src/HOL/Ln.thy
author huffman
Thu Aug 18 19:53:03 2011 -0700 (2011-08-18)
changeset 44289 d81d09cdab9c
parent 43336 05aa7380f7fc
child 44305 3bdc02eb1637
permissions -rw-r--r--
optimize some proofs
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(*  Title:      HOL/Ln.thy
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    Author:     Jeremy Avigad
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*)
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header {* Properties of ln *}
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theory Ln
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imports Transcendental
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begin
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lemma exp_first_two_terms: "exp x = 1 + x + suminf (%n. 
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  inverse(fact (n+2)) * (x ^ (n+2)))"
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proof -
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  have "exp x = suminf (%n. inverse(fact n) * (x ^ n))"
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    by (simp add: exp_def)
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  also from summable_exp have "... = (SUM n::nat : {0..<2}. 
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      inverse(fact n) * (x ^ n)) + suminf (%n.
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      inverse(fact(n+2)) * (x ^ (n+2)))" (is "_ = ?a + _")
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    by (rule suminf_split_initial_segment)
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  also have "?a = 1 + x"
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    by (simp add: numeral_2_eq_2)
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  finally show ?thesis .
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qed
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lemma exp_tail_after_first_two_terms_summable: 
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  "summable (%n. inverse(fact (n+2)) * (x ^ (n+2)))"
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proof -
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  note summable_exp
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  thus ?thesis
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    by (frule summable_ignore_initial_segment)
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qed
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lemma aux1: assumes a: "0 <= x" and b: "x <= 1"
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    shows "inverse (fact ((n::nat) + 2)) * x ^ (n + 2) <= (x^2/2) * ((1/2)^n)"
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proof (induct n)
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  show "inverse (fact ((0::nat) + 2)) * x ^ (0 + 2) <= 
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      x ^ 2 / 2 * (1 / 2) ^ 0"
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    by (simp add: real_of_nat_Suc power2_eq_square)
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next
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  fix n :: nat
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  assume c: "inverse (fact (n + 2)) * x ^ (n + 2)
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       <= x ^ 2 / 2 * (1 / 2) ^ n"
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  show "inverse (fact (Suc n + 2)) * x ^ (Suc n + 2)
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           <= x ^ 2 / 2 * (1 / 2) ^ Suc n"
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  proof -
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    have "inverse(fact (Suc n + 2)) <= (1/2) * inverse (fact (n+2))"
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    proof -
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      have "Suc n + 2 = Suc (n + 2)" by simp
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      then have "fact (Suc n + 2) = Suc (n + 2) * fact (n + 2)" 
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        by simp
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      then have "real(fact (Suc n + 2)) = real(Suc (n + 2) * fact (n + 2))" 
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        apply (rule subst)
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        apply (rule refl)
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        done
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      also have "... = real(Suc (n + 2)) * real(fact (n + 2))"
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        by (rule real_of_nat_mult)
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      finally have "real (fact (Suc n + 2)) = 
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         real (Suc (n + 2)) * real (fact (n + 2))" .
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      then have "inverse(fact (Suc n + 2)) = 
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         inverse(Suc (n + 2)) * inverse(fact (n + 2))"
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        apply (rule ssubst)
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        apply (rule inverse_mult_distrib)
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        done
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      also have "... <= (1/2) * inverse(fact (n + 2))"
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        apply (rule mult_right_mono)
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        apply (subst inverse_eq_divide)
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        apply simp
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        apply (rule inv_real_of_nat_fact_ge_zero)
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        done
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      finally show ?thesis .
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    qed
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    moreover have "x ^ (Suc n + 2) <= x ^ (n + 2)"
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      by (simp add: mult_left_le_one_le mult_nonneg_nonneg a b)
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    ultimately have "inverse (fact (Suc n + 2)) *  x ^ (Suc n + 2) <=
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        (1 / 2 * inverse (fact (n + 2))) * x ^ (n + 2)"
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      apply (rule mult_mono)
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      apply (rule mult_nonneg_nonneg)
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      apply simp
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      apply (subst inverse_nonnegative_iff_nonnegative)
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      apply (rule real_of_nat_ge_zero)
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      apply (rule zero_le_power)
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      apply (rule a)
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      done
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    also have "... = 1 / 2 * (inverse (fact (n + 2)) * x ^ (n + 2))"
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      by simp
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    also have "... <= 1 / 2 * (x ^ 2 / 2 * (1 / 2) ^ n)"
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      apply (rule mult_left_mono)
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      apply (rule c)
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      apply simp
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      done
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    also have "... = x ^ 2 / 2 * (1 / 2 * (1 / 2) ^ n)"
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      by auto
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    also have "(1::real) / 2 * (1 / 2) ^ n = (1 / 2) ^ (Suc n)"
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      by (rule power_Suc [THEN sym])
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    finally show ?thesis .
