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F−1 a0 f−1 2b0 +c0 F−3 a1 f−2 2b1 +c1 . . 59). The E proof to find the explicit form of RF is similar. 2. 7) K+ (u) ⊗ C T 1/2 K− v + 3n + 23 −i−1 du i n≥0 i≥0 where K± (u) = ln h± (u), C(q) = q+1+q −1 , T is the shift operator: Tf (u) = f (u+1) and (ψ(u))i = ψi for any function ψ(u) = i ψi u−i−1 . Proof. The proof is inspired by the results exposed in [12]. 9) −1 . 10) Therefore one obtains < T −1 + T 1/2 − T − T −1/2 −1 K− (u), d −1 K+ (v) > = . 11) The formal inversion of the operator (T −1 + T 1/2 − T − T −1/2 )−1 is given by T −1 + T 1/2 − T − T −1/2 −1 = T 3n+2 + T 3n+3/2 + T 3n+1 .
Let θ, θ1 ∈ (0, 1) and suppose that I = [tan ω0 , tan ω1 ] is a subinterval of [0, 1] of size |I | εθ , and Q is an integer such that Q = cosεω0 + O(ε −θ ). 1, and we denote cI = I dt = ω1 − ω0 , 1 + t2 1 Eθ,θ1 ,δ (ε) = εmin(2θ, 2 −2θ1 −δ,θ+θ1 −δ) . It is convenient to write H˜ ε,I,Q (t) = ω ; tan ω ∈ I, l 1 (ω) cos ω > Q t cos ω ε . 1) We also define for each interval I ⊆ [0, 1] and each integer Q ≥ 1 the quantity H˜ I,Q (t) = ω ; tan ω ∈ I, l 1 (ω) cos ω > tQ . Q Let Q− and Q+ be two integers such that Q− ≤ cos ω1 cos ω0 ≤ ≤ Q+ .
If the second element belongs to F + , the assertion for p + 1 is proven. 16). 47) and fm ap−1 , hm+k ap−1 are ordered thanks to the hypothesis on p. This ends the induction on p and (i) is proven for A+ . For A− , the proof is almost similar. However, an additional difficulty appears because of exchange relations between e−k and h−1 . 23) allows us to order eˆ−l ≡ 2e−l − e−l−1 − e−l−2 (l ≥ 1) and h−1 . Fortunately, e−k can be expressed in terms of eˆ−l : ∀k ≥ 1, e−k = 1 3 +∞ 1− − l=k 1 2 l−k+1 eˆ−l .