src/HOL/Taylor.thy
 author paulson Mon May 23 15:33:24 2016 +0100 (2016-05-23) changeset 63114 27afe7af7379 parent 61954 1d43f86f48be child 63569 7e0b0db5e9ac permissions -rw-r--r--
Lots of new material for multivariate analysis
```     1 (*  Title:      HOL/Taylor.thy
```
```     2     Author:     Lukas Bulwahn, Bernhard Haeupler, Technische Universitaet Muenchen
```
```     3 *)
```
```     4
```
```     5 section \<open>Taylor series\<close>
```
```     6
```
```     7 theory Taylor
```
```     8 imports MacLaurin
```
```     9 begin
```
```    10
```
```    11 text \<open>
```
```    12 We use MacLaurin and the translation of the expansion point \<open>c\<close> to \<open>0\<close>
```
```    13 to prove Taylor's theorem.
```
```    14 \<close>
```
```    15
```
```    16 lemma taylor_up:
```
```    17   assumes INIT: "n>0" "diff 0 = f"
```
```    18   and DERIV: "(\<forall> m t. m < n & a \<le> t & t \<le> b \<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t))"
```
```    19   and INTERV: "a \<le> c" "c < b"
```
```    20   shows "\<exists>t::real. c < t & t < b &
```
```    21     f b = (\<Sum>m<n. (diff m c / (fact m)) * (b - c)^m) + (diff n t / (fact n)) * (b - c)^n"
```
```    22 proof -
```
```    23   from INTERV have "0 < b-c" by arith
```
```    24   moreover
```
```    25   from INIT have "n>0" "((\<lambda>m x. diff m (x + c)) 0) = (\<lambda>x. f (x + c))" by auto
```
```    26   moreover
```
```    27   have "ALL m t. m < n & 0 <= t & t <= b - c --> DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)"
```
```    28   proof (intro strip)
```
```    29     fix m t
```
```    30     assume "m < n & 0 <= t & t <= b - c"
```
```    31     with DERIV and INTERV have "DERIV (diff m) (t + c) :> diff (Suc m) (t + c)" by auto
```
```    32     moreover
```
```    33     from DERIV_ident and DERIV_const have "DERIV (%x. x + c) t :> 1+0" by (rule DERIV_add)
```
```    34     ultimately have "DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1+0)"
```
```    35       by (rule DERIV_chain2)
```
```    36     thus "DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)" by simp
```
```    37   qed
```
```    38   ultimately obtain x where
```
```    39         "0 < x & x < b - c &
```
```    40         f (b - c + c) = (\<Sum>m<n. diff m (0 + c) / (fact m) * (b - c) ^ m) +
```
```    41           diff n (x + c) / (fact n) * (b - c) ^ n"
```
```    42      by (rule Maclaurin [THEN exE])
```
```    43   then have "c<x+c & x+c<b \<and> f b = (\<Sum>m<n. diff m c / (fact m) * (b - c) ^ m) +
```
```    44     diff n (x+c) / (fact n) * (b - c) ^ n"
```
```    45     by fastforce
```
```    46   thus ?thesis by fastforce
```
```    47 qed
```
```    48
```
```    49 lemma taylor_down:
```
```    50   fixes a::real
```
```    51   assumes INIT: "n>0" "diff 0 = f"
```
```    52   and DERIV: "(\<forall> m t. m < n & a \<le> t & t \<le> b \<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t))"
```
```    53   and INTERV: "a < c" "c \<le> b"
```
```    54   shows "\<exists> t. a < t & t < c &
```
```    55     f a = (\<Sum>m<n. (diff m c / (fact m)) * (a - c)^m) + (diff n t / (fact n)) * (a - c)^n"
```
```    56 proof -
```
```    57   from INTERV have "a-c < 0" by arith
```
```    58   moreover
```
