(* Author: John Harrison and Valentina Bruno
Ported from "hol_light/Multivariate/complexes.ml" by L C Paulson
*)
section \<open>Polynomial Functions: Extremal Behaviour and Root Counts\<close>
theory Poly_Roots
imports Complex_Main
begin
subsection\<open>Basics about polynomial functions: extremal behaviour and root counts\<close>
lemma sub_polyfun:
fixes x :: "'a::{comm_ring,monoid_mult}"
shows "(\i\n. a i * x^i) - (\i\n. a i * y^i) =
(x - y) * (\<Sum>j<n. \<Sum>k= Suc j..n. a k * y^(k - Suc j) * x^j)"
proof -
have "(\i\n. a i * x^i) - (\i\n. a i * y^i) =
(\<Sum>i\<le>n. a i * (x^i - y^i))"
by (simp add: algebra_simps sum_subtractf [symmetric])
also have "... = (\i\n. a i * (x - y) * (\j
by (simp add: power_diff_sumr2 ac_simps)
also have "... = (x - y) * (\i\n. (\j
by (simp add: sum_distrib_left ac_simps)
also have "... = (x - y) * (\ji=Suc j..n. a i * y^(i - Suc j) * x^j))"
by (simp add: sum.nested_swap')
finally show ?thesis .
qed
lemma sub_polyfun_alt:
fixes x :: "'a::{comm_ring,monoid_mult}"
shows "(\i\n. a i * x^i) - (\i\n. a i * y^i) =
(x - y) * (\<Sum>j<n. \<Sum>k<n-j. a (j+k+1) * y^k * x^j)"
proof -
{ fix j
have "(\k = Suc j..n. a k * y^(k - Suc j) * x^j) =
(\<Sum>k <n - j. a (Suc (j + k)) * y^k * x^j)"
by (rule sum.reindex_bij_witness[where i="\i. i + Suc j" and j="\i. i - Suc j"]) auto }
then show ?thesis
by (simp add: sub_polyfun)
qed
lemma polyfun_linear_factor:
fixes a :: "'a::{comm_ring,monoid_mult}"
shows "\b. \z. (\i\n. c i * z^i) =
(z-a) * (\<Sum>i<n. b i * z^i) + (\<Sum>i\<le>n. c i * a^i)"
proof -
{ fix z
have "(\i\n. c i * z^i) - (\i\n. c i * a^i) =
(z - a) * (\<Sum>j<n. (\<Sum>k = Suc j..n. c k * a^(k - Suc j)) * z^j)"
by (simp add: sub_polyfun sum_distrib_right)
then have "(\i\n. c i * z^i) =
(z - a) * (\<Sum>j<n. (\<Sum>k = Suc j..n. c k * a^(k - Suc j)) * z^j)
+ (\<Sum>i\<le>n. c i * a^i)"
by (simp add: algebra_simps) }
then show ?thesis
by (intro exI allI)
qed
lemma polyfun_linear_factor_root:
fixes a :: "'a::{comm_ring,monoid_mult}"
assumes "(\i\n. c i * a^i) = 0"
shows "\b. \z. (\i\n. c i * z^i) = (z-a) * (\i
using polyfun_linear_factor [of c n a] assms
by simp
lemma adhoc_norm_triangle: "a + norm(y) \ b ==> norm(x) \ a ==> norm(x + y) \ b"
by (metis norm_triangle_mono order.trans order_refl)
proposition polyfun_extremal_lemma:
fixes c :: "nat \ 'a::real_normed_div_algebra"
assumes "e > 0"
shows "\M. \z. M \ norm z \ norm(\i\n. c i * z^i) \ e * norm(z) ^ Suc n"
proof (induction n)
case 0
show ?case
by (rule exI [where x="norm (c 0) / e"]) (auto simp: mult.commute pos_divide_le_eq assms)
next
case (Suc n)
then obtain M where M: "\z. M \ norm z \ norm (\i\n. c i * z^i) \ e * norm z ^ Suc n" ..
