Theory Polya_Vinogradov
section ‹The Pólya--Vinogradov Inequality›
theory Polya_Vinogradov
imports
Gauss_Sums
"Dirichlet_Series.Divisor_Count"
begin
unbundle no_vec_lambda_notation
subsection ‹The case of primitive characters›
text ‹
We first prove a stronger variant of the Pólya--Vinogradov inequality for primitive characters.
The fully general variant will then simply be a corollary of this. First, we need some bounds on
logarithms, exponentials, and the harmonic numbers:
›
lemma ln_add_one_self_less_self:
fixes x :: real
assumes "x > 0"
shows "ln (1 + x) < x"
proof -
have "0 ≤ x" "0 < x" "exp x > 0" "1+x > 0" using assms by simp+
have "1 + x < 1 + x + x⇧2 / 2"
using ‹0 < x› by auto
also have "… ≤ exp x"
using exp_lower_Taylor_quadratic[OF ‹0 ≤ x›] by blast
finally have "1 + x < exp (x)" by blast
then have "ln (1 + x) < ln (exp (x))"
using ln_less_cancel_iff[OF ‹1+x > 0› ‹exp(x) > 0›] by auto
also have "… = x" using ln_exp by blast
finally show ?thesis by auto
qed
lemma exp_1_bounds:
assumes "x > (0::real)"
shows "exp 1 > (1 + 1 / x) powr x" and "exp 1 < (1 + 1 / x) powr (x+1)"
proof -
have "ln (1 + 1 / x) < 1 / x"
using ln_add_one_self_less_self assms by simp
thus "exp 1 > (1 + 1 / x) powr x" using assms
by (simp add: field_simps powr_def)
next
have "1 < (x + 1) * ln ((x + 1) / x)" (is "_ < ?f x")
proof (rule DERIV_neg_imp_decreasing_at_top[where ?f = ?f])
fix t assume t: "x ≤ t"
have "(?f has_field_derivative (ln (1 + 1 / t) - 1 / t)) (at t)"
using t assms by (auto intro!: derivative_eq_intros simp:divide_simps)
moreover have "ln (1 + 1 / t) - 1 / t < 0"
using ln_add_one_self_less_self[of "1 / t"] t assms by auto
ultimately show "∃y. ((λt. (t + 1) * ln ((t + 1) / t)) has_real_derivative y) (at t) ∧ y < 0"
by blast
qed real_asymp
thus "exp 1 < (1 + 1 / x) powr (x + 1)"
using assms by (simp add: powr_def field_simps)
qed
lemma harm_aux_ineq_1:
fixes k :: real
assumes "k > 1"
shows "1 / k < ln (1 + 1 / (k - 1))"
proof -
have "k-1 > 0" ‹k > 0› using assms by simp+
from exp_1_bounds(2)[OF ‹k-1 > 0›]
have "exp 1 < (1 + 1 / (k - 1)) powr k" by simp
then have n_z: "(1 + 1 / (k - 1)) powr k > 0"
using assms not_exp_less_zero by auto
have "(1::real) = ln (exp(1))" using ln_exp by auto
also have "… < ln ((1 + 1 / (k - 1)) powr k)"
using ln_less_cancel_iff[of "exp(1)",simplified,OF ‹(1 + 1 / (k - 1)) powr k > 0›]
exp_1_bounds[OF ‹k - 1 > 0›] by simp
also have "… = k * ln (1 + 1 / (k - 1))"
using ln_powr n_z by simp
finally have "1 < k * ln (1 + 1 / (k - 1))"
by blast
then show ?thesis using assms by (simp add: field_simps)
qed
lemma harm_aux_ineq_2_lemma:
assumes "x ≥ (0::real)"
shows "1 < (x + 1) * ln (1 + 2 / (2 * x + 1))"
proof -
have "0 < ln (1+2/(2*x+1)) - 1 / (x + 1)" (is "_ < ?f x")
proof (rule DERIV_neg_imp_decreasing_at_top[where ?f = ?f])
fix t assume t: "x ≤ t"
from assms t have "3 + 8 * t + 4 * t^2 > 0"
by (intro add_pos_nonneg) auto
hence *: "3 + 8 * t + 4 * t^2 ≠ 0"
by auto
have "(?f has_field_derivative (-1 / ((1 + t)^2 * (3 + 8 * t + 4 * t ^ 2)))) (at t)"
apply (insert assms t *, (rule derivative_eq_intros refl | simp add: add_pos_pos)+)
apply (auto simp: divide_simps)
apply (auto simp: algebra_simps power2_eq_square)
done
moreover have "-1 / ((1 + t)^2 * (3 + 8 * t + 4 * t^2)) < 0"
using t assms by (intro divide_neg_pos mult_pos_pos add_pos_nonneg) auto
ultimately show "∃y. (?f has_real_derivative y) (at t) ∧ y < 0"
by blast
qed real_asymp
thus "1 < (x + 1) * ln (1+2/(2*x+1))"
using assms by (simp add: field_simps)
qed
lemma harm_aux_ineq_2:
fixes k :: real
assumes "k ≥ 1"
shows "1 / (k + 1) < ln (1 + 2 / (2 * k + 1))"
proof -
have "k > 0" using assms by auto
have "1 < (k + 1) * ln (1 + 2 / (2 * k + 1))"
using harm_aux_ineq_2_lemma assms by simp
then show ?thesis
by (simp add: ‹0 < k› add_pos_pos mult.commute mult_imp_div_pos_less)
qed
lemma nat_0_1_induct [case_names 0 1 step]:
assumes "P 0" "P 1" "⋀n. n ≥ 1 ⟹ P n ⟹ P (Suc n)"
