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 Should tetration be a multivalued function? sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 01/09/2016, 06:20 AM (01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically: $\vspace{15}{(^{r+1}a)^{^{-(r+1)}a}=(^{-r}a)^{^{r}a}}$ ok, first lets define y and z as follows: $y=\; ^{-(r+1)}a \;\;\;$ $z=\; ^{r}a \;\;\;$ Then substitute these values of y and z into the Op's equation above, noting that $^{(r+1)}a = a^z\;\;\;\; ^{-r}a = a^y$ $(a^z)^{y}=(a^y)^z\;\;\;$ This is the Op's equations with the substitutions $(a^z)^{y}=(a^y)^z=a^{(y\cdot z)}\;\;\;$ this equation holds for all values of a,y,z - Sheldon marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/09/2016, 05:14 PM (This post was last modified: 01/11/2016, 06:25 PM by marraco.) (01/09/2016, 06:20 AM)sheldonison Wrote: (01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically: $\vspace{15}{(^{r+1}a)^{^{-(r+1)}a}=(^{-r}a)^{^{r}a}}$ ok, first lets define y and z as follows: $y=\; ^{-(r+1)}a \;\;\;$ $z=\; ^{r}a \;\;\;$ Then substitute these values of y and z into the Op's equation above, noting that $^{(r+1)}a = a^z\;\;\;\; ^{-r}a = a^y$ $(a^z)^{y}=(a^y)^z\;\;\;$ This is the Op's equations with the substitutions $(a^z)^{y}=(a^y)^z=a^{(y\cdot z)}\;\;\;$ this equation holds for all values of a,y,z^^ Good. That's far more elegant that my reasoning, which used the commutative x^ln(y) and goofy induction. Maybe is the key to calculate the derivative at the origin? (taking limit of r → 0) Here is a graphic illustrating the equality. The exponentiation of the blue arrow is equal to the one in the red arrow. The expression inside the rectangles are equal. By induction, if we invert the red arrow, it is equal to a larger arrow; I mean $\vspace{20}{(^{r+(n+1)}a)^{^{-(r+(n+1))}a}=(^{-(r+n)}a)^{^{r+n}a}}$. I wrote it thinking that n was entire, but since r is real, and °a is arbitrary, it is obvious that is for any real n. But is even more general, because this is valid on all branches, and for any definition of °a. Graphically, you can move the arrows to the left or right, and the equality remains valid. For example, $\vspace{20}{^2a\,^{^3a}\,=\,^4a\,^{^1a}}$ and also $\vspace{20}{^1a\,^{^4a}\,=\,^5a\,^{^0a}}$, $\vspace{20}{^0a\,^{^5a}\,=\,^6a\,^{^{-1}a}}$, etc. I have the result, but I do not yet know how to get it. marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/09/2016, 05:56 PM (This post was last modified: 01/09/2016, 06:17 PM by marraco.) The general equation is $\vspace{15}{(^xa)^{^{y}a}=(^{y+1}a)^{^{x-1}a}}$ Because $\vspace{20}{(^xa)^{^{y}a}\,=\,(a^{^{x-1}a})^{^{y}a}}\,=\,a^{^{x-1}a.\,^{y}a}\,=\,(a^{^{y}a})^{^{x-1}a}\,=\,(^{y+1}a)^{^{x-1}a }$ It is somewhat trivial. I have the result, but I do not yet know how to get it. tommy1729 Ultimate Fellow Posts: 1,676 Threads: 368 Joined: Feb 2009 01/10/2016, 03:37 AM (This post was last modified: 01/10/2016, 03:38 AM by tommy1729.) Hold on we can do a bit better. For base sqrt(2) = a using sheldon's y and z. a^^0 is defined by a^^0 ^ a^^0 = 2^4 = 4^2 = 16. ( 2 and 4 are a^^oo or the fixpoints ) So a^^0 = ssqrt(16). From sheldon we then get a^(y z) = 16. subst a = sqrt(2) [ for example , this idea works for other bases too ] => y z = 8. Now notice r = -(r+1) for r = -1/2. So a^^(-1/2) ^ 2 = 8. So a^^(-1/2) = sqrt( 8 ) = 2 sqrt(2). Since we have a^^0 and a^^(-1/2) we actually have a^^(K/2) for all integer K. Maybe this idea can be extended to a unique analytic solution ... ¯\_(ツ)_/¯ Regards tommy1729 The master marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/11/2016, 09:28 PM (01/10/2016, 03:37 AM)tommy1729 Wrote: Hold on we can do a bit better.I need more clarification. You propose $\vspace{15}{^0a\,^{^0a}=c_1\,^{c_2}=c_2\,^{c_1}=a\,^{c_1*c_2}}$, where the c's are the asymptotic values. (01/10/2016, 03:37 AM)tommy1729 Wrote: From sheldon we then get a^(y z) = 16.I do not understand from where comes that 16. y and z are defined by Sheldonison as functions of r. I have the result, but I do not yet know how to get it. