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Half-iteration of x^(n^2) + 1
#1
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
Reply
#2
(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.

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

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()={
default(format,"g0.4");
}

D(z,n)={for(i=0,n-1,z=z');return(z)}

Car(f,dim)={return(subst(matrix(dim,dim,k,j,D(f^(j-1),k-1)/(k-1)!),x,0))}

Decar(M,dim)={
f=0;for(i=1,dim,f+=M[i,2]*x^(i-1));
return(f);
}

Msqrt(B,dim,prec)={
A=matid(dim);
for(i=0,prec,A=(B*A^-1+A)/2);
return(A);
}

I hope it helps you, and you can develope this code. If you can, please share it with me.
Xorter Unizo
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#3
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.
Reply
#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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