shows "P n"
proof (induction n rule: less_induct)
case (less n)
show ?case
using assms(3)[OF _ less.IH[of "n - 1"]]
by (cases "n ≤ 1")
(insert assms(1-2),auto simp: eval_nat_numeral le_Suc_eq)
qed
lemma harm_less_ln:
fixes m :: nat
assumes "m > 0"
shows "harm m < ln (2 * m + 1)"
using assms
proof (induct m rule: nat_0_1_induct)
case 0
then show ?case by blast
next
case 1
have "harm 1 = (1::real)" unfolding harm_def by simp
have "harm 1 < ln (3::real)"
by (subst ‹harm 1 = 1›,subst ln3_gt_1,simp)
then show ?case by simp
next
case (step n)
have "harm (n+1) = harm n + 1/(n+1)"
by ((subst Suc_eq_plus1[symmetric])+,subst harm_Suc,subst inverse_eq_divide,blast)
also have "… < ln (real (2 * n + 1)) + 1/(n+1)"
using step(1-2) by auto
also have "… < ln (real (2 * n + 1)) + ln (1+2/(2*n+1))"
proof -
from step(1) have "real n ≥ 1" by simp
have "1 / real (n + 1) < ln (1 + 2 / real (2 * n + 1))"
using harm_aux_ineq_2[OF ‹1 ≤ (real n)›] by (simp add: add.commute)
then show ?thesis by auto
qed
also have "… = ln ((2 * n + 1) * (1+2/(2*n+1)))"
by (rule ln_mult[symmetric],simp,simp add: field_simps)
also have "… = ln (2*(n+1)+1)"
proof -
have "(2 * n + 1) * (1+2/(2*n+1)) = 2*(n+1)+1"
by (simp add: field_simps)
then show ?thesis by presburger
qed
finally show ?case by simp
qed
text‹Theorem 8.21›
theorem (in primitive_dchar) polya_vinogradov_inequality_primitive:
fixes x :: nat
shows "norm (∑m=1..x. χ m) < sqrt n * ln n"
proof -
define τ :: complex where "τ = gauss_sum 1 div sqrt n"
have τ_mod: "norm τ = 1" using fourier_primitive(2)
by (simp add: τ_def)
{
fix m
have "χ m = (τ div sqrt n) * (∑k = 1..n. (cnj (χ k)) * unity_root n (-m*k))"
using fourier_primitive(1)[of m] τ_def by blast}
note chi_expr = this
have "(∑m = 1..x. χ(m)) = (∑m = 1..x. (τ div sqrt n) * (∑k = 1..n. (cnj (χ k)) * unity_root n (-m*k)))"
by(rule sum.cong[OF refl]) (use chi_expr in blast)
also have "… = (∑m = 1..x. (∑k = 1..n. (τ div sqrt n) * ((cnj (χ k)) * unity_root n (-m*k))))"
by (rule sum.cong,simp,simp add: sum_distrib_left)
also have "… = (∑k = 1..n. (∑m = 1..x. (τ div sqrt n) * ((cnj (χ k)) * unity_root n (-m*k))))"
by (rule sum.swap)
also have "… = (∑k = 1..n. (τ div sqrt n) * (cnj (χ k) * (∑m = 1..x. unity_root n (-m*k))))"
by (rule sum.cong,simp,simp add: sum_distrib_left)
also have "… = (∑k = 1..<n. (τ div sqrt n) * (cnj (χ k) * (∑m = 1..x. unity_root n (-m*k))))"
using n by (intro sum.mono_neutral_right) (auto intro: eq_zero)
also have "… = (τ div sqrt n) * (∑k = 1..<n. (cnj (χ k) * (∑m = 1..x. unity_root n (-m*k))))"
by (simp add: sum_distrib_left)
finally have "(∑m = 1..x. χ(m)) = (τ div sqrt n) * (∑k = 1..<n. (cnj (χ k) * (∑m = 1..x. unity_root n (-m*k))))"
by blast
hence eq: "sqrt n * (∑m=1..x. χ(m)) = τ * (∑k=1..<n. (cnj (χ k) * (∑m=1..x. unity_root n (-m*k))))"
by auto
define f where "f = (λk. (∑m = 1..x. unity_root n (-m*k)))"
hence "(sqrt n) * norm(∑m = 1..x. χ(m)) = norm(τ * (∑k=1..<n. (cnj (χ k) * (∑m = 1..x. unity_root n (-m*k)))))"
proof -
have "norm(sqrt n * (∑m=1..x. χ(m))) = norm (sqrt n) * norm((∑m = 1..x. χ(m)))"
by (simp add: norm_mult)
also have "… = (sqrt n) * norm((∑m = 1..x. χ(m)))"
by simp
finally have 1: "norm((sqrt n) * (∑m = 1..x. χ(m))) = (sqrt n) * norm((∑m = 1..x. χ(m)))"
by blast
then show ?thesis using eq by algebra
qed
also have "… = norm (∑k = 1..<n. (cnj (χ k) * (∑m = 1..x. unity_root n (-m*k))))"
by (simp add: norm_mult τ_mod)
also have "… ≤ (∑k = 1..<n. norm (cnj (χ k) * (∑ m = 1..x. unity_root n (-m*k))))"
using norm_sum by blast
also have "… = (∑k = 1..<n. norm (cnj (χ k)) * norm((∑ m = 1..x. unity_root n (-m*k))))"
by (rule sum.cong,simp, simp add: norm_mult)
also have "… ≤ (∑k = 1..<n. norm((∑m = 1..x. unity_root n (-m*k))))"
proof -
show ?thesis
proof (rule sum_mono)
fix k
assume "k ∈ {1..<n}"
define sum_aux :: real where "sum_aux = norm (∑m=1..x. unity_root n (- int m * int k))"
have "sum_aux ≥ 0" unfolding sum_aux_def by auto
