# Theory Taylor

Up to index of Isabelle/HOL

theory Taylor
imports MacLaurin
`(*  Title:      HOL/Taylor.thy    Author:     Lukas Bulwahn, Bernhard Haeupler, Technische Universitaet Muenchen*)header {* Taylor series *}theory Taylorimports MacLaurinbegintext {*We use MacLaurin and the translation of the expansion point @{text c} to @{text 0}to prove Taylor's theorem.*}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. c < t & t < b &     f b = setsum (%m. (diff m c / real (fact m)) * (b - c)^m) {0..<n} +      (diff n t / real (fact n)) * (b - c)^n"proof -  from INTERV have "0 < b-c" by arith  moreover   from INIT have "n>0" "((λm x. diff m (x + c)) 0) = (λx. f (x + c))" by auto  moreover  have "ALL 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)    thus "DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)" by simp  qed  ultimately   have EX:"EX t>0. t < b - c &     f (b - c + c) = (SUM m = 0..<n. diff m (0 + c) / real (fact m) * (b - c) ^ m) +      diff n (t + c) / real (fact n) * (b - c) ^ n"     by (rule Maclaurin)  show ?thesis  proof -    from EX obtain x where       X: "0 < x & x < b - c &         f (b - c + c) = (∑m = 0..<n. diff m (0 + c) / real (fact m) * (b - c) ^ m) +          diff n (x + c) / real (fact n) * (b - c) ^ n" ..    let ?H = "x + c"    from X have "c<?H & ?H<b ∧ f b = (∑m = 0..<n. diff m c / real (fact m) * (b - c) ^ m) +      diff n ?H / real (fact n) * (b - c) ^ n"      by fastforce    thus ?thesis by fastforce  qedqedlemma taylor_down:  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 = setsum (% m. (diff m c / real (fact m)) * (a - c)^m) {0..<n} +      (diff n t / real (fact n)) * (a - c)^n" proof -  from INTERV have "a-c < 0" by arith  moreover   from INIT have "n>0" "((λm x. diff m (x + c)) 0) = (λx. f (x + c))" by auto  moreover  have "ALL 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)    thus "DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)" by simp  qed  ultimately   have EX: "EX t>a - c. t < 0 &    f (a - c + c) = (SUM m = 0..<n. diff m (0 + c) / real (fact m) * (a - c) ^ m) +      diff n (t + c) / real (fact n) * (a - c) ^ n"     by (rule Maclaurin_minus)  show ?thesis  proof -    from EX obtain x where X: "a - c < x & x < 0 &      f (a - c + c) = (SUM m = 0..<n. diff m (0 + c) / real (fact m) * (a - c) ^ m) +        diff n (x + c) / real (fact n) * (a - c) ^ n" ..    let ?H = "x + c"    from X have "a<?H & ?H<c ∧ f a = (∑m = 0..<n. diff m c / real (fact m) * (a - c) ^ m) +      diff n ?H / real (fact n) * (a - c) ^ n"      by fastforce    thus ?thesis by fastforce  qedqedlemma taylor:  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 & t < c) else (c < t & t < x)) &    f x = setsum (% m. (diff m c / real (fact m)) * (x - c)^m) {0..<n} +      (diff n t / real (fact n)) * (x - c)^n" proof (cases "x<c")  case True  note INIT  moreover from DERIV and INTERV  have "∀m t. m < n ∧ x ≤ t ∧ t ≤ b --> DERIV (diff m) t :> diff (Suc m) t"    by fastforce  moreover note True  moreover from INTERV have "c ≤ b" by simp  ultimately have EX: "∃t>x. t < c ∧ f x =    (∑m = 0..<n. diff m c / real (fact m) * (x - c) ^ m) +      diff n t / real (fact n) * (x - c) ^ n"    by (rule taylor_down)  with True show ?thesis by simpnext  case False  note INIT  moreover from DERIV and INTERV  have "∀m t. m < n ∧ a ≤ t ∧ t ≤ x --> DERIV (diff m) t :> diff (Suc m) t"    by fastforce  moreover from INTERV have "a ≤ c" by arith  moreover from False and INTERV have "c < x" by arith  ultimately have EX: "∃t>c. t < x ∧ f x =    (∑m = 0..<n. diff m c / real (fact m) * (x - c) ^ m) +      diff n t / real (fact n) * (x - c) ^ n"     by (rule taylor_up)  with False show ?thesis by simpqedend`