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) ε(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 ε(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, ε(ε(x)), since it uses composition with d a series with constant term 1.) First, note that dx ε(x) = ε(x). Then, for example, we can prove that ε(x)ε(−x) = 1, since ε(x)ε(−x) has constant term ε(0)ε(−0) = 1, and d (ε(x)ε(−x)) = ε(x)ε(−x) − ε(x)ε(−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 − ε(−x))) = (ε(−x))−1 ε(−x) = 1, dx by the chain rule, and λ(1 − ε(−0)) = 0, and we conclude that λ(1 − ε(−x)) = x, by uniqueness of solution of differential equations. (The series 1 − ε(−x) has constant term 0, C&O 631 NOTES 3 so the composition λ(1 − ε(−x)) is valid.) Similarly, we prove that ε(λ(x)) = (1 − x)−1, using ε(λ(0)) = 1, and d ((1 − x)ε(λ(x))) = −ε(λ(x)) + (1 − x)ε(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 ε(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 ε(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. Then define −1 −1 k −1 A(x) = a−kx B(x) . d P i−1 We define differentiation in the same way as for formal power series, by dx A(x) = i≥−k iaix . The formal residue of a formal Laurent series is given by the coefficient of x−1. This has a nice invariance property, given in the following result.

P i Theorem 1.1. Suppose that A(x) is a formal Laurent series, and that B(x) = i≥m bix is a formal power series with m a positive integer, and bm invertible. Then 1 d [x−1]A(x) = [x−1]A(B(x)) B(x). m dx (Note, that for formal Laurent series A(x), the composition A(B(x)) is well-formed only when B(x) is a formal power series with constant term 0.)

P n Proof. Let A(x) = n≥−k anx , so on the right hand side of the result we have 1 X d [x−1] a B(x)n B(x). m n dx n≥−k For n ≥ 0, note that B(x)nB0(x) is a formal power series in x, so we have 1 [x−1]B(x)nB0(x) = 0 m in this case. For −k < n < −1, we have 1 1 1 [x−1]B(x)nB0(x) = [x−1] B(x)n+10 = 0, m m n + 1 since [x−1]f 0(x) = 0 for any formal Laurent series f(x). C&O 631 NOTES 5

m Finally, for n = −1, let B(x) = bmx H(x), so H(x) is a formal power series with H(0) = 1. Then we have 0 m−1 1 −1 n 0 1 −1 bmxmH (x) + mbmx H(x) [x ]B(x) B (x) = [x ] m m m bmx H(x) 1 H0(x) = [x−1] + mx−1H(x) m H(x) 1 = [x−1] (lnH(x))0 + mx−1H(x) m 1 = (0 + m) = 1. m The result follows by combining the results for these cases.

Before we apply Theorem 1.1 to prove Lagrange’s Theorem, we prove that the formal P i power series F (x) = i≥1 fix has a unique compositional inverse whenever f1 is invertible. P i Suppose the compositional inverse is given by G(x) = i≥1 gix . Then equate coeﬃcients of xn in F (G(x)) = x, to obtain the equation, n ≥ 1, n X X fi gj1 ··· gji = δn,1,

i=1 j1+...+ji=n (where, on the right hand side, we obtain 1 if n = 1 and 0 otherwise.) When n = 1, this −1 gives g1 = f1 , and when n > 1, this gives n −1 X X gn = −f1 fi gj1 ··· gji , i=2 j1+...+ji=n which allows us to recursively obtain each coeﬃcient gn ﬁnitely in terms of f1, . . ., fn. Lagrange’s Implicit Function Theorem. We have the functional equation w = tφ(w), for some formal power series φ with an invertible constant term. Rewrite this functional equation as Φ(w) = t, where Φ(w) = w/φ(w), and let Ψ be the compositional inverse of Φ, so we have w = Ψ(t). Now consider any formal Laurent series f. Then we have, for n 6= 0, [tn]f(w) = [t−1]t−n−1f(w) = [t−1]t−n−1f(Ψ(t)), and we can apply Theorem 1.1, with B = Φ (for which we have m = 1), to obtain 1 [tn]f(w) = [w−1]Φ(w)−n−1f(w)Φ0(w) = − [w−1]f(w) Φ−n(w)0 . n But [w−1]f(w)g0(w) = −[w−1]f 0(w)g(w) for any formal Laurent series f, g, since [w−1](f(w)g(w))0 = 0, so we have 1 1 1 (5) [tn]f(w) = [w−1]f 0(w)Φ−n(w) = [w−1]f 0(w)φn(w)w−n = [wn−1]f 0(w)φn(w), n n n for n ≥ 1, where w = tφ(w), which is the usual statement of Lagrange’s Implicit Function Theorem. For an analytic treatment of Lagrange’s Theorem, see Section 5.1 of Wilf’s book Generatingfunctionology. 6 C&O 631 NOTES

2. Lecture of May 1 We now turn to symmetric functions. For ﬁxed positive integer n, we consider formal power series in x1, . . ., xn. A formal power series A(x1, . . ., xn) is symmetric if A(xσ(1), . . ., xσ(n)) = A(x1, . . ., xn) for all permutations σ of {1, . . ., n} (so the word “symmetric” here refers to the symmetric group S(n), which consists of all permutations of {1, . . ., n}). We begin with the simplest example. Let λ = (λ1, . . ., λn), where λ1 ≥ ... ≥ λn ≥ 0, so λ is a partition with at most n parts (the number of parts in λ is given by the number of positive λi’s). Then the monomial symmetric function mλ = mλ(x1, . . ., xn) is deﬁned to be the

λ1 λn (6) mλ = x1 ···xn + . . .,

λ1 λn in which we add the minimal set of monomials that, together with the monomial x1 ···xn , n! creates a symmetric function. In general, mλ is a sum of |autλ| distinct monomials, where |autλ| is the number of permutations that ﬁx the n-tuple (λ1, . . ., λn). For example, we have 2 2 2 m(2,2,2)(x1, x2, x3) = x1x2x3, and 3 2 3 2 2 3 3 2 2 3 2 3 m(3,2,0)(x1, x2, x3) = x1x2 + x1x3 + x1x2 + x2x3 + x1x3 + x2x3.

