Now for the second condition. The second derivative is positive for t>0. This means that the first derivative (which we just found) is strictly increasing. We don't actually care what the second derivative is, so long as we can prove that the first derivative is strictly increasing. In other words, we're simply making a qualitative analysis, not a quantitative analysis (at this point, anyway).
In order to simplify notation, and without loss of generality, let's drop the bases from the notation. Note that this doesn't mean we're assuming "default" bases; we're simply not displaying them.
\(
\begin{eqnarray}
\text{D}_z\left[\mathcal{E}^{'}(z)\right] & > & 0 \\
\\[10pt]
\\
\lim_{h \to 0^{\small +}} \frac{\mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z)}{\Large h} & > & 0
\end{eqnarray}
\)
The h in denominator is now unimportant, since 0*h is still 0. Notice we can only do this if h is positive; otherwise, the ">" would reverse.
\(
\begin{eqnarray}
\lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z) & > & 0 \\
\\[10pt]
\\
\lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h) & > & \mathcal{E}^{'}(z) \\
\end{eqnarray}
\)
Okay, time to plug in the formula for E' we found above. Notice that I can only safely extract h because \( \mathcal{T} \) is strictly increasing, which in turn implies that \( \mathcal{T}^{\small-1} \) is strictly increasing. Again, we're looking for a qualititative analysis, so the exact value of epsilon after extraction is not important, so long as it's positive.
\(
\begin{eqnarray}
\lim_{h \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z+h)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z+h)\right)}
& > &
\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}
& > &
\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}
& > &
\frac{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\end{eqnarray}
\)
At this point, it's time to take a logarithm. Without loss of generality, we can use any base greater than 1, but for simplicity we can pretend it's base e, if it helps to understand.
\(
\begin{eqnarray}
\lim_{\epsilon \to 0^{+}} \left[ \log\left(\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}\right)\right.
& > &
\left. \log\left(\frac{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)}\right) \right] \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \left[ \log\left(\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)\right)-\log\left(\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)\right) \right.
& > &
\left. \log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)\right) \right]\\
\end{eqnarray}
\)
Now we divide through by epsilon, and lo and behold, we now have a proper derivative again!
\(
\begin{eqnarray}
\lim_{\epsilon \to 0^{+}} \left[ \frac{\log\left(\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)\right)-\log\left(\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)\right)}{\Large \epsilon} \right.
& > &
\left. \frac{\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)\right)}{\Large \epsilon} \right] \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \left[ \frac{\log\left(\mathcal{T}^{'}\left(w+t+\epsilon\right)\right)-\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)}{\Large \epsilon} \right.
& > &
\left. \frac{\log\left(\mathcal{T}^{'}\left(w+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left(w\right)\right)}{\Large \epsilon} \right] \\
\end{eqnarray}
\)
A quick substitution to simplify notation:
\(
\begin{eqnarray}
\text{D}_{w+t} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\text{D}_w \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\\[10pt]
\\
\frac{\text{d}}{\text{d}w} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\frac{\text{d}}{\text{d}w} \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\\[10pt]
\\
\frac{\text{d}z}{\text{d}w}\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\frac{\text{d}z}{\text{d}w}\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\end{eqnarray}
\)
Note that \( \frac{\text{d}z}{\text{d}w} = \left(\frac{\text{d}w}{\text{d}z}\right)^{-1} \). w is strictly increasing, so dw/dz is positive, so dz/dw is positive. Therfore, we can eliminate it from both sides without affecting the ">" comparison. And let's extract t. The derivative T' is strictly increasing, so when we extract a positive t, we'll get a positive s. The value will be uninmportant, so long as we know it's positive.
