(* Title: HOL/MacLaurin.thy
Author: Jacques D. Fleuriot, 2001 University of Edinburgh
Author: Lawrence C Paulson, 2004
Author: Lukas Bulwahn and Bernhard Häupler, 2005
*)
section \<open>MacLaurin and Taylor Series\<close>
theory MacLaurin
imports Transcendental
begin
subsection \<open>Maclaurin's Theorem with Lagrange Form of Remainder\<close>
text \<open>This is a very long, messy proof even now that it's been broken down
into lemmas.\<close>
lemma Maclaurin_lemma:
"0 < h \
\<exists>B::real. f h = (\<Sum>m<n. (j m / (fact m)) * (h^m)) + (B * ((h^n) /(fact n)))"
by (rule exI[where x = "(f h - (\m
lemma eq_diff_eq': "x = y - z \ y = x + z"
for x y z :: real
by arith
lemma fact_diff_Suc: "n < Suc m \ fact (Suc m - n) = (Suc m - n) * fact (m - n)"
by (subst fact_reduce) auto
lemma Maclaurin_lemma2:
fixes B
assumes DERIV: "\m t. m < n \ 0\t \ t\h \ DERIV (diff m) t :> diff (Suc m) t"
and INIT: "n = Suc k"
defines "difg \
(\<lambda>m t::real. diff m t -
((\<Sum>p<n - m. diff (m + p) 0 / fact p * t ^ p) + B * (t ^ (n - m) / fact (n - m))))"
(is "difg \ (\m t. diff m t - ?difg m t)")
shows "\m t. m < n \ 0 \ t \ t \ h \ DERIV (difg m) t :> difg (Suc m) t"
proof (rule allI impI)+
fix m t
assume INIT2: "m < n \ 0 \ t \ t \ h"
have "DERIV (difg m) t :> diff (Suc m) t -
((\<Sum>x<n - m. real x * t ^ (x - Suc 0) * diff (m + x) 0 / fact x) +
real (n - m) * t ^ (n - Suc m) * B / fact (n - m))"
by (auto simp: difg_def intro!: derivative_eq_intros DERIV[rule_format, OF INIT2])
moreover
from INIT2 have intvl: "{.. and "0 < n - m"
unfolding atLeast0LessThan[symmetric] by auto
have "(\x
(\<Sum>x<n - Suc m. real (Suc x) * t ^ x * diff (Suc m + x) 0 / fact (Suc x))"
unfolding intvl by (subst sum.insert) (auto simp: sum.reindex)
moreover
have fact_neq_0: "\x. (fact x) + real x * (fact x) \ 0"
by (metis add_pos_pos fact_gt_zero less_add_same_cancel1 less_add_same_cancel2
less_numeral_extra(3) mult_less_0_iff of_nat_less_0_iff)
have "\x. (Suc x) * t ^ x * diff (Suc m + x) 0 / fact (Suc x) = diff (Suc m + x) 0 * t^x / fact x"
by (rule nonzero_divide_eq_eq[THEN iffD2]) auto
moreover
have "(n - m) * t ^ (n - Suc m) * B / fact (n - m) = B * (t ^ (n - Suc m) / fact (n - Suc m))"
using \<open>0 < n - m\<close> by (simp add: field_split_simps fact_reduce)
ultimately show "DERIV (difg m) t :> difg (Suc m) t"
unfolding difg_def by (simp add: mult.commute)
qed
lemma Maclaurin:
assumes h: "0 < h"
and n: "0 < n"
and diff_0: "diff 0 = f"
and diff_Suc: "\m t. m < n \ 0 \ t \ t \ h \ DERIV (diff m) t :> diff (Suc m) t"
shows
"\t::real. 0 < t \ t < h \
f h = sum (\<lambda>m. (diff m 0 / fact m) * h ^ m) {..<n} + (diff n t / fact n) * h ^ n"
proof -
from n obtain m where m: "n = Suc m"
by (cases n) (simp add: n)
from m have "m < n" by simp
obtain B where f_h: "f h = (\m
using Maclaurin_lemma [OF h] ..
