src/HOL/Ln.thy
 author huffman Thu Aug 18 19:53:03 2011 -0700 (2011-08-18) changeset 44289 d81d09cdab9c parent 43336 05aa7380f7fc child 44305 3bdc02eb1637 permissions -rw-r--r--
optimize some proofs
1 (*  Title:      HOL/Ln.thy
2     Author:     Jeremy Avigad
3 *)
5 header {* Properties of ln *}
7 theory Ln
8 imports Transcendental
9 begin
11 lemma exp_first_two_terms: "exp x = 1 + x + suminf (%n.
12   inverse(fact (n+2)) * (x ^ (n+2)))"
13 proof -
14   have "exp x = suminf (%n. inverse(fact n) * (x ^ n))"
15     by (simp add: exp_def)
16   also from summable_exp have "... = (SUM n::nat : {0..<2}.
17       inverse(fact n) * (x ^ n)) + suminf (%n.
18       inverse(fact(n+2)) * (x ^ (n+2)))" (is "_ = ?a + _")
19     by (rule suminf_split_initial_segment)
20   also have "?a = 1 + x"
21     by (simp add: numeral_2_eq_2)
22   finally show ?thesis .
23 qed
25 lemma exp_tail_after_first_two_terms_summable:
26   "summable (%n. inverse(fact (n+2)) * (x ^ (n+2)))"
27 proof -
28   note summable_exp
29   thus ?thesis
30     by (frule summable_ignore_initial_segment)
31 qed
33 lemma aux1: assumes a: "0 <= x" and b: "x <= 1"
34     shows "inverse (fact ((n::nat) + 2)) * x ^ (n + 2) <= (x^2/2) * ((1/2)^n)"
35 proof (induct n)
36   show "inverse (fact ((0::nat) + 2)) * x ^ (0 + 2) <=
37       x ^ 2 / 2 * (1 / 2) ^ 0"
38     by (simp add: real_of_nat_Suc power2_eq_square)
39 next
40   fix n :: nat
41   assume c: "inverse (fact (n + 2)) * x ^ (n + 2)
42        <= x ^ 2 / 2 * (1 / 2) ^ n"
43   show "inverse (fact (Suc n + 2)) * x ^ (Suc n + 2)
44            <= x ^ 2 / 2 * (1 / 2) ^ Suc n"
45   proof -
46     have "inverse(fact (Suc n + 2)) <= (1/2) * inverse (fact (n+2))"
47     proof -
48       have "Suc n + 2 = Suc (n + 2)" by simp
49       then have "fact (Suc n + 2) = Suc (n + 2) * fact (n + 2)"
50         by simp
51       then have "real(fact (Suc n + 2)) = real(Suc (n + 2) * fact (n + 2))"
52         apply (rule subst)
53         apply (rule refl)
54         done
55       also have "... = real(Suc (n + 2)) * real(fact (n + 2))"
56         by (rule real_of_nat_mult)
57       finally have "real (fact (Suc n + 2)) =
58          real (Suc (n + 2)) * real (fact (n + 2))" .
59       then have "inverse(fact (Suc n + 2)) =
60          inverse(Suc (n + 2)) * inverse(fact (n + 2))"
61         apply (rule ssubst)
62         apply (rule inverse_mult_distrib)
63         done
64       also have "... <= (1/2) * inverse(fact (n + 2))"
65         apply (rule mult_right_mono)
66         apply (subst inverse_eq_divide)
67         apply simp
68         apply (rule inv_real_of_nat_fact_ge_zero)
69         done
70       finally show ?thesis .
71     qed
72     moreover have "x ^ (Suc n + 2) <= x ^ (n + 2)"
73       by (simp add: mult_left_le_one_le mult_nonneg_nonneg a b)
74     ultimately have "inverse (fact (Suc n + 2)) *  x ^ (Suc n + 2) <=
75         (1 / 2 * inverse (fact (n + 2))) * x ^ (n + 2)"
76       apply (rule mult_mono)
77       apply (rule mult_nonneg_nonneg)
78       apply simp
79       apply (subst inverse_nonnegative_iff_nonnegative)
80       apply (rule real_of_nat_ge_zero)
81       apply (rule zero_le_power)
82       apply (rule a)
83       done
84     also have "... = 1 / 2 * (inverse (fact (n + 2)) * x ^ (n + 2))"
85       by simp
86     also have "... <= 1 / 2 * (x ^ 2 / 2 * (1 / 2) ^ n)"
87       apply (rule mult_left_mono)
88       apply (rule c)
89       apply simp
90       done
91     also have "... = x ^ 2 / 2 * (1 / 2 * (1 / 2) ^ n)"
92       by auto
93     also have "(1::real) / 2 * (1 / 2) ^ n = (1 / 2) ^ (Suc n)"
94       by (rule power_Suc [THEN sym])
95     finally show ?thesis .
