08/12/2009, 04:49 PM

(08/12/2009, 03:47 PM)Gottfried Wrote:Yes, this is it! This is much faster than what I was using before!(08/12/2009, 02:29 PM)jaydfox Wrote: Actually, since I'm trying to find the Taylor series, I don't worry about the factorials. So I work directly with the matrix above, where the first row is [1 -2 4 -8 16].

Inverting this and multiplying by 4!, or (n-1)! in general, I get back a matrix of integer coefficients. For this 5x5 matrix, I get:

Code:`[ 0 0 24 0 0]`

[ 2 -16 0 16 -2]

[ -1 16 -30 16 -1]

[ -2 4 0 -4 2]

[ 1 -4 6 -4 1]

Yes, exactly that's wat XI is for.

Using

I get ScI (= Sc^-1 without general matrix-inversion!)Code:`´`

dim=5

ScI = XI *FacI*PI *(dim-1)!

Code:`´`

0 0 24 0 0

2 -16 0 16 -2

-1 16 -30 16 -1

-2 4 0 -4 2

1 -4 6 -4 1

~ Jay Daniel Fox