User talk:Edward Z. Yang

Thank you for your understanding and happy editing :) Editing on behalf of

Hi! I am the developer of a WikiPraise UserScript. May I have a look into the sources of your extension? It may be very interesting for further research. What is the format of the data your extension provides? -- NetAction (talk) 13:38, 19 September 2011 (UTC)[reply]

See the attachments to https://bugzilla.wikimedia.org/show_bug.cgi?id=639 . Though it probably makes more sense to take advantage of existing annotation engines like those seen in Git. — Edward Z. Yang^(Talk) 01:18, 20 September 2011 (UTC)[reply]

Dear Edward Z. Yang,

My name is Jonathan Obar user:Jaobar, I'm a professor in the College of Communication Arts and Sciences at Michigan State University and a Teaching Fellow with the Wikimedia Foundation's Education Program. This semester I've been running a little experiment at MSU, a class where we teach students about becoming Wikipedia administrators. Not a lot is known about your community, and our students (who are fascinated by wiki-culture by the way!) want to learn how you do what you do, and why you do it. A while back I proposed this idea (the class) to the communityHERE, where it was met mainly with positive feedback. Anyhow, I'd like my students to speak with a few administrators to get a sense of admin experiences, training, motivations, likes, dislikes, etc. We were wondering if you'd be interested in speaking with one of our students.

So a few things about the interviews:

• Interviews will last between 15 and 30 minutes.
• Interviews can be conducted over skype (preferred), IRC or email. (You choose the form of communication based upon your comfort level, time, etc.)
• All interviews will be completely anonymous, meaning that you (real name and/or pseudonym) will never be identified in any of our materials, unless you give the interviewer permission to do so.
• All interviews will be completely voluntary. You are under no obligation to say yes to an interview, and can say no and stop or leave the interview at any time.
• The entire interview process is being overseen by MSU's institutional review board (ethics review). This means that all questions have been approved by the university and all students have been trained how to conduct interviews ethically and properly.

Bottom line is that we really need your help, and would really appreciate the opportunity to speak with you. If interested, please send me an email at [email protected] (to maintain anonymity) and I will add your name to my offline contact list. If you feel comfortable doing so, you can post your nameHERE instead.

If you have questions or concerns at any time, feel free to email me at [email protected]. I will be more than happy to speak with you.

Thanks in advance for your help. We have a lot to learn from you.

Sincerely,

Jonathan Obar --Jaobar (talk) — Preceding unsigned comment added by Chlopeck (talk • contribs) 23:16, 14 February 2012 (UTC)[reply]

I saw the wiki page, but I couldn't find any examples using actual numbers evaluating the formula. Could you give some examples of convolution, please? Mathijs Krijzer (talk) 22:14, 9 March 2013 (UTC)[reply]

Definition

The convolution of f and g is written f∗g, using an asterisk or star. It is defined as the integral of the product of the two functions after one is reversed and shifted. As such, it is a particular kind of integral transform:

${\displaystyle (f*g)(t)\ \ \,}$	${\displaystyle {\stackrel {\mathrm {def} }{=}}\ \int _{-\infty }^{\infty }f(\tau )\,g(t-\tau )\,d\tau }$
	${\displaystyle =\int _{-\infty }^{\infty }f(t-\tau )\,g(\tau )\,d\tau .}$       (commutativity)

Domain of definition

The convolution of two complex-valued functions on R^d

${\displaystyle (f*g)(x)=\int _{\mathbf {R} ^{d}}f(y)g(x-y)\,dy}$

is well-defined only if f and g decay sufficiently rapidly at infinity in order for the integral to exist. Conditions for the existence of the convolution may be tricky, since a blow-up in g at infinity can be easily offset by sufficiently rapid decay in f. The question of existence thus may involve different conditions on f and g.

Circular discrete convolution

When a function g_N is periodic, with period N, then for functions, f, such that f∗g_N exists, the convolution is also periodic and identical to:

${\displaystyle (f*g_{N})[n]\equiv \sum _{m=0}^{N-1}\left(\sum _{k=-\infty }^{\infty }{f}[m+kN]\right)g_{N}[n-m].\,}$

Circular convolution

When a function g_T is periodic, with period T, then for functions, f, such that f∗g_T exists, the convolution is also periodic and identical to:

${\displaystyle (f*g_{T})(t)\equiv \int _{t_{0}}^{t_{0}+T}\left[\sum _{k=-\infty }^{\infty }f(\tau +kT)\right]g_{T}(t-\tau )\,d\tau ,}$

where t_o is an arbitrary choice. The summation is called a periodic summation of the function f.

Discrete convolution

For complex-valued functions f, g defined on the set Z of integers, the discrete convolution of f and g is given by:

${\displaystyle (f*g)[n]\ {\stackrel {\mathrm {def} }{=}}\ \sum _{m=-\infty }^{\infty }f[m]\,g[n-m]}$

${\displaystyle =\sum _{m=-\infty }^{\infty }f[n-m]\,g[m].}$       (commutativity)

When multiplying two polynomials, the coefficients of the product are given by the convolution of the original coefficient sequences, extended with zeros where necessary to avoid undefined terms; this is known as the Cauchy product of the coefficients of the two polynomials.

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Hi,
You appear to be eligible to vote in the current

Hello, Edward Z. Yang. Voting in the

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If you wish to participate in the 2016 election, please review the candidates' statements and submit your choices on the voting page. Mdann52 (talk) 22:08, 21 November 2016 (UTC)[reply]

Hello, Edward Z. Yang. Voting in the

2016 Arbitration Committee elections

is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016.

The

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describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2016 election, please review the candidates' statements and submit your choices on the voting page. MediaWiki message delivery (talk) 22:08, 21 November 2016 (UTC)[reply]

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