claimroutineinferred_dep

Combined error margins contradict determinant comparison

Total error

E_{\rm ar}+E_{\rm an}<\varepsilon_0

, so determinant comparison gives a contradiction

finishes the upper bound by contradiction

Claim·§4 Choice of parameters and proof of Theorem~thm:main·inside thmIrrationality exponent of pi is 2

Statement (paper)

With the dimension, weights and centers chosen as above, the total error satisfies

E_{\rm ar}+E_{\rm an}=\frac{\nu}{F_0}+\frac{\Lambda F_0m+2\log2}{v_0}+\frac{100K}{w_0}+\Lambda\sum_i\frac1{w_i}+\frac{\theta+\log4+\log(2K)+\nu+\log(200K)}{w_*}<\varepsilon_0

. In particular

E_{\rm ar}+E_{\rm an}<g

and

cL>2>1+E_{\rm ar}+E_{\rm an}

, so all hypotheses of Proposition hold, which yields a contradiction.

In plain English

With the chosen

F_0

m

K

w_0

v_0

and approximants

p_i/q_i

, the total error

E_{\rm ar}+E_{\rm an}

is less than

\varepsilon_0

. Hence

E_{\rm ar}+E_{\rm an}<g

and

cL>2>1+E_{\rm ar}+E_{\rm an}

, so the determinant comparison hypotheses all hold and give a contradiction.

Why it's here

The total error splits into three pieces, each bounded by a third of

\varepsilon_0

by the earlier parameter choices: the

\nu/F_0

term, the

m

K

w_0

v_0

terms, and the weight terms. Since

\varepsilon_0=\min(g/2,1/2)

, the sum is below

g

and below

1/2

, so

1+E_{\rm ar}+E_{\rm an}<2<cL

. This meets the budget conditions of the determinant comparison proposition, which rules out the assumed approximations and so finishes the proof by contradiction.

Classified routine:

This is the final bookkeeping step: the earlier parameter choices bound the error terms, and the resulting inequalities meet the hypotheses of the determinant comparison proposition, yielding the contradiction.

Selected node

Statement (paper)

In plain English

Why it's here