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Let f(x) be a non-constant real-analytic function and for real x it satisfies :
f(2^x) = f(4^x + 2^(x+1) + 2) - f(4^x + 1)
This should be solvable.
regards
tommy1729
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04/07/2011, 01:06 PM
(This post was last modified: 04/07/2011, 01:12 PM by bo198214.)
(03/17/2011, 12:11 AM)tommy1729 Wrote: f(2^x) = f(4^x + 2^(x+1) + 2) - f(4^x + 1)
Isnt that equivalent to
f(y) = f(y^2+2y+2) - f(y^2 +1)
?
Posts: 1,924
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not quite.
remember that an exponential function a^z = 0 has no finite solution z for a =/= 0.
it is strongly related of course.
informally speaking , we want to avoid certain fixpoints and zero's on the reals ...
tommy1729