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  qed
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qed
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lemma aux2: "(%n. (x::real) ^ 2 / 2 * (1 / 2) ^ n) sums x^2"
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proof -
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  have "(%n. (1 / 2::real)^n) sums (1 / (1 - (1/2)))"
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    apply (rule geometric_sums)
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    by (simp add: abs_less_iff)
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  also have "(1::real) / (1 - 1/2) = 2"
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    by simp
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  finally have "(%n. (1 / 2::real)^n) sums 2" .
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  then have "(%n. x ^ 2 / 2 * (1 / 2) ^ n) sums (x^2 / 2 * 2)"
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    by (rule sums_mult)
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  also have "x^2 / 2 * 2 = x^2"
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    by simp
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  finally show ?thesis .
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qed
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lemma exp_bound: "0 <= (x::real) ==> x <= 1 ==> exp x <= 1 + x + x^2"
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proof -
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  assume a: "0 <= x"
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  assume b: "x <= 1"
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  have c: "exp x = 1 + x + suminf (%n. inverse(fact (n+2)) * 
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      (x ^ (n+2)))"
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    by (rule exp_first_two_terms)
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  moreover have "suminf (%n. inverse(fact (n+2)) * (x ^ (n+2))) <= x^2"
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  proof -
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    have "suminf (%n. inverse(fact (n+2)) * (x ^ (n+2))) <=
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        suminf (%n. (x^2/2) * ((1/2)^n))"
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      apply (rule summable_le)
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      apply (auto simp only: aux1 a b)
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      apply (rule exp_tail_after_first_two_terms_summable)
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      by (rule sums_summable, rule aux2)  
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    also have "... = x^2"
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      by (rule sums_unique [THEN sym], rule aux2)
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    finally show ?thesis .
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  qed
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  ultimately show ?thesis
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    by auto
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qed
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lemma aux4: "0 <= (x::real) ==> x <= 1 ==> exp (x - x^2) <= 1 + x" 
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proof -
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  assume a: "0 <= x" and b: "x <= 1"
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  have "exp (x - x^2) = exp x / exp (x^2)"
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    by (rule exp_diff)
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  also have "... <= (1 + x + x^2) / exp (x ^2)"
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    apply (rule divide_right_mono) 
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    apply (rule exp_bound)
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    apply (rule a, rule b)
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    apply simp
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    done
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  also have "... <= (1 + x + x^2) / (1 + x^2)"
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    apply (rule divide_left_mono)
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    apply (auto simp add: exp_ge_add_one_self_aux)
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    apply (rule add_nonneg_nonneg)
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    using a apply auto
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    apply (rule mult_pos_pos)
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    apply auto
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    apply (rule add_pos_nonneg)
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    apply auto
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    done
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  also from a have "... <= 1 + x"
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    by (simp add: field_simps add_strict_increasing zero_le_mult_iff)
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  finally show ?thesis .
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qed
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lemma ln_one_plus_pos_lower_bound: "0 <= x ==> x <= 1 ==> 
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    x - x^2 <= ln (1 + x)"
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proof -
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  assume a: "0 <= x" and b: "x <= 1"
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  then have "exp (x - x^2) <= 1 + x"
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    by (rule aux4)
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  also have "... = exp (ln (1 + x))"
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  proof -
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    from a have "0 < 1 + x" by auto
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    thus ?thesis
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      by (auto simp only: exp_ln_iff [THEN sym])
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  qed
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  finally have "exp (x - x ^ 2) <= exp (ln (1 + x))" .
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  thus ?thesis by (auto simp only: exp_le_cancel_iff)
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qed
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lemma ln_one_minus_pos_upper_bound: "0 <= x ==> x < 1 ==> ln (1 - x) <= - x"
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proof -
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  assume a: "0 <= (x::real)" and b: "x < 1"
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  have "(1 - x) * (1 + x + x^2) = (1 - x^3)"
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    by (simp add: algebra_simps power2_eq_square power3_eq_cube)
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  also have "... <= 1"
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    by (auto simp add: a)
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  finally have "(1 - x) * (1 + x + x ^ 2) <= 1" .
