src/HOL/Taylor.thy
 author hoelzl Fri Feb 19 13:40:50 2016 +0100 (2016-02-19) changeset 62378 85ed00c1fe7c parent 61954 1d43f86f48be child 63569 7e0b0db5e9ac permissions -rw-r--r--
generalize more theorems to support enat and ennreal
1 (*  Title:      HOL/Taylor.thy
2     Author:     Lukas Bulwahn, Bernhard Haeupler, Technische Universitaet Muenchen
3 *)
5 section \<open>Taylor series\<close>
7 theory Taylor
8 imports MacLaurin
9 begin
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>
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
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
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
115 end