Exact and Unique solution for base e^(1/e)

[jaydfox]

Ok, step by step we continue. Now we want to have g also differentiable for other bases b. It looks as if you change the initial interval for this purpose (though keeping the initial function f linear) in a way that provides for differentiability of g.

It came to me while I was traveling (flew from Dallas to Fresno). For a base b > eta, there's a very good reason that the "critical interval" (v-2, v-1] should be defined such that \( ^v {\Large b} = e \):

\(
\begin{eqnarray}
g(v-1) & = & {\Large b}^{g(v-2)} \\
g'(v-1) & = & ln(b) \left({\Large b}^{g(v-2)}\right) g'(v-2) \\
g'(v-1) & = & ln(b) \left( g(v-1) \right) g'(v-2)\end{eqnarray}
\)

Notice that when \( g(v-1)\ =\ log_b(e),\text{ then }g'(v-1)\ =\ ln(b) \left(log_b(e)\right) g'(v-2) = 1 \times g'(v-2) \), no matter what the funciton g(x) might look like. This is why I consider this particular unit interval the "critical" interval. Somewhere in that interval, there must be an inflection point, unless the interval is truly linear, which can't happen for b > eta. This means that the first derivative has a local minimum in this interval, and intuitively, we can claim this to be a global minimum for the first derivative. This in turn implies that the second derivative has a zero in this interval. It should also be clear that this will be the only zero for the second derivative.

Now, for bases that are integer superroots of e, the critical interval will have well-defined endpoints. For example, 1.601075..., which is third superroot of e, the endpoints are 1 and 2, exactly. However, for base 2, for example, we know the value of the function at v-2 and v-1, but we don't know what v is. It might be 1.484386863..., but it could be 1.47 or 1.50. Without further investigation, it would only be a guess at this point to make a claim one way or the other.