Concrete mathematics : a foundation for computer science

is somehow involved. So we steer toward that kind of sum:
we have
G(z) = l+t(n+l)g,M+’
= 1 + t ng, zn+l + x gn zn+’
n
= 1 +z’G’(z)+zG(z).
7.3 SOLVING RECURRENCES 333
Let’s check this equation, using the values of g,, for small n. Since
G = 1 +z+2z2 + 6z3 +24z4 + ... ,
G’ = 1+42 +18z2+96z3+-.,
z2G’ zz z2+4z3+18z4+96z5+.-,
zG = z+z2 +2z3 + 6z4 +24z5 + ... ,
1 = 1.
“This will be ouick.”
These three lines add up to G, so we’re fine so far. Incidentally, we often find
it convenient to write ‘G’ instead of ‘G(z)‘; the extra ‘(2)’ just clutters up the
formula when we aren’t changing z.
Step 3 is next, and it’s different from what we’ve done before because we
have a differential equation to solve. But this is a differential equation that
we can handle with the hypergeometric series techniques of Section 5.6; those
techniques aren’t too bad. (Readers who are unfamiliar with hypergeometrics
needn’t worrv- this will be quick.)
That’s what the
doctor said just
First we must
both sides:
get rid of the constant ‘l’, so we take the derivative of
before he stuck me
with that needle.
Come to think of it,
“hypergeometric”
sounds a lot like
“hypodermic.”
G’ = @‘G’S zG + 1)’ = (2zG’+z’G”)+(G +zG’)
= z2G”+3zG’+G.
The theory in Chapter 5 tells us to rewrite this using the 4 operator, and we
know from exercise 6.13 that
9G = zG’, B2G = z2G” +zG’.
Therefore the desired form of the differential equation is
4G = ~9~G+224G+zG = z(9+1)‘G.
According to (5.1og), the solution with go = 1 is the hypergeometric series
F(l,l;;z).