@SQB Hmm, that does look weird, since you only added information already provided by OP in a comment
I don't really agree with editing the question at all though. It is 4 years old and it seems that nobody really cared about it, so not much point in bumping it.
Let $f,g:\Bbb R\to\Bbb R$ be two differentiable functions. Suppose that $f'(x)>g'(x)>0$ for $x>0$. Then $f(x)-f(0)\ge g(x)-g(0)$ for all $x>0$.
I tried this: Suppose $f(x)-f(0)< g(x)-g(0)$ for some $x>0$. I applied MVT, so, there is $\psi\in(0,x)$ such that $f'(\psi)=\frac{f(x)}x$ and $\gamma\in(0,x)$ such that$g'(\gamma)=\frac{g(x)}x$. But I can't show that $f'(c)\le g'(c)$ for some $c\in(0,x)$.
I am working through the proof of the following theorem: Let $A$ be a commutative ring. Then $A^n \simeq A^m$ iff $m=n$...I 'understand' the proof of the forward direction, but I am having trouble confirming some of the details. Since $A$ is commutative, it has at least one maximal ideal $\mathfrak{m}$. Now, in order to show that $A^n/\mathfrak{m}A^n \simeq A^m/\mathfrak{m}A^m$, don't I first need to show $\mathfrak{m}A^n \simeq \mathfrak{m}A^m$? How do I do that?
Let $f$ denotes the isomorphism from $A^n$ to $A^m$. I tried to show $f(\mathfrak{m}A^n) = \mathfrak{m}A^m$, but every attempt seemed circular.
@user193319 beware: $M \cong M'$ and $M \supset N \cong N' \subset M'$ doesn't imply $M/N \cong M'/N'$
Consider e.g. $M=N=M'=\Bbb Z$ and $N'=2\Bbb Z$
@user193319 what part of $f(\mathfrak{m}A^n)=\mathfrak{m}A^m$ gives you trouble?
@user193319 let's show in general that if $f: M \to N$ is a $R$-linear map and $I \subset R$ is an ideal and $X \subset M$ is a submodule, then $f(IX)=If(X)$. $IX$ is generated by elements of the form $ix$ with $i \in I$, $x \in X$. Thus $f(IX)$ will be the submodule generated by elements of the form $f(ix)=if(x)$ with $i \in I$ and $x \in X$, or to put it differently, $f(IX)$ will be generated by elements of the form $iy$ with $i \in I$ and $y \in f(X)$. Thus $f(IX)=If(X)$
I wonder if it might help to express the $f$ in the integral in terms of $w_{xx}$ and $w_{tt}$ of the wave equation, cause I suspect there's a lapacian somewhere
@MatheinBoulomenos Ah! That's what I was looking for! I trying to figure out the more general theorem of which my claim would be a corollary. Sometimes they're tricky to identify, but they're almost always easier to prove. Thanks for the help!
@MatheinBoulomenos Actually, by your first remark, I cannot conclude that $A^n/\mathfrak{m}A^n \simeq A^m/\mathfrak{m}A^m$ from $\mathfrak{m}A^n \simeq \mathfrak{m}A^n$. So how do I conclude that the quotients are isomorphic?
I am trying to understand that what could be the possible outcomes of this following expression as $n \to \infty$ and as $|t| \to \infty$ where $t \in \mathbb{Z}$ .
$$\sum_{k=1}^{n}\dfrac{(\prod_{m=1}^{t}k)^{-i}}{\sqrt{k}}$$
Considering the numerator first:
$$(\prod_{m=1}^{t}k)^{-i}=\prod_{m=1}...
Hold on! I think I figured it out. Let $f : A^n \to A^m$ be an $A$-module isomorphism, and let $\pi : A^m \to A^m/\mathfrak{m}A^m$ be the canonical projection. Then $\pi \circ f$ is a surjective homomorphism and therefore $A^n/\ker (\pi \circ f) \simeq \pi(f(A^n)) = A^m/\mathfrak{m}A^m$ or $A^n/\mathfrak{m}A^n \simeq A^m/\mathfrak{m}A^m$, because $\ker (\pi \circ f) = f^{-1}(\pi^{-1}(0)) = f^{-1}(\mathfrak{m}A^m) = \mathfrak{m}A^n$.
A ring $R$ (not necessarily commutative) is called stably finite if for each $n \in \Bbb N$, the matrix ring $M_n(R)$ has the property that $AB=1$ implies $BA=1$ for $A,B \in M_n(R)$.
Claim Any commutative ring $R$ is stably finite.
Proof: Suppose $AB=1$ for $A,B \in M_n(R)$, then $\det(A)\...
