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 paradox, accurate taylor series half iterate of eta not analytic at e sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/02/2011, 04:58 PM (This post was last modified: 06/03/2011, 02:12 PM by sheldonison.) For base exp(1/e), there have been many posts that the half-iterate of $\exp_\eta(z)$ is not analytic at z=e. Below I am posting a paradoxical accurate 30 term Taylor series, for that non-analytic half iterate of $\exp_\eta(z)$, developed at z=e. This series is double precision accurate, out to a radius of around 1. An example, with z=e-0.6, and this series puts out 2.1507815747789682, which is the correct result for the half iterate, generated via $\text{sexp}_\eta(\text{sexp}_\eta^{-1}(e-0.6)+0.5)=\text{sexp}_\eta(5.5342534224332571+0.5)$. The puzzle, is how is it possible to develop such a paradoxical accurate Taylor series, for a function which is not even analytic? A further complication, is that this Taylor series is required to seamlessly stitch together two different functions, the half iterate generated using $\text{sexp}_\eta(z)$, and the half iterate generated using the upper entire superfunction, $\text{cheta}(z)$. This is also the explanation for how it is possible. For a small enough series radius, it turns out the two half iterates can be seamlessly stitched together, with very little discontinuity, since as imag(z) increases, $\text{sexp}_\eta(z)$ exponentially converges to $\text{cheta}(z+k)$. And the imag(z) stitching value gets larger as the Taylor series radius gets smaller. I sampled this series at a radius of 1. $y=\text{sexp}^{-1}_\eta(e+i)=-3.3628841572099938 + 4.9027399771826196i$. At imag(y)=4.9, the two functions half iterates are already consistent to an accuracy of approximately 15 digits, which allows for the merged Taylor series. For real(z)=e, we use the half iterate generated from $\text{cheta}(z)$. The singularity and misbehavior occurs for the half iterate of cheta(z), when real(z)<=e, and similar misbehavior occurs for the half iterate of $\text{sexp}_\eta(z)$ for real(z)>=e. At e itself, both functions have singularities, but both functions agree that the half iterate of e=e. For real(z)=e, at smaller values of imag(z), paradoxically, the stitch is occurring for larger values of imag(y), where the stitch becomes more and more seamless. For the half iterate of $y=\text{sexp}^{-1}_\eta(e+0.5i)=-3.5937424498587124+10.344418312596874i$, where the two functions are consistent to 31 decimal digits! But at a larger sampling radius of r=1.5, the two half iterates are only consistent to approximately 11 decimal digits. So, within acceptable accuracy limits, it turns out it is possible to develop a Taylor series for the half iterate of $\text{sexp}_\eta(z)$, at e, where the function is not even analytic. Code:a0=   2.7182818284590452 a1=   1.0000000000000000 a2=   0.091969860292860588 a3=   0.0028194850674294418 a4=  -8.0085798390301461 E-18 a5=   0.0000047696976272632850 a6=  -0.00000051177982848104645 a7=   0.0000000038423117936633581 a8=   0.000000014046730882359691 a9=  -0.0000000030441687473700918 a10= -1.0220786581293779 E-11 a11=  1.6377605039633389 E-10 a12= -2.8311894064410302 E-11 a13= -8.1773414171527632 E-12 a14=  4.0125808678932662 E-12 a15=  3.0797537350253160 E-13 a16= -5.8574107605901771 E-13 a17=  3.8771237610746716 E-14 a18=  1.0016443539438077 E-13 a19= -2.2477472152925032 E-14 a20= -2.0211321002183050 E-14 a21=  8.7613198763649902 E-15 a22=  4.7138857780993590 E-15 a23= -3.4998961878228463 E-15 a24= -1.2239509050193108 E-15 a25=  1.5383321198806298 E-15 a26=  3.3098494797748605 E-16 a27= -7.5476712927521906 E-16 a28= -7.8877856873676488 E-17 a29=  4.1188557816475547 E-16 a30=  3.5020724883949410 E-18 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/02/2011, 05:48 PM Hey Sheldon, its not that paradoxical. These series are called asymptotic powerseries. One says a function on a domain D has an asymptotic development $\sum_{n=0}^\infty a_n (z-z_0)^n$ at $z_0\in\partial D$ (boundary of D) if it satisfies: $\frac{f(z)-\sum_{n=0}^N a_n (z-z_0)^n}{z^N} \to 0$ for $z\to z_0$. Or equivalently $f(z)=\sum_{n=0}^N a_n (z-z_0)^n + o(z^N)$ In our case the half-iterate has this kind of asymptotic development. So if we glue the left and right half-iterate together at e, then all derivatives exists and are finite at e, but still the function is not analytic at e, i.e. the convergence radius is 0. (Generally for a powerseries $f(z)=z+a_{m+1}z^{m+1}+a_{m+2}z^{m+2}+\dots$ there is exactly one formal powerseries of the fractional iterate $f^t$. This means all the 2m fractional iterates $f^t$ have the same asymptotic powerseries and hence the same derivations at 0.) However such divergent series are often used in physics, to still get good values (though not to unlimited precision). There is a certain point of truncation where the series returns the best and often quite precise value. I think its where the root-test $\frac{1}{\left|a_n\right|^{1/n}}$ has its maximum, i.e. where one - informally speaking - has the biggest radius of convergence. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 06/02/2011, 07:06 PM (This post was last modified: 06/02/2011, 07:07 PM by JmsNxn.) Since we're dealing with half-iterates of $\text{sexp}_\eta(z)$, does this tell us anything about pentation? If $\text{pent}_\eta(z)$ is pentation base $\eta$, if $\text{sexp}_\eta^{\frac{1}{2}}(z) = \text{pent}_\eta(\text{pent}_\eta^{-1}(z) + \frac{1}{2})$ holds up, I would assume since $\text{sexp}_\eta^{\frac{1}{2}}(z)$ isn't analytic about e, this says $\text{pent}_\eta(z)$ and $\text{pent}_\eta^{-1}(z)$ aren't analytic about e also right? I'm not sure, but the composition of two analytic functions is always an analytic function right? so if $\text{sexp}_\eta^{\frac{1}{2}}(z)$ isn't analytic about e, they probably aren't. I guess this glueing together has consequences for all hyperoperators base $\eta$, because if $\text{pent}_\eta(z)$ isn't analytic about e $\text{pent}_\eta^{\frac{1}{2}}(z)$ isn't analytic about e, so the same is true about hexation, etc etc.. Still though, it's very interesting that you can already develop half-iterates of tetration. Very fascinating. I guess using some sort of complicated recursive formula, I'm pretty sure you can theoretically evaluate any value for all hyperoperators base $\eta$. That's pretty incredible. At least, using the cheta function method. With this knowledge, I know it's possible for me to extend logarithmic semi-operators base $\eta$ to domain C. This would give a complex extension of the Ackerman function. This thread was all very beautiful. Thank you for posting it. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/02/2011, 07:17 PM (This post was last modified: 06/02/2011, 09:32 PM by sheldonison.) (06/02/2011, 07:06 PM)JmsNxn Wrote: Since we're dealing with half-iterates of $\text{sexp}_\eta(z)$, does this tell us anything about pentation? If $\text{pent}_\eta(z)$ is pentation base $\eta$, if $\text{sexp}_\eta^{\frac{1}{2}}(z) = \text{pent}_\eta(\text{pent}_\eta^{-1}(z) + \frac{1}{2})$ holds up, I would assume since $\text{sexp}_\eta^{\frac{1}{2}}(z)$ isn't analytic about e, this says $\text{pent}_\eta(z)$ and $\text{pent}_\eta^{-1}(z)$ aren't analytic about e also right? ..... This thread was all very beautiful. Thank you for posting it.you're welcome, but fyi, the wording may have been misleading, in that the half iterate I was generating was $\exp_\eta^{[0.5]}(z)$, where $\eta$=1/e. This is the half iterate of an exponential function, not the half iterate of a superexponential function. But one way to calculate the half iterate is this equation, which works for other bases than $\eta$ as well. $\exp_\eta^{[0.5]}(z)=\text{sexp}^_\eta(\text{sexp}^{-1}_\eta(z)+0.5)$. Also, your equation for the half iterate of sexp(z) in terms of pent(z) is correct. But notice, that we're only saying that the half iterate of eta^z is not analytic at a single point, when z=e. At other points, it is analytic, and at z=e, all of its derivatives are apparently defined, though with a zero radius of convergence. As you may have noticed from other posts, the two superfunctions of $\eta^z$ are a bit of an oddball, and this is just one more example. As a consequence, I don't think anyone has ever thought about or calculated pentation for base eta. Also the lower sexp(z) for eta never grows super-exponentially. The upper superfunction for eta^z does grow superexponentially, but never gets smaller than e. So that might take away some of the appeal for doing pentation base eta. - Sheldon tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/02/2011, 09:12 PM (06/02/2011, 04:58 PM)sheldonison Wrote: For base exp(1/e), there have been many posts that the half-iterate of $\text{sexp}_\eta(z)$ is not analytic at z=e. wow euh ... for starters i think you meant something else from what you actually said. i do not believe you consider half-iterates of any superfunction. if i get it correctly : the uppersuperfunction of eta^z is called cheta(z) the lowersuperfunction of eta^z is called sexp(z) you claim both cheta(z) and sexp(z) are not analytic at z = e. that is because e is the fixpoint of eta^z and it has f ' (e) = 1. since f(z) is not analytic at e , its taylor series is only an approximation. just like with e^z - 1. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/02/2011, 09:23 PM (This post was last modified: 06/02/2011, 09:25 PM by bo198214.) (06/02/2011, 09:12 PM)tommy1729 Wrote: if i get it correctly : the uppersuperfunction of eta^z is called cheta(z) the lowersuperfunction of eta^z is called sexp(z) you claim both cheta(z) and sexp(z) are not analytic at z = e. No, not the superfunctions, of course is sexp analytic at e, its analytic on $(-2,\infty)$ dont you remember? We talk about the half iterate of $\eta^x$, i.e. the function $f$ such that $f(f(x))=\eta^x$. Though Sheldon's wording was indeed a bit misleading. And even the explanation $\eta^{0.5}(z)$ is not really clear, because $\eta$ is not a function but a constant. Better may be the expression $\exp_\eta ^{0.5}(z)$ where $\exp_\eta(z)=\eta^z$. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/02/2011, 09:44 PM (06/02/2011, 09:23 PM)bo198214 Wrote: (06/02/2011, 09:12 PM)tommy1729 Wrote: if i get it correctly : the uppersuperfunction of eta^z is called cheta(z) the lowersuperfunction of eta^z is called sexp(z) you claim both cheta(z) and sexp(z) are not analytic at z = e. No, not the superfunctions, of course is sexp analytic at e, its analytic on $(-2,\infty)$ dont you remember? We talk about the half iterate of $\eta^x$, i.e. the function $f$ such that $f(f(x))=\eta^x$. Though Sheldon's wording was indeed a bit misleading. And even the explanation $\eta^{0.5}(z)$ is not really clear, because $\eta$ is not a function but a constant. Better may be the expression $\exp_\eta ^{0.5}(z)$ where $\exp_\eta(z)=\eta^z$.Sorry for the confusing wording. It seems like I made a few serious typos. I tried to fix the mistakes. As far as cheta as the upper superfunction for eta, that is a shorthand name that Jay invented. I try to occasionally throw in a detailed description. Then sexp for base eta follows the definition of sexp(0)=1. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 06/02/2011, 10:52 PM (06/02/2011, 09:23 PM)bo198214 Wrote: (06/02/2011, 09:12 PM)tommy1729 Wrote: if i get it correctly : the uppersuperfunction of eta^z is called cheta(z) the lowersuperfunction of eta^z is called sexp(z) you claim both cheta(z) and sexp(z) are not analytic at z = e. No, not the superfunctions, of course is sexp analytic at e, its analytic on $(-2,\infty)$ dont you remember? We talk about the half iterate of $\eta^x$, i.e. the function $f$ such that $f(f(x))=\eta^x$. Though Sheldon's wording was indeed a bit misleading. And even the explanation $\eta^{0.5}(z)$ is not really clear, because $\eta$ is not a function but a constant. Better may be the expression $\exp_\eta ^{0.5}(z)$ where $\exp_\eta(z)=\eta^z$. lol , yes thats what i meant. i didnt write what i wanted either. sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/03/2011, 04:22 PM (06/02/2011, 05:48 PM)bo198214 Wrote: Hey Sheldon, its not that paradoxical. These series are called asymptotic powerseries. ..... In our case the half-iterate has this kind of asymptotic development. So if we glue the left and right half-iterate together at e, then all derivatives exists and are finite at e, but still the function is not analytic at e, i.e. the convergence radius is 0. ....Thanks for the information! So does this mean, the infinite series will diverge, no matter how small abs(z-e) is? But the truncated finite series may be fairly accurate, depending on how many terms of the series are included, and abs(z-e)? I'm convinced that the series terms, $a_n(z-e)^n$, start getting larger for terms beyond a40 or so, and the series begins to diverge for |z-e|=1. - Sheldon bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 06/03/2011, 05:56 PM (06/03/2011, 04:22 PM)sheldonison Wrote: So does this mean, the infinite series will diverge, no matter how small abs(z-e) is?Yes. Quote: But the truncated finite series may be fairly accurate, depending on how many terms of the series are included, and abs(z-e)? Yes. But there is a trick you can use. Take a truncated powerseries of at least degree 2 (or in general case $f(x)=a+x+c(x-a)^{m+1}+\dots$ of degree $m+1$) of the intended by $t$ iterated powerseries, i.e. a polynomial $\pi$, of the form $\pi(x)=a+x+tc(x-a)^{m+1}+\dots$. Then one knows that $f^{\pm n}(\pi(f^{ \mp n}(z)))$ converges to the precise value of the fractional iteration (for $n\to\infty$), where you choose the opposing signs so that the inner $f^{\mp n}(z)$ converges towards the fixpoint $a$ (and the outer $f^{\pm n}$ then takes it back). Its because the series gets more precise - so to say - the closer the value is at the fixpoint a. And because $f^{\pm n}\circ f^t \circ f^{\mp n}=f^t$. « Next Oldest | Next Newest »

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