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 using sinh(x) ? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/02/2010, 02:58 AM (06/01/2010, 10:58 PM)tommy1729 Wrote: but is that equivalent to my solution ? I wasnt aware that you featured a solution, my impression was rather that of a vague idea or as you put it "something like that I think". So I dunno: if you provide some values for your solution we can compare with Lévy's solution. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/02/2010, 04:47 PM (This post was last modified: 06/03/2010, 03:57 AM by sheldonison.) (03/09/2010, 01:31 PM)tommy1729 Wrote: i wrote sinh(x) but i meant 2*sinh(x) !! tommy1729I'm trying to understand Tommy's suggestions here, so I'm restating what you guys already know. It seems like a different version of sexp, but it might be interesting to graph it. Tommy suggests calculating a superfunction of $f(z) = 2\sinh(z) = e^z - e^{-z}$ The fixed point of 2sinh(z) is z=0. The behavior near the fixed point of z=0 is f(z)=2*z. Then the superfunction would be real valued at the real axis, which would lead to a straightforward real valued superfunction, based on iterating from the fixed point/function of $2^z$, although its not clear if Tommy was suggesting generating the limits from the fixed point. It seems to work in an excel spreadsheet, and the super function would be generated from the limit: $\operatorname{superfunc}(z)=\lim_{n \to \infty} \operatorname{2sinh}^{[n]} ( 2^{z-n} )$ Once this super function starts growing, it grows super-exponentially. I'm assuming Tommy is suggested modifying this into a $\operatorname{sexp}_e$ by iterating natural logarithms. This is similar to what Bo is referring to when using the Abel function (inverse of the super function) to convert from the Abel function of exp(z)-1 to the Abel function of exp(z), which is also the "base change" equation. Here the Abel function is for 2sinh, so its a different sexp_e equation than all of the other sexp_e definitions we've seen so far. Later edit: By the way, Bo, thanks for the info on Levy's Abel function equation, that you posted earlier in this thread! I now understand Abel function notation! So the "change of base" idea has been around for a long time .... bo198214 Wrote:Actually Paul Lévy [1] showed how to obtain an iteration of e^x if we have an iteration of e^x-1. Say $\beta$ is an Abel function of e^x-1, then $\alpha(x)-\alpha(x_0)=\lim_{n\to\infty} \beta(\exp^{[n]}(x)) - \beta(\exp^{[n]}(x_0))$ Continuing on, we take the natural logarithm "m" times of the superfunc(z+m+k). The constant k has to be calculated from the inverse superfunction (Abel function) to normalize the equation so that sexp_e(0)=1. Then we have something like: $\operatorname{TommySexp_e}(z)= \lim_{m \to \infty } \ln^{[m]} ( \lim_{n \to \infty} \operatorname{2sinh}^{[n]} ( 2^{z-n+m+k} ) )$ Anyway, the limits appear to work, and this may be what Tommy or Bo have in mind. Because the e^-z term in 2sinh(z) decays rapidly to zero, this leaves us with an exponential with the same base as the iterated logarithms. So one could hope that this definition would not fall prey to the "base change" version of tetration, which gives a smooth function that is infinitely differentiable but seems to be nowhere analytic. On the other hand, it may fall into the same smooth but nowhere analytic class that the base change definition falls into, due to an infinite number of singularities generated from iterated logarithms in the complex plane. Either way, it could be difficult to make sense of the infinitely dense thicket of superexponential windings and logarithmic singularities. But it would be interesting to graph the superfunction of 2sinh in the complex plane, which is well defined. I believe it would be periodic in the imaginary direction. Is the period i*2pi/ln(2)? Does it grow superexponentially negative at imag(z)=i*pi/ln(2)? Could one extend this 2sinh superfunction definition to other real bases? Which ones? My quick initial guess is that it would be limited to bases greater than e^(0.5)? (06/02/2010, 02:58 AM)bo198214 Wrote: .... if you provide some values for your solution we can compare with Lévy's solution.Back to converting this superfunction