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/13/2016, 08:24 AM @marraco, @tommy That certainly is a cool equation, even if it is easily provable. @everyone Also, I think I can express my earlier comment in different words now. Tetration is defined as the 1-initialized superfunction of exponentials. The previous functions discussed earilier are 3-initialized and 5-initialized, which makes them, not tetration, by definition. However, if there is an analytic continuation of the 1-initialized superfunction that overlaps with the 3-initialized superfunction, AND if on the overlap f(0) = 3, then they can be considered branches of the same function. But until that is proven, I don't think it's accurate to say that they're all "tetration". They are, however, iterated exponentials in the sense that they extend $\exp_b^n(3)$ to non-integer n. And so I would probably write these functions as $\exp_b^n(1),\, \exp_b^n(3),\, \exp_b^n(5)$ instead of saying that ${}^{n}b$ is a multivalued function that returns all three. sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 01/13/2016, 01:37 PM (This post was last modified: 01/13/2016, 04:59 PM by sheldonison.) (01/13/2016, 08:24 AM)andydude Wrote: ...The previous functions discussed earilier are 3-initialized and 5-initialized, which makes them, not tetration, by definition. However, if there is an analytic continuation of the 1-initialized superfunction that overlaps with the 3-initialized superfunction, AND if on the overlap f(0) = 3, then they can be considered branches of the same function. But until that is proven, I don't think it's accurate to say that they're all "tetration". They are, however, iterated exponentials in the sense that they extend $\exp_b^n(3)$ to non-integer n. And so I would probably write these functions as $\exp_b^n(1),\, \exp_b^n(3),\, \exp_b^n(5)$ instead of saying that ${}^{n}b$ is a multivalued function that returns all three. ok, I was thinking of posting something similar, but I wasn't sure if it was off topic or not. From my perspective, the definition of tetration begins with real valued b>eta, where the base b $b>\eta \;\;\; b>\exp(\frac{1}{e})$. These bases require a Kneser mapping, to get a unique real valued function at the real axis, so that: $\text{tet}_b(z) =\;^z b= \exp_b^z(1)\;\;\; \text{tet}_b(0)=1, \; \text{tet}_b(1)=b;\;\;\;$ and tet(z) grows superexponentially as z gets larger The Op is working with iterated exponentials using b=sqrt(2). These iterated functions are valid super-exponentials. But it turns out they are not Kneser tetration, in the above sense. Maybe that's not an important distinction. Anyway, it turns out, we can extend tetration to complex bases, and experiment with how the function behaves as we vary the base around $\eta$. I didn't know any of this when I wrote kneser.gp for real valued tetration. But I have since written two different pari-gp programs posted on this site to experiment with complex base tetration. One is called tetcomplex.gp, and the other more recent effort calculates the Abel/slog function for arbitrary complex bases, fatou.gp The two pari-gp programs agree with each other. And they both agree that when you rotate 180 degrees around eta, the function you get is no longer real valued at the real axis! So there is such a thing as a well defined tetration $b=\sqrt{2}\; \text{tet}_b(z)\;\;$ that is an analytic continuation of Kneser type tetration. But guess what. Its not real valued at the real axis. Its some wierd function, that is almost real valued. In the upper half of the complex plane, the function gets arbitrarily close to behaving like the attracting fixed point of 2 function whose period is ~=17.1i , and in the lower half of the complex plane, it gets arbitrarily close to behaving like the repelling fixed point function, whose period is ~=19.2i. At the real axis, $\text{tet}_{\sqrt{2}}(0.5) =$1.24362162766852 - 1.18899611608401 E-48*I It is much easier to see the effect with a base like b=1.25, where the default settings and precision are sufficient for the calculations. Code:\r \fatou.gp sexpinit(1.25); sexp(0.5) 1.16359536196501 - 2.01428234045867 E-16*I - Sheldon marraco Fellow Posts: 100 Threads: 12 Joined: Apr 2011 01/14/2016, 04:24 AM (01/09/2016, 05:56 PM)marraco Wrote: The general equation is $\vspace{15}{(^xa)^{^{y}a}=(^{y+1}a)^{^{x-1}a}}$ Because $\vspace{20}{(^xa)^{^{y}a}\,=\,(a^{^{x-1}a})^{^{y}a}}\,=\,a^{^{x-1}a.\,^{y}a}\,=\,(a^{^{y}a})^{^{x-1}a}\,=\,(^{y+1}a)^{^{x-1}a }$ It is somewhat trivial. I just noticed that the equation remains valid between branches, and "branches" include the asymptotes. I have the result, but I do not yet know how to get it. « Next Oldest | Next Newest »

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