have "norm (cnj (χ k)) ≤ 1" using norm_le_1[of k] by simp
then have "norm (cnj (χ k)) * sum_aux ≤ 1 * sum_aux"
using ‹sum_aux ≥ 0› by (simp add: mult_left_le_one_le)
then show " norm (cnj (χ k)) *
norm (∑m = 1..x. unity_root n (- int m * int k))
≤ norm (∑m = 1..x. unity_root n (- int m * int k))"
unfolding sum_aux_def by argo
qed
qed
also have "… = (∑k = 1..<n. norm(f k))"
using f_def by blast
finally have 24: "(sqrt n) * norm(∑m = 1..x. χ(m)) ≤ (∑k = 1..<n. norm(f k))"
by blast
{
fix k :: int
have "f(n-k) = cnj(f(k))"
proof -
have "f(n-k) = (∑m = 1..x. unity_root n (-m*(n-k)))"
unfolding f_def by blast
also have "… = (∑m = 1..x. unity_root n (m*k))"
proof (rule sum.cong,simp)
fix xa
assume "xa ∈ {1..x}"
have "(k * int xa - int n * int xa) mod int n = (k * int xa - 0) mod int n"
by (intro mod_diff_cong) auto
thus "unity_root n (-int xa * (int n - k)) = unity_root n (int xa * k)"
unfolding ring_distribs by (intro unity_root_cong) (auto simp: cong_def algebra_simps)
qed
also have "… = cnj(f(k))"
proof -
have "cnj(f(k)) = cnj (∑m = 1..x. unity_root n (- int m * k))"
unfolding f_def by blast
also have "cnj (∑m = 1..x. unity_root n (- int m * k)) =
(∑m = 1..x. cnj(unity_root n (- int m * k)))"
by (rule cnj_sum)
also have "… = (∑m = 1..x. unity_root n (int m * k))"
by (intro sum.cong) (auto simp: unity_root_uminus)
finally show ?thesis by auto
qed
finally show "f(n-k) = cnj(f(k))" by blast
qed
hence "norm(f(n-k)) = norm(cnj(f(k)))" by simp
hence "norm(f(n-k)) = norm(f(k))" by auto
}
note eq = this
have 25:
"odd n ⟹ (∑k = 1..n - 1. norm (f (int k))) ≤
2 * (∑k = 1..(n-1) div 2. norm (f (int k)))"
"even n ⟹ (∑k = 1..n - 1. norm (f (int k))) ≤
2 * (∑k = 1..(n-2) div 2. norm (f (int k))) + norm(f(n div 2))"
proof -
assume "odd n"
define g where "g = (λk. norm (f k))"
have "(n-1) div 2 = n div 2" using ‹odd n› n
using div_mult_self1_is_m[OF pos2,of "n-1"]
odd_two_times_div_two_nat[OF ‹odd n›] by linarith
have "(∑i=1..n-1. g i) = (∑i∈{1..n div 2}∪{n div 2<..n-1}. g i)"
using n by (intro sum.cong,auto)
also have "… = (∑i∈{1..n div 2}. g i) + (∑i∈{n div 2<..n-1}. g i)"
by (subst sum.union_disjoint,auto)
also have "(∑i∈{n div 2<..n-1}. g i) = (∑i∈{1..n - (n div 2 + 1)}. g (n - i))"
by (rule sum.reindex_bij_witness[of _ "λi. n - i" "λi. n - i"],auto)
also have "… ≤ (∑i∈{1..n div 2}. g (n - i))"
by (intro sum_mono2,simp,auto simp add: g_def)
finally have 1: "(∑i=1..n-1. g i) ≤ (∑i=1..n div 2. g i + g (n - i))"
by (simp add: sum.distrib)
have "(∑i=1..n div 2. g i + g (n - i)) = (∑i=1..n div 2. 2 * g i)"
unfolding g_def
apply(rule sum.cong,simp)
using eq int_ops(6) by force
also have "… = 2 * (∑i=1..n div 2. g i)"
by (rule sum_distrib_left[symmetric])
finally have 2: "(∑i=1..n div 2. g i + g (n - i)) = 2 * (∑i=1..n div 2. g i)"
by blast
from 1 2 have "(∑i=1..n-1. g i) ≤ 2 * (∑i=1..n div 2. g i)" by algebra
then show "(∑n = 1..n - 1. norm (f (int n))) ≤ 2 * (∑n = 1..(n-1) div 2. norm (f (int n)))"
unfolding g_def ‹(n-1) div 2 = n div 2› by blast
next
assume "even n"
define g where "g = (λn. norm (f (n)))"
have "(n-2) div 2 = n div 2 - 1" using ‹even n› n by simp
have "(∑i=1..n-1. g i) = (∑i∈{1..<n div 2}∪ {n div 2} ∪ {n div 2<..n-1}. g i)"
using n by (intro sum.cong,auto)
also have "… = (∑i∈{1..<n div 2}. g i) + (∑i∈{n div 2<..n-1}. g i) + g(n div 2)"
by (subst sum.union_disjoint,auto)
also have "(∑i∈{n div 2<..n-1}. g i) = (∑i∈{1..n - (n div 2+1)}. g (n - i))"
by (rule sum.reindex_bij_witness[of _ "λi. n - i" "λi. n - i"],auto)
also have "… ≤ (∑i∈{1..<n div 2}. g (n - i))"
proof (intro sum_mono2,simp)
have "n - n div 2 = n div 2" using ‹even n› n by auto
then have "n - (n div 2 + 1) < n div 2"
using n by (simp add: divide_simps)
then show "{1..n - (n div 2 + 1)} ⊆ {1..<n div 2}" by fastforce
qed auto
finally have 1: "(∑i=1..n-1. g i) ≤ (∑i=1..<n div 2. g i + g (n - i)) + g(n div 2)"
by (simp add: sum.distrib)