into sexp_e .... using the limit equation I just posted above, I was able to use Tommy's 2*sinh solution to generate the half iterate. I calculate: TommySexp_e(-0.5)=0.49874336 The published value from Dimitrii's taylor series coefficients is DimitriiSexp_e(-0.5)=0.49856329 As might be expected, Tommy's sexp gives different values. The function also has different values than the base change solution, discussed earlier in this forum, which Bo points out is equivalent to Lévy's solution. So, it seems we have yet another super exponential, different from all the others. - Sheldon tommy1729 Ultimate Fellow Posts: 1,372 Threads: 336 Joined: Feb 2009 06/02/2010, 10:13 PM (This post was last modified: 06/07/2010, 12:22 AM by tommy1729.) (06/02/2010, 04:47 PM)sheldonison Wrote: (03/09/2010, 01:31 PM)tommy1729 Wrote: i wrote sinh(x) but i meant 2*sinh(x) !! tommy1729I'm trying to understand Tommy's suggestions here, so I'm restating what you guys already know. It seems like a different version of sexp, but it might be interesting to graph it. Tommy suggests calculating a superfunction of $f(z) = 2\sinh(z) = e^z - e^{-z}$ The fixed point of 2sinh(z) is z=0. The behavior near the fixed point of z=0 is f(z)=2*z. Then the superfunction would be real valued at the real axis, which would lead to a straightforward real valued superfunction, based on iterating from the fixed point/function of $2^z$, although its not clear if Tommy was suggesting generating the limits from the fixed point. It seems to work in an excel spreadsheet, and the super function would be generated from the limit: $\operatorname{superfunc}(z)=\lim_{n \to \infty} \operatorname{2sinh}^{[n]} ( 2^{z-n} )$ Once this super function starts growing, it grows super-exponentially. I'm assuming Tommy is suggested modifying this into a $\operatorname{sexp}_e$ by iterating natural logarithms. This is similar to what Bo is referring to when using the Abel function (inverse of the super function) to convert from the Abel function of exp(z)-1 to the Abel function of exp(z), which is also the "base change" equation. Here the Abel function is for 2sinh, so its a different sexp_e equation than all of the other sexp_e definitions we've seen so far. Continuing on, we take the natural logarithm "m" times of the superfunc(z+m+k). The constant k has to be calculated from the inverse superfunction (Abel function) to normalize the equation so that sexp_e(0)=1. Then we have something like: $\operatorname{TommySexp_e}(z)= \lim_{m \to \infty } \ln^{[m]} ( \lim_{n \to \infty} \operatorname{2sinh}^{[n]} ( 2^{z-n+m+k} ) )$ Anyway, the limits appear to work, and this may be what Tommy or Bo have in mind. Because the e^-z term in 2sinh(z) decays rapidly to zero, this leaves us with an exponential with the same base as the iterated logarithms. So one could hope that this definition would not fall prey to the "base change" version of tetration, which gives a smooth function that is infinitely differentiable but seems to be nowhere analytic. On the other hand, it may fall into the same smooth but nowhere analytic class that the base change definition falls into, due to an infinite number of singularities generated from iterated logarithms in the complex plane. Either way, it could be difficult to make sense of the infinitely dense thicket of superexponential windings and logarithmic singularities. But it would be interesting to graph the superfunction of 2sinh in the complex plane, which is well defined. I believe it would be periodic in the imaginary direction, and would be interesting to graph. Is the period i*2pi/ln(2)? Does it grow superexponentially negative at imag(z)=i*pi/ln(2)? (06/02/2010, 02:58 AM)bo198214 Wrote: .... if you provide some values for your solution we can compare with Lévy's solution.Using the limit equation I just posted above, I was able to use Tommy's 2*sinh solution to generate the half iterate. I calculate: TommySexp_e(-0.5)=0.49874336 The published value from Dimitrii's taylor series coefficients is DimitriiSexp_e(-0.5)=0.49856329 As might be expected, Tommy's sexp gives different values. The function also has different values than the base change solution, discussed earlier in this forum, which Bo points out is equivalent to Lévy's