Metamath Proof Explorer


Theorem basellem4

Description: Lemma for basel . By basellem3 , the expression P ( ( cot x ) ^ 2 ) = sin ( N x ) / ( sin x ) ^ N goes to zero whenever x = n _pi / N for some n e. ( 1 ... M ) , so this function enumerates M distinct roots of a degree- M polynomial, which must therefore be all the roots by fta1 . (Contributed by Mario Carneiro, 28-Jul-2014)

Ref Expression
Hypotheses basel.n N = 2 M + 1
basel.p P = t j = 0 M ( N 2 j ) 1 M j t j
basel.t T = n 1 M tan n π N 2
Assertion basellem4 M T : 1 M 1-1 onto P -1 0

Proof

Step Hyp Ref Expression
1 basel.n N = 2 M + 1
2 basel.p P = t j = 0 M ( N 2 j ) 1 M j t j
3 basel.t T = n 1 M tan n π N 2
4 1 basellem1 M n 1 M n π N 0 π 2
5 tanrpcl n π N 0 π 2 tan n π N +
6 4 5 syl M n 1 M tan n π N +
7 2z 2
8 znegcl 2 2
9 7 8 ax-mp 2
10 rpexpcl tan n π N + 2 tan n π N 2 +
11 6 9 10 sylancl M n 1 M tan n π N 2 +
12 11 rpcnd M n 1 M tan n π N 2
13 1 2 basellem3 M n π N 0 π 2 P tan n π N 2 = sin N n π N sin n π N N
14 4 13 syldan M n 1 M P tan n π N 2 = sin N n π N sin n π N N
15 elfzelz n 1 M n
16 15 adantl M n 1 M n
17 16 zred M n 1 M n
18 pire π
19 remulcl n π n π
20 17 18 19 sylancl M n 1 M n π
21 20 recnd M n 1 M n π
22 2nn 2
23 nnmulcl 2 M 2 M
24 22 23 mpan M 2 M
25 24 peano2nnd M 2 M + 1
26 1 25 eqeltrid M N
27 26 adantr M n 1 M N
28 27 nncnd M n 1 M N
29 27 nnne0d M n 1 M N 0
30 21 28 29 divcan2d M n 1 M N n π N = n π
31 30 fveq2d M n 1 M sin N n π N = sin n π
32 sinkpi n sin n π = 0
33 16 32 syl M n 1 M sin n π = 0
34 31 33 eqtrd M n 1 M sin N n π N = 0
35 34 oveq1d M n 1 M sin N n π N sin n π N N = 0 sin n π N N
36 20 27 nndivred M n 1 M n π N
37 36 resincld M n 1 M sin n π N
38 37 recnd M n 1 M sin n π N
39 27 nnnn0d M n 1 M N 0
40 38 39 expcld M n 1 M sin n π N N
41 sincosq1sgn n π N 0 π 2 0 < sin n π N 0 < cos n π N
42 4 41 syl M n 1 M 0 < sin n π N 0 < cos n π N
43 42 simpld M n 1 M 0 < sin n π N
44 43 gt0ne0d M n 1 M sin n π N 0
45 27 nnzd M n 1 M N
46 38 44 45 expne0d M n 1 M sin n π N N 0
47 40 46 div0d M n 1 M 0 sin n π N N = 0
48 14 35 47 3eqtrd M n 1 M P tan n π N 2 = 0
49 1 2 basellem2 M P Poly deg P = M coeff P = n 0 ( N 2 n ) 1 M n
50 49 simp1d M P Poly
51 plyf P Poly P :
52 ffn P : P Fn
53 50 51 52 3syl M P Fn
54 53 adantr M n 1 M P Fn
55 fniniseg P Fn tan n π N 2 P -1 0 tan n π N 2 P tan n π N 2 = 0
56 54 55 syl M n 1 M tan n π N 2 P -1 0 tan n π N 2 P tan n π N 2 = 0
57 12 48 56 mpbir2and M n 1 M tan n π N 2 P -1 0
58 57 3 fmptd M T : 1 M P -1 0
59 fveq2 k = m T k = T m
60 fveq2 k = x T k = T x
