Do $\alpha: [0,2\pi]\rightarrow \mathbb{R}^2, \alpha(t) = (\cos(t), \sin(t))$ and $β:[0,4π]→\mathbb{R}^2,β(t)=(cos(t),sin(t))$ represent the same curve?
Let $(X_n; n \geq 1)$ be a collection of independent positive identically distributed random variables, with density $f(x)$. They are inspected in order from $n = 1$. An observer conjectures that $X_1$ will be greater than all the subsequent $X_n, n \geq 2$. Show that this conjecture will be proved wrong with probability 1.
Let $f : M \to N$ be a smooth map. Think of $C^\infty(M, N)$ as an infinite dimensional manifold; then $T_f C^\infty(M, N)$ is the space of sections of $f^* TN$ over $M$
@LeakyNun Ideas: It seems suspect that $\text{Homeo}(\Bbb R^n)$ even has finite homological dimension for high $n$ to me. $H_k(\text{Homeo}(\Bbb R^n))$ can be recovered as maps from $k$-stratifolds $X \to \text{Homeo}(\Bbb R^n)$ upto $(k+1)$-stratifold cobordism.
A map $X \to \text{Homeo}(\Bbb R^n)$ gives rise to a map $\Sigma X \to B\text{Homeo}(\Bbb R^n)$ and since $X$ was a stratifold, so is $\Sigma X$. Cobordism classes of maps from $k$-stratifolds to $B\text{Homeo}(\Bbb R^n)$ correspond to topological $n$-plane bundles on $k$-stratifold upto concordance, and we're asking if this is trivial for large $k$.
On the other hand, why should this be different for $\mathrm{Diffeo}(\Bbb R^n)$, which we know has homotopy type of a manifold, $O(n)$? For high $k$, is it easy to see rank $n$ vector bundles become trivial upto concordance over $k$-stratifolds?
Maybe this is because of stable triviality of vector bundles.
That has to be it
So we want examples of topological $n$-plane bundles which are not "stably trivial" - what would that mean?
$E \to X$ be a topological $n$-plane bundle. Define $E \oplus \Bbb R$ to be the topological $(n+1)$-plane bundle given by defining the transition functions $\varphi_{ij}' : U_{ij} \to \text{Homeo}(\Bbb R^{n+1})$ from the original transition functions $\varphi_{ij} : U_{ij} \to \text{Homeo}(\Bbb R^n)$ by using the natural inclusion $\text{Homeo}(\Bbb R^n) \to \text{Homeo}(\Bbb R^{n+1})$, taking product of a self-homeo of $\Bbb R^n$ with the identity map on $\Bbb R$.
What's an example of a topological $n$-plane bundle which doesn't become trivial in this process? There has to be one
@LeakyNun You might find the top answer by Mike here useful, by the way. Gives examples of non-isomorphic vector bundles which are isomorphic as topological plane bundles.
I forgot I asked that question a long time ago lol
Also the stable triviality thing is garbage, there are vector bundles which are not stably trivial. There is always a complement, but that doesn't help; there are complements for topological plane bundles as well
I have no clue how to see that rank $n$ vector bundles on $m$-manifolds are all equivalent upto concordance, if $m$ is very large.
Maybe I am confused
I suppose because you can "localize" the locus of nontriviality to some small subset of the manifold, which you can surger out and replace by something on which the bundle trivializes
Maybe
Not exactly sure where I should be using the property of being a vector bundle
$\pi_1 : E_1 \to B_1$ and $\pi_2 : E_2 \to B_2$ be rank $k$ bundles on $n$-manifolds. I call a concordance between them to be a rank $k$-bundle $\pi : E \to B$ where $B$ is a cobordism between $B_1$ and $B_2$ such that $\pi$ restricts to $\pi_1$ and $\pi_2$ on the two respective components of $\partial B = B_1 \sqcup B_2$
The argument was: H_n(Homeo(k)) <=> Cobordism classes of maps from n-stratifolds to Homeo(k) <=> Cobordism classes of maps from (n+1)-stratifolds to BHomeo(k) <=> Concordance classes of k-plane bundles on (n+1)-stratifolds, therefore if Homeo(k) has finite homological dimension concordance classes of k-plane bundles on m-stratifolds should be trivial for large m
Ok nevermind I am not an idiot (yet). I just don't see why this story is different from H_n(Diffeo(k)) = H_n(O(k))
If $v \neq 0$ is $(A-\alpha I)$-cyclic, with period $r$, then $\{ v, (A-\alpha I)v, \dots, (A-\alpha I)^{r-1}v\}$ is linearly independent. My proof consists of repeatedly composing by $A^{r-1}$, $A^{r}$... My textbook however writes the linear dependence relation as $f(A-\alpha I)v=0$ where $f$ is a polynomial, and straight out talks about the greatest common divisor of $f(t)$ and $g(t)=t^r$.