96   qed
97 qed
99 lemma aux2: "(%n. (x::real) ^ 2 / 2 * (1 / 2) ^ n) sums x^2"
100 proof -
101   have "(%n. (1 / 2::real)^n) sums (1 / (1 - (1/2)))"
102     apply (rule geometric_sums)
103     by (simp add: abs_less_iff)
104   also have "(1::real) / (1 - 1/2) = 2"
105     by simp
106   finally have "(%n. (1 / 2::real)^n) sums 2" .
107   then have "(%n. x ^ 2 / 2 * (1 / 2) ^ n) sums (x^2 / 2 * 2)"
108     by (rule sums_mult)
109   also have "x^2 / 2 * 2 = x^2"
110     by simp
111   finally show ?thesis .
112 qed
114 lemma exp_bound: "0 <= (x::real) ==> x <= 1 ==> exp x <= 1 + x + x^2"
115 proof -
116   assume a: "0 <= x"
117   assume b: "x <= 1"
118   have c: "exp x = 1 + x + suminf (%n. inverse(fact (n+2)) *
119       (x ^ (n+2)))"
120     by (rule exp_first_two_terms)
121   moreover have "suminf (%n. inverse(fact (n+2)) * (x ^ (n+2))) <= x^2"
122   proof -
123     have "suminf (%n. inverse(fact (n+2)) * (x ^ (n+2))) <=
124         suminf (%n. (x^2/2) * ((1/2)^n))"
125       apply (rule summable_le)
126       apply (auto simp only: aux1 a b)
127       apply (rule exp_tail_after_first_two_terms_summable)
128       by (rule sums_summable, rule aux2)
129     also have "... = x^2"
130       by (rule sums_unique [THEN sym], rule aux2)
131     finally show ?thesis .
132   qed
133   ultimately show ?thesis
134     by auto
135 qed
137 lemma aux4: "0 <= (x::real) ==> x <= 1 ==> exp (x - x^2) <= 1 + x"
138 proof -
139   assume a: "0 <= x" and b: "x <= 1"
140   have "exp (x - x^2) = exp x / exp (x^2)"
141     by (rule exp_diff)
142   also have "... <= (1 + x + x^2) / exp (x ^2)"
143     apply (rule divide_right_mono)
144     apply (rule exp_bound)
145     apply (rule a, rule b)
146     apply simp
147     done
148   also have "... <= (1 + x + x^2) / (1 + x^2)"
149     apply (rule divide_left_mono)
151     apply (rule add_nonneg_nonneg)
152     using a apply auto
153     apply (rule mult_pos_pos)
154     apply auto
155     apply (rule add_pos_nonneg)
156     apply auto
157     done
158   also from a have "... <= 1 + x"
159     by (simp add: field_simps add_strict_increasing zero_le_mult_iff)
160   finally show ?thesis .
161 qed
163 lemma ln_one_plus_pos_lower_bound: "0 <= x ==> x <= 1 ==>
164     x - x^2 <= ln (1 + x)"
165 proof -
166   assume a: "0 <= x" and b: "x <= 1"
167   then have "exp (x - x^2) <= 1 + x"
168     by (rule aux4)
169   also have "... = exp (ln (1 + x))"
170   proof -
171     from a have "0 < 1 + x" by auto
172     thus ?thesis
173       by (auto simp only: exp_ln_iff [THEN sym])
174   qed
175   finally have "exp (x - x ^ 2) <= exp (ln (1 + x))" .
176   thus ?thesis by (auto simp only: exp_le_cancel_iff)
177 qed
179 lemma ln_one_minus_pos_upper_bound: "0 <= x ==> x < 1 ==> ln (1 - x) <= - x"
180 proof -
181   assume a: "0 <= (x::real)" and b: "x < 1"
182   have "(1 - x) * (1 + x + x^2) = (1 - x^3)"
183     by (simp add: algebra_simps power2_eq_square power3_eq_cube)
184   also have "... <= 1"
185     by (auto simp add: a)
186   finally have "(1 - x) * (1 + x + x ^ 2) <= 1" .
187   moreover have "0 < 1 + x + x^2"
188     apply (rule add_pos_nonneg)
189     using a apply auto
190     done
191   ultimately have "1 - x <= 1 / (1 + x + x^2)"
192     by (elim mult_imp_le_div_pos)
193   also have "... <= 1 / exp x"
194     apply (rule divide_left_mono)
195     apply (rule exp_bound, rule a)
196     using a b apply auto
197     apply (rule mult_pos_pos)
198     apply (rule add_pos_nonneg)
199     apply auto
200     done
201   also have "... = exp (-x)"
202     by (auto simp add: exp_minus divide_inverse)
203   finally have "1 - x <= exp (- x)" .