define g where [abs_def]: "g t =
f t - (sum (\<lambda>m. (diff m 0 / fact m) * t^m) {..<n} + B * (t^n / fact n))" for t
have g2: "g 0 = 0" "g h = 0"
by (simp_all add: m f_h g_def lessThan_Suc_eq_insert_0 image_iff diff_0 sum.reindex)
define difg where [abs_def]: "difg m t =
diff m t - (sum (\<lambda>p. (diff (m + p) 0 / fact p) * (t ^ p)) {..<n-m} +
B * ((t ^ (n - m)) / fact (n - m)))" for m t
have difg_0: "difg 0 = g"
by (simp add: difg_def g_def diff_0)
have difg_Suc: "\m t. m < n \ 0 \ t \ t \ h \ DERIV (difg m) t :> difg (Suc m) t"
using diff_Suc m unfolding difg_def [abs_def] by (rule Maclaurin_lemma2)
have difg_eq_0: "\m
by (auto simp: difg_def m Suc_diff_le lessThan_Suc_eq_insert_0 image_iff sum.reindex)
have isCont_difg: "\m x. m < n \ 0 \ x \ x \ h \ isCont (difg m) x"
by (rule DERIV_isCont [OF difg_Suc [rule_format]]) simp
have differentiable_difg: "\m x. m < n \ 0 \ x \ x \ h \ difg m differentiable (at x)"
using difg_Suc real_differentiable_def by auto
have difg_Suc_eq_0:
"\m t. m < n \ 0 \ t \ t \ h \ DERIV (difg m) t :> 0 \ difg (Suc m) t = 0"
by (rule DERIV_unique [OF difg_Suc [rule_format]]) simp
have "\t. 0 < t \ t < h \ DERIV (difg m) t :> 0"
using \<open>m < n\<close>
proof (induct m)
case 0
show ?case
proof (rule Rolle)
show "0 < h" by fact
show "difg 0 0 = difg 0 h"
by (simp add: difg_0 g2)
show "continuous_on {0..h} (difg 0)"
by (simp add: continuous_at_imp_continuous_on isCont_difg n)
qed (simp add: differentiable_difg n)
next
case (Suc m')
then have "\t. 0 < t \ t < h \ DERIV (difg m') t :> 0"
by simp
then obtain t where t: "0 < t" "t < h" "DERIV (difg m') t :> 0"
by fast
have "\t'. 0 < t' \ t' < t \ DERIV (difg (Suc m')) t' :> 0"
proof (rule Rolle)
show "0 < t" by fact
show "difg (Suc m') 0 = difg (Suc m') t"
using t \<open>Suc m' < n\<close> by (simp add: difg_Suc_eq_0 difg_eq_0)
have "\x. 0 \ x \ x \ t \ isCont (difg (Suc m')) x"
using \<open>t < h\<close> \<open>Suc m' < n\<close> by (simp add: isCont_difg)
then show "continuous_on {0..t} (difg (Suc m'))"
by (simp add: continuous_at_imp_continuous_on)
qed (use \<open>t < h\<close> \<open>Suc m' < n\<close> in \<open>simp add: differentiable_difg\<close>)
with \<open>t < h\<close> show ?case
by auto
qed
then obtain t where "0 < t" "t < h" "DERIV (difg m) t :> 0"
by fast
with \<open>m < n\<close> have "difg (Suc m) t = 0"
by (simp add: difg_Suc_eq_0)
show ?thesis
proof (intro exI conjI)
show "0 < t" by fact
show "t < h" by fact
show "f h = (\m
using \<open>difg (Suc m) t = 0\<close> by (simp add: m f_h difg_def)
qed
qed
lemma Maclaurin2:
fixes n :: nat
and h :: real
assumes INIT1: "0 < h"
and INIT2: "diff 0 = f"
and DERIV: "\m t. m < n \ 0 \ t \ t \ h \ DERIV (diff m) t :> diff (Suc m) t"
shows "\t. 0 < t \ t \ h \ f h = (\m
proof (cases n)
case 0
with INIT1 INIT2 show ?thesis by fastforce
next
case Suc
then have "n > 0" by simp
from INIT1 this INIT2 DERIV
have "\t>0. t < h \ f h = (\m
by (rule Maclaurin)
then show ?thesis by fastforce
qed
lemma Maclaurin_minus:
fixes n :: nat and h :: real
assumes "h < 0" "0 < n" "diff 0 = f"
and DERIV: "\m t. m < n \ h \ t \ t \ 0 \ DERIV (diff m) t :> diff (Suc m) t"
shows "\t. h < t \ t < 0 \ f h = (\m
proof -
txt \<open>Transform \<open>ABL'\<close> into \<open>derivative_intros\<close> format.\<close>
note DERIV' = DERIV_chain'[OF _ DERIV[rule_format], THEN DERIV_cong]
let ?sum = "\t.
