src/HOL/Ln.thy
 author huffman Sat, 31 Mar 2012 20:09:24 +0200 changeset 47242 1caeecc72aea parent 44305 3bdc02eb1637 child 47244 a7f85074c169 permissions -rw-r--r--
tuned proof

(*  Title:      HOL/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: numeral_2_eq_2)
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 -
have "2 * 2 ^ n \<le> fact (n + 2)"
by (induct n, simp, simp)
hence "real ((2::nat) * 2 ^ n) \<le> real (fact (n + 2))"
by (simp only: real_of_nat_le_iff)
hence "2 * 2 ^ n \<le> real (fact (n + 2))"
by simp
hence "inverse (fact (n + 2)) \<le> inverse (2 * 2 ^ n)"
by (rule le_imp_inverse_le) simp
hence "inverse (fact (n + 2)) \<le> 1/2 * (1/2)^n"
by (simp add: inverse_mult_distrib power_inverse)
hence "inverse (fact (n + 2)) * (x^n * x\<twosuperior>) \<le> 1/2 * (1/2)^n * (1 * x\<twosuperior>)"
by (rule mult_mono)
(rule mult_mono, simp_all add: power_le_one a b mult_nonneg_nonneg)
thus ?thesis
unfolding power_add by (simp add: mult_ac del: fact_Suc)
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 a b)
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)
using a apply 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 add_strict_increasing zero_le_mult_iff)
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)
using a apply 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)
using a b apply 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 a b by simp
hence e: "-x - 2 * x^2 <= - x / (1 - x)"
using mult_right_le_one_le[of "x*x" "2*x"] a b
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 x1: "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 x x1 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 a: "-(1 / 2) <= x"
assume b: "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)
using a apply 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)
using a b apply 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 x: "exp 1 <= x" "x <= y"
moreover have "0 < exp (1::real)" by simp
ultimately have a: "0 < x" and b: "0 < y"
by (fast intro: less_le_trans order_trans)+
have "x * ln y - x * ln x = x * (ln y - ln x)"
by (simp add: algebra_simps)
also have "... = x * ln(y / x)"
by (simp only: ln_div a b)
also have "y / x = (x + (y - x)) / x"
by simp
also have "... = 1 + (y - x) / x"
using x a 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)
using x a apply 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 fact
apply (rule a)
apply (rule x)
using x apply 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

lemma ln_le_minus_one:
"0 < x \<Longrightarrow> ln x \<le> x - 1"
using exp_ge_add_one_self[of "ln x"] by simp

lemma ln_eq_minus_one:
assumes "0 < x" "ln x = x - 1" shows "x = 1"
proof -
let "?l y" = "ln y - y + 1"
have D: "\<And>x. 0 < x \<Longrightarrow> DERIV ?l x :> (1 / x - 1)"
by (auto intro!: DERIV_intros)

show ?thesis
proof (cases rule: linorder_cases)
assume "x < 1"
from dense[OF `x < 1`] obtain a where "x < a" "a < 1" by blast
from `x < a` have "?l x < ?l a"
proof (rule DERIV_pos_imp_increasing, safe)
fix y assume "x \<le> y" "y \<le> a"
with `0 < x` `a < 1` have "0 < 1 / y - 1" "0 < y"
by (auto simp: field_simps)
with D show "\<exists>z. DERIV ?l y :> z \<and> 0 < z"
by auto
qed
also have "\<dots> \<le> 0"
using ln_le_minus_one `0 < x` `x < a` by (auto simp: field_simps)
finally show "x = 1" using assms by auto
next
assume "1 < x"
from dense[OF `1 < x`] obtain a where "1 < a" "a < x" by blast
from `a < x` have "?l x < ?l a"
proof (rule DERIV_neg_imp_decreasing, safe)
fix y assume "a \<le> y" "y \<le> x"
with `1 < a` have "1 / y - 1 < 0" "0 < y"
by (auto simp: field_simps)
with D show "\<exists>z. DERIV ?l y :> z \<and> z < 0"
by blast
qed
also have "\<dots> \<le> 0"
using ln_le_minus_one `1 < a` by (auto simp: field_simps)
finally show "x = 1" using assms by auto
qed simp
qed

end