@MikeMiller LOL, I trust you implicitly. But it's interesting that we all seem to have fallen in this trap. I remember struggling for quite a while and being so proud of the solution I finally found (30+ years ago).
Let A be an abelian group, let y be an element of maximal order n, and let x be an element of order m with m > d = (m,n) > 1. Consider the d-torsion subgroup B of A, the set of elements with d*b = 0. Then in the group A/B, we have ord(x+B) = m/d and ord(y+B) = n/d, again maximal.
Because the correct statement is that gcd(m/d, n/d) = 1, we have reduced to the case that we have an element of order coprime to the maximal order.
Did you see my approach above? Instead of dividing by $d$, you split $d$ up in such a way that $m/r$ and $n/s$ have to be relatively prime (with $rs=d$). I sorta liked that.
Hi everyone! Random question: what is the maximal interval of existence for a solution of a Cauchy problem with a separable ODE and an initial condition?
For example, in a problem I got $y(t)=\frac{-2}{t^2-2}$. In this case the solution is defined on the whole of R, except for $+\sqrt2$ and $-\sqrt2$, or we should consider only the maximal neighbourhood?
The initial datum was $1=y(0)$
So should I find the maximal interval of existence between $-\sqrt2$ and $+\sqrt2$ or is such interval the whole domain of the solution?
Because the local existence and uniqueness theorem states that such system has a unique solution $y(t)$ defined in a neighborhood of $t_0$...
@Ted I don't think it's that complicated, if $x$ and $y$ have different orders in an abelian group, then $xy$ has order $\operatorname{lcm}(\operatorname{ord}(x),\operatorname{ord}(y))$
Hi guys, can I have a suggestion on how to check if the following series is converges? $\sum_{n=1}^{\infty}\frac{(-1)^n}{4\sqrt{n}}\frac{(16n-4)}{\sqrt{16n+64}}$
But there's no boundary to use any maximum principle (we didn't talk about it though, all I'm supposed to know about harmonic functions at this point is the definition and the fact that Green's identities work on any Riemannian manifold)
@BalarkaSen Oh I see now what you meant. Now I have to convince myself that if $f:M\to\Bbb R$ is harmonic also its coordinate representation $\Bbb R^n\to\Bbb R$ is
@AlessandroCodenotti Any harmonic function $f$ on a Riemannian manifold $(M, g)$ satisfies $f(a) = \int_{B_{\varepsilon}(a)} f d\text{vol}_g$ where $B_{\varepsilon}(a)$ is a Riemannian ball of radius $\varepsilon$ around $a$, you know that, right?
You cannot localize to usual harmonic functions on $\Bbb R^n$ when you look at a chart, the metric is different.
In any case if you can prove this, let $a$ be the maximum. Then $f(a) = \int_{B_\varepsilon(a)} fd\text{vol}_g$ implies $f$ is constant $f(a)$ on $B_{\varepsilon}(a)$
Then you argue by looking at the set $\{x \in M : f(x) = f(a)\} \subset M$, say it's open (by above), then closed (direct), etc.
I remember you can switch between the surface and volume mean estimates with a polar coordinates+Fubini argument (this was also discussed in chat recently)
@Alessandro I googled around and it seems the mean value property becomes more subtle when you try to average over a geodesic sphere instead of on the ball
@StupidQuestionsInc I'm studying the mechanics of deformable bodies and there we got the equation of motion of a point in volume element (under consideration), so for the x coordinate of that point we have something like this $$ x' = x'_0 +\frac{\partial x'}{\partial x} + \frac{\partial y'}{\partial y} + \frac{\partial z'}{\partial z} +...$$
@adeshmishra every point on a rigid body, to be more precise. (And this ignores the kinetic theory of such a solid, ie, you don’t expect every atom of the solid to move exactly the same)
@StupidQuestionsInc I'm studying the mechanics of deformable bodies and there we got the equation of motion of a point in volume element (under consideration), so for the x coordinate of that point we have something like this $$ x' = x'_0 +\frac{\partial x'}{\partial x} + \frac{\partial y'}{\partial y} + \frac{\partial z'}{\partial z} +...$$
So what my book meant when it said "the first part of right hand side represents the translational motion of the particle in volume element (under consideration)"
Absent the next page, I’d suppose that what they mean is: Suppose there’s no deformation / rotation involved. Let O and P then be the coordinates of some chosen point, as the entire body moves
@Alessandro I was unsatisfied with the polar coordinates approach back then. I think the perhaps better justification of "integration over ball is integration over spheres and radius" is given by the coarea formula.
