| ( | n k | ) | + | ( | n k+1 | ) | = | ( | n+1 k+1 | ) | , denn | n(n-1)...(n-k+1)k! | + | n(n-1)...(n-k-1+1)(k+1)! | = | n(n-1)...(n+1-(k+1)+1)(k+1+n-k)k!(k+1) |
| an | an-1 | an-2 | ... | a2 | a1 | a0 |
| b0,n | b0,n-1 | b0,n-2 | ... | b0,2 | b0,1 | b0,0 |
| b1,n | b1,n-1 | b1,n-2 | ... | b1,2 | b1,1 | |
| ... | ... | ... | ... | ... | ||
| bn-1,n | bn-1,n-1 | |||||
| bn,n |
x0 := 1 für alle x
bi,n-k = x×bi,n-k+1 + bi-1,n-k für k > 0
und i > -1
bi,n-k = bi-1,n-k=an-k für k = 0 und i > -1
bi,n-k = x×bi,n-k+1 + an-k für k > -1 und i = 0
| Behauptung: bi,n-k =
|
(3.1) |
= an.
und
=
. | a) |
bt,n-(s+1) = x×bt,n-(s+1)+1 + bt-1,n-(s+1) bt,n-(s+1) = bt,n-(s+1) bt,n-(s+1) bt,n-(s+1) bt,n-(s+1) bt,n-(s+1)
|
| b) | bt+1,n-s = x×bt+1,n-s+1 + bt,n-s bt+1,n-s = bt+1,n-s bt+1,n-s bt+1,n-s bt+1,n-s bt+1,n-s
|
QED
q=0
(q+1)aqxq = folgt
f (i)(x) =
n-i
q=0
(q+1)(q+2)...(q+i)aq+ixq.
Also ist i! bi,i = f (i)(x).