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 Uniqueness Criterion for Tetration jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/09/2007, 07:01 AM (This post was last modified: 08/09/2007, 07:31 AM by jaydfox.) In Andrew Robbins paper [1], he mentions three properties, two of which are applicable to tetration: Property 1. Iterated exponential property $^{y}x=x^{\left({ ^{y-1} x }\right)}$ for all real y. Property 3. Infinite differentiability property $\large f(x)\ \text{is}\ C^\infty\ \equiv\ D^k f(x)$ exists for all integer k. At the bottom of page three, Andrew goes on to say: Quote:It is the goal of this paper, however, to show that these properties are sufficient to find such an extension, and that the extension found will be unique. So, I set out to work on a hunch I had (see my posts on sci.math.research for the line of study I was on when I had this hunch), and came up with the following formula: $T(x,\ y,\ n) = \left{ \begin{eqnarray} \alpha_0\ +\ y\ +\ x^{T(x,\ y-1,\ n-1)} & , & n\ >\ 0 \\ \alpha_0\ +\ y & , & n\ =\ 0 \end{eqnarray} \right. \\ \ \\ \ \\ \ \\ {\Large ^y x}\ =\ \lim_{m,n\to\infty}{ln^{\small (m)}T(x,m+y,m+n)}$ Here, $\alpha_0$ is a constant that shifts the function left or right so that it's in the "correct" place. $ln^{\small (m)}(z)$ means iterating the logarithm m times. Of course, due to the nature of the formula, even if you can find a suitable $\alpha_0$ for a particular value (e.g., x^^0=1), the formula won't precisely satisfy property 1 unless you take the limits all the way to infinity. But it can get close enough with a relatively small m so that it exceeds the precision of any possible physical computer. But the point is, the function is infinitely differentiable, satisying property 3. And, as m and n are increased sufficiently to the point where machine precision is exceeded, property 1 is satisfied as well. However, a quick check shows that the values calculated by my formula aren't remotely close to Andrews. I started to panic at this point. Andrew's method aims to satisfy both properties 1 and 3, and I was fairly certain (without writing a very long and esoteric proof) that my formula does as well, and yet we get very different answers. How could this be? And then it hit me. Think of the iterated multiplication formula (you know, "exponentiation"). It satisfies the following three criteria (the first to give us a known reference point): $E(a,\ 0)\ =\ 1\\ \ \\ E(a,\ z+1)\ =\ a{E(a,\ z)}\\ \ \\ E\ \text{is}\ C^{\infty}$ Okay, so that's sufficient to define the Exponentiation function uniquely, right? Well, what about this formula: $F(a,\ z)\ =\ E\left({a,\ z+0.1\sin{\left({2\pi z}\right)}}\right)$ Is Property 1b (iterated multiplication property) still valid? $F(a,\ z+1)\ \overset{\tiny \text{?}}{=}\ a F(a,\ z) \\ E\left({a,\ (z+1)+0.1\sin{\left({2\pi (z+1)}\right)}}\right)\ \overset{\tiny \text{?}}{=}\ a E\left(a,\ {z+0.1\sin{\left({2\pi z}\right)}}\right) \\ E\left({a,\ (z+1)+0.1\sin{\left({2\pi z}\right)}}\right)\ \overset{\tiny \text{?}}{=}\ a E\left({a,\ z+0.1\sin{\left({2\pi z}\right)}}\right) \\ E\left({a,\ \left[(z+0.1\sin{\left({2\pi z}\right)}\right]+1}\right)\ \overset{\tiny \text{?