I don't know. i have the two vectors, one of them is $(X+1, \frac{1}{X-1})$ and another one and the rest is "show that they are linear independent in the vector space of $\mathbb{R}(X)$ which is $\mathbb{R}(X)^2$
wakaranai is more like "I don't get it" or "I don't understand", while shiranai is simply that you don't know (or aren't familiar with, depending on the context)
Just something I knew for a while that $[0,1]/(0,1]$ is the Sierpiński space
In fact all non-Hausdorff spaces are quotients of Hausdorff spaces, if I remember right
Ooh, here's a neat one
Define $f$ from $\Bbb Q$ to the natural numbers with the cofinite topology
by $f(x)$ equals the denominator of $x$
(Integers including $0$ are said to have denominator $1$)
(Write $x$ in lowest terms first)
And this is a quotient map as well (that is, $\Bbb Q/{\sim}$, where $x\sim y$ iff they have the same denominator in lowest form, is the naturals with the cofinite topology)
Hate this... I wanted to prove by contradiction... The claim is that we can't have a filter on $\omega$ that is countably closed.
So what I'm thinking is that there exists an uncountable sequence and if there can't be a filter then it's an ideal but that also doesn't make sense either because ideals are small subsets of X. Bddkmabs
Whatever you can prove without contradiction, you can prove by contradiction. Just assume the negation of the consequence, prove that it is true, contradiction. In general though, if you can prove by contradiction, you don't necessarily can prove without contradiction.
Maybe $\omega$ is countably closed if it's an ideal... Since ideals are small subsets of X and countably closed may possibly mean that there aren't that many elements
Maybe I can prove it directly since $\omega$ is countably closed (which by the way $\omega$ is the set of all natural numbers)...it can't be a filter because filters are usually large and ideals are small. Hmm
@usukidoll You can have one, as long as it's principal
For a nonprincipal filter $F$ you know that $\omega\setminus\{n\}$ is in $F$ since every nonprincipal filter contains the cofinite one, but if $F$ were countably closed their intersection, which is empty, would be in $F$, a contradiction
The same argument shows that no nonprincipal filter on $\kappa$ is closed under intersections of size $\kappa$
(and whether there exist a $\kappa$ with an $F$ closed under all intersections of size less than $\kappa$ is the statement that $\kappa$ is a measurable cardinal)
Suppose that X, Y and Z are independent random variables, each uniformly distributed on (0, 1). How can you prove that (XY )^Z is also uniformly distributed on (0, 1).
If $F(D) = \Pi_{i = 1}^n (D - m_i) $, and Differential equation is $F(D) y = r(x) $ , then prove that, the particular solution of the differential equation would be $$ y = \sum_{i = i}^{n} A_i e^{m_i x} \int e^{-m_i x} r(x) dx$$
After I sit down and I compute the character table for $D_4$: $$\begin{bmatrix}D_4&(1)&(r)&(r^2)&(sr)&(sr^2)\\\tau_1&1&1&1&1&1\\\tau_2&1&1&1&-1&-1\\\tau_3&1&-1&1&1&-1\\\tau_4&1&-1&1&-1&1\\\tau_5&2&0&-2&0&0\end{bmatrix}$$
I know that $\tau_5$ is an irrep of degree $2$, so $\rho(g)$ is a $2$ by $2$ matrix for each $g\in D_4$, and I know its trace, and I know since these matrices are finite order, they are diagonalisable, and their eigenvalues are roots of unity (k^th roots if it's order k), but since $D_4$ isn't abelian, I can't simultaneously diagonalise these matrices, so how then do I pick matrix representatives for $\tau_5$?
Being an essential epi doesn't require us check only projective modules, where indecomposable probably rules them out since they can't be direct summands of an indecomposable guy
Assume $c$ is a column vector. What mathematical operation or expression can produce a diagonal matrix with entries of that of $c$ in same order. Does such an expression exist?
Basically I know that the sum of entries of $c$ is zero. I want to write it in an matrix expression/equation. I am hopi...
