532 CHAPTER 19. HILBERT SPACES
Proof: By Theorem 19.3.4
d2 =G( fp1 , · · · , fpn , fm)
G( fp1 , · · · , fpn).
(fpi , fp j
)=∫ 1
0xpixp j dx =
11+ pi + p j
Therefore,
d2 =
∣∣∣∣∣∣∣∣∣∣∣∣
11+p1+p1
11+p1+p2
· · · 11+p1+pn
11+m+p1
11+p2+p1
11+p2+p2
· · · 11+p2+pn
11+m+p2
......
......
11+pn+p1
11+pn+p2
· · · 11+pn+pn
11+pn+m
11+m+p1
11+m+p2
· · · 11+m+pn
11+m+m
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
11+p1+p1
11+p1+p2
· · · 11+p1+pn
11+p2+p1
11+p2+p2
· · · 11+p2+pn
......
...1
1+pn+p11
1+pn+p2· · · 1
1+pn+pn
∣∣∣∣∣∣∣∣∣∣Now from the Cauchy identity, letting ai = pi +
12 and b j =
12 + p j with pn+1 = m, the
numerator of the above equals
∏ j<i≤n+1 (pi− p j)(pi− p j)
∏i, j≤n+1 (pi + p j +1)
=∏
nk=1 (m− pk)
2∏ j<i≤n (pi− p j)
2
∏ni=1 (m+ pi +1)∏
nj=1 (m+ p j +1)∏i, j≤n (pi + p j +1)(2m+1)
=∏
nk=1 (m− pk)
2∏ j<i≤n (pi− p j)
2
∏ni=1 (m+ pi +1)2
∏i, j≤n (pi + p j +1)(2m+1)
while the denominator equals∏ j<i≤n (pi− p j)
2
∏i, j≤n (pi + p j +1)
Therefore,
d2 =
(∏
nk=1(m−pk)
2∏ j<i≤n(pi−p j)
2
∏ni=1(m+pi+1)2
∏i, j≤n(pi+p j+1)(2m+1)
)(
∏ j<i≤n(pi−p j)2
∏i, j≤n(pi+p j+1)
)=
∏nk=1 (m− pk)
2
∏ni=1 (m+ pi +1)2 (2m+1)