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  moreover have "0 < 1 + x + x^2"
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    apply (rule add_pos_nonneg)
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    using a apply auto
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    done
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  ultimately have "1 - x <= 1 / (1 + x + x^2)"
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    by (elim mult_imp_le_div_pos)
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  also have "... <= 1 / exp x"
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    apply (rule divide_left_mono)
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    apply (rule exp_bound, rule a)
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    using a b apply auto
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    apply (rule mult_pos_pos)
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    apply (rule add_pos_nonneg)
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    apply auto
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    done
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  also have "... = exp (-x)"
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    by (auto simp add: exp_minus divide_inverse)
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  finally have "1 - x <= exp (- x)" .
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  also have "1 - x = exp (ln (1 - x))"
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  proof -
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    have "0 < 1 - x"
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      by (insert b, auto)
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    thus ?thesis
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      by (auto simp only: exp_ln_iff [THEN sym])
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  qed
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  finally have "exp (ln (1 - x)) <= exp (- x)" .
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  thus ?thesis by (auto simp only: exp_le_cancel_iff)
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qed
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lemma aux5: "x < 1 ==> ln(1 - x) = - ln(1 + x / (1 - x))"
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proof -
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  assume a: "x < 1"
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  have "ln(1 - x) = - ln(1 / (1 - x))"
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  proof -
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    have "ln(1 - x) = - (- ln (1 - x))"
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      by auto
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    also have "- ln(1 - x) = ln 1 - ln(1 - x)"
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      by simp
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    also have "... = ln(1 / (1 - x))"
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      apply (rule ln_div [THEN sym])
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      by (insert a, auto)
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    finally show ?thesis .
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  qed
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  also have " 1 / (1 - x) = 1 + x / (1 - x)" using a by(simp add:field_simps)
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  finally show ?thesis .
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qed
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lemma ln_one_minus_pos_lower_bound: "0 <= x ==> x <= (1 / 2) ==> 
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    - x - 2 * x^2 <= ln (1 - x)"
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proof -
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  assume a: "0 <= x" and b: "x <= (1 / 2)"
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  from b have c: "x < 1"
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    by auto
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  then have "ln (1 - x) = - ln (1 + x / (1 - x))"
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    by (rule aux5)
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  also have "- (x / (1 - x)) <= ..."
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  proof - 
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    have "ln (1 + x / (1 - x)) <= x / (1 - x)"
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      apply (rule ln_add_one_self_le_self)
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      apply (rule divide_nonneg_pos)
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      by (insert a c, auto) 
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    thus ?thesis
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      by auto
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  qed
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  also have "- (x / (1 - x)) = -x / (1 - x)"
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    by auto
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  finally have d: "- x / (1 - x) <= ln (1 - x)" .
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  have "0 < 1 - x" using a b by simp
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  hence e: "-x - 2 * x^2 <= - x / (1 - x)"
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    using mult_right_le_one_le[of "x*x" "2*x"] a b
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    by (simp add:field_simps power2_eq_square)
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  from e d show "- x - 2 * x^2 <= ln (1 - x)"
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    by (rule order_trans)
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qed
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lemma exp_ge_add_one_self [simp]: "1 + (x::real) <= exp x"
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  apply (case_tac "0 <= x")
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  apply (erule exp_ge_add_one_self_aux)
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  apply (case_tac "x <= -1")
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  apply (subgoal_tac "1 + x <= 0")
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  apply (erule order_trans)
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  apply simp
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  apply simp
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  apply (subgoal_tac "1 + x = exp(ln (1 + x))")
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  apply (erule ssubst)
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  apply (subst exp_le_cancel_iff)
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  apply (subgoal_tac "ln (1 - (- x)) <= - (- x)")
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  apply simp
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  apply (rule ln_one_minus_pos_upper_bound) 
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  apply auto
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done
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lemma ln_add_one_self_le_self2: "-1 < x ==> ln(1 + x) <= x"
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  apply (subgoal_tac "x = ln (exp x)")
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  apply (erule ssubst)back
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  apply (subst ln_le_cancel_iff)
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  apply auto
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done
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lemma abs_ln_one_plus_x_minus_x_bound_nonneg:
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    "0 <= x ==> x <= 1 ==> abs(ln (1 + x) - x) <= x^2"
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proof -
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  assume x: "0 <= x"
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  assume x1: "x <= 1"
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  from x have "ln (1 + x) <= x"
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    by (rule ln_add_one_self_le_self)
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  then have "ln (1 + x) - x <= 0" 
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    by simp
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  then have "abs(ln(1 + x) - x) = - (ln(1 + x) - x)"
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    by (rule abs_of_nonpos)
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  also have "... = x - ln (1 + x)" 
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    by simp
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  also have "... <= x^2"
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  proof -
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    from x x1 have "x - x^2 <= ln (1 + x)"
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      by (intro ln_one_plus_pos_lower_bound)
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    thus ?thesis
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      by simp
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  qed
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  finally show ?thesis .