204   also have "1 - x = exp (ln (1 - x))"
205   proof -
206     have "0 < 1 - x"
207       by (insert b, auto)
208     thus ?thesis
209       by (auto simp only: exp_ln_iff [THEN sym])
210   qed
211   finally have "exp (ln (1 - x)) <= exp (- x)" .
212   thus ?thesis by (auto simp only: exp_le_cancel_iff)
213 qed
215 lemma aux5: "x < 1 ==> ln(1 - x) = - ln(1 + x / (1 - x))"
216 proof -
217   assume a: "x < 1"
218   have "ln(1 - x) = - ln(1 / (1 - x))"
219   proof -
220     have "ln(1 - x) = - (- ln (1 - x))"
221       by auto
222     also have "- ln(1 - x) = ln 1 - ln(1 - x)"
223       by simp
224     also have "... = ln(1 / (1 - x))"
225       apply (rule ln_div [THEN sym])
226       by (insert a, auto)
227     finally show ?thesis .
228   qed
229   also have " 1 / (1 - x) = 1 + x / (1 - x)" using a by(simp add:field_simps)
230   finally show ?thesis .
231 qed
233 lemma ln_one_minus_pos_lower_bound: "0 <= x ==> x <= (1 / 2) ==>
234     - x - 2 * x^2 <= ln (1 - x)"
235 proof -
236   assume a: "0 <= x" and b: "x <= (1 / 2)"
237   from b have c: "x < 1"
238     by auto
239   then have "ln (1 - x) = - ln (1 + x / (1 - x))"
240     by (rule aux5)
241   also have "- (x / (1 - x)) <= ..."
242   proof -
243     have "ln (1 + x / (1 - x)) <= x / (1 - x)"
244       apply (rule ln_add_one_self_le_self)
245       apply (rule divide_nonneg_pos)
246       by (insert a c, auto)
247     thus ?thesis
248       by auto
249   qed
250   also have "- (x / (1 - x)) = -x / (1 - x)"
251     by auto
252   finally have d: "- x / (1 - x) <= ln (1 - x)" .
253   have "0 < 1 - x" using a b by simp
254   hence e: "-x - 2 * x^2 <= - x / (1 - x)"
255     using mult_right_le_one_le[of "x*x" "2*x"] a b
256     by (simp add:field_simps power2_eq_square)
257   from e d show "- x - 2 * x^2 <= ln (1 - x)"
258     by (rule order_trans)
259 qed
261 lemma exp_ge_add_one_self [simp]: "1 + (x::real) <= exp x"
262   apply (case_tac "0 <= x")
263   apply (erule exp_ge_add_one_self_aux)
264   apply (case_tac "x <= -1")
265   apply (subgoal_tac "1 + x <= 0")
266   apply (erule order_trans)
267   apply simp
268   apply simp
269   apply (subgoal_tac "1 + x = exp(ln (1 + x))")
270   apply (erule ssubst)
271   apply (subst exp_le_cancel_iff)
272   apply (subgoal_tac "ln (1 - (- x)) <= - (- x)")
273   apply simp
274   apply (rule ln_one_minus_pos_upper_bound)
275   apply auto
276 done
278 lemma ln_add_one_self_le_self2: "-1 < x ==> ln(1 + x) <= x"
279   apply (subgoal_tac "x = ln (exp x)")
280   apply (erule ssubst)back
281   apply (subst ln_le_cancel_iff)
282   apply auto
283 done
285 lemma abs_ln_one_plus_x_minus_x_bound_nonneg:
286     "0 <= x ==> x <= 1 ==> abs(ln (1 + x) - x) <= x^2"
287 proof -
288   assume x: "0 <= x"
289   assume x1: "x <= 1"
290   from x have "ln (1 + x) <= x"
291     by (rule ln_add_one_self_le_self)
292   then have "ln (1 + x) - x <= 0"
293     by simp
294   then have "abs(ln(1 + x) - x) = - (ln(1 + x) - x)"
295     by (rule abs_of_nonpos)
296   also have "... = x - ln (1 + x)"
297     by simp
298   also have "... <= x^2"
299   proof -
300     from x x1 have "x - x^2 <= ln (1 + x)"
301       by (intro ln_one_plus_pos_lower_bound)
302     thus ?thesis
303       by simp
304   qed
305   finally show ?thesis .
