“…If s ∈ 9 2 , 17 2 , then 2s+3 8 > 3 2 , and therefore h 0 (0), h 0 (0) are both well-defined, but z x (0, 0) = 0, and hence we must have h 0 (0) = 0 in addition to h 0 (0) = 0. More generally, if s ∈ 1 2 + 4(j − 1), 1 2 + 4j for some j ≥ 1, then using also the main equation, we deduce that ∂ k t h 0 (0) = 0 is a necessary condition for all 0 ≤ k ≤ j − 1. By similar arguments, we deduce that if s ∈ 3 2 + 4(l − 1), 3 2 + 4l for some l ≥ 1, then ∂ k t h 1 (0) = 0 is a necessary condition for all 0 ≤ k ≤ l − 1.…”