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Half-iteration of x^(n^2) + 1
#4
(03/09/2017, 01:28 PM)tommy1729 Wrote: 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.

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.
Xorter Unizo
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Messages In This Thread
Half-iteration of x^(n^2) + 1 - by tommy1729 - 02/13/2017, 12:12 PM
RE: Half-iteration of x^(n^2) + 1 - by Xorter - 03/04/2017, 11:43 AM
RE: Half-iteration of x^(n^2) + 1 - by tommy1729 - 03/09/2017, 01:28 PM
RE: Half-iteration of x^(n^2) + 1 - by Xorter - 03/09/2017, 10:02 PM

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