solution. So, it seems we have yet another super exponential, different from all the others. - Sheldon thanks for the reply. that is intresting. however i dont think it is the same , i do it differently , although the end result may be the same ; for real x and y , both >=0 : let g(x,y) be the y'th iterate of exp(x) evaluated at x. let f(x,y) be the y'th iterate of 2 sinh(x) evaluated at x. then g(x,y) = lim k -> oo log log log ... (k times ) [f( g(x,k) ,y)] regards tommy1729 btw g(x,y) satisfies ( derivative with respect to x ) g'(x,y) / g'(exp(x),y) = exp(x)/g(exp(x),y) this shows directly that it - my solution g(x,y) - commutes with exp(x). but i guess you already knew that. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/02/2010, 10:56 PM (This post was last modified: 06/02/2010, 11:50 PM by sheldonison.) (06/02/2010, 10:13 PM)tommy1729 Wrote: thanks for the reply. that is intresting. however i dont think it is the same , i do it differently , although the end result may be the same ; for real x and y , both >=0 : let g(x,y) be the y'th iterate of exp(x) evaluated at x. let f(x,y) be the y'th iterate of 2 sinh(x) evaluated at x.How do you calculate f(x,y)? I would drop the "x" term entirely... I calculate f(y) using the superfunction equation I gave. If f(z)=2^z is a small number, then it approximates the behavior of the superfunction for 2sinh. Do you have a different equation, and could you share it? $\operatorname{superfunc}(z)=\lim_{n \to \infty} \operatorname{2sinh}^{[n]} ( 2^{z-n} )$ After that, do you agree that converting from your f(y) to g(y) by iterating natural logarithms would appear to be exactly the same? Here, the superfunction <=> f(y), and TommySexp <=> g(y). The only reason I threw in the "k" constant was so that the resulting sexp_e(z) can be normalized so that TommySexp_e(0)=1. Otherwise, with the k constant=0, I get TommySexp(0)=0.92715... The value I'm using for k is 0.067838366. $\operatorname{TommySexp_e}(z)= \lim_{n \to \infty } \ln^{[n]} ( \operatorname{superfunc}(z+n+k))$ Quote:then g(x,y) = lim k -> oo log log log ... (k times ) [f( g(x,k) ,y)] regards tommy1729 btw g(x,y) satisfies ( derivative with respect to x ) g'(x,y) / g'(exp(x),y) = exp(x)/F(exp(x) but i guess you already knew that. Nope, certainly didn't, nor do I understand it But if it holds in the complex plane, that would really be something, since g and f are such different functions. f is the super function of 2sinh, and g is the iterated natural logarithm (n times) of f(z+n). Do you think g(x) is analytic (as n=>oo)? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/03/2010, 08:00 AM Sheldon, you are really a blessing! Nice to have you on the forum. I couldnt have it better explained. (06/02/2010, 04:47 PM)sheldonison Wrote: So the "change of base" idea has been around for a long time .... Exactly! Seems Paul Lévy being an old Fox (in the double meaning haha). Quote:But it would be interesting to graph the superfunction of 2sinh in the complex plane, which is well defined. I believe it would be periodic in the imaginary direction. Is the period i*2pi/ln(2)? Ya every regular superfunction at a hyperbolic fixed point $z_0$ has the period $2\pi i/\kappa$, $\kappa=\ln(f'(z_0))$. This is because it can be expressed as $\sigma(z)=z_0+\eta(e^{\kappa z})$, $\eta$ is the inverse Schröder function. You can find details in [1] Kouznetsov, D., & Trappmann, H. (2010). Portrait of the four regular super-exponentials to base sqrt(2). Math. Comp., 79, 1727–1756. Quote: Does it grow superexponentially negative at imag(z)=i*pi/ln(2)? Yes, by the above representation of the superfunction: $\sigma(z+\frac{\pi i}{\kappa})=z_0+\eta(-e^{\kappa z})$ $\eta$ provides the superexponential growth on the real axis, so $\eta(-x)$ falls superexponentially. Again details in [1], at every fixed point we have these *two* regular superfunctions. Quote: Could one extend this 2sinh superfunction definition to other real bases? Which ones? My quick initial guess is that it would be limited to bases greater than e^(0.5)? You mean base $b$ in $b^x-b^{-x}$ which then would approach $b^x$ for $x\to\infty$? Quote:As might be expected, Tommy's sexp gives different values. The function also has different values than