320 Appendix
The polynomial
ψmn(x)=
m
∑
r=0
(
m
r
)
αn+rx
r
satisfies the relations
ψ
′
mn=mψm−^1 ,n+1,
ψmn−ψm− 1 ,n=xψ
′
mn
=mxψm− 1 ,n+1.
Exercises
1.Prove that
φm(x−h)=
m
∑
r=0
(
m
r
)
(−h)
r
φm−r(x)
=∆
m
hφ^0.
2.If
Sm(x)=
∑
r+s=m
φrφs,
Tm(x)=
∑
r+s+t=m
φrφsφt,
prove that
S
′
m
=(m+1)Sm− 1 ,
Sm(x+h)=
m
∑
r=0
(
m+1
r
)
h
r
Sm−r(x),
T
′
m=(m+2)Tm−^1 ,
Tm(x+h)=
m
∑
r=0
(
m+2
r
)
h
r
Tm−r(x).
3.Prove that
φ
− 1
m =
1
αm
∞
∑
n=0
(−1)
n
cmnx
n
,
where
cm 0 =1,
cmn=
1
α
n
m
∣
∣
∣
∣
(
m
m−i+j− 1
)
αm−i+j− 1
∣
∣
∣
∣
n
,n≥ 1.
This determinant is of Hessenberg form, is symmetric about its sec-
ondary diagonal, and contains no more than (m+ 1) nonzero diagonals
parallel to and including the principal diagonal.