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qed
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lemma abs_ln_one_plus_x_minus_x_bound_nonpos:
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    "-(1 / 2) <= x ==> x <= 0 ==> abs(ln (1 + x) - x) <= 2 * x^2"
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proof -
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  assume a: "-(1 / 2) <= x"
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  assume b: "x <= 0"
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  have "abs(ln (1 + x) - x) = x - ln(1 - (-x))" 
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    apply (subst abs_of_nonpos)
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    apply simp
avigad@16959
   316
    apply (rule ln_add_one_self_le_self2)
wenzelm@41550
   317
    using a apply auto
avigad@16959
   318
    done
avigad@16959
   319
  also have "... <= 2 * x^2"
avigad@16959
   320
    apply (subgoal_tac "- (-x) - 2 * (-x)^2 <= ln (1 - (-x))")
nipkow@29667
   321
    apply (simp add: algebra_simps)
avigad@16959
   322
    apply (rule ln_one_minus_pos_lower_bound)
wenzelm@41550
   323
    using a b apply auto
nipkow@29667
   324
    done
avigad@16959
   325
  finally show ?thesis .
avigad@16959
   326
qed
avigad@16959
   327
avigad@16959
   328
lemma abs_ln_one_plus_x_minus_x_bound:
avigad@16959
   329
    "abs x <= 1 / 2 ==> abs(ln (1 + x) - x) <= 2 * x^2"
avigad@16959
   330
  apply (case_tac "0 <= x")
avigad@16959
   331
  apply (rule order_trans)
avigad@16959
   332
  apply (rule abs_ln_one_plus_x_minus_x_bound_nonneg)
avigad@16959
   333
  apply auto
avigad@16959
   334
  apply (rule abs_ln_one_plus_x_minus_x_bound_nonpos)
avigad@16959
   335
  apply auto
avigad@16959
   336
done
avigad@16959
   337
avigad@16959
   338
lemma ln_x_over_x_mono: "exp 1 <= x ==> x <= y ==> (ln y / y) <= (ln x / x)"  
avigad@16959
   339
proof -
wenzelm@41550
   340
  assume x: "exp 1 <= x" "x <= y"
huffman@44289
   341
  moreover have "0 < exp (1::real)" by simp
huffman@44289
   342
  ultimately have a: "0 < x" and b: "0 < y"
huffman@44289
   343
    by (fast intro: less_le_trans order_trans)+
avigad@16959
   344
  have "x * ln y - x * ln x = x * (ln y - ln x)"
nipkow@29667
   345
    by (simp add: algebra_simps)
avigad@16959
   346
  also have "... = x * ln(y / x)"
huffman@44289
   347
    by (simp only: ln_div a b)
avigad@16959
   348
  also have "y / x = (x + (y - x)) / x"
avigad@16959
   349
    by simp
huffman@44289
   350
  also have "... = 1 + (y - x) / x"
huffman@44289
   351
    using x a by (simp add: field_simps)
avigad@16959
   352
  also have "x * ln(1 + (y - x) / x) <= x * ((y - x) / x)"
avigad@16959
   353
    apply (rule mult_left_mono)
avigad@16959
   354
    apply (rule ln_add_one_self_le_self)
avigad@16959
   355
    apply (rule divide_nonneg_pos)
wenzelm@41550
   356
    using x a apply simp_all
avigad@16959
   357
    done
nipkow@23482
   358
  also have "... = y - x" using a by simp
nipkow@23482
   359
  also have "... = (y - x) * ln (exp 1)" by simp
avigad@16959
   360
  also have "... <= (y - x) * ln x"