show ?case
proof (rule exI [where x="max 1 (max M ((e + norm(c(Suc n))) / e))"], clarify)
fix z::'a
assume "max 1 (max M ((e + norm (c (Suc n))) / e)) \ norm z"
then have norm1: "0 < norm z" "M \ norm z" "(e + norm (c (Suc n))) / e \ norm z"
by auto
then have norm2: "(e + norm (c (Suc n))) \ e * norm z" "(norm z * norm z ^ n) > 0"
apply (metis assms less_divide_eq mult.commute not_le)
using norm1 apply (metis mult_pos_pos zero_less_power)
done
have "e * (norm z * norm z ^ n) + norm (c (Suc n) * (z * z ^ n)) =
(e + norm (c (Suc n))) * (norm z * norm z ^ n)"
by (simp add: norm_mult norm_power algebra_simps)
also have "... \ (e * norm z) * (norm z * norm z ^ n)"
using norm2
using assms mult_mono by fastforce
also have "... = e * (norm z * (norm z * norm z ^ n))"
by (simp add: algebra_simps)
finally have "e * (norm z * norm z ^ n) + norm (c (Suc n) * (z * z ^ n))
\<le> e * (norm z * (norm z * norm z ^ n))" .
then show "norm (\i\Suc n. c i * z^i) \ e * norm z ^ Suc (Suc n)" using M norm1
by (drule_tac x=z in spec) (auto simp: intro!: adhoc_norm_triangle)
qed
qed
lemma norm_lemma_xy: assumes "\b\ + 1 \ norm(y) - a" "norm(x) \ a" shows "b \ norm(x + y)"
proof -
have "b \ norm y - norm x"
using assms by linarith
then show ?thesis
by (metis (no_types) add.commute norm_diff_ineq order_trans)
qed
proposition polyfun_extremal:
fixes c :: "nat \ 'a::real_normed_div_algebra"
assumes "\k. k \ 0 \ k \ n \ c k \ 0"
shows "eventually (\z. norm(\i\n. c i * z^i) \ B) at_infinity"
using assms
proof (induction n)
case 0 then show ?case
by simp
next
case (Suc n)
show ?case
proof (cases "c (Suc n) = 0")
case True
with Suc show ?thesis
by auto (metis diff_is_0_eq diffs0_imp_equal less_Suc_eq_le not_less_eq)
next
case False
with polyfun_extremal_lemma [of "norm(c (Suc n)) / 2" c n]
obtain M where M: "\z. M \ norm z \
norm (\<Sum>i\<le>n. c i * z^i) \<le> norm (c (Suc n)) / 2 * norm z ^ Suc n"
by auto
show ?thesis
unfolding eventually_at_infinity
proof (rule exI [where x="max M (max 1 ((\B\ + 1) / (norm (c (Suc n)) / 2)))"], clarsimp)
fix z::'a
assume les: "M \ norm z" "1 \ norm z" "(\B\ * 2 + 2) / norm (c (Suc n)) \ norm z"
then have "\B\ * 2 + 2 \ norm z * norm (c (Suc n))"
by (metis False pos_divide_le_eq zero_less_norm_iff)
then have "\B\ * 2 + 2 \ norm z ^ (Suc n) * norm (c (Suc n))"
by (metis \<open>1 \<le> norm z\<close> order.trans mult_right_mono norm_ge_zero self_le_power zero_less_Suc)
then show "B \ norm ((\i\n. c i * z^i) + c (Suc n) * (z * z ^ n))" using M les
apply auto
apply (rule norm_lemma_xy [where a = "norm (c (Suc n)) * norm z ^ (Suc n) / 2"])
apply (simp_all add: norm_mult norm_power)
done
qed
qed
qed
proposition polyfun_rootbound:
fixes c :: "nat \ 'a::{comm_ring,real_normed_div_algebra}"
assumes "\k. k \ n \ c k \ 0"
shows "finite {z. (\i\n. c i * z^i) = 0} \ card {z. (\i\n. c i * z^i) = 0} \ n"
using assms
