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5.1 Determinants Which Represent Particular Polynomials 173
1
φ n =
(1 − x) n+1
1 − x
x
1 1 − x
0
1/2! 1 1 − x
0
1/3! 1/2! 1 1 − x 0 (5.1.8)
.....................................................
1/(n − 1)! 1/(n − 2)! .............. 1 1 − x 0
1/n! 1/(n − 1)! .............. 1/2! 1 0
n+1
After expanding the determinant by the single nonzero element in the last
column, the theorem follows from (5.1.2) and (5.1.4).
Exercises
Prove that
α 0 α 1 α 2 α 3 ··· α n−1 α n
−y x
n
−y x
1. α r x n−r r .
−y x
y =
r=0
.................
−y x
n+1
1 x x 2 x 3 ··· x n−1 x n
2
−1 y xy x y ··· x n−2 y x n−1
y
−1 y xy ··· x y x
n−3 n−2
y
n .
−1 y ··· x n−4 y x n−3
2. (x + y) =
y
...........................
−1 y
n+1
x b
n
3. (−b) 2 F 0 , −n; −
a a
b
−c 1
a −c 2 b
2a b
−c 3
3a −c 4 ,
=
··· ··· ···
−c n−1
b
(n − 1)a −c n n
where
c r =(r − 1)a + b + x. (Frost and Sackfield)
and 2 F 0 is the generalized hypergeometric function.