bo198214 Wrote:However I would guess that the claimed uniqueness for a solution satisfying 1' and 2' is not guarantied. We can use different approximations, for example for a given constant we can consider the equation systems, resulting from letting for . Because interestingly the sum converges to where e is the Euler constant and B_k are the Bell numbers. So by setting for the remaining sum converges and merely introduces an additive constant in the linear equation system. The obtained solutions are different from the solution obtained by c=0.

However I didnt verify yet the convergence properties of these alternative solutions.

I will do some testing, but my hunch is that the series will still converge on the "correct" solution, despite your attempt to break the solution. Because the modification you made is convergent, it has no radius of convergence, and hence it does not introduce a "false" singularity. Therefore, the solutions as n grows should still converge on the correct solution (since it does have a radius of convergence, and hence the terms will be much larger in magnitude), even if that convergence is slower.

If you made a modification that introduced a false singularity, I think that could possibly break the convergence, especially if the false singularity were closer to the origin than the true singularity, i.e. gave a root test (for terms k>n) greater than abs(1/c_0).

I already have code in place to remove the "correct" singularity, and I could just modify it to instead introduce a false set of coefficients for k>n, and see what happens. As I think more about what a partial series (only terms k>n) of a singularity would look like as a function, i.e., an increasingly insignificant singularity, I'm actually going to guess that the solution will still converge on the correct one, even if only very, very slowly. (And convergence with v_k=0, k>n, is already very slow.)

~ Jay Daniel Fox