Second, the elementary symmetric functions ek = ek(x1, . . ., xn), k ≥ 0, are deﬁned by ek = m(1,...,1,0,...,0), where (1, . . ., 1, 0, . . ., 0) has k 1’s and n − k 0’s. Equivalently, using a generating function, we have n X k Y (7) ekt = (1 + xjt). k≥0 j=1 n Thus ek is the sum of k monomials with exponents 0 or 1, each of which corresponds to a k-subset of {1. . ., n}, the k-subset specifying which of the indeterminates have exponent 1. Third, the homogeneous (or complete) symmetric functions hk = hk(x1, . . ., xn), k ≥ 0, are deﬁned by X hk = mλ,

where the summation is over all λ = (λ1, . . ., λn) with λ1 + ··· + λn = k. But this means that hk is the sum of all monomials in x1, . . ., xn of (total) degree k, so, equivalently, using a generating function, we have n n X k Y −1 Y 2 2 (8) hkt = (1 − xjt) = (1 + xjt + xj t + ...). k≥0 j=1 j=1 n+k−1 Thus hk is the sum of all of the k monomials with non-negative exponents totalling k. Fourth, the power sum symmetric functions pk = pk(x1, . . ., xn), k ≥ 0, are deﬁned by pk = m(k,0,...,0), so we have p0 = 1 and n X k (9) pk = xj , k ≥ 1. j=1 In terms of generating functions, we obtain n n n X tk X X tk X X tk X X p = xk = xk = ln(1 − x t)−1 = ln h ti. k k j k j k j i k≥1 k≥1 j=1 j=1 k≥1 j=1 i≥0 C&O 631 NOTES 7

As a ﬁfth example, we now consider a less familiar example of symmetric functions, the Schur functions. We shall make extensive use of determinants in this presentation, for which it will be useful to recall the deﬁnition as a summation over the symmetric group: n X Y det (ai,j)i,j=1,...,n = sgn (ρ) ai,ρ(i). ρ∈S(n) i=1

Then for λ = (λ1, . . ., λn) with λ1 ≥ ... ≥ λn ≥ 0, the Schur symmetric function sλ = sλ(x1, . . ., xn) is deﬁned by

λj +n−j det xi (10) s = i,j=1,...,n . λ n−j det xi i,j=1,...,n This is symmetric because sgn(σ) s (x , . . ., x ) = s (x , . . ., x ), λ σ(1) σ(n) sgn(σ) λ 1 n where sgn(σ) is the sign of the permutation σ (since in both the numerator and denominator determinants of sλ, to evaluate at xσ(1), . . ., xσ(n), we simply permute the rows by σ). The denominator determinant in (10) is well-known, as the Vandermonde determinant. Call it V (x1, . . ., xn). Note that V is a polynomial in x1, . . ., xn, and has value 0 if xk = x` for any k 6= ` (since this makes rows k and ` identical). Therefore V is divisible by Y (11) (xk − x`). 1≤k<`≤n n But V is homogeneous of total degree (n−1)+...+1+0 = 2 , and (11) is also homogeneous n of total degree 2 , so we conclude that Y V (x1, . . ., xn) = c · (xk − x`), 1≤k<`≤n n−1 1 0 for some constant c. Now, the monomial x1 ··· xn−1xn appears in V with coefficient 1, since it is produced uniquely in the expansion of the determinant as the product of the entries on the main diagonal (for which ρ is the identity permutation, with sgn +1); this monomial also appears in the expansion of (11) with coefficient 1, since it is produced uniquely in the expansion of the product by selecting the xk from each xk − x`. This implies that c = 1, giving Y (12) V (x1, . . ., xn) = (xk − x`). 1≤k<`≤n Now the same argument as we used for the denominator proves that the numerator determinant in (10) is a polynomial divisible by (11), which implies that sλ is a symmetric polynomial in x1, . . ., xn (not just a symmetric rational function), since the denominator in (10) perfectly divides the numerator. Moreover, since the numerator of (10) is a homo- n geneous polynomial of total degree λ1 + ... + λn + 2 , we conclude that the polynomial sλ itself is homogeneous, of total degree λ1 + ... + λn. We now give a second, combinatorial definition of the polynomial sλ (and will then prove that these definitions are consistent). The Ferrers graph of the partition λ is an array of boxes (called cells), with λi cells in row i (indexed from the top), for each positive λi, aligned at 8 C&O 631 NOTES the left. For example, the Ferrers graph of (5, 3, 2) is given on the left hand side of Figure 1.

1 1 3 3 5

2 3 5

4 4

Figure 1. A Ferrers graph and tableau.

A tableau of shape λ is obtained by placing positive integers into the cells of the Ferrers graph of λ (one in each cell), subject to the condition that the integers weakly increase from left to right across each row, and strictly increase from top to bottom down each column. An example of a tableau of shape (5, 3, 2) is given on the right hand side of Figure 1. Our combinatorial deﬁnition of sλ(x1, . . ., xn) is now given by n X Y nj (T ) (13) sλ = wt (T ), wt (T ) = xj , T j=1 where the summation is over all tableaux T of shape λ, and nj(T ) is the number of times j appears in a cell of T . For example, the tableau given in Figure 1 contributes the monomial 2 3 2 2 x1x2x3x4x5 to s(5,3,2)(x1, . . ., xn) (for any n ≥ 5).