```    59   from INIT have "n>0" "((\<lambda>m x. diff m (x + c)) 0) = (\<lambda>x. f (x + c))" by auto
```
```    60   moreover
```
```    61   have "ALL m t. m < n & a-c <= t & t <= 0 --> DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)"
```
```    62   proof (rule allI impI)+
```
```    63     fix m t
```
```    64     assume "m < n & a-c <= t & t <= 0"
```
```    65     with DERIV and INTERV have "DERIV (diff m) (t + c) :> diff (Suc m) (t + c)" by auto
```
```    66     moreover
```
```    67     from DERIV_ident and DERIV_const have "DERIV (%x. x + c) t :> 1+0" by (rule DERIV_add)
```
```    68     ultimately have "DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c) * (1+0)" by (rule DERIV_chain2)
```
```    69     thus "DERIV (%x. diff m (x + c)) t :> diff (Suc m) (t + c)" by simp
```
```    70   qed
```
```    71   ultimately obtain x where
```
```    72          "a - c < x & x < 0 &
```
```    73       f (a - c + c) = (\<Sum>m<n. diff m (0 + c) / (fact m) * (a - c) ^ m) +
```
```    74         diff n (x + c) / (fact n) * (a - c) ^ n"
```
```    75      by (rule Maclaurin_minus [THEN exE])
```
```    76   then have "a<x+c & x+c<c \<and> f a = (\<Sum>m<n. diff m c / (fact m) * (a - c) ^ m) +
```
```    77       diff n (x+c) / (fact n) * (a - c) ^ n"
```
```    78     by fastforce
```
```    79   thus ?thesis by fastforce
```
```    80 qed
```
```    81
```
```    82 lemma taylor:
```
```    83   fixes a::real
```
```    84   assumes INIT: "n>0" "diff 0 = f"
```
```    85   and DERIV: "(\<forall> m t. m < n & a \<le> t & t \<le> b \<longrightarrow> DERIV (diff m) t :> (diff (Suc m) t))"
```
```    86   and INTERV: "a \<le> c " "c \<le> b" "a \<le> x" "x \<le> b" "x \<noteq> c"
```
```    87   shows "\<exists> t. (if x<c then (x < t & t < c) else (c < t & t < x)) &
```
```    88     f x = (\<Sum>m<n. (diff m c / (fact m)) * (x - c)^m) + (diff n t / (fact n)) * (x - c)^n"
```
```    89 proof (cases "x<c")
```
```    90   case True
```
```    91   note INIT
```
```    92   moreover from DERIV and INTERV
```
```    93   have "\<forall>m t. m < n \<and> x \<le> t \<and> t \<le> b \<longrightarrow> DERIV (diff m) t :> diff (Suc m) t"
```
```    94     by fastforce
```
```    95   moreover note True
```
```    96   moreover from INTERV have "c \<le> b" by simp
```
```    97   ultimately have "\<exists>t>x. t < c \<and> f x =
```
```    98     (\<Sum>m<n. diff m c / (fact m) * (x - c) ^ m) + diff n t / (fact n) * (x - c) ^ n"
```
```    99     by (rule taylor_down)
```
```   100   with True show ?thesis by simp
```
```   101 next
```
```   102   case False
```
```   103   note INIT
```
```   104   moreover from DERIV and INTERV
```
```   105   have "\<forall>m t. m < n \<and> a \<le> t \<and> t \<le> x \<longrightarrow> DERIV (diff m) t :> diff (Suc m) t"
```
```   106     by fastforce
```
```   107   moreover from INTERV have "a \<le> c" by arith
```
```   108   moreover from False and INTERV have "c < x" by arith
```
```   109   ultimately have "\<exists>t>c. t < x \<and> f x =
```
```   110     (\<Sum>m<n. diff m c / (fact m) * (x - c) ^ m) + diff n t / (fact n) * (x - c) ^ n"
```
```   111     by (rule taylor_up)
```
```   112   with False show ?thesis by simp
```
```   113 qed
```
```   114
```
```   115 end
```