Si a, b ∈ ℝ tales que a ≤ b, entonces log(1 + eᵃ) ≤ log(1 + eᵇ)
Demostrar que si a, b ∈ ℝ tales que a ≤ b, entonces log(1 + eᵃ) ≤ log(1 + eᵇ).
Para ello, completar la siguiente teoría de Lean:
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import analysis.special_functions.log.basic open real variables a b : ℝ example (h : a ≤ b) : log (1 + exp a) ≤ log (1 + exp b) := sorry |
Soluciones con Lean
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import analysis.special_functions.log.basic open real variables a b : ℝ -- 1ª demostración -- =============== example (h : a ≤ b) : log (1 + exp a) ≤ log (1 + exp b) := begin have h₀ : 0 < 1 + exp a, { apply add_pos, exact one_pos, apply exp_pos, }, have h₁ : 0 < 1 + exp b, { apply add_pos, exact one_pos, apply exp_pos }, apply (log_le_log h₀ h₁).mpr, apply add_le_add, apply le_refl, apply exp_le_exp.mpr h, end -- 2ª demostración -- =============== example (h : a ≤ b) : log (1 + exp a) ≤ log (1 + exp b) := begin have h₀ : 0 < 1 + exp a := add_pos one_pos (exp_pos a), have h₁ : 0 < 1 + exp b := add_pos one_pos (exp_pos b), exact (log_le_log h₀ h₁).mpr (add_le_add rfl.ge (exp_le_exp.mpr h)) end -- 3ª demostración -- =============== lemma aux : 0 < 1 + exp a := add_pos one_pos (exp_pos a) example (h : a ≤ b) : log (1 + exp a) ≤ log (1 + exp b) := begin have h₀ : 0 < 1 + exp a := aux a, have h₁ : 0 < 1 + exp b := aux b, exact (log_le_log h₀ h₁).mpr (add_le_add rfl.ge (exp_le_exp.mpr h)) end -- 4ª demostración -- =============== example (h : a ≤ b) : log (1 + exp a) ≤ log (1 + exp b) := (log_le_log (aux a) (aux b)).mpr (add_le_add rfl.ge (exp_le_exp.mpr h)) |
Se puede interactuar con la prueba anterior en esta sesión con Lean.
Referencias
- J. Avigad, K. Buzzard, R.Y. Lewis y P. Massot. Mathematics in Lean, p. 17.