$A(x, y) = (x + cy, cx + y) = (1 + c)(x, y)$ implies $x + cy = x + cx$ and $cx + y = y + cy$ so $x = y$. $(1, 1)$ is an eigenvector
Big brain time
The other one is $(1, -1)$
So an orthonormal pair is $1/\sqrt{2}(e_1 + e_2)$ and $1/\sqrt{2}(e_1 - e_2)$. The dual basis is some scaling of $dx + dy$ and $dx - dy$, let's figure out what
$(dx + dy)(1/\sqrt{2}(e_1 + e_2)) = 2/\sqrt{2} = \sqrt{2}$. So the same scaling
Alright, so $\omega_1 = (dx + dy)/\sqrt{2}$ and $\omega_2 = (dx - dy)/\sqrt{2}$. Let's say $\omega_{12} = \alpha \omega_1 + \beta \omega_2$
Yeah what the hell am I doing. $A = (1, c|c, 1)$ where $c = \cos(\theta)$. Let $e_1 = (1, 1)/\sqrt{2}$, $e_2 = (1, -1)/\sqrt{2}$. Then $e_1^T A e_1 = (1 + c)$ and $e_2^T A e_2 = (1 - c)$ and $e_1^T A e_2 = 0$. I have found a pair of vectors orthogonal wrt the inner product given by $A$, I need to scale by $1/\sqrt{1\pm c}$ each to make it an orthonormal basis for $A$.
Fine, so $e_1 = \frac{(1, 1)}{\sqrt{2(1 + c)}}$ and $e_2 = \frac{(1, -1)}{\sqrt{2(1 - c)}}$ are the pair of orthonormal vectors, pointwise.
I need $\omega_1, \omega_2$ such that $\omega_i(e_j) = \langle e_i, e_j \rangle_A$. So $\omega_1 = E e^1 + F e^2 = e^1 + \cos(\theta) e^2$ and $\omega_2 = F e^1 + G e^2 = \cos(\theta) e^1 + e^2$
Compute curvature of $dx^2 - 2\cos(\theta) dx dy + dy^2$
I kind of see that it has to be $- \theta_{xy}/\sin(\theta)$
Because the coordinate curves form something called a Chebyshev net, if you draw a small rectangle of coordinate curves, opposite sides have equal length, and the two independent coordinate directions always make an angle of $\theta$ between them. So if you parallel transport the tangent vector to the coordinate curves around a coordinate square, you'll expect to get $-\theta_{xy}$ as the holonomy
And curvature is limit of holonomy by area. Area of the coordinate square is $\sin(\theta)$ in the limit
I'm trying to figure out why all these different L^p spaces and continuously differentiable function spaces have the norms that they do because they seem somewhat arbitrary. Do norms that make a particular function space complete maybe have the property that they are unique? Because that would explain it.
Then we have some polynomial $P$ such that $P(x,s)=t$ and we need to solve for $x$
@Masterphile We knew that algebraic expressions of any algebraic number is closed in the algebraic numbers, so you cannot for example, get $\pi$ from applying roots, sums, mutliplications to an algebraic number
Well, there are transcendentals whose product are algebraic, and as long you get one algebraic, it seems you can get all of them as you are not limited to use only a subset of transcendental
Problem is whether $st$ is transcendental given $s,t$ transcendental is not solved yet in the general case, it is only known transcendentals are not closed under products
something like $\pi e$ is still unsolved for example
yes, but the operations should be on non-trivial pairs to be interesting, not like the one i mentioned, so the problem could be: given mutually different transcendentals $t_1,...,t_n$ can we with a finite number of operations reach algebraic $a(t_1,...,t_n)$, and it needs to be defined which operations are allowable
yeah I am wondering about that as well, I don't recall seeing many nontrivial examples other than something like $2^{\sqrt{2}}$ is transcendental or something
I think the question is, do we know a lot about equivalence class of transcendentals so far. I think we so far only knew about $e^a$ where $a$ algebraic
you can state the problem like this: is it true that for every algebraic $a$ there exist transcendentals $t_1,...t_n$ such that $t_i$ is not obtained from $t_j$ by a finite number of trivial operations with algebraic numbers if $i \neq j$ and such that it is possible to have that $a$ is equal to some finite combination of $t_1,...,t_n$ obtained by a finite number of operations (but even now not all is well and not very strictly defined)
Yeah I think $t_i$ not obtainable from $t_j$ when $i\neq j$ at least will help establish a criteria for two equivalence classes of transcendentals, and if in the end there is only one equivalence class, there will be a contradiction like $t_i=t_j$
I should investigate more in that direction, hmm...
