let x, r be Real; for n being Element of NAT
for Z being open Subset of REAL st x in Z holds
((diff (r (#) sin ),Z) . n) . x = r * (sin . (x + ((n * PI ) / 2)))
let n be Element of NAT ; for Z being open Subset of REAL st x in Z holds
((diff (r (#) sin ),Z) . n) . x = r * (sin . (x + ((n * PI ) / 2)))
let Z be open Subset of REAL ; ( x in Z implies ((diff (r (#) sin ),Z) . n) . x = r * (sin . (x + ((n * PI ) / 2))) )
assume A1:
x in Z
; ((diff (r (#) sin ),Z) . n) . x = r * (sin . (x + ((n * PI ) / 2)))
A2:
sin is_differentiable_on n,Z
by TAYLOR_2:21;
per cases
( n > 0 or n = 0 )
;
suppose
n > 0
;
((diff (r (#) sin ),Z) . n) . x = r * (sin . (x + ((n * PI ) / 2)))then
0 < 0 + n
;
then
1
<= n
by NAT_1:19;
then reconsider i =
n - 1 as
Element of
NAT by INT_1:18;
A3:
(diff sin ,Z) . i is_differentiable_on Z
by A2, TAYLOR_1:def 6;
dom ((diff sin ,Z) . n) =
dom ((diff sin ,Z) . (i + 1))
.=
dom (((diff sin ,Z) . i) `| Z)
by TAYLOR_1:def 5
.=
Z
by A3, FDIFF_1:def 8
;
then A4:
x in dom (r (#) ((diff sin ,Z) . n))
by A1, VALUED_1:def 5;
((diff (r (#) sin ),Z) . n) . x =
(r (#) ((diff sin ,Z) . n)) . x
by A2, Th21
.=
r * (((diff sin ,Z) . n) . x)
by A4, VALUED_1:def 5
.=
r * (sin . (x + ((n * PI ) / 2)))
by A1, Th11
;
hence
((diff (r (#) sin ),Z) . n) . x = r * (sin . (x + ((n * PI ) / 2)))
;
verum end; end;