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6:18 AM
The tag is gone.
5 hours later…
11:25 AM
I don't know about this tag, but it seems a bit weird and unnecessary.
@MartinSleziak is this tag unnecessary?
11:55 AM
is this the R programming language?
12:12 PM
As pointed out above, a new tag was created.
Q: Simulating Brownian motion in R. Is this correct?

ParsevalI have a Weiner process $\{W(t)\}_{t\ge0}$ with $\sigma^2=\text{Var}(W(1))=1$. For a real constant $\epsilon>0$ consider the differential ratio process $\Delta_\epsilon=\{\Delta_\epsilon(t)\}_{r>0}$ given by \begin{equation} \Delta_\epsilon(t) = \frac{W(t+\epsilon)-W(t)}{\epsilon}, \quad \text{fo...

R is a programming language and free software environment for statistical computing and graphics supported by the R Foundation for Statistical Computing. The R language is widely used among statisticians and data miners for developing statistical software and data analysis. Polls, data mining surveys, and studies of scholarly literature databases show substantial increases in popularity; as of September 2020, R ranks 9th in the TIOBE index, a measure of popularity of programming languages.A GNU package, the official R software environment is written primarily in C, Fortran, and R itself (thus,...
Certainly, if a tag for R (programming language) is needed, it would be better to choose a more descriptive name for the tag.
@MartinSleziak oh okay
1 hour later…
1:22 PM
Some time ago there was suggestion that a synonym $\to$ would be a better one than the existing one $\to$ . And the tag (gradient) is mentioned also in the question which was bumped recently on meta.
A: Tag management 2019

Travis WillseProposal: Add the tag grad-curl-div; make gradient, curl and divergence aliases thereof. Edit 2: I've revised the proposal statement to Alexander Gruber's (in my view superior) suggestion in the comments. Implementing this proposal poses some advantages over the status quo: It would (productivel...

Q: Where did [gradient] go?

Rodrigo de AzevedoAt the moment, Mathematics SE has tags hessian-matrix and jacobian. However, tag gradient is a synonym for tag vector-analysis. Why? Motivation Since tag gradient does not exist, tags gradient-descent and gradient-flows are often used instead, which is unfortunate. Two tags are being corrupted b...

I wonder what the moderators think about this - after all, only moderator can remove/change a synonym which already exists.
@AlexanderGruber Since you took part in the discussion on meta, I hope it is ok that I pinged you about this.
@quid AFAICT among the moderators, you are the one who deals with the tags most frequently. So I tried notifying you, too.
Although I suppose that the moderators have noticed the question about which was recently bumped.
This has already been discussed also here in chat:
Dec 5 '19 at 8:13, by Travis Willse
I would be content with that outcome, except that I strongly recommend that the tag list the operator names in the canonical order: .
Dec 5 '19 at 16:25, by Alexander Gruber
My view is we're pretty much ready to implement the change, I don't think a meta post is necessary, but if someone writes one I'll wait for the upvotes before making the synonyms.
Dec 5 '19 at 16:33, by Alexander Gruber
I'll wait a day or so to finalize in case of any unforseen objections or input on word order

(gradient), (curl) and (divergence)