have "(∑i=1..<n div 2. g i + g (n - i)) = (∑i=1..<n div 2. 2 * g i)"
unfolding g_def
apply(rule sum.cong,simp)
using eq int_ops(6) by force
also have "… = 2 * (∑i=1..<n div 2. g i)"
by (rule sum_distrib_left[symmetric])
finally have 2: "(∑i=1..<n div 2. g i + g (n - i)) = 2 * (∑i=1..<n div 2. g i)"
by blast
from 1 2 have 3: "(∑i=1..n-1. g i) ≤ 2 * (∑i=1..<n div 2. g i) + g(n div 2)" by algebra
then have "(∑i=1..n-1. g i) ≤ 2 * (∑i=1..(n-2) div 2. g i) + g(n div 2)"
proof -
have "{1..<n div 2} = {1..(n-2) div 2}" by auto
then have "(∑i=1..<n div 2. g i) = (∑i=1..(n-2) div 2. g i)"
by (rule sum.cong,simp)
then show ?thesis using 3 by presburger
qed
then show "(∑k = 1..n - 1. norm (f (int k))) ≤ 2 * (∑n = 1..(n-2) div 2. norm (f (int n))) + g(n div 2)"
unfolding g_def by blast
qed
{fix k :: int
assume "1 ≤ k" "k ≤ n div 2"
have "k ≤ n - 1"
using ‹k ≤ n div 2› n by linarith
define y where "y = unity_root n (-k)"
define z where "z = exp (-(pi*k/n)* 𝗂)"
have "z^2 = exp (2*(-(pi*k/n)* 𝗂))"
unfolding z_def using exp_double[symmetric] by blast
also have "… = y"
unfolding y_def unity_root_conv_exp by (simp add: algebra_simps)
finally have z_eq: "y = z^2" by blast
have z_not_0: "z ≠ 0"
using z_eq by (simp add: z_def)
then have "y ≠ 1"
using unity_root_eq_1_iff_int ‹1 ≤ k› ‹k ≤ n - 1› not_less
unity_root_eq_1_iff_int y_def zdvd_not_zless by auto
have "f(k) = (∑m = 1..x . y^m)"
unfolding f_def y_def
by (subst unity_root_pow,rule sum.cong,simp,simp add: algebra_simps)
also have sum: "… = (∑m = 1..<x+1 . y^m)"
by (rule sum.cong,fastforce,simp)
also have "… = (∑m = 0..<x+1 . y^m) - 1"
by (subst (2) sum.atLeast_Suc_lessThan) auto
also have "… = (y^(x+1) - 1) div (y - 1) - 1"
using geometric_sum[OF ‹y ≠ 1›, of "x+1"] by (simp add: atLeast0LessThan)
also have "… = (y^(x+1) - 1 - (y-1)) div (y - 1)"
proof -
have "y - 1 ≠ 0" using ‹y ≠ 1› by simp
show ?thesis
using divide_diff_eq_iff[OF ‹y - 1 ≠ 0›, of "(y^(x+1) - 1)" 1] by auto
qed
also have "… = (y^(x+1) - y) div (y - 1)"
by (simp add: algebra_simps)
also have "… = y * (y^x - 1) div (y - 1)"
by (simp add: algebra_simps)
also have "… = z^2 * ((z^2)^x - 1) div (z^2 - 1)"
unfolding z_eq by blast
also have "… = z^2 * (z^(2*x) - 1) div (z^2 - 1)"
by (subst power_mult[symmetric, of z 2 x],blast)
also have "… = z^(x+1)*((z ^x -inverse(z^x))) / (z - inverse(z))"
proof -
have "z^x ≠ 0" using z_not_0 by auto
have 1: "z ^ (2 * x) - 1 = z^x*(z ^x -inverse(z^x))"
by (simp add: semiring_normalization_rules(36) right_inverse[OF ‹z^x ≠ 0›] right_diff_distrib')
have 2: "z⇧2 - 1 = z*(z - inverse(z))"
by (simp add: right_diff_distrib' semiring_normalization_rules(29) right_inverse[OF ‹z ≠ 0›])
have 3: "z⇧2 * (z^x / z) = z^(x+1)"
proof -
have "z⇧2 * (z^x / z) = z⇧2 * (z^x * inverse z)"
by (simp add: inverse_eq_divide)
also have "… = z^(x+1)"
by (simp add: algebra_simps power2_eq_square right_inverse[OF ‹z ≠ 0›])
finally show ?thesis by blast
qed
have "z⇧2 * (z ^ (2 * x) - 1) / (z⇧2 - 1) =
z⇧2 * (z^x*(z ^x -inverse(z^x))) / (z*(z - inverse(z)))"
by (subst 1, subst 2,blast)
also have "… = (z⇧2 * (z^x / z)) * ((z ^x -inverse(z^x))) / (z - inverse(z))"
by simp
also have "… = z^(x+1) *((z ^x -inverse(z^x))) / (z - inverse(z))"
by (subst 3,simp)
finally show ?thesis by simp
qed
finally have "f(k) = z^(x+1) *((z ^x -inverse(z^x))) / (z - inverse(z))" by blast
then have "norm(f(k)) = norm(z^(x+1) * (((z ^x -inverse(z^x))) / (z - inverse(z))))" by auto
also have "… = norm(z^(x+1)) * norm(((z ^x -inverse(z^x))) / (z - inverse(z)))"
using norm_mult by blast
also have "… = norm(((z ^x -inverse(z^x))) / (z - inverse(z)))"
proof -
have "norm(z) = 1"
unfolding z_def by auto
have "norm(z^(x+1)) = 1"
by (subst norm_power,simp add: ‹norm(z) = 1›)
then show ?thesis by simp
qed
also have "… = norm((exp (-(x*pi*k/n)* 𝗂) - exp ((x*pi*k/n)* 𝗂)) div
(exp (-(pi*k/n)* 𝗂) - exp ((pi*k/n)* 𝗂)))"
proof -
have 1: "z ^ x = exp (-(x*pi*k/n)* 𝗂)"
unfolding z_def
