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Half-iteration of x^(n^2) + 1 - tommy1729 - 02/13/2017
I was intrested in the half-iteration of f_n(x) = x^(n^2) + 1 for Large x. For instance g_n(x) = f_n ^[1/2](x) - x^n. H_n(x) = f_n ^[1/2](x) / x^n. For Large x : Is abs g_n(x) increasing or decreasing with n ? Is abs H_n(x) decreasing ? Probably abs g_n is increasing and abs H_n decreasing. The focus is on integer n and branch structure. But also if n is real , are these functions analytic in n ? Perturbation Theory suggests this. I wonder how these functions look like on the complex plane , especially with resp to n. Regards Tommy1729 RE: Half-iteration of x^(n^2) + 1 - Xorter - 03/04/2017
(02/13/2017, 12:12 PM)tommy1729 Wrote: I was intrested in the half-iteration of f_n(x) = x^(n^2) + 1 for Large x. Okey, I got some Taylor series of the half -iteration of f_n(x) = x^(n^2)+1 by a PARI/gp programme code: I know these are not the best results, but this is that I could get from my programme. Here is the code: Code: `init()={` I hope it helps you, and you can develope this code. If you can, please share it with me. RE: Half-iteration of x^(n^2) + 1 - tommy1729 - 03/09/2017
Thank u for your reply. However i have questions 1) your coëfficiënt 1/2048 occurs twice !? Are you sure about that. 2) also the coëfficiënts : i noticed all of them ( though truncated ) are positive. Does this pattern remain ? Are they correct ? 3) not sure how you computed it. I assume no fixpoint but a kind of carleman matrix method ? If you used a fixpoint , which one ? 4) im intrested in using the fix with largest real part. Regards Tommy1729. RE: Half-iteration of x^(n^2) + 1 - Xorter - 03/09/2017
(03/09/2017, 01:28 PM)tommy1729 Wrote: Thank u for your reply. I did not use fixpoint, because by the Carleman matrix it can be calculated, too. You can see above, how I computed. Just save it in gp and open it with gp.exe and enter this code: Decar(Msqrt(Car(x^4+1,20),20,5),20)*1.0 Where Car makes a 20x20 Carleman matrix from x^4+1, Msqrt get its square root and Decar gets the Taylor series of the function from the matrix. It is simple, because: M[f]M[g]=M[fog] thus sqrt M[f] = M[f^o0.5], right? Of course! N root of M[f] = M[f^o1÷N] If you check the code above, you can see it has a lot of (infinity) part with negative sign. Naturally, it is not perfect, the bigger Carleman matrices you use, the better the results are. |