"It is not an integral in the Lebesgue sense; it is called integration because it has analogous properties and since it is used in physics as a sum over histories for fermions, an extension of the path integral. "
Hi. I have a little question about limits. I want to prove by $\epsilon-\delta$ definition, the following limit is 0: $$\displaystyle\lim_{(x,y)\to(0,0)} \left(\left(x^2+7y^4\right)sin\left(\frac{1}{xy}\right)\right)$$. I thought in using the result of sum of limits.
I.e: consider $\lim\displaystyle \:_{\left(x,y\right)\to \:\left(0,0\right)\:}\left(x^2sin\left(\frac{1}{xy}\right)\right)$ and $\lim \:_{\left(x,y\right)\to \:\left(0,0\right)\:}\left(7y^4sin\left(\frac{1}{xy}\right)\right)$
I don't think is correct since I don't know that the desired limit exists, so I can't apply it, right?
Yeah, I mean, I proved that $\lim\displaystyle \:_{\left(x,y\right)\to \:\left(0,0\right)\:}\left(x^2sin\left(\frac{1}{xy}\right)\right) = 0$ and the other one too. But I don't think I can conclude.
Yeah, I did. I proved this limit $\lim\displaystyle \:_{\left(x,y\right)\to \:\left(0,0\right)\:}\left(x^2sin\left(\frac{1}{xy}\right)\right)$ exists using the definition, with that domain.
Proof of the shoe-sock theorem: You can't homotope $\text{Sock}_\text{on}$ into $\text{Sock}_\text{in drawer}$ over the set $\text{Earth} \setminus (\text{Shoes} \cup \text{Foot})$
So the function is not defined on a punctured neighborhood of the origin. Therefore, unless you write the problem differently, the limit does not exist.
In particular I want to read more Spanish literature since they seem to be a compromise between the extreme dark symbolism, surrealist and chaotic things I like and the "aesthetic-surrealism" that I'm a big fan of
@MikeMiller It seems the argument they use is that the closure of the saturated is set is saturated by 11, so you look at a transversal to all those leaves. Those hit in something which is invariant under the holonomy group, so you use Zorn's lemma to find a minimal subset of that invariant under the holonomy group
Here is a problem if someone is bored: Consider the funcion $f : \mathbb{N} \to \mathbb{Z}$ defined as follows: $$f(n) = \begin{cases} -f\left(\frac{n}{2}\right) & \text{si }n\text{ is even}\\f(n-1) + 1 & \text{si } n \text{ is odd}\end{cases}$$ for $n\geq 0$. Prove that $f(n)$ is multiple of $3$ if and only if $n$ is multiple of $3$. Compute the smallest number that holds $f(n) = 2017$.
write $(uv^T)^T=-uv^T$, rewrite left side as $vu^T$, notice both sides as operators must have the same kernels, which are $u^\perp$ and $v^\perp$ respectively, so $u$ and $v$ are parallel. write $u=\lambda v$, get $\lambda vv^T=-\lambda vv^T$, so $u=\lambda v=0$
Statistics problem: Say that the parameter $\theta$ can only that on the value $\theta=1$ or $\theta=2$. The density function of a single observation, $X$, is $$f(x)=f(x;\theta)=\theta x^{\theta-1}$$ for $0<x<1$. Assume that $H_0: \theta=1$ will be rejected in favor of $H_1:\theta=2$ if $X>K$ (so critical region is C={$X>K$}. Evaluate $K$ for the critical region of size $\alpha$.
My attempt: $\alpha=P_{\theta_0}(X>K)$ which is same as $1-\alpha=P_{\theta_0}(X\le K)$. But $P_{\theta_0}(X\le K)$ is just the CDF of $X$ so $$1-\alpha=\int_{0}^K 1 dx = x\Big|_0^K \implies K=1-\alpha$$
I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?
I tried to play around with it:
$x = (c - by)/a$, hence $a|(c - by)$.
$a$, $c$ and $b$ are known. So to obtain integer solutio...
if gcd(a,b) doesn't divide c, then no solutions. if gcd(a,b) divides c, then divide equation by gcd(a,b) so without loss of generality gcd(a,b)=1. solutions are then of the form (x,y)=(u+nb,v-na) where u,v are such that au+bv=1
"So the function is not defined on a punctured neighborhood of the origin. Therefore, unless you write the problem differently, the limit does not exist."
That is what Ted said me a few hours ago.
And the same thing happens here.
Yeah @Daminark, but note the function isn't defined in $B_r((0,0),\epsilon)$
So let's say $A\unlhd A^*$ and $B\unlhd B^*$ are all subgroups of some $G$. Now the first thing it does is to say that $A \unlhd A^* \cap B^*$, but why do we know that $A$ is even a subset of that intersection?
But it proceeds to say "and so $A\cap B^* = A\cap (A^* \cap B^*) \unlhd A^* \cap B^*$", which suggests an implication, but I'm not seeing why this uses that $A\cap B \unlhd A^* \cap B^*$
If it were true that $A\unlhd A^* \cap B^*$ then there would be this implication
If I am one of 100 people in a room, and 10 people are picked (at random) sequentially, what are the chances I'll be picked? I managed to get two answers but don't know which is correct. Is it 1/10 or 1/9.55?
But I was proving that the Hilbert's cube $Q$ is compact. So first I showed it is closed: if $x \in \ell^2 \setminus Q$ defining $\epsilon = |x_n| - 1/n >0$ we have $B_{\epsilon}$ doesn't 'meet' $Q$.
> You plan to test all pairwise comparisons among 4 means. What is the critical value after a Bonferroni adjustment needed to maintain an experiment-wise Type I error rate of 0.05?
I wonder how one calculates the number of comparisons among 4 means.
How can we find the height of hcp unit cell ?
1. Each basal plane has nearest neighbor atoms making equilateral triangles. So, a=2R (where R is the sphere radius). #2. Each atom at height c/2 above the basal plane is positioned directly above the centroid of the triangles in the base plane. Fo...
@MikeMiller Damn, I can't get it to work. For example $SL(2,\mathbb{R}) \times \mathbb{R} \to SL(3,\mathbb{R})$ via $(A,t) \mapsto \begin{pmatrix}e^t A & 0 \\ 0 & e^{1-t}\end{pmatrix}$ just doesn't work to give an additive homomorphism.
I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?
I tried to play around with it:
$x = (c - by)/a$, hence $a|(c - by)$.
$a$, $c$ and $b$ are known. So to obtain integer solutio...
How can we find the height of hcp unit cell ?
1. Each basal plane has nearest neighbor atoms making equilateral triangles. So, a=2R (where R is the sphere radius). #2. Each atom at height c/2 above the basal plane is positioned directly above the centroid of the triangles in the base plane. Fo...
