Borel summation

[bo198214]

No. It is just a (parametrizable) summation matrix,

But in the end it is regular iteration, I think.

I have no such matrix M for the Borel-summation because then another parameter (the "x" from the description above) must be taken in respect too ... and too much fiddling ...

Yeah, exactly. For the integral method, I numerically searched for the minimum of the corresponding function. But even there you have to give a good initial guess in which interval that minimum is to find.

Using the functional equation with \( z_{-40} = f°^{-40}(1) \approx 0.048759544221234150945 \)
then I can even use the simpler Euler-sum matrix with little order 0.25 partsums = ESum(0.25,64) * dV(z_40,64) * Mat( G[1..64] )
and the retransformation, iterating each of this partial sums 40 times forward  by   [b]  Mat( vectorv ( 64,r,  it ( partsums [r,1] , 40 ))) 

But isn't this quite \(f^{\circ 40}(h_{64}(f^{\circ -40}(x)))\), except that you Euler-summed \(h_{64}\) instead of just truncating the powerseries? This is what I meant - you actually don't need to care much about the power series, but you can get arbitrary precision via the iterations, i.e. just increase the 40, instead of to make efforts about the summation of the 64.