Page 348 - Determinants and Their Applications in Mathematical Physics

√
2 2n
A.10 Formulas Related to the Function (x + 1+ x ) 333
2
Proof. Replace x by −x −1 in (A.10.1), multiply by x , and put x = z.
2n
The result is

√ 1
n
n
(−1+ 1+ z) 2n = z + λ ni z n−i − (1 + z) 2 µ ni z n−i
n
i=1 i=1
n+1 n
j−1 1 j−1
= λ n,n−j+1 z − (1 + z) 2 µ n,n−j+1 z .
j=1 j=1
Rearrange, multiply by (1 + z) −1/2 and apply (A.10.4):
∞ n+1 n
j−1 j−1 −1/2 √
ν i z i λ n,n−j+1 z = µ n,n−j+1 z +(1+z) (−1+ 1+ z) .
2n
i=0 j=1 j=1
In some detail,
2
(1 + ν 1 z + ν 2 z + ···)(λ nn + λ n,n−1 z + ··· + λ n1 z n−1 + λ n0 z )
n
z 2n −1/2 1
1 2
2n
n−1
=(µ nn + µ n,n−1 z + ··· + µ n1 z )+ (1 + z) 1 − z + ··· .
2 4
Note that there are no terms containing z , z n+1 ,...,z 2n−1 on the right-
n
hand side and that the coefficient of z 2n is 2 −2n . Hence, equating coefficients
of z n−1+i ,1 ≤ i ≤ n +1,
n+1
0, 1 ≤ i ≤ n

λ n,j−1 ν i+j−2 =
2 −2n ,i = n +1.
j=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 functions λ nr and µ nr for an application
in Section 4.13.1, which, in turn, is applied in Section 6.10.5 on the Einstein
and Ernst equations.
If n is a positive integer,
+ +
2
(x + 1+ x ) = g n + h n 1+ x , (A.10.6)
2 2n
where
n

g n = λ nr (2x) , an even function,
2r
r=0
g 0 =1;
n
2r−1
h n (x)= µ nr (2x) , an odd function,
r=1
h 0 =0. (A.10.7)