Theory Ln (Isabelle2005: October 2005)

# Theory Ln

Up to index of Isabelle/HOL/HOL-Complex

theory Ln
imports Transcendental
begin

    (*  Title:      Ln.thy
        Author:     Jeremy Avigad
        ID:         $Id: Ln.thy,v 1.3 2005/08/03 12:48:42 avigad Exp $
    *)

    header {* Properties of ln *}

    theory Ln

    imports Transcendental
    begin

    lemma exp_first_two_terms: "exp x = 1 + x + suminf (%n.
      inverse(real (fact (n+2))) * (x ^ (n+2)))"
    proof -
      have "exp x = suminf (%n. inverse(real (fact n)) * (x ^ n))"
        by (unfold exp_def, simp)
      also from summable_exp have "... = (SUM n : {0..<2}.
          inverse(real (fact n)) * (x ^ n)) + suminf (%n.
          inverse(real (fact (n+2))) * (x ^ (n+2)))" (is "_ = ?a + _")
        by (rule suminf_split_initial_segment)
      also have "?a = 1 + x"
        by (simp add: numerals)
      finally show ?thesis .
    qed

    lemma exp_tail_after_first_two_terms_summable:
      "summable (%n. inverse(real (fact (n+2))) * (x ^ (n+2)))"
    proof -
      note summable_exp
      thus ?thesis
        by (frule summable_ignore_initial_segment)
    qed

    lemma aux1: assumes a: "0 <= x" and b: "x <= 1"
        shows "inverse (real (fact (n + 2))) * x ^ (n + 2) <= (x^2/2) * ((1/2)^n)"
    proof (induct n)
      show "inverse (real (fact (0 + 2))) * x ^ (0 + 2) <=
          x ^ 2 / 2 * (1 / 2) ^ 0"
        apply (simp add: power2_eq_square)
        apply (subgoal_tac "real (Suc (Suc 0)) = 2")
        apply (erule ssubst)
        apply simp
        apply simp
        done
    next
      fix n
      assume c: "inverse (real (fact (n + 2))) * x ^ (n + 2)
           <= x ^ 2 / 2 * (1 / 2) ^ n"
      show "inverse (real (fact (Suc n + 2))) * x ^ (Suc n + 2)
               <= x ^ 2 / 2 * (1 / 2) ^ Suc n"
      proof -
        have "inverse(real (fact (Suc n + 2))) <=
            (1 / 2) *inverse (real (fact (n+2)))"
        proof -
          have "Suc n + 2 = Suc (n + 2)" by simp
          then have "fact (Suc n + 2) = Suc (n + 2) * fact (n + 2)"
            by simp
          then have "real(fact (Suc n + 2)) = real(Suc (n + 2) * fact (n + 2))"
            apply (rule subst)
            apply (rule refl)
            done
          also have "... = real(Suc (n + 2)) * real(fact (n + 2))"
            by (rule real_of_nat_mult)
          finally have "real (fact (Suc n + 2)) =
             real (Suc (n + 2)) * real (fact (n + 2))" .
          then have "inverse(real (fact (Suc n + 2))) =
             inverse(real (Suc (n + 2))) * inverse(real (fact (n + 2)))"
            apply (rule ssubst)
            apply (rule inverse_mult_distrib)
            done
          also have "... <= (1/2) * inverse(real (fact (n + 2)))"
            apply (rule mult_right_mono)
            apply (subst inverse_eq_divide)
            apply simp
            apply (rule inv_real_of_nat_fact_ge_zero)
            done
          finally show ?thesis .
        qed
        moreover have "x ^ (Suc n + 2) <= x ^ (n + 2)"
          apply (simp add: mult_compare_simps)
          apply (simp add: prems)
          apply (subgoal_tac "0 <= x * (x * x^n)")
          apply force
          apply (rule mult_nonneg_nonneg, rule a)+
          apply (rule zero_le_power, rule a)
          done
        ultimately have "inverse (real (fact (Suc n + 2))) *  x ^ (Suc n + 2) <=
            (1 / 2 * inverse (real (fact (n + 2)))) * x ^ (n + 2)"
          apply (rule mult_mono)
          apply (rule mult_nonneg_nonneg)
          apply simp
          apply (subst inverse_nonnegative_iff_nonnegative)
          apply (rule real_of_nat_fact_ge_zero)
          apply (rule zero_le_power)
          apply assumption
          done
        also have "... = 1 / 2 * (inverse (real (fact (n + 2))) * x ^ (n + 2))"
          by simp
        also have "... <= 1 / 2 * (x ^ 2 / 2 * (1 / 2) ^ n)"
          apply (rule mult_left_mono)
          apply (rule prems)
          apply simp
          done
        also have "... = x ^ 2 / 2 * (1 / 2 * (1 / 2) ^ n)"
          by auto
        also have "(1::real) / 2 * (1 / 2) ^ n = (1 / 2) ^ (Suc n)"
          by (rule realpow_Suc [THEN sym])
        finally show ?thesis .
      qed
    qed

