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12:05 AM
12:33 AM
If $X$ is a Hausdorff space, so is $X\times X$. Is the converse true?
What do you think?
I'm really new to topology. I'm guessing the answer is yes, but I'm trying to prove it
Here’s a hint: Is a subspace of a Hausdorff space Hausdorff?
I thought of that, but $X$ isn't really a subspace of $X\times X$, right? it's not a subset.
There are many ways you can make it be :)
12:40 AM
I am allowed to fix an arbitrary element, say $X\times x$, and treat this space like $X$?
They’re homeomorphic. Prove it :)
Thank you!
You’re welcome.
12:55 AM
Given a sequence $(a_n)_{n\in\mathbb{N}}$, if I prove that for a fixed $k_0 \in \mathbb{N}$ it is $a_{n-k_0} \ge 0$ for each $n \in \mathbb{N}$ can I deduce that $a_n \ge 0$ for each $n\in\mathbb{N}$? I think that this is true, because if $a_{n-k_0}$ for each $n \in \mathbb{N}$ then given an arbitrary $i\in\mathbb{N}$ this holds in particular for $i+k_0 \in \mathbb{N}$; that is, $0 \le a_{i+k_0-k_0}=a_i$ for each $i\in\mathbb{N}$. Is this a correct proof?
1:06 AM
It doesn’t make sense, actually. $n-k_0$ can be negative, and then the term isn’t defined.
But otherwise, yes, $i=(i+k_0)-k_0$ is correct.
What was the actual exercise you were given?
Does anyone know how to prove, Taylor's Theorem with Lagrange's form of remainder using Cauchy Mean Value Theorem?
I don't know, how to begin...
Look at Spivak later editions. I had him put that proof in there.
@TedShifrin You talkin 'bout this : Michael Spivak - Calculus.pdf - The Swiss Bay theswissbay.ch/pdf/Gentoomen%20Library/Maths/Calculus/… ?
Or this : Calculus Spivak : Free Download, Borrow, and Streaming archive.org/details/CalculusSpivak ?
No I don't find them here. Maybe these are old editions.
Needs to be 4th edition, I think. Possibly 3rd.
You just apply Cauchy repeatedly. It’s completely intuitive.
1:22 AM
@TedShifrin The problem is constructing the two functions with satisfies the conditions of Cauchy Mean Value Theorem.
The denominator is a power of $x-a$. The numerator is $f-P$.
1:34 AM
@TedShifrin actually, it was just a thought of mine; no exercise. You are right, I should have assumed that $n-k_0$ is nonnegative! So, assuming that, the reasoning works? Thank you.
1:50 AM
Yup, otherwise fine.
Topology is making me feel really dumb
2:08 AM
@TedShifrin Thanks! I was able to solve it.
2:27 AM
I dont get why the change in condition of this theorem is valid?
2:44 AM
@DavidRaveh You’ve had some abstract algebra and analysis already?
@X4J $$a_k=\frac1{(n+1)^24^n}\quad\text{where $4^n\le k\lt4^{n+1}\quad$and $n\ge0$}$$
I'm taking all three simultaneously. But the other courses are much more intuitive for me so far
Bad idea. Do topology later. I advised students for close to 40 years. You need plenty of experience and topology generalizes what you learn in analysis. Take it later.
Did your faculty adviser approve this?
I'm really a physics major; I've never officially met with a math advisor
@Thomas Try writing the result with multiplications, not divisions.
2:48 AM
But other math majors are taking all 3 simultaneously
Bad idea. None of my students ever did that.
I’m done making comments.
I'm just really interested in these abstract maths, and I'm not sure if I will get the chance in grad school
2 hours later…
4:29 AM
I recently wrote a Taylor Series Expansion of $e^{a\arcsin x}$ but then a question says to find the Taylor expansion of $\arcsin x$ from the expansion of $e^{a\arcsin x}$ and I have no idea on how to do it?
@SoumikMukherjee and @leslietownes Can you please help me with this? I am stuck with this problem badly.
4:54 AM
@Jakobian I have to think about this. Thank you for your effort.
@ThomasFinley Differentiate with respect to ……
5:27 AM
Does anyone have any idea about how to find closed form of this integral $\int_0^\infty \frac{\ln(1+x^2)}{e^x - 1} dx$ ?
@leslietownes yeah but I don't want to see the solution directly :(
Q: Evaluating $\int_0^\infty\frac{\ln(1+x^2)}{e^{\pi x}-1}dx$

