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 Personal Scratchpad jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 08:10 PM I'll start by solving for the fractional iterate of exponentiation itself. I.e., the base b and the iteration t are to be assumed constant, so that we can solve for, e.g., the half-iterate of base e $\left(\exp_e^{\circ {\tiny 1/2}}(z)\right)$, or the quarter-iterate of base 2 $\left(\exp_{\small 2}^{\circ {\tiny 1/4}}(z)\right)$, etc. First, we'll mix notations, noting that we must ensure that the bases are constant. In this case, the base z for T is not constant, so we must pick a constant base and then use an inverse function to allow for the variable z: $ \begin{eqnarray} {\Large \mathcal{E}_{\normalsize [b,t]}(z)} & = & {\Large \mathcal{T}_{\normalsize [b,z]}(t)} \\ & = & {\Large \mathcal{T}_{\normalsize [b,p]}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right)} \end{eqnarray}$ Note that if you set p=1, this becomes sexp(slog(z)+t). For bases between 1 and eta, other p values allow us to extend our analysis. Let's attempt to define some desirable characteristics of the fractional iterates of exponentiation. First, for bases above eta (and probably above 1 in general), they should be strictly increasing. This includes negative iterates as well. Second, for bases above eta, they should be convex. For positive iterates, this would mean that the second derivative is positive, while for negative iterates it would be negative. This looks likes: 1. ${\Large \mathcal{E}_{\normalsize [b,t]}^{'}(z)}\ >\ 0$ 2. ${\Large \mathcal{E}_{\normalsize [b,t]}^{''}(z)}\ >\ 0,\ t>0 \\ {\Large \mathcal{E}_{\normalsize [b,t]}^{''}(z)}\ <\ 0,\ t<0 \\$ At this point, it's time to start taking derivatives of T and see what happens. For now, we'll assume t positive, so we can control the direction of the ">". $ \begin{eqnarray} {\Large \text{D}_z\left[\mathcal{E}_{\normalsize [b,t]}(z)\right]} & > & 0 \\ \vspace{10}\\ {\Large \text{D}_z\left[\mathcal{T}_{\normalsize [b,p]}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right)\right]} & > & 0 \\ \vspace{10}\\ {\Large \mathcal{T}_{\normalsize [b,p]}^{'}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right) \text{D}_z\left[ \mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right]} & > & 0 \\ \vspace{10}\\ {\Large \mathcal{T}_{\normalsize [b,p]}^{'}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right) \frac{\Large 1}{ \mathcal{T}_{\normalsize [b,p]}^{'}\left( \mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)\right)}} & > & 0 \\ \vspace{10}\\ {\Large \frac{\mathcal{T}_{\normalsize [b,p]}^{'}\left(\mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)+t\right)}{ \mathcal{T}_{\normalsize [b,p]}^{'}\left( \mathcal{T}_{\normalsize [b,p]}^{\small-1}(z)\right)}} & > & 0 \\ \end{eqnarray}$ This last line implies that the first derivative of a tetration solution must be positive, i.e., $\mathcal{T}$ is strictly increasing. Not a difficult condition to meet, as this would be necessary for the sexp function to be invertible. So hardly any help has been gotten by this exercise. But it's a start. ~ Jay Daniel Fox « Next Oldest | Next Newest »