# Documentation

Mathlib.Algebra.ContinuedFractions.ContinuantsRecurrence

# Recurrence Lemmas for the continuants Function of Continued Fractions. #

## Summary #

Given a generalized continued fraction g, for all n ≥ 1, we prove that the continuants function indeed satisfies the following recurrences:

• Aₙ = bₙ * Aₙ₋₁ + aₙ * Aₙ₋₂, and
• Bₙ = bₙ * Bₙ₋₁ + aₙ * Bₙ₋₂.
theorem GeneralizedContinuedFraction.continuantsAux_recurrence {K : Type u_1} {g : } {n : } [] {ppred : } {pred : } (nth_s_eq : Stream'.Seq.get? g.s n = some gp) (nth_conts_aux_eq : ) (succ_nth_conts_aux_eq : = pred) :
= { a := gp.b * pred.a + gp.a * ppred.a, b := gp.b * pred.b + gp.a * ppred.b }
theorem GeneralizedContinuedFraction.continuants_recurrenceAux {K : Type u_1} {g : } {n : } [] {ppred : } {pred : } (nth_s_eq : Stream'.Seq.get? g.s n = some gp) (nth_conts_aux_eq : ) (succ_nth_conts_aux_eq : = pred) :
= { a := gp.b * pred.a + gp.a * ppred.a, b := gp.b * pred.b + gp.a * ppred.b }
theorem GeneralizedContinuedFraction.continuants_recurrence {K : Type u_1} {g : } {n : } [] {ppred : } {pred : } (succ_nth_s_eq : Stream'.Seq.get? g.s (n + 1) = some gp) (nth_conts_eq : ) (succ_nth_conts_eq : = pred) :
= { a := gp.b * pred.a + gp.a * ppred.a, b := gp.b * pred.b + gp.a * ppred.b }

Shows that Aₙ = bₙ * Aₙ₋₁ + aₙ * Aₙ₋₂ and Bₙ = bₙ * Bₙ₋₁ + aₙ * Bₙ₋₂.

theorem GeneralizedContinuedFraction.numerators_recurrence {K : Type u_1} {g : } {n : } [] {ppredA : K} {predA : K} (succ_nth_s_eq : Stream'.Seq.get? g.s (n + 1) = some gp) (nth_num_eq : = ppredA) (succ_nth_num_eq : = predA) :
= gp.b * predA + gp.a * ppredA

Shows that Aₙ = bₙ * Aₙ₋₁ + aₙ * Aₙ₋₂.

theorem GeneralizedContinuedFraction.denominators_recurrence {K : Type u_1} {g : } {n : } [] {ppredB : K} {predB : K} (succ_nth_s_eq : Stream'.Seq.get? g.s (n + 1) = some gp) (nth_denom_eq : ) (succ_nth_denom_eq : = predB) :
= gp.b * predB + gp.a * ppredB

Shows that Bₙ = bₙ * Bₙ₋₁ + aₙ * Bₙ₋₂.