(\<Sum>m<n. (- 1) ^ m * diff m (- 0) / (fact m) * (- h) ^ m) +
(- 1) ^ n * diff n (- t) / (fact n) * (- h) ^ n"
from assms have "\t>0. t < - h \ f (- (- h)) = ?sum t"
by (intro Maclaurin) (auto intro!: derivative_eq_intros DERIV')
then obtain t where "0 < t" "t < - h" "f (- (- h)) = ?sum t"
by blast
moreover have "(- 1) ^ n * diff n (- t) * (- h) ^ n / fact n = diff n (- t) * h ^ n / fact n"
by (auto simp: power_mult_distrib[symmetric])
moreover
have "(\mm
by (auto intro: sum.cong simp add: power_mult_distrib[symmetric])
ultimately have "h < - t \ - t < 0 \
f h = (\<Sum>m<n. diff m 0 / (fact m) * h ^ m) + diff n (- t) / (fact n) * h ^ n"
by auto
then show ?thesis ..
qed
subsection \<open>More Convenient "Bidirectional" Version.\<close>
lemma Maclaurin_bi_le:
fixes n :: nat and x :: real
assumes "diff 0 = f"
and DERIV : "\m t. m < n \ \t\ \ \x\ \ DERIV (diff m) t :> diff (Suc m) t"
shows "\t. \t\ \ \x\ \ f x = (\m
(is "\t. _ \ f x = ?f x t")
proof (cases "n = 0")
case True
with \<open>diff 0 = f\<close> show ?thesis by force
next
case False
show ?thesis
proof (cases rule: linorder_cases)
assume "x = 0"
with \<open>n \<noteq> 0\<close> \<open>diff 0 = f\<close> DERIV have "\<bar>0\<bar> \<le> \<bar>x\<bar> \<and> f x = ?f x 0"
by auto
then show ?thesis ..
next
assume "x < 0"
with \<open>n \<noteq> 0\<close> DERIV have "\<exists>t>x. t < 0 \<and> diff 0 x = ?f x t"
by (intro Maclaurin_minus) auto
then obtain t where "x < t" "t < 0"
"diff 0 x = (\m
by blast
with \<open>x < 0\<close> \<open>diff 0 = f\<close> have "\<bar>t\<bar> \<le> \<bar>x\<bar> \<and> f x = ?f x t"
by simp
then show ?thesis ..
next
assume "x > 0"
with \<open>n \<noteq> 0\<close> \<open>diff 0 = f\<close> DERIV have "\<exists>t>0. t < x \<and> diff 0 x = ?f x t"
by (intro Maclaurin) auto
then obtain t where "0 < t" "t < x"
"diff 0 x = (\m
by blast
with \<open>x > 0\<close> \<open>diff 0 = f\<close> have "\<bar>t\<bar> \<le> \<bar>x\<bar> \<and> f x = ?f x t" by simp
then show ?thesis ..
