11/05/2007, 07:42 AM

By the way, looking at it from this point of view helps make it clear why solving directly for the sexp was doomed to failure. You see, solving for the slog means composing F(z) with exp(z), to get F(exp(z)).

On the other hand, for a sexp function T, solving it would require the composition exp(T(z)). Exponentiating T necessarily means creating column vectors of powers of T, which means we are no longer dealing with a linear system.

And this is probably a general situation: finding G(F(x)), given the known power series for G and an unknown power series F, is going to lead to a non-linear system. However, F(G(x)) is still a linear system, because G is known.

On the other hand, for a sexp function T, solving it would require the composition exp(T(z)). Exponentiating T necessarily means creating column vectors of powers of T, which means we are no longer dealing with a linear system.

And this is probably a general situation: finding G(F(x)), given the known power series for G and an unknown power series F, is going to lead to a non-linear system. However, F(G(x)) is still a linear system, because G is known.

~ Jay Daniel Fox