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 (#) exp_R),Z)) . n) . x = r * (exp_R . x)
let n be Element of NAT ; for Z being open Subset of REAL st x in Z holds
((diff ((r (#) exp_R),Z)) . n) . x = r * (exp_R . x)
let Z be open Subset of REAL; ( x in Z implies ((diff ((r (#) exp_R),Z)) . n) . x = r * (exp_R . x) )
assume A1:
x in Z
; ((diff ((r (#) exp_R),Z)) . n) . x = r * (exp_R . x)
A2:
exp_R is_differentiable_on n,Z
by TAYLOR_2:10;
per cases
( n > 0 or n = 0 )
;
suppose
n > 0
;
((diff ((r (#) exp_R),Z)) . n) . x = r * (exp_R . x)then
0 < 0 + n
;
then
1
<= n
by NAT_1:19;
then reconsider i =
n - 1 as
Element of
NAT by INT_1:5;
A3:
(diff (exp_R,Z)) . i is_differentiable_on Z
by A2;
dom ((diff (exp_R,Z)) . n) =
dom ((diff (exp_R,Z)) . (i + 1))
.=
dom (((diff (exp_R,Z)) . i) `| Z)
by TAYLOR_1:def 5
.=
Z
by A3, FDIFF_1:def 7
;
then A4:
x in dom (r (#) ((diff (exp_R,Z)) . n))
by A1, VALUED_1:def 5;
((diff ((r (#) exp_R),Z)) . n) . x =
(r (#) ((diff (exp_R,Z)) . n)) . x
by Th21, TAYLOR_2:10
.=
r * (((diff (exp_R,Z)) . n) . x)
by A4, VALUED_1:def 5
.=
r * (exp_R . x)
by A1, TAYLOR_2:7
;
hence
((diff ((r (#) exp_R),Z)) . n) . x = r * (exp_R . x)
;
verum end; end;