let n be Element of NAT ; for h being Real
for f being Function of REAL ,REAL st f is constant holds
for x being Real holds ((cdif f,h) . (n + 1)) . x = 0
let h be Real; for f being Function of REAL ,REAL st f is constant holds
for x being Real holds ((cdif f,h) . (n + 1)) . x = 0
let f be Function of REAL ,REAL ; ( f is constant implies for x being Real holds ((cdif f,h) . (n + 1)) . x = 0 )
defpred S1[ Element of NAT ] means for x being Real holds ((cdif f,h) . ($1 + 1)) . x = 0 ;
assume A1:
f is constant
; for x being Real holds ((cdif f,h) . (n + 1)) . x = 0
A2:
for x being Real holds (f . (x + (h / 2))) - (f . (x - (h / 2))) = 0
A4:
S1[ 0 ]
A5:
for k being Element of NAT st S1[k] holds
S1[k + 1]
proof
let k be
Element of
NAT ;
( S1[k] implies S1[k + 1] )
assume A6:
for
x being
Real holds
((cdif f,h) . (k + 1)) . x = 0
;
S1[k + 1]
let x be
Real;
((cdif f,h) . ((k + 1) + 1)) . x = 0
A7:
((cdif f,h) . (k + 1)) . (x - (h / 2)) = 0
by A6;
A8:
(cdif f,h) . (k + 1) is
Function of
REAL ,
REAL
by Th19;
((cdif f,h) . (k + 2)) . x =
((cdif f,h) . ((k + 1) + 1)) . x
.=
(cD ((cdif f,h) . (k + 1)),h) . x
by Def8
.=
(((cdif f,h) . (k + 1)) . (x + (h / 2))) - (((cdif f,h) . (k + 1)) . (x - (h / 2)))
by A8, Th5
.=
0
by A6, A7
;
hence
((cdif f,h) . ((k + 1) + 1)) . x = 0
;
verum
end;
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A4, A5);
hence
for x being Real holds ((cdif f,h) . (n + 1)) . x = 0
; verum