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+--- Thread: f(2^x) = f(4^x + 2^(x+1) + 2) - f(4^x + 1) (/showthread.php?tid=610)
f(2^x) = f(4^x + 2^(x+1) + 2) - f(4^x + 1) - tommy1729 - 03/17/2011
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
RE: f(2^x) = f(4^x + 2^(x+1) + 2) - f(4^x + 1) - bo198214 - 04/07/2011
(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)
?
RE: f(2^x) = f(4^x + 2^(x+1) + 2) - f(4^x + 1) - tommy1729 - 04/07/2011
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