Niven polynomials #
The auxiliary polynomials Fₚ = Xᵖ⁻¹ Tᵖ and basic facts about their degree, used
to build the integer that drives the contradiction in Niven's proof.
Fₚ = Xᵖ⁻¹ Tᵖ.
Instances For
If T ≠ 0, then deg(Fₚ) = (p - 1) + p deg(T).
If T ≠ 0, then deg(Fₚ⁽ⁱ⁾) ≤ (p - 1) + p deg(T) - i.
The (p - 1)-th coefficient of Fₚ is T(0)ᵖ.
Fₚ⁽ᵖ⁻¹⁾(0) = (p - 1)! * T(0)ᵖ.
P(0) is the constant coefficient of P.
If T ≠ 0, then multₐ(Tᵖ) = p * multₐ(T).
If T ≠ 0, then multₐ(Fₚ) = p - 1 + p * multₐ(T) if a = 0 and p * multₐ(T) otherwise.
Sₚ,ᵢ is defined so that Fₚ⁽ⁱ⁾ = i! • Sₚ,ᵢ.
Equations
Instances For
The proof that Fₚ⁽ⁱ⁾ = i! • Sₚ,ᵢ.
If multₐ(T) ≥ m, then F_{p,d}(a) is expressed in terms of Fₚ⁽ⁱ⁾(a).
The definition of ∑ᵢ₌ₚᵈ Fₚ⁽ⁱ⁾ with d = deg(Fₚ).
Equations
- sumStartpDerivFp T p = ∑ i ∈ Finset.Icc p (Fp T p).natDegree, (⇑Polynomial.derivative)^[i] (Fp T p)
Instances For
If a ≠ 0 and a is a root of T, then F_{p,d} = ∑ᵢ₌ₚᵈ Fₚ⁽ⁱ⁾(a).
If 0 < p and T ≠ 0, then F_{p,d}(0) = (p - 1)! T(0)ᵖ + ∑ᵢ₌ₚᵈ Fₚ⁽ⁱ⁾(0).
∑ᵢ bᵢ exp(aᵢ) ∫₀¹ aᵢ * exp(-(t aᵢ)) * T(t aᵢ) dt is equal to
F_{p,d}(0) * ∑ᵢ bᵢ exp(aᵢ) - ∑ᵢ bᵢ F_{p,d}(0)(aᵢ).
∑ᵢ₌ₚᵈ Fₚ⁽ⁱ⁾ = p! * Gₚ.