For a function to be real analytic it must satisfy two properties. (1) it must be infinitely differentiable (iterated real derivatives exist) for all points in the domain, or in other words, there exists a Taylor series expansion, and (2) the Taylor series expansion must converge, or in other words, the radius of convergence must be nonzero. Sometimes there is also a requirement that the function the Taylor series converges to is also the function being differentiated, but this is hard to find a good counterexample for.
The general iterate of the function \( e^x-1 \) has a Taylor series expansion, thus it is infinitely differentiable, but since the radius of convergence is 0 (see this thread), it is not real analytic. The reason why the radius of convergence is 0 is because the root test has no bound. For there to be a nonzero radius of convergence, the root test must be bounded. Actually, to be more specific, its unbounded for non-integer t, for integer t the root test is bounded.
For a function to be complex analytic (or holomorphic) it must satisfy essentially the same properties, but over complex derivatives, not real derivatives. In order for a complex derivative to exist a function must satisfy the Cauchy-Riemann equations which are a little more strict than a real derivative.
Since the primary difference between real analytic and complex analytic is the kind of derivatives used, there is a tendency to drop the real/complex part and just refer to analycity. In both cases the requirements are (1) infinite differentiability and (2) Taylor series has nonzero radius of convergence for all points in the domain. A special terminology is reserved for cases in which this is almost true except for a finite number of points. A meromorphic function is a function that is holomorphic for all except a finite number of points in the domain.
It may be that the iteration of \( e^x-1 \) (written \( DE^t(x) \)) is analytic or holomorphic everywhere except x=0. I don't know, but we do know that the general iterate is not analytic at x=0. Also, to be more specific, \( DE^t(x) \) is analytic at x=0 for integer t, but is not analytic at x=0 for non-integer t, thus it is not analytic for all (t, x) in the domain \( T \times X \). If this is too confusing, then it may be because the coefficients of x in the Taylor series are finite polynomials in t which we don't need to worry about convergence, since they're always "convergent" in some sense. So the Taylor series in x is the only one we need to worry about.
In some sense this kind of thinking can also be applied to hyperbolic iteration as well, since the coefficients of x in the Taylor series are functions of t that are representable in closed form.
Andrew Robbins
The general iterate of the function \( e^x-1 \) has a Taylor series expansion, thus it is infinitely differentiable, but since the radius of convergence is 0 (see this thread), it is not real analytic. The reason why the radius of convergence is 0 is because the root test has no bound. For there to be a nonzero radius of convergence, the root test must be bounded. Actually, to be more specific, its unbounded for non-integer t, for integer t the root test is bounded.
For a function to be complex analytic (or holomorphic) it must satisfy essentially the same properties, but over complex derivatives, not real derivatives. In order for a complex derivative to exist a function must satisfy the Cauchy-Riemann equations which are a little more strict than a real derivative.
Since the primary difference between real analytic and complex analytic is the kind of derivatives used, there is a tendency to drop the real/complex part and just refer to analycity. In both cases the requirements are (1) infinite differentiability and (2) Taylor series has nonzero radius of convergence for all points in the domain. A special terminology is reserved for cases in which this is almost true except for a finite number of points. A meromorphic function is a function that is holomorphic for all except a finite number of points in the domain.
It may be that the iteration of \( e^x-1 \) (written \( DE^t(x) \)) is analytic or holomorphic everywhere except x=0. I don't know, but we do know that the general iterate is not analytic at x=0. Also, to be more specific, \( DE^t(x) \) is analytic at x=0 for integer t, but is not analytic at x=0 for non-integer t, thus it is not analytic for all (t, x) in the domain \( T \times X \). If this is too confusing, then it may be because the coefficients of x in the Taylor series are finite polynomials in t which we don't need to worry about convergence, since they're always "convergent" in some sense. So the Taylor series in x is the only one we need to worry about.
In some sense this kind of thinking can also be applied to hyperbolic iteration as well, since the coefficients of x in the Taylor series are functions of t that are representable in closed form.
Andrew Robbins