src/HOL/Ln.thy
author huffman
Sat Mar 31 20:09:24 2012 +0200 (2012-03-31)
changeset 47242 1caeecc72aea
parent 44305 3bdc02eb1637
child 47244 a7f85074c169
permissions -rw-r--r--
tuned proof
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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 -
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  have "2 * 2 ^ n \<le> fact (n + 2)"
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    by (induct n, simp, simp)
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  hence "real ((2::nat) * 2 ^ n) \<le> real (fact (n + 2))"
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    by (simp only: real_of_nat_le_iff)
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  hence "2 * 2 ^ n \<le> real (fact (n + 2))"
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    by simp
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  hence "inverse (fact (n + 2)) \<le> inverse (2 * 2 ^ n)"
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    by (rule le_imp_inverse_le) simp
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  hence "inverse (fact (n + 2)) \<le> 1/2 * (1/2)^n"
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    by (simp add: inverse_mult_distrib power_inverse)
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  hence "inverse (fact (n + 2)) * (x^n * x\<twosuperior>) \<le> 1/2 * (1/2)^n * (1 * x\<twosuperior>)"
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    by (rule mult_mono)
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      (rule mult_mono, simp_all add: power_le_one a b mult_nonneg_nonneg)
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  thus ?thesis
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    unfolding power_add by (simp add: mult_ac del: fact_Suc)
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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
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    apply (rule ln_add_one_self_le_self2)
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    using a apply auto
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    done
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  also have "... <= 2 * x^2"
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    apply (subgoal_tac "- (-x) - 2 * (-x)^2 <= ln (1 - (-x))")
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    apply (simp add: algebra_simps)
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    apply (rule ln_one_minus_pos_lower_bound)
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    using a b apply auto
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    done
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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:
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    "abs x <= 1 / 2 ==> abs(ln (1 + x) - x) <= 2 * x^2"
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  apply (case_tac "0 <= x")
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  apply (rule order_trans)
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  apply (rule abs_ln_one_plus_x_minus_x_bound_nonneg)
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  apply auto
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  apply (rule abs_ln_one_plus_x_minus_x_bound_nonpos)
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  apply auto
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done
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lemma ln_x_over_x_mono: "exp 1 <= x ==> x <= y ==> (ln y / y) <= (ln x / x)"  
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proof -
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  assume x: "exp 1 <= x" "x <= y"
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  moreover have "0 < exp (1::real)" by simp
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  ultimately have a: "0 < x" and b: "0 < y"
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    by (fast intro: less_le_trans order_trans)+
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  have "x * ln y - x * ln x = x * (ln y - ln x)"
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    by (simp add: algebra_simps)
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  also have "... = x * ln(y / x)"
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    by (simp only: ln_div a b)
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  also have "y / x = (x + (y - x)) / x"
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    by simp
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  also have "... = 1 + (y - x) / x"
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    using x a by (simp add: field_simps)
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  also have "x * ln(1 + (y - x) / x) <= x * ((y - x) / x)"
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    apply (rule mult_left_mono)
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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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    using x a apply simp_all
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    done
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  also have "... = y - x" using a by simp
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  also have "... = (y - x) * ln (exp 1)" by simp
avigad@16959
   314
  also have "... <= (y - x) * ln x"
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   315
    apply (rule mult_left_mono)
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   316
    apply (subst ln_le_cancel_iff)
huffman@44289
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    apply fact
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   318
    apply (rule a)
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   319
    apply (rule x)
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   320
    using x apply simp
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   321
    done
avigad@16959
   322
  also have "... = y * ln x - x * ln x"
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   323
    by (rule left_diff_distrib)
avigad@16959
   324
  finally have "x * ln y <= y * ln x"
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   325
    by arith
wenzelm@41550
   326
  then have "ln y <= (y * ln x) / x" using a by (simp add: field_simps)
wenzelm@41550
   327
  also have "... = y * (ln x / x)" by simp
wenzelm@41550
   328
  finally show ?thesis using b by (simp add: field_simps)
avigad@16959
   329
qed
avigad@16959
   330
hoelzl@43336
   331
lemma ln_le_minus_one:
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   332
  "0 < x \<Longrightarrow> ln x \<le> x - 1"
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   333
  using exp_ge_add_one_self[of "ln x"] by simp
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   334
hoelzl@43336
   335
lemma ln_eq_minus_one:
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   336
  assumes "0 < x" "ln x = x - 1" shows "x = 1"
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   337
proof -
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   338
  let "?l y" = "ln y - y + 1"
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   339
  have D: "\<And>x. 0 < x \<Longrightarrow> DERIV ?l x :> (1 / x - 1)"
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   340
    by (auto intro!: DERIV_intros)
hoelzl@43336
   341
hoelzl@43336
   342
  show ?thesis
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   343
  proof (cases rule: linorder_cases)
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   344
    assume "x < 1"
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   345
    from dense[OF `x < 1`] obtain a where "x < a" "a < 1" by blast
hoelzl@43336
   346
    from `x < a` have "?l x < ?l a"
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   347
    proof (rule DERIV_pos_imp_increasing, safe)
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   348
      fix y assume "x \<le> y" "y \<le> a"
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   349
      with `0 < x` `a < 1` have "0 < 1 / y - 1" "0 < y"
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   350
        by (auto simp: field_simps)
hoelzl@43336
   351
      with D show "\<exists>z. DERIV ?l y :> z \<and> 0 < z"
hoelzl@43336
   352
        by auto
hoelzl@43336
   353
    qed
hoelzl@43336
   354
    also have "\<dots> \<le> 0"
hoelzl@43336
   355
      using ln_le_minus_one `0 < x` `x < a` by (auto simp: field_simps)
hoelzl@43336
   356
    finally show "x = 1" using assms by auto
hoelzl@43336
   357
  next
hoelzl@43336
   358
    assume "1 < x"
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   359
    from dense[OF `1 < x`] obtain a where "1 < a" "a < x" by blast
hoelzl@43336
   360
    from `a < x` have "?l x < ?l a"
hoelzl@43336
   361
    proof (rule DERIV_neg_imp_decreasing, safe)
hoelzl@43336
   362
      fix y assume "a \<le> y" "y \<le> x"
hoelzl@43336
   363
      with `1 < a` have "1 / y - 1 < 0" "0 < y"
hoelzl@43336
   364
        by (auto simp: field_simps)
hoelzl@43336
   365
      with D show "\<exists>z. DERIV ?l y :> z \<and> z < 0"
hoelzl@43336
   366
        by blast
hoelzl@43336
   367
    qed
hoelzl@43336
   368
    also have "\<dots> \<le> 0"
hoelzl@43336
   369
      using ln_le_minus_one `1 < a` by (auto simp: field_simps)
hoelzl@43336
   370
    finally show "x = 1" using assms by auto
hoelzl@43336
   371
  qed simp
hoelzl@43336
   372
qed
hoelzl@43336
   373
avigad@16959
   374
end