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The symmetrized Fermi function and its transforms

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6530 D W L Sprung <strong>and</strong> J Martorell<br />

<strong>and</strong> α = λπ = πqd (in this section, z is real). Using now D = d/dz, equation (15)<br />

becomes<br />

q 2 v(q) =<br />

αD H(z). (30)<br />

sin αD<br />

Exp<strong>and</strong>ing in powers of α = πqd, <strong>and</strong> taking the derivatives, one finds for the first few<br />

terms<br />

q 2 v(q) = qRJ 1 (qR) − (πqd)2 [qRJ 1 (qR) − J 0 (qR)]<br />

6<br />

[<br />

+ 7(πqd)4 qRJ 1 (qR) − 2J 0 (qR) + J ]<br />

1(qR)<br />

360<br />

qR<br />

− 31(πqd)6<br />

360(42)<br />

[<br />

qRJ 1 (qR) − 3J 0 (qR) + 3<br />

qR J 1(qR) − 3<br />

(qR) 2 J 2(qR)<br />

]<br />

+···.<br />

(31)<br />

As noted in appendix A, this series diverges for large |α|, so one would like to know <strong>its</strong><br />

sum. Further study reveals that the coefficients involve <strong>function</strong>s<br />

y p (z) ≡ (−) p+1 (2p − 1)!! J p(z)<br />

(32)<br />

z p<br />

with y 0 =−J 0 (z) <strong>and</strong> y −1 ≡ zJ 1 (z). Differentiation is facilitated by the relation<br />

y p ′′ =−y p−y p+1 . (33)<br />

For example, one has<br />

(zJ 1 (z)) ′′ =−zJ 1 (z) + J 0 =−y −1 −y 0 . (34)<br />

At large z, the J 0 term is of relative order 1/z. If we could neglect it altogether, then we<br />

could replace the derivative operator in equation (30) by i since only even powers of D<br />

occur in the expansion. <strong>The</strong> result would be zJ 1 (z)α/ sinh α, as in equation (27).<br />

Taking the derivatives systematically leads to the tableau<br />

−(zJ 1 ) (2) = zJ 1 + y 0<br />

(zJ 1 ) (4) = zJ 1 + 2y 0 + y 1<br />

−(zJ 1 ) (6) = zJ 1 + 3y 0 + 3y 1 + y 2<br />

(35)<br />

(zJ 1 ) (8) = zJ 1 + 4y 0 + 6y 1 + 4y 2 + y 3<br />

−(zJ 1 ) (10) = zJ 1 + 5y 0 + 10y 1 + 10y 2 + 5y 3 + y 4 .<br />

One sees that the rule for carrying out the derivatives generates coefficients according to<br />

Pascal’s triangle, so they are the binomial coefficients l C p . <strong>The</strong> general case is therefore<br />

l∑<br />

(−) l D 2l zJ 1 (z) = lC p y p−1 . (36)<br />

p=0<br />

Following equation (A.3), with c 0 = 1 , we have<br />

2<br />

αD<br />

∞∑<br />

sinh αD = 2 (−) l c 2l α 2l D 2l . (37)<br />

l=0<br />

This gives<br />

∞∑<br />

l∑<br />

q 2 v(q) = 2 (−) l c 2l α 2l<br />

lC p y p−1 . (38)<br />

l=0<br />

p=0

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