While graphing the continuous iterations of the just discussed \( F(x)=e^{x/e} \), they looked quite divergent. Of course the numerics can be quite errournous for our rapidly increasing functions. But now indeed I found a proof of the divergence of \( F^{\circ t}(x) \) for nearly any t.
Proof: Assume that \( F^{\circ t}(x) \) has a convergence radius \( >0 \) for some t developed at the fixed point e.
Then consider the linear transformation \( \tau(x)=(x+1)e \) and its inverse \( \tau^{-1}(x)=\frac{x}{e}-1 \) and define
Further it is clear that \( G^{\circ t}=\tau^{-1}\circ F^{\circ t}\circ \tau \). If \( F^{\circ t}(x) \) has a convergence radius at \( e=\tau(0) \) then \( G^{\circ t}(x) \) has a convergence radius at \( 0 \). However it is well known (see [1]) that the (unique) continuous iteration \( G^{\circ t}(x) \) of \( e^x-1 \) has a convergence radius at 0 merely for integer \( t \), see [1].
[1] I. N. Baker, Zusammensetzungen ganzer Funktionen, Math. Z 69, 1958, 121-163.
Proof: Assume that \( F^{\circ t}(x) \) has a convergence radius \( >0 \) for some t developed at the fixed point e.
Then consider the linear transformation \( \tau(x)=(x+1)e \) and its inverse \( \tau^{-1}(x)=\frac{x}{e}-1 \) and define
\( G:=\tau^{-1}\circ F\circ \tau \),
i.e.\( G(x)=\frac{\left(e^{\frac{1}{e}}\right)^{(x+1)e}}{e}-1=\frac{e^{x+1}}{e}-1=e^x-1 \)
Further it is clear that \( G^{\circ t}=\tau^{-1}\circ F^{\circ t}\circ \tau \). If \( F^{\circ t}(x) \) has a convergence radius at \( e=\tau(0) \) then \( G^{\circ t}(x) \) has a convergence radius at \( 0 \). However it is well known (see [1]) that the (unique) continuous iteration \( G^{\circ t}(x) \) of \( e^x-1 \) has a convergence radius at 0 merely for integer \( t \), see [1].
[1] I. N. Baker, Zusammensetzungen ganzer Funktionen, Math. Z 69, 1958, 121-163.