Kemono ChenI want to evaluate $$\int_0^\infty\frac{\ln(1+x^2)}{e^{\pi x}-1}\,\mathrm dx$$ I tried to let $$I(a)=\int_0^\infty\frac{\ln(1+a^2x^2)}{e^{\pi x}-1}\,\mathrm dx,$$ and then $$ \begin{align} I'(a)&=\int_0^\infty\frac{2ax^2}{(1+a^2x^2)(e^{\pi x}-1)}\,\mathrm dx\\[25pt] &=2a\int_0^\infty\mat...

none of these are complete solutions
@TedShifrin Differentiate what?
all purpose no-spoiler ideas include 'try residue methods over some cleverly chosen something' 'try introducing a parameter and differentiating with respect to it' etc.
it seems to be difficult, or the people who regularly jump on these things would have jumped on the first linked problem above
5:33 AM
Leslie, I have one question which is "Is there anything bad in seeing a solution? Or when to give up on a problem? Cuz sometimes when I see solution "I often feel like oh gosh I was almost there.
@leslietownes yeah!
lucky i dunno, with something like "finding a 'closed form' for [some random definite integral]" i don't see what would be spoiled, exactly, by seeing a fully worked out answer. computing nice forms for definite integrals is a gigantic bag of tricks, and you add to your bag by looking at what others have done.
i see concern about spoilers arising, if at all, in a situation where maybe seeing routine things worked out by other people prevents you from ever bothering to learn basic techniques yourself. but this is deep enough in the weeds that i don't see that popping up
in the interest of complete honesty i also don't know why anyone gives an ess aych eye tee about definite integrals like that
so learn 'em, don't learn 'em, do 'em, don't do 'em, it's all kinda the same to me
@leslietownes yeah! You are right. But I don't know why "my mind doesn't allow me to see the solution." I think I should solve it by myself. When I have full worked out solution I am allowed to see what other people have done. lol,
@leslietownes Yeah,
i guess part of my background apathy is, i regard this subject as one in which clever people can come up with simple-looking problems of essentially arbitrarily high "difficulty" (however you measure it)
so i guess i see it as mostly just a puzzle thing, and you get to make your own rules. i wouldn't look at partial solutions when i work out a crossword puzzle, something like that. not good or bad, just how i puzzle.
@leslietownes great, so do you also see solutions? Are you a professor or anything like that?
i also wouldn't generally expect there to be a "nice" closed form, but that's another thing
the people who do those problems on MSE will be like "oh wow! here is a nice closed form for your integral it is twice the octo-logarithm of the apery constant plus 5 euler's gamma" and i wouldn't regard that as "nice", but they do
@LuckyChouhan i taught for a while but certainly not this kind of stuff
i wonder how often people have taught topics courses where you just do definite integrals all day. not for any contest, just to do them
5:43 AM
@leslietownes people like using fancy symbols. It gives them feeling that they are doing some high level math :)
@LuckyChouhan Nowadays he sells lesliecoins.
@leslietownes yeah! I will see the solution but with some questions in mind.
prithu: we "mint" them. and it's not me who mints them, it's really the community minting them together
@leslietownes we need teachers like you.
@TedShifrin when you can't solve a math problem. When do you see solutions after that? I mean what is the thinking process going on in your mind.
@leslietownes I had an obsession of 'minting' fake money for myself using a pen and a lot of A4 paper.
5:48 AM
When stuck on advanced/research problems, there are no solutions to look at.
@TedShifrin Exactly! So what do you do in that situation?
Keep trying. Try to think of other approaches. See if there is any literature on related matters.
I'm curious how Thomas was supposed to "know"/"find" the Taylor expansion of $e^{a\arcsin x}$.
So basically, READ AND READ A LOT!
"I recently wrote a Taylor expansion ..." Hmm. You mean someone gave it to you?
@TedShifrin yeah! I would write it as \sum_{n=0}^\infty \frac{(a\arcsin{x})^n}{n!} Now I am too lazy to do long long differentiation
@TedShifrin Please send here...
5:53 AM
ted: you can get a fairly simple recursion out a differential equation it satisfies. the top voted answer in math.stackexchange.com/questions/4157490/… works it out.
Seems odd that Thomas could make his way through that and has to ask me "differentiate what?"
I wouldn't think of that solution to the original question, unless someone told me what to do.
well if it's anything like the others of these, it's probably on page 123 from some textbook from 1880 and the author does something similar with another problem on page 122.
that's often how these monsters are created.
Yes, another 1880 textbook.
Manifolds question: is it true that if $M$ is an $n$-manifold and $F\subseteq M$ is a finite set, then there is an open $U\subseteq M$ with $F\subseteq U$ and $U\cong\Bbb R^n$?
You guys are too funny
5:56 AM
Not the usual statement, Alessandro, but if $\dim M\ge 2$ the isotopy lemma says you can isotope the manifold and move the finite set anywhere you want.
Ah of course, manifolds are $n$-transitive
The proof I like is just a connectedness argument along with flowing by a vector field.
I confused myself by thinking about manifolds with boundary and then missed that I knew the answer to my own question after reducing it to one about manifolds without boundaries!
Oh, yeah, I haven't thought about what's true with a boundary.
If the points are not on the boundary all is good
6:00 AM
LOL, so remove the boundary in the first place :P
Because $M\setminus\partial M$ is an honest manifold, and you can move finitely many points wherever you want with a map that becomes the identity at infinity
Oh, yeah, I should have said the isotopy was compactly supported.
@TedShifrin well the points are not on the boundary in the case I'm interested in :P
Actually @Ted the situation I'm working with is the following: I have a closed disk $D'$ from which I've removed the interior of finitely many pairwise disjoint closed disks $C_i$ to obtain a manifold with boundary $D$. I have a homeomorphism $h\colon\bigcup\partial C_i\to\bigcup\partial C_i$ so that, for every $j$, $h\upharpoonright C_j$ is either orientation preserving or orientation reversing. I want to extend it to $D$. Now @Astyx swears it can be done, but I'd like a reference
Do you happen to have one?
@AkivaWeinberger Hi. Can you help me with this problem:
18 hours ago, by Prithu Biswas
@SoumikMukherjee Is there a way to prove that "Every subspace of a finite dimensional vector space is finite dimensional" using induction?
6:20 AM
A: Total space of vector bundle deformation retracts onto 0-section of base space

