Taylor expansion and double roots
TheoremFirst-order Taylor expansion of a polynomial#✓fully proved
Let \(R\) be a commutative ring, let \(f \in R[X]\) and \(a \in R\), and let \(f'\) be the formal derivative of \(f\). There is a unique polynomial \(g \in R[X]\) such that
theorem existsUnique_taylor_remainder (f : R[X]) (a : R) :
∃! g : R[X],
f = C (f.eval a) + C (f.derivative.eval a) * (X - C a) +
g * (X - C a) ^ 2
Used by 1
TheoremDouble root criterion#✓fully proved
Let \(R\) be a commutative ring, let \(f \in R[x]\) and \(a \in R\), and let \(f'\) be the formal derivative of \(f\). Then \(f(a) = 0\) and \(f'(a) = 0\) if and only if \(f = (x - a)^2 g\) for some \(g \in R[x]\).
public theorem eval_and_derivative_eval_eq_zero_iff (f : R[X]) (a : R) :
f.eval a = 0 ∧ f.derivative.eval a = 0 ↔
∃ g : R[X], f = (X - C a) ^ 2 * g
Sources
- ATLAS NumberTheoryI, Corollary 9.13.
Uses 1 · Used by 1
TheoremDouble root criterion in divisibility form#✓fully proved
Let \(R\) be a commutative ring, let \(f \in R[x]\) and \(a \in R\), and let \(f'\) be the formal derivative of \(f\). Then \((x - a)^2\) divides \(f\) in \(R[x]\) if and only if \(f(a) = 0\) and \(f'(a) = 0\).
public theorem sq_dvd_iff_eval_and_derivative_eval_eq_zero (f : R[X])
(a : R) :
(X - C a) ^ 2 ∣ f ↔ f.eval a = 0 ∧ f.derivative.eval a = 0
Sources
- ATLAS NumberTheoryI, Corollary 9.13.
Uses 1 · Used by 1
TheoremA nonvanishing derivative excludes a double root#✓fully proved
Let \(R\) be a commutative ring, let \(f \in R[x]\) and \(a \in R\), and let \(f'\) be the formal derivative of \(f\). If \(f'(a) \neq 0\), then \((x - a)^2\) does not divide \(f\) in \(R[x]\).
No assumption on \(f(a)\) is needed.
public theorem not_sq_dvd_of_derivative_eval_ne_zero
(f : R[X]) (a : R) (hf' : f.derivative.eval a ≠ 0) :
¬ (X - C a) ^ 2 ∣ f
Sources
- ATLAS NumberTheoryI, consequence of Corollary 9.13.
Uses 1
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