lemmaroutine

Choice of hierarchy parameters K, L, k

Lemma·bud:parameters·§5.1 Parameters, rare levels, and histories

Statement (paper)

There are integers

K,L,k\to\infty

such that

\begin{aligned} K\log L&=o(\log t),& D/K&=o(1),\\ kK\log L&=o(\log t),& k^{20}\left((D/K)^{1/4}+L^{-1}+S^{-1}+d_{Q,\min}^{-1}\right)&=o(1). \end{aligned}

Set

\epsilon=k^{-1/16}

and

e_t=(D/K)^{1/4}+L^{-1}

. Then

k^3/S=o(1)

k^{3+1/16}e_t=o(1)

, and

k=t^{o(1)}

In plain English

There are integers

K,L,k\to\infty

with

K\log L=o(\log t)

D/K=o(1)

kK\log L=o(\log t)

, and

k^{20}\big((D/K)^{1/4}+L^{-1}+S^{-1}+d_{Q,\min}^{-1}\big)=o(1)

. With

\epsilon=k^{-1/16}

and

e_t=(D/K)^{1/4}+L^{-1}

, this gives

k^3/S=o(1)

k^{3+1/16}e_t=o(1)

and

k=t^{o(1)}

Why it's here

This fixes the parameters once and for all, so later arguments can treat

k

K

L

as growing slowly enough to absorb error terms. The tradeoff is that

K

L

and

k

grow, but only slowly relative to

\log t

, while

k^{20}

times the error terms

(D/K)^{1/4}

L^{-1}

S^{-1}

and

d_{Q,\min}^{-1}

still tends to 0. The budget estimates and the popular-circle deletion use this to show their error terms are

o(1)

o(S)

Classified routine:

Standard diagonal choice of slowly growing parameters: since D, S, d_{Q,min} and log t are given sequences with the appropriate growth, one picks K, L, k tending to infinity slowly enough that each listed quantity is o(1). This is pure bookkeeping with no new idea.

Uses

Nothing else in the paper

Used by

propUpper score budget for pair experimentslemPopular unit circles carry few wedgesclaimEqual-endpoint collisions contribute o(S)

Sources

PaperOpen

Section: §5.1 Parameters, rare levels, and histories
Page: —
Label: bud:parameters
Line: sections/04-budgets.tex:27

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