Andrew D. HwangScalar multiplication $v \mapsto (1 - t)v$ commutes with arbitrary linear transformations, in particular with transition functions. Consequently, if $E$ denotes the total space of your vector bundle, $x$ denotes a local coordinate on the base, and $v$ denotes a local coordinate in the fibres, th...

what is local coordinate?
@PrithuBiswas Have you seen the answer sketch by Jackobian?
@SoumikMukherjee Yea. He himself said that it might be wrong.
I also don't follow some of the lines.
6:32 AM
Suppose that is true for all vector spaces of dimension $<n$, now take an $n$ dimensional vector space $V$ and consider a subspace $V_1$ of $V$. Now $V_1$ is contained in some maximal subspace $V_2$. Apply the hypothesis on $V_2$
But I am not sure whether this qualifies as a proof by induction, also I don't think everything can be shown by induction
@SoumikMukherjee What is a maximal subspace?
A subspace that is not contained in any other subspace except the whole space
7:20 AM
@SoumikMukherjee How do you prove that V1 is contained in some maximal subspace V2.
Sorry if this is something trivial. Axler doesn't seem to talk about maximal subspaces =(
7:39 AM
Suppose there is some element $x$ that is not in $V_1$, adjoin it to $V_1$. If the resulting subspace is the total space then $V_1$ is itself a maximal subspace. If the resulting subspace is maximal then you are done, otherwise keep adjoining elements. This process should terminate after finite steps as the whole space is finite dimensional
8:03 AM
@SoumikMukherjee but doesn't showing that the process terminates after finitely many steps assume that a subspace of a finite dimensional space is finite dimensional, making the argument circular?
what does 'dimension' mean? it can certainly be circular if that core notion isn't pinned down in the right way. a common way of confronting the key issue is to prove that if a space is spanned by even a single list of m vectors, then any list of more than m vectors in that space is linearly dependent.
and you build your definition of dimension somehow around that.
if that happens to be what axler does, the question of whether an 'inductive' proof exists is maybe an objection to this idea of proofs that work by inductive constructions that produce increasingly long lists of vectors, instead of inductive constructions that reduce somehting about dimension n to something about dimension n-1.
if for example dim V := max {|S|: S is a linearly independent subset of V} it is trivial that a subspace of a finite dimensional space is finite dimensional and it is not something that would naturally lend itself to a proof by induction. but you still need probably need all of those lemmas about lists of vectors, removing things from linearly dependent spanning sets without changing the span, etc. to operationalize the definition.
Axler did something similar to Asaf.
A: Can we prove without the axiom of choice that subspaces of finite dimensional vector spaces are finite dimensional as well?

