05/27/2014, 12:23 PM
Well we get exp(x) replaced by exp(f(x)) and ln(x) replaced by f^[-1](ln(x)).
Hence we arrive at f ( f^[-1](a) + f^[-1](b) ) = a + b
( with f at least C^3 )
now differentiate with respect to a :
f ' (f^[-1](a) + f^[-1](b)) / f ' (f^[-1](a)) = 1
substitute X = f(a) , Y = f^[-1](b) :
f ' ( a + Y) = f ' (a)^2
Now rewrite f ' = G and differentiate with respect to Y :
G ' (Y + a) = 0
Hence G is constant and thus f = lineair.
Since f(0) = 0 we conclude f(x) = C x.
thus it only holds for r = 0,1
QED
regards
tommy1729
Hence we arrive at f ( f^[-1](a) + f^[-1](b) ) = a + b
( with f at least C^3 )
now differentiate with respect to a :
f ' (f^[-1](a) + f^[-1](b)) / f ' (f^[-1](a)) = 1
substitute X = f(a) , Y = f^[-1](b) :
f ' ( a + Y) = f ' (a)^2
Now rewrite f ' = G and differentiate with respect to Y :
G ' (Y + a) = 0
Hence G is constant and thus f = lineair.
Since f(0) = 0 we conclude f(x) = C x.
thus it only holds for r = 0,1
QED
regards
tommy1729

