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∂F = 2[(1 ÷e)(1 -β )2÷ χ]3 u
du <⅛1 -β )4(1+<≈)3
(B.3)
(B.3) is positive.
(6) Demonstration that dF > 0
dX
The first derivative of F with respect to χ is given by
3[(1+ε)(1 -β ) + χ]2U2
σ2∕1 -β )4(1+ε)3
(B.4)
It can easily be checked that (B.4) is positive.
∂F
(7) Demonstration that < 0.
dσ2
The first derivative of F with respect to σμ2 is given by
∂F _ -[(1 +ε)(1 -β)2 + χ]3U2
∂σ2μ [σ2μ(1 -β )2]2(1 + ε)3
(B.5)
(B.5) is negative.
∂F
(8) Demonstration that > 0.
The first derivative of F with respect to (1-β)-1 is given by
∂F = 2U2[2χ-(1+ε)(1 -β )2][(1 +ɛ)ɑ-β )2÷ χ]2
d(1 -β )"1 σ2(1 -β )3(1+ε)3
(B.6)
(B.6) is positive if χ > (1+g)(1 β) .