(08/25/2009, 08:12 PM)bo198214 Wrote:Sorry, I was misremembering some old posts of yours, where you show that the Abel function of x+c is x/c, or something like that.(08/25/2009, 07:56 PM)jaydfox Wrote: Well, when in doubt, test with something we know the answer for. If I recall correctly, this method can be adapted to find the Abel function of cx, namely the logarithm base c?This is still an open question whether intuitive slog of cx is log_c!
The method is too complicated to get grip of it theoretically.
Edit: Hmm, experimenting with f(x) = e*(x+1)-1 seems to be solving to the power series of log(x+1). So it does indeed seem to be solvable by this approach. Obviously we can't solve it at x=0.
The first 10 terms of a 250-term solution are 0.999350826516384*x - 0.498825775056848*x^2 + 0.336537863799800*x^3 - 0.258262994469142*x^4 + 0.203808987395232*x^5 - 0.157090948593760*x^6 + 0.121920358237637*x^7 - 0.103864057515499*x^8 + 0.101694333870558*x^9 - 0.108441429864967*x^10
~ Jay Daniel Fox