@Knight I have seen your situation and I'm not happy about it. There's no humility, no empathy, only selfishness and injustice. Come on TeX.SE and there are a lot of nice and extraordinary people also from a human point of view. You are welcome for me on TeX.SE.
Let $p$ be an odd prime, and let $a$ be an odd integer such that $p \nmid a .$ Prove that
$$
a^{p-1} \equiv 1 \quad(\bmod 2 p)
$$
I thought about Fermat Little Theorem could be useful but for that we need prime in modulo. But her we have 2p so it is an even integer. Can somebody give me hint for...
Reading Baker atm to see what else can be used, otherwise I think there is an intuition behind Gelfond–Schneider theorem that explains how can transcendence be reached from algebraic numbers
Using the example of $2^{\sqrt{2}}$ as a guide, I noticed something. Let $x$ be an integer. Then we want to multiply $x$ by some $y$ to make it transcendental
Hi there, can anybody help me with this seemingly easy (intuitively at least) fact: Let $V,W$ be two subspaces of $\mathbb{R^n}$. If there exists a $x\in \mathbb{R}^n \setminus\{0\}$ such that $
x\in V\cap W$, then it must hold that either $V\subseteq W$ or $W\subseteq V$.
Martin: Are you sure? Already in $\Bbb{R}^3$ both the xy and yz planes are subspaces of it, but they are not subseteq to each other since any such x lies on the intersection
yeah that's what I am thinking. If $1+a+b$ irrational, $x^{1+a+b}$ is transcendental by Gelfond–Schneider
If we express $a,b$ as infinite sums, it means we are effectively having an infinite product of $x$ raised to the rational power
Thus if $a+b$ is rational instead, it means countably many terms in the infinite sum cancels out, thus turning the infinite product involving $x$ into a finite one, and thus ensuring it is algebraic
1+a+b is irrational if and only if a+b is irrational, and if a and b are irrational then for a+b to be irrational is it enough that b is not shifted by some rational r with respect to a, for example b=r-a
Martin: then it means there are some vectors that span the same subspace within the intersection of $V$ and $W$. Still not enough to ensure they properly contain each other
It takes at least $\Bbb{R}^4$ to see that though, two intersecting dim 3 subspaces at a dim 2 subspace for example (e.g. xyz and yzw volume)
@Secret Maybe it help with the concrete problem I'm working on. Let P we a orthogonal projection onto R(A), where R(A) and R(Z) denote the range of a linear transformations A and Z onto R^n. If R(Z)^\perp \cap R(PZ)P\perp contains a non-zero element, then R(Z) \subset R(P). This is at least what is claimed in an article im reading, but I can't see why this holds.
