[AIS] (alternating) Iteration series: Half-iterate using the AIS?

[Gottfried]

Here is some explanation in terms of Pari/GP-code, how the serial alternating iteration sums asum(x) can be expressed/computed with the help of power series.

Preliminaries: Globale variable for base, log of the base and dim for dimension. Then the stdcall for the exponential/logarithm with a variable (but integer) heightparameter:

Code:

[tb = 1.3, tl=log(tb)]
dim=64    \\ for power series expansion to "dim" terms, and matrix-size
\ps 64

  \\ for direct access to the iteration height
exph(x,h=1) = for(k=1,h, x= exp(x*tl)-1);for(k=1,-h,x=log(1+x)/tl);return(x);

glx = 1.0    \\ for sequential access to the iteration by one step
{nextx(x,h=0)= if (h==0, glx=x;return(glx));
    glx=if(h>0,exph(glx,1)
              ,exph(glx,-1));
  return(glx); }

First, the formula for the alternating iteration series; this is "asum(x)". This is simply the summa-tion of the iterated exponentials/logarithms.

Because we want to compare this with a true power series solution (via the Carleman-matrix-concept) where we have to separate this in two parts, we do this here too; the alternating iteration series towards fp0 is asump(x), and that towards the fp1 it is asumn(x). After adding both two segments we have to reduce the joint sum by x because it was doubly accounted for:

Code:

asump(x)= sumalt(k=0,(-1)^k*nextx(x, k))
asumn(x)= sumalt(k=0,(-1)^k*nextx(x,-k))
asum(x) = asump(x)+asumn(x) - x
\\ -------------------------------------------------------------

Now we create Carleman-matrices for that sums, that means we shall get power series for that sums. First we need the Carleman-matrices for tb^x-1 developed around fp0 and that developed around fp1. These are the "Carl0" and the "Carl1"-matrices.

Then the matrices, which provide the coefficients for the power-series for the alternating sums, are created using the Neumann-series-expression for that Carlemanmatrices (CarlAsp,CarlAsn)
Note the small difference in the formula for the CarlAsn compared to that of CarlAsp.

\\ Series and Carleman-matrix related to fixpoint fp0 (= 0)

Code:

coeffs0 = polcoeffs(exp(tl*(x+fp0))-1  - fp0, dim);coeffs0[1]=0;
   print(    coeffs0)   \\ that statement to display coeffs0 in the gp-dialogue
Carl0   = matfromser(coeffs0);           \\ Carlemanmatrix for tb^x-1 by powerseries around fp0 (=0)
CarlAsp = (matid(dim) + Carl0)^-1      \\ Newtonseries-matrix for Carl0 giving series-coefficients for "asump" (= the altern. series towards fp0)
coeffs_asp =CarlAsp[,2]                    \\ that coefficients in a vector,  for computation of "asp_mat(x-fp0)+fp0" = "asp_mat(x)" because fp0=0

\\ Series and Carleman-matrix related to fixpoint fp1
coeffs1 = polcoeffs(exp(tl*(x+fp1))-1  - fp1, dim);coeffs1[1]=0;
   print(    coeffs1 )  \\ that statement to display coeffs1 in the gp-dialogue
Carl1= matfromser(coeffs1);             \\ Carlemanmatrix for tb^(x+fp1)-1-fp1 by powerseries around fp1 (=???)
CarlAsn = (matid(dim) + Carl1^-1)^-1     \\ Newtonseries-matrix for Carl1^-1 giving coefficients for "asumn" (= the altern. series towards fp1)
coeffs_asn = CarlAsn[,2]               \\ that coefficients in a vector, for computation of "asn_mat (x-fp1)+fp1"

{asump_mat(x)=local(h,su);
   while(abs(x-fp0)>0.5,    \\ move the argument x for the powerseries towards fp0
           su=su+x; x=exph(x,1);
           su=su-x; x=exph(x,1);
    );
   x = x-fp0;
   su= su + sum(k=0,dim-1,x^k*coeffs_asp[1+k]) ;  \\ evaluate asp_mat at x_h
   su= su + fp0/2;
   return(su); }

