summary |
shortlog |
changelog |
graph |
tags |
bookmarks |
branches |
files |
changeset |
file |
latest |
revisions |
annotate |
diff |
comparison |
raw |
help

src/HOL/Ln.thy

author | nipkow |

Wed, 01 Dec 2010 20:59:29 +0100 | |

changeset 40864 | 4abaaadfdaf2 |

parent 36777 | be5461582d0f |

child 41550 | efa734d9b221 |

permissions | -rw-r--r-- |

moved activation of coercion inference into RealDef and declared function real a coercion.
Made use of it in theory Ln.

(* 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