avigad@16959
   361
    apply (rule mult_left_mono)
avigad@16959
   362
    apply (subst ln_le_cancel_iff)
huffman@44289
   363
    apply fact
avigad@16959
   364
    apply (rule a)
wenzelm@41550
   365
    apply (rule x)
wenzelm@41550
   366
    using x apply simp
avigad@16959
   367
    done
avigad@16959
   368
  also have "... = y * ln x - x * ln x"
avigad@16959
   369
    by (rule left_diff_distrib)
avigad@16959
   370
  finally have "x * ln y <= y * ln x"
avigad@16959
   371
    by arith
wenzelm@41550
   372
  then have "ln y <= (y * ln x) / x" using a by (simp add: field_simps)
wenzelm@41550
   373
  also have "... = y * (ln x / x)" by simp
wenzelm@41550
   374
  finally show ?thesis using b by (simp add: field_simps)
avigad@16959
   375
qed
avigad@16959
   376
hoelzl@43336
   377
lemma ln_le_minus_one:
hoelzl@43336
   378
  "0 < x \<Longrightarrow> ln x \<le> x - 1"
hoelzl@43336
   379
  using exp_ge_add_one_self[of "ln x"] by simp
hoelzl@43336
   380
hoelzl@43336
   381
lemma ln_eq_minus_one:
hoelzl@43336
   382
  assumes "0 < x" "ln x = x - 1" shows "x = 1"
hoelzl@43336
   383
proof -
hoelzl@43336
   384
  let "?l y" = "ln y - y + 1"
hoelzl@43336
   385
  have D: "\<And>x. 0 < x \<Longrightarrow> DERIV ?l x :> (1 / x - 1)"
hoelzl@43336
   386
    by (auto intro!: DERIV_intros)
hoelzl@43336
   387
hoelzl@43336
   388
  show ?thesis
hoelzl@43336
   389
  proof (cases rule: linorder_cases)
hoelzl@43336
   390
    assume "x < 1"
hoelzl@43336
   391
    from dense[OF `x < 1`] obtain a where "x < a" "a < 1" by blast
hoelzl@43336
   392
    from `x < a` have "?l x < ?l a"
hoelzl@43336
   393
    proof (rule DERIV_pos_imp_increasing, safe)
hoelzl@43336
   394
      fix y assume "x \<le> y" "y \<le> a"
hoelzl@43336
   395
      with `0 < x` `a < 1` have "0 < 1 / y - 1" "0 < y"
hoelzl@43336
   396
        by (auto simp: field_simps)
hoelzl@43336
   397
      with D show "\<exists>z. DERIV ?l y :> z \<and> 0 < z"
hoelzl@43336
   398
        by auto
hoelzl@43336
   399
    qed
hoelzl@43336
   400
    also have "\<dots> \<le> 0"
hoelzl@43336
   401
      using ln_le_minus_one `0 < x` `x < a` by (auto simp: field_simps)
hoelzl@43336
   402
    finally show "x = 1" using assms by auto
hoelzl@43336
   403
  next
hoelzl@43336
   404
    assume "1 < x"
hoelzl@43336
   405
    from dense[OF `1 < x`] obtain a where "1 < a" "a < x" by blast
hoelzl@43336
   406
    from `a < x` have "?l x < ?l a"
hoelzl@43336
   407
    proof (rule DERIV_neg_imp_decreasing, safe)
hoelzl@43336
   408
      fix y assume "a \<le> y" "y \<le> x"
hoelzl@43336
   409
      with `1 < a` have "1 / y - 1 < 0" "0 < y"
hoelzl@43336
   410
        by (auto simp: field_simps)
hoelzl@43336
   411
      with D show "\<exists>z. DERIV ?l y :> z \<and> z < 0"
hoelzl@43336
   412
        by blast
hoelzl@43336
   413
    qed
hoelzl@43336
   414
    also have "\<dots> \<le> 0"
hoelzl@43336
   415
      using ln_le_minus_one `1 < a` by (auto simp: field_simps)
hoelzl@43336
   416
    finally show "x = 1" using assms by auto
hoelzl@43336
   417
  qed simp
hoelzl@43336
   418
qed
hoelzl@43336
   419
avigad@16959
   420
end