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I'm looking for literature on fractional iterates of $x^2+c$, where c>0. For c=0, generating the half iterate is trivial.
$$h(h(x))=x^2$$ $$h(x)=x^{\sqrt{2}}$$

The question is, for $c>0,$ and $x>1$, when is the half iterate of $x^2+c$ smaller than the half iterate of $x^2$? We know that the full iterate is always larger, since $x^2+c>x^2$, for $c>0$, and $x>1$. Intuitively, one would think that the half iterate of $x^2+c$ would also always be larger, but I believe I have found some counter examples.

In examining the parabolic case for $c=0.25$, I believe $x=800000000$ is a counter example. $800000000^{\sqrt{2}} \approx 3898258249628$, but I calculate the half iterate of $f(x)=x^2+0.25$, $h_{x^2+0.25}(800000000) \approx 3898248180100$, which is smaller.

For $c=0$, this is the equation for the superfunction which can be used to calculate fractional iterations. $f(x)=x^2$, and $g(x) = f^{o x}$, $g(z) = 2^{2^z}$. For $c=0.25$, this is the parabolic case, which has been studied a great deal in understanding the mandelbrot set, and the superfunction is entire, and I presume there is a uniqueness criteria. For $c>0.25$, the problem becomes trickier because $x^2+c$ has complex fixed points, and I am also looking for any literature on unique solutions to calculating real valued fractional iterates for $c>0.25$.

What I am also interested in is the abel function of $x^2$, which is $\text{abel}(z) = \log_2(\log_2(z))$. I am interested in the abel function of $x^2$ composed with the superfunction of $x^2+c$. $$\theta(z)=\text{abel}_{x^2}(\text{superfunction}_{x^2+c}(z))-z$$

As real $z$ increases, if $\theta$ converges to a $1$-cyclic function, as opposed to a constant, then there are counter examples like the one I gave, and sometimes the superfunction is growing slower than $2^{2^z}$, and othertimes it is growing faster, with the two function intersecting each other an infinite number of times. I'm also wondering if $\theta$ converge to an analytic function? Any relevant links would be appreciated.
- Sheldon

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    Hi Shel - that's really surprising and interesting! Just for the record: I reproduced your counterexample to more digits precision and the same result (Pari/GP, 800 digits precision, fixpoint 0.5, squareroot of the 64x64 triangular carlemanmatrix for g(x)=x+x^2 which occurs by recentering the polynomial at the fixpoint)2012-10-08
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    Hey Gottfried, thanks for your comments2012-10-08
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    Hey Gottfried, thanks for your comments. I posted this because of the link to mick's post on fractional iterations for exponentials, which show some of the same characteristics. I figured that the similar problem for iterations of x^2+c was much simpler. Much to my surprise, the partial iterates of x^2+c show some of the same behaviors, with different values of "c" having both bigger and smaller fractional iterates, depending on the value of x. Iterates of x^+c have been studied much more than tetration, so relevant material should be published someplace ...2012-10-08
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    Have you tried to compare the half-iterates of $x^2$ when you solve by the two versions of regular iteration according to the two fixpoints (0 and 1) of that *same* function? I suspect, we'll have "wobbling" even between that methods of halfiteration of the same function...2012-10-08
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    x^2 shows double exponential growth from the repelling fixed point of 1. I don't know how to get double exponential growth from the attracting fixed point of zero. btw, I had a typo in my original calculation. I also actually calculated the half iterate of x^2+x, for x=800000000, instead of the half iterate of x^2+0.25. To shift between the two equivalent functions, you half to add or subtract 1/2.2012-10-08
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    Sorry to interrupt this conversation but when $f:x\mapsto x^2+\frac14$, how do you DEFINE the function $h$?2012-10-08
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    @did:Let $g(x)=x^2+x$ then $f(x)=g(x-0.5)+0.5$ because $(x-0.5)^2+(x-0.5)+0.5 = (x^2-2*0.5x+0.25)+(x-0.5)+0.5 = x^2 + 0.25$. The function $g(x)$ has now no constant term and a linear term, of which we can get the square root. This allows a power series representation for a half-iterate of $g(x)$ , lets call this $d(x)$ with $d(d(x))=g(x)$ then $h(x) = d(x-0.5)+0.5$ , We need more considerations, for instance how to evaluate the series of $d(x)$ which might have convergence-radius zero... but we can get approximations...2012-10-08
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    Right, $f$ and $g$ are conjugate hence solving $d\circ d=g$ and solving $h\circ h=f$ are equivalent. Now, there is a unique sequence $(a_n)_{n\geqslant2}$ such that, for every $n\geqslant2$, $d_n(x)=x+a_2x^2+\cdots+a_nx^n$ is such that $d_n\circ d_n(x)=x+x^2+o(x^{n+1})$ when $x\to0$. But, as you say, this might not be enough to define a function $d$... What makes you think the procedure works in this specific case, for example in the sense that $(d_n)_{n\geqslant2}$ converges pointwise?2012-10-08
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    @sheldonison How did you proceed to *generate the superfunction* of $f(x)=x^2+0.25$? Note that superfunctions are mentioned in your post, but only after you declare that you defined and manipulated $h$.2012-10-08
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    Note: my last comment refers to a now-disappeared comment by @sheldonison.2012-10-08
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    Did,The parabolic fixed point of x^2+0.25 is 0.5, so the superfunction is well defined and entire and gives accurate results for polynomial interpolation for a sequence of values a little bit bigger than 0.5. I interpolated 25 points on either side of 0.512, for a total of 51 interpolation points. This is accurate to 67 decimal digits, which is the precision I was using in pari-gp. Then I iterate f inverse of z=800000000 until the value is near 0.512, calculate the inverse of the superfunction, add 1/2 and then calculate the superfunction. Then iterate f, to generate the half iterate of z.2012-10-08
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    @did: not sure, whether I understand the intention of your question. a) the sequence of $d_n$ can uniquely be determined by solving finite linear equations up to rational values. b) I assume the radius of convergence for the power series $d(x)$ is zero. What makes me confident, that the method produces meaningful values is, that computing $x_1=d(x_0)$ and $x_2=d(x_1)$ approximates $g(x_0) $ well and seemingly the approximation can be improved by extending the powerseries of *d* to more terms and the help of non-fractional iterations. However, this is no proof for the solution to be exact...2012-10-08
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    @did:(...cont...) