deffunc H1( Element of NAT ) -> Element of COMPLEX = (tau to_power $1) / (sqrt 5);
let F be Real_Sequence; :: thesis: ( ( for n being Element of NAT holds F . n = (Fib (n + 1)) / (Fib n) ) implies ( F is convergent & lim F = tau ) )
consider f being Real_Sequence such that
A1: for n being Element of NAT holds f . n = Fib n from SEQ_1:sch 1();
 set f2 = f ^\ 2;
set f1 = f ^\ 1;
A2: (f ^\ 1) ^\ 1 = f ^\ (1 + 1) by NAT_1:49
.= f ^\ 2 ;
A3: for n being Element of NAT holds (f ^\ 2) . n <> 0
proof
let n be Element of NAT ; :: thesis: (f ^\ 2) . n <> 0
(f ^\ 2) . n = f . (n + 2) by NAT_1:def 3
.= Fib ((n + 1) + 1) by A1 ;
hence (f ^\ 2) . n <> 0 by Lm3, NAT_1:5; :: thesis: verum
end;
A4: for n being Nat holds ((f ^\ 2) /" (f ^\ 2)) . n = 1
proof
let n be Nat; :: thesis: ((f ^\ 2) /" (f ^\ 2)) . n = 1
A5: n in NAT by ORDINAL1:def 13;
then ((f ^\ 2) /" (f ^\ 2)) . n = ((f ^\ 2) . n) * (((f ^\ 2) . n) ") by Th11
.= ((f ^\ 2) . n) * (1 / ((f ^\ 2) . n))
.= 1 by A3, A5, XCMPLX_1:107 ;
hence ((f ^\ 2) /" (f ^\ 2)) . n = 1 ; :: thesis: verum
end;
then A6: (f ^\ 2) /" (f ^\ 2) is V23() by VALUED_0:def 18;
A7: (f /" f) ^\ 2 = (f ^\ 2) /" (f ^\ 2) by SEQM_3:43;
then A8: f /" f is convergent by A6, SEQ_4:35;
((f ^\ 2) /" (f ^\ 2)) . 0 = 1 by A4;
then lim ((f ^\ 2) /" (f ^\ 2)) = 1 by A6, SEQ_4:40;
then A9: lim (f /" f) = 1 by A6, A7, SEQ_4:36;
ex g being Real_Sequence st
for n being Element of NAT holds g . n = H1(n) from SEQ_1:sch 1();
 then consider g being Real_Sequence such that
A10: for n being Element of NAT holds g . n = H1(n) ;
set g1 = g ^\ 1;
A11: for n being Element of NAT holds g . n <> 0
proof
let n be Element of NAT ; :: thesis: g . n <> 0
A12: (sqrt 5) " <> 0 by SQUARE_1:85, SQUARE_1:95, XCMPLX_1:203;
A13: tau |^ n <> 0 by Lm12, PREPOWER:12;
g . n = (tau to_power n) / (sqrt 5) by A10
.= (tau to_power n) * ((sqrt 5) ")
.= (tau |^ n) * ((sqrt 5) ") by POWER:46 ;
hence g . n <> 0 by A13, A12, XCMPLX_1:6; :: thesis: verum
end;
then A14: g is non-zero by SEQ_1:7;
then A15: (f ^\ 2) /" (f ^\ 1) = ((f ^\ 2) /" (g ^\ 1)) (#) ((g ^\ 1) /" (f ^\ 1)) by Th10;
set g2 = (g ^\ 1) ^\ 1;
for n being Element of NAT holds (f ^\ 1) . n <> 0
proof
let n be Element of NAT ; :: thesis: (f ^\ 1) . n <> 0
(f ^\ 1) . n = f . (n + 1) by NAT_1:def 3
.= Fib (n + 1) by A1 ;
hence (f ^\ 1) . n <> 0 by Lm6; :: thesis: verum
end;
then A16: f ^\ 1 is non-zero by SEQ_1:7;
for n being Element of NAT holds (((g ^\ 1) ^\ 1) /" (f ^\ 2)) . n <> 0
proof
let n be Element of NAT ; :: thesis: (((g ^\ 1) ^\ 1) /" (f ^\ 2)) . n <> 0
A17: ((g ^\ 1) ^\ 1) . n <> 0 by A14, SEQ_1:7;
A18: (((g ^\ 1) ^\ 1) /" (f ^\ 2)) . n = (((g ^\ 1) ^\ 1) . n) * (((f ^\ 2) . n) ") by Th11;
((f ^\ 2) . n) " <> 0 by A16, A2, SEQ_1:7, XCMPLX_1:203;
hence (((g ^\ 1) ^\ 1) /" (f ^\ 2)) . n <> 0 by A17, A18, XCMPLX_1:6; :: thesis: verum
end;
then A19: ((g ^\ 1) ^\ 1) /" (f ^\ 2) is non-zero by SEQ_1:7;
(g ^\ 1) ^\ 1 = g ^\ (1 + 1) by NAT_1:49;
then A20: ((g ^\ 1) ^\ 1) /" (f ^\ 2) = (g /" f) ^\ 2 by SEQM_3:43;
