Manuscript Hawk07-B@1 · Hawk07-B.tex · section · b9785538

Fixed residual dimension

Payload-resource tradeoffs before a positive cutoff › Fixed residual dimension · lines 680–703

Hawk07-B@1#eq:exacth

At a cut with specified residual dimension $b$ , (eq:momentbound) becomes

K = \frac{d ^{2} f ( b ^{2} - 1 ) ( d ^{2} - 1 )}{4 ( d ^{2} h ^{2} f ^{2} - 1 )} .

Thus its exact second-moment sufficient condition is

h^{2} \geq \frac{1}{d ^{2} f ^{2}} + \frac{( b ^{2} - 1 ) ( d ^{2} - 1 )}{4 p ζ ^{2} f} .

With $ζ = ε^{2} /8$ , this reads

h^{2} \geq \frac{1}{d ^{2} f ^{2}} + \frac{16 ( b ^{2} - 1 ) ( d ^{2} - 1 )}{p ε ^{4} f} .

These are design allowances, not a converse capacity theorem.

Displayed formulas in this passage

Hawk07-B@1#eq:exacth · line 690
$h^{2} \geq \frac{1}{d ^{2} f ^{2}} + \frac{( b ^{2} - 1 ) ( d ^{2} - 1 )}{4 p ζ ^{2} f} .$
unlabeled · line 684
$K = \frac{d ^{2} f ( b ^{2} - 1 ) ( d ^{2} - 1 )}{4 ( d ^{2} h ^{2} f ^{2} - 1 )} .$
unlabeled · line 697
$h^{2} \geq \frac{1}{d ^{2} f ^{2}} + \frac{16 ( b ^{2} - 1 ) ( d ^{2} - 1 )}{p ε ^{4} f} .$