Let a = L + 1− Lτ and b = τ, then according to (3.28), ¯g0 = a1 > 0 and ¯gi =
−ab2(1−ab)i−1 < 0 for i = 1, 2,· · · , L, and
¯
g0+ ¯g1+ ¯g2+· · · + ¯gl = 1a− ab2 − ab2(1− ba)− · · · −ab2(1− ba)l−1
= 1a− ab2 · 1·[1−(1−1−(1−bab)l] a)
= 1a− 1a[1− (1 −ab)l]
= 1a(1− ba)l Hence
m22 = ¯g02+ (¯g0 + ¯g1)2+· · · + (¯g0+ ¯g1+· · · + ¯gL)2
= [1a]2+ [1a(1− ba)]2+· · · + [1a(1− ab)L]2
= a12[1 + (1− ba)2+· · · + (1 − ba)2L]
= a12 · 1·[1−(1−1−(1−ab)b2(L+1)] a)2
= 1−(1−ab)2(L+1)
a2·[1−1−b2a2+2ba]
= 1−(1−2ab−bab)2(L+1)2
= 1−(1−
τ
L+1−Lτ)2(L+1) 2(L+1−Lτ)τ−τ2
and
d
dτm22 = [2(L+1−Lτ)τ−τ2]·[−2(L+1)(1−L+1−Lττ )2L+1]·[−dτd(L+1−Lττ )]
[2(L+1−Lτ)τ−τ2]2 − [1−(1−L+1−Lττ )2(L+1)]·(−4Lτ−2τ) [2(L+1−Lτ)τ−τ2]2
= [(1−τ)τ(2L+1)+τ]·[2(L+1)(1−L+1−Lττ )2L+1]·[(L+1−Lτ)2L+1 ]
[2(L+1−Lτ)τ−τ2]2 + [1−(1−
τ
L+1−Lτ)2(L+1)]·(4Lτ+2τ) [2(L+1−Lτ)τ−τ2]2
Because 0 < 1− τ < 1 and 0 < (1 −L+1−Lττ ) < 1 for 0 < τ < 1, dτdm22 > 0 for 0 < τ < 1.
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