2(x)/log_2(e) ...
You just provide an approximation of log_2(e)
And then you can use exp to compute any exponential once you find a way to compute exp using one of its identities.
So it would be silly, really, to use series or recursions
Also I just learned you can use cosh(x) to compute any exponential just as well by using a constant of ln(N) as cosh(x*ln(N)) although I don't know the exact relationship between a^b and cosh(x * ln(N)) yet (would need to do a bit more digging).
Wait, an approximation of $\log_2(e)$? How does that fit in with "no error" Each approximation will have to be made more accurate. That seems like using more terms in a series, and I though that was what you were getting away from.
log_2(e) is a constant. That form log_2(x)/log_2(e) is exactly the sort of form in principle that I'm looking for with regards to circular functions.
If I want to increase the precision, I just provide a more accurate approximation of the constant log_2(e) :)
I don't have to add more terms
e is transcendental, so there'll never be an exact form apart from using the more efficient series definition of e as the limit of sum k=0,infinity 1/k!
$\log_2(x)$ is not a constant, and even in binary, you need a series or recursion or something similar to compute it. Even with $\log_2(e)$, I thought you wanted something that you just needed to plug in a bigger $n$ and it would give more precise approximation without series or recursion or the like.
Even thought it is a constant, $\log_2(e)$ to an arbitrary precision is a task that will probably require something that I believe you were trying to avoid.
I hope you find what you're looking for, but as far as I've seen, I don't see how it's possible.
I honestly don't get your point, and you can't assume that I'm using floating point math either. log_2(x) is trivially implemented as N - lzcnt(x) (count leading zeros) where N is the number of bits used to represent the integer x. I won't be using floating point as I'll be making my own rational number encoding that uses integer math only.
And I don't think you quite understand what I'm saying with regards to log_2(e) as a constant with arbitrary precision.
could I have a hint for the following problem? Show that if T is locally compact hausdorff, and H \subset T is locally compact, then H = A \ B for some A and B closed in T
@AMDG Well, you seem to know enough about my rational math package, and I don't seem to be getting where you're going, so I am going to bow out from this discussion.
I don't understand. I just want to know what methods of analysis that I don't necessarily know about might exist for me to be able to get my error down to zero for my cosine approximation displayed there in desmos.
@robjohn I don't know anything about your rational math package, but on the contrary, you can't assume that mine will be like yours. You don't know what my encoding will be, and therefore the behaviors and constraints of what it will be.
@sai-kartik Ha! it says "Server side rendering: KaTeX produces the same output regardless of browser or environment, so you can pre-render expressions using Node.js and send them as plain HTML." Obviously, that is not true.
@robjohn not to be disrespectful, but that just went over my head. Can you please point out where exactly is the problem and if you have any suggestions to fix it?
I just worry that AMDG is wasting time on his quest, but it's his quest, so I am trying not to interfere. I said my 200 cents worth and now I'm shutting the hell up.
@robjohn now I'm actually curious to know what your rational math package is like.
What I meant by "not FP" is that it isn't standard IEEE spec. My current idea to avoid ever growing spatial requirements for ratios is to use custom FP for numerators and denominators.
You'd just have some bits to specify where in the word the decimal place is. I could otherwise just use fixed point as well to simplify things even more.
@AMDG It started out as an arbitrary size integer math package (bigNum), then I added a rational type (bigFrac) and then built polynomials on top of that and specialized code for fixed point math using a common denominator. I also added strings for displaying things.
The only use I have for mathematica is complex number plots because nothing else really does that for some reason except for giving some more or less useless "2D pretty plots".
@AMDG I have lots of things I've written for Mathematica. The two most useful it turns out have been an Euler-Maclaurin Sum Formula implementation and a LaTeX rendering implementation.
