jaydfox Wrote:Now comes the interesting part. Since F(e^x) = F(x)+1, we can solve for F(e^x)-F(x) = 1Jay, an additional remark.

Given P_F as the power series for F, we can find F(e^x) by multiplying , and then subtract , where I is the identity matrix. Set this equal to a column vector of [1, 0, 0, ..]~, and then solve for :

On the rhs of the equation as well as the right term on the lhs you have only vectors, and it seems even truncated vectors(row 0 missing). If I approach such questions I usually try to generalize also the rhs and the right term to have square matrices; in this case, to have a full identity matrix on the result-side of the equation. The advantage of this is then, that all relations are also expressible in their inverted view (or it is shown, that this is impossible, or only conditionally possible, if such an inversion is impossible/ not meaningful in a context).

I would like to know, what a matrix of [P_F0,P_F1,P_F2,...] looks like, if the [0,1,0,0,...] vector is also expanded (into identity matrix), and P_F1 is your P_F with additional leading row.

Do you know something about it (perhaps it is the eigensystem)? (or did I simply overlook some obvious?)

Gottfried

Gottfried Helms, Kassel