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 Personal Scratchpad jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 04:18 PM My scratchpad Notation for iteration of exponentiation: $\exp_b^{\circ t}(z)$ Iterated logarithm is then shorthand: $\log_b^{\circ t}(z)\ \equiv\ \exp_b^{\circ ({\small -}t)}(z)$ Tetration is a special case: ${}^{t} b\ \equiv\ \exp_b^{\circ t}(1)$ "Cleaner" notations to allow "primed" derivative notation: $\mathcal{T}_{[b,z]}(t)\ \equiv\ \exp_b^{\circ t}(z)$ $ \begin{eqnarray} \mathcal{T}_{b}(t) & = & \mathcal{T}_{[b,1]}(t) \\ & \equiv & {}^{t} b \\ \vspace{5} \\ & \equiv & \exp_b^{\circ t}(1) \end{eqnarray}$ $\mathcal{E}_{[b,t]}(z)\ \equiv\ \exp_b^{\circ t}(z)$ $ \begin{eqnarray} \mathcal{E}_{b}(z) & = & \mathcal{E}_{[b,1]}(z) \\ & \equiv & b^z \\ \vspace{5} \\ & \equiv & \exp_b^{\circ 1}(z) \end{eqnarray}$ $\mathcal{B}_{[t,z]}(b)\ \equiv\ \exp_b^{\circ t}(z)$ $ \begin{eqnarray} \mathcal{B}_{t}(b) & = & \mathcal{B}_{[t,1]}(b) \\ & \equiv & {}^{t} b \\ \vspace{5} \\ & \equiv & \exp_b^{\circ t}(1) \end{eqnarray}$ Generalizing what was discussed earlier, differentiation with respect to t: $\mathcal{T}_{[b,z]}^{'}(t)\ =\ \ln(b)\mathcal{T}_{[b,z]}(t)\mathcal{T}_{[b,z]}^{'}(t-1)$ $\mathcal{T}_{[b,z]}(t)\ =\ \frac{\mathcal{T}_{[b,z]}^{'}(t)}{\ln(b)\mathcal{T}_{[b,z]}^{'}(t-1)}$ ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 04:19 PM (This post was last modified: 09/01/2007, 04:24 PM by jaydfox.) Now for E(z): $ \begin{eqnarray} \exp_b^{\circ t}(z) & = & \exp_b^{\circ (t{\small +1})}(\log_b(z)) \\ & = & \exp_b\left(\exp_b^{\circ t}(\log_b(z))\right) \\ \mathcal{E}_{[b,t]}(z) & = & \exp_b\left(\mathcal{E}_{[b,t]}(\log_b(z))\right) \end{eqnarray}$ Derivative of E with respect to z: $ \begin{eqnarray} \text{D}_z\left(\mathcal{E}_{[b,t]}(z)\right) & = & \text{D}_z\left[\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right)\right] \\ \mathcal{E}_{[b,t]}^{'}(z) & = & \ln(b)\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right] \\ & = & \ln(b)\exp_b\left(\mathcal{E}_{[b,t]}\left(\log_b(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)\text{D}_z\left[\log_b(z)\right] \\ & = & \mathcal{E}_{[b,t]}(z)\left(\frac{\mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)}{\Large z}\right) \end{eqnarray}$ Transforming: $\mathcal{E}_{[b,t]}(z)\ =\ \frac{z\mathcal{E}_{[b,t]}^{'}(z)}{\mathcal{E}_{[b,t]}^{'}\left(\log_b(z)\right)}$ Going the other way: $ \begin{eqnarray} \exp_b^{\circ t}(z) & = & \exp_b^{\circ (t{\small -1})}(\exp_b(z)) \\ & = & \log_b\left(\exp_b^{\circ t}(\exp_b(z))\right) \\ \mathcal{E}_{[b,t]}(z) & = & \log_b\left(\mathcal{E}_{[b,t]}(\exp_b(z))\right) \end{eqnarray}$ And the derivative: $ \begin{eqnarray} \text{D}_z\left(\mathcal{E}_{[b,t]}(z)\right) & = & \text{D}_z\left[\log_b\left(\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)\right] \\ \mathcal{E}_{[b,t]}^{'}(z) & = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right] \\ & = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)\text{D}_z\left[\exp_b(z)\right] \\ & = & \left(\ln(b)\mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\right)^{-1} \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)\exp_b(z)\ln(b) \\ & = & \frac{\mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right) \exp_b(z)}{ \mathcal{E}_{[b,t]}\left(\exp_b(z)\right)} \\ \end{eqnarray}$ Finally, transforming: $ \mathcal{E}_{[b,t]}\left(\exp_b(z)\right)\ =\ \frac{\exp_b(z) \mathcal{E}_{[b,t]}^{'}\left(\exp_b(z)\right)}{\mathcal{E}_{[b,t]}^{'}(z)}$ ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 05:05 PM Inverse: $ \begin{eqnarray} z & = & \exp_b^{\circ ({\small -}t)}\left(\exp_b^{\circ t}(z)\right) \\ z & = & \mathcal{E}_{[b,-t]}\left(\mathcal{E}_{[b,t]}(z)\right) \end{eqnarray}$ Differentiate: $ \begin{eqnarray} \text{D}_z\left[ z \right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small -}t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right] \\ 1 & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] \\ & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\ \end{eqnarray}$ Transforming: $\mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right)\ =\ \frac{1}{\mathcal{E}_{[b,t]}^{'}(z)}$ Self-extraction: $ \begin{eqnarray} \exp_b^{\circ t}(z) & = & \exp_b^{\circ (t{\small -}t)}\left(\exp_b^{\circ t}(z)\right) \\ & = & \log_b^{\circ t}\left(\exp_b^{\circ t}\left(\exp_b^{\circ t}(z)\right) \right) \\ \mathcal{E}_{[b,t]}(z) & = & \mathcal{E}_{[b,-t]}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \end{eqnarray}$ Differentiate: $ \begin{eqnarray} \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small -}t]}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right)\right] \\ \mathcal{E}_{[b,t]}^{'}(z) & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right] \\ & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,t]}(z)\right] \\ & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,t]}\left(\mathcal{E}_{[b,t]}(z)\right)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\ & = & \mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,2t]}(z)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right) \mathcal{E}_{[b,t]}^{'}(z) \\ \end{eqnarray}$ Cancelling $\mathcal{E}_{[b,t]}^{'}(z)$ from both sides: $\mathcal{E}_{[b,{\small -}t]}^{'}\left(\mathcal{E}_{[b,2t]}(z)\right) \mathcal{E}_{[b,t]}^{'}\left(\mathcal{E}_{[b,t]}(z)\right)\ =\ 1$ In hindsight, that was obvious from the inverse. ~ Jay Daniel Fox bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/01/2007, 06:04 PM jaydfox Wrote:Notation for iteration of exponentiation: $\exp_b^{\circ t}(z)$ Tetration is a special case: ${}^{t} b\ \equiv\ \exp_b^{\circ t}(1)$ "Cleaner" notations to allow "primed" derivative notation: $\mathcal{T}_{[b,z]}(t)\ \equiv\ \exp_b^{\circ t}(z)$ $\mathcal{E}_{[b,t]}(z)\ \equiv\ \exp_b^{\circ t}(z)$ $\mathcal{B}_{[t,z]}(b)\ \equiv\ \exp_b^{\circ t}(z)$ I rather would additionally introduce:  \begin{align*} \text{sexp}_b (t) &={}^tb = \mathcal{T}_{[b,1]}(t)\\ \text{spow}_t (b) &={}^tb = \mathcal{B}_{[t,1]}(b)\\ \text{slog}_b&=\text{sexp}_b^{\circ -1} \end{align*} And we can recover the full 3 variable expression by means of $\text{sexp}_b$ and its inverse $\text{slog}_b$: $\exp_b^{\circ t}(z) = \text{sexp}_b(\text{slog}_b(z) + t)$ jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 08:25 PM (This post was last modified: 09/01/2007, 08:28 PM by jaydfox.) The half-iterate: $ \begin{eqnarray} \exp_b(z) & = & \exp_b^{\circ {\small \frac{1}{2}}}\left(\exp_b^{\circ {\small \frac{1}{2}}}(z)\right) \\ \exp_b(z) & = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \end{eqnarray}$ Differentiate: $ \begin{eqnarray} \text{D}_z\left[ \exp_b(z) \right] & = & \text{D}_z\left[\mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right)\right] \\ \ln(b) \exp_b(z) & = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \text{D}_z\left[\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right] \\ & = & \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \end{eqnarray}$ Which leads to: $ \begin{eqnarray} \exp_b(z) & = & \frac{1}{\ln(b)} \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \\ \mathcal{E}_{[b,{\small \frac{1}{2}}]}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) & = & \frac{1}{\ln(b)} \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}(z)\right) \mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z) \\ \mathcal{E}_{[b,{\small \frac{1}{2}}]}(z) & = & \frac{1}{\ln(b)} \left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}(z)\right) \left(\mathcal{E}_{[b,{\small \frac{1}{2}}]}^{'}\left( \mathcal{E}_{[b,{\small -\frac{1}{2}}]}(z) \right)\right) \end{eqnarray}$ (Yes, I'm on a fishing expedition.) ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 10:28 PM bo198214 Wrote:jaydfox Wrote:Notation for iteration of exponentiation: $\exp_b^{\circ t}(z)$ Tetration is a special case: ${}^{t} b\ \equiv\ \exp_b^{\circ t}(1)$ "Cleaner" notations to allow "primed" derivative notation: $\mathcal{T}_{[b,z]}(t)\ \equiv\ \exp_b^{\circ t}(z)$ $\mathcal{E}_{[b,t]}(z)\ \equiv\ \exp_b^{\circ t}(z)$ $\mathcal{B}_{[t,z]}(b)\ \equiv\ \exp_b^{\circ t}(z)$ I rather would additionally introduce:  \begin{align*} \text{sexp}_b (t) &={}^tb = \mathcal{T}_{[b,1]}(t)\\ \text{spow}_t (b) &={}^tb = \mathcal{B}_{[t,1]}(b)\\ \text{slog}_b&=\text{sexp}_b^{\circ -1} \end{align*} And we can recover the full 3 variable expression by means of $\text{sexp}_b$ and its inverse $\text{slog}_b$: $\exp_b^{\circ t}(z) = \text{sexp}_b(\text{slog}_b(z) + t)$Those notations are useful for bases above eta. However, for bases between 1 and eta, you cannot recover the full exp function, unless you extend the definition of slog to include the 2 additional domains (only 1 additional for eta). While I'm in favor of such an extension, I'm still working on defining the correct z value to start from, just as 1 is the "correct" value to start from for the principal domain. Until we define good starting points, a full analysis of fractional iterates will require the more general "T" notation, where we can start from a location other than 1 and leave it at that. For example, for base sqrt(2), we could start at sqrt( or e, etc., to solve for the corridor between the asymptotes at 2 and 4, and we could start at 4.1 or 5 or 16, etc. to solve above the asymptote at 4. (BTW, 16 seems to me the best place. Essentially, exponentiate the upper and lower asymptotes. The order doesn't matter: 2^4=4^2, and this is true for any base between 1 and eta, so it seems pretty "unique" to me.) ~ Jay Daniel Fox jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/01/2007, 10:36 PM jaydfox Wrote:Those notations are useful for bases above eta. However, for bases between 1 and eta, you cannot recover the full exp function, unless you extend the definition of slog to include the 2 additional domains (only 1 additional for eta). While I'm in favor of such an extension, I'm still working on defining the correct z value to start from, just as 1 is the "correct" value to start from for the principal domain. Until we define good starting points, a full analysis of fractional iterates will require the more general "T" notation, where we can start from a location other than 1 and leave it at that. For example, for base sqrt(2), we could start at sqrt( or e, etc., to solve for the corridor between the asymptotes at 2 and 4, and we could start at 4.1 or 5 or 16, etc. to solve above the asymptote at 4. (BTW, 16 seems to me the best place. Essentially, exponentiate the upper and lower asymptotes. The order doesn't matter: 2^4=4^2, and this is true for any base between 1 and eta, so it seems pretty "unique" to me.)Ach, what am I saying? Even if we extended slog to include all 2 or three domains, the sexp function would cease to be a function if we tried to take the inverse of the slog. In some ways, this would indicate that the slog function may be more fundamental than the sexp function. Given that Andrew's solution is based on solving the slog, this is interesting... ~ Jay Daniel Fox 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 jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 08:11 PM (This post was last modified: 09/02/2007, 08:15 PM by jaydfox.) 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 \\ \vspace{10}\\ \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 \\ \vspace{10}\\ \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)} \\ \vspace{10}\\ \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)} \\ \vspace{10}\\ \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] \\ \vspace{10}\\ \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] \\ \vspace{10}\\ \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] \\ \vspace{10}\\ \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] \\ \vspace{10}\\ \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] \\ \vspace{10}\\ \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 jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/02/2007, 08:34 PM (This post was last modified: 09/02/2007, 08:40 PM by jaydfox.) Hmm, I assumed I could pull the t out of $\text{D}_{w+t}$, since t is a constant, but now I'm not so sure. My calculus is pretty rusty. $ \frac{\text{d}}{\text{d}\left(w+t\right)}\ =\ \frac{\text{d}w}{\text{d}\left(w+t\right)}\frac{\text{d}}{\text{d}w}$ If my understanding is correct, dw/d(w+t) should equal 1. Yes? ~ Jay Daniel Fox « Next Oldest | Next Newest »

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