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Lecture 16 MATH1904 Generating Functions When faced with a difficult counting problem, such as finding a formula for the n-th Catalan number cn, one of the most powerful techniques is to consider the numbers c0, c1, c2, . . . simultaneously and to bundle them up into a single mathematical object called a generating function (otherwise known as a formal power series). Definition Given a sequence a0 , a 1 , a 2 , a 3 , . . . , its generating function is the power series G(z) = a0 + a1z + a2z 2 + a3z 3 + · · · . In Σ-notation we write this as G(z) = ∞ X ak z k . k=0 1 Example. The generating function of the Catalan numbers is C(z) = ∞ X cn z n = n=0 1 + z + 2z 2 + 5z 3 + 14z 4 + 42z 5 + · · · The algebra of generating functions Generating functions can be added and multiplied just like polynomials. That is, if G1(z) = a0 + a1z + a2z 2 + a3z 3 + · · · and G2(z) = b0 + b1z + b2z 2 + b3z 3 + · · · then G1 (z) + G2 (z) = (a0 + b0 ) + (a1 + b1 )z + (a2 + b2 )z 2 + · · · and G1 (z)G2 (z) = c0 + c1 z + c2 z 2 + c3 z 3 + · · · , where ck = a0 bk + a1 bk−1 + a2 bk−2 + · · · + ak b0 . 2 Example. The Vandermonde convolution From the binomial theorem we have w X w k w (1 + z) = z k k=0 and (1 + z)m = m X m k z k=0 k In this the case the generating functions really are polynomials. In any case, when we multiply them together we get (1 + z)w (1 + z)m = (1 + z)w+m On comparing coefficients of z n we see that n X w m k=0 k n−k = w + m n . 3 Closed forms Generating functions are power series which usually involve infinitely many terms. We would like to find simpler expressions for these series, using only a finite number of mathematical operations. These simpler expressions are called closed forms. If the terms of the sequence an are 0 from some point on, the generating function G(z) is really an ordinary polynomial and it is already a closed form, as was the case for the previous example. Fortunately, for many sequences, even those with infinitely many non-zero terms, we can still find closed forms. Example. The generating function of 1, 1, 1, 1, 1, 1, . . . is G(z) = 1 + z + z 2 + · · · . Multiplying by z and subtracting from G(z) we find that G(z) − zG(z) = 1 and this allows us to write 1 . 1−z This is a closed form for G(z). G(z) = 4 Negative powers of 1 − z It will often be useful to know the coefficient of z m in (1 − z)−n. From the previous example we have (1 − z)−1 = 1 + z + z 2 + z 3 + · · · and therefore, for n ≥ 1, (1 − z)−n = (1 + z + z 2 + z 3 + · · · ) . . . (1 + z + z 2 + z 3 + · · · ). | {z } n factors The coefficient of z m in this product is equal to the number of ways one can form z m by multiplying together n terms, one from each factor. That is, it is the number of choices of d1 , d2 , . . . , dn such that z d1 z d2 . . . z dn = z m . This is the number of solutions to the equation d1 + d2 + · · · + dn = m, where each di is a non-negative integer. This is just the number of ways to select m things from n things, allowing repetition and therefore the number is m + n − 1 . m Putting these observations together gives the formula (1 − z) −n ∞ X m + n − 1 m = z . m m=0 5 Some special cases When n = 2 we have m + n − 1 m + 1 = = m + 1, m m and so (1 − z)−2 = 1 + 2z + 3z 2 + 4z 3 + · · · . When n = 3 we have m + n − 1 m + 2 = = (m + 2)(m + 1)/2, m m and so (1 − z)−3 = 1 + 3z + 6z 2 + 10z 3 + · · · . Derivatives The derivative of G(z) = a0 + a1z + a2z 2 + a3z 3 + · · · is G0(z) = a1 + 2a2z + 3a3z 2 + · · · . Using the derivative and mathematical induction we can derive anew the formula for (1 − z)−n. 6