154 4. Particular Determinants
The theorem follows from (4.12.2) and (4.12.3) after replacing xby
x
− 1
.
In the next theorem,Pm(x) is the Legendre polynomial.
Theorem 4.55.
|Pm(x)|n=2
−(n−1)
2
(x
2
−1)
n(n−1)/ 2
.
0 ≤m≤ 2 n− 2
First Proof.Let
φm(x)=(1−x
2
)
−m/ 2
Pm(x).
Then
φ
′
m(x)=mF φm−^1 (x)
where
F=(1−x
2
)
− 3 / 2
φ 0 =P 0 (x)=1. (4.12.4)
Hence, ifA=|φm(x)|n, thenA
′
= 0 andA=|φm(0)|n.
|Pm(x)|n=
∣
∣(1−x^2 )m/^2 φ
m(x)
∣
∣
n
, 0 ≤m≤ 2 n− 2
=(1−x
2
)
n(n−1)/ 2
|φm(x)|n
=(1−x
2
)
n(n−1)/ 2
|φm(0)|n
=(1−x
2
)
n(n−1)/ 2
|Pm(0)|n.
The formula
|Pm(0)|n=(−1)
n(n−1)/ 2
2
−(n−1)
2
is proved in Theorem 4.50 in Section 4.11.3 on determinants with binomial
and factorial elements. The theorem follows.
Other functions which contain orthogonal polynomials and which satisfy
the Appell equation are given by Carlson.
The second proof, which is a modified and detailed version of a proof
outlined by Burchnall with an acknowledgement to Chaundy, is preceded
by two lemmas.
Lemma 4.56. The Legendre polynomialPn(x)is equal to the coefficient
oft
n
in the polynomial expansion of[(u+t)(v+t)]
n
, whereu=
1
2
(x+1)
andv=
1
2
(x−1).
Proof. Applying the Rodrigues formula forPn(x) and the Cauchy
integral formula for thenth derivative of a function,
Pn(x)=
1
2
n
n!
D
n
(x
2
−1)
n