the base change solution, discussed earlier in this forum, which Bo points out is equivalent to Lévy's solution. So, it seems we have yet another super exponential, different from all the others. Ya and we can construct lots of others by choosing instead of $2\sinh(x)$ any function $f$ that has similar growth like $e^x$ for $x\to\infty$! sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/03/2010, 01:47 PM (This post was last modified: 06/03/2010, 11:52 PM by sheldonison.) (06/03/2010, 08:00 AM)bo198214 Wrote: (06/02/2010, 04:47 PM)sheldonison Wrote: Could one extend this 2sinh superfunction definition to other real bases? Which ones? My quick initial guess is that it would be limited to bases greater than e^(0.5)?You mean base $b$ in $b^x-b^{-x}$ which then would approach $b^x$ for $x\to\infty$?Yeah, exactly that. I like that $b^x-b^{-x}$ is on odd function with a fixed point of zero. The function grows exponentially both in the positive and negative directions. This seems to lead to a nice super function definition that shares some of the characteristics of exponentiation in the complex plane. Also, I think the superfunction of 2sinh is entire; it has no singularities. The math is much easier than the complex fixed point of base e, followed by a Riemann mapping... My secret wish was that for the $b^x-b^{-x}$ superfunction, the base change function would work without a wobble. I generated the superfunction for $f(x)=2^x-2^{-x}$ with a base change back to the superfunction for 2sinh, $f(x)=e^x-e^{-x}$. As expected, there was a small (magnitude of +/-0.02%) 1-cyclic wobble... - Sheldon bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/04/2010, 03:28 AM (06/03/2010, 01:47 PM)sheldonison Wrote: Also, I think the superfunction of 2sinh is entire; it has no singularities.Yes, generally the regular superfunction is entire at a repelling fixed point (because the inverse Schröder function is). Quote: The math is much easier than the complex fixed point of base e, followed by a Riemann mapping... Well but for the derived superfunction of b^x we dont know even whether it is analytic. While we know that for the Kneser-solution. Quote:My secret wish was that for the $b^x-b^{-x}$ superfunction, the base change function would work without a wobble. I generated the superfunction for $f(x)=2^x-2^{-x}$ with a base change back to the superfunction for 2sinh, $f(x)=e^x-e^{-x}$. As expected, there was a small (magnitude of +/-0.02%) 1-cyclic wobble... Hm, so even there! Actually I found out that you can do the basechange also for bases smaller $\eta$: The limit $\mu_{a,b}(x)=\lim_{n\to\infty} \log_a^{[n]}(\exp_b^{[n]}(x))$ exists for all $1 and $x\in (-\infty,\infty)$ and satisfies $\mu_{a,b}\circ\exp_b = \exp_a\circ\mu_{a,b}$. However it is constant for $x\le a^+$ (the upper fixed point of of $a^x$). BUT it is strictly increasing for $x>a^+$! This means we can use the change of base to change bases of regular tetration at the *upper* fixed point. But even then: If $\sigma_a$ and $\sigma_b$ are the regular superfunctions to base $a$ and $b$ respectively at the *upper* fixed point, then $\mu_{a,b}\circ \sigma_b$ is a superfunction to base $a$ but $\mu_{a,b}\circ \sigma_b \neq \sigma_a$. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/04/2010, 04:48 AM (This post was last modified: 06/04/2010, 04:52 AM by sheldonison.) I think I know what Tommy may have in mind, based on these posts, and his other posts on half iterates. tommy1729 Wrote:... if we take the half-iterate of 2sinh(x) by using taylor series , we get a good approximation of the half-iterate of exp(x) for x large. if this good approximation is analytic in say [e^e,e^e^e^e] we could use that interval and take logs or exp of it to compute the half-iterate for [-oo,+oo] up to a relatively high precision. (06/02/2010, 10:13 PM)tommy1729 Wrote: ...i dont think it is the same , i do it differently , although the end result may be the same ; for real x and y , both >=0 : let g(x,y) be the y'th iterate of exp(x) evaluated at x. let f(x,y) be the y'th iterate of 2 sinh(x) evaluated at x. then g(x,y) = lim k -> oo log log log ... (k times ) [f( g(x,k) ,y)] For small values of x, 2sinh(x) = 2x, (the derivative at x=0), so the "half