Well, I think I'm not clear on the argument that makes it immediate that indecomposable implies $k[G]$ is a projective cover of the trivial $k[G]$-module (although for irreducible there seems to be an argument)
Question: let $M_k$ be the space of weight $k$ entire modular forms and $S_k$ the subspace of cusp forms. If $f \in M_k\setminus S_k$.. does that mean $f$ vanishes at NO cusp (by which I mean $s \in \Bbb Q \cup \lbrace \infty\rbrace$) or just that there are cusps at which $f$ does not vanish, or are these equivalent, since there's only one equivalence class of cusps for the action of $\operatorname{SL}_2(\Bbb Z)$ on the upper half plane?
I mean if $f(s) = 0$ for some $s \in \Bbb Q\cup \lbrace \infty \rbrace$ then there is a $\gamma \in \operatorname{SL}_2(\Bbb Z)$ such that $\gamma \infty = s$ so that $f(s) = j_\gamma(\infty)^kf(\infty) = 0$
and in fact $f(\gamma z) = j_\gamma(z)^kf(z)$ for all $\gamma \in \operatorname{SL}_2(\Bbb Z)$ so this is gonna hold for whichever $\gamma$ I pick, since I can just move them around with the action of SL_2(Z)
In mathematics, a matrix of ones or all-ones matrix is a matrix where every element is equal to one. Examples of standard notation are given below:
J
2
=
(
1
1
1
1
)...
When people say $p(e)$ is a probability density, what is $e$ here? Is it a random variable, a variable, or something else? In an ML paper, I am reading, the authors say: "let $e$ be a random variable with probability density p(e)". Now, usually, random variables are written in upper case, so this is already confusing. Then they say "let $t(\theta, e)$ be a deterministic function".
Furthermore, sometimes, this notation and terminology confuse me even more, because people sometimes use p(E=e) or P(E=e) or P(E) or P(e) or p(E). What the hell are the differences? I've already read several questions on Stats, but it is still confusing
For example, in p(E=e), I suppose that E is a random variable and e is the actual observation or realization of the random variable.
ML people are so ignorant when it comes to notation, terminology and actual probability theory
In this same paper, the author says "suppose further that the marginal probability density of w, $p(w \mid \theta)$".
$p(w \mid \theta)$ should be a conditional prob. density, not a marginal
Of course, I know that, in this case, $\theta$ is considered a parameter, rather than a random variable, but how stupid is this?
Can someone tell me which calculus rule is being used here?
I understand that the derivative is first exchanged with the expectation. So, in the first line, we have the derivative of an expectation, in the second line, we have the expectation of derivatives.
I understand that this can be done, under certain conditions, which I assume are satisfied in this case
But the partial derivative of f(w, theta) with respect to theta, should just be the partial derivative of f with respect to w multiplied by the partial derivative of w with respect to theta. No?
So I just conjectured something in an answer to my own question that turns out is false (forwards implication holds, backwards does not). Does etiquette say I should delete it or not?
@ÍgjøgnumMeg Great for me at least. If privileges continue to go hand in hand with reputation points, then I am going to get all kinds of undue influence that I can misuse.
Or maybe not so much on this site. The next reputation privilege is probably several points ahead for me.
@Leaky I just multiplied $\theta(z)$ by $\sqrt{\frac{z}{i}}$ and naively ripped the root apart to show that $\sqrt{\frac{z}{i}}\theta(z) = \theta\left(\frac{-1}{z} \right)$ rofl
@LeakyNun What is the difference between the one you sent me and the one I sent you? They both seem to specify that x and y also belong to the same sets.
Have any of you read "Road to Reality" by Penrose?
I wonder if you guys have any resources for better understanding one-forms and p-forms? I guess I have some understanding of it, but I would like to improve my geometrical intuition of it.
Also, any resources on Clifford and Grasmann algebras? Those sections in Road to Reality were a little confusing
Their geometrical intuition. like what they mean geometrically/physically. I can "see" a vector and a vector field, but it is harder to do so with a one-form and p-forms
I have heard that for a one-form, since the scalar product of a one-form with a vector is a scalar (that's the definition after all), the one-form can be represented by a plate that the vector crosses to result in a scalar... is this similar to the interpretation you are describing here?
Are you describing the piercing one? That is, that the 1-form measures how many "level surfaces" the vector punctures/crosses/intersects?
Like this?
This is not what I am describing exactly.
This one is often used in physics books, but is problematic if you take it too far.
I had this example in some notes I wrote, but I took it out explicitly because I am now of the belief it is kind of harmful to focus on drawing a picture of differential forms, rather than just viewing how they are used.