}}{=}\ a E\left(a,\ \left[{z+0.1\sin{\left({2\pi z}\right)}}\right]\right) \\ E\left({a,\ Z+1}\right)\ \overset{\tiny \surd}{=}\ a E\left(a,\ Z\right)$ The sine function has a period of 1, so this function still has property 1. And this trivial change obviously hasn't altered the infinite differentiability property either. So as you can see, properties 1 and 3, unfortunately, are not sufficient to define a unique solution. Note: I made a similar claim about the Gamma function in a personal correspondence with Andrew Robbins, but after thinking it through, I realize I was mistaken in that particular case. Because the Gamma function is fixed relative to its input values, its inputs cannot be shifted around. Iterated multiplication (i.e., exponentiation) does not suffer this drawback. Of course, this makes defining the "correct" formula for iterated multiplication a little tricky. Pretend you know nothing about exponentiation, and try to figure out the "correct" formula. One can interpolate with square roots, cube roots, etc., which only gives you values for rational inputs. Using Cauchy sequences, one could prove that this extends to real answers, but until one comes up with a series expansion or defines a limit for repeated multiplication by $\small {1+\epsilon}$, for example, how can one be sure their solution is correct? So, never fear. My formula can satisfy both of these properties and still be "wrong". I'm not upset about it, because now I have a new pursuit: trying to define a third uniqueness criterion (fourth if you count a reference point as a criterion). I'd be interested if anyone knows of any resources that give a good such criterion, including a decent explanation of why that criterion is a good one. [1] Robbins, Andrew (2005) Solving for the Analytic Piecewise Extension of Tetration and the Super-logarithm link: http://tetration.itgo.com/paper.html PS: I've never used TeX before, so apologies if I went overboard. Edit: Moved postscript and updated LaTeX formats (I'm still learning how to use this). bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/09/2007, 09:05 AM (This post was last modified: 08/09/2007, 09:07 AM by bo198214.) Welcome Jayd, yes you observed correctly: the 3 properties 1. ${}^1 b = b$ 2. ${}^{x+1} b = b^{{}^{x}b}$ 3. and infinite differentiability (even analyticity) do not suffice to uniquely determine an extension of tetration. See also my post about Andrew's solution where I describe modified solutions also following from the approach of defining piecewise an infinite differentiable solution. They suffice neither for the Gamma function nor for exponentiation nor for multiplication. The criterion that makes the Gamma function unique is logarithmic convexity and the criterion that makes exponentiation and multiplication unique is the translation equation (see the FAQ for a description of both). So - as you said - for uniqueness there shall be found a suitable criterion, (however I beleave there exists nothing suitable XD). For the fractional iteration of functions there are uniqueness criterions at hand if the function has a fixed point. Unfortunately this is not the case for exp, indeed there is no uniqueness criterion for continuous iteration of exp too. (For the relationship between iteration of exp and tetration see also the FAQ.) Can you explain your formula $T(x,\ y,\ n) = \left{ \begin{eqnarray} \alpha_0\ +\ y\ +\ x^{T(x,\ y-1,\ n-1)} & , & n\ >\ 0 \\ \alpha_0\ +\ y & , & n\ =\ 0 \end{eqnarray} \right. \\ \ \\ \ \\ \ \\ {\Large ^y x}\ =\ \lim_{m,n\to\infty}{ln^{\small (m)}T(x,m+y,m+n)}$ in a bit more detail? PS: For not using TeX before your post was amazing jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/09/2007, 03:58 PM (This post was last modified: 08/19/2007, 01:17 PM by bo198214.) bo198214 Wrote:They suffice neither for the Gamma function nor for exponentiation nor for multiplication. The criterion that makes the Gamma function unique is logarithmic convexity and the criterion that makes exponentiation and multiplication unique is the translation equation (see the FAQ for a description of both). So - as you said - for uniqueness there shall