90
W. Herr and E. Forest
c · (q 0 , q 1 , q 2 , . . . q N ) = (c · q 0 , c · q 1 , c · q 2 , . . . c · q N )
(3.149)
(q 0 , q 1 , q 2 , . . . q N ) · (r 0 , r 1 , r 2 , . . . r N ) = (s 0 , s 1 , s 2 , . . . s N )
(3.150)
with:
s i =
i
k=0
i!
k!(i − k)!
q k r i−k
(3.151)
If we had started with:
x = (a, 1, 0, 0, 0 . . .)
(3.152)
we would get:
f (x) = ( f (a), f
(a), f
(a), f
(a), . . . f
(n) (a) )
(3.153)
Some special cases are:
(x, 0, 0, 0, ..)
n
= (x
n , 0, 0, 0, ..)
(3.154)
(0, 1, 0, 0, ..)
n
= (0, 0, 0, ..,
n+1
n! , 0, 0, ..)
(3.155)
(x, 1, 0, 0, ..)
2
= (x
2 , 2x, 2, 0, ..)
(3.156)
(x, 1, 0, 0, ..)
3
= (x
3 , 3x
2 , 6x, 6, ..)
(3.157)
As another exercise one can consider the function f (x) = x −3
f (x) → f (x, 1, 0, 0, ..) =
(x, 1, 0, 0, ..)
−3
= (f 0 , f
f
, f
, ..)
next : multiply both sides with (x, 1, 0, 0) 3
(1, 0, 0, ..) = (x, 1, 0, 0, ..) 3 · (f 0 , f , f , f , . . .)
(1, 0, 0, ..) = (x 3 , 3x 2 , 6x, 6, ..) · (f 0 , f , f , f , . . .) using (3.157)
(1, 0, 0, ..) = (x
3
· f 0
q 0 = 1
, 3x
2
· f 0 + x
3
·f
q 1 = 0
, 6x · f 0 + 2 · 3x
2
· f
+ x
3
·f
q 2 = 0
, . . .)
This can easily be solved by forward substitution:
1 = x 3 · f 0
→ f 0 = x −3
0 = 3x 2 · f 0 + x 3 · f
→ f = − 3x −4
0 = 6x · f 0 + 2 · 3x 2 · f + x 3 · f
→ f = 12x −5
....
(3.158)