(*  Title:      Ln.thy

     Author:     Jeremy Avigad

 *)

 header {* Properties of ln *}

 theory Ln

 imports Transcendental

 begin

 lemma exp_first_two_terms: "exp x = 1 + x + suminf (%n. 

   inverse(fact (n+2)) * (x ^ (n+2)))"

 proof -

   have "exp x = suminf (%n. inverse(fact n) * (x ^ n))"

     by (simp add: exp_def)

   also from summable_exp have "... = (SUM n::nat : {0..<2}. 

       inverse(fact n) * (x ^ n)) + suminf (%n.

       inverse(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(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 (fact ((n::nat) + 2)) * x ^ (n + 2) <= (x^2/2) * ((1/2)^n)"

 proof (induct n)

   show "inverse (fact ((0::nat) + 2)) * x ^ (0 + 2) <= 

       x ^ 2 / 2 * (1 / 2) ^ 0"

     by (simp add: real_of_nat_Suc power2_eq_square)

 next

   fix n :: nat

   assume c: "inverse (fact (n + 2)) * x ^ (n + 2)

        <= x ^ 2 / 2 * (1 / 2) ^ n"

   show "inverse (fact (Suc n + 2)) * x ^ (Suc n + 2)

            <= x ^ 2 / 2 * (1 / 2) ^ Suc n"

   proof -

     have "inverse(fact (Suc n + 2)) <= (1/2) * inverse (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(fact (Suc n + 2)) = 

          inverse(Suc (n + 2)) * inverse(fact (n + 2))"

         apply (rule ssubst)

         apply (rule inverse_mult_distrib)

         done

       also have "... <= (1/2) * inverse(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 (fact (Suc n + 2)) *  x ^ (Suc n + 2) <=

         (1 / 2 * inverse (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_ge_zero)

       apply (rule zero_le_power)

       apply (rule a)

       done

     also have "... = 1 / 2 * (inverse (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 power_Suc [THEN sym])

     finally show ?thesis .

   qed

 qed

 lemma aux2: "(%n. (x::real) ^ 2 / 2 * (1 / 2) ^ n) sums x^2"

 proof -

   have "(%n. (1 / 2::real)^n) sums (1 / (1 - (1/2)))"

     apply (rule geometric_sums)

     by (simp add: abs_less_iff)

   also have "(1::real) / (1 - 1/2) = 2"

     by simp

   finally have "(%n. (1 / 2::real)^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::real) ==> 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(fact (n+2)) * 

       (x ^ (n+2)))"

     by (rule exp_first_two_terms)

   moreover have "suminf (%n. inverse(fact (n+2)) * (x ^ (n+2))) <= x^2"

   proof -

     have "suminf (%n. inverse(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 aux4: "0 <= (x::real) ==> 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(simp add:field_simps zero_compare_simps)

   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: algebra_simps power2_eq_square power3_eq_cube)

   also have "... <= 1"

     by (auto 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 divide_inverse)

   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)" using a by(simp add:field_simps)

   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 "0 < 1 - x" using prems by simp

   hence e: "-x - 2 * x^2 <= - x / (1 - x)"

     using mult_right_le_one_le[of "x*x" "2*x"] prems

     by(simp add:field_simps power2_eq_square)

   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::real) <= 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

 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 x: "0 <= x"

   assume "x <= 1"

   from x 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: algebra_simps)

     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 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::real)")

     apply arith

     apply auto

     apply (subgoal_tac "0 < exp (1::real)")

     apply arith

     apply auto

     done

   have "x * ln y - x * ln x = x * (ln y - ln x)"

     by (simp add: algebra_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" using a prems by(simp add:field_simps)

   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" using a by 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" using a by(simp add:field_simps)

   also have "... = y * (ln x / x)"  by simp

   finally show ?thesis using b by(simp add:field_simps)

 qed

 end