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products/Sources/formale Sprachen/Isabelle/HOL/ (Beweissystem Isabelle Version 2025-1^©) Datei vom 16.11.2025 mit Größe 24 kB

Quellcode-Bibliothek MacLaurin.thy Sprache: Isabelle

(*  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 obtain t where t: "0 < t" "t < h" "DERIV (difg m') t :> 0"
      by force
    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" "difg (Suc m) t = 0"
    using \<open>m < n\<close> difg_Suc_eq_0 by force
  show ?thesis
  proof (intro exI conjI)
    show "0 < t" "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 Maclaurin [OF INIT1 this INIT2 DERIV]
  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> show ?thesis by force
  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 show ?thesis by force
next
  assume "x > 0"
  with assms have "\t>0. t < x \ f x = ?f x t"
    by (intro Maclaurin) auto
  then show ?thesis by force
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
    using DERIV INIT Maclaurin_bi_le by auto
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)"
  by (smt (verit) ln_eq_minus_one ln_le_minus_one)

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)"
      using DERIV_shift sin_expansion_lemma by fastforce
  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)"
      using DERIV_shift sin_expansion_lemma by fastforce
  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)"
      using cos_expansion_lemma
      by (intro allI derivative_eq_intros | simp)+
  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
    apply (subst t2)
    apply (rule sin_bound_lemma)
     apply (rule sum.cong[OF refl])
    apply (simp add: diff_m_0 sin_coeff_def)
    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 show ?thesis
    by (smt (verit) sum.cong)
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 show "DERIV (\x. diff m (x + c)) t :> diff (Suc m) (t + c)"
      using DERIV_chain2 DERIV_shift by blast
  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
  ultimately show ?thesis
    using True INTERV Taylor_down 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
  ultimately show ?thesis
    using Taylor_up INTERV False by simp
qed

end

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