14.8. THE CAUCHY FORMULA 359
Consider 14.21 which involves identifying the an in terms of the derivatives of f . This isobvious if k = 0. Suppose it is true for k. Then for small h ∈ C,
1h
(f (k) (z+h)− f (k) (z)
)=
1h
∞
∑n=k
n(n−1) · · ·(n− k+1)an
((z+h− z0)
n−k− (z− z0)n−k)
=1h
∞
∑n=k+1
n(n−1) · · ·(n− k+1)an
∑n−kj=0
(n− k
j
)h j (z− z0)
(n−k)− j
−(z− z0)n−k
=
∞
∑n=k+1
n(n−1) · · ·(n− k+1)an
(n−k
∑j=1
(n− k
j
)h j−1 (z− z0)
(n−k)− j
)
=∞
∑n=k+1
n(n−1) · · ·(n− k+1)(n− k)an (z− z0)(n−k)−1
+h∞
∑n=k+1
n(n−1) · · ·(n− k+1)an
(n−k
∑j=2
(n− k
j
)h j−2 (z− z0)
(n−k)− j
)
By what was shown earlier, limsupn→∞ ∥an∥1/n |z− z0|< 1. Consider the norm of the partin the above which multiplies h, |h|< 1.∥∥∥∥∥n(n−1) · · ·(n− k+1)an
(n−k
∑j=2
(n− k
j
)h j−2 (z− z0)
(n−k)− j
)∥∥∥∥∥≤ n(n−1) · · ·(n− k+1)∥an∥|z− z0|(n−k)−2
n−k
∑j=2
(n− k
j
)(|h||z− z0|
) j−2
(mj
)=
m(m−1) · · ·(m− j+1)j!
=m(m−1) · · ·(m− j+3)(m− j+2)(m− j+1)
j ( j−1)( j−2)!
=
(m
j−2
)(m− j+2)(m− j+1)
j ( j−1)≤(
mj−2
)(m)(m−1)
2
Thus the above is no more than
≤ n(n−1) · · ·(n− k+1)∥an∥|z− z0|(n−k)−2 ·(n− k)((n− k)−1)
2
n−k
∑j=2
(n− kj−2
)(h
|z− z0|
) j−2
≤ n(n−1) · · ·(n− k+1)(n− k)((n− k)−1)∥an∥(
1+|h||z− z0|
)n−k