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house price is consequently CH = 4(4an1b)a.
Composition of the regularizer with the site-to-color row cancels = 1. Its trace-ideal price is
and its ordinary price 4
. The shear row has ordinary price 4
, so the corrected row has price
. The two reflection products have combined trace-ideal price
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CR = 4rt + 4(4p)r.
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For the orbit term let si bound the fixed stage row in direction i, and put S(z) = P3 i=0 4(4siz)si. Its unweighted trace price is μ2S(B). The two changes to the dual massive frames cost at most e4μ−1 each, yielding CO = e4S(B)e4.
The motive film factors through the fixed-dimensional space, with inclusions of ordinary price one; its trace-ideal price is 4
. All widths and derivative orders have been fixed before
, and each cancellation of a block power has been displayed. These four estimates supply (DP10). Composing a difference factor with an ordinary bounded factor keeps (DP10). In particular the house film has price
; the two reflection products have combined price
has price 16
, including its two film-composition factors; and the motive film has price 4
Loc
. Thus one may use exactly
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K∗= CH + CR + 16CO + 4| Loc0 |. (DP11)
Each price here is the absolute weighted column sum of its stated local factor with its zero-, one-and two-field placements, not a norm of a full trace. The analytic row and localized column estimates bound those sums by the fixed stage and strip prices. They use the same128, branch bounds and common stationary rows as (DP3), and are uniform in
. The inverse identity gives difference price
. For a two-field determinant frame, the direct term difference costs
. Telescoping its three-factor bubble gives (
Taking its trace through (DP10) gives precisely (DP2) and (DP4). The factor 1
2 in the coreminus-rank determinant may be discarded when increasing this bound. This proves the relative projection estimate while retaining the
factor in the unrelated individual estimates (DP3). The stationary carrier change has zero kinetic reading:
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αa(Mact −Mcore) = 0. (DP5)
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To see this, write its diagonal film as 1 2(W(p) + W(−p) −V (p)/2). The stationary tax row has zero jets below order two. Its mixed carrier row has zero jets below order one, by the stationary graph identity. Hence each summand of W, a product of these two rows, vanishes to order at least three. Every summand of V contains two tax rows and vanishes to order at least four. These films are analytic on the common strip, so their external degree-two coefficient is zero. Linearity proves (DP5). For an analytic scalar function bounded by A on width r0, two successive Cauchy derivative estimates give
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∥D2f(0)∥≤ A r0/2 1 r0/4.
For the normalized reading (DP1), the source keeps the slightly weaker explicit price
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B(r0, A) = A r0/4 1 r0/8∥axes(e0)∥2. (DP6)
The factor two in (DP1) cancels the Taylor factorial. Apply this estimate to (DP4), with and use (DP5). The exact projection bound is
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|αa(Mact) −αa(Mmrank)| ≤B(h/2, R∗). (DP7)
No inverse power of the lattice period and no multiplier occurs in (DP7).
There is one more comparison, which is an equality rather than a bounded error. Let Π be the analytic film of the massive pole test on the fine lattice, with
. Its finite-period analytic tensors approximate that test with error cents(
, where
0, and the rank film approximates the same test pulled back through the stationary uniform map with error cents(
, where
for fixed
. The constants and analytic strip in these approximation statements may depend on
. They are used for identification only, not as the uniform bound in (DP7).
Fourier evaluation of the stationary pullback gives the exact alias formula