Asaf KaragilaThe usual proof (which you mentioned) doesn't use the axiom of choice at all. You can prove this directly using induction. Pick some non-zero $w_1\in F$. Suppose that we chose $w_1,\ldots,w_n$ and $F_n$ is their span. If $F_n=F$ then $F$ is finite dimensional. Otherwise, $F_n$ is a proper sub...

Hi, does it the function bounded in L^2([a,b]) implies function bounded in L^infty([a,b])?
Something about "You don't need AOC because you are only making finitely many choices". So thought maybe the algorithm can be turned into some kind of inductive or recursive argument.
But I don't really know how to do it =/
8:22 AM
AP: what does 'bounded in X' mean here, where X is a normed space? it isn't necessarily true, for example, that if a function has a finite L^2 norm, then its L^infty norm also has to be finite. e.g. 1/x on [0,1] has finite L^2 norm but is not in L^infty.
Oh, that is a nice comment. Just I was writing bounded in X as, there exists M such that ||f||<M with ||..|| norm defined on X.
I was trying show that f_n=sin^(2)(k^nt) is bounded in C[0,2pi] with the norm uniform with k natural
Then I was trying majoree it as like as ||f_n||_infty<=||f_n|| <=M with some norm in L^p but then I suppose it does not work
because L^infty is strictly contained in L^p then we have ||f||_p<=||f||_infty but not >=
Since sin(…) is bounded is it sufficient to write: |f_n|<=1 implies ||f_n||_infty <=1 and so bounded (f_n)?
prithu: if V has a linearly independent spanning set {v_1, ..., v_n} and w_1, ..., w_{n+1} are n+1 vectors in V, by writing each w_j as a linear combination of the v's and using fairly concrete scalar math (that a homogeneous system of n+1 equations in n unknowns has a nonzero solution, findable by gaussian elimination and such) you can expressly write one of the w's as a linear combination of the others.
and that's enough to rule out [any subspace of] V containing n+1 linearly independent vectors, let alone arbitrarily long lists of linearly independent vectors.
i don't know that you can remove the difficulty of, in general, constructively doing this with lists of vectors will involve the equivalent of matrix computation, and that if n is large this could be unwieldy to do by hand, or unwieldy numerically, or something. but it should be clear that you don't need some new principle of set theory to do it.
AP: yes, if |f_n(t)| <= 1 for all n and t then each f_n is in L^oo, and ||f_n||_oo <= 1
[i meant n equations in n+1 unknowns above, if that wasn't clear.]
8:44 AM
@leslietownes thank you!
8:54 AM
@leslietownes My brain feels foggy right now. But thank you so much for your response.
prithu, you might compare your feelings about the abstract vector space situation with your feelings about whether you really believe that something like row echelon form of a matrix actually exists in general, or has just been shown to exist in small examples
i don't mean that as a joke, i think it's fundamentally the same kind of indeterminacy where as a logical matter, of course it exists, but actually doing it (even in fairly concrete cases) is enough of a pain in the ass that you might wonder
9:07 AM
@AlessandroCodenotti Oh right, it is circular
@leslietownes I think this is the case, inductive construction instead of induction on the dimension
9:23 AM
Just a soft question: If $X$ is a metrizable space, then can we always find a complete metric that induces the same underlying topology on $X$?
i don't think so. there's certainly a separate adjective for such spaces ("completely metrizable"), and there probably wouldn't be if it were coextensive with metrizable
is Q completely metrizable?
[rationals with usual metric]
oh there's a notion of completely metrizable? I didn't know that.
Aug 22 at 16:32, by Sourav Ghosh
A subset of a complete metric space is completely metrizable iff it is a $G_{\delta}$ set.
jakobian also talks about this concept if you search the chat
Ah, Jacobian, of course.
9:55 AM
@onepotatotwopotato no, complete metric spaces without isolated points are uncountable by the Baire category theorem, so it fails for Q for example
10:24 AM
Is the area to the left of Y-axis negative?
I mean the signed area?
10:55 AM
If a function on average is greater than another over the same interval, this always implies that the integral of the bigger function is greater, The converse not true right? if the function had a jump discontinuity?
But what is both the functions are continuous without jump discontinuities , will the converse hold?
2 hours later…
12:29 PM
@TedShifrin No, I derived that using the differential equation $(1-x^2)y_2=xy_1+a^2y$
1:00 PM
Q: Deduce, the Taylor series expansion of $\arcsin x$ from the Taylor Series expansion of $e^{a\arcsin x}$