    lemma aux2: "(%n. x ^ 2 / 2 * (1 / 2) ^ n) sums x^2"
    proof -
      have "(%n. (1 / 2)^n) sums (1 / (1 - (1/2)))"
        apply (rule geometric_sums)
        by (simp add: abs_interval_iff)
      also have "(1::real) / (1 - 1/2) = 2"
        by simp
      finally have "(%n. (1 / 2)^n) sums 2" .
      then have "(%n. x ^ 2 / 2 * (1 / 2) ^ n) sums (x^2 / 2 * 2)"
        by (rule sums_mult)
      also have "x^2 / 2 * 2 = x^2"
        by simp
      finally show ?thesis .
    qed

    lemma exp_bound: "0 <= x ==> x <= 1 ==> exp x <= 1 + x + x^2"
    proof -
      assume a: "0 <= x"
      assume b: "x <= 1"
      have c: "exp x = 1 + x + suminf (%n. inverse(real (fact (n+2))) *
          (x ^ (n+2)))"
        by (rule exp_first_two_terms)
      moreover have "suminf (%n. inverse(real (fact (n+2))) * (x ^ (n+2))) <= x^2"
      proof -
        have "suminf (%n. inverse(real (fact (n+2))) * (x ^ (n+2))) <=
            suminf (%n. (x^2/2) * ((1/2)^n))"
          apply (rule summable_le)
          apply (auto simp only: aux1 prems)
          apply (rule exp_tail_after_first_two_terms_summable)
          by (rule sums_summable, rule aux2)
        also have "... = x^2"
          by (rule sums_unique [THEN sym], rule aux2)
        finally show ?thesis .
      qed
      ultimately show ?thesis
        by auto
    qed

    lemma aux3: "(0::real) <= x ==> (1 + x + x^2)/(1 + x^2) <= 1 + x"
      apply (subst pos_divide_le_eq)
      apply (simp add: zero_compare_simps)
      apply (simp add: ring_eq_simps zero_compare_simps)
    done

    lemma aux4: "0 <= x ==> x <= 1 ==> exp (x - x^2) <= 1 + x"
    proof -
      assume a: "0 <= x" and b: "x <= 1"
      have "exp (x - x^2) = exp x / exp (x^2)"
        by (rule exp_diff)
      also have "... <= (1 + x + x^2) / exp (x ^2)"
        apply (rule divide_right_mono)
        apply (rule exp_bound)
        apply (rule a, rule b)
        apply simp
        done
      also have "... <= (1 + x + x^2) / (1 + x^2)"
        apply (rule divide_left_mono)
        apply (auto simp add: exp_ge_add_one_self_aux)
        apply (rule add_nonneg_nonneg)
        apply (insert prems, auto)
        apply (rule mult_pos_pos)
        apply auto
        apply (rule add_pos_nonneg)
        apply auto
        done
      also from a have "... <= 1 + x"
        by (rule aux3)
      finally show ?thesis .
    qed

    lemma ln_one_plus_pos_lower_bound: "0 <= x ==> x <= 1 ==>
        x - x^2 <= ln (1 + x)"
    proof -
      assume a: "0 <= x" and b: "x <= 1"
      then have "exp (x - x^2) <= 1 + x"
        by (rule aux4)
      also have "... = exp (ln (1 + x))"
      proof -
        from a have "0 < 1 + x" by auto
        thus ?thesis
          by (auto simp only: exp_ln_iff [THEN sym])
      qed
      finally have "exp (x - x ^ 2) <= exp (ln (1 + x))" .
      thus ?thesis by (auto simp only: exp_le_cancel_iff)
    qed

