claimroutineinferred_dep

Two-adic loss of E_t(0) is at most 6Dh

Claim·§D.4.4 Tapered Hermite recovery·inside lemTapered Hermite recovery of F(0)

Statement (paper)

For t∈S∖{0}t\in S\setminus\{0\} with i=v2(t)i=v_2(t) and D=Dt=L/2iD=D_t=L/2^i, one has v2(Et(0))=∑q≠0,tqq(v2(q)−v2(t−q))=D∑j=1a−ij2j(q(i+j)−q(i))⩾−6Dh.v_2(E_t(0))=\sum_{q\ne0,t}q_q\bigl(v_2(q)-v_2(t-q)\bigr)=D\sum_{j=1}^{a-i}\frac{j}{2^j}\bigl(q(i+j)-q(i)\bigr)\ge -6Dh.

In plain English

For t∈S∖{0}t\in S\setminus\{0\} with i=v2(t)i=v_2(t) and D=Dt=L/2iD=D_t=L/2^i, the 2-adic valuation of Et(0)E_t(0) equals ∑q≠0,tqq (v2(q)−v2(t−q))\sum_{q\ne0,t}q_q\,(v_2(q)-v_2(t-q)), which can be rewritten as D∑j=1a−ij2j(q(i+j)−q(i))D\sum_{j=1}^{a-i}\frac{j}{2^j}(q(i+j)-q(i)) and is at least −6Dh-6Dh.

Why it's here

This bounds how much 2-adic precision is lost when the Hermite formula multiplies by Et(0)E_t(0), so Et(0)E_t(0) has valuation no worse than −6Dh-6Dh. The following claim uses this bound, together with the extra 8hDt8hD_t in ptp_t, to show the stated precisions ptp_t are enough to recover F(0)F(0) modulo 2p2^p. The proof computes the valuation as a sum over the other points weighted by the tapered multiplicities, regroups it by 2-adic level of tt, and bounds the regrouped sum using the shape of q(i)=hmax⁡(0,u−i)q(i)=h\max(0,u-i).
Classified routine:
A direct 2-adic valuation calculation on the Hermite interpolation coefficients. It regroups the sum by 2-adic level and bounds it using the explicit form q(i)=h max(0,u-i), which is standard bookkeeping following the earlier claim's setup.

Sources

PaperOpen
Section
§D.4.4 Tapered Hermite recovery
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quantitative-contraction.tex:301
Leannot aligned

Phase 2 aligns this node to a Lean declaration.

Friction

  • info
    inferred_dep
    outline: "Two-adic loss of E_t(0) at most 6Dh" rests on claim-lem-tapered-hermite-1