sqrt(exp) Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 10/29/2008, 04:49 AM (This post was last modified: 11/23/2008, 01:00 AM by Kouznetsov.)     Hello, I did not find figuires of power of exponential. So, I suggest few for $f=\exp^c(z)$. levels $\Re(f)=-3,-2,-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ and $\Im(f)=-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14$ are shown with thick lines in the comples $z$ plane for values $c=1, 0.9, 0.5, 0.1$ (left) and $c=-1, -0.9, -0.5, -0.1$ (right). One of levels $\Im(f)=0$ coincides with the real axis, but at $c<0$ one of cuts occupies some of the negative part of the real axis. Some intermediate levels are shown with thin lines. Also, levels $\Re(f)=\Re(L)$ and $\Im(f)=\Im(L)$ are shown with green curves. These levels cross at the branchpoint $z=L\approx 0.31813150520476413 + 1.3372357014306895 {\rm i}$ , which is fixed point of logarithm. The thick pink lines indicate the cuts of the complex plane. Graphics are symmetric with respect to the real axis; at real $c$, $\exp^c(z^*)=\exp^c(z)^*$ ;so, I show only upper half of the complex plane. Below I incert the graphic $y=\exp^c(x)$ for real valuees of $x$, and various values of $c$: $\exp(\exp(x))$, $\exp(x)$, $\exp^{0.9}(x)$, ... $\log(\log(x))$     ----- After to post the pics, I got the crytics (see the first comment). bo198214 suggests that I describe the method. I expected, that the method is obvious from the name of the forum... $\exp^c(z)=F\!\Big(c+F^{-1}(z) \Big)$ where $F$ is holomorpic solution of $F(z+1)=\exp(F(z))$ The entire soluiton of this equation is described by Kneser [1]; he used it to construct $\sqrt{\exp}$, although it is not the only aplication. Past century, there were no computers, so, Kneser could not plot his solution, and only in this century the plotting becomes possible. I have implemented tetration "tet", id est, the function $F$, that is holomorphic at least in $\{ z\in \mathbb{C} ~:~ \Re(z)>-2 \}$ and bounded at least in $\{ z\in \mathbb{C} ~:~ |\Re(z)| \le 1 \}$, satisfying conditions $F(z+1)=\exp\big(F(z)\big)~ \forall z\in \mathbb{C} ~:~ \Re(z)> -2$ $F(0)=1$ $F\big(z^*\big)=F(z)^* ~ \forall z\in \mathbb{C} ~:~ \Re(z)> -2$ My tetation [2] is holomorphic in the whole complex $z$ plane except $z\le -2$. I estimate, my algorithm returns at least 14 correct decimal digits; at least, while the argument is of order of unity. Then, I have implemented, with similar precision, its derivative $F^{\prime}$ and its interse $G=\mathrm{slog}$; $F(G(z))=z \forall z \in \mathbb{C} \backslash \{z~:~ z\!\le\! -\!2 \}$. Such inverse function has two barnchpoints at fixed points $L$ and $L^*$ of logarithm, which are solutions of equation $L=\log(L)$. I put the cutlines horizontally, along the halflines $\Re(z)<\Re(L)$, $\Im(z)=\pm \Im(L)$. One of these lines is seen in the figures for $\exp^c$ at non-ineger $c$. In order to understand, how does the inverse function work, I plot the image of the upper halfplane with function kslog.     Images of gridlines $\Re(z)=-1,0,\Re(L),1,2,3$ and $\Im(z)=1,\Im(L),2,3,4$ are shown. For $z$ in the shaded region, $\mathrm{kslog}(\mathrm{tet}(z))=z$ References: 1. H.Kneser. Reelle analytische Losungen der Gleichung $\varphi(\varphi(x))={\rm e}^{x}$ und verwandter Funktionalgleichungen''. Journal fur die reine und angewandte Mathematik, v.187 (1950), 56-67. 2. D.Kouznetsov. Solution of $F(z+1)=\exp(F(z))$ in complez z-plane. Mathematics Of Computations, in press. The up-to-last version is available at my homepage http://www.ils.uec.ac.jp/~dima/PAPERS/analuxp99.pdf 3. http://en.citizendium.org/wiki/Tetration bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 10/29/2008, 08:37 PM Hey Dmitrii, thanks for the cool pictures. However you didnt tell the community by what method you computed the values! I mean, I know. But its not only me on the forum. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 11/18/2008, 12:24 PM I would expect singularities for $\exp^{1/2}$ at the other fixed points of $\exp$ is that true? For example at $L_2=2.062277729598284 + 7.588631178472513*I$. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/20/2008, 02:50 AM I do not expect any other singularities. There are only 2 branchpoints and only two cutlines. