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C&O 631 NOTES 1. Formal Power Series the Series That We Shall Use

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C&O 631 NOTES 1. Formal Power Series the Series That We Shall Use C&O 631 NOTES 1. Formal power series The series that we shall use in this course are formal power series, not the power series of real variables that have been studied in calculus courses. A formal power series is given by P i i i A(x) = i≥0 aix , where ai = [x ]A(x), the coefficient of x , is a complex number, for i ≥ 0. P i The basic rule for A(x) is that ai is determined finitely for each finite i. Let B(x) = i≥0 bix . Then A(x) = B(x) if and only if ai = bi for all i ≥ 0, and we define sum and product by i ! X i X X i A(x) + B(x) = (ai + bi)x ;A(x)B(x) = ajbi−j x ; i≥0 i≥0 j=0 P i and a special case of product is the scalar product c A(x) = i≥0(c ai)x , for a complex number c. We write A(0) = a0, and unless A is a polynomial, this is the only evaluation we allow. If b0 = 0, then we define the composition X i X X n A(B(x)) = aiB(x) = aibj1 : : :bji x ; i≥0 n≥0 i≥0;j1;:::;ji≥1 j1+:::+ji=n and note that the summations above are finite. Note that we only allow substitutions of this type - where we substitute a constant-free B(x) for x. Now suppose A(0) = 1. Then if B(x) is a multiplicative inverse of A(x), we have (since multiplication of complex numbers is commutative, so is multiplication of A(x) with B(x), P i P j so there is no difference between a left-inverse and a right-inverse) i≥0 aix j≥0 bjx = 1, and equating coefficients of xn on both sides, for n ≥ 0, we obtain b0 = 1 a1b0 + b1 = 0 a2b0 + a1b1 + b2 = 0; where the nth equation is anb0 +an−1b1 +:::+b0 = 0, n ≥ 1. But this gives b0 = 1, and allows us to determine bn uniquely in terms of b0; : : :; bn−1, for each n ≥ 1, so, by induction on n, B(x) is unique. Applying this process to obtain the multiplicative inverse of A(x) = 1 − x, −1 P i we obtain bn = 1, n ≥ 0, by induction on n, or (1 − x) = i≥0 x . But substitution into this, for an arbitrary A(x) with A(0) = 1, gives −1 X i A(x)−1 = (1 − (1 − A(x))) = 1 + (1 − A(x)) ; i≥1 which is therefore the unique multiplicative inverse of A(x). We define differentiation and integration operators by d X X ai A(x) = ia xi−1;I A(x) = xi+1: dx i x i + 1 i≥1 i≥0 Date: May 2, 2017. 1 2 C&O 631 NOTES d d Now note that we have uniqueness for solution of differential equations: if dx A(x) = dx B(x) and A(0) = B(0), then A(x) = B(x). Now d X X X d d (A(x) + B(x)) = i(a + b )xi−1 = ia xi−1 + ib xi−1 = A(x) + B(x); dx i i i i dx dx i≥1 i≥1 i≥1 so the differentiation satisfies the sum rule, and i d X X (A(x)B(x)) = ia b xi−1 dx j i−j i≥1 j=0 i X X i−1 = (j + i − j)ajbi−jx i≥1 j=0 d d = A(x) B(x) + A(x) B(x) ; dx dx d n n−1 d and differentiation satisfies the product rule. Induction on n then gives dx B(x) = nB(x) dx B(x) for positive integers n, which allows us to prove the chain rule: d d A(B(x)) = A0(B(x)) B(x): dx dx To differentiate A(x)−1, where A(0) = 1, we apply the product rule to A(x) · A(x)−1 = 1, to obtain d d (1) A(x)−1 A(x) + A(x) A(x)−1 = 0; dx dx and thus conclude that d d A(x)−1 = −A(x)−2 A(x): dx dx We now define three special series X 1 X 1 X a(a − 1):::(a − n + 1) (2) "(x) = xn; λ(x) = xn;B (x) = xn; n! n a n! n≥0 n≥1 n≥0 where a is a complex number parameter in Ba(x). Our object is to show that "(x); λ(x);Ba(x) have the properties of the familiar functions ex; ln(1 − x)−1; (1 + x)a, respectively. (Except that we will NOT be able to consider, for example, "("(x)), since it uses composition with d a series with constant term 1.) First, note that dx "(x) = "(x). Then, for example, we can prove that "(x)"(−x) = 1, since "(x)"(−x) has constant term "(0)"(−0) = 1, and