{asumn_mat(x)=local(h,su);
   while(abs(x-fp1)>0.5,          \\ move the argument x for the powerseries towards fp1
           su=su+x; x=exph(x,-1);
           su=su-x; x=exph(x,-1);
    );
   x = x-fp1;
   su= su + sum(k=0,dim-1,x^k*coeffs_asn[1+k]) ; \\ evaluate asn_mat at x_(-h)
   su= su + fp1/2;
   return(su); }

asum_mat(x)=asump_mat(x)+asumn_mat(x)-x
\\ ==============================================

Let us ignore the integer-height iterations to shift an argument x sufficiently towards the fixpoints, then the formulae for the power-series-solutions are

\( \text{asump}(x) = \sum_{k=0}^{\infty} p_k \cdot x^k \)

where the coefficients p are taken from the coeffs_asp-array.

\( \text{asumn}(x) =\frac{t}{2}+ \sum_{k=0}^{\infty} n_k \cdot (x-t)^k \)

where the coefficients n are taken from the coeffs_asn-array and t denotes the upper fixpoint fp1.


Now we test these functions:

Code:

\\ test that at x0=1.0
x0=1.0
[aser=asump(x0),amat=asump_mat(x0),err=aser-amat]
[aser=asumn(x0),amat=asumn_mat(x0),err=aser-amat]
[aser=asum (x0),amat=asum_mat (x0),err=aser-amat]

\\         serial           matrix-based        difference
%1269 = [0.764698042872,   0.764698042872,   7.30924528016 E-132]
%1270 = [0.245204215222,   0.245204215222,   1.43545677556 E-118]
%1271 = [0.00990225809450, 0.00990225809450, 1.43545677556 E-118]

The errors with power series truncated to 64 terms are far smaller than 1e-100. So we should accept this as very good approximation and I think it is a reasonable hypothese to assume it a true analyti-cal solution.

However, for me it is somehow unusual to have two power series to include in the computation, and also the initial "shift" of the x-argument towards the resp. fixpoints by integer iterations, such that the power series converge. I have no idea, how, for instance, could this process then be inverted and get a power series (even if only a formal power series) assigned to...

---

Here are the first few coefficients for the power series for reference for you, if you want to check this. Column 1 and 2 for the b^x-1 resp log(1+x)/log(b)-expressions, then column 3 for asump_mat and column 4 for asumn_mat .

\( \small \begin{array} {r|r|r|r}
tb^x-1 & tb^{(x+fp_1)}-1-fp_1 & coeffsasp(x) & coeffsasn(x) \\
\hline \\
0 & 0 & 0 & 0 \\
0.262364264467 & 2.52769252337 & 0.792164376122 & 0.716528582529 \\
0.0344175036348 & 0.331588094847 & -0.0255084462139 & 0.0127206425725 \\
0.00300997434198 & 0.0289989555369 & -0.00188959012829 & -0.000764040945147 \\
0.000197427426075 & 0.00190207240994 & -0.0000721182735413 & 0.0000509718272678 \\
0.0000103595802856 & 0.0000998071657596 & 0.00000283330292899 & -0.00000353974018266 \\
0.000000452997276970 & 0.00000436430560552 & 0.000000868177178099 & 0.000000246153936735 \\
0.0000000169786139111 & 0.000000163576832872 & 0.000000108306933352 & -0.0000000164875359311 \\
0.000000000556822693809 & 0.00000000536458943004 & 0.0000000106042541295 & 0.000000000995691627932 \\
1.62322640556E-11 & 1.56386284443E-10 & 0.000000000903757476731 & -4.44139406950E-11 \\
4.25876601958E-13 & 4.10301724906E-12 & 6.72094524757E-11 & -4.06144047569E-13 \\
1.01577092206E-14 & 9.78622820588E-14 & 3.96472142050E-12 & 4.80559320387E-13 \\
2.22084992361E-16 & 2.13963047096E-15 & 9.51525791516E-14 & -8.90297750191E-14 \\
4.48208966694E-18 & 4.31817365188E-17 & -2.18803878944E-14 & 1.27988586030E-14 \\
\ldots & \ldots & \ldots & \ldots
\end{array} \)