306 qed
308 lemma abs_ln_one_plus_x_minus_x_bound_nonpos:
309     "-(1 / 2) <= x ==> x <= 0 ==> abs(ln (1 + x) - x) <= 2 * x^2"
310 proof -
311   assume a: "-(1 / 2) <= x"
312   assume b: "x <= 0"
313   have "abs(ln (1 + x) - x) = x - ln(1 - (-x))"
314     apply (subst abs_of_nonpos)
315     apply simp
316     apply (rule ln_add_one_self_le_self2)
317     using a apply auto
318     done
319   also have "... <= 2 * x^2"
320     apply (subgoal_tac "- (-x) - 2 * (-x)^2 <= ln (1 - (-x))")
321     apply (simp add: algebra_simps)
322     apply (rule ln_one_minus_pos_lower_bound)
323     using a b apply auto
324     done
325   finally show ?thesis .
326 qed
328 lemma abs_ln_one_plus_x_minus_x_bound:
329     "abs x <= 1 / 2 ==> abs(ln (1 + x) - x) <= 2 * x^2"
330   apply (case_tac "0 <= x")
331   apply (rule order_trans)
332   apply (rule abs_ln_one_plus_x_minus_x_bound_nonneg)
333   apply auto
334   apply (rule abs_ln_one_plus_x_minus_x_bound_nonpos)
335   apply auto
336 done
338 lemma ln_x_over_x_mono: "exp 1 <= x ==> x <= y ==> (ln y / y) <= (ln x / x)"
339 proof -
340   assume x: "exp 1 <= x" "x <= y"
341   moreover have "0 < exp (1::real)" by simp
342   ultimately have a: "0 < x" and b: "0 < y"
343     by (fast intro: less_le_trans order_trans)+
344   have "x * ln y - x * ln x = x * (ln y - ln x)"
345     by (simp add: algebra_simps)
346   also have "... = x * ln(y / x)"
347     by (simp only: ln_div a b)
348   also have "y / x = (x + (y - x)) / x"
349     by simp
350   also have "... = 1 + (y - x) / x"
351     using x a by (simp add: field_simps)
352   also have "x * ln(1 + (y - x) / x) <= x * ((y - x) / x)"
353     apply (rule mult_left_mono)
354     apply (rule ln_add_one_self_le_self)
355     apply (rule divide_nonneg_pos)
356     using x a apply simp_all
357     done
358   also have "... = y - x" using a by simp
359   also have "... = (y - x) * ln (exp 1)" by simp
360   also have "... <= (y - x) * ln x"
361     apply (rule mult_left_mono)
362     apply (subst ln_le_cancel_iff)
363     apply fact
364     apply (rule a)
365     apply (rule x)
366     using x apply simp
367     done
368   also have "... = y * ln x - x * ln x"
369     by (rule left_diff_distrib)
370   finally have "x * ln y <= y * ln x"
371     by arith
372   then have "ln y <= (y * ln x) / x" using a by (simp add: field_simps)
373   also have "... = y * (ln x / x)" by simp
374   finally show ?thesis using b by (simp add: field_simps)
375 qed
377 lemma ln_le_minus_one:
378   "0 < x \<Longrightarrow> ln x \<le> x - 1"
379   using exp_ge_add_one_self[of "ln x"] by simp
381 lemma ln_eq_minus_one:
382   assumes "0 < x" "ln x = x - 1" shows "x = 1"
383 proof -
384   let "?l y" = "ln y - y + 1"
385   have D: "\<And>x. 0 < x \<Longrightarrow> DERIV ?l x :> (1 / x - 1)"
386     by (auto intro!: DERIV_intros)
388   show ?thesis
389   proof (cases rule: linorder_cases)
390     assume "x < 1"
391     from dense[OF `x < 1`] obtain a where "x < a" "a < 1" by blast
392     from `x < a` have "?l x < ?l a"
393     proof (rule DERIV_pos_imp_increasing, safe)
394       fix y assume "x \<le> y" "y \<le> a"
395       with `0 < x` `a < 1` have "0 < 1 / y - 1" "0 < y"
396         by (auto simp: field_simps)
397       with D show "\<exists>z. DERIV ?l y :> z \<and> 0 < z"
398         by auto
399     qed
400     also have "\<dots> \<le> 0"
401       using ln_le_minus_one `0 < x` `x < a` by (auto simp: field_simps)
402     finally show "x = 1" using assms by auto
403   next
404     assume "1 < x"
405     from dense[OF `1 < x`] obtain a where "1 < a" "a < x" by blast
406     from `a < x` have "?l x < ?l a"
407     proof (rule DERIV_neg_imp_decreasing, safe)
408       fix y assume "a \<le> y" "y \<le> x"
409       with `1 < a` have "1 / y - 1 < 0" "0 < y"
410         by (auto simp: field_simps)
411       with D show "\<exists>z. DERIV ?l y :> z \<and> z < 0"
412         by blast
413     qed
414     also have "\<dots> \<le> 0"
415       using ln_le_minus_one `1 < a` by (auto simp: field_simps)
416     finally show "x = 1" using assms by auto
417   qed simp
418 qed
420 end