Martin Rubey wrote:
>
> Waldek Hebisch <[EMAIL PROTECTED]> writes:
>
> > 3) I tried guessExpRat to force the result to be polynomial, but guessExpRat
> > seems to a lot of time (I was too impatient to actually see any answer). I
> > wonder what method is used in guessExpRat?
>
> Also described in the article. Yes, this method actually doesn't really work
> for lists longer than about 10 items, unless you have a supercomputer, that
> is.
>
> > guessRec seem to reasonably quickly find difference equation satisfied by a
> > function. Once we know f(n+1)/f(n) asymptotic at infinity should immediatly
> > give us exponential factor.
>
> The hard part is to get the asymptotic at infinity :-) As far as I know,
> functions of the form
>
> (a+b*n)^n*q(n)
>
> (a,b are fractions, q a rational function) does not necessarily satisfy a
> difference equation.
>
Hmm, if f(n) = (a+b*n)^nW_1(n)/W_2(n) and W_2(n+1) and W_2(n) are non-zero,
then we have
W_2(n+1)*W_1(n)*f(n+1) = (a+n*n)*W_2(n)*W_1(n+1)*f(n) (**)
If W_2(n+1) or W_2(n) is zero, then corresponding value of f is
undefined, so the equation is valid when ever it makes sense.
I guess that you are saying the following: given degree bound
d_1 for W_1 and d_2 for W_2 we may try to rewrite the equation in
linear form:
P_1*f(n+1) = P_2*f(n) (***)
with deg(P_1) = d_1 + d2 and deg(P_2) = 1 + d_1 + d_2. To
have unique solition we need 2*(d_1 + d_2 + 1) + 1 equations, so
2*(d_1 + d_2) + 3 data points. In nonliner equation (**) we
have d_1 + d_2 + 3 unknowns, so d_1 + d_2 + 4 data points
should be enough. Indeed, if I gave guessExpRat smaller
number of points, then it worked.
--
Waldek Hebisch
[EMAIL PROTECTED]
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