The example mentioned above [exp(x)/loggamma(1/x)] is typical for the
kind of expansion we do in limits, in this case we need two terms. This
takes about 0.5 seconds currently.
This seems to me can be done with a bivariate series
in x and w, where w = 1/y = 1/log(x)
Maybe this was a bad example. My point is not the log(x) - this should
be treated as an independent symbol, yes. My point is can your fast
taylor function cooperate with the ordinary series expansion code for
things it does not know how to expand?
multivariate series are not yet in lpoly, but I plan to
translate them from rmpoly.
Using rmpoly
import sys
from time import time
from rmpoly import *
from sympy import *
rp, x,w = rgens('x,w',8,sympify)
hx = 4
hw = 3
t0 = time()
p1 = 1 - x/2 + w - x*w*log(2*pi)/2 - x**2*w/12
p1 = p1.pow_trunc(-1,['x','w'],[hx,hw-1]).mul_trunc(w,'w',hw)
p2 = -x.exp('x',hx).mul_trunc(x,'x',hx)
p = p2.mul_trunc(p1,'x',hx)
t1 = time()
print p1
print '%.2f' %(t1-t0)
The output is
+(-5/12 + 3*log(2*pi)/8)*w^2*x^3 +(-2/3 + log(2*pi)/2)*w^2*x^2 +(-1 +
log(2*pi)/2)*w^2*x +(-1)*w^2 +(1/8)*w*x^3 +(1/4)*w*x^2 +(1/2)*w*x
+(1)*w
0.01
Is this the result expected? If I have misunderstood the
problem please tell me.
with hw=10 it takes 0.04s
I guess that in lpoly it would be something like
p1 = multi_taylor(-w*(x*exp(x)*(1-x/2+w -x*w*log(2*pi)/2 -x**2*w/
12)**-1, [x,y],[hx,hw])
and that it would be 2x or 3x slower.
Mario
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