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 Functional super-iteration and hierarchy of functional hyper-iterations bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/04/2009, 09:01 PM (This post was last modified: 05/04/2009, 09:06 PM by bo198214.) (05/04/2009, 08:57 PM)Tetratophile Wrote: bo, you needed to see my explanation. 1. Evaluate g(x) at c first. 2. Hyper-n-iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. For n=1 (iteration), to evaluate this expression at any given natural x=c: 1. Evaluate g(x) at c first. 2. Iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. The set of all ordered pairs resulting from this evaluation is {(x, [f It_1 g(x)] (x))}. I know, but I dont know what you want to say with that *questioningly look*. We agree on the meaning, the question is more how to write it down properly. Edit: Oh ok I see you extended your article. Yes and that computer program can be expressed with the 3 lines (1),(2),(3) that I gave. Perhaps the difficulty of mutual understanding results from that you have no experience in functional programming. Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/04/2009, 09:06 PM (This post was last modified: 05/04/2009, 09:10 PM by Base-Acid Tetration.) (05/04/2009, 09:01 PM)bo198214 Wrote: (05/04/2009, 08:57 PM)Tetratophile Wrote: bo, you needed to see my explanation. 1. Evaluate g(x) at c first. 2. Hyper-n-iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. For n=1 (iteration), to evaluate this expression at any given natural x=c: 1. Evaluate g(x) at c first. 2. Iterate f to the OUTPUT of Step 1. 3. Evaluate the resulting function at c. The set of all ordered pairs resulting from this evaluation is {(x, [f It_1 g(x)] (x))}. I know, but I dont know what you want to say with that *questioningly look*. We agree on the meaning, the question is more how to write it down properly. Ok, I now see that you were only trying to clarify my notation. I would agree that your notation emphasizes that hyper-iterations are actually: function * OUTPUT of a function -> function. tongue in cheek edit: all this editing posts only to see it was too late biz, i know... wastes such a h**k of time... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/04/2009, 09:12 PM (05/04/2009, 09:06 PM)Tetratophile Wrote: tongue in cheek edit: all this editing posts only to see it was too late biz, i know... wastes such a h**k of time... Haha, see I edited my previous too, after you replied to it Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/12/2009, 02:29 AM (This post was last modified: 07/06/2009, 12:54 AM by Base-Acid Tetration.) To test the usefulness of this concept, I am now trying to define the hyper-operations in terms of hyper-iteration of the successor operation x+1. to try to see if the levels of hyper operations correspond to the hyper-iterations. if S(a) := a+1, than iteration of this function is addition: $a+b = [S \operatorname{It}_1 b] (a).$ Problem is, $[S \operatorname{It}_2 b] (a)$ produces ab+1 instead of ab, so it needs to be written as $[S \operatorname{It}_2 (b)] (a)-1$. So to define exponentiation etc... How do I put it in It_3? Don't know how to do this! Dammit! bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/12/2009, 07:11 AM (05/12/2009, 02:29 AM)Tetratophile Wrote: So to define exponentiation etc... How do I put it in It_3? Don't know how to do this! Dammit! I dont think they are compatible: Roughly your hyper-iteration ladder does: $f_{n+1} (x) = f^{[x]}(x)$ while the hyper-operations ladder does: $f_{n+1}(x) = f^{[x]}( c)$ where $c=0$ or $c=1$. « Next Oldest | Next Newest »

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