be found a suitable criterion, (however I beleave there exists nothing suitable XD). For the fractional iteration of functions there are uniqueness criterions at hand if the function has a fixed point. Unfortunately this is not the case for exp, indeed there is no uniqueness criterion for continuous iteration of exp too. (For the relationship between iteration of exp and tetration see also the FAQ.) Well, I'm not sure if we can define a uniqueness criterion, but we can impose certain limits. For example, it seems very desirable to me that fractional exponential functions should be convex. For example, if we define F(F(x)) = e^x, then it seems sensible to me that F(x) should be convex. Ditto for G(G(G(x))) = e^x, and so on. Well, as I'm sure we're all aware, for qth iterate of e^x, we can define the function as parameterized curve: $R(t)\ =\ < ^{t} e,\ ^{t+q} e > \\ \ \text{or} \\ R(z)\ =\ tet_e(tet_e^{\tiny -1}(z)+q)$ Pushing forward to real iterates (rather than just the "well-defined" rational iterates), let's use r instead of q. Differentiation then tells us that: $\begin{eqnarray}R^\prime(z) & = & tet_e^\prime(tet_e^{\tiny -1}(z)+r)\ \times\ \left(D_z\ tet_e^{\tiny -1}(z) \right) \\ & = & tet_e^\prime(tet_e^{\tiny -1}(z)+r)\ \times\ \left(\frac {1}{tet_e^\prime(tet_e^{\tiny -1}(z))} \right)\\ & = & \frac {tet_e^\prime(tet_e^{\tiny -1}(z)+r)}{tet_e^\prime(tet_e^{\tiny -1}(z))}\\ & = & \frac {tet_e^\prime(Z+r)}{tet_e^\prime(Z)}\end{eqnarray}$ My calculus is a bit rusty, so someone spot-check the conversion from: $D_z\ tet_e^{\tiny -1}(z) \\ \ \text{to} \\ \frac {1}{D_z\ tet_e(tet_e^{\tiny -1}(z))$ (I'm pretty sure that's an identity, but please let me know if I'm wrong.) 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. It turns out that my formula meets this criterion, so it's not a uniqueness constraint, but it can help sift out the really "wrong" solutions. Additionally, while the first derivative of my function is log-convex, a look at the second derivative shows that it's "wavy". (See attached image, which is for my formula that you quoted below.)     (Question to forum experts: Is there a way to attach a full-sized image, or perhaps enlarge the thumbnail that displays for attachments?) Quote:Can you explain your formula $T(x,\ y,\ n) = \left{ \begin{eqnarray} \alpha_0\ +\ y\ +\ x^{T(x,\ y-1,\ n-1)} & , & n\ >\ 0 \\ \alpha_0\ +\ y & , & n\ =\ 0 \end{eqnarray} \right. \\ \ \\ \ \\ \ \\ {\Large ^y x}\ =\ \lim_{m,n\to\infty}{ln^{\small (m)}T(x,m+y,m+n)}$ in a bit more detail? PS: For not using TeX before your post was amazing I'll get back to this question after work tonight. Edit by bo198214: I inlined the image. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/10/2007, 06:19 PM (This post was last modified: 08/19/2007, 01:16 PM by bo198214.) jaydfox Wrote:(Question to forum experts: Is there a way to attach a full-sized image, or perhaps enlarge the thumbnail that displays for attachments?) Generally if there is an image on some server, say at blafoo.com/img.jpg then you can inline it in your text via Code:[img]http://blafoo.com/img.jpg[/img](there is also a clickable icon that performs this for you in the posting form) If you however merely attached an image, then there is a trick. After a first post you get the url of the attachement - for example yours is http://math.eretrandre.org/tetrationforu....php?aid=3 - and then you edit your post to include the image via the img tag: Code:[attachment=3] UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 08/12/2007, 10:52 PM [quote=jaydfox] In Andrew Robbins paper [1], he mentions three properties, two of which are applicable to tetration: Property 1. Iterated exponential property $^{y}x=x^{\left({ ^{y-1} x }\right)}$ for all real y. Property 3. Infinite differentiability property $\large f(x)\ \text{is}\ C^\infty\ \equiv\ D^k f(x)$ exists for all integer k. At the bottom of page three, Andrew goes on to say: [QUOTE]It is the goal of this paper, however, to show that these properties are sufficient to find such an extension, and that the extension found will be unique.