User talk:AJRobbins
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[edit] Talk
Hello everybody! Welcome to my talk page!
[edit] Welcome
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[edit] Not so many capitals
Hello. Please see my recent edits to Carleman matrix and Bell polynomials. Under Wikipedia:Manual of Style, you shouldn't capitalize the initial letter of a word merely because it's in a section heading. Thus See also, not See Also.
Also, it's usually a good idea to tell the non-mathematician reader at the beginning that mathematics is what the article is about. Otherwise in some cases they think it's just incomrehensible symbols that someone put there as graffiti. Michael Hardy (talk) 21:09, 19 November 2007 (UTC)
Thanks, I'm trying my best, but I'm still learning how to Wiki... AJRobbins (talk) 00:50, 20 November 2007 (UTC)
[edit] Comment on Talk:Tetration
I wasn't exactly sure what you meant by your recent addition to Talk:Tetration, but I thought I would explain how to create a new page.
- Type the name of the page you wish to create into the search box on the left sidebar.
- The page you arrive at will have the text "You searched for ____" in small light-gray type.
- Note that the article name you typed in is a red link (meaning no such article exists yet). Click on the red link, and it will send you to a page where you can type the text of the new article.
Please post on my talk page if you have any questions or if this isn't clear enough. I've been with Wikipedia for a while now, and I am willing to help if I can, but I'm definitely not an expert. --Whiteknox (talk) 04:55, 23 November 2007 (UTC)
- Thanks, I know how to make a page now. I made the super-logarithm page. Let me know what you think. AJRobbins (talk) 06:38, 25 November 2007 (UTC)
[edit] rational arguments for hyper-operator
Cant hyper-operator be defined such a way: hyper (a,n,b/c) = hroot_n(hyper (a,n,b),c) ?--Dojarca (talk) 10:15, 14 December 2007 (UTC)
- No, not with the standard hyper-operators. I think that anything that includes (n=1 as addition, n=2 as multiplication) can be called a "hyper-operator" of sorts. But for the standard hyper-operators, n=4 gives tetration, which would have to satisfy that equation, but it doesn't. Plugging in n=4 and b=1 gives which boils down the the assumption that reciprocal hyper-exponents will always give hyper-roots, which is false. The reason why it is false is because there is a well-known counter-example. The counter-example is which means if the formula were to hold, then we would get a contraction. Thus it is false. AJRobbins (talk) 21:40, 17 December 2007 (UTC)