I'm not much of an abstract algebra nor arithmetic person so please tell me why would you or anyone else think like this?
I mean, it is useful to see that a $k$-linear map $T: V \to V$ where $V$ is a vector space over $k$ defines a $k[t]$-module structure on $V$ by having $t$ act as $T$
then we're just taking the annihilator of $v$, which is an ideal in $k[t]$
and $k[t]$ is a Euclidean domain hence PID, so it makes sense to consider greatest common divisor
@FuzzyPixelz It makes sense to do that, though, right? Write a linear dependence relation of $(A - \alpha I)^i v$ out for $0 \leq i \leq r-1$, and cancel coefficients one by one.
I would start with the last equation and apply $A - \alpha I$ to iteratively conclude every coefficient from the left is zero, instead of starting from the first and applying $(A - \alpha I)^{r-1}$ like you are doing
It's much easier (try it!)
@LeakyNun He doesn't need to know this right now, I think :)
Well, this particular linear independence trash is a manipulation trick. You can put it in the $k[A]$-module wrapper, but it doesn't actually help too much
But it does give more context to the whole setup, I agree
@LeakyNun $f : (M, x) \to (N, y)$ be a germ of a smooth map, and let $A$ be a finitely generated $C^\infty_x(M)$-module. Then $A$ is a finitely generated $C^\infty_y(N)$-module if and only if $A/f^* \mathfrak{m}_y A$ is a finite dimensional vector space over $\Bbb R$.
This is called the Malgrange preparation theorem. Full algebra
so you're saying that if I have a map $f: X \times \Bbb R^m \to X \times \Bbb R^n$ with constant rank then $\ker f = X \times \ker f_p$ for a given $p \in X$
let's say $X$ is connected
where $=$ means "homeomorphic" or something similar
in words this means the kernel of a family of maps varies smoothly
so let's even say $m=2$ and $n=1$
then $f_p: \Bbb R^2 \to \Bbb R^1$ is taking dot product with some vector $v_p$
Take $TS^2$. $TS^2 \oplus \underline{\Bbb R} = \underline{\Bbb R}^3$, so $TS^2$ is kernel of a map $\underline{\Bbb R}^3 \to \underline{\Bbb R}$ of bundles of constant rank. Not globally trivial
$\pi_1 : E_1 \to B_1$ and $\pi_2 : E_2 \to B_2$ be rank $k$ bundles on $n$-manifolds. I call a concordance between them to be a rank $k$-bundle $\pi : E \to B$ where $B$ is a cobordism between $B_1$ and $B_2$ such that $\pi$ restricts to $\pi_1$ and $\pi_2$ on the two respective components of $\partial B = B_1 \sqcup B_2$
The argument was: H_n(Homeo(k)) <=> Cobordism classes of maps from n-stratifolds to Homeo(k) <=> Cobordism classes of maps from (n+1)-stratifolds to BHomeo(k) <=> Concordance classes of k-plane bundles on (n+1)-stratifolds, therefore if Homeo(k) has finite homological dimension concordance classes of k-plane bundles on m-stratifolds should be trivial for large m
In repeating confidence interval experiments, are we allowed to take samples of different size every time?
Because confidence interval of 95% means that if the sampling process is repeated infinite times, 95% of all the intervals obtained will have the parameter of interest.
So wrt this process of repeating infinite times - do we have to repeat with the same sample size?
I think I can reduce this to $\varphi: [0,1] \to \Bbb R^{n \times m}$ a constant-rank path, then we can define $\ker \varphi(0) \to \ker \varphi(1)$ by sending $v$ to...???