qed
qed
lemma Maclaurin_all_lt:
fixes x :: real
assumes INIT1: "diff 0 = f"
and INIT2: "0 < n"
and INIT3: "x \ 0"
and DERIV: "\m x. DERIV (diff m) x :> diff(Suc m) x"
shows "\t. 0 < \t\ \ \t\ < \x\ \ f x =
(\<Sum>m<n. (diff m 0 / fact m) * x ^ m) + (diff n t / fact n) * x ^ n"
(is "\t. _ \ _ \ f x = ?f x t")
proof (cases rule: linorder_cases)
assume "x = 0"
with INIT3 show ?thesis ..
next
assume "x < 0"
with assms have "\t>x. t < 0 \ f x = ?f x t"
by (intro Maclaurin_minus) auto
then obtain t where "t > x" "t < 0" "f x = ?f x t"
by blast
with \<open>x < 0\<close> have "0 < \<bar>t\<bar> \<and> \<bar>t\<bar> < \<bar>x\<bar> \<and> f x = ?f x t"
by simp
then show ?thesis ..
next
assume "x > 0"
with assms have "\t>0. t < x \ f x = ?f x t"
by (intro Maclaurin) auto
then obtain t where "t > 0" "t < x" "f x = ?f x t"
by blast
with \<open>x > 0\<close> have "0 < \<bar>t\<bar> \<and> \<bar>t\<bar> < \<bar>x\<bar> \<and> f x = ?f x t"
by simp
then show ?thesis ..
qed
lemma Maclaurin_zero: "x = 0 \ n \ 0 \ (\m
for x :: real and n :: nat
by simp
lemma Maclaurin_all_le:
fixes x :: real and n :: nat
assumes INIT: "diff 0 = f"
and DERIV: "\m x. DERIV (diff m) x :> diff (Suc m) x"
shows "\t. \t\ \ \x\ \ f x = (\m
(is "\t. _ \ f x = ?f x t")
proof (cases "n = 0")
case True
with INIT show ?thesis by force
next
case False
show ?thesis
proof (cases "x = 0")
case True
with \<open>n \<noteq> 0\<close> have "(\<Sum>m<n. diff m 0 / (fact m) * x ^ m) = diff 0 0"
by (intro Maclaurin_zero) auto
with INIT \<open>x = 0\<close> \<open>n \<noteq> 0\<close> have " \<bar>0\<bar> \<le> \<bar>x\<bar> \<and> f x = ?f x 0"
by force
then show ?thesis ..
next
case False
with INIT \<open>n \<noteq> 0\<close> DERIV have "\<exists>t. 0 < \<bar>t\<bar> \<and> \<bar>t\<bar> < \<bar>x\<bar> \<and> f x = ?f x t"
by (intro Maclaurin_all_lt) auto
then obtain t where "0 < \t\ \ \t\ < \x\ \ f x = ?f x t" ..
then have "\t\ \ \x\ \ f x = ?f x t"
by simp
then show ?thesis ..