Hello! I was solving this simple equation: $|\sin(x)|=1/2$. The answer according to the book is $\pm \frac{\pi/18}{18}+(\pi \cdot n)/3$ which I doubt, isn't it $(-1)^n \pm \frac{\pi/18}{18}+(\pi \cdot n)/3$
Oh, sorry 1st expression should be $\pm \frac{\pi}{18}+(\pi \cdot n)/3$ and mine is $(-1)^n \pm \frac{\pi}{18}+(\pi \cdot n)/3$
@Tug'Tegin wouldn't you solve these types of equations like $\sin^2(x) = 1/4$ then $\sin(x) = 1/2$ and $\sin(x) = -1/2$. Calculate the basic angle which is $\pi/6$. Answers would be, $2\pi.n+\pi/6$,$2\pi.n+5\pi/6$,$2\pi.n+7\pi/6$,$2\pi.n-2\pi/6$ ?
Could someone draw the Poincaré model for $\mathbb{H}^3$ with this $P_t$? I can not find $P_t$ orthogonal to $\partial B^3$ and perpendicular to the line that contains $\overrightarrow{v}$.
While it is quite common to use piecewise constant functions to describe reality, e.g. the optical properties of a layered system, or the Fermi–Dirac statistics at (the impossible to reach exactly) $T=0$, I wonder if in a fundamental theory such as QFT some statement on the analyticity of the fie...
I think nonanalytic events is one way we can have indeterminism in classical phenomenon because then derivatives breaks down and events before the abrupt change no longer matter
Dan has a pair of glasses and a monocle which in total cost $300. At which point is getting surgery to remove his extra eye less financially wise than just breaking his glasses apart into monocles and putting them in a rock tumbler until they become diamonds?
Ed visits his mother every day, his grandmother every month, and his great grandmother every year. How often will he visit his daughter?
@TobiasKildetoft is it better to read something like Lam "A first course in noncommutative algebra" before Assem-Simson-Skowronski or is that independent?
we skeched a proof that $H^2(\operatorname{Gal}(K^{sep}/K),{K^{sep}}^\times) = Br(K)$ in a seminar, but we used some heavy stuff like double-centralizer or Jacobson density which we didn't prove
@TobiasKildetoft if $B$ is a simple subalgebra of a finite-dimensional centrally simple algebra $A$, then $C_A(C_A(B))=B$ and some equality of dimensions I don't remember right now
@MatheinBoulomenos I see. I wonder if that follows in any way from the results of Double centralizer properties, dominant dimension, and tilting modules
@gateprep There does not seem to be any links there to anything hosted on researchgate
@TobiasKildetoft I'm not sure I'm at a point yet where I can comfortably read research papers. I'm in a seminar right now that is a based on a paper in the annals and it's pretty hard
@MatheinBoulomenos That one is not that hard to read actually. It gives a more generalized approach to the classical Schur-Weyl duality, seen as a double centralizer.
@TobiasKildetoft I think I asked this recently, so sorry if I'm bothering you, but you do know a simple proof that groups of order 720 are not simple? All proofs I could find were quite involved
Not to mention the increased number of possibilities for the individual Sylow subgroups
So it makes sense that there is either some very obvious and slick proof, or any proof will be full of nasty details
I just recalled that a similar order had caused me to include an exception in an earlier paper. But that was order $768$ instead of $720$.
And there it was just because that was the order $< 1000$ which was not a nilpotent number but for which there were too many groups to deal with using GAP
@TobiasKildetoft one thing I was wondering: if $G$ is a finite group we have purely-group theoretic description of the number of conjugacy classes of irreducible representations of $G$ with coefficients in $\Bbb C$, $\Bbb R$ and $\overline{\Bbb F_p}$. What about other fields, e.g. $\Bbb Q_p$ (which is one of the most frequently used coefficient fields for representations at least in number theory)?
It's a well-known fact that
$$\lim_{n\to\infty} (H_n-\log(n))=\gamma$$
Now, if I change things a bit and use the fact that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder
if it's possible to compute the following limit in a closed-form...
Straight integration seems pretty tedious and difficult, and I suppose that the symmetry might possibly open some new ways of which I'm not aware. What would your idea be?
$$\int_0^1 \int_0^1 \int_0^1 \frac{x^2}{\sqrt{x^2+1} \left(x^2-y^2\right) \left(x^2-z^2\right)}+\frac{y^2}{\sqrt{y^2+1} \lef...
@MatheinBoulomenos I think they would be the same as for $\Bbb C$, right? As $\Bbb Q_p$ is an algebraically closed field of characteristic zero and cardinality the continuum, it is abstractly isomorphic to $\Bbb C$
@Waiting Do you by any chance know the true value of integral? The sum/integral forms at the end of that answers feels like it's on the cusp of a solution.
