src/HOL/Ln.thy
 author huffman Thu, 18 Aug 2011 19:53:03 -0700 changeset 44289 d81d09cdab9c parent 43336 05aa7380f7fc child 44305 3bdc02eb1637 permissions -rw-r--r--
optimize some proofs
```
(*  Title:      HOL/Ln.thy
*)

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))"
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"
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"
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)"
by (simp add: mult_left_le_one_le mult_nonneg_nonneg a b)
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 c)
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)
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)
using a apply auto
apply (rule mult_pos_pos)
apply auto
apply auto
done
also from a have "... <= 1 + x"
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"
finally have "(1 - x) * (1 + x + x ^ 2) <= 1" .
moreover have "0 < 1 + x + x^2"
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 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 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
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 (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"
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
using a apply auto
done
also have "... <= 2 * x^2"
apply (subgoal_tac "- (-x) - 2 * (-x)^2 <= ln (1 - (-x))")
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)"
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 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
```