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192 CHAPTER 7. CALCULUS
Putting these together proves the following result.
Lemma 12 ([Ris69]) α ≤ max(min(γ − 1, γ − β), F/p ′ ), where the last term
only applies when β = 1, and when it gives rise to a positive integer.
In fact, it is not necessary to factorise the denominators into irreducible polynomials.
It is enough to find square-free polynomials p i , relatively prime in
pairs, and non-negative integers β i and γ i such that den(f) = ∏ p βi
i and
den(g) = ∏ p γi
i . When β = 1, we have, in theory, to factorise p completely,
but it is enough to find the integral roots of Res x (F − zp ′ , p), by an argument
similar to Trager’s algorithm for calculating the logarithmic part of the integral
of a rational expression.
We have, therefore, been able to bound the denominator of y by D = ∏ p αi
i ,
so that y = Y/D with Y polynomial. So it is possible to suppress the denominators
in our equation, and to find an equation
RY ′ + SY = T. (7.48)
Let α, β, γ and δ be the degrees of Y , R, S and T . Then (7.48) becomes
RY
}{{}
′ +
}{{}
SY
α+β−1 α+γ
There are again three possibilities 8 .
= T
}{{}
δ
. (7.49)
1. β − 1 > γ. In this case, the terms of degree α + β − 1 must cancel out the
terms of degree δ, therefore α = δ + 1 − β.
2. β − 1 < γ. In this case, the terms of degree α + γ must cancel out the
terms of degree δ, therefore α = δ − γ.
3. β − 1 = γ. In this case, the terms of degree α + β − 1 on the left may
cancel. If not, the previous analysis still holds, and α = δ + 1 − β. To
analyse the cancellation, we write Y = ∑ α
i=0 y ix i , R = ∑ β
i=0 r ix i and
S = ∑ γ
i=0 s ix i . The coefficients of the terms of degree α+β −1 are αr β y α
and s γ y α . The cancellation is equivalent to α = −s γ /r β .
Lemma 13 ([Ris69]) α ≤ max(min(δ − γ, δ + 1 − β), −s γ /r β ), where the last
term is included only when β = γ + 1, and only when it gives rise to a positive
integer.
Determining the coefficients y i of Y is a problem of linear algebra. In fact, the
system of equations is triangular, and is easily solved.
Example 11 Let us consider y ′ − 2xy = 1, i.e. f = −2x, g = 1. Since neither
f nor g have any denominator, y does not, i.e. it is purely polynomial, of degree
α, and R = 1, S = −2x and T = 1 in (7.48). Comparing degrees gives us
y ′
}{{}
α
+ −2xy
} {{ }
α+1
= 1
}{{}
0
, (7.50)
8 This is not an accidental similarity, but we refer to [Dav84] for a unifying exposition.