    lemma ln_one_minus_pos_upper_bound: "0 <= x ==> x < 1 ==> ln (1 - x) <= - x"
    proof -
      assume a: "0 <= (x::real)" and b: "x < 1"
      have "(1 - x) * (1 + x + x^2) = (1 - x^3)"
        by (simp add: ring_eq_simps power2_eq_square power3_eq_cube)
      also have "... <= 1"
        by (auto intro: zero_le_power simp add: a)
      finally have "(1 - x) * (1 + x + x ^ 2) <= 1" .
      moreover have "0 < 1 + x + x^2"
        apply (rule add_pos_nonneg)
        apply (insert a, auto)
        done
      ultimately have "1 - x <= 1 / (1 + x + x^2)"
        by (elim mult_imp_le_div_pos)
      also have "... <= 1 / exp x"
        apply (rule divide_left_mono)
        apply (rule exp_bound, rule a)
        apply (insert prems, auto)
        apply (rule mult_pos_pos)
        apply (rule add_pos_nonneg)
        apply auto
        done
      also have "... = exp (-x)"
        by (auto simp add: exp_minus real_divide_def)
      finally have "1 - x <= exp (- x)" .
      also have "1 - x = exp (ln (1 - x))"
      proof -
        have "0 < 1 - x"
          by (insert b, auto)
        thus ?thesis
          by (auto simp only: exp_ln_iff [THEN sym])
      qed
      finally have "exp (ln (1 - x)) <= exp (- x)" .
      thus ?thesis by (auto simp only: exp_le_cancel_iff)
    qed

    lemma aux5: "x < 1 ==> ln(1 - x) = - ln(1 + x / (1 - x))"
    proof -
      assume a: "x < 1"
      have "ln(1 - x) = - ln(1 / (1 - x))"
      proof -
        have "ln(1 - x) = - (- ln (1 - x))"
          by auto
        also have "- ln(1 - x) = ln 1 - ln(1 - x)"
          by simp
        also have "... = ln(1 / (1 - x))"
          apply (rule ln_div [THEN sym])
          by (insert a, auto)
        finally show ?thesis .
      qed
      also have " 1 / (1 - x) = 1 + x / (1 - x)"
      proof -
        have "1 / (1 - x) = (1 - x + x) / (1 - x)"
          by auto
        also have "... = (1 - x) / (1 - x) + x / (1 - x)"
          by (rule add_divide_distrib)
        also have "... = 1 + x / (1-x)"
          apply (subst add_right_cancel)
          apply (insert a, simp)
          done
        finally show ?thesis .
      qed
      finally show ?thesis .
    qed

    lemma ln_one_minus_pos_lower_bound: "0 <= x ==> x <= (1 / 2) ==>
        - x - 2 * x^2 <= ln (1 - x)"
    proof -
      assume a: "0 <= x" and b: "x <= (1 / 2)"
      from b have c: "x < 1"
        by auto
      then have "ln (1 - x) = - ln (1 + x / (1 - x))"
        by (rule aux5)
      also have "- (x / (1 - x)) <= ..."
      proof -
        have "ln (1 + x / (1 - x)) <= x / (1 - x)"
          apply (rule ln_add_one_self_le_self)
          apply (rule divide_nonneg_pos)
          by (insert a c, auto)
        thus ?thesis
          by auto
      qed
      also have "- (x / (1 - x)) = -x / (1 - x)"
        by auto
      finally have d: "- x / (1 - x) <= ln (1 - x)" .
      have e: "-x - 2 * x^2 <= - x / (1 - x)"
        apply (rule mult_imp_le_div_pos)
        apply (insert prems, force)
        apply (auto simp add: ring_eq_simps power2_eq_square)
        apply (subgoal_tac "- (x * x) + x * (x * (x * 2)) = x^2 * (2 * x - 1)")
        apply (erule ssubst)
        apply (rule mult_nonneg_nonpos)
        apply auto
        apply (auto simp add: ring_eq_simps power2_eq_square)
        done
      from e d show "- x - 2 * x^2 <= ln (1 - x)"
        by (rule order_trans)
    qed

    lemma exp_ge_add_one_self [simp]: "1 + x <= exp x"
      apply (case_tac "0 <= x")
      apply (erule exp_ge_add_one_self_aux)
      apply (case_tac "x <= -1")
      apply (subgoal_tac "1 + x <= 0")
      apply (erule order_trans)
      apply simp
      apply simp
      apply (subgoal_tac "1 + x = exp(ln (1 + x))")
      apply (erule ssubst)
      apply (subst exp_le_cancel_iff)
      apply (subgoal_tac "ln (1 - (- x)) <= - (- x)")
      apply simp
      apply (rule ln_one_minus_pos_upper_bound)
      apply auto
      apply (rule sym)
      apply (subst exp_ln_iff)
      apply auto
    done

