src/HOL/Taylor.thy

misc tuning and modernization;

(*  Title:      HOL/Taylor.thy
    Author:     Lukas Bulwahn, Bernhard Haeupler, Technische Universitaet Muenchen
*)

section \<open>Taylor series\<close>

theory Taylor
imports MacLaurin
begin

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: "\<forall>m t. m < n \<and> a \<le> t \<and> t \<le> b \<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t)"
    and INTERV: "a \<le> c" "c < b"
  shows "\<exists>t::real. c < t \<and> t < b \<and>
    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" "(\<lambda>m x. diff m (x + c)) 0 = (\<lambda>x. f (x + c))"
    by auto
  moreover
  have "\<forall>m t. m < n \<and> 0 \<le> t \<and> t \<le> b - c \<longrightarrow> DERIV (\<lambda>x. diff m (x + c)) t :> diff (Suc m) (t + c)"
  proof (intro strip)
    fix m t
    assume "m < n \<and> 0 \<le> t \<and> t \<le> 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 (\<lambda>x. x + c) t :> 1 + 0"
      by (rule DERIV_add)
    ultimately have "DERIV (\<lambda>x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1 + 0)"
      by (rule DERIV_chain2)
    then show "DERIV (\<lambda>x. diff m (x + c)) t :> diff (Suc m) (t + c)"
      by simp
  qed
  ultimately obtain x where
    "0 < x \<and> x < b - c \<and>
      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 \<and> x + c < b \<and> 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: "(\<forall>m t. m < n \<and> a \<le> t \<and> t \<le> b \<longrightarrow> DERIV (diff m) t :> diff (Suc m) t)"
    and INTERV: "a < c" "c \<le> b"
  shows "\<exists>t. a < t \<and> t < c \<and>
    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" "(\<lambda>m x. diff m (x + c)) 0 = (\<lambda>x. f (x + c))"
    by auto
  moreover
  have "\<forall>m t. m < n \<and> a - c \<le> t \<and> t \<le> 0 \<longrightarrow> DERIV (\<lambda>x. diff m (x + c)) t :> diff (Suc m) (t + c)"
  proof (rule allI impI)+
    fix m t
    assume "m < n \<and> a - c \<le> t \<and> t \<le> 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 (\<lambda>x. x + c) t :> 1 + 0"
      by (rule DERIV_add)
    ultimately have "DERIV (\<lambda>x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1 + 0)"
      by (rule DERIV_chain2)
    then show "DERIV (\<lambda>x. diff m (x + c)) t :> diff (Suc m) (t + c)"
      by simp
  qed
  ultimately obtain x where
    "a - c < x \<and> x < 0 \<and>
      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 \<and> x + c < c \<and>
    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: "\<forall>m t. m < n \<and> a \<le> t \<and> t \<le> b \<longrightarrow> DERIV (diff m) t :> diff (Suc m) t"
    and INTERV: "a \<le> c " "c \<le> b" "a \<le> x" "x \<le> b" "x \<noteq> c"
  shows "\<exists>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 "\<forall>m t. m < n \<and> x \<le> t \<and> t \<le> b \<longrightarrow> DERIV (diff m) t :> diff (Suc m) t"
    using DERIV and INTERV by fastforce
  moreover note True
  moreover from INTERV have "c \<le> b"
    by simp
  ultimately have "\<exists>t>x. t < c \<and> 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 "\<forall>m t. m < n \<and> a \<le> t \<and> t \<le> x \<longrightarrow> DERIV (diff m) t :> diff (Suc m) t"
    using DERIV and INTERV by fastforce
  moreover from INTERV have "a \<le> c"
    by arith
  moreover from False and INTERV have "c < x"
    by arith
  ultimately have "\<exists>t>c. t < x \<and> 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