Manuscript X-L-ST@1 · X-L-ST.tex · section · 20a32a22

Order-by-order compatibility

Conditional corrections and quantum-presentation tests › Order-by-order compatibility · lines 1206–1226

X-L-ST@1#eq:MCorders

Write

d = d_{0} + κ d_{1} + \dots + κ^{N} d_{N} (mod κ^{N + 1}), d_{0} = s_{0} .

Square-zero requires

i + j = n \sum d_{i} d_{j} = 0 (mod J), 0 \leq n \leq N .

For example,

s_{0} d_{1} + d_{1} s_{0} = 0, s_{0} d_{2} + d_{2} s_{0} + d_{1}^{2} = 0.

These are additional equations. A first-order linear solve does not discharge the second equation.

Displayed formulas in this passage

X-L-ST@1#eq:MCorders · line 1214
$i + j = n \sum d_{i} d_{j} = 0 (mod J), 0 \leq n \leq N .$
unlabeled · line 1209
$d = d_{0} + κ d_{1} + \dots + κ^{N} d_{N} (mod κ^{N + 1}), d_{0} = s_{0} .$
unlabeled · line 1220
$s_{0} d_{1} + d_{1} s_{0} = 0, s_{0} d_{2} + d_{2} s_{0} + d_{1}^{2} = 0.$