Thomas FinleyFind the Taylor Series expansion of $e^{a\arcsin x}$ and hence deduce, the Taylor series expansion of $\arcsin x$. I could find the Taylor Series expansion of $e^{a\arcsin x}$ as $$1+ax+\frac{a^2x^2}{2}+\frac{a^3+a}{6}x^3+\cdots $$ However, I have no idea how to deduce the series for $\arcsin x$ ...

Need some help with this.
1:39 PM
Q: List of textbooks on Abstract Algebra in the order of time

SensebeI am knowing Abstract Algebra things; I am searching aims of Abstract Algebra and origins of parts of Abstract Algebra. I thought original initial textbooks have explicit links to aims and origins of the parts of Abstract Algebra. Can we list major textbooks on Abstract Algebra in the order of ti...

2:04 PM
@ThomasFinley What is $\frac{\partial}{\partial a}e^{a\arcsin(x)}$ at $a=0$? apply that to the power series you've found.
If $G=GL_2(\Bbb{R})$, $H=\{g\in G: g_{21}=0\}$ and $X=\Bbb{R}\cup \{\infty\}$. How can I find a map $f:G/H\rightarrow X$ s.t. $f(gy)=gf(y)$ for $y\in G/H$? I'm a bit confused since $y=g'H$ for some $g'\in G$ but then the map should end into the real line. The beginning of the exercise we showed that $g\cdot x=\frac{ax+b}{cx+d}$ where $g$ is the matrix with entries $a,b,c,d$ is a group action so I think I need to do somthing with it
can someone help me?
3:01 PM
Rudin's Functional analysis is more readable than I expected. That's good.
maybe I should try RCA later.
The thing is that I don't know what would be a good time distribution between self-studying topics I'm interested in and coursework. I used to spend most of my time on the former but maybe I need to care about coursework too.
3:22 PM
Q: Find the values of $p$ and $q$ such that $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13.$ Assume, that L' Hospital's rule is applicable.

Thomas FinleyFind the values of $p$ and $q$ such that $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13.$ Assume, that L' Hospital's rule is applicable. I tried solving the problem as follows: Given, $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13$ and according to the given information, that L'...