qed
qed
lemma Maclaurin_all_le_objl:
"diff 0 = f \ (\m x. DERIV (diff m) x :> diff (Suc m) x) \
(\<exists>t::real. \<bar>t\<bar> \<le> \<bar>x\<bar> \<and> f x = (\<Sum>m<n. (diff m 0 / fact m) * x ^ m) + (diff n t / fact n) * x ^ n)"
for x :: real and n :: nat
by (blast intro: Maclaurin_all_le)
subsection \<open>Version for Exponential Function\<close>
lemma Maclaurin_exp_lt:
fixes x :: real and n :: nat
shows
"x \ 0 \ n > 0 \
(\<exists>t. 0 < \<bar>t\<bar> \<and> \<bar>t\<bar> < \<bar>x\<bar> \<and> exp x = (\<Sum>m<n. (x ^ m) / fact m) + (exp t / fact n) * x ^ n)"
using Maclaurin_all_lt [where diff = "\n. exp" and f = exp and x = x and n = n] by auto
lemma Maclaurin_exp_le:
fixes x :: real and n :: nat
shows "\t. \t\ \ \x\ \ exp x = (\m
using Maclaurin_all_le_objl [where diff = "\n. exp" and f = exp and x = x and n = n] by auto
corollary exp_lower_Taylor_quadratic: "0 \ x \ 1 + x + x\<^sup>2 / 2 \ exp x"
for x :: real
using Maclaurin_exp_le [of x 3] by (auto simp: numeral_3_eq_3 power2_eq_square)
corollary ln_2_less_1: "ln 2 < (1::real)"
proof -
have "2 < 5/(2::real)" by simp
also have "5/2 \ exp (1::real)" using exp_lower_Taylor_quadratic[of 1, simplified] by simp
finally have "exp (ln 2) < exp (1::real)" by simp
thus "ln 2 < (1::real)" by (subst (asm) exp_less_cancel_iff) simp
qed
subsection \<open>Version for Sine Function\<close>
lemma mod_exhaust_less_4: "m mod 4 = 0 \ m mod 4 = 1 \ m mod 4 = 2 \ m mod 4 = 3"
for m :: nat
by auto
text \<open>It is unclear why so many variant results are needed.\<close>
lemma sin_expansion_lemma: "sin (x + real (Suc m) * pi / 2) = cos (x + real m * pi / 2)"
by (auto simp: cos_add sin_add add_divide_distrib distrib_right)
lemma Maclaurin_sin_expansion2:
"\t. \t\ \ \x\ \
sin x = (\<Sum>m<n. sin_coeff m * x ^ m) + (sin (t + 1/2 * real n * pi) / fact n) * x ^ n"
proof (cases "n = 0 \ x = 0")
case False
let ?diff = "\n x. sin (x + 1/2 * real n * pi)"
have "\t. 0 < \t\ \ \t\ < \x\ \ sin x =
(\<Sum>m<n. (?diff m 0 / fact m) * x ^ m) + (?diff n t / fact n) * x ^ n"
proof (rule Maclaurin_all_lt)
show "\m x. ((\t. sin (t + 1/2 * real m * pi)) has_real_derivative
sin (x + 1/2 * real (Suc m) * pi)) (at x)"
by (rule allI derivative_eq_intros | use sin_expansion_lemma in force)+
qed (use False in auto)
then show ?thesis
apply (rule ex_forward, simp)
apply (rule sum.cong[OF refl])
apply (auto simp: sin_coeff_def sin_zero_iff elim: oddE simp del: of_nat_Suc)
done
qed auto
lemma Maclaurin_sin_expansion:
"\t. sin x = (\m
using Maclaurin_sin_expansion2 [of x n] by blast
lemma Maclaurin_sin_expansion3:
assumes "n > 0" "x > 0"
shows "\t. 0 < t \ t < x \
sin x = (\<Sum>m<n. sin_coeff m * x ^ m) + (sin (t + 1/2 * real n * pi) / fact n) * x ^ n"
proof -
let ?diff = "\n x. sin (x + 1/2 * real n * pi)"
have "\t. 0 < t \ t < x \ sin x = (\m
proof (rule Maclaurin)
show "\m t. m < n \ 0 \ t \ t \ x \
((\<lambda>u. sin (u + 1/2 * real m * pi)) has_real_derivative
sin (t + 1/2 * real (Suc m) * pi)) (at t)"
apply (simp add: sin_expansion_lemma del: of_nat_Suc)
apply (force intro!: derivative_eq_intros)
done
qed (use assms in auto)
then show ?thesis
apply (rule ex_forward, simp)
apply (rule sum.cong[OF refl])
apply (auto simp: sin_coeff_def sin_zero_iff elim: oddE simp del: of_nat_Suc)
done
qed
lemma Maclaurin_sin_expansion4:
assumes "0 < x"