@LeakyNun Let $\Sigma_r$ be the subset of rank $r$ matrices of $M_{m \times n}(\Bbb R)$. Any matrix in $\Sigma_r$ has a distinguished minor $A_{r \times r}$ which is invertible; repositioning $A$ to be the top-left block, break the rest of the elements into blocks $B_{r \times (n-r)}$, $C_{(m-r) \times r}$ and $D_{(m-r) \times (n-r)}$. Then you can show $D = C A^{-1} B$, in which case the kernel is everything of the form $(I_{r \times (n-r)} - A^{-1} B)v$ where $v \in \Bbb R^{n-r}$.
For any matrix $M \in \Sigma_r$ you can choose a neighborhood of $M$ in which the position of the block $A$ is the same, so $A, B, C, D$ are globally defined on that neighborhood
In general if $E \to F$ is a bundle homomorphism of vector bundles, the kernel is something I would call a stratified bundle, i.e., you can find a stratification of the base manifold $M$ such that the kernel is locally trivial on each stratum
@MikeMiller Yeah you cooked up some proof of your own I recall
I think I just got used to this
@LeakyNun The annoying thing is that kernel of bundle homs escape the category of bundles and land into stratibundle but also kernel of stratibundle homs escape the category of stratibundles lol
It's a clusterfuck and Thom came up with the right condition to fix this
Thom, what a legend
Need to get back to reading singularity theory pray for me
@Thorgott Okay I'm finally feeling like I have a better grasp of this! I was just writing up a summary to iron everything out and I was considering the example where f(x,y) = sinx/x over the unbounded rectangle where x goes from 0 to infinity and y goes from 0 to 1. Now the double integral of the absolute value of sinx/x over that region is infinite, so technically Fubini's theorem shouldn't work. And yet, reversing the order of integration doesn't matter - both iterated integrals give pi/2.
Is this one of those cases where you would need to consider HK integrability instead?
@Thorgott Hmm actually Wolfram can't actually calculate the double integral of the absolute value of sinx/x from x = 0 to infinity and y = 0 to 1, so I don't actually know if it's finite or not. :/
Suppose that I have a polynomial of large degree with integer coefficients which I know is a product of linear and irreducible quadratic factors. I'm looking for a procedure to factor such a polynomial. The linear factors can be separated using the rational root theorem, but is there any way to find the quadratic factors?
Additionally, all of the factors are monic, which should simplify this problem.
@Thorgott Ok so I've now confirmed that the double integral of the absolute value of sinx/x from x = 0 to infinity and y = 0 to 1 does indeed diverge. So my earlier question still remains!
From Wikipedia https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test#Time_complexity it is not obvious to me how first step leads to second $= 110000_2 \mod{2^5−1}$ $= (10000_2 + 1_2) \mod{2^5−1}$
Can anyone tell me the definition of derivative other than that limit one? I’m searching for something like series, can we write $f’(x)$ as a sum of $f(x)$ , $x$ or something like that
This follows from the definition of the derivative:$$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)\right)$$ Or so I was told by Lucia at mathoverflow.
@LeakyNun Let $f : [0,1] \rightarrow \mathbb R$ be a continuous function on its domain and differentialable on $(0,1)$ with the property that $ f(0) = f(1) =0$. Prove that $f(x) = f’(x) $ for some $x$ belonging to $(0,1)$
Let $g(x) = e^{-x} f(x)$ which yields $g(0) = e^{-0} f(0) = 0 = e^{-1}f(1)$. $g$ is also continuous on $[0,1]$ and differentiable on $(0,1)$, so Rolle's Theorem applies. Hence there exists an $x \in (0,1)$ such that $g'(0) = 0$. This implies that $-e^{-x} f(x) + e^{-x} f'(x) = 0 \implies e^{-x} (f'(x) - f(x)) = 0$ and since $e^{-x} \neq 0$ for all $x \in \mathbb{R}$ then $f'(x) - f(x) = 0 \implies f'(x) = f(x)$ for that choice of $x$.
I'm trying to compute the contour integral
$$\frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \zeta^2(\omega) \frac{8^\omega}{\omega} \ d \omega$$
where $c > 1$, $\zeta(s)$ is the Riemann zeta function.
Using Perron's Formula and defining $D(x) = \sum_{k \leq x} \sigma_0(n)$, where $\sigma_0...
This is how I found it: $$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \cdot \zeta(c)}{\zeta(c + s - 1)}$$
@MikeMiller Consider the action of $\mathrm{Diff}(M) \times \mathrm{Diff}(N)$ on $C^\infty(M, N)$. If for a smooth map $f : M\to N$ the orbit of $f$ is open, we call $f$ structurally stable.
