Consider the generating function g(x) for the distribution of component sizes where the
coefficient of x k is the probability that a vertex chosen at random is in a component of
size k

g ( x)   kak x k
k 1
From the formula for the a i ' s , we will derive
g  2xg'2xgg' x
and then use the equation for g to determine the value of  at which the phase transition
for the appearance of the giant component occurs.
Derivation of g(x)
From
a1 
and
ak 
we derive the equations

1
1  2
k 1
1  2k
 j (k  j )a a
j k j
j 1
a1 1  2  1  0
and
k 1
ak 1  2k     j (k  j )a j ak  j
j 1
The generating function is formed by multiplying the kth equation by kx k and summing
over all k. This gives


k 1
k 1

k 1
k 1
j 1
 x   kak x k  2 x ak k 2 x k 1    kx k  j (k  j )a j ak  j


k 1
k 1
Note that g ( x)   kak x k , and g '( x)   ak k 2 x k 1 .
Thus

k 1
k 1
j 1
 x  g ( x)  2 xg '( x)    kx k  j (k  j )a j ak  j
Working with the right hand side

k 1
 k 1
k 1
j 1
k 1 j 1
  kxk  j (k  j )a j ak  j   x j (k  j )( j  k  j ) x k 1a j ak  j
Now breaking the j+k-j into two sums gives
 k 1
 k 1
k 1 j 1
k 1 j 1
 x j 2 a j x j 1 (k  j )ak  j x k  j   x ja j x j (k  j )2 ak  j x k  j 1
Notice that the second sum is obtained from the first by substituting k-j for j and that both
terms are  xg ' g . Thus
 x  g ( x)  2 xg '( x)  2 xg '( x) g ( x)
Hence
g
1
x.
g' 
2 1  g
1