224 5. Further Determinant Theory
=
∞
∑m=0zm
um+1m!∞
∑r=0zr
Frr!. (5.7.13)
Equating coefficients ofz
n
,Fn+1n!=
n
∑r=0ur+1Fn−rr!(n−r)!,
Fn+1=n
∑r=0(
nr)
ur+1Fn−r. (5.7.14)This recurrence relation inFnis identical in form to the recurrence relation
inEngiven in (5.7.7). Furthermore,
E 1 =F 1 =u 1 ,E 2 =F 2 =u2
1
+u 2.Hence,
En=Fnwhich proves the theorem.
Second proof.Express the lemma in the form∞
∑i=0z
ii!H
i
(f, g)=f(x+z)g(x−z). (5.7.15)Hence,
H
i
(f, g)=[
D
i
z
{f(x+z)g(x−z)}]
z=0. (5.7.16)
Put
f(x)=eF(x)
,g(x)=eG(x)
,w=F(x+z)+G(x−z).Then,
H
i
(eF
,eG
)=[
D
i
z(ew
)]
z=0=[
D
i− 1
z
(ew
wz)]
z=0=
i− 1
∑j=0(
i− 1j)
[
D
i−j
z
(w)Dj
z
(ew
)]
z=0=
i− 1
∑j=0(
i− 1j)
ψi−jHj
(eF
,eG
),i≥ 1 , (5.7.17)where
ψr=[
D
r
z
(w)]
z=0