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the distributive property
#10
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
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Messages In This Thread
the distributive property - by tommy1729 - 09/25/2010, 12:05 AM
RE: the distributive property - by tommy1729 - 09/28/2010, 12:22 PM
RE: the distributive property - by tommy1729 - 10/13/2010, 11:07 PM
RE: the distributive property - by tommy1729 - 05/23/2014, 11:12 PM
RE: the distributive property - by MphLee - 05/24/2014, 09:46 AM
RE: the distributive property - by tommy1729 - 05/25/2014, 12:16 AM
RE: the distributive property - by MphLee - 05/25/2014, 09:35 AM
RE: the distributive property - by tommy1729 - 05/26/2014, 11:37 PM
RE: the distributive property - by MphLee - 05/27/2014, 07:43 AM
RE: the distributive property - by tommy1729 - 05/27/2014, 12:23 PM
RE: the distributive property - by MphLee - 05/27/2014, 12:46 PM
RE: the distributive property - by tommy1729 - 05/27/2014, 10:49 PM
RE: the distributive property - by andydude - 05/27/2014, 07:45 PM
RE: the distributive property - by MphLee - 05/27/2014, 08:22 PM
RE: the distributive property - by tommy1729 - 05/27/2014, 10:55 PM
RE: the distributive property - by MphLee - 06/06/2014, 09:56 AM
RE: the distributive property - by tommy1729 - 06/06/2014, 11:56 AM
RE: the distributive property - by MphLee - 06/06/2014, 12:09 PM

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