Statement
Notation as on the
Statement (I) page;
the nonnegative zeros of pn are j=0,1,…,n and those of qn are
j=0,1,…,n+1.
Slopes at the zeros (Section 4, pp. 297--298). With 0!=1,
∣pn′(j)∣=(n+j)!(n−j)!,j=0,1,…,n,(16)
∣qn′(j)∣=(n+j)!(n+1−j)!,j=0,1,…,n+1.(18)
Hence (p. 297, (17))
∣pn′(j+1)∣−∣pn′(j)∣=(2j+1)(n+j)!(n−j−1)!>0,j=0,1,…,n−1,
and (p. 298, (19))
∣qn′(j+1)∣−∣qn′(j)∣=(2j)(n+j)!(n−j)!>0,j=1,…,n.
So ∣pn′(j)∣ increases for j=0,…,n, and ∣qn′(j)∣ increases for
j=1,…,n+1, while ∣qn′(0)∣=∣qn′(1)∣=n!(n+1)!, for each fixed
n.
Statement (III) (p. 298). The sequences {∣pn′(j)∣},
j=0,…,n, and {∣qn′(j)∣}, j=1,…,n+1, are each absolutely
monotonic.
Here a sequence {aj} is absolutely monotonic when every defined
difference is non-negative: Δmaj≥0 for all m and j, where
Δ0aj=aj and Δm+1aj=Δmaj+1−Δmaj
(p. 298). For pn this is the inequality
Δm{(n+j)!(n−j)!}≥0,j=0,…,n;m=0,…,n−j,(20)
and the paper shows it with strict inequality. Remark (i) (p. 299) says
that similar results hold for these sequences taken at fixed j as n
varies.
Read depth. Claims checked: (16)--(20) and (III) were read on the page
images of pp. 297--299. The proofs were followed but not checked step by
step.
Proof pointer
Pp. 297--299. (16) comes from the product rule applied to (1): at an
interior zero j only one term survives, and the two remaining products
evaluate to (2j)! and ∏k=j+1n(k2−j2), whose product is
(n+j)!(n−j)!; the cases j=0,1,n are checked directly. (18) follows from
qn′=(x−n−1)pn′+pn, with qn′(n+1)=pn(n+1)=(2n+1)!. For (20) the
paper proves by induction that Δm{(n+j)!(n−j)!} equals
(n+j)!(n−j−m)! times a polynomial in j,m,n with non-negative integer
coefficients, not all zero; the case of qn is said to follow in the same
fashion. Remark (ii) (p. 299) reports D. J. Newman's observation that a beta
function integral for these differences makes (20) obvious.
Dependencies
The definitions (1) and (2).
Bears on
None recorded.