@Masterphile I think the general case is more complicated than a shift. There is an answer here that tries to unravel what is going on by generalising this into a group: math.stackexchange.com/questions/157245/…
the converse, if a+b is rational so a+b=r, if a is irrational then b is irrational and if b is irrational then a is irrational, or either a and b are both rational, and if a is not a shift of b by a rational r, then a=c-b, where c is irrational so a+b=b+(c-b)=c, an irrational, a contradiction
@Martin Ok, the key here is P is an orthogonal projection, so it only takes those elements in Z that are parallel to Z, thus the range of projections has to be larger since many elements of the domain of P is the kernel in Z if you get what I am trying to say in this imprecise terminologies here as it has been getting rusty for me
@Secret if you are interested at least slightly, i am trying to obtain at least something of when for a natural $n=\displaystyle \prod_{i=1}^r {p_i}^{a_i}$ is the number $$\dfrac{1}{\prod_{i=1}^r \dfrac {{p_i}^{a_i+1}-{p_i}^{a_i}}{{p_i}^{a_i+1}-1}+ \prod_{i=1}^r \dfrac {{p_i}^{a_i+1}-2{p_i}^{a_i}+{p_i}^{a_i-1}}{{p_i}^{a_i+1}-1}}$$ a natural number? this could be hard
so the denominator needs to cancel out into a number of the form $\frac{1}{m}$
and what we have here is something like:
$\prod$ a finite sum of fractions that may or may not have a prime denominator, adding to another more complicated $\prod$
The way the numerator of the "produtum" (I don't know what is the product version of summand) seemed to remind a bit of polynomials, like the second one is definitely "2nd order" in terms of the indices
I wonder if there is a known result on how you get composite numbers from a finite combination of primes that is more specific than the fundemental theorem of algebra
as well a finite combination of prime fractions to get a composite fraction
that´s it, now, for example, you can research what happens if all the a_i are equal to 1, that is, when the number n is the product of just r different primes all raised to the power of one
$$\dfrac{\prod_{i=1}^r ({p_i}^{2}-1)}{\prod_{i=1}^r ({p_i}^{2}-{p_i}^{1})+ \prod_{i=1}^r ({p_i}^{2}-2{p_i}^{1}+{p_i}^{0})}$$ and this further simplifies to:
@Knight how? what happened? downvotes or? i think that you are studying very efficiently and i see that you seek for help asking for clarifications of some points in the texts that you do not understand, so there is no need to be discouraged if "some folks got in the way", you seem to be a rather clever student
@Secret the term with (p_i-1) will have some negative terms but it seems that´s not enough to cancel the contribution of the term with p_i only
yes, basically, the numerator grows almost as fast (a little faster) as the term $\prod_{i=1}^r p_i$ and the two terms in the denominator grow almost as exactly as $\prod_{i=1}^r p_i$, of course $\prod_{i=1}^r p_i$ grows exactly as $\prod_{i=1}^r p_i$ since it is the same term, so n=2 because of that should be the only solution, i think
@Masterphile Thanks for the kind comments. No, it was not downvotes that caused problem, there were some users who collectively attacked us and then suddenly they grew in number. You can check Physics.Meta.stackexchange and you will find so many controversies going on, if you want to see something live go to the h bar and you will find some animals roaming there. By the way, are you a student or a teacher or a self researcher :-) ?
@Knight self-researcher, but i have enough experience with similar problems, i were on a college once so i was a student also, and i was also teaching math to some people, preparing them for exams and similar stuff
@Secret yes, "same" here
@Knight by the way, do not call folks there "animals", it is not nice to do so
@Knight i can help, but limitedly, some folks here know much about some branches i even did not try to self-research appropriately, so they can also help, if they feel they really need to
@Knight i understand, but you are going further down if you misbehave towards them as they are misbehaving towards you, think about that
@Knight i got much downvoted here, on MSE, yes right here, so i decided i will not ask much questions on the main site, i think even none in the future, but am instead in some chatrooms, and i am trying to collaborate with some folks here,sometimes i ask questions in this chatroom and some folks are being nice and they really give nice advices and are being helpful, that´s pretty much that, i am also often in this chatroom if you need some advices:
Hi all. Is there any reason that you would be able to generally assume that the design matrix/covariate matrix X or whatever you want to call it is invertible?
Or if it is a random matrix composed of i.i.d. standard normal entries?