    lemma ln_add_one_self_le_self2: "-1 < x ==> ln(1 + x) <= x"
      apply (subgoal_tac "x = ln (exp x)")
      apply (erule ssubst)back
      apply (subst ln_le_cancel_iff)
      apply auto
    done

    lemma abs_ln_one_plus_x_minus_x_bound_nonneg:
        "0 <= x ==> x <= 1 ==> abs(ln (1 + x) - x) <= x^2"
    proof -
      assume "0 <= x"
      assume "x <= 1"
      have "ln (1 + x) <= x"
        by (rule ln_add_one_self_le_self)
      then have "ln (1 + x) - x <= 0"
        by simp
      then have "abs(ln(1 + x) - x) = - (ln(1 + x) - x)"
        by (rule abs_of_nonpos)
      also have "... = x - ln (1 + x)"
        by simp
      also have "... <= x^2"
      proof -
        from prems have "x - x^2 <= ln (1 + x)"
          by (intro ln_one_plus_pos_lower_bound)
        thus ?thesis
          by simp
      qed
      finally show ?thesis .
    qed

    lemma abs_ln_one_plus_x_minus_x_bound_nonpos:
        "-(1 / 2) <= x ==> x <= 0 ==> abs(ln (1 + x) - x) <= 2 * x^2"
    proof -
      assume "-(1 / 2) <= x"
      assume "x <= 0"
      have "abs(ln (1 + x) - x) = x - ln(1 - (-x))"
        apply (subst abs_of_nonpos)
        apply simp
        apply (rule ln_add_one_self_le_self2)
        apply (insert prems, auto)
        done
      also have "... <= 2 * x^2"
        apply (subgoal_tac "- (-x) - 2 * (-x)^2 <= ln (1 - (-x))")
        apply (simp add: compare_rls)
        apply (rule ln_one_minus_pos_lower_bound)
        apply (insert prems, auto)
        done
      finally show ?thesis .
    qed

    lemma abs_ln_one_plus_x_minus_x_bound:
        "abs x <= 1 / 2 ==> abs(ln (1 + x) - x) <= 2 * x^2"
      apply (case_tac "0 <= x")
      apply (rule order_trans)
      apply (rule abs_ln_one_plus_x_minus_x_bound_nonneg)
      apply auto
      apply (rule abs_ln_one_plus_x_minus_x_bound_nonpos)
      apply auto
    done

    lemma DERIV_ln: "0 < x ==> DERIV ln x :> 1 / x"
      apply (unfold deriv_def, unfold LIM_def, clarsimp)
      apply (rule exI)
      apply (rule conjI)
      prefer 2
      apply clarsimp
      apply (subgoal_tac "(ln (x + xa) + - ln x) / xa + - (1 / x) =
          (ln (1 + xa / x) - xa / x) / xa")
      apply (erule ssubst)
      apply (subst abs_divide)
      apply (rule mult_imp_div_pos_less)
      apply force
      apply (rule order_le_less_trans)
      apply (rule abs_ln_one_plus_x_minus_x_bound)
      apply (subst abs_divide)
      apply (subst abs_of_pos, assumption)
      apply (erule mult_imp_div_pos_le)
      apply (subgoal_tac "abs xa < min (x / 2) (r * x^2 / 2)")
      apply force
      apply assumption
      apply (simp add: power2_eq_square mult_compare_simps)
      apply (rule mult_imp_div_pos_less)
      apply (rule mult_pos_pos, assumption, assumption)
      apply (subgoal_tac "xa * xa = abs xa * abs xa")
      apply (erule ssubst)
      apply (subgoal_tac "abs xa * (abs xa * 2) < abs xa * (r * (x * x))")
      apply (simp only: mult_ac)
      apply (rule mult_strict_left_mono)
      apply (erule conjE, assumption)
      apply force
      apply simp
      apply (subst diff_minus [THEN sym])+
      apply (subst ln_div [THEN sym])
      apply arith
      apply (auto simp add: ring_eq_simps add_frac_eq frac_eq_eq
        add_divide_distrib power2_eq_square)
      apply (rule mult_pos_pos, assumption)+
      apply assumption
    done