by (subst exp_of_nat_mult[symmetric],simp add: algebra_simps)
have "inverse (z ^ x) = inverse (exp (-(x*pi*k/n)* 𝗂))"
using ‹z ^ x = exp (-(x*pi*k/n)* 𝗂)› by auto
also have "… = (exp ((x*pi*k/n)* 𝗂))"
by (simp add: exp_minus)
finally have 2: "inverse(z^x) = exp ((x*pi*k/n)* 𝗂)" by simp
have 3: "inverse z = exp ((pi*k/n)* 𝗂)"
by (simp add: exp_minus z_def)
show ?thesis using 1 2 3 z_def by simp
qed
also have "… = norm((sin (x*pi*k/n)) div (sin (pi*k/n)))"
proof -
have num: "(exp (-(x*pi*k/n)* 𝗂) - exp ((x*pi*k/n)* 𝗂)) = (-2*𝗂* sin((x*pi*k/n)))"
proof -
have 1: "exp (-(x*pi*k/n)* 𝗂) = cos(-(x*pi*k/n)) + 𝗂 * sin(-(x*pi*k/n))"
"exp ((x*pi*k/n)* 𝗂) = cos((x*pi*k/n)) + 𝗂 * sin((x*pi*k/n))"
using Euler Im_complex_of_real Im_divide_of_nat Im_i_times Re_complex_of_real
complex_Re_of_int complex_i_mult_minus exp_zero mult.assoc mult.commute by force+
have "(exp (-(x*pi*k/n)* 𝗂) - exp ((x*pi*k/n)* 𝗂)) =
(cos(-(x*pi*k/n)) + 𝗂 * sin(-(x*pi*k/n))) -
(cos((x*pi*k/n)) + 𝗂 * sin((x*pi*k/n)))"
using 1 by argo
also have "… = -2*𝗂* sin((x*pi*k/n))" by simp
finally show ?thesis by blast
qed
have den: "(exp (-(pi*k/n)* 𝗂) - exp ((pi*k/n)* 𝗂)) = -2*𝗂* sin((pi*k/n))"
proof -
have 1: "exp (-(pi*k/n)* 𝗂) = cos(-(pi*k/n)) + 𝗂 * sin(-(pi*k/n))"
"exp ((pi*k/n)* 𝗂) = cos((pi*k/n)) + 𝗂 * sin((pi*k/n))"
using Euler Im_complex_of_real Im_divide_of_nat Im_i_times Re_complex_of_real
complex_Re_of_int complex_i_mult_minus exp_zero mult.assoc mult.commute by force+
have "(exp (-(pi*k/n)* 𝗂) - exp ((pi*k/n)* 𝗂)) =
(cos(-(pi*k/n)) + 𝗂 * sin(-(pi*k/n))) -
(cos((pi*k/n)) + 𝗂 * sin((pi*k/n)))"
using 1 by argo
also have "… = -2*𝗂* sin((pi*k/n))" by simp
finally show ?thesis by blast
qed
have "norm((exp (-(x*pi*k/n)* 𝗂) - exp ((x*pi*k/n)* 𝗂)) div
(exp (-(pi*k/n)* 𝗂) - exp ((pi*k/n)* 𝗂))) =
norm((-2*𝗂* sin((x*pi*k/n))) div (-2*𝗂* sin((pi*k/n))))"
using num den by presburger
also have "… = norm(sin((x*pi*k/n)) div sin((pi*k/n)))"
by (simp add: norm_divide)
finally show ?thesis by blast
qed
also have "… = norm((sin (x*pi*k/n))) div norm((sin (pi*k/n)))"
by (simp add: norm_divide)
also have "… ≤ 1 div norm((sin (pi*k/n)))"
proof -
have "norm((sin (pi*k/n))) ≥ 0" by simp
have "norm (sin (x*pi*k/n)) ≤ 1" by simp
then show ?thesis
using divide_right_mono[OF ‹norm (sin (x*pi*k/n)) ≤ 1› ‹norm((sin (pi*k/n))) ≥ 0›]
by blast
qed
finally have 26: "norm(f(k)) ≤ 1 div norm((sin (pi*k/n)))"
by blast
{
fix t
assume "t ≥ 0" "t ≤ pi div 2"
then have "t ∈ {0..pi div 2}" by auto
have "convex_on {0..pi/2} (λx. -sin x)"
by (rule convex_on_realI[where f' = "λx. - cos x"])
(auto intro!: derivative_eq_intros simp: cos_monotone_0_pi_le)
from convex_onD_Icc'[OF this ‹t ∈ {0..pi div 2}›] have "sin(t) ≥ (2 div pi)*t" by simp
}
note sin_ineq = this
have sin_ineq_inst: "sin ((pi*k) / n) ≥ (2 * k) / n"
proof -
have "pi / n ≥ 0" by simp
have 1: "(pi*k) / n ≥ 0" using ‹1 ≤ k› by auto
have "(pi*k)/n = (pi / n) * k" by simp
also have "… ≤ (pi / n) * (n / 2)"
using mult_left_mono[of "k" "n / 2" "pi / n"]
‹k ≤ n div 2› ‹0 ≤ pi / real n› by linarith
also have "… ≤ pi / 2"
by (simp add: divide_simps)
finally have 2: "(pi*k)/n ≤ pi / 2" by auto
have "(2 / pi) * (pi * k / n) ≤ sin((pi * k) / n)"
using sin_ineq[OF 1 2] by blast
then show "sin((pi * k) / n) ≥ (2*k) / n"
by auto
qed
from 26 have "norm(f(k)) ≤ 1 div abs((sin (pi*k/n)))" by simp
also have "… ≤ 1 / abs((2*k) / n)"
proof -
have "sin (pi*k/n) ≥ (2*k) / n" using sin_ineq_inst by simp
moreover have "(2*k) / n > 0" using n ‹1 ≤ k› by auto
ultimately have "abs((sin (pi*k/n))) ≥ abs((2*k)/n)" by auto
have "abs((2*k)/n) > 0" using ‹(2*k)/n > 0› by linarith
then show "1 div abs((sin (pi*k/n))) ≤ 1 / abs(((2*k)/n))"
using ‹abs((2*k)/n) > 0› ‹abs((sin (pi*k/n))) ≥ abs(((2*k)/n))›
by (intro frac_le) auto
qed
also have "… = n / (2*k)" using ‹k ≥ 1› by simp
finally have "norm(f(k)) ≤ n / (2*k)" by blast
}
note ineq = this
have "sqrt n * norm (sum χ {1..x}) < n * ln n"