@Secret for small $r$, should it be appropriate to assume that $$\dfrac{\prod_{i=1}^r (p_i+1)}{\prod_{i=1}^r {p_i} +\prod_{i=1}^r (p_i-1)}=w$$ and $w \geq 2$ and somehow try to derive a contradiction?
ok it turns out I overthink about $p_i+1$. It is always even since 2 is the only even prime to make $p_i+1$ odd. So the numerator has a factor of $2$
Likewise, $\prod_{i=1}^r (p_i-1)$ is also even, as the smallest odd value can be get is 1, which is neither prime nor composite and is the identity, so it does not contribute
My math basics are very patchy. I´m only remembering now that an invertible matrix is always a quadratic matrix. So if I´m not mistaken the answer to my question above should be an obvious "no" because X is a pxn matrix with p being smaller than n in my case
since the smallest prime is 2, it means the denominator's evenness cannot match up to cancel out (as it is always too big), or the numerator is even and denominator is odd, so for $a_i=1$, there is no integer unless $r=1$ and $p_i=2$
I'm studyin,g for the first time, the Tensor Product of Vector Spaces. After the answer of a particular question, $[1]$, I think that I grasped the key point of the role of Quotient Vector Space; the answer is:
It's literally immediately from the definition of quotient. If $V/W$ is a quotient...
I just ran some quick ppppppp + (p-1)(p-1)(p-1)... at the back of the envelope and it pretty much checks out. If $r$ is odd and huge, the second product can get very small, and lead to the $p_i^{2r}$ condition to fail
but I have not checked whether it fail just the right way to divide the numerator
that can be difficult since I have no intuition on what this map $\prod x \mapsto \prod (x-1)$ does other than it can get very small
@Secret I got a very small doubt in physics, it won’t take much time. In book (context is Fluid Mechanics) it is written “The free surface, characterised by $p=0$”. I want to know what is free surface?
@EdwardEvans Do working people getting their salaries? (Because for some sectors it is not possible to work from home, my cousin works in Heidelberg Cements in Logistic Deoartment)
Yeah. It contains the main ideas behind Freedman's proof of topological 4-dimensional Poincare conjecture and uses nontrivial point set topology ideas of Bing
I have vague memories of a class of groups for which it is open whether it is the class of all groups or not. I thought it was sofic groups, but that's something else, but it might have a similar name. Am I going crazy or does this sound familiar to anyone?
Would it be ethical to allow an AI to make life-or-death medical decisions?
For instance, where there an insufficient number of ventilators during a respiratory pandemic, not every patient can have one. It seems like a straight forward question, but before you answer, consider:
Human decision...
"A drop of water falls onto a football and rolls down, following the path of steepest descent; that is, it moves in the direction tangent to the football most nearly vertically downward. Find the path the water drop follows if the surface of the football is ellipsoidal and given by the equation $$4x^2 +y^2 +4z^2 =9$$ and the drop starts at the point $$(1,1,1)$$"
What I did is: I'v found the gradient at that point, from that I've found a vector that is tangent to the level curve
$(1,-4,0)$ then I used the dot product with the vector $(x-1,y-1,z-1)$ to find the plane that intersects the football in the trajectory that I believe the drop would take from $(1,1,1)$ down.
@Simone Not quite right. When you talk about the gradient, you want to work with $z=f(x,y)$ and work only in the $xy$-plane. You then want to find the curve whose tangent line at each point is parallel to the gradient. This gives you a differential equation to solve.
if it is the case that the solution is non-trivial then MO is really more appropriate for that question, as it is part of a research, although from a naive standpoint
Obviously, since I have to reduce the equation to one with only 2 variables and since the drop of waters falls on the top half of the football, I can work with only that top half, that is: $f(x,y)=+(-x^2 -1/4 y^2 +9/4)^{1/2}$
I'm studyin,g for the first time, the Tensor Product of Vector Spaces. After the answer of a particular question, $[1]$, I think that I grasped the key point of the role of Quotient Vector Space; the answer is:
It's literally immediately from the definition of quotient. If $V/W$ is a quotient...
Would it be ethical to allow an AI to make life-or-death medical decisions?
For instance, where there an insufficient number of ventilators during a respiratory pandemic, not every patient can have one. It seems like a straight forward question, but before you answer, consider:
Human decision...
Here, $n-1$ is even and $n>5$.
P.S. I have don't have much knowledge about the Binomial Theorem.
