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Because power series can be differentiated term by term, Eq. (7.87)
gives:

∞
∞
+
(12− xt t 2 ) ∑ lP l ()x t − l 1 + ( −t x ) ∑ P l () =x t l 0 (7.88)
= l 0 = l 0
Since this equation should hold true for all values of t, this means
that all coefficients of any power of t should be zero; therefore:

(l + 1 ) ()P x − 2 lxP () (x + l − 1 )P ()x + P ()x − xP ()x = 0 (7.89)
l l l−1 l−1 l

or collecting terms, this can be written as:

(l + 1 ) () (P x − 2 l + 1 )xP ()x + lP ()x = 0 (7.90)
l l l−1

This is the recursion relation of Pb. 2.25.
4. By substitution in the explicit expression of the generating function,
we can also verify that:

+
)
(12− xt t 2 ∂ G − tG = 0 (7.91)
x ∂
which leads to:

∞
∞
+
(12− xt t 2 ) ∑ dP l ()x − ∑ Px + l 1 = 0 (7.92)
()t
l
= l 0 dx = l 0
Again, looking at the coefficients of the same power of t permits
us to obtain another recursion relation:

x
x
dP () dP x() dP ()
l+1 − 2 x l + l−1 − Px = 0() (7.93)
dx dx dx l
dP ()
x
l−1
Differentiating Eq. (7.90), we first eliminate and then
dx
dP x()
l from the resulting equation, and use Eq. (7.93) to obtain
dx
two new recursion relations:

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