proof (cases "even n")
case True
have "norm (f(n div 2)) ≤ 1"
proof -
have "int (n div 2) ≥ 1" using n ‹even n› by auto
show ?thesis
using ineq[OF ‹int (n div 2) ≥ 1›] True n by force
qed
from 24 have "sqrt n * norm (sum χ {1..x})
≤ (∑k = 1..<n. norm (f (int k)))" by blast
also have "… = (∑k = 1..n-1. norm (f (int k)))"
by (intro sum.cong) auto
also have "… ≤ 2 * (∑k = 1..(n - 2) div 2. norm (f (int k))) + norm(f(n div 2))"
using 25(2)[OF True] by blast
also have "… ≤ real n * (∑k = 1..(n - 2) div 2. 1 / k) + norm(f(n div 2))"
proof -
have "(∑k = 1..(n - 2) div 2. norm (f (int k))) ≤ (∑k = 1..(n - 2) div 2. real n div (2*k))"
proof (rule sum_mono)
fix k
assume "k ∈ {1..(n - 2) div 2}"
then have "1 ≤ int k" "int k ≤ n div 2" by auto
show "norm (f (int k)) ≤ real n / (2*k)"
using ineq[OF ‹1 ≤ int k› ‹int k ≤ n div 2›] by auto
qed
also have "… = (∑k = 1..(n - 2) div 2. (real n div 2) * (1 / k))"
by (rule sum.cong,auto)
also have "… = (real n div 2) * (∑k = 1..(n - 2) div 2. 1 / k)"
using sum_distrib_left[symmetric] by fast
finally have "(∑k = 1..(n - 2) div 2. norm (f (int k))) ≤
(real n div 2) * (∑k = 1..(n - 2) div 2. 1 / k)"
by blast
then show ?thesis by argo
qed
also have "… = real n * harm ((n - 2) div 2) + norm(f(n div 2))"
unfolding harm_def inverse_eq_divide by simp
also have "… < n * ln n"
proof (cases "n = 2")
case True
have "real n * harm ((n - 2) div 2) + norm (f (int (n div 2))) ≤ 1"
using ‹n = 2› ‹norm (f (int (n div 2))) ≤ 1›
unfolding harm_def by simp
moreover have "real n * ln (real n) ≥ 4 / 3"
using ‹n = 2› ln2_ge_two_thirds by auto
ultimately show ?thesis by argo
next
case False
have "n > 3" using n ‹n ≠ 2› ‹even n› by auto
then have "(n-2) div 2 > 0" by simp
then have "harm ((n - 2) div 2) < ln (real (2 * ((n - 2) div 2) + 1))"
using harm_less_ln by blast
also have "… = ln (real (n - 1))"
using ‹even n› ‹n > 3› by simp
finally have 1: "harm ((n - 2) div 2) < ln (real (n - 1))"
by blast
then have "real n * harm ((n - 2) div 2) < real n * ln (real (n - 1))"
using n by simp
then have "real n * harm ((n - 2) div 2) + norm (f (int (n div 2)))
< real n * ln (real (n - 1)) + 1"
using ‹norm (f (int (n div 2))) ≤ 1› by argo
also have "… = real n * ln (real (n - 1)) + real n * 1 / real n"
using n by auto
also have "… < real n * ln (real (n - 1)) + real n * ln (1 + 1 / (real n - 1))"
proof -
have "real n > 1" "real n > 0" using n by simp+
then have "real n * (1 / real n) < real n * ln (1 + 1 / (real n - 1))"
by (intro mult_strict_left_mono harm_aux_ineq_1) auto
then show ?thesis by auto
qed
also have "… = real n * ( ln (real (n - 1)) + ln (1 + 1 / (real n - 1)))"
by argo
also have "… = real n * ( ln (real (n - 1) * (1 + 1 / (real n - 1))))"
proof -
have "real (n - 1) > 0" "1 + 1 / (real n - 1) > 0"
using n by (auto simp add: add_pos_nonneg)
show ?thesis
by (subst ln_mult [OF ‹real (n - 1) > 0› ‹1 + 1 / (real n - 1) > 0›,symmetric],blast)
qed
also have "… = real n * ln n"
using n by (auto simp add: divide_simps)
finally show ?thesis by blast
qed
finally show ?thesis by blast
next
case False
from 24 have "sqrt n * norm (sum χ {1..x}) ≤ (∑k= 1..<n. norm (f (int k)))"
by blast
also have "… = (∑k= 1..n-1. norm (f (int k)))"
by (intro sum.cong) auto
also have "… ≤ 2 * (∑k = 1..(n - 1) div 2. norm (f (int k)))"
using 25(1)[OF False] by blast
also have "… ≤ real n * (∑k = 1..(n - 1) div 2. 1 / k)"
proof -
have "(∑k = 1..(n - 1) div 2. norm (f (int k))) ≤ (∑k = 1..(n - 1) div 2. real n div (2*k))"
proof (rule sum_mono)
fix k
assume "k ∈ {1..(n - 1) div 2}"
then have "1 ≤ int k" "int k ≤ n div 2" by auto
show "norm (f (int k)) ≤ real n / (2*k)"
using ineq[OF ‹1 ≤ int k› ‹int k ≤ n div 2›] by auto
qed
also have "… = (∑k = 1..(n - 1) div 2. (n / 2) * (1 / k))"
by (rule sum.cong,auto)
also have "… = (n / 2) * (∑k = 1..(n - 1) div 2. 1 / k)"
using sum_distrib_left[symmetric] by fast
finally have "(∑k = 1..(n - 1) div 2. norm (f (int k))) ≤
