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 Cyclic complex functions and uniqueness bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/22/2008, 11:02 AM (This post was last modified: 04/22/2008, 11:04 AM by bo198214.) I dont know what you mean, Gottfried, the definition $\exp^0(t)=t$ $\exp^{n+1}(t)=\exp(\exp^n(t))$ uniquely defines $\exp^n$ for any natural $n$. This definition is equivalent to if you substitute in the second line $\exp^{n+1}(t)=\exp^n(\exp(t))$. So the finite iterations are equal and hence also the limit for $n\to\infty$!? But lets continue this discussion of Dmitrii's article in the other thread. As this thread is about cyclic complex functions. Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 04/24/2008, 11:25 AM (This post was last modified: 04/24/2008, 11:44 AM by Gottfried.) bo198214 Wrote:I dont know what you mean, Gottfried, the definition $\exp^0(t)=t$ $\exp^{n+1}(t)=\exp(\exp^n(t))$ uniquely defines $\exp^n$ for any natural $n$. This definition is equivalent to if you substitute in the second line $\exp^{n+1}(t)=\exp^n(\exp(t))$. So the finite iterations are equal and hence also the limit for $n\to\infty$!? But lets continue this discussion of Dmitrii's article in the other thread. As this thread is about cyclic complex functions.Hmm, Henryk - I don't know how to say more than in the previous msg. In lim n->oo $\exp^{n}(t)=\exp(\exp^{n-1}(t))$ we have $\exp^{\infty}(t)=\exp(\exp^{\infty}(t))$ and we have to evaluate the limit to evaluate the limit... Perhaps another view makes my problem better visible. Look at D.F.Barrow's (*1) illustration     This images suggests a sequence of parameters, indexed from 0 to n, but which we had to evaluate beginning at n instead of 0 according to right associativity of the operation. However, the convention of partial evaluation to the approximate result would be a0, a0^a1, a0^a1^a2,... Again hmm, I cannot say more. If it isn't sufficient to explain my concern, then I'm rather helpless, and am apparently facing another mystery of math from outside (of the game)... Gottfried (*1) D.F.Barrow; Infinite Exponentials; The American Mathematical Monthly, Vol. 43, no. 3 (Mar., 1936), 150-160 (Sorry, I still appended my reply still in this thread - seemed a bit strange to change the place. You may reorder as you think it is appropriate) Gottfried Helms, Kassel bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/24/2008, 08:50 PM (This post was last modified: 04/24/2008, 08:54 PM by bo198214.) Gottfried, if the definitions agree on finite arguments then they also agree in the limit, this is a triviality. Everything else is a matter of taste. If we define $\exp^0(x)=x$ $\exp^{n+1}(x)=\exp^n(\exp(x))$ then of course also $\exp^{n+1}(x)=\exp(\exp^n(x))$ and vice versa. If you want, you can prove that by induction. « Next Oldest | Next Newest »

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