Question about infinite tetrate tommy1729 Ultimate Fellow Posts: 1,918 Threads: 414 Joined: Feb 2009 04/01/2009, 09:53 PM (This post was last modified: 04/06/2009, 03:58 AM by andydude.) probably a stupid question the complex function h(z) = z^z^z^z^z^.... converges in a heart shaped region on the complex plane. what is the area of that region ? sorry andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 04/06/2009, 08:28 AM tommy1729 Wrote:what is the area of that region? That's not a stupid question, its actually a good question, and I don't know the answer. But let's see if we can answer that using what is known about this region. Galidakis (1) and I (2) both call it the Shell-Thron region since these two authors have both investigated this region in great detail. Shell and Thron note that ${}^{\infty}b$ converges where $b=h^{1/h}$ and $|\ln(h)| \le 1$ (a result which they both attribute to Barrow). So if we want the outer path, then we change the less-than sign to an equals sign, and this should give us the answer. If $|\ln(h)| = 1$, then we can parameterize this as $\ln(h) = e^{it}$ where $i=\sqrt{-1}$. This means that $h = e^{e^{it}}$, and putting this back in the relationship with b gives the parameterization $f(t) = {\left(e \uparrow e^{it}\right)} \uparrow {\left(e \uparrow {-e^{it}}\right)}$ where $a \uparrow b = a^b$. There are some interesting points that can be expressed with this function $\begin{tabular}{rrl} f(0) = & 1.4446678610097661337 & = e^{1/e} \\ f(1.4488307492834293737) = & 2.0477905274644031305 & + i\ 0.842045503530840715 \\ f(1.927907601568660839 = & 0.8152027425068848021 & + i\ 2.0166402199122300356 \\ f(2.316910654383280043 = & -0.380979728647791984 & + i\ 0.8997001955459000918 \\ f(\pi) = & 0.0659880358453125371 & = e^{-e} \end{tabular}$ While we could use these points to integrate each section of the region, we could also use the parametric integration formula $A = \int_{x=\alpha}^{x=\beta} y dx = \int_{t=a}^{t=b} y(t) x'(t) dt$ where $x(t) = \text{Re}(f(t))$ and $y(t) = \text{Im}(f(t))$. Using numerical integration we find that $\begin{tabular}{rl} A & = \int_{t=\pi}^{t=0} \text{Im}(f(t)) \left(\text{Re}(f(t))\right)^{'} dt \\ & = \frac{i}{2} \int_{h=1/e}^{h=e} h^{(1/h - 2)} \left(h^{1/h} - e^{1/(e^{1/\ln(h)} \ln(h))}\right) (\ln(h)-1) dh \\ & \approx 4.02546664 \end{tabular}$ which is only the top half of the region, and since A = 4.02546664046975481171259768713, then 2A = 8.05093328093950962342519537425 should be the area of the whole region. Andrew Robbins (1) I.N.Galidakis The Birth of the Infinite Tetration Fractal. (2) A.Robbins and H.Trappmann Tetration Reference, page 37. andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 04/06/2009, 08:51 PM (This post was last modified: 04/06/2009, 08:53 PM by andydude.) I also made a pretty graph:     where the points are expressed with the previous parameterization: f(2.8104771698906146985) = 0 + (...)i f(2.316910654383280043 -- minimum real part f(2.2893404841466060077) = (...) +i f(2.1025947819458660569) = 0 + (...)i f(1.9681078867187501493) = (...) + 2i f(1.927907601568660839 -- maximum imaginary part f(1.8909946364512036029) = 1 + (...)i f(1.8869709745663253961) = (...) + 2i f(1.5600433372731975189) = 2 + (...)i f(1.5064318596452087082) = (...) + i f(1.4488307492834293737) -- maximum real part f(1.3099015915373765040) = 2 + (...)i from this graph you can almost see how the area is almost 4, because it takes up almost 4 full squares, then it has some over-hand outside of these 4 squares, and the overhang looks like it could fill up the remaining area... Andrew Robbins tommy1729 Ultimate Fellow Posts: 1,918 Threads: 414 Joined: Feb 2009 04/15/2009, 04:19 PM andydude Wrote:tommy1729 Wrote:what is the area of that region? That's not a stupid question, its actually a good question, and I don't know the answer. But let's see if we can answer that using what is known about this region. Galidakis (1) and I (2) both call it the Shell-Thron region since these two authors have both investigated this region in great detail. Shell and Thron note that ${}^{\infty}b$ converges where $b=h^{1/h}$ and $|\ln(h)| \le 1$ (a result which they both attribute to Barrow). So if we want the outer path, then we change the less-than sign to an equals sign, and this should give us the answer. If $|\ln(h)| = 1$, then we can parameterize this as $\ln(h) = e^{it}$ where $i=\sqrt{-1}$. This means that $h = e^{e^{it}}$, and putting this back in the relationship with b gives the parameterization $f(t) = {\left(e \uparrow e^{it}\right)} \uparrow {\left(e \uparrow {-e^{it}}\right)}$ where $a \uparrow b = a^b$. hmmm i was thinking about contour integration to arrive at a closed form .... andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 04/17/2009, 03:33 AM tommy1729 Wrote:hmmm i was thinking about contour integration to arrive at a closed form .... Well if you can find a closed form, then more power to you! I give up. Andrew Robbins « Next Oldest | Next Newest »

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