Dec 3 '19 at 8:26, 7 days total – 210 messages, 7 users, 13 stars

Bookmarked Sep 6 at 23:53 by Martin Sleziak

1:42 PM
@MartinSleziak I don't have a fixed opinion on this. I think there is no good reason that is a synonym while others are not (and arguably to all synonmise them is too much). The proposed solution of the triple-tag is alright. Another could be to put gradient and jacobian (matrix) together. I also see not big problem with keeping them all apart. Or at least keeping gradient separate from divergence-curl.
The point being that there are likely plenty of people that know the notion gradient yet not the other two.
I would guess that keeping separately would be the closest to what @RodrigodeAzevedo wanted to achieve by the post on meta. (Although he did not make any specific proposal in the post on meta, basically just stated that the current situation is not very good.)
In any case, thanks for the response.
We'll see what Alexander Gruber has to say on this. Clearly, his message from last December (that he plans to "wait a day or so to finalize in case of any unforseen objections") shows that he leaned towards some solution at the time.
2:18 PM
@quid I find your view incomprehensible. Tags are cheap. One's time is not. Or should not be. The more tags, the easier it is to find duplicates and prevent duplicates not being flagged as duplicates. What is the point of answering the exact same question 100s of times?
1 hour later…
3:30 PM
@RodrigodeAzevedo what exactly are you objecting to? I'm against the existing synonym. And I said "I also see not big problem with keeping them all apart."
To say "The more tags, the easier it is to find duplicates and prevent duplicates not being flagged as duplicates." is of course an oversimplification.
If there are redundant tags that's also a problem.
There is also a limit on the number of tags per question.
This was (maybe still is) a considerable problem on MO.
No one cared much about "cohesion" of tags and this created things like "primes" and "prime-numbers" existing in parallel.
I might agree that at times on math.se we suffer from the opposite problem that too many things are subsumed in one not so telling tag.
Having two tags for the same thing isn't that unusual on Mathematics, either. We have both and not too long ago. (Although one of them was removed relatively quickly, this was just an example I could remember - since it was quite recent.)
3:59 PM
@quid 5 tags is not enough. To make matters worse, there is no tag hierarchy, unlike on Quora. In an ideal world, all tags of the form [#-matrices] would be "daughters" of tag [matrices].
@RodrigodeAzevedo Of course, this isn't something which cannot be change at the level of the site, it would have to be changed for the whole network. (So the local mods cannot really change this.)
It seems unlikely that this limit changes. And while the limit exists, we have to keep it in mind when deciding which tags to create (and other tag-related stuff).
You can find many related discussions on Meta Stack Exchange, such as: Why is there a limit to the number of tags? or Could we make tags imply other tags? (I did not look much intot the details of five tags limit - but if you're curious, you can have a look.)
5 hours later…
9:23 PM
A new tag was created and added to 7 questions.
Q: Factoring a given rank-$1$ matrix

MichthanSuppose you have a $n \times 1$ column vector $$a=\begin{bmatrix}a_1\\{a_2}\\ \vdots\\{a_n}\end{bmatrix}$$ and a $1 \times m$ row vector $$\quad b=\begin{bmatrix}b_1 & b_2 & \ldots & b_m\end{bmatrix}$$ If you then multiply these $$A=ab=\begin{bmatrix} a_{1}b_1 & a_{1}b_2 & \ldots & a_{1}b_m\...

Q: Does $\det(I+A) = 1 + \mbox{tr}(A)$ hold if $A$ is a rank-$1$ complex matrix?

BAYMAX If $A$ is a complex $n \times n$ matrix of rank $1$, then $$\det(I+A) = 1 + \mbox{tr}(A)$$ How to approach this problem? Rank-$1$ matrices have special properties. Also, thinking about the determinant of a matrix as the product of its eigenvalues and the trace of the matrix as the sum of it...

Q: What properties does a rank-$1$ matrix have?

MatthewI have seen a lot of papers mentioning that a certain matrix is rank-$1$. What properties does a rank-$1$ matrix have? I know that if a matrix is rank-$1$ then there are no independent columns or rows in that matrix.

Q: $\det(I+A)=1+\operatorname{Tr}(A)$ if $\operatorname{rank}(A)=1$

user598858 Let $A$ be a complex matrix of rank $1$. Show that $$\det (I+A) = 1 + \operatorname{Tr}(A)$$ where $\det(X)$ denotes the determinant of $X$ and $\operatorname{Tr}(X)$ denotes the trace of $X$. Any hint, please. I do not get how to combine the ideas of rank, determinant and trace. Thank you.

Q: How can I solve "average" best rank-1 approximation?

Thomas ArildsenAssume I want to minimise this: $$ \min_{x,y} \|A - x y^T\|_F^2$$ then I am finding best rank-1 approximation of A in the squared-error sense and this can be done via the SVD, selecting $x$ and $y$ as left and right singular vectors corresponding to the largest singular value of A. Now instead, ...

Q: Given rank-$1$ matrix $A$, how to compute $A^{100}$?

abhishek tyagi $$A = \begin{bmatrix} 6 & 4\\ -6 & -4\end{bmatrix}$$ Find $A^{100}$. I tried to find it using diagonalization, but as it is a singular matrix so one of eigenvectors came out zero. How $A^{100}$ can be calculated of same matrix?

Q: Is the product of a diagonal matrix and a rank-$1$ matrix still rank-$1$?

J. GoeFirst let me make two statements to give my question the proper context. Consider $D$ a diagonal matrix and $u v^T$ a rank 1 matrix. From my current knowledge there exist computationally cheap algorithms to diagonalize $D+uv^T$ as discussed here: maximum eigenvalue of a diagonal plus rank-one ...


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