The centroid and second origin identities #
Two of the four identities of the blog post's Lemma 1 come from comparing the two
factorizations of Sendov.Counterexample.Factor coefficient by coefficient, or by evaluating
them at a point. Neither needs any integration.
- the centroid identity, from the coefficient of
X^{n-2}inp', equivalently fromp.coeff (n-1)andp'.coeff (n-2); - the second origin identity, from
p'(0).
Everything is stated division-free. The blog post writes the second origin identity as
(-1)^{n-1} (∏ zⱼ) (1 + a ∑ⱼ 1/zⱼ) = (n / ∏ qⱼ) F(1),
with a convention that singularities are removed when some zⱼ vanishes. Multiplying out,
(∏ zⱼ)(∑ⱼ 1/zⱼ) is ∑ⱼ ∏_{k≠j} z_k, which is defined whether or not any zⱼ is zero, and
∏ qⱼ clears the other denominator. So the convention becomes unnecessary rather than being
formalized: no junk value of 0⁻¹ is ever evaluated.
Main statements #
Sendov.sumEraseProd:∑ⱼ ∏_{k≠j} sₖ, the division-free form of(∏ s)(∑ 1/sⱼ);Sendov.centroid_identity;Sendov.second_origin_identity.
∑ⱼ ∏_{k≠j} sₖ. This is (∏ s) · (∑ⱼ 1/sⱼ) with the denominators cleared, and unlike
that expression it is well defined when some sⱼ vanishes.
Equations
- Sendov.sumEraseProd s = (Multiset.map (fun (j : ℂ) => (s.erase j).prod) s).sum
Instances For
Two rewritings of the factorizations #
The factorization of p, with a put back into the multiset.
The centroid identity #
The centroid identity. (n-1)(a + ∑ zⱼ) = n ∑ⱼ (a - 1/qⱼ): the centroid of the
zeroes equals the centroid of the critical points.
The second origin identity #
Splitting off one element, exactly as Multiset.esymm does at the top index.
∏ⱼ (X - zⱼ) at the origin.
The derivative of ∏ⱼ (X - zⱼ) at the origin. Proved by induction rather than through
Polynomial.derivative_prod, which would need the evaluation of a multiset sum.
The second origin identity, division-free. The blog post's form
(-1)^{n-1}(∏ zⱼ)(1 + a ∑ 1/zⱼ) = (n/∏ qⱼ) F(1) after clearing both denominators.
The integral representation #
The integrand of the representation below is continuous.
The integral representation. p(w) = (w-a) ∫₀¹ p'(a + t(w-a)) dt when a is a root
of p. This is the fundamental theorem of calculus along the segment from a to w,
parametrized by a real variable; it replaces the blog post's p(z) = (z-a)∫₀¹ n ∏ⱼ (t(z-a) + 1/qⱼ) dt, which is the same statement with p' already factored.
The first origin identity #
The first origin identity, division-free. The blog post's
(-1)^{n-1} ∏ zⱼ = (n / ∏ qⱼ) ∫₀¹ F(t) dt with the denominator cleared, where
F(t) = ∏ⱼ (1 - a t qⱼ).
The polar identity #
p'(a) two ways. (∏ⱼ qⱼ) ∏ⱼ (a - zⱼ) = n. This is what makes the polar identity
usable without dividing: it is the denominator of the blog post's
∏ⱼ (1-azⱼ)/(a-zⱼ), expressed as a product.
The polar identity, division-free. The blog post's
∏ⱼ (1-azⱼ)/(a-zⱼ) = ∫₀¹ ∏ⱼ (t(1-a²)qⱼ + a) dt with the denominator cleared using
Sendov.prod_sub_mul_prod.