2.4 Formal Power Series

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2.4 Formal Power Series 52 CHAPTER 2. ADVANCED COUNTING AND GENERATING FUNCTIONS 2.4 Formal Power Series In this chapter, we will first try to develop the theory of generating functions by getting closed form expressions for some known recurrence relations. These ideas will be used later to get some binomial identities. To do so, we first recall from Page 41 that for all n Q and k Z, k 0, the binomial ∈ n(n 1)(∈n 2) ≥ (n k + 1) coefficients, n , are well defined, using the idea that n = − − · · · − . We k k k! now start with the definition of “formal power series” over Q and study its properties in some detail. n Definition 2.4.1 (Formal power series). An algebraic expression of the form f(x)= anx , n≥0 where a Q for all n 0, is called a formal power series in the indeterminate x overPQ. n ∈ ≥ The set of all formal power series in the indeterminate x, with coefficients from Q will be denoted by (x). P Remark 2.4.2. 1. Given a sequence of numbers an Q : n = 0, 1, 2,... , one associates { ∈ n } n x n two formal power series, namely, anx and an . The expression anx is called n≥0 n≥0 n! n≥0 P P xn P the generating function and the expression an is called the exponential generating n≥0 n! function, for the numbers a : n 0 . P { n ≥ } n n 2. Let f(x) = anx be a formal power series. Then the coefficient of x , for n 0, in n≥0 ≥ f(x) is denotedP by [xn]f(x). That is, a = [x0]f(x) and a = [xn]f(x), for n 1. 0 n ≥ n 3. One thinks of anx as an algebraic expression. In general, one is interested only in n≥0 computing the coefficientP of certain power of x and not in evaluating them for any value of x. But, if at all there is a need to evaluate it at a point, say x0, then one needs to determine its “radius of convergence” and then evaluate it if x0 lies within that radius. We need the following definition to proceed further. n Definition 2.4.3 (Equality of two formal power series). Two elements f(x) = anx and n≥0 n g(x)= bnx of (x) are said to be equal if an = bn, for all n 0. P n≥0 P ≥ P We are now ready to define the algebraic rules: n n Definition 2.4.4. Let f(x)= anx , g(x)= bnx (x). Then their n≥0 n≥0 ∈ P P P 1. sum/addition is defined by n n n f(x)+ g(x)= anx + bnx = (an + bn)x . nX≥0 nX≥0 Xn≥0 2.4. FORMAL POWER SERIES 53 2. product is defined by n f(x) g(x)= a xn b xn = c xn, where c = a b , for n 0. · n · n n n k n−k ≥ Xn≥0 nX≥0 nX≥0 Xk=0 This product is also called the Cauchy product. xn Remark 2.4.5. 1. In case of exponential power series, the product of f(x)= an and n≥0 n! n n n x x n P g(x)= bn equals dn , where dn = k akbn−k, for n 0. n≥0 n! n≥0 n! k=0 ≥ P P x P 2. Note that the expression ee −1 is a well defined formal power series as the definition ey = n x n y x (e 1) implies that ee −1 = − and hence n≥0 n! n≥0 n! P P x n m x n x (e 1) (e 1) [xm]ee −1 = [xm] − = [xm] − . (2.1) n! n! n=0 nX≥0 X x That is, for each m 0, [xm]ee −1 is a sum of a finite number of real numbers. Where x≥ x as the expression ee is not a formal power series as the computation of [xm]ee , for all m 0, will indeed require an infinite sum. ≥ Thus, under the algebraic operations defined above, it can be checked that the set (x) forms P a Commutative Ring with identity, where the identity element is given by the formal power series n f(x) = 1. In this ring, the element f(x) = anx is said to have a reciprocal if there exists n≥0 n another element g(x)= bnx (x) suchP that f(x) g(x) = 1. So, the question arises, under n≥0 ∈ P · what conditions on the coefficientsP of f(x), can we find g(x) (x) such that f(x)g(x) = 1. ∈ P The answer to this question is given in the following proposition. n Proposition 2.4.6. Let f(x) = anx (x). Then there exists g(x) (x) satisfying n≥0 ∈ P ∈ P f(x) g(x) = 1 if and only if a = 0P. · 0 6 n Proof. Let g(x)= bnx (x). Then, by the definition of Cauchy product, f(x)g(x)= n≥0 ∈ P n n P cnx , where cn = akbn−k, for all n 0. Therefore, using the definition of equality of two n≥0 k=0 ≥ powerP series, we see thatP f(x)g(x) = 1 if and only if c = 1 and c = 0, for all n 1. 0 n ≥ Therefore, if a0 = 0 then c0 = 0 and hence the Cauchy product f(x)g(x) can never equal 1. However, if a = 0, then the coefficients b ’s can be recursively obtained as follows: 0 6 n 1 b0 = as 1= c0 = a0b0. a0 1 b1 = − (a1b0)as0= c1 = a0b1 + a1b0. a0 · 54 CHAPTER 2. ADVANCED COUNTING AND GENERATING FUNCTIONS 1 b2 = − (a2b0 + a1b1)as 0= c2 = a0b2 + a1b1 + a2b0. And in general, if we have already a0 · computed the values of b , for k r, then k ≤ 1 br+1 = − (ar+1b0 + arb1 + + a1br)as0= cr+1 = ar+1b0 + arb1 + + a1br + a0br+1. a0 · · · · · · · Note that if the coefficients a ’s come from , a commutative ring with unity, then one needs n R b to be an invertible element of for Proposition 2.4.6 to hold true. Let us now look at the 0 R n n composition of two formal power series. Recall that, if f(x)= anx , g(x)= bnx (x) n≥0 n≥0 ∈ P n m n then the composition (f g)(x) = f(g(x)) = an(g(x)) =P an( bmx P) may not be ◦ n≥0 n≥0 m≥0 defined (just to compute the constant term ofP the compositionP, one mayP have to look at an infinite sum). For example, let f(x) = ex and g(x) = x + 1. Note that g(0) = 1 = 0. Here, 6 (f g)(x) = f(g(x)) = f(x +1) = ex+1. So, as function f g is well defined, but there is no ◦ ◦ x+1 k x+1 formal procedure to write e as akx (x) (i.e., with ak Q) and hence e is not a k≥0 ∈ P ∈ formal power series over Q. P The next result gives the condition under which the composition (f g)(x) is well defined. ◦ n n Proposition 2.4.7. Let f(x)= anx and g(x)= bnx be two formal power series. Then n≥0 n≥0 the composition (f g)(x) is wellP defined if either f isP a polynomial or b = 0. ◦ 0 n Moreover, suppose that a0 = 0. Then, there exists g(x) = bnx , with b0 = 0, such that n≥0 (f g)(x)= x. Furthermore, (g f)(x) is well defined and (g Pf)(x)= x. ◦ ◦ ◦ n Proof. Let (f g)(x) = f(g(x)) = cnx and suppose that either f is a polynomial or ◦ n≥0 b = 0. Then to compute c = [xk] (f Pg)(x), for k 0, one just needs to consider the terms 0 k ◦ ≥ a + a g(x)+ a (g(x))2 + + a (g(x))k. Hence, each c is a real number and (f g)(x) is well 0 1 2 · · · k k ◦ defined. This completes the proof of the first portion. The proof of the other part is left to the readers. We now define the formal differentiation of elements of (x) and state a result without proof. P n Definition 2.4.8 (Differentiation). Let f(x) = anx (x). Then the formal differentia- n≥0 ∈ P tion of f(x), denoted Df(x), is defined by P Df(x)= a + 2a x + + na xn−1 + = na xn−1. 1 2 · · · n · · · n nX≥1 n Proposition 2.4.9. Let f(x) = anx (x). Then f(x) = a0, a constant, whenever n≥0 ∈ P x Df(x) = 0. Also, f(x)= a0e wheneverP Df(x)= f(x)..
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