shows "\t. 0 < t \ t \ x \ sin x = (\m
proof -
let ?diff = "\n x. sin (x + 1/2 * real n * pi)"
have "\t. 0 < t \ t \ x \ sin x = (\m
proof (rule Maclaurin2)
show "\m t. m < n \ 0 \ t \ t \ x \
((\<lambda>u. sin (u + 1/2 * real m * pi)) has_real_derivative
sin (t + 1/2 * real (Suc m) * pi)) (at t)"
apply (simp add: sin_expansion_lemma del: of_nat_Suc)
apply (force intro!: derivative_eq_intros)
done
qed (use assms in auto)
then show ?thesis
apply (rule ex_forward, simp)
apply (rule sum.cong[OF refl])
apply (auto simp: sin_coeff_def sin_zero_iff elim: oddE simp del: of_nat_Suc)
done
qed
subsection \<open>Maclaurin Expansion for Cosine Function\<close>
lemma sumr_cos_zero_one [simp]: "(\m
by (induct n) auto
lemma cos_expansion_lemma: "cos (x + real (Suc m) * pi / 2) = - sin (x + real m * pi / 2)"
by (auto simp: cos_add sin_add distrib_right add_divide_distrib)
lemma Maclaurin_cos_expansion:
"\t::real. \t\ \ \x\ \
cos x = (\<Sum>m<n. cos_coeff m * x ^ m) + (cos(t + 1/2 * real n * pi) / fact n) * x ^ n"
proof (cases "n = 0 \ x = 0")
case False
let ?diff = "\n x. cos (x + 1/2 * real n * pi)"
have "\t. 0 < \t\ \ \t\ < \x\ \ cos x =
(\<Sum>m<n. (?diff m 0 / fact m) * x ^ m) + (?diff n t / fact n) * x ^ n"
proof (rule Maclaurin_all_lt)
show "\m x. ((\t. cos (t + 1/2 * real m * pi)) has_real_derivative
cos (x + 1/2 * real (Suc m) * pi)) (at x)"
apply (rule allI derivative_eq_intros | simp)+
using cos_expansion_lemma by force
qed (use False in auto)
then show ?thesis
apply (rule ex_forward, simp)
apply (rule sum.cong[OF refl])
apply (auto simp: cos_coeff_def cos_zero_iff elim: evenE simp del: of_nat_Suc)
done
qed auto
lemma Maclaurin_cos_expansion2:
assumes "x > 0" "n > 0"
shows "\t. 0 < t \ t < x \
cos x = (\<Sum>m<n. cos_coeff m * x ^ m) + (cos (t + 1/2 * real n * pi) / fact n) * x ^ n"
proof -
let ?diff = "\n x. cos (x + 1/2 * real n * pi)"
have "\t. 0 < t \ t < x \ cos x = (\m
proof (rule Maclaurin)
show "\m t. m < n \ 0 \ t \ t \ x \
((\<lambda>u. cos (u + 1 / 2 * real m * pi)) has_real_derivative
cos (t + 1 / 2 * real (Suc m) * pi)) (at t)"
by (simp add: cos_expansion_lemma del: of_nat_Suc)
qed (use assms in auto)
then show ?thesis
apply (rule ex_forward, simp)
apply (rule sum.cong[OF refl])
apply (auto simp: cos_coeff_def cos_zero_iff elim: evenE)
done
qed
lemma Maclaurin_minus_cos_expansion:
assumes "n > 0" "x < 0"
shows "\t. x < t \ t < 0 \
cos x = (\<Sum>m<n. cos_coeff m * x ^ m) + ((cos (t + 1/2 * real n * pi) / fact n) * x ^ n)"
proof -
let ?diff = "\n x. cos (x + 1/2 * real n * pi)"
have "\t. x < t \ t < 0 \ cos x = (\m
proof (rule Maclaurin_minus)
show "\m t. m < n \ x \ t \ t \ 0 \
((\<lambda>u. cos (u + 1 / 2 * real m * pi)) has_real_derivative
cos (t + 1 / 2 * real (Suc m) * pi)) (at t)"
by (simp add: cos_expansion_lemma del: of_nat_Suc)
qed (use assms in auto)
then show ?thesis
apply (rule ex_forward, simp)
apply (rule sum.cong[OF refl])
apply (auto simp: cos_coeff_def cos_zero_iff elim: evenE)
done
qed
(* Version for ln(1 +/- x). Where is it?? *)
lemma sin_bound_lemma: "x = y \ \u\ \ v \ \(x + u) - y\ \ v"
for x y u v :: real
by auto
lemma Maclaurin_sin_bound: "\sin x - (\m \ inverse (fact n) * \x\ ^ n"
proof -
have est: "x \ 1 \ 0 \ y \ x * y \ 1 * y" for x y :: real
by (rule mult_right_mono) simp_all
let ?diff = "\(n::nat) (x::real).
