08/30/2010, 11:14 PM
ahh
i think ....
f(z) is NOT periodic , ONLY in the interval [z,z^^2] IF f(z) = 0.
meaning that not only f(z) = f(z^^2) = 0 but also
f(z+(z^^2-z)/2) = = f(z/2 + z^^2 /2) = f(z) = 0.
hmm.
f(z) = f(z^^2). for all z.
f(z) = - f(z^z) right ?
but if f(z) = f(z^^2) = 0 then - f(z^z) must be 0 but z^z is not in the middle of z and z^^2.
i.e. z<> z^z => z^z =/= (z + z^^2)/2.
or ... is that the equation to solve for f(z) = 0 ??
man your f(z) is weird !
i think ....
f(z) is NOT periodic , ONLY in the interval [z,z^^2] IF f(z) = 0.
meaning that not only f(z) = f(z^^2) = 0 but also
f(z+(z^^2-z)/2) = = f(z/2 + z^^2 /2) = f(z) = 0.
hmm.
f(z) = f(z^^2). for all z.
f(z) = - f(z^z) right ?
but if f(z) = f(z^^2) = 0 then - f(z^z) must be 0 but z^z is not in the middle of z and z^^2.
i.e. z<> z^z => z^z =/= (z + z^^2)/2.
or ... is that the equation to solve for f(z) = 0 ??
man your f(z) is weird !