iterate" of 2sinh(x) would be $\sqrt{2}*x$. Using this "half iterate" of 2sinh(x), along with Tommy's equations for converting back and forth between f and g (the same as the base change equations, and the same as Levy's equations!), we can calculate TommySexp(0.5) and, yes, the results are the same as what I posted earlier. Perhaps Tommy can verify that this is what he means when he says he does it differently. The idea of using the 2sinh(x) half iterate limit of $\sqrt{2}*x$ is attractive due to its simplicity. That the TommySexp(0.5) is different than the Knesser sexp(0.5) by a small amount is another example of the base change function almost working, but not quite. It would be nice to graph the 1-cyclic wobble between TommySexp(x) and Kouznetsov's published sexp. Anyway, for large enough numbers, if we iterate exponents enough times, the inverse superfunctions (for any function with exponential growth, for any base) all agree as exactly as desired on what a full iterate means, but they all seem to disagree on what a fractional iterate means! I see Henryk has posted something on base changes for bases between 0 and eta, and now I will go read that, and get some sleep! - Shel sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/04/2010, 12:23 PM (This post was last modified: 06/04/2010, 12:32 PM by sheldonison.) (06/04/2010, 03:28 AM)bo198214 Wrote: Hm, so even there! Actually I found out that you can do the basechange also for bases smaller $\eta$: The limit $\mu_{a,b}(x)=\lim_{n\to\infty} \log_a^{[n]}(\exp_b^{[n]}(x))$ exists for all $1 and $x\in (-\infty,\infty)$ and satisfies $\mu_{a,b}\circ\exp_b = \exp_a\circ\mu_{a,b}$. However it is constant for $x\le a^+$ (the upper fixed point of of $a^x$). BUT it is strictly increasing for $x>a^+$! This means we can use the change of base to change bases of regular tetration at the *upper* fixed point.I followed most of that, what is the constant value? the upper fixed point of a? Also, what is the circle $\circ$ notation, $\mu_{a,b}\circ\exp_b = \exp_a\circ\mu_{a,b}$? Is it the same as $\mu_{a,b}(\exp_b) = \exp_a(\mu_{a,b})$? No... Quote:But even then: If $\sigma_a$ and $\sigma_b$ are the regular superfunctions to base $a$ and $b$ respectively at the *upper* fixed point, then $\mu_{a,b}\circ \sigma_b$ is a superfunction to base $a$ but $\mu_{a,b}\circ \sigma_b \neq \sigma_a$. Yeah, I would expect all of the same problems of iterating complex logarithms, with super exponential windings, and singularities. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/05/2010, 07:56 AM (06/04/2010, 12:23 PM)sheldonison Wrote: I followed most of that, what is the constant value? the upper fixed point of a?Yes I write $a^+$ for the upper fixed point of $a^x$ and $a^-$ for the lower fixed point of $a^x$ if $1. Quote: Also, what is the circle $\circ$ notation, $\mu_{a,b}\circ\exp_b = \exp_a\circ\mu_{a,b}$? Is it the same as $\mu_{a,b}(\exp_b) = \exp_a(\mu_{a,b})$? No... Quote:Yes $f\circ g$ is the function given by $(f\circ g)(x)=f(g(x))$. $\circ$ is called the composition operator. Thatswhy you see sometimes in literature $f^{\circ n}$, which means the $n$-th power of $f$, but not the power for the multiplication operation, but the power for the composition operation, i.e. $f^{\circ n}=f\circ \dots\circ f$. The notation $f(g)$ is formally a bit arguable because $g$ is not from the domain of definition of f, like it is if you write $f(x)$. [quote][quote] But even then: If $\sigma_a$ and $\sigma_b$ are the regular superfunctions to base $a$ and $b$ respectively at the *upper* fixed point, then $\mu_{a,b}\circ \sigma_b$ is a superfunction to base $a$ but $\mu_{a,b}\circ \sigma_b \neq \sigma_a$. Yeah, I would expect all of the same problems of iterating complex logarithms, with super exponential windings, and singularities. I just wanted to point out that regular iteration at $\eta=e^{1/e}$ basechange to $e$ does not give Kouznetsov's (but wobbles) regular iteration at $a<\eta$ basechange to $b<\eta$ does not give regular iteration (but wobbles). (though regular iteration must be considered always at the upper fixed point, not as we are used to at the lower fixed point.) « Next Oldest | Next Newest »

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