I need a little help with this.
If $G$ is a topological group and $H$ a open subgroup, why does the qotient topology of $G/H$ is equal to the discrete topology
its points are all open
@Thorgott I mean that $\tau_{G/H}\subset \tau_{d}$ where $\tau_{d}$ is the discrete topology is clear. But I mean If I take $U\in \tau_{d}$ , i.e. $U$ is any subset of $G$. Then to show that it is open in $G/H$ we need to show that $\pi^{-1}(U)$ is open in $G$ right?
3:38 PM
sure, you can also do it like that
But I mean can I immediately say that $\pi^{-1}(U)$ is open in $G$ since $G$ has the discrete topology
3:51 PM
@Thorgott Because it seems a bit strange to me since I know $H$ needs to be open and I don't used it
4:06 PM
since when does $G$ have the discrete topology? you didn't mention that as a hypothesis
A naive question: we know that the composition of two bijections is a bijection. Why should I not be surprised that one can form a bijection out of two functions, neither of which is a bijection?
This comes up in the following context:
**Lemma:** If a set $X$ injects into $\mathbb{N}$, then it's countable.

*Proof.* If $X$ finite we are done, so let $X$ be infinite. Let $f : X \to \mathbb{N}$ be the relevant injection injection.

Being a subset of $\mathbb{N}$, the set $f(X) = \{ f(x)\,:\, x \in X \}$ is well-ordered, and in particular it has a least element, a 2nd-least element, a 3rd-least element, and so on. Define $g : f(X) \to \mathbb{N}$ by sending the $n^{\text{th}}$-least element of $f(X)$ to $n \in \mathbb{N}$. Then the composite $g \circ f : X \to f(X) \to \mathbb{N}$ is i
None of them might not be bijections but I assume at least the first one has to be an injective function
@robjohn Thanks, I've just seen it tho
The somewhat confusing bit is that $h:= g \circ f$ is a bijection, but $f$ is not. I think $g$ may be a bijection to by its definition. So my confusion is that (1) we get a bijection out of an injection and a bijection (i guess this is OK?) and (2) if $f$ is not a bijection, we can still write $f = g^{-1} \circ g \circ f = g^{-1} \circ h$, which shows $f$ is a bijection? Where am I going wrong?
4:21 PM
Rob, do you know where I can find a bunch of advanced exercises at infinite series for real analysis 2? The university I study at doesn't provide much and most of what I see in books are ones that are very common
Not sure this is the flavor you're looking for, but Counterexamples in Analysis by Gelbaum could be of interest
Ah I missed exercises
4:38 PM
@Thorgott Ah no sorry it does not have the discrete topology. but how do I show that $\pi^{-1}(U)$ is open in $G$
@X4J The ratios you are looking at vary between $1$ and $1/4$, thus the limit does not exist.
But the series converges to $\frac{\pi^2}2$
@X4J I'm sorry, I don't know where you can find such a list.
4:59 PM
@X4J I found Calculus II - Series & Sequences on the web. There are a few problems (with solutions) for each different concept.
@robjohn Oh this is exactly what I was looking for, thanks
Great. I just Googled "problems in infinite series"
5:30 PM
@user123234 try following my hint, do it for $U$ a point
Hi, loosely speaking, if I had a ball with a dot on it, then deciding which direction that dot should point could be thought of as choosing an element of SO(3), right?
5:47 PM
cool, thank you
6:47 PM
Is it true that a vector-valued function is Lipschitz continuous if and only if all of its components are? If so, is there any formula for the Lipschitz constant of the function over a given set in terms of the Lipschitz constants of the components?
If the above is true, it seems like it should be a proposition/theorem in every other lecture notes on Lipschitz continuity, but I haven't found this in any sources I've been looking at.
7:02 PM
Actually, my question is a bit strange, since the Lipschitz constant is not unique, but I was wondering if there is a formula, say $L=L_1+L_2$, such that $L$ is a Lipschitz constant of the function when $L_1$ and $L_2$ are for the respective components.
@sunny yeah, its true
but there should be no such formula
cool, and shame at the same time
you can compare the $\ell^1$ and $\ell^2$ norm in $\mathbb{R}^n$ for example
7:43 PM
