Faster exponentials of power series

We describe a new algorithm for computing exp f where f is a power series in CJxK. If M(n) denotes the cost of multiplying polynomials of degree n, the new algorithm costs (2.1666 . . . + o(1))M(n) to compute exp f to order n. This improves on the previous best result, namely (2.333 . . . + o(1))M(n). The author recently gave new algorithms for computing the square root and reciprocal of power series in CJxK, achieving better running time constants than those previously known [Har09]. In this paper we apply similar techniques to the problem of computing exp f for a power series f ∈ CJxK. Previously, the best known algorithm was that of van der Hoeven [vdH06, p. 6], computing g = exp(f) mod x in time (7/3+ o(1))M(n), where M(n) denotes the cost of multiplying polynomials of degree n. We give a new algorithm that performs the same task in time (13/6+ o(1))M(n). Van der Hoeven’s algorithm works by decomposing f into blocks, and solving g = f g by operating systematically with FFTs of blocks. Our starting point is the observation that his algorithm computes too much, in the sense that at the end of the algorithm, the FFT of every block of g is known. Our new algorithm uses van der Hoeven’s algorithm to compute the first half of g, and then extends the approximation to the target precision using a Newton iteration due to Brent [Bre76] (see also [HZ04] or [Ber04] for other exponential algorithms based on a similar iteration). At the end of the algorithm, only the FFTs of the blocks of the first half of g are known. In fact, the reduction in running time relative to van der Hoeven’s algorithm turns out to be equal to the cost of these ‘missing’ FFTs. We freely use notation and complexity assumptions introduced in [Har09]. Briefly: ‘running time’ always means number of ring operations in C. The Fourier transform of length n is denoted by Fn(g), and its cost by T (n). We assume that T (2n) = (1/3+o(1))M(n) for a sufficiently dense set of integers n. For Proposition 1 below, we fix a block size m, and for any f ∈ CJxK we write f = f[0] + f[1]X+ f[2]X + · · · where X = x and deg f[i] < m. The key technical tool is [Har09, Lemma 1], which asserts that if f, g ∈ CJxK, k ≥ 0, and if F2m(f[i]) and F2m(g[i]) are known for 0 ≤ i ≤ k, then (fg)[k] may be computed in time T (2m) +O(m(k + 1)). We define a differential operator δ by δf = xf (x), and we set δkf = X δ(Xf). In particular δ(f[0] + f[1]X + · · · ) = (δ0f[0]) + (δ1f[1])X + · · · .