RE: [math-fun] factoring formal power series
Neil asked (excerpt): << [W]hat about [factoring] power series with integer coefficients? There are two versions, depending on whether you work to order n, or with the full power series. Version 1: Given 2 3 4 5 6 7 8 t3 := 1 + 3 x + 7 x + 17 x + 42 x + 113 x + 321 x + 976 x + 3109 x + 9 10 11 12 10258 x + 34698 x + 119554 x + O(x ), find t1 and t2 such that t1*t2 = t3 (mod x^12). Answer:
t1; 2 3 4 5 6 7 8 9 1 + x + 2 x + 5 x + 14 x + 42 x + 132 x + 429 x + 1430 x + 4862 x
10 11 + 16796 x + 58786 x + ...
t2; 2 3 4 5 6 7 8 9 10 11 1 + 2 x + 3 x + 5 x + 7 x + 11 x + 13 x + 17 x + 19 x + 23 x + 29 x + 31 x + ...
Version 2 (harder): Factor: 1 + 2*Sum_{m >= 1} x^(m^2) = 1 + 2x + 2x^4 + 2x^9 + 2x^16 + ... Answer: Product_{n >= 1} (1-x^2n)(1+x^(2n-1))^2
Since the group of units is exactly the set of power series beginning with constant term +-1, finding the equivalence classes up to multiplication by a unit seems worthwhile. Suppose the lowest term of a power series S(x) in Z[[x]] is +-N*(x^d), with d >= 0 and N > 0. Then there is a unique unit U_S(x) such that (*) U_S(x)S(x) = x^d * (N + Sum_{k >= 1} (L_k x^k)), where for all k >= 1 we have 0 <= L_k < N. Thus each equivalence class up to multiplication by a unit has a unique representative of the form [RHS of (*)]. (In fact the cofficients L_k are just the original coefficients of x^(d+k) reduced mod N.) This last fact allows us to multiply such representatives of each equivalence class and figure out the corresponding representative of the product. But still unanswered is the crucial question: What are the representatives of the irreducible elements of Z[[x]] ? --Dan
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