Alternating binomial sums of rational functions
TheoremThe factor \((x - 1)^n\) of an alternating binomial sum#●ready to prove
For every real number \(a\) and every natural number \(n\), the polynomial \((x - 1)^n\) divides, in \(\mathbb{R}[x]\), the polynomial
For \(a \ne -1\) (the paper's condition), \(H_n(x; a)\) is the numerator of \(G_n(x; a) = \sum_{i=0}^{n} (-1)^i \binom{n}{i} / (x^i + a)\) over the common denominator \(\prod_{k=0}^{n} (x^k + a)\); at \(a = -1\) the factor \(x^0 + a\) is zero and only the polynomial statement remains.
public theorem_wanted X_sub_one_pow_dvd_alternating_sum_prod_X_pow_add_C (n : ℕ) (a : ℝ) :
(X - 1 : ℝ[X]) ^ n ∣ ∑ i ∈ Finset.range (n + 1),
(-1 : ℝ[X]) ^ i * (n.choose i : ℝ[X]) *
∏ k ∈ (Finset.range (n + 1)).erase i, (X ^ k + C a)
Sources
- Takashi Agoh, An Intrinsic Factor of an Alternating Sum Involving Certain Rational Functions, INTEGERS 26 (2026), Article A9, Theorem 1.1.
TheoremThe factor \((x - 1)^n\) for polynomials with distinct real roots#●ready to prove
Let \(c\) be a real number, let \(s\) be a finite set of real numbers, and let \(f(x) = c \prod_{r \in s} (x - r)\) in \(\mathbb{R}[x]\). For every natural number \(n\), the polynomial \((x - 1)^n\) divides the polynomial
For \(c \ne 0\) and \(1 \notin s\) (the paper's conditions), this polynomial is the numerator of \(Q_n(x; f) = \sum_{i=0}^{n} (-1)^i \binom{n}{i} / f(x^i)\) over the common denominator \(\prod_{k=0}^{n} f(x^k)\); for \(c = 0\) or \(1 \in s\) some \(f(x^k)\) is zero and only the polynomial statement remains.
public theorem_wanted X_sub_one_pow_dvd_alternating_sum_prod_comp_X_pow (n : ℕ) (c : ℝ)
(s : Finset ℝ) :
(X - 1 : ℝ[X]) ^ n ∣ ∑ i ∈ Finset.range (n + 1),
(-1 : ℝ[X]) ^ i * (n.choose i : ℝ[X]) *
∏ k ∈ (Finset.range (n + 1)).erase i, (C c * ∏ r ∈ s, (X - C r)).comp (X ^ k)
Sources
- Takashi Agoh, An Intrinsic Factor of an Alternating Sum Involving Certain Rational Functions, INTEGERS 26 (2026), Article A9, Theorem 3.1.
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