Can You Edit Wikipedia Articles as a New User?

  • Thread starter jackmell
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In summary: I have to say that I completely agree with arildno.In summary, experts who want to contribute to Wikipedia should first make an account, and should be aware of the policies and guidelines of the site. They should also be prepared to argue their points on the associated talk page, and should be aware that not all of their edits will be seen immediately.
  • #36
arildno said:
Read WAnnabeNewton's comment.
And, I strongly suggest that you make a thread in the Maths&Science Materials section here at PF.
There, you can go really deep into the matter, and it most certainly will be an appreciated contribution.
Afterwards, you can crystallize that article into a Wikipedia entry.

Ok, I didn't see that comment before making the previous one. I can do that arildno. As I stated though, will be a while as I'm still studying the concept.
 
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  • #37
Now, I've made a few Wikipedia articles myself, on strictly historical issues.
The one I made which has reached "Good Article" status concerns the historical execution method known as impalement. It is shock full of references, and is an overview article over periods and techniques of impalement, but is designedly short of analyzing how, for example, such execution methods fugues with other mentalities/attitudes within the society in which it appear.
That would be the work of a HISTORIAN to deal with such broad social studies and interpretations, rather than a "just the facts" approach.
 
  • #38
jackmell said:
Ok, I didn't see that comment before making the previous one. I can do that arildno. As I stated though, will be a while as I'm still studying the concept.
I'm sure you can get a mentor to accommodate you in opening a thread that you can work on here at PF WHILE you are studying this, as an "in progress" project.
:smile:
 
  • #39
jackmell said:
Ok, no offense to jgens as you seem to be very knowledgeable but I don't have a clue what that means and I suspect most layman won't either.

None taken. Basically the homology group H1 consists of formal algebraic sums of loops modulo some relations. So if we have a loop/chain, then homologous to zero just means that it is zero in H1. Since the Pochhammer Contour can be written as the formal algebraic sum A+B-A-B = 0 we see that it is homologous to zero.

jackmell said:
But they do go into great detail with some subjects

Yes some articles go into great detail. If your vision for Wikipedia is as a giant pedagogical tool or an open source textbook, then these sorts of articles are fine. But there is also value in short articles that highlight the important details and direct you to other sources for the nitty-gritty stuff.

I suspect the reason it's not being done with this article is that few people really understand what's going on and we're only waiting for someone like me to elaborate.

I suspect not. There are probably plenty of people who understand this much better than me or you. This might be exactly how they like the article.
 
  • #40
jgens said:
None taken. Basically the homology group H1 consists of formal algebraic sums of loops modulo some relations. So if we have a loop/chain, then homologous to zero just means that it is zero in H1. Since the Pochhammer Contour can be written as the formal algebraic sum A+B-A-B = 0 we see that it is homologous to zero.

Ok, you're just playing around with me now jgens. That's ok. I don't mind you being smarter than me. Lots of people here are. :)
 
  • #41
jackmell said:
Ok, you're just playing around with me now jgens. That's ok. I don't mind you being smarter than me. Lots of people here are. :)

That honestly was not the point and I do not know nor care who is smarter. Basically the point is that your Pochhammer Contour can be written as a chain [itex]\Gamma = A+B-A-B[/itex] and that this particular chain is obviously homologous to zero. There is no need to draw nice pictures or anything to see this fact. It follows simply from the contour being defined as the commutator ABA-1B-1 of loops.
 
  • #42
jgens, please don't write wiki articles with that attitude.
 
  • #43
Pythagorean said:
jgens, please don't write wiki articles with that attitude.

I would ask the same of those who like extensive Wikipedia articles :rolleyes:
 
  • #44
jgens said:
I would ask the same of those who like extensive Wikipedia articles :rolleyes:

Seriously. That's what textbooks are for. An encyclopedia is meant to be a summary of information about a topic, not a detailed exposition of said topic.
 
  • #47
jackmell said:
I wish the article to simply be more practical so that readers can actually use the information. For example, the published integral expression is correct for only one path through the function. There is a different integral expressions corresponding to beginning the integration over each covering of the function or there is a possibility I'm not understanding what particular path is implied in the article. For example, here are the six (numeric) values of the pochhammer integral for the function [itex]w[/itex] above:

$$
\left(
\begin{array}{c}
-1.37669-0.794831 i \\
1.37669\, +0.794831 i \\
-\text{2.480233703139323$\grave{ }$*${}^{\wedge}$-8}-1.58966 i \\
-1.37669+0.794831 i \\
1.37669\, -0.794831 i \\
\text{2.480233429746903$\grave{ }$*${}^{\wedge}$-8}+1.58966 i \\
\end{array}
\right)
$$

only one of which can be used in the integral expression stated in that article and every other article about the matter that I've written.

And I hope no one reading this is the author of that article cus' now they're mad at me.

The values for the third and sixth entries seem a little garbled.
 
  • #48
SteamKing said:
The values for the third and sixth entries seem a little garbled.

Hi, I didn't take the time to format it nicely but rather just cut and pasted from Mathematica. Keep in mind I'm numerically integrating

[tex]\int_P z^{1/2}(1-z)^{1/3}dz[/tex]

over six different versions of that rainbow-colored contour I posted above, one version for each determination of the function I begin the integration on. The actual values for the third and sixth path are likely pure imaginary. I'll be working on this problem in much greater detail in the thread I initially created about it several weeks ago:

https://www.physicsforums.com/showthread.php?t=718609

so if you like, you can check there for more information about it as I add to the thread.
 
Last edited:
  • #49
The entries are all imaginary, at least in the numerical sense, and there are three pairs of conjugates.

Still, if you want to write an article on your work for whatever forum, at least take the time to proof the text and results.
 
  • #50
It has been suggested that this thread be closed.
 

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