"liE

In fact, a straightforward but cumbersome calculation shows that, for fixed n, we have

Assuming without loss of generality that s ~ t we get from (14.57)

Taking into account that all terms in (14.56) are zero for k > s and

(14.58)

a.+! (a) = a.(a) we further obtain •

L)an(a) - an-l(a)J[bn(,8) - bn-l (,8)] n=1

• • - ~ an(a)bn - l (,8) +~ an-l (a)bn - 1(,8)

• • = ~[2an(a) - an-l(a) - an+!(a)]bn(,8) = ~ An(a)bn(,8).