01/18/2021, 10:24 PM
Consider the equation
\( f(x+1) = \exp(x + f(x)) \)
such that f maps the reals to a subset of the reals.
The solution \( \phi(s+c) \) for a real constant \( c \) seems to be the unique entire solution such that also
\( f(x) = f(x+2\pi i) \)
To see why notice that for a real 1-period function \( g(s) \) all the solutions are probably
\( \phi(s + g(s)) \)
But if our solution is \( 2\pi i \) periodic than our \( g(s) \) should be as well.
But if \( g(s) \) is double periodic than by the theory of (analytic) double periodic functions , \( g(s) \) can not be an entire function.
regards
tommy1729
\( f(x+1) = \exp(x + f(x)) \)
such that f maps the reals to a subset of the reals.
The solution \( \phi(s+c) \) for a real constant \( c \) seems to be the unique entire solution such that also
\( f(x) = f(x+2\pi i) \)
To see why notice that for a real 1-period function \( g(s) \) all the solutions are probably
\( \phi(s + g(s)) \)
But if our solution is \( 2\pi i \) periodic than our \( g(s) \) should be as well.
But if \( g(s) \) is double periodic than by the theory of (analytic) double periodic functions , \( g(s) \) can not be an entire function.
regards
tommy1729