Thinking of everything as Frechet manifolds if necessary, take the derivative at identity of the map $\text{Diff}(M) \times \text{Diff}(N) \to C^\infty(M, N)$, $(g, h) \mapsto h \circ f \circ g$. The derivative is a map $\Gamma_M(TM) \times \Gamma_N(TN) = T_e(\text{Diff}(M) \times \text{Diff}(N)) \to T_f C^\infty(M, N) = \Gamma_M(f^* TN)$, which can be checked to be $(X, Y) \mapsto df(X) + Y \circ f$
If $f$ is stable, then for any 1-parameter variation $f_t$ of $f$, the section $s = \partial_t f_t$ of $f^*TN$ over $M$ can be written as $s = df(X) + Y \circ f$ for some vector fields $X, Y$ on $M, N$ respectively, because $f_t$ goes lies inside the orbit of $f$ for some time
The condition that $\Gamma_M(TM) \times \Gamma_N(TN) \to \Gamma_M(f^* TN)$, $(X, Y) \mapsto df(X) + Y \circ f$ is surjective is thus called infinitesimal stability. Stable maps are infinitisimally stable; the converse is true by a theorem of Mather.
Guillemin-Golubitsky mentions if implicit function theorem was true for Frechet manifolds we'd have the converse immediately, but it's not true in Frechet manifolds
I assume he uses the charts somehow to construct a map $M \to Sym^2(M)$, the space of unordered (distinct) pairs --- and further I guess one has that $m \not \in f(m)$.
After possibly taking a double cover this is a map $M \to M \times M$
Write $U$ for a neighborhood of the diagonal $M \times M$ which deformation retracts to the diagonal; this should produce a map $M \to U \setminus \Delta$
Maybe you should be able to parse the proof idea as the codimension 1 foliation breaking the classifying map M -> BTOP(n) of the tangent microbundle as M -> BTOP(n-1) x BTOP(1), so the tangent microbundle is germinally equivalent to a rank n-1 microbundle direct sum a trivial rank 1 microbundle
Then taking a nonzero section of the trivial microbundle gives a perturbation of the diagonal in M x M which intersects the diagonal trivially
Clearly A must be non-unital; and once we realize we allow non-unital algebras, this is easy: take any abelian group with standard addition and multiplication x*y = 0.
For a semigroup $(S,\cdot)$ and a subset $A\subset S$ we ask the sufficient and necessary condition for the quotient $S/A$ is well-defined as a semigroup, where here $S/A$ is taken to be the set of all left(or right) cosets of $A$ in $S$.
Of course when $A$ has the property that $sA=As$ for all $s\in S$ the quotient is well-defined as a semigroup.
But I am looking for an example of such subset $A\in S$ which does not satisfy the property $sA=As$ but still make $S/A$ into a well-defined semigroup.
If $g_{n}(t):=\dfrac{f(k^{n}t)-f(k^{n+1}t)}{k^{n}t}$ then is true that $\dfrac{f(t)-f(0)}{t}=g_{0}(t)+kg_{1}(t)+\dots+k^{n-1}g_{n}(t)+\dfrac{f(k^{n+1})-f(0)}{t}$? I think is not because this term: $\dfrac{f(t)-f(0)}{t}=g_{0}(t)+kg_{1}(t)+\dots+{\bf {k^{n-1}}g_{n}}(t)+\dfrac{f(k^{n+1})-f(0)}{t}$ should be $k^n$
@VJ123 yes, it is HK-integrable (and not Lebesgue-integrable, even the iterated integral does not make sense as Lebesgue integral). anything that is improperly Riemann-integrable is also HK-integrable. in fact, there is a theorem called Hake's theorem which roughly states there is no such thing as an "improper HK integral" (in the sense that anything improperly HK-integrable is already HK-integrable).
First, it reduces to the case where $A$ is nilpotent (Primary Decomposition Th. and adding $\lambda Id$)
Then it follows by induction on the dimension of $V$, proving that there is always a basis of eigenvectors $\{v_1, \dots, v_n\}$ s.t. $Av_i$ is either $v_{i+1}$ or $0$
@TedShifrin I want to prove that it is possible for a function which is continuous on $(a,b)$ that $$ \int_{c}^{d} f(x) dx = f(d)$$ That is the area under $f(x)$ between two points can be equal to the value of the function at the upper limit
