A.10 Formulas Related to the Function (x+
√
1+x^2 )
2 n
333
Proof. Replacexby−x
− 1
in (A.10.1), multiply byx
2 n
, and putx
2
=z.
The result is
(−1+
√
1+z)
2 n
=
[
z
n
+
n
∑
i=1
λniz
n−i
]
−(1 +z)
1
2
n
∑
i=1
μniz
n−i
=
n+1
∑
j=1
λn,n−j+1z
j− 1
−(1 +z)
1
2
n
∑
j=1
μn,n−j+1z
j− 1
.
Rearrange, multiply by (1 +z)
− 1 / 2
and apply (A.10.4):
∞
∑
i=0
νiz
i
n+1
∑
j=1
λn,n−j+1z
j− 1
=
n
∑
j=1
μn,n−j+1z
j− 1
+(1+z)
− 1 / 2
(−1+
√
1+z)
2 n
.
In some detail,
(1 +ν 1 z+ν 2 z
2
+···)(λnn+λn,n− 1 z+···+λn 1 z
n− 1
+λn 0 z
n
)
=(μnn+μn,n− 1 z+···+μn 1 z
n− 1
)+
(
z
2
) 2 n
(1 +z)
− 1 / 2
(
1 −
1
4
z+···
) 2 n
.
Note that there are no terms containingz
n
,z
n+1
,...,z
2 n− 1
on the right-
hand side and that the coefficient ofz
2 n
is 2
− 2 n
. Hence, equating coefficients
ofz
n−1+i
,1≤i≤n+1,
n+1
∑
j=1
λn,j− 1 νi+j− 2 =
{
0 , 1 ≤i≤n
2
− 2 n
,i=n+1.
The theorem appears when n is replaced by (n−1) and is ap-
plied in Section 4.11.3 in connection with a determinant with binomial
elements.
It is convenient to redefine the functionsλnrandμnrfor an application
in Section 4.13.1, which, in turn, is applied in Section 6.10.5 on the Einstein
and Ernst equations.
Ifnis a positive integer,
(x+
√
1+x
2
)
2 n
=gn+hn
√
1+x
2
, (A.10.6)
where
gn=
n
∑
r=0
λnr(2x)
2 r
, an even function,
g 0 =1;
hn(x)=
n
∑
r=1
μnr(2x)
2 r− 1
, an odd function,
h 0 =0. (A.10.7)