11/06/2007, 01:58 PM

bo198214 Wrote:Sorry, it's how I picture it in my head and hence how I describe it. If you substituted the terms for the power series of 1/(z-3) for k>n, then you will have introduced a singularity to the solution. The system you solve would NOT just consist of the n terms you calculated, but an infinite number of terms, the first n of which are determined by the solution to the system, the remaining for k>n being given as an initial condition. Therefore, it would have a singularity at z=3, a simple pole in fact. This is a "false singularity", because it was introduced artificially, i.e., it was not predicted by the convergence of the finite systems. In fact, I would predict that convergence would nonetheless be to a series with singularities at the upper and lower fixed points (for base e).jaydfox Wrote: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.I am not sure what you mean by false singularity and false function, and how you would introduce a singularity somewhere. Despite I am very interested in the results you will come up with.

Quote:I also would suggest to stop calling one solution as being "correct" (as it doesnt fit the meaning of the word), I rather would suggest to call the solution gotten by Andrew's method as being "natural".I call it correct in the sense that the system should converge on the same solution, regardless of whatever initial conditions we specify for k>n, so long as those initial conditions do not attempt to reduce the radius of convergence below that of the system solved with 0's for k>n. But we can call it "natural" as well.

~ Jay Daniel Fox