(real n div 2) * (∑k = 1..(n - 1) div 2. 1 / k)"
by blast
then show ?thesis by argo
qed
also have "… = real n * harm ((n - 1) div 2)"
unfolding harm_def inverse_eq_divide by simp
also have "… < n * ln n"
proof -
have "n > 2" using n ‹odd n› by presburger
then have "(n-1) div 2 > 0" by auto
then have "harm ((n - 1) div 2) < ln (real (2 * ((n - 1) div 2) + 1))"
using harm_less_ln by blast
also have "… = ln (real n)" using ‹odd n› by simp
finally show ?thesis using n by simp
qed
finally show ?thesis by blast
qed
then have 1: "sqrt n * norm (sum χ {1..x}) < n * ln n"
by blast
show "norm (sum χ {1..x}) < sqrt n * ln n"
proof -
have 2: "norm (sum χ {1..x}) * sqrt n < n * ln n"
using 1 by argo
have "sqrt n > 0" using n by simp
have 3: "(n * ln n) / sqrt n = sqrt n * ln n"
using n by (simp add: field_simps)
show "norm (sum χ {1..x}) < sqrt n * ln n"
using mult_imp_less_div_pos[OF ‹sqrt n > 0› 2] 3 by argo
qed
qed
subsection ‹General case›
text ‹
We now first prove the inequality for the general case in terms of the divisor function:
›
theorem (in dcharacter) polya_vinogradov_inequality_explicit:
assumes nonprincipal: "χ ≠ principal_dchar n"
shows "norm (sum χ {1..x}) < sqrt conductor * ln conductor * divisor_count (n div conductor)"
proof -
write primitive_extension ("Φ")
write conductor ("c")
interpret Φ: primitive_dchar c "residue_mult_group c" primitive_extension
using primitive_primitive_extension nonprincipal by metis
have *: "k ≤ x div b ⟷ b * k ≤ x" if "b > 0" for b k
by (metis that antisym_conv div_le_mono div_mult_self1_is_m
less_imp_le not_less times_div_less_eq_dividend)
have **: "a > 0" if "a dvd n" for a
using n that by (auto intro!: Nat.gr0I)
from nonprincipal have "(∑m=1..x. χ m) = (∑m | m ∈ {1..x} ∧ coprime m n. Φ m)"
by (intro sum.mono_neutral_cong_right) (auto simp: eq_zero_iff principal_decomposition)
also have "… = (∑m=1..x. Φ m * (∑d | d dvd gcd m n. moebius_mu d))"
by (subst sum_moebius_mu_divisors', intro sum.mono_neutral_cong_left)
(auto simp: coprime_iff_gcd_eq_1 simp del: coprime_imp_gcd_eq_1)
also have "… = (∑m=1..x. ∑d | d dvd gcd m n. Φ m * moebius_mu d)"
by (simp add: sum_distrib_left)
also have "… = (∑m=1..x. ∑d | d dvd m ∧ d dvd n. Φ m * moebius_mu d)"
by (intro sum.cong) auto
also have "… = (∑(m, d)∈(SIGMA m:{1..x}. {d. d dvd m ∧ d dvd n}). Φ m * moebius_mu d)"
using n by (subst sum.Sigma) auto
also have "… = (∑(d, q)∈(SIGMA d:{d. d dvd n}. {1..x div d}). moebius_mu d * Φ (d * q))"
by (intro sum.reindex_bij_witness[of _ "λ(d,q). (d * q, d)" "λ(m,d). (d, m div d)"])
(auto simp: * ** Suc_le_eq)
also have "… = (∑d | d dvd n. moebius_mu d * Φ d * (∑q=1..x div d. Φ q))"
using n by (subst sum.Sigma [symmetric]) (auto simp: sum_distrib_left mult.assoc)
finally have eq: "(∑m=1..x. χ m) = …" .
have "norm (∑m=1..x. χ m) ≤
(∑d | d dvd n. norm (moebius_mu d * Φ d) * norm (∑q=1..x div d. Φ q))"
unfolding eq by (intro sum_norm_le) (simp add: norm_mult)
also have "… < (∑d | d dvd n. norm (moebius_mu d * Φ d) * (sqrt c * ln c))"
(is "sum ?lhs _ < sum ?rhs _")
proof (rule sum_strict_mono_ex1)
show "∀d∈{d. d dvd n}. ?lhs d ≤ ?rhs d"
by (intro ballI mult_left_mono less_imp_le[OF Φ.polya_vinogradov_inequality_primitive]) auto
show "∃d∈{d. d dvd n}. ?lhs d < ?rhs d"
by (intro bexI[of _ 1] mult_strict_left_mono Φ.polya_vinogradov_inequality_primitive) auto
qed (use n in auto)
also have "… = sqrt c * ln c * (∑d | d dvd n. norm (moebius_mu d * Φ d))"
by (simp add: sum_distrib_left sum_distrib_right mult_ac)
also have "(∑d | d dvd n. norm (moebius_mu d * Φ d)) =
(∑d | d dvd n ∧ squarefree d ∧ coprime d c. 1)"
using n by (intro sum.mono_neutral_cong_right)
(auto simp: moebius_mu_def Φ.eq_zero_iff norm_mult norm_power Φ.norm)
also have "… = card {d. d dvd n ∧ squarefree d ∧ coprime d c}"
by simp
also have "card {d. d dvd n ∧ squarefree d ∧ coprime d c} ≤ card {d. d dvd (n div c)}"
proof (intro card_mono; safe?)