\(
\begin{eqnarray}
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}(w+t)\right)\right]
& > &
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}(w)\right) \right] \\
\\[10pt]
\\
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}(w)+s\right)\right]
& > &
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}(w)\right) \right] \\
\end{eqnarray}
\)
To satisfy this equation for all positive s, we want the first derivative (of the logarithm of the first derivative) to be strictly increasing. so we really want the second derivative to be positive:
\(
\begin{eqnarray}
\frac{\text{d}^{\tiny 2}}{\text{d}z^{\tiny 2}} \left[\log\left(\mathcal{T}_^{'}(w)+s\right)\right]
& > & 0
\end{eqnarray}
\)
And this, finally, means that we want T' to be log-convex. I mentioned this here, providing only the briefest sketch of a proof:
http://math.eretrandre.org/tetrationforu...d=11#pid11
As I mentioned:
Formalizing that wasn't "easy". It was the long way there, and someone may know a quicker/cleaner way to get there. But that's where we get. In order for all the partial iterates of exponentiation (for bases b>1) to be convex, the first derivative of T' must be log-convex. (What I now call T, I previously called R, if that clears up any confusion.)
Now, this is the general case. For the special case of tetration, we fix z at 1, and now we know that a tetration solution must have its first derivative be log-convex, in order to get convex partial iterates based on the solution.
My solution meets this new criterion, as does Andrew's. So we still need further criteria.
My hunch is that if we impose convexity requirements on the (odd?) derivatives of the partial iterates, we'll impose log-convexity requirements on higher derivatives of the tetration solution. We know these requirements exist for the integer iterates (at least for bases above eta, but I assume for bases above 1 in general), so it doesn't seem too much to ask this of the fractional iterates.
These log-convexity requirements would in turn require good old-fashioned convexity of the odd derivatives of a tetration solution, as we observed in Andrew's solution. If my hunch is correct, then we can find a more "solid" reason to believe that Andrew's solution is "the" solution.
In order to simplify notation, and without loss of generality, let's drop the bases from the notation. Note that this doesn't mean we're assuming "default" bases; we're simply not displaying them.
\(
\begin{eqnarray}
\text{D}_z\left[\mathcal{E}^{'}(z)\right] & > & 0 \\
\\[10pt]
\\
\lim_{h \to 0^{\small +}} \frac{\mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z)}{\Large h} & > & 0
\end{eqnarray}
\)
The h in denominator is now unimportant, since 0*h is still 0. Notice we can only do this if h is positive; otherwise, the ">" would reverse.
\(
\begin{eqnarray}
\lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h)-\mathcal{E}^{'}(z) & > & 0 \\
\\[10pt]
\\
\lim_{h \to 0^{+}} \mathcal{E}^{'}(z+h) & > & \mathcal{E}^{'}(z) \\
\end{eqnarray}
\)
Okay, time to plug in the formula for E' we found above. Notice that I can only safely extract h because \( \mathcal{T} \) is strictly increasing, which in turn implies that \( \mathcal{T}^{\small-1} \) is strictly increasing. Again, we're looking for a qualititative analysis, so the exact value of epsilon after extraction is not important, so long as it's positive.
\(
\begin{eqnarray}
\lim_{h \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z+h)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z+h)\right)}
& > &
\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}
& > &
\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}
& > &
\frac{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)} \\
\end{eqnarray}
\)
At this point, it's time to take a logarithm. Without loss of generality, we can use any base greater than 1, but for simplicity we can pretend it's base e, if it helps to understand.
\(
\begin{eqnarray}
\lim_{\epsilon \to 0^{+}} \left[ \log\left(\frac{\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)}{\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)}\right)\right.
& > &
\left. \log\left(\frac{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)}{ \mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)}\right) \right] \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \left[ \log\left(\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)\right)-\log\left(\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)\right) \right.
& > &
\left. \log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)\right) \right]\\
\end{eqnarray}
\)
Now we divide through by epsilon, and lo and behold, we now have a proper derivative again!
\(
\begin{eqnarray}
\lim_{\epsilon \to 0^{+}} \left[ \frac{\log\left(\mathcal{T}^{'}\left(\mathcal{T}^{\tiny-1}(z)+\epsilon+t\right)\right)-\log\left(\mathcal{T}_^{'}\left(\mathcal{T}^{\tiny-1}(z)+t\right)\right)}{\Large \epsilon} \right.