d ("(x)"(−x)) = "(x)"(−x) − "(x)"(−x) = 0; dx where we have used the product rule and chain rule. The result follows by the uniqueness of solution of differential equations ( since 1 also has constant term 1 and derivative 0). Also, d P n −1 we have dx λ(x) = n≥0 x = (1 − x) , so d (λ(1 − "(−x))) = ("(−x))−1 "(−x) = 1; dx by the chain rule, and λ(1 − "(−0)) = 0, and we conclude that λ(1 − "(−x)) = x, by uniqueness of solution of differential equations. (The series 1 − "(−x) has constant term 0, C&O 631 NOTES 3 so the composition λ(1 − "(−x)) is valid.) Similarly, we prove that "(λ(x)) = (1 − x)−1, using "(λ(0)) = 1, and d ((1 − x)"(λ(x))) = −"(λ(x)) + (1 − x)"(λ(x))(1 − x)−1 = 0; dx using the product rule and chain rule. Also, if A(0) = 1, the chain rule together with (1) gives d d d (3) λ(1 − A(x)−1) = A(x) 1 − A(x)−1 = A(x)−1 A(x): dx dx dx d For the series Ba(x), we have dx Ba(x) = a Ba−1(x), and we omit further details of these computations. In summary, we have demonstrated above that the series "(x); λ(x);Ba(x) defined in (2) have most of the properties of the familiar functions ex; ln(1 − x)−1; (1 + x)a, respectively, and we shall write X 1 X 1 X a(a − 1):::(a − n + 1) (4) ex = xn; ln(1 − x)−1 = xn; (1 + x)a = xn; n! n n! n≥0 n≥1 n≥0 where, as usual, the only substitutions that we allow for x are constant-free formal power series. For example, in terms of these familiar functions, (3) becomes the familiar logarithmic differentiation rule d d lnA(x) = A(x)−1 A(x); dx dx where A(0) = 1. As another example, consider the problem of finding the nth root of a formal power series: begin by noting that it is easy to verify that there are no zero divisors for formal power series, and this fact allows us to establish that nth roots are unique, at least with given constant term, as follows. Suppose A(0) = B(0) = 1, and A(x)n = B(x)n, for some positive integer n. Then we have 0 = A(x)n − B(x)n = (A(x) − B(x))(A(x)n−1 + A(x)n−2B(x) + ::: + B(x)n−1): Now the constant term in the second factor is A(0)n−1 + A(0)n−2B(0) + ::: + B(0)n−1 = n 6= 0, so we conclude that A(x) − B(x) = 0, since there are no zero divisors, which gives A(x) = B(x), as required. But we can determine the nth root of A(x) with A(0) = 1 by 1 substitution in the binomial series B 1 (x) = (1 + x) n , to obtain n 1 1 X 1 ( 1 − 1):::( 1 − i + 1) A(x) n = (1 + (A(x) − 1)) n = 1 + n n n (A(x) − 1)i; i! i≥1 which is therefore the unique nth root with constant term 1. We introduce trigonometric series by defining x3 x5 x2 x4 sin(x) = x − + − : : :; cos(x) = 1 − + − : : :; 3! 5! 2! 4! and then proving the properties of these series from properties of the series "(x) = ex, by eix − e−ix eix + e−ix sin(x) = ; cos(x) = ; 2i 2 4 C&O 631 NOTES so, for example, we have eix − e−ix 2 eix + e−ix 2 sin(x)2 + cos(x)2 = + 2i 2 1 2ix −2ix 2ix −2ix = 4 −(e − 2 + e ) + (e + 2 + e ) = 1: Then, noting that cos(x) has constant term 1, so it is invertible, we define sin(x) x3 2x5 2x3 16x5 tan(x) = = x + + + ::: = x + + + : : :; cos(x) 3 15 3! 5! 1 x2 5x4 x2 5x4 sec(x) = = 1 + + + ::: = 1 + + + : : :: cos(x) 2 24 2! 4! Various special functions are of interest combinatorially. For example, the coefficients in these formal power series for sec and tan, scaled by the factorial as in the above expressions, give the number of permutations of a special kind, called alternating permutations. P i In a similar way, consider formal Laurent series, given by A(x) = i≥−k aix , for some finite negative integer −k. We proceed as for formal power series above, with little modifi- −k cation required. To handle multiplicative inverses, suppose that A(x) = a−kx B(x), where a−k is invertible, and B(x) is a formal power series with B(0) = 1.
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