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 changing terminology (was: overview paper co-author invitation) andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/13/2009, 07:53 PM (08/13/2009, 11:20 AM)bo198214 Wrote: So then the terms super-exponential, super-logarithm and superfunction would still be in effect (?). Yes. I think so. Just because there are systematic names the superfunctions and their inverses, doesn't mean existing terms stop being used. The term "superlogarithm" and exponential-logarithm are synonyms now (I also support "tetralogarithm" but have been using "superlogarithm" to distinguish from the 4th polylogarithm). Also, take ethylene for example, its systematic name is "ethene" but no respectable chemist uses the systematic name. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/13/2009, 09:21 PM (08/13/2009, 11:20 AM)bo198214 Wrote: I would rather name it !-exponential and !-logarithm, as short form for functional exponential/logarithm of the factorial.I don't see the difference. "FactorialExp" is the Mathematica spelling, and "!-exponential" is the linguistic spelling. They are both pronounced "factorial-exponential", so they are basically the same word. bo198214 Administrator Posts: 1,395 Threads: 91 Joined: Aug 2007 08/13/2009, 10:17 PM (08/13/2009, 09:21 PM)andydude Wrote: I don't see the difference. "FactorialExp" is the Mathematica spelling, and "!-exponential" is the linguistic spelling. They are both pronounced "factorial-exponential", so they are basically the same word. Just to keep the confusion low, perhaps the computer term should be better FactorialFexp and FactorialFlog, for "functional logarithm of the factorial". andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/14/2009, 09:34 AM (This post was last modified: 08/14/2009, 09:42 AM by andydude.) (08/13/2009, 10:17 PM)bo198214 Wrote: Just to keep the confusion low, perhaps the computer term should be better FactorialFexp and FactorialFlog, for "functional logarithm of the factorial". Here we go again... did you know that in English flog means "to beat with a rod or whip"? Though, slog means "to tramp heavily", so it doesn't matter... Also, I think it would increase confusion, since Mathematica's conventions would be FactorialFLog (note the uppercase L), and so a lowercase l would cause people typing the name to get lots of spelling errors. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 08/22/2009, 05:50 PM (This post was last modified: 08/22/2009, 11:57 PM by Base-Acid Tetration.) Uh Oh I saw a problem with the f-logarithm terminology. What if it can mean a function/operator being plugged into the exp's Taylor series (powers meaning iteration)? Someone might call THAT a functional exponential, in analog to the matrix exponential... consider this example (just an example, don't say it's circular because i use exp's taylor series) this statement of analyticity of a function: Iff the following statement holds true for all $x, x_0 \in U:$ $f(x) = e^{(x-x_0)\frac{d}{dx}}[f](x_0),$ then the function is analytic in $U.$ Can you call such a usage an operator exponential? Annotation: if you're confused about my example: (x-x0) is scalar multiplication by x-x0, and d/dx is the differentiation operator ... so $e^{(x-x_0)\frac{d}{dx}} = \sum^\infty_{n=0}\frac{1}{n!} (x-x_0)^n \frac{d^n}{dx^n}$ Apply that to f, and you get: $e^{(x-x_0)\frac{d}{dx}}[f] = \sum^\infty_{n=0}\frac{(x-x_0)^n}{n!} \frac{d^n f}{dx^n}$ "Plug in" x_0 to this resulting function: $e^{(x-x_0)\frac{d}{dx}}[f](x_0) = \sum^\infty_{n=0}\frac{(x-x_0)^n}{n!} f^{(n)} (x_0)= \sum^\infty_{n=0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n$. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/23/2009, 12:46 AM What you are talking about only makes sense with a descriptive definition, a definition that attempts to capture current usage and the like. What I think we were doing was making a stipulative definition, one that we hold true no matter what. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 08/23/2009, 02:07 PM (This post was last modified: 08/24/2009, 04:39 PM by Base-Acid Tetration.) I mean, what would we call "exponentiating" a function by plugging it into the taylor series, if we were to call f's superfuntion an f-exponential? I suggest the term "f-iteration series of g" for the function/operator g plugged into f's taylor series. So for a function/operator f, "exp-iteration series of f" would mean the series: $\sum_{n=0}^\infty \frac{f^{\circ n}(x)}{n!}$ The taylor's theorem will say: for a smooth function f, f approximately equals f plugged into the exp-iteration series of (x-x0)D around x0. "sin-iteration series of f" will mean: $\sum_{n=0}^\infty \frac{-^n f^{\circ 2n+1}(x)}{(2n+1)!}$ etc. etc. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 09/02/2009, 05:01 PM (This post was last modified: 09/09/2009, 04:05 AM by Base-Acid Tetration.) Let's stick with iter- and abel- f. i suggest "f-iterational" for "f-exponential", but does that mean "functional power" in the literature? andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/04/2009, 03:24 AM But back to the original question. If superfunction is something everyone is comfortable with, then what prefix would make Abel function into a single word? Did anyone recommend the obvious: subfunction? Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 11/05/2009, 02:06 AM (This post was last modified: 11/05/2009, 02:15 AM by Base-Acid Tetration.) people may want "f is a subfunction of g" to mean the same as "g is a superfunction of f", as in "exponentiation is the subfunction of tetration" rather than "slog is a subfunction of exponentation" as for the abel function thing, there's no special prefix needed. just say "abel function". (though i prefer "f-iterational" to "superfunction of f.") « Next Oldest | Next Newest »

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