& > &
\left. \frac{\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left( \mathcal{T}^{\tiny-1}(z)\right)\right)}{\Large \epsilon} \right] \\
\\[10pt]
\\
\lim_{\epsilon \to 0^{+}} \left[ \frac{\log\left(\mathcal{T}^{'}\left(w+t+\epsilon\right)\right)-\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)}{\Large \epsilon} \right.
& > &
\left. \frac{\log\left(\mathcal{T}^{'}\left(w+\epsilon\right)\right)-\log\left(\mathcal{T}^{'}\left(w\right)\right)}{\Large \epsilon} \right] \\
\end{eqnarray}
\)
A quick substitution to simplify notation:
\(
\begin{eqnarray}
\text{D}_{w+t} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\text{D}_w \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\\[10pt]
\\
\frac{\text{d}}{\text{d}w} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\frac{\text{d}}{\text{d}w} \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\\[10pt]
\\
\frac{\text{d}z}{\text{d}w}\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}\left(w+t\right)\right)\right]
& > &
\frac{\text{d}z}{\text{d}w}\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}\left(w\right)\right) \right] \\
\end{eqnarray}
\)
Note that \( \frac{\text{d}z}{\text{d}w} = \left(\frac{\text{d}w}{\text{d}z}\right)^{-1} \). w is strictly increasing, so dw/dz is positive, so dz/dw is positive. Therfore, we can eliminate it from both sides without affecting the ">" comparison. And let's extract t. The derivative T' is strictly increasing, so when we extract a positive t, we'll get a positive s. The value will be uninmportant, so long as we know it's positive.
\(
\begin{eqnarray}
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}(w+t)\right)\right]
& > &
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}(w)\right) \right] \\
\\[10pt]
\\
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}_^{'}(w)+s\right)\right]
& > &
\frac{\text{d}}{\text{d}z} \left[\log\left(\mathcal{T}^{'}(w)\right) \right] \\
\end{eqnarray}
\)
To satisfy this equation for all positive s, we want the first derivative (of the logarithm of the first derivative) to be strictly increasing. so we really want the second derivative to be positive:
\(
\begin{eqnarray}
\frac{\text{d}^{\tiny 2}}{\text{d}z^{\tiny 2}} \left[\log\left(\mathcal{T}_^{'}(w)+s\right)\right]
& > & 0
\end{eqnarray}
\)
And this, finally, means that we want T' to be log-convex. I mentioned this here, providing only the briefest sketch of a proof:
http://math.eretrandre.org/tetrationforu...d=11#pid11
As I mentioned:
jaydfox Wrote:From here, we need to guarantee that R'(z) is always increasing. To do this, it suffices to show that ln(D_z tet_e(z)) is convex (I don't have time to post the explanation, I'll get back to that after work). Which in turn simply means that:
\( D_z^2\ ln(tet_e^\prime(z))\ \ge \ 0 \)
This in turn is equivalent to saying that the first derivative of tet_e(z) is log-convex.
Formalizing that wasn't "easy". It was the long way there, and someone may know a quicker/cleaner way to get there. But that's where we get. In order for all the partial iterates of exponentiation (for bases b>1) to be convex, the first derivative of T' must be log-convex. (What I now call T, I previously called R, if that clears up any confusion.)
Now, this is the general case. For the special case of tetration, we fix z at 1, and now we know that a tetration solution must have its first derivative be log-convex, in order to get convex partial iterates based on the solution.
My solution meets this new criterion, as does Andrew's. So we still need further criteria.
My hunch is that if we impose convexity requirements on the (odd?) derivatives of the partial iterates, we'll impose log-convexity requirements on higher derivatives of the tetration solution. We know these requirements exist for the integer iterates (at least for bases above eta, but I assume for bases above 1 in general), so it doesn't seem too much to ask this of the fractional iterates.
These log-convexity requirements would in turn require good old-fashioned convexity of the odd derivatives of a tetration solution, as we observed in Andrew's solution. If my hunch is correct, then we can find a more "solid" reason to believe that Andrew's solution is "the" solution.
~ Jay Daniel Fox