Abel's binomial theorem The following was assigned as homework ...

n−1 �
� �
n − 1
= n
x(x+kz)
k
k=0
k−1 [(y+z)+((n−1)−k)z] (n−1)−k = nfn−1(x, y+z,z).
(9)
It follows from the induction hypothesis that
nfn−1(x, y + z,z) =ngn−1(x, y + z,z),
and this finishes the proof of (5). We turn to the proof of (6). This will be proved
much like in the manner of the proof to (5). Since obviously gn(x, −x−nz, z) =
0, we have to prove fn(x, −x − nz, z) = 0. We are going to do it by induction
on n. The base case is trivial again. We have to prove the following two facts:
Let us start with
To verify (10),
fn(x, −x − nz, z) =
∂
∂z fn(x, −x − nz, z) = 0; (10)
fn(x, −x, 0) = 0. (11)
n�
k=0
∂
∂z fn(x, −x − nz, z)=(n−1)
(using k � � � � n n−1
k = n k−1 for k>0)
= n(n − 1)
j=0
n�
k=1
� �
n
(−1)
k
n−k x(x + kz) n−1 .
n�
k=0
� �
n
xk(−1)
k
n−k (x + kz) n−2
� �
n − 1
x(−1)
k − 1
n−k (x + kz) n−2
n−1 �
� �
n − 1
= n(n − 1)
x(−1)
j
n−1−j (x +(j+1)z) n−2 . (12)
We have to show that (12) is equal to zero. Use the hypothesis
to prove
0=fn−1(x, −x, −(n − 1)z,z)
n−1 �
0= (−1) n−1−j
� �
n − 1
x(x + jz)
j
n−2 . (13)
j=0
Substitute in (13) x + z to the place of x and multiply the resulting formula by
x/(x + z) to obtain that (12) is equal to zero.
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