Exact degrees
The final summand of 𝐴𝑛 has
strictly larger degree than all the others. We use it to determine the
leading coefficient as well as the degree. After cancellation, this will
give the least common power of the base denominator that clears the
pair.
Let 𝑑𝑛,𝑘 =deg(𝑐𝑛,𝑘(𝑋)𝑋𝑎0𝑘).
The Gaussian factors are monic of degrees (𝑎1 −1)(𝑘 −𝑎1) and (𝛽𝑛 −𝑘 −1)(𝑘 −𝑎2). Adding these to the
monomial exponent gives
𝑑𝑛,𝑘=𝑒𝑛,𝑘+𝑎0𝑘+(𝑎1−1)(𝑘−𝑎1)+(𝛽𝑛−𝑘−1)(𝑘−𝑎2)=−𝑘2+80𝑘𝑛+3𝑘−340𝑛2−26𝑛2.
Taking the difference
at consecutive indices, we obtain
𝑑𝑛,𝑘+1−𝑑𝑛,𝑘=40𝑛+1−𝑘>0(𝑎2≤𝑘≤𝛽𝑛−2),
the inequality because 𝑘 ≤𝛽𝑛 −2 =27𝑛 and 27𝑛 <40𝑛 +1. The summand of top degree
is therefore the one at 𝑘 =𝛽𝑛 −1
alone, no cancellation occurs there, and
𝐾𝑛:=deg𝐴𝑛=𝑑𝑛,𝛽𝑛−1=1091𝑛2+81𝑛+22,𝑊𝑛:=deg𝑈𝑛=𝐾𝑛−𝑀𝑛+∑𝑙≤𝑁𝜑(𝑙)−∑2≤𝑙≤𝑁𝜈𝑛,𝑙𝜑(𝑙),(8)
the second equality because deg𝐷𝑁 =∑𝑙≤𝑁𝜑(𝑙) and
degΩ𝑛 =∑2≤𝑙≤𝑁𝜈𝑛,𝑙𝜑(𝑙), and because 𝑈𝑛 is a polynomial by (7),
so subtracting 𝑀𝑛 from the degree
of 𝐷𝑁𝐴𝑛/Ω𝑛 is legitimate.
In particular 𝑈𝑛 ≠0. There is
also no integer leading-coefficient factor to cancel: the top summand
has leading coefficient ( −1)𝑎1+𝑎2+𝛽𝑛 =( −1)𝑛, and both
Gaussian factors and 𝐷𝑁/Ω𝑛
are monic. Thus the leading coefficient of 𝑈𝑛 is ( −1)𝑛.
The net decrease in degree is
𝐾𝑛−𝑊𝑛=𝑀𝑛−deg(𝐷𝑁/Ω𝑛).
Dividing
by 𝑋𝑀𝑛 lowers the degree by
𝑀𝑛, but multiplying by the
remaining cyclotomic factors increases it by deg(𝐷𝑁/Ω𝑛). This is why the
remainder estimate below involves 𝐾𝑛 −𝑊𝑛, rather than 𝑀𝑛 alone.
For a reduced fraction 𝑎/𝑏, the
unit leading coefficient also gives the exact denominator of 𝑈𝑛(𝑎/𝑏), not just an upper bound. For
every prime 𝑝 ∣𝑏,
𝑏𝑊𝑛𝑈𝑛(𝑎/𝑏)≡(−1)𝑛𝑎𝑊𝑛≢0(mod𝑝).
No prime dividing 𝑏 cancels from the cleared numerator, so
the reduced denominator is exactly 𝑏𝑊𝑛. In particular every common
integer divisor of the cleared row is coprime to 𝑏. This excludes cancelling a factor of
𝑏 from this row, not cancellation
at other primes or a saving from a different pair.
The formal arithmetic checks cover the value
of 2𝑀𝑛 from the source’s
formula (16), the degree
difference 𝑑𝑛,𝑘+1 −𝑑𝑛,𝑘 =40𝑛 +1 −𝑘, its positivity
for 𝑎2 ≤𝑘 ≤𝛽𝑛 −2, and
the top
degree 2𝐾𝑛 =1091𝑛2 +81𝑛 +2.
These statements check the polynomial degree calculation, not the
analytic estimates used later.
For the second coefficient, fix 𝑛 and let 𝑥 →∞. For 𝑥 ≥2, every finite product in (2)
lies between ∏𝑗≥1(1 −2−𝑗) >0 and 1, independently of 𝑡. Also ∑𝑡≥0𝑥−𝑎0𝑡 ≤2. These bounds
give 𝐻𝑛(𝑥) =𝑂(1) for the whole sum,
not just each summand. Also 𝐹(𝑥) =𝑥−1 +𝑂(𝑥−2), because 𝜏(1) =1 and ∑𝑚≥2𝑚𝑥−𝑚 =𝑂(𝑥−2). The monic
normalising factor at infinity gives Λ𝑛(𝑥) =𝑂(𝑥𝑊𝑛−𝐾𝑛). Since 𝐾𝑛 ≥2 and 𝑈𝑛 has leading coefficient ( −1)𝑛,
𝑉𝑛(𝑥)=𝑈𝑛(𝑥)𝐹(𝑥)−Λ𝑛(𝑥)=(−1)𝑛𝑥𝑊𝑛−1+𝑂(𝑥𝑊𝑛−2).
The degree formula gives
𝑊𝑛 ≥𝐾𝑛 −𝑀𝑛 =(559𝑛2 +13𝑛)/2 >0.
Polynomial integrality was established earlier, so the displayed
asymptotic proves the exact statement
deg𝑉𝑛=𝑊𝑛−1,lc𝑉𝑛=(−1)𝑛.(9)
In particular 𝑉𝑛 is not the zero polynomial. Applying
the same prime-by-prime reduction at its own degree shows that 𝑉𝑛(𝑎/𝑏) has reduced denominator exactly
𝑏𝑊𝑛−1. Thus 𝑏𝑊𝑛 is the least common clearing
denominator, although the second coordinate alone needs one fewer power
of 𝑏. In the common normalisation,
gcd(𝑏𝑊𝑛𝑉𝑛(𝑎/𝑏),𝑏𝑊𝑛)=𝑏.
Indeed 𝑏𝑊𝑛𝑉𝑛(𝑎/𝑏) is 𝑏 times an integer coprime to 𝑏. These facts concern the fixed source
pair, not all approximants. The degree calculation fixes 𝑛 and varies 𝑥; the decay calculation fixes 𝑥 and varies 𝑛.