[/QUOTE] It doesn't seem to be, tho. To complicate matters, I fixed the piecewise differentiability problem with the frac extension and it turns out that the original construction can be made $C^\infty$, when the base is e. I also posted a note on sci.math.research Details here jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 08/13/2007, 02:58 AM (This post was last modified: 08/13/2007, 03:02 AM by jaydfox.) I'm not sure I follow why your solution is $C^\infty$ The function is continuous, of course: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} F(x,y, 0) & = & x^y \\ & = & x^1 \\ \vspace{10} \\ & = & x \\ \vspace{15} \\ \lim_{y \to {\small 1^{+}}} F(x,y-1, 1) & = & x^{x^{y-1}} \\ & = & x^{x^0} \\ \vspace{5} \\ & = & x^1 \\ \vspace{10} \\ & = & x \end{eqnarray}$ And for base e, the first derivative is continuous as well: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} D_y F(x,y, 0) & = & D_y x^y \\ & = & x^y \ln(x) \\ & = & x^1 \ln(x) \\ & = & x \ln(x) \\ \vspace{10} \\ \lim_{y \to {\small 1^{+}}} D_y F(x,y-1, 1) & = & D_y \left(x^{x^{y-1}}\right) \\ & = & \left(x^{x^{y-1}}\right) \ln(x) D_x (x^{y-1}) \\ & = & \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^2 \\ & = & \left(x^{x^0}\right) (x^0) \left(\ln(x)\right)^2 \\ & = & x \left(\ln(x)\right)^2 \end{eqnarray}$ However, even at the second derivative, we begin to see problems: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} D_y^2 F(x,y, 0) & = & D_y^2 x^y \\ & = & D_y \left( x^y\right) \ln(x) \\ & = & x^y \left(\ln(x)\right)^2 \\ & = & x^1 \left(\ln(x)\right)^2 \\ & = & x \left(\ln(x)\right)^2 \\ \vspace{10} \\ \lim_{y \to {\small 1^{+}}} D_y^2 F(x,y-1, 1) & = & D_y^2 \left(x^{x^{y-1}}\right) \\ & = & D_y \left( \left(x^{x^{y-1}}\right)\left( x^{y-1}\right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[D_y \left(x^{x^{y-1}}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left[D_y \left( x^{y-1}\right)\right]\right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[x^{x^{y-1}} \ln(x) D_y \left(x^{y-1}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[x^{x^{y-1}} \ln(x) \left(x^{y-1}\right)\ln(x)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\ & = & \left( \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^3 \right) \left(\left(x^{y-1}\right)\ln(x) + 1\right) \\ & = & \left( \left(x^{x^0}\right) \left(x^0\right) \left(\ln(x)\right)^3 \right) \left(\left(x^0\right)\ln(x) + 1\right) \\ & = & \left( \left(x^1\right) \left(1\right) \left(\ln(x)\right)^3 \right) \left(\left(1\right)\ln(x) + 1\right) \\ & = & \left( x \left(\ln(x)\right)^3 \right) \left(\ln(x) + 1\right) \\ \end{eqnarray}$ For base e, we get a second derivative of 1 from the left, but 2 from the right. Due to the product rule, the second and higher derivatives will get quite complex, preventing a simple solution. UVIR Junior Fellow Posts: 16 Threads: 2 Joined: Aug 2007 08/13/2007, 01:33 PM jaydfox Wrote:I'm not sure I follow why your solution is $C^\infty$ The function is continuous, of course: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} F(x,y, 0) & = & x^y \\ & = & x^1 \\ \vspace{10} \\ & = & x \\ \vspace{15} \\ \lim_{y \to {\small 1^{+}}} F(x,y-1, 1) & = & x^{x^{y-1}} \\ & = & x^{x^0} \\ \vspace{5} \\ & = & x^1 \\ \vspace{10} \\ & = & x \end{eqnarray}$ And for base e, the first derivative is continuous as well: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} D_y F(x,y, 0) & = & D_y x^y \\ & = & x^y \ln(x) \\ & = & x^1 \ln(x) \\ & = & x \ln(x) \\ \vspace{10} \\ \lim_{y \to {\small 1^{+}}} D_y F(x,y-1, 1) & = & D_y \left(x^{x^{y-1}}\right) \\ & = & \left(x^{x^{y-1}}\right) \ln(x) D_x (x^{y-1}) \\ & = & \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^2 \\ & = & \left(x^{x^0}\right) (x^0) \left(\ln(x)\right)^2 \\ & = & x \left(\ln(x)\right)^2 \end{eqnarray}$ However, even at the second derivative, we begin to see problems: $\begin{eqnarray} \lim_{y \to {\small 1^{-}}} D_y^2 F(x,y, 0) & = & D_y^2 x^y \\ & = & D_y \left( x^y\right) \ln(x) \\ & = & x^y \left(\ln(x)\right)^2 \\ & = & x^1 \left(\ln(x)\right)^2 \\ & = & x \left(\ln(x)\right)^2 \\ \vspace{10} \\ \lim_{y \to {\small 1^{+}}} D_y^2 F(x,y-1, 1) & = & D_y^2 \left(x^{x^{y-1}}\right) \\ & = & D_y \left( \left(x^{x^{y-1}}\right)\left( x^{y-1}\right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[D_y \left(x^{x^{y-1}}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left[D_y \left( x^{y-1}\right)\right]\right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[x^{x^{y-1}} \ln(x) D_y \left(x^{y-1}\right)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\ & = & \left(\left( \left[x^{x^{y-1}} \ln(x) \left(x^{y-1}\right)\ln(x)\right]\left( x^{y-1}\right)\right)+\left( \left(x^{x^{y-1}}\right) \left( x^{y-1}\right) \ln(x) \right)\right) \left(\ln(x)\right)^2 \\ & = & \left( \left(x^{x^{y-1}}\right) \left(x^{y-1}\right) \left(\ln(x)\right)^3 \right) \left(\left(x^{y-1}\right)\ln(x) + 1\right) \\ & = & \left( \left(x^{x^0}\right) \left(x^0\right) \left(\ln(x)\right)^3 \right) \left(\left(x^0\right)\ln(x) + 1\right) \\ & = & \left( \left(x^1\right) \left(1\right) \left(\ln(x)\right)^3 \right) \left(\left(1\right)\ln(x) + 1\right) \\ & = & \left( x \left(\ln(x)\right)^3 \right) \left(\ln(x) + 1\right) \\ \end{eqnarray}$ For base e, we get a second derivative of 1 from the left, but 2 from the right. Due to the product rule, the second and higher derivatives will get quite complex, preventing a simple solution. I see. I guess you are right. Thanks for the correction. tommy1729 Ultimate Fellow Posts: 1,359 Threads: 331 Joined: Feb 2009 04/27/2014, 08:17 PM Im a bit confused about this one. First you cannot simple take two variables and say they both go to (+)infinity. Limits do not work that simple. Generally m and n depend on eachother or otherwise the limit is undefined. Maybe this is a special case and the relations between m and n really does not matter. In that case my apologies but plz show me why. Secondly, why does this work ? And is this not equivalent to the base change method ?? Seems alot like ackermann's function at first sight. Im going to some older threads here, because there is alot of stuff written before I registered here too. Digging for gold in old threads. regards tommy1729 Gottfried Ultimate Fellow Posts: 759 Threads: 117 Joined: Aug 2007 05/01/2014, 09:07 PM (04/27/2014, 08:17 PM)tommy1729 Wrote: (...) In that case my apologies but plz show me why.Tommy, "Uvir" (whom I know by his real name) seems to have completely resigned from math a couple of years ago and also Jay D. Fox has not appeared here for three (?) years because of rising a family. Perhaps, however, there are other ones active to step in. Gottfried Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,359 Threads: 331 Joined: Feb 2009 05/01/2014, 10:21 PM Maybe the other one is you Gottfried ? regards tommy1729 « Next Oldest | Next Newest »

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