Consider $F(x)=(x,\sqrt{1+x^2})$. Is this function globally Lipschitz on $\mathbb R$? Using the above result, its components are Lipschitz continuous for any $x,y\in\mathbb R$ and hence the function is, however, I can not determine whether this is global or local Lipschitz continuity.
8:01 PM
@Novice No. An element of $SO(3)$ tells you where an orthonormal basis maps, not just a single vector.
Neat :)
No, what I said is wrong. You’re using the usual Euclidean norm on the target?
If so, then it’ll be the square root of the sums of the squares.
Just write down the obvious formula.
@TedShifrin Hey Ted. Couple of days ago, you said to me that if a sequence na_n approaches zero at infinity then it roughly means the terms decay 1\n faster. I think I did not understand it, do you know what can help me to intuitively grasp it?
8:16 PM
I meant by comparison with having just $\sum a_n$ converge. I’m just saying that $a_n = (na_n)/n$.
Ah I see
In order for $na_n$ to converge, the $a_n$ must be smaller than without the $n$, right?
@TedShifrin $$\begin{align}\left \lVert f_1(x)\mathbf{e_1}+f_2(x)\mathbf{e_2}-f_1(y)\mathbf{e_1}-f_2(y)\mathbf{e_2} \right \rVert &\leq \left \lVert f_1(x)-f_1(y) \right \rVert +\left \lVert f_2(x)-f_2(y) \right \rVert \\ &\leq L_1\left \lVert x-y \right \lVert+L_2\left \lVert x-y \right \rVert \\ &\leq (L_1+L_2)\left \lVert x-y \right \rVert\end{align}$$
Use the actual definition of the norm, not triangle inequality?
But that’s OK. Just what I said is a “better” constant.
:) ok
8:23 PM
spilled soup all over my paper
I'm not doing the homework again. Made me lose all my motivation
That’ll show them !
Exactly. No one messes with Hades
@TedShifrin yeah I see, thanks
8:29 PM
@TedShifrin Thank you, so I guess for my ball, I would be looking at subgroup of $\text{SO}(3)$ (I think)
Why is it a group?
Can you explain exactly what you’re doing?
There is something called reinforcement learning wherein an "agent" must take an "action" at every discrete time step. Decisions for how to take an action are made via function that takes in an element of the state space and gives a probability distribution over action space. If an agent is controlling, say, a camera and has to decide where to look, then I thought this might connect with rotation groups (and ultimately, probability on groups).
But as you might be able to tell, I am a little over my skis
(for extra clarity: the "action space" would be a group, e.g. $\textit{SO}(3)$, although I'm apparently wrong about that)
Well, the group is the group of isometries of the unit sphere, but there’s a whole circle worth of choices of group element that take the north pole to any given point of the unit sphere. That’s a lot of ambiguity. But maybe you’re ok with that.
I don’t see why you don’t just use the unit sphere itself.
8:46 PM
Well, the idea is in very early gestation and might well be nonsense or merely not useful.
I just thought that there might be a connection with group theory, but I guess I have to think about it more, and learn more group theory too, when I find the time. Thanks for your help
9:32 PM
@Thorgott how to show that $CP^n\times S^{2m+1}$ and $CP^m\times S^{2n+1}$ are not homotopic?
m is not equal to n.
Looking it up made me learn about Kunneth formula.
But I don't want to use that.
and what is the meaning of a cw pair (X,A) of dimension n?
10:23 PM
nvm, I found it.
10:40 PM
musing aloud: I guess if I forgot about the groups and just thought of the action space as the unit sphere, then I would have a probability distribution over the sphere (a manifold)
@Novice That makes good sense to me. :)
And you know standard coordinates on the sphere easily ….
Q: Elementary inequality related to twin prime conjecture.

MathCrackExchangeDefine $m = \sum_{d \mid p_n\#} \frac{(-1)^{\omega(d)}2^{\omega(d) -(2\mid d)}}{d}$. Where $(2\mid d) = 1 \iff 2 \mid d, \text{ else } 0$ is shorthand. Conjecture. $$ m(p_{n + 2}^2 - p_{n + 1}^2) \geq \log_{p_{n + 1}}(p_{n + 2}^2 - 2) $$ for all $n \geq 1$. Is it true? Is there a way to prove t...

It's an inequality involving that slope $m$ of a line that lower bounds the twin prime average counter
and prime numbers themselve
11:21 PM
When the math just ain't mathing anymore
11:58 PM
Still spilling soup?

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