There's also a smooth coarea formula that works over Riemannian manifolds, which may be useful to your problem, but I don't really know about that.
Assume we have the following stochastic process:
$$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$
where $(B)_{t\geq 0}$ is a Brownian Motion.
I have to show that $X_t$ is not a martingale.
I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process ...
@LukasHeger I'm not sure about details with what you've said. Only when the orders of $a$ and $b$ are relatively prime is the order of $ab$ necessarily the lcm.
@LukasHeger We probably have the same construction. I'm thinking of it in terms of writing the gcd as a product $rs$ so that $m/r$ and $n/s$ are relatively prime. Although I tried last night to get $r$ and $s$ from the Euclidean algorithm, that doesn't work, and I have to use prime factorization to get them.
yeah, I'm not exactly sure how to approach the problem though. An example would be: "Let $B^2$ be the closed unit disk in $\mathbb{R}^2$. Let $\sim$ be the equivalence relation on $B^2$ generated by $(x,y)\sim (-x,y)$ for all $(x,y)$ in the boundary of $B^2$"
I know what it means for $\sim$ to be the equivalence relation generated by another relation $R$. But those two don't seem the same.
If $R$ is a relation on a set $X$ then there is a smallest equivalence relation $\sim$ such that $xRy\implies x\sim y$. That I know. $\sim$ is called the equivalence relation generated by $R$. But, the problem with the closed unit disk is talking about an equivalence relation generated by $(x,y)\sim (-x,y)$ , which is not a relation, and as a consequence I'm not sure what that means to generate the equivalence relation in that way.
Yes, I know that. I meant when he sais Let $\sim$ be the equivalence relation on $B^2$ generated by $(x,y)\sim (-x,y)$ for all $(x,y)$ in the boundary, does he mean the equivalence relation on $B^2$ generated by the relation,$\{$ (x,y) in boundary : (x,y)\sim (-x,y) $\}$?
No, you yourself told me that a relation needs to be a subset of the cartesian product of the set and itself.
I think you're being way too formal about mathematics here. Do you seriously not understand what he's doing? He's leaving the interior of the ball alone and gluing on the boundary by reflecting across the $y$-axis.
@TedShifrin for that ellipse problem, the person grading my hw said eigenvectors were unnecessary :'-) i wanted to be like "well that's a good attitude if you don't actually care about learning how to use the math we learn" but i kept my opinion to myself.
Is the half iterate of $2\sinh(x)$ - expanded from fixed point $0$ - analytic near the real line $\mathbb{R}$?
I know this function has 2 other fixed points apart from $0$, so I'm not sure.
Also does analytic at some place imply it satisfies its functional equation(being half-iterate of $2\sinh...
Assume we have the following stochastic process:
$$X_t=\int_0^t e^{B(s)^2}dB(s)\, ,0\leq t \leq 1$$
where $(B)_{t\geq 0}$ is a Brownian Motion.
I have to show that $X_t$ is not a martingale.
I know that if $t< \frac 1 4$ then $\int_0^t \mathbb E(e^{2B(s)^2})ds < \infty $ and then the process ...
Is the half iterate of $2\sinh(x)$ - expanded from fixed point $0$ - analytic near the real line $\mathbb{R}$?
I know this function has 2 other fixed points apart from $0$, so I'm not sure.
Also does analytic at some place imply it satisfies its functional equation(being half-iterate of $2\sinh...