61 fveq2 k = y T k = T y
62 15 zred n 1 M n
63 62 ssriv 1 M
64 11 rpred M n 1 M tan n π N 2
65 64 3 fmptd M T : 1 M
66 65 ffvelrnda M k 1 M T k
67 simplr M k < m k 1 M m 1 M k < m
68 63 sseli k 1 M k
69 68 ad2antrl M k < m k 1 M m 1 M k
70 63 sseli m 1 M m
71 70 ad2antll M k < m k 1 M m 1 M m
72 18 a1i M k < m k 1 M m 1 M π
73 pipos 0 < π
74 73 a1i M k < m k 1 M m 1 M 0 < π
75 ltmul1 k m π 0 < π k < m k π < m π
76 69 71 72 74 75 syl112anc M k < m k 1 M m 1 M k < m k π < m π
77 67 76 mpbid M k < m k 1 M m 1 M k π < m π
78 remulcl k π k π
79 69 18 78 sylancl M k < m k 1 M m 1 M k π
80 remulcl m π m π
81 71 18 80 sylancl M k < m k 1 M m 1 M m π
82 26 ad2antrr M k < m k 1 M m 1 M N
83 82 nnred M k < m k 1 M m 1 M N
84 82 nngt0d M k < m k 1 M m 1 M 0 < N
85 ltdiv1 k π m π N 0 < N k π < m π k π N < m π N
86 79 81 83 84 85 syl112anc M k < m k 1 M m 1 M k π < m π k π N < m π N
87 77 86 mpbid M k < m k 1 M m 1 M k π N < m π N
88 neghalfpirx π 2 *
89 pirp π +
90 rphalfcl π + π 2 +
91 rpge0 π 2 + 0 π 2
92 89 90 91 mp2b 0 π 2
93 halfpire π 2
94 le0neg2 π 2 0 π 2 π 2 0
95 93 94 ax-mp 0 π 2 π 2 0
96 92 95 mpbi π 2 0
97 iooss1 π 2 * π 2 0 0 π 2 π 2 π 2
98 88 96 97 mp2an 0 π 2 π 2 π 2
99 1 basellem1 M k 1 M k π N 0 π 2
100 99 ad2ant2r M k < m k 1 M m 1 M k π N 0 π 2
101 98 100 sseldi M k < m k 1 M m 1 M k π N π 2 π 2
102 1 basellem1 M m 1 M m π N 0 π 2
103 102 ad2ant2rl M k < m k 1 M m 1 M m π N 0 π 2
104 98 103 sseldi M k < m k 1 M m 1 M m π N π 2 π 2
105 tanord k π N π 2 π 2 m π N π 2 π 2 k π N < m π N tan k π N < tan m π N
106 101 104 105 syl2anc M k < m k 1 M m 1 M k π N < m π N tan k π N < tan m π N
107 87 106 mpbid M k < m k 1 M m 1 M tan k π N < tan m π N
108 tanrpcl k π N 0 π 2 tan k π N +
109 100 108 syl M k < m k 1 M m 1 M tan k π N +
110 tanrpcl m π N 0 π 2 tan m π N +
111 103 110 syl M k < m k 1 M m 1 M tan m π N +
112 rprege0 tan k π N + tan k π N 0 tan k π N
113 rprege0 tan m π N + tan m π N 0 tan m π N
114 lt2sq tan k π N 0 tan k π N tan m π N 0 tan m π N tan k π N < tan m π N tan k π N 2 < tan m π N 2
115 112 113 114 syl2an tan k π N + tan m π N + tan k π N < tan m π N tan k π N 2 < tan m π N 2
116 109 111 115 syl2anc M k < m k 1 M m 1 M tan k π N < tan m π N tan k π N 2 < tan m π N 2
117 107 116 mpbid M k < m k 1 M m 1 M tan k π N 2 < tan m π N 2
118 rpexpcl tan k π N + 2 tan k π N 2 +
119 109 7 118 sylancl M k < m k 1 M m 1 M tan k π N 2 +
120 rpexpcl tan m π N + 2 tan m π N 2 +
121 111 7 120 sylancl M k < m k 1 M m 1 M tan m π N 2 +
122 119 121 ltrecd M k < m k 1 M m 1 M tan k π N 2 < tan m π N 2 1 tan m π N 2 < 1 tan k π N 2
123 117 122 mpbid M k < m k 1 M m 1 M 1 tan m π N 2 < 1 tan k π N 2
124 oveq1 n = m n π = m π
125 124 fvoveq1d n = m tan n π N = tan m π N
126 125 oveq1d n = m tan n π N 2 = tan m π N 2
127 ovex tan m π N 2 V
128 126 3 127 fvmpt m 1 M T m = tan m π N 2
129 128 ad2antll M k < m k 1 M m 1 M T m = tan m π N 2
130 111 rpcnd M k < m k 1 M m 1 M tan m π N
131 2nn0 2 0
132 expneg tan m π N 2 0 tan m π N 2 = 1 tan m π N 2
133 130 131 132 sylancl M k < m k 1 M m 1 M tan m π N 2 = 1 tan m π N 2
134 129 133 eqtrd M k < m k 1 M m 1 M T m = 1 tan m π N 2
135 oveq1 n = k n π = k π
136 135 fvoveq1d n = k tan n π N = tan k π N
137 136 oveq1d n = k tan n π N 2 = tan k π N 2
138 ovex tan k π N 2 V
139 137 3 138 fvmpt k 1 M T k = tan k π N 2
140 139 ad2antrl M k < m k 1 M m 1 M T k = tan k π N 2
141 109 rpcnd M k < m k 1 M m 1 M tan k π N
142 expneg tan k π N 2 0 tan k π N 2 = 1 tan k π N 2
143 141 131 142 sylancl M k < m k 1 M m 1 M tan k π N 2 = 1 tan k π N 2
144 140 143 eqtrd M k < m k 1 M m 1 M T k = 1 tan k π N 2
145 123 134 144 3brtr4d M k < m k 1 M m 1 M T m < T k
146 145 an32s M k 1 M m 1 M k < m T m < T k
147 146 ex M k 1 M m 1 M k < m T m < T k
148 59 60 61 63 66 147 eqord2 M x 1 M y 1 M x = y T x = T y
149 148 biimprd M x 1 M y 1 M T x = T y x = y
150 149 ralrimivva M x 1 M y 1 M T x = T y x = y
151 dff13 T : 1 M 1-1 P -1 0 T : 1 M P -1 0 x 1 M y 1 M T x = T y x = y
152 58 150 151 sylanbrc M T : 1 M 1-1 P -1 0
153 49 simp2d M deg P = M
154 nnne0 M M 0
155 153 154 eqnetrd M deg P 0
156 fveq2 P = 0 𝑝 deg P = deg 0 𝑝
157 dgr0 deg 0 𝑝 = 0
158 156 157 syl6eq P = 0 𝑝 deg P = 0
159 158 necon3i deg P 0 P 0 𝑝
160 155 159 syl M P 0 𝑝
161 eqid P -1 0 = P -1 0
162 161 fta1 P Poly P 0 𝑝 P -1 0 Fin P -1 0 deg P
163 50 160 162 syl2anc M P -1 0 Fin P -1 0 deg P
164 163 simpld M P -1 0 Fin
165 f1domg P -1 0 Fin T : 1 M 1-1 P -1 0 1 M P -1 0
166 164 152 165 sylc M 1 M P -1 0
167 163 simprd M P -1 0 deg P
168 nnnn0 M M 0
169 hashfz1 M 0 1 M = M
170 168 169 syl M 1 M = M
171 153 170 eqtr4d M deg P = 1 M
172 167 171 breqtrd M P -1 0 1 M
173 fzfid M 1 M Fin
174 hashdom P -1 0 Fin 1 M Fin P -1 0 1 M P -1 0 1 M
175 164 173 174 syl2anc M P -1 0 1 M P -1 0 1 M
176 172 175 mpbid M P -1 0 1 M
177 sbth 1 M P -1 0 P -1 0 1 M 1 M P -1 0
178 166 176 177 syl2anc M 1 M P -1 0
179 f1finf1o 1 M P -1 0 P -1 0 Fin T : 1 M 1-1 P -1 0 T : 1 M 1-1 onto P -1 0
180 178 164 179 syl2anc M T : 1 M 1-1 P -1 0 T : 1 M 1-1 onto P -1 0
181 152 180 mpbid M T : 1 M 1-1 onto P -1 0