330 Appendix
=−
1
n+1[
n+1
∑r=0(−1)
r(
n+1r)
x2 r
−n+1
∑r=0(−1)
r(
n+1r)
]
=−
1
n+1[(1−x2
)n+1
−0]S(0) =−
1
n+1.
The result follows. It is applied withc= 1 in Section 4.10.4 on a particular
case of the Yamazaki–Hori determinant.
Example A.4. If
ψm=∞
∑r=1rm
xr
,then
∆
m
ψ 0 =xψm.ψmis the generalized geometric series (Appendix A.6).
Proof.
(r−1)m
=m
∑s=0(−1)
m−s(
ms)
rs
.Multiply both sides byx
r
and sum overrfrom 1 to∞. (In the sum on theleft, the first term is zero and can therefore be omitted.)
x∞
∑r=2(r−1)m
xr− 1
=∞
∑r=1xrm
∑s=0(−1)
m−s(
ms)
rs
,x∞
∑s=1sm
xs
=m
∑s=0(−1)
m−s(
ms) ∞
∑
r=1rs
xr
,xψm=m
∑s=0(−1)
m−s(
ms)
ψs=∆
m
ψ 0.This result is applied in Section 5.1.2 to prove Lawden’s theorem.
A.9 The Euler and Modified Euler Theorems on
Homogeneous Functions
The two theorems which follow concern two distinct kinds of homogeneity
of the function
f=f(x 0 ,x 1 ,x 2 ,...,xn). (A.9.1)