if n mod 4 = 0 then sin x
else if n mod 4 = 1 then cos x
else if n mod 4 = 2 then - sin x
else - cos x"
have diff_0: "?diff 0 = sin" by simp
have "DERIV (?diff m) x :> ?diff (Suc m) x" for m and x
using mod_exhaust_less_4 [of m]
by (auto simp: mod_Suc intro!: derivative_eq_intros)
then have DERIV_diff: "\m x. DERIV (?diff m) x :> ?diff (Suc m) x"
by blast
from Maclaurin_all_le [OF diff_0 DERIV_diff]
obtain t where t1: "\t\ \ \x\"
and t2: "sin x = (\m
by fast
have diff_m_0: "?diff m 0 = (if even m then 0 else (- 1) ^ ((m - Suc 0) div 2))" for m
using mod_exhaust_less_4 [of m]
by (auto simp: minus_one_power_iff even_even_mod_4_iff [of m] dest: even_mod_4_div_2 odd_mod_4_div_2)
show ?thesis
unfolding sin_coeff_def
apply (subst t2)
apply (rule sin_bound_lemma)
apply (rule sum.cong[OF refl])
apply (subst diff_m_0, simp)
using est
apply (auto intro: mult_right_mono [where b=1, simplified] mult_right_mono
simp: ac_simps divide_inverse power_abs [symmetric] abs_mult)
done
qed
section \<open>Taylor series\<close>
text \<open>
We use MacLaurin and the translation of the expansion point \<open>c\<close> to \<open>0\<close>
to prove Taylor's theorem.
\<close>
lemma Taylor_up:
assumes INIT: "n > 0" "diff 0 = f"
and DERIV: "\m t. m < n \ a \ t \ t \ b \ DERIV (diff m) t :> (diff (Suc m) t)"
and INTERV: "a \ c" "c < b"
shows "\t::real. c < t \ t < b \
f b = (\<Sum>m<n. (diff m c / fact m) * (b - c)^m) + (diff n t / fact n) * (b - c)^n"
proof -
from INTERV have "0 < b - c" by arith
moreover from INIT have "n > 0" "(\m x. diff m (x + c)) 0 = (\x. f (x + c))"
by auto
moreover
have "\m t. m < n \ 0 \ t \ t \ b - c \ DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c)"
proof (intro strip)
fix m t
assume "m < n \ 0 \ t \ t \ b - c"
with DERIV and INTERV have "DERIV (diff m) (t + c) :> diff (Suc m) (t + c)"
by auto
moreover from DERIV_ident and DERIV_const have "DERIV (\x. x + c) t :> 1 + 0"
by (rule DERIV_add)
ultimately have "DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1 + 0)"
by (rule DERIV_chain2)
then show "DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c)"
by simp
qed
ultimately obtain x where
"0 < x \ x < b - c \
f (b - c + c) =
(\<Sum>m<n. diff m (0 + c) / fact m * (b - c) ^ m) + diff n (x + c) / fact n * (b - c) ^ n"
by (rule Maclaurin [THEN exE])
then have "c < x + c \ x + c < b \ f b =
(\<Sum>m<n. diff m c / fact m * (b - c) ^ m) + diff n (x + c) / fact n * (b - c) ^ n"
by fastforce
then show ?thesis by fastforce
qed
lemma Taylor_down:
fixes a :: real and n :: nat
