Formal Power Series
Consider the set N0 C = (a0, a1, a2,...): an ∈ C for all n of all sequences of complex numbers. On this set, we introduce operations of addition, scalar multiplication, and multiplication, via
(a0, a1,...) + (b0, b1,...) := (a0 + b0, a1 + b1,...)
a(a0, a1,...) := (aa0, aa1,...)
(a0, a1,...)(b0, b1,...) := (c0, c1,...), where n X cn := akbn−k. k=0 Exercise 1.
(i) Verify that, with respect to these operations, CN0 simultaneously carries the structure of a commutative ring with identity element, and of a vector space over C. (ii) Show that the ring of formal power series C[[z]] over C (as defined in the course) and CN0 are isomorphic as C-algebras.
We note that the ring of polynomials C[z] is a subring of C[[z]]. It is the above product rule, which accounts for the wide applicability of power series in combinatorial problems. Indeed, frequently we can construct all an of the objects of type n in some family of objects by choosing an object of type k and an object of type n − k, and stitching them together to obtain an object of type n. In this situation, clearly n X an = akan−k, k=0 so that the Cauchy product has a natural combinatorial interpretation.
Exercise 2. Prove from the multiplication rule that (1 − z)(1 + z + z2 + z3 + ··· ) = 1, and deduce that the series 1 − z has a reciprocal, and that this reciprocal is the geometric P n series n≥0 z (and, of course, the other way round, too).
P n More generally, a formal power series f(z) = n≥0 anz has a reciprocal, if there exists a P n power series g(z) = n≥0 bnz such that f(z)g(z) = 1. The series g is then called the reciprocal of f, and vice versa.
P n Exercise 3. Show the following: A formal power series f(z) = n≥0 anz has a reciprocal if, and only if, a0 6= 0. In this case, the reciprocal is unique. In the ring of formal power series, there are further operations defined which reflect operations of functional calculus, but make no use of limiting processes. One of these operations is 1 2
P n differentiation. By definition, the derivative of the formal power series f(z) = n≥0 anz is the series 0 X n−1 Dzf(z) = f (z) = nanz . n≥1
Exercise 4. Show that differentiation of formal power series follows the usual rules of P n P n calculus. Specifically, show that, for series f(z) = n≥0 anz and g(z) = n≥0 bnz ,
(i) (f + g)0 = f 0 + g0, (ii) (fg)0 = f 0g + fg0, (iii) (f k)0 = kf k−1f 0, k ≥ 1, f 0 f 0g − fg0 (iv) = , provided that b 6= 0. g g2 0
Exercise 5. Show the following:
0 (i) If f = 0, then f = a0 is a constant. (ii) If f 0 = f, then f = cez, where c is a constant, and X ez = zn/n! n≥0 is the (series expansion of the) exponential function. (iii) If f 00 + f = 0, then f = a sin(z) + b cos(z), where a, b are constants, and X sin(z) = (−1)nz2n+1/(2n + 1)! n≥0 X cos(z) = (−1)nz2n/(2n)!. n≥0
P i Let fn(z) = i≥0 cniz (n = 0, 1, 2,...) be a sequence of elements from C[[z]] with the property that
∀ i ∃ ni : n > ni ⇒ cni = 0. (1) Then we can formally define
∞ ∞ ni X X X i fn(z) = cni z . n=0 i=0 n=0 P n This definition allows us to introduce formal substitution of a power series g(z) = n bnz P n for the variable z of a formal power series f(z) = n anz . If b0 = 0, then the powers n n g (z) := (g(z)) satisfy the summability condition (1) with ni = i, that is ∞ X n f(g(z)) := ang (z) n=0 makes sense as a formal power series. Note in particular that, in this sense, eez−1 is a well- defined formal series, whereas eez is not definable via the above composition of series. 3 Exercise 6. (i) Show that, if fn(z) n≥0 is a summable sequence of formal power series, 0 then the sequence of derivatives fn(z) n≥0 is also summable, and we have ∞ 0 ∞ X X 0 fn(z) = fn(z). n=0 n=0 (ii) Prove the chain rule f(g(z))0 = f 0(g(z))g0(z).