A21: for n being Element of NAT holds (f ^\ 1) . n = Fib (n + 1)
proof
let n be Element of NAT ; :: thesis: (f ^\ 1) . n = Fib (n + 1)
(f ^\ 1) . n = f . (n + 1) by NAT_1:def 3
.= Fib (n + 1) by A1 ;
hence (f ^\ 1) . n = Fib (n + 1) ; :: thesis: verum
end;
assume A22: for n being Element of NAT holds F . n = (Fib (n + 1)) / (Fib n) ; :: thesis: ( F is convergent & lim F = tau )
for n being Element of NAT holds F . n = ((f ^\ 1) /" f) . n
proof
let n be Element of NAT ; :: thesis: F . n = ((f ^\ 1) /" f) . n
((f ^\ 1) /" f) . n = ((f ^\ 1) . n) / (f . n) by Th11
.= (Fib (n + 1)) / (f . n) by A21
.= (Fib (n + 1)) / (Fib n) by A1 ;
hence F . n = ((f ^\ 1) /" f) . n by A22; :: thesis: verum
end;
then F = (f ^\ 1) /" f by FUNCT_2:113;
then A23: (f ^\ 2) /" (f ^\ 1) = F ^\ 1 by A2, SEQM_3:43;
A24: ((g ^\ 1) ^\ 1) /" (g ^\ 1) = ((g ^\ 1) /" g) ^\ 1 by SEQM_3:43;
A25: for n being Nat holds ((g ^\ 1) /" g) . n = tau
proof
let n be Nat; :: thesis: ((g ^\ 1) /" g) . n = tau
A26: n in NAT by ORDINAL1:def 13;
then A27: g . n = (tau to_power n) / (sqrt 5) by A10
.= (tau to_power n) * ((sqrt 5) ")
.= (tau |^ n) * ((sqrt 5) ") by POWER:46 ;
A28: g . n <> 0 by A11, A26;
(g ^\ 1) . n = g . (n + 1) by NAT_1:def 3
.= (tau to_power (n + 1)) / (sqrt 5) by A10
.= (tau to_power (n + 1)) * ((sqrt 5) ")
.= (tau |^ (n + 1)) * ((sqrt 5) ") by POWER:46
.= (tau * (tau |^ n)) * ((sqrt 5) ") by NEWTON:11
.= tau * (g . n) by A27 ;
then ((g ^\ 1) /" g) . n = (tau * (g . n)) * ((g . n) ") by A26, Th11
.= tau * ((g . n) * ((g . n) "))
.= tau * 1 by A28, XCMPLX_0:def 7
.= tau ;
hence ((g ^\ 1) /" g) . n = tau ; :: thesis: verum
end;
tau in REAL by XREAL_0:def 1;
then A29: (g ^\ 1) /" g is V23() by A25, VALUED_0:def 18;
A30: for x being real number st 0 < x holds
ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs (((f ") . m) - 0) < x
proof
let x be real number ; :: thesis: ( 0 < x implies ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs (((f ") . m) - 0) < x )
assume 0 < x ; :: thesis: ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs (((f ") . m) - 0) < x
then consider k being Element of NAT such that
A31: k > 0 and
0 < 1 / k and
A32: 1 / k <= x by Th3;
for m being Element of NAT st k + 2 <= m holds
abs (((f ") . m) - 0) < x
proof
let m be Element of NAT ; :: thesis: ( k + 2 <= m implies abs (((f ") . m) - 0) < x )
k + 2 = (k + 1) + 1 ;
then A33: Fib (k + 2) >= k + 1 by Lm3;
assume k + 2 <= m ; :: thesis: abs (((f ") . m) - 0) < x
then Fib (k + 2) <= Fib m by Lm5;
then k + 1 <= Fib m by A33, XXREAL_0:2;
then A34: k + 1 <= f . m by A1;
then 0 < f . m by NAT_1:5, XXREAL_0:2;
then A35: 0 <= (f . m) " by XREAL_1:124;
k + 0 < k + 1 by XREAL_1:8;
then A36: 1 / (k + 1) < 1 / k by A31, XREAL_1:90;
A37: abs (((f ") . m) - 0) = abs ((f . m) ") by VALUED_1:10
.= (f . m) " by A35, ABSVALUE:def 1
.= 1 / (f . m) ;
1 / (f . m) <= 1 / (k + 1) by A34, NAT_1:5, XREAL_1:87;
then 1 / (f . m) < 1 / k by A36, XXREAL_0:2;