@MikeMiller I think that's the completion of the algebraic closure of Q_p under the p-adic metric, as bar{Q_p} is not a complete metric space under the p-adic metric
@BalarkaSen @MikeMiller suppose that I have a Lie group structure on a sphere $S^n$, does this give rise to a normed real division algebra of dimension $n+1$? The other direction is easy and we know that the answer is actually yes, because only $n=1$ and $n=3$ are possible, but I don't see how one could define a multiplication that is distributive
Hurwitz's theorem on classification of real normed division algebras is really easy though, isn't it? The classification of H-space structures on spheres, on the other hand, is really hard.
Maybe classifying the Lie groups is not that hard, I don't remember.
I heard that using some relatively basic differential geometry, you can show that the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$. My friend who told me this thought that it involved de Rham cohomology, but I don't really know anything about the cohomology of Lie groups so this...
If $G$ is a Lie group I can give it a bi-invariant metric $g$, which ends up satisfying the identity $g([x, y], z) = g(x, [y, z])$, with which you can prove $\omega = g([-, -], -)$ is a 3-form on $G$.
That's somehow a closed but not exact form. I don't remember why.
We know actually that since $S^1$ and $S^3$ have trivial outer automorphism groups, that the product defined by $v \cdot w = \|v\| \|w\| (\frac{v}{\|v\|} \frac{w}{\|w\|})$ will always be assocative
Riemann be like "ugh, the Dirichlet function is crap" and Lebesgue be like "bruh, $\displaystyle \int_0^1 1_{\Bbb Q} \ \mathrm dx = \lambda (\Bbb Q \cap [0,1]) = 0$"
@Symposium I'm pretty deep in the matter I'm interested in (not sure if that sounds fine in English). However, there is always something one can learn from the others, and that's fine.
@LeakyNun the Bolzano-Weierstraß theorem depends on the topology of $\Bbb R$, dominated convergence holds for general measure spaces, so I doubt there's a connection
@LeakyNun the problem is that if you have a bonded sequence, a priori nothing converges. Dominated convergence does assume you already have one type of convergence, namely pointwise, and want to pass to L^1
@MatheinBoulomenos Right, so the correct statement should be this. Take the space of all Lebesgue-integrable functions on [0, 1]. The Lebesgue integral operator is continuous on this space under the dumb product topology.
It's a well-known fact that
$$\lim_{n\to\infty} (H_n-\log(n))=\gamma$$
Now, if I change things a bit and use the fact that $\displaystyle \Gamma \left( \displaystyle \frac{1}{ n}\right) \approx n$ when $n$ is large, then I wonder
if it's possible to compute the following limit in a closed-form...
Straight integration seems pretty tedious and difficult, and I suppose that the symmetry might possibly open some new ways of which I'm not aware. What would your idea be?
$$\int_0^1 \int_0^1 \int_0^1 \frac{x^2}{\sqrt{x^2+1} \left(x^2-y^2\right) \left(x^2-z^2\right)}+\frac{y^2}{\sqrt{y^2+1} \lef...
The most general version I know is that if you have $f_n\to f$ and you have $|f_n|\le g_n$, where $g_n\to g$ and $\int g_n \to \int g$, then you have $\int f_n \to \int f$
@Waiting What you said made sense (your English is excellent). I just wasn't sure what it was in reference to (I asked whether you knew the closed form answer for the integral, perhaps through a symbolic calculator or something).
@Symposium I also like to share sometimes some of my creations. I know the community enjoy such questions, but on the other side, it's against the community rules to post question where you already have solutions (I think). So, it's complicated.
@MatheinBoulomenos That's a natural question but I have never known the answer. I would anticipate one approach passes through the Lie algebra and then constructs an algebra with an extra unit vector (like quaternions out of imaginary quarternions)
I am trying to understand that what could be the possible outcomes of this following expression as $n \to \infty$ and as $|t| \to \infty$ where $t \in \mathbb{Z}$ .
$$\sum_{k=1}^{n}\dfrac{(\prod_{m=1}^{t}k)^{-i}}{\sqrt{k}}$$
Considering the numerator first:
$$(\prod_{m=1}^{t}k)^{-i}=\prod_{m=1}...
A ring $R$ (not necessarily commutative) is called stably finite if for each $n \in \Bbb N$, the matrix ring $M_n(R)$ has the property that $AB=1$ implies $BA=1$ for $A,B \in M_n(R)$.
Claim Any commutative ring $R$ is stably finite.
Proof: Suppose $AB=1$ for $A,B \in M_n(R)$, then $\det(A)\...