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 11/20/2008, 08:38 AM Kouznetsov Wrote:I do not expect any other singularities. There are only 2 branchpoints and only two cutlines. Hm perhaps then they lie on other branches. Those singularities need not to be branch points, they can also be isolated singularities. But why do there have to be singularities? First one would expect that a half iterate has the same fixed points as the function. (Is this true for your branch of $\exp^{1/2}$, i.e. is $L_2$ a fixed point?) But if so, a non-integer iterate can usually be holomorphic at most at one fixed point. This is the case for regular iteration at that fixed point. As your $\exp^{1/2}$ is not regular at any fixed point (regular iteration yields non-real values at the real axis for $\exp$) it must be singular at any fixed point. However if it is non-singular at $L_2$ then this may be due to $L_2$ being not a fixed point of $\exp^{1/2}$ in the branch that you chose. Can you please verify this? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/22/2008, 01:46 AM bo198214 Wrote:Kouznetsov Wrote:I do not expect any other singularities. There are only 2 branchpoints and only two cutlines. Hm perhaps then they lie on other branches. Those singularities need not to be branch points, they can also be isolated singularities. But why do there have to be singularities? First one would expect that a half iterate has the same fixed points as the function. (Is this true for your branch of $\exp^{1/2}$, i.e. is $L_2$ a fixed point?) But if so, a non-integer iterate can usually be holomorphic at most at one fixed point. This is the case for regular iteration at that fixed point. As your $\exp^{1/2}$ is not regular at any fixed point (regular iteration yields non-real values at the real axis for $\exp$) it must be singular at any fixed point. However if it is non-singular at $L_2$ then this may be due to $L_2$ being not a fixed point of $\exp^{1/2}$ in the branch that you chose. Can you please verify this?Henryk: There is unigue tetration $f$, holomorphic in $\mathbb{C}\{ x\in \mathbb{R~:~} x\le -2 \}$. There is inverse function $g$ such that $f(g(z))=z \forall z\in F$. Function $g$ has unavoidable branchpoints at eigenvalues $L \approx 0.318131505204764135312654+1.33723570143068940890116 \rm i$ abe $L^*$ of logarithm. I choose the cutline in such a way that function $g$ is holomorphic in the range $F=\mathbb{C}\{ z\in \mathbb{C~:~} \Re(z)\le \Re(L),~ |\Im(z)|=\Im(L)\}$. In the set $F$, function $g$ takes each value just once, except fixed points of logarithm, i.e., $L\approx 0.318131505204764135312654+1.33723570143068940890116 ~\rm i$ and its conjugation. This function preserves the signum of the imaginary part. (While the imaginary part of the argument is positive, the imaginary part of the function is also positive). Define square root of the exponential as $\sqrt{\exp}(z)=f(0.5+g(z))$ for all $z\in F$. As function $f$ has no singularities ourside the real axis, funciton $\sqrt{\exp}$ also is not allowed to have singularities which would not be singulatities of function $g$. Theregore function $\sqrt{\exp}$ is holomorphic in $F$. I attach the extended figure of $f=sqrt{exp} (z)$ in the upper part of the complex $z$. It shows the branchpoint at eigenvalue $L \approx 0.3181315052+1.33723570143 ~\rm i$ of logarithm. It is holomorphic and has no cuts in vicinity of the "first" branchpoint $L_{{\rm e, 1}}\approx 2.062277729598284 + 7.5886311784725127 ~\rm i$ of exponential. (at elast while the imaginary part is equal to 8.     The grid covers the range $-4 \le Re(z) \le 4$, $0 \le Im(z) \le 0$ with unity step. Levels $\Re(f)f=-3,-2,-1$ are shown with thick red curves. Levels $\Re(f)f=1,2,.. 14$ are shown with thick blue curves. Level $\Re(f)f=0$ is shown with thick black curve. Level $\Re(f)f=\Re(L)$ is shown with thick green curve. Levels $\Im(f)f=\pm 0.1, \pm 0.2, .. \pm 0.9$ are shown with thin green curves. Level $\Im(f)f=\Im(L)$ is shown with thick green curve. Level $\Im(f)f=-1,-2$ is shown with thick red curves. Levels $\Im(f)f=1,2,.. 14$ are shown with thick blue curves. Levels $\Im(f)f= -0.1, -0.2, .. -0.9$ are shown with thin red curves. The cut $\{z \in \mathbb{C~:~} \Im(z)=\Im(L)~,~ \Re(z)<\Re(L)\}$ is shown with thick pink horizontal line. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 11/22/2008, 04:22 PM It seems we dont speak about the same thing. So my first question is: $\sqrt{\exp}(L_{e,1})=L_{e,1}$? where $L_{e,1}\approx 2.062 + 7.589i$ is the fixed point of $\exp$ on the upper halfplane that has the second lowest distance to the real line. I mean one could expect that a half iterate has the same fixed points as the function itself. Unfortunately I can not verify the above question from the picture. So if it turns out that $\sqrt{\exp}(L_{e,1})\neq L_{e,1}$ on the given domain of definition, then I would expect that there is some branch$\sqrt{\exp}_k$ such that $\sqrt{\exp}_k(L_{e,1})=L_{e,1}$. And further if there is such a branch (made up at the branch point $L$) then I would expect that $\sqrt{\exp}_k$ has a (maybe isolated) singularity at $L_{e,1}$. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/23/2008, 12:58 AM bo198214 Wrote:It seems we dont speak about the same thing. So my first question is: $\sqrt{\exp}(L_{e,1})=L_{e,1}$? where $L_{e,1}\approx 2.062 + 7.589i$ is the fixed point of $\exp$ on the upper halfplane that has the second lowest distance to the real line.My first answer is NO. $\sqrt{\exp}( 2.062277729598284 + 7.5886311784725127 ~\mathrm{i}) \approx -17.11069793592735 + 5.77820343698599 ~\mathrm{i}$. Looking at the graphic, I see, that the $\Re\Big(\sqrt{\exp}(L_{\mathrm{e},1})\Big)$ is negative. bo198214 Wrote:I mean one could expect that a half iterate has the same fixed points as the function itself.No, One knows that if $z^2=1$, then it does not imply that $z=1$. bo198214 Wrote:Unfortunately I can not verify the above question from the picture.I can. It is difficult to see that the real part is of order of $-17$, but it is easy to see that it is negative. Do you want me to draw more levels for negative values of the real part of $\sqrt{\exp}$? bo198214 Wrote:So if it turns out that $\sqrt{\exp}(L_{e,1})\neq L_{e,1}$ on the given domain of definition, then I would expect that there is some branch$\sqrt{\exp}_k$ such that $\sqrt{\exp}_k(L_{e,1})=L_{e,1}$.Yes, and you will have to determine somehow the positions of all new cutlines you create in such a way. bo198214 Wrote:And further if there is such a branch (made up at the branch point $L$) then I would expect that $\sqrt{\exp}_k$ has a (maybe isolated) singularity at $L_{e,1}$.I doubt about "isolated". There should be a branchpoint. If you want to keep $\mathrm{slog}\big(z^*\big)=\mathrm{slog}(z)^*$, two additional branchpoints. I expect, you will get a pair of new branchpoints per each turn. You may combine the cutlines at the real axis; then the $\sqrt{\exp}$ will not be defined at the real axis. bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 11/23/2008, 09:02 AM Now we have found consent Kouznetsov Wrote:bo198214 Wrote:I mean one could expect that a half iterate has the same fixed points as the function itself.No, One knows that if $z^2=1$, then it does not imply that $z=1$. Well, modification: one would expect that for each fixed point of the function the half iterate has a branch with the same fixed point. Quote:bo198214 Wrote:And further if there is such a branch (made up at the branch point $L$) then I would expect that $\sqrt{\exp}_k$ has a (maybe isolated) singularity at $L_{e,1}$. I doubt about "isolated". There should be a branchpoint. So thats something left to find out Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/25/2008, 04:28 AM (This post was last modified: 11/25/2008, 04:33 AM by Kouznetsov.) bo198214 Wrote:.. Kouznetsov Wrote:I doubt about "isolated". There should be a branchpoint.So thats something left to find out I work with slog. I have rotated the cutlines, and I found out the $2\pi \mathrm{i}$ periodicity and set of logarithmic singularities (with set of cutlines, of course) at the left hand side from the cutline     I do not copypast here desctiption from the previous figure; the only changes: I have rorated the cutline for 90 degrees, and I add levels $\Re(\sqrt{\exp}(z))=-3,-4,..-9$ « Next Oldest | Next Newest »

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