18a9042b3fc5b02fe3d57fea87d6992f-Supplemental.pdf

Projecting this differential equation on the last coordinate givesdHe+1t = dt, that is, He+1t = t. Finally,let (a(n))n N beaCauchysequencein T . Straightforward calculations yield the equality,valid for any x R, tanh(x)=2σ(2x) 1. But,foranyn 1, Next, it is clear that the signature of a constant path is equal to1 = (1,0,...,0,...) which is the nullelementinT . More precisely, fork = 1, C(1;0) = 1 00 = 1 and C(1;1) = 0 01 = 0. Assume that the formula is true at orderk. Then, at order k + 1, there are two cases.