assumes INIT: "n > 0" "diff 0 = f"
and DERIV: "(\m t. m < n \ a \ t \ t \ b \ DERIV (diff m) t :> diff (Suc m) t)"
and INTERV: "a < c" "c \ b"
shows "\t. a < t \ t < c \
f a = (\<Sum>m<n. (diff m c / fact m) * (a - c)^m) + (diff n t / fact n) * (a - c)^n"
proof -
from INTERV have "a-c < 0" by arith
moreover from INIT have "n > 0" "(\m x. diff m (x + c)) 0 = (\x. f (x + c))"
by auto
moreover
have "\m t. m < n \ a - c \ t \ t \ 0 \ DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c)"
proof (rule allI impI)+
fix m t
assume "m < n \ a - c \ t \ t \ 0"
with DERIV and INTERV have "DERIV (diff m) (t + c) :> diff (Suc m) (t + c)"
by auto
moreover from DERIV_ident and DERIV_const have "DERIV (\x. x + c) t :> 1 + 0"
by (rule DERIV_add)
ultimately have "DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1 + 0)"
by (rule DERIV_chain2)
then show "DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c)"
by simp
qed
ultimately obtain x where
"a - c < x \ x < 0 \
f (a - c + c) =
(\<Sum>m<n. diff m (0 + c) / fact m * (a - c) ^ m) + diff n (x + c) / fact n * (a - c) ^ n"
by (rule Maclaurin_minus [THEN exE])
then have "a < x + c \ x + c < c \
f a = (\<Sum>m<n. diff m c / fact m * (a - c) ^ m) + diff n (x + c) / fact n * (a - c) ^ n"
by fastforce
then show ?thesis by fastforce
qed
theorem Taylor:
fixes a :: real and n :: nat
assumes INIT: "n > 0" "diff 0 = f"
and DERIV: "\m t. m < n \ a \ t \ t \ b \ DERIV (diff m) t :> diff (Suc m) t"
and INTERV: "a \ c " "c \ b" "a \ x" "x \ b" "x \ c"
shows "\t.
(if x < c then x < t \<and> t < c else c < t \<and> t < x) \<and>
f x = (\<Sum>m<n. (diff m c / fact m) * (x - c)^m) + (diff n t / fact n) * (x - c)^n"
proof (cases "x < c")
case True
note INIT
moreover have "\m t. m < n \ x \ t \ t \ b \ DERIV (diff m) t :> diff (Suc m) t"
using DERIV and INTERV by fastforce
moreover note True
moreover from INTERV have "c \ b"
by simp
ultimately have "\t>x. t < c \ f x =
(\<Sum>m<n. diff m c / (fact m) * (x - c) ^ m) + diff n t / (fact n) * (x - c) ^ n"
by (rule Taylor_down)
with True show ?thesis by simp
next
case False
note INIT
moreover have "\m t. m < n \ a \ t \ t \ x \ DERIV (diff m) t :> diff (Suc m) t"
using DERIV and INTERV by fastforce
moreover from INTERV have "a \ c"
by arith
moreover from False and INTERV have "c < x"
by arith
ultimately have "\t>c. t < x \ f x =
(\<Sum>m<n. diff m c / (fact m) * (x - c) ^ m) + diff n t / (fact n) * (x - c) ^ n"
by (rule Taylor_up)
with False show ?thesis by simp
qed
end
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