show "finite {d. d dvd (n div c)}"
using dvd_div_eq_0_iff[of c n] n conductor_dvd by (intro finite_divisors_nat) auto
next
fix d assume d: "d dvd n" "squarefree d" "coprime d c"
hence "d > 0" by (intro Nat.gr0I) auto
show "d dvd (n div c)"
proof (rule multiplicity_le_imp_dvd)
fix p :: nat assume p: "prime p"
show "multiplicity p d ≤ multiplicity p (n div c)"
proof (cases "p dvd d")
assume "p dvd d"
with d ‹d > 0› p have "multiplicity p d = 1"
by (auto simp: squarefree_factorial_semiring' in_prime_factors_iff)
moreover have "p dvd (n div c)"
proof -
have "p dvd c * (n div c)"
using ‹p dvd d› ‹d dvd n› conductor_dvd by auto
moreover have "¬(p dvd c)"
using d p ‹p dvd d› coprime_common_divisor not_prime_unit by blast
ultimately show "p dvd (n div c)"
using p prime_dvd_mult_iff by blast
qed
hence "multiplicity p (n div c) ≥ 1"
using n p conductor_dvd dvd_div_eq_0_iff[of c n]
by (intro multiplicity_geI) (auto intro: Nat.gr0I)
ultimately show ?thesis by simp
qed (auto simp: not_dvd_imp_multiplicity_0)
qed (use ‹d > 0› in simp_all)
qed
also have "card {d. d dvd (n div c)} = divisor_count (n div c)"
by (simp add: divisor_count_def)
finally show "norm (sum χ {1..x}) < sqrt c * ln c * divisor_count (n div c)"
using conductor_gr_0 by (simp add: mult_left_mono)
qed
text ‹
Next, we obtain a suitable upper bound on the number of divisors of ‹n›:
›
lemma divisor_count_upper_bound_aux:
fixes n :: nat
shows "divisor_count n ≤ 2 * card {d. d dvd n ∧ d ≤ sqrt n}"
proof (cases "n = 0")
case False
hence n: "n > 0" by simp
have *: "x > 0" if "x dvd n" for x
using that n by (auto intro!: Nat.gr0I)
have **: "real n = sqrt (real n) * sqrt (real n)"
by simp
have ***: "n < x * sqrt n ⟷ sqrt n < x" "x * sqrt n < n ⟷ x < sqrt n" for x
by (metis ** n of_nat_0_less_iff mult_less_iff1 real_sqrt_gt_0_iff)+
have "divisor_count n = card {d. d dvd n}"
by (simp add: divisor_count_def)
also have "{d. d dvd n} = {d. d dvd n ∧ d ≤ sqrt n} ∪ {d. d dvd n ∧ d > sqrt n}"
by auto
also have "card … = card {d. d dvd n ∧ d ≤ sqrt n} + card {d. d dvd n ∧ d > sqrt n}"
using n by (subst card_Un_disjoint) auto
also have "bij_betw (λd. n div d) {d. d dvd n ∧ d > sqrt n} {d. d dvd n ∧ d < sqrt n}"
using n by (intro bij_betwI[of _ _ _ "λd. n div d"])
(auto simp: Real.real_of_nat_div real_sqrt_divide field_simps * ***)
hence "card {d. d dvd n ∧ d > sqrt n} = card {d. d dvd n ∧ d < sqrt n}"
by (rule bij_betw_same_card)
also have "… ≤ card {d. d dvd n ∧ d ≤ sqrt n}"
using n by (intro card_mono) auto
finally show "divisor_count n ≤ 2 * …" by simp
qed auto
lemma divisor_count_upper_bound:
fixes n :: nat
shows "divisor_count n ≤ 2 * nat ⌊sqrt n⌋"
proof (cases "n = 0")
case False
have "divisor_count n ≤ 2 * card {d. d dvd n ∧ d ≤ sqrt n}"
by (rule divisor_count_upper_bound_aux)
also have "card {d. d dvd n ∧ d ≤ sqrt n} ≤ card {1..nat ⌊sqrt n⌋}"
using False by (intro card_mono) (auto simp: le_nat_iff le_floor_iff Suc_le_eq intro!: Nat.gr0I)
also have "… = nat ⌊sqrt n⌋" by simp
finally show ?thesis by simp
qed auto
lemma divisor_count_upper_bound':
fixes n :: nat
shows "real (divisor_count n) ≤ 2 * sqrt n"
proof -
have "real (divisor_count n) ≤ 2 * real (nat ⌊sqrt n⌋)"
using divisor_count_upper_bound[of n] by linarith
also have "… ≤ 2 * sqrt n"
by simp
finally show ?thesis .
qed
text ‹
We are now ready to prove the `regular' Pólya--Vinogradov inequality.
Apostol formulates it in the following way (Theorem 13.15, notation adapted):
`If ‹χ› is any nonprincipal character mod ‹n›, then for all ‹x ≥ 2› we have
$\sum_{m\leq x} \chi(m) = O(\sqrt{n}\log n)$.'
The precondition ‹x ≥ 2› here is completely unnecessary. The `Big-O' notation is somewhat
problematic since it does not make explicit in what way the variables are quantified
(in particular the ‹x› and the ‹χ›). The statement of the theorem in this way (for a fixed
character ‹χ›) seems to suggest that ‹n› is fixed here, which would make the use of `Big-O'
completely vacuous, since it is an asymptotic statement about ‹n›.
We therefore decided to formulate the inequality in the following more explicit way,
even giving an explicit constant factor:
›
theorem (in dcharacter) polya_vinogradov_inequality:
assumes nonprincipal: "χ ≠ principal_dchar n"
shows "norm (∑m=1..x. χ m) < 2 * sqrt n * ln n"
proof -
have "n div conductor > 0"
using n conductor_dvd dvd_div_eq_0_iff[of conductor n] by auto
have "norm (∑m=1..x. χ m) < sqrt conductor * ln conductor * divisor_count (n div conductor)"
using nonprincipal by (rule polya_vinogradov_inequality_explicit)
also have "… ≤ sqrt conductor * ln conductor * (2 * sqrt (n div conductor))"
using conductor_gr_0 ‹n div conductor > 0›
by (intro mult_left_mono divisor_count_upper_bound') (auto simp: Suc_le_eq)
also have "sqrt (n div conductor) = sqrt n / sqrt conductor"
using conductor_dvd by (simp add: Real.real_of_nat_div real_sqrt_divide)
also have "sqrt conductor * ln conductor * (2 * (sqrt n / sqrt conductor)) =
2 * sqrt n * ln conductor"
using conductor_gr_0 n by (simp add: algebra_simps)
also have "… ≤ 2 * sqrt n * ln n"
using conductor_le_modulus conductor_gr_0 by (intro mult_left_mono) auto
finally show ?thesis .
qed
unbundle vec_lambda_notation
end