The region ππ <π and the
base 31/4
The rational-base calculation compares a gain in the remainder with
the cost of making its coefficients integral. If the original
coefficient has degree πΎπ and the
cancelled coefficient has degree ππ, the leading balance is
πππ(π/π)β(πΎπβππ)=ππΎππβ(πΎπβππ).(1)
The polynomial identities and remainder
estimates below justify this balance, including its lower-order factors.
With πΎπ/π2 βπΆ1 and (πΎπ βππ)/π2 βπΆ0, the cleared
remainder tends to zero when πΆ1logβ‘π <πΆ0logβ‘π. The degree drop has two effects: it improves
the real remainder and changes the denominator needed to make its
coefficients integral. Both effects concern the same cancelled form.
This is why one must calculate the degree before rational evaluation,
rather than borrow a denominator bound from the uncancelled family.
We evaluate Zudilinβs 2004 forms at π/π after cancelling their common
polynomial factors. His direction (14,12,14;27) [8] gives the constants below; the
comparison between decay and the remaining degree determines the
irrationality region:
πΆ1=(14+12+14)27β12(122+142+272)=10912,πΆ0=266β3π2(225βπ½),
π½=13βπ=1(π1(π’π)βπ1(π£π)),π1(π₯)=βπβ₯01(π+π₯)2,
where π1 is the trigamma function, used
only for positive arguments, and the thirteen half-open intervals [π’π,π£π) are [1/14,1/12), [1/7,1/6), [3/14,1/4), [2/7,1/3), [5/14,2/5), [3/7,7/15), [1/2,8/15), [4/7,3/5), [9/14,2/3), [5/7,11/15), [11/14,4/5), [6/7,13/15), [13/14,14/15). The intervals are disjoint
and lie in [1/14,1), so 0 β€π½ β€π1(1/14) βπ1(1) <196:
in the defining series, π1(1/14) =196 +βπβ₯1(π +1/14)β2 <196 +π1(1).
Since π >3, it follows that
191 <πΆ0 <266 <πΆ1/2. In
particular, the reciprocal constants below are positive without using
decimal approximations. The accompanying rational certificate gives
77.94318445520543<π½<77.94318449473569,221.30008815898543<πΆ0<221.30008817100119,0.40568302137302<πβ:=πΆ0/πΆ1<0.40568302139506.
Put π =1/πβ. These parameter ratios,
intervals and constants, including πΆ1/πΆ0 =2.46497868β¦, occur in [8]; that ratio is the
integer-base exponent bound in its TheoremΒ 1, not a new optimised
constant. The displayed interval endpoints are rounded outwards from
rational bounds; no digits beyond their certified accuracy are
required.
The proof below needs π½ in its
combinatorial form as well. Put
π(π)=max{0,β14πβ+β13πβββ12πβββ15πβ,2β14πβββ13πβββ15πβ},
the weight that [8]
attaches to the six-tuple π =(13,14,12,14,15,13) for the chosen
parameter ratios; it gives the exponents ππ =π(π/π) of the sourceβsΒ (22)
and enters its limitΒ (26). Zudilin lists the support of π in the same place; the next lemma
verifies that list by exact evaluation. Since 14 +13 =12 +15 and 2 β
14 =13 +15, each floor difference is
unchanged by π β¦π +1. Thus
π has period 1, and the lemma also determines every
value π(π/π) needed for the
cyclotomic exponents.
Lemma 2.1 (the weight is an indicator). On [0,1) the function π takes only the values 0 and 1, and π =1 exactly on the union of the
thirteen half-open intervals listed above.
Proof. Each of the three expressions inside the maximum is
constant on every interval between consecutive elements of {π/π :π β{12,13,14,15},Β 0 β€π β€π}, since those are the only points at which one of the four
floor functions changes. There are 48 such intervals in [0,1); evaluating π at the midpoint of each gives the
value 1 on the thirteen listed
half-open intervals and 0
elsewhere, and the thirteen listed intervals are unions of consecutive
cells. Each floor function is right-continuous, so a cell has the same
value at its left endpoint as at its midpoint. This also verifies the
half-open endpoint convention. All evaluations use exact rational
arithmetic.Β β»
LemmaΒ 2.1 and
termwise integration of the positive series βπβ²1(π) =βπβ₯02(π +π)β3
give the two forms of π½ used below,
π½=β«10π(π)βπβ₯02(π+π)3ππ=13βπ=1(π1(π’π)βπ1(π£π)),
the
first being the integral β«10π π( βπ1) of [8]. Since π vanishes near 0, both are finite.
For π =1 the logarithmic ratio is
zero, so the condition includes every integer base. For π >1 it requires π >π2.46497868β¦, substantially
more than π >π. There are still
infinitely many admissible coprime numerators for each fixed
denominator. The base 31/4 is
included, whereas 3/2 is excluded.
Taking positive integral powers does not change the ratio. At equality
with the cutoff, the estimates give no conclusion.
Theorem 2.2 (rational-base region). Let π >π β₯1 be coprime integers with
ππ<π,equivalentlylogβ‘πlogβ‘π<πβ=0.40568302138406054β¦
Then πΉ(π/π) =βπβ₯1((π/π)π β1)β1 is irrational.
The proof has three steps. We cancel the common polynomial factors
using Zudilinβs integrality argument [8]. We then compute the degree of the
remaining polynomials, using his cyclotomic limits and the
reciprocal-interval method of [7]. Finally, we evaluate at π/π and show that the positive remainder
still decays after multiplication by the required power of π. These steps occupy SectionsΒ 2.1β2.4. The
construction, direction and constants are Zudilinβs; the rational
specialisation and its degree and remainder estimates are proved
below.
The polynomial forms and their common factors
Exact degrees
The cyclotomic limit
Positive linear forms with integer coefficients
The rational bracket around πβΞΈβ
Earlier criteria and changes of direction
Finite calculations