The name is absent

Pι(rι)xi^υ(rι) = exp[I>ι(rι) - βV₁^ext(rβ)] exp(⅜ι).           (4.52)

Now using this result in eqn. 4.9 for (neglecting the 1 in the denominator in
comparison to the second term which contains the bonding energy and ɛɑ ~÷ ∞) ^can
be rewritten as

_________________________________1_________________________________

exp(βμ₁) ʃ drɪ exp[A(rι) - ,z3V₁^ea=ⁱ(rɪ)]ʌ(ɪ'²)(r_l, r₂) '

Substituting for Xg∖r₂) in the Euler - Lagrange equation for the second segment leads
to an expression for p2(^r2)^∖²∖^r2), from which an expression for X^∖r₃) similar to
eqn. 4.53 can be obtained. Repeating this procedure through the last segment in the
chain, lead to accurate expressions for Xa and Xb for all the segments in the chain.

ʌ-b)/ɪ, ∖ __________________________________________________________________________________________

^{a j} exp(β(μ_j+1 + μ_j+2 + .. + μ_m)) ʃ .. ʃ dr_j+1dr_j+2..dr_m exp[D_j+l(r_j+1) - βVffir_j+β)
+D_j+2(r_j+2) - βV⅝(r_j+2) + .. + D_ro(r_m) - βV^(r_m)]Δ^+V(r_j, r_j+ι)..∆(-i--)(r_m-₁, r_m) ’

(4.54)

and

X^)(_rl =_______________________________________________________

^{b 3} exp(β(μ₁ +μ₂ +.. + μ_j-₁')') f .. f dr_idr₂..dr_j-₁exp[D₁(r₁) - βV^^xt(r₁)

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