| I doubt that this is what you're really after but here's one way that
will work, at least for 0 < x < e^(1/e):
Fix x in the range above and form a sequence of values of f(x,n), n>0.
Note that this sequence is geometrically convergent to a fixed point,
say f(x). Now in some small neighborhood of the fixed point,
the error e[n] = |f(x)-f(x,n)| will be proportional to some a^-n,
a > 1. Thus, you can define a way of 'interpolating' between
iterations in the vicinity of the fixed point, with as much accuracy
as you desire, by taking n great enough.
Suppose you want to interpolate the value of f(x,t) where k < t < k+1.
Note that f(x,k) < f(x,t) < f(x,k+1). Simply iterate f(x,k) into the
neighborhood of the fixed point, geometrically interpolate between
the images of f(x,k) and f(x,k+1) there, namely, f(x,n) and f(x,n+1),
and work backwards to infer the value of f(x,t).
I believe this can be shown to lead to a well defined analytic
continuation of the function away from the integers, but don't
know off-hand how to extend this past the upper limit. But if
this does lead to any analytic function, it is the official one.
It may also be possible to expand the function in a power series
about x = 1, since f(1,n) is identically 1, and you could perturb
this in a series.
- Jim
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