Tetration series for integer exponent. Can you find the pattern?

[marraco]

You are great, Sheldonison.

(I had lot of work, so I was unable to answer for like a week.)

I sampled 1000 points around the point r=1.1, with a radius R=.01.

this is the code:

Code:

\r fatou.gp
\p 100
/*Integration path over a semicircle*/

r=1.1            /*center of calculation*/
R=(r-1)/10       /*Radius of integration*/
Ndq=1000         /*number of angles*/
dq=2*Pi/N°dq     /*differential of angle*/


/*Calculation points/path z of integral*/
/*[z0,z1,z2,...,z(Ndq)=z0]*/ /*last point==first*/
z=vector(1+Ndq,n,  r+R*exp(I*(n-1)*dq)  )


/*(z-a) average over the integration interval*/
/*[0_1,1_2,2_3,...,Ndq_0]*/
z_a=vector(Ndq,n,  R*exp(I*(n*dq-dq/2))   )


/*differential vector*/
/*[d1_0,d2_1,d3_1,...,d0_Ndq]*/
dz=vector(Ndq,n,  dq*(z_a[n]*I) )

/*start integrand numerator*/
    /*calculation of tetrations*/
    TetZ = z*0  /*Same size as z, and initializated=0*/
    /*[Tz0,Tz1,Tz2,...,Tz(Ndq)=Tz0]*/ /*last point==first*/
    {
    for(n=1,1+Ndq,
       if(imag(z[n])<0,
          sexpinit(conj(z[n]));
          TetZ[n]=conj(sexp(1.5));
          ,
          sexpinit(z[n]);
          TetZ[n]=sexp(1.5);
        );
    )};
    
    /*Calculation of numerator of integrand*/
    /*Same size as z_a*/
    /*[I0_1,I1_2,I2_3,...,INdq_0]*/
    Integrand_num=vector(length(z_a),n, (TetZ[n]+TetZ[n+1])/2 *  dz[n]  )
/*end integrand numerator*/

/*Cauchy formula*/
jDerivative=vector(50,j,(j-1)!/(2*Pi*I)*sum(n=1,length(Integrand_num),Integrand_num[n]/z_a[n]^(j)))

(I need the Taylor centered at 1, so I cannot avoid calculating inside the shelltron region)

It took days to run, but I only got 7 converging derivatives (I know that the Taylor series (around r) should be equal to 1 when evaluated at 1, so I count the converged derivatives as the ones that give a result closest to the correct one.

I also was forced to pick only the real part of the coefficients.

Anyways, I got those bj:

[0.998577152804501, 2.89169703329950, -9.32293754538342, 410.790219668510, -8400.01108400142, 77233.5910778554, 1116744.84553185]

Unfortunately, it is not much better than the last attempt:

Code:

%38 = [1.000152704474621237210741951681248733595287259524430467296, 2.861320494145353285475346429548911736678675196129994723805, -13.6475978846314412502384523327046630578, 756.92186972158813546692266102007258863]

maybe using a fourier series to interpolate fatou.gp results improve the integral.

I will attempt to implement a conjugate gradient on the red blue equation, use knesser.gp to get a very good first guess, and increase precision.
That also may lead nowhere, since the red blue eq is mined with local minima.