    lemma ln_x_over_x_mono: "exp 1 <= x ==> x <= y ==> (ln y / y) <= (ln x / x)"
    proof -
      assume "exp 1 <= x" and "x <= y"
      have a: "0 < x" and b: "0 < y"
        apply (insert prems)
        apply (subgoal_tac "0 < exp 1")
        apply arith
        apply auto
        apply (subgoal_tac "0 < exp 1")
        apply arith
        apply auto
        done
      have "x * ln y - x * ln x = x * (ln y - ln x)"
        by (simp add: ring_eq_simps)
      also have "... = x * ln(y / x)"
        apply (subst ln_div)
        apply (rule b, rule a, rule refl)
        done
      also have "y / x = (x + (y - x)) / x"
        by simp
      also have "... = 1 + (y - x) / x"
        apply (simp only: add_divide_distrib)
        apply (simp add: prems)
        apply (insert a, arith)
        done
      also have "x * ln(1 + (y - x) / x) <= x * ((y - x) / x)"
        apply (rule mult_left_mono)
        apply (rule ln_add_one_self_le_self)
        apply (rule divide_nonneg_pos)
        apply (insert prems a, simp_all)
        done
      also have "... = y - x"
        by (insert a, simp)
      also have "... = (y - x) * ln (exp 1)"
        by simp
      also have "... <= (y - x) * ln x"
        apply (rule mult_left_mono)
        apply (subst ln_le_cancel_iff)
        apply force
        apply (rule a)
        apply (rule prems)
        apply (insert prems, simp)
        done
      also have "... = y * ln x - x * ln x"
        by (rule left_diff_distrib)
      finally have "x * ln y <= y * ln x"
        by arith
      then have "ln y <= (y * ln x) / x"
        apply (subst pos_le_divide_eq)
        apply (rule a)
        apply (simp add: mult_ac)
        done
      also have "... = y * (ln x / x)"
        by simp
      finally show ?thesis
        apply (subst pos_divide_le_eq)
        apply (rule b)
        apply (simp add: mult_ac)
        done
    qed

    end

lemma exp_first_two_terms:

      exp x = 1 + x + (∑n. inverse (real (fact (n + 2))) * x ^ (n + 2))

lemma exp_tail_after_first_two_terms_summable:

      summable (%n. inverse (real (fact (n + 2))) * x ^ (n + 2))

lemma aux1:

      [| 0 ≤ x; x ≤ 1 |]
      ==> inverse (real (fact (n + 2))) * x ^ (n + 2) ≤ x² / 2 * (1 / 2) ^ n

lemma aux2:

      (%n. x² / 2 * (1 / 2) ^ n) sums x²

lemma exp_bound:

      [| 0 ≤ x; x ≤ 1 |] ==> exp x ≤ 1 + x + x²

lemma aux3:

      0 ≤ x ==> (1 + x + x²) / (1 + x²) ≤ 1 + x

lemma aux4:

      [| 0 ≤ x; x ≤ 1 |] ==> exp (x - x²) ≤ 1 + x

lemma ln_one_plus_pos_lower_bound:

      [| 0 ≤ x; x ≤ 1 |] ==> x - x² ≤ ln (1 + x)

lemma ln_one_minus_pos_upper_bound:

      [| 0 ≤ x; x < 1 |] ==> ln (1 - x) ≤ - x

lemma aux5:

      x < 1 ==> ln (1 - x) = - ln (1 + x / (1 - x))

lemma ln_one_minus_pos_lower_bound:

      [| 0 ≤ x; x ≤ 1 / 2 |] ==> - x - 2 * x² ≤ ln (1 - x)

lemma exp_ge_add_one_self:

      1 + x ≤ exp x

lemma ln_add_one_self_le_self2:

      -1 < x ==> ln (1 + x) ≤ x

lemma abs_ln_one_plus_x_minus_x_bound_nonneg:

      [| 0 ≤ x; x ≤ 1 |] ==> ¦ln (1 + x) - x¦ ≤ x²

lemma abs_ln_one_plus_x_minus_x_bound_nonpos:

      [| - (1 / 2) ≤ x; x ≤ 0 |] ==> ¦ln (1 + x) - x¦ ≤ 2 * x²

lemma abs_ln_one_plus_x_minus_x_bound:

      ¦x¦ ≤ 1 / 2 ==> ¦ln (1 + x) - x¦ ≤ 2 * x²

lemma DERIV_ln:

      0 < x ==> DERIV ln x :> 1 / x

lemma ln_x_over_x_mono:

      [| exp 1 ≤ x; x ≤ y |] ==> ln y / y ≤ ln x / x