hence abs (((f ") . m) - 0) < x by A32, A37, XXREAL_0:2; :: thesis: verum
end;
hence ex n being Element of NAT st
for m being Element of NAT st n <= m holds
abs (((f ") . m) - 0) < x ; :: thesis: verum
end;
then A38: f " is convergent by SEQ_2:def 6;
then A39: lim (f ") = 0 by A30, SEQ_2:def 7;
deffunc H2( Element of NAT ) -> Element of COMPLEX = (tau_bar to_power $1) / (sqrt 5);
ex h being Real_Sequence st
for n being Element of NAT holds h . n = H2(n) from SEQ_1:sch 1();
 then consider h being Real_Sequence such that
A40: for n being Element of NAT holds h . n = H2(n) ;
A41: for n being Element of NAT holds f . n = (g . n) - (h . n)
proof
let n be Element of NAT ; :: thesis: f . n = (g . n) - (h . n)
f . n = Fib n by A1
.= ((tau to_power n) - (tau_bar to_power n)) / (sqrt 5) by Th7
.= ((tau to_power n) / (sqrt 5)) - ((tau_bar to_power n) / (sqrt 5))
.= (g . n) - ((tau_bar to_power n) / (sqrt 5)) by A10
.= (g . n) - (h . n) by A40 ;
hence f . n = (g . n) - (h . n) ; :: thesis: verum
end;
for n being Element of NAT holds g . n = (f . n) + (h . n)
proof
let n be Element of NAT ; :: thesis: g . n = (f . n) + (h . n)
f . n = (g . n) - (h . n) by A41;
hence g . n = (f . n) + (h . n) ; :: thesis: verum
end;
then g = f + h by SEQ_1:11;
then A42: g /" f = (f /" f) + (h /" f) by SEQ_1:57;
for n being Element of NAT holds abs (h . n) < 1
proof
let n be Element of NAT ; :: thesis: abs (h . n) < 1
h . n = (tau_bar to_power n) / (sqrt 5) by A40;
hence abs (h . n) < 1 by Lm16; :: thesis: verum
end;
then A43: h is bounded by SEQ_2:15;
f " is convergent by A30, SEQ_2:def 6;
then A44: h /" f is convergent by A43, A39, SEQ_2:39;
then A45: g /" f is convergent by A8, A42, SEQ_2:19;
((g ^\ 1) /" g) . 0 = tau by A25;
then lim ((g ^\ 1) /" g) = tau by A29, SEQ_4:40;
then A46: lim (((g ^\ 1) ^\ 1) /" (g ^\ 1)) = tau by A29, A24, SEQ_4:33;
A47: (g ^\ 1) /" (f ^\ 1) = (g /" f) ^\ 1 by SEQM_3:43;
lim (h /" f) = 0 by A43, A38, A39, SEQ_2:40;
then A48: lim (g /" f) = 1 + 0 by A44, A8, A9, A42, SEQ_2:20
.= 1 ;
then A49: lim (((g ^\ 1) ^\ 1) /" (f ^\ 2)) = 1 by A45, A20, SEQ_4:33;
then (((g ^\ 1) ^\ 1) /" (f ^\ 2)) " is convergent by A45, A20, A19, SEQ_2:35;
then A50: (f ^\ 2) /" ((g ^\ 1) ^\ 1) is convergent by SEQ_1:48;
A51: (f ^\ 2) /" (g ^\ 1) = ((f ^\ 2) /" ((g ^\ 1) ^\ 1)) (#) (((g ^\ 1) ^\ 1) /" (g ^\ 1)) by A14, Th10;
then A52: (f ^\ 2) /" (g ^\ 1) is convergent by A29, A50, A24, SEQ_2:28;
then A53: (f ^\ 2) /" (f ^\ 1) is convergent by A45, A47, A15, SEQ_2:28;
hence F is convergent by A23, SEQ_4:35; :: thesis: lim F = tau
lim ((((g ^\ 1) ^\ 1) /" (f ^\ 2)) ") = 1 " by A45, A20, A49, A19, SEQ_2:36
.= 1 ;
then lim ((f ^\ 2) /" ((g ^\ 1) ^\ 1)) = 1 by SEQ_1:48;
then A54: lim ((f ^\ 2) /" (g ^\ 1)) = 1 * tau by A29, A51, A50, A24, A46, SEQ_2:29
.= tau ;
lim ((g ^\ 1) /" (f ^\ 1)) = 1 by A45, A48, A47, SEQ_4:33;
then lim ((f ^\ 2) /" (f ^\ 1)) = tau * 1 by A45, A54, A52, A47, A15, SEQ_2:29;
hence lim F = tau by A53, A23, SEQ_4:36; :: thesis: verum