Suppose I have a nonempty simply connected space $X$ (which is not necessarily a CW Complex), does there exist a continuous map $f : X \to K(\pi_2(X), 2)$ such the $f_* = 1_{\pi_2(X)}$?
The answer is true if $X$ is a CW-Complex
But I am not sure if it holds if $X$ is not a CW-Complex
The reason being is that $X$ is the domain of some map, say $f$, and we turn $f$ into a fibration, and fibrations are well-defined up to homotopy equivalence and not weak homotopy equivalence I think
$H_n(X)$ is the homology of $H_{n+1}(X^{n+1}, X^n) \to H_n(X^n, X^{n-1}) \to H_{n-1}(X^{n-1}, X^{n-2})$
for this case we have $X/X^{n-1}$ instead
so the right hand side becomes $0$ and the middle is $\bigoplus_\alpha \Bbb Z$
oh $[S^n, X/X^{n-1}]$ should be equal to $[S^n,X^{n+1}/X^{n-1}]$ instead
and so they are the same, because they're just spelling out the presentations
or rather I should prove that $\pi_n(X^{n+1}/X^{n-1}) = \langle \alpha \mid f(\beta) \rangle$ where $\alpha$ indexes the $n$-cells and $\beta$ the $n+1$-cells
this fact is used in the construction of K(G,n)...
@gustaffIR suppose $P_1$ divides $\mathfrak p_1S$ as many times as $v$, then $P_1$ divides $IS$ as many times as $ve_1$
So, what propeties does a space need to have in order to ensure that every element has at least one square root? I imagine it needs to be complete, to start.
Suppose I have a power series with a term-wise index $n$, and my power series has factorial one higher than usual, so usually the taylor expansion reads $$ \sum_0^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$, now I have $(n+1)!$ instead of $n!$. Is there any way to know how the closed form expression changes?
A family of positive convex function $\Psi_{\alpha} : \mathbb{R}^N \to \mathbb{R}$ is defined as $$\Psi_{\alpha}(x) = A(x) + B(x) + \alpha D(x)$$ with paramter $\alpha \in (0,\infty)$. $A(x),B(x),C(x)$ are all positive,quadratic and convex functions. (edit), Also $A,B,D$ are linearly independent....
Am I misunderstanding something or is the Wikipedia article wrong?
Never mind. I calculated the derivative of the first polynomial wrongly. The pieces on [0,1] and [1,2] indeed join in value and first derivative, at t=1, in the first example.
Note that $\frac{d}{dx}(x-a)\frac{(x-a)^n}{(n+1)!} = \frac{(x-a)^n}{n!}$
So to turn your desired series into your old one, you'd multiply by $(x-a)$ and differentiate. Conversely, your desired series should be found by computing the antiderivative and dividing by $(x-a)$
So $f(x)\to \frac{1}{x-a}\int_a^x f(t)\,dt$ where $f(x)$ is your series. (the reason I'm integrating from $t=a$ to $x$ is to avoid a spurious integration constant)
So that makes that the new function will be the average value of $f(t)$ for $t\in [a,x]$. That's sorta cute.
@Semiclassical Can you look at my question? It seems like, in the first example, the derivative of the 2nd and 3rd polynomials of S have the same value at t=2, so this would be a C1 smooth function
But, in that section of the article, they say that the 2nd and 3rd polynomials only join in value and not in the first derivative
Never mind. I again made a wrong calculation. In fact, $P_1'(t) = 2$ and $P_2'(t) = -1 + 2t$, so $P_1'(2) = 2$ and $P_2'(2) = 3$, so it is not C1
Hi all, I have somewhat of a basic question. Consider of a seqence $\{f_n\}$ such that there is a $g$ with $|f_n| < g$. Suppose $f_n \rightarrow f$ pointwise. Will it follow that $|f| < g$? And how to prove this?
I was thinking this: $|f| = |\lim f_n| = \lim |f_n| < \lim g = g$ but im not sure
Text below copied from here
The Blue-Eyed Islander problem is one of my favorites. You can read
about it here on Terry Tao's website, along with some discussion.
I'll copy the problem here as well.
There is an island upon which a tribe resides. The tribe consists of
1000 people, w...
So effectively I can resum and get to $$ \sum_0^\infty f^{(n+2)}(a) * (x-a)^{n+2} / (n+2)!$$, so this would imply I could factor out an $(x-a)^2$, redefine $g\equiv f^{(2)}$ then do two integrations, it would imply I get back f(x), but that doesn't make sense to me :/
What do you guys do when you sort of "hit a wall" while thinking about a specific problem.
Do you try for a few hours and then check the solution or you carry on thinking?
This has been bugging me for quite a while. I can't think fast enough and once my thinking stalls and I can't proceed any further no matter how differently I try. This has been bugging me for quite a while.
Linear algebra question with an obvious answer (I think): Is every inner product on $\mathbb{R}^n$ of the form $\langle x,y\rangle=x^\top A y$ for positive definite A?
Any idea how that goes over in infinite dimensions? My knowledge of Fourier series makes me inclined to say it does, but presumably there’s some subtleties I’m forgetting
@Semiclassical I realize that such a number must always either end in 3 or 5 and any number ending in 5 (besides 5) cannot be prime, but I'm not sure how to prove that any number ending in 3 cannot be prime either
@TedShifrin Trying to figure out why 4^a - 1 can never be prime. I wrote out the first few numbers and realize their digits always add up to a number that's divisible by 3, therefore they're all divisible by 3.
one thing that generalizes some part of this is the following lemma I learned in functional analysis. Let $(V,\langle -,-\rangle)$ be a real Hilbert space and let $s:V \times V \to \Bbb R$ be a bounded bilinear form in the sense that $|s(v,w)| \leq C \|v\|\|w\|$ for some $C$ independent of $v$ and $w$, then there is a unique bounded operator $A:V \to V$ such that $s(v,w)=\langle v,Aw\rangle$ If $s$ is actually an inner product, then we get that $A$ is orthogonal and positive definite
DogAteMy: I would probably have stuck you in something more advanced, but you didn't really learn everything in my book thoroughly. What book are they using now?
Text below copied from here
The Blue-Eyed Islander problem is one of my favorites. You can read
about it here on Terry Tao's website, along with some discussion.
I'll copy the problem here as well.
There is an island upon which a tribe resides. The tribe consists of
1000 people, w...