proof (induction n arbitrary: c)
case (Suc n) show ?case
proof (cases "{z. (\i\Suc n. c i * z^i) = 0} = {}")
case False
then obtain a where a: "(\i\Suc n. c i * a^i) = 0"
by auto
from polyfun_linear_factor_root [OF this]
obtain b where "\z. (\i\Suc n. c i * z^i) = (z - a) * (\i< Suc n. b i * z^i)"
by auto
then have b: "\z. (\i\Suc n. c i * z^i) = (z - a) * (\i\n. b i * z^i)"
by (metis lessThan_Suc_atMost)
then have ins_ab: "{z. (\i\Suc n. c i * z^i) = 0} = insert a {z. (\i\n. b i * z^i) = 0}"
by auto
have c0: "c 0 = - (a * b 0)" using b [of 0]
by simp
then have extr_prem: "\ (\k\n. b k \ 0) \ \k. k \ 0 \ k \ Suc n \ c k \ 0"
by (metis Suc.prems le0 minus_zero mult_zero_right)
have "\k\n. b k \ 0"
apply (rule ccontr)
using polyfun_extremal [OF extr_prem, of 1]
apply (auto simp: eventually_at_infinity b simp del: sum.atMost_Suc)
apply (drule_tac x="of_real ba" in spec, simp)
done
then show ?thesis using Suc.IH [of b] ins_ab
by (auto simp: card_insert_if)
qed simp
qed simp
corollary
fixes c :: "nat \ 'a::{comm_ring,real_normed_div_algebra}"
assumes "\k. k \ n \ c k \ 0"
shows polyfun_rootbound_finite: "finite {z. (\i\n. c i * z^i) = 0}"
and polyfun_rootbound_card: "card {z. (\i\n. c i * z^i) = 0} \ n"
using polyfun_rootbound [OF assms] by auto
proposition polyfun_finite_roots:
fixes c :: "nat \ 'a::{comm_ring,real_normed_div_algebra}"
shows "finite {z. (\i\n. c i * z^i) = 0} \ (\k. k \ n \ c k \ 0)"
proof (cases " \k\n. c k \ 0")
case True then show ?thesis
by (blast intro: polyfun_rootbound_finite)
next
case False then show ?thesis
by (auto simp: infinite_UNIV_char_0)
qed
lemma polyfun_eq_0:
fixes c :: "nat \ 'a::{comm_ring,real_normed_div_algebra}"
shows "(\z. (\i\n. c i * z^i) = 0) \ (\k. k \ n \ c k = 0)"
proof (cases "(\z. (\i\n. c i * z^i) = 0)")
case True
then have "\ finite {z. (\i\n. c i * z^i) = 0}"
by (simp add: infinite_UNIV_char_0)
with True show ?thesis
by (metis (poly_guards_query) polyfun_rootbound_finite)
next
case False
then show ?thesis
by auto
qed
theorem polyfun_eq_const:
fixes c :: "nat \ 'a::{comm_ring,real_normed_div_algebra}"
shows "(\z. (\i\n. c i * z^i) = k) \ c 0 = k \ (\k. k \ 0 \ k \ n \ c k = 0)"
proof -
{fix z
have "(\i\n. c i * z^i) = (\i\n. (if i = 0 then c 0 - k else c i) * z^i) + k"
by (induct n) auto
} then
have "(\z. (\i\n. c i * z^i) = k) \ (\z. (\i\n. (if i = 0 then c 0 - k else c i) * z^i) = 0)"
by auto
also have "... \ c 0 = k \ (\k. k \ 0 \ k \ n \ c k = 0)"
by (auto simp: polyfun_eq_0)
finally show ?thesis .
qed
end
¤ Dauer der Verarbeitung: 0.20 Sekunden
(vorverarbeitet)
¤
|
Haftungshinweis
Die Informationen auf dieser Webseite wurden
nach bestem Wissen sorgfältig zusammengestellt. Es wird jedoch weder Vollständigkeit, noch Richtigkeit,
noch Qualität der bereit gestellten Informationen zugesichert.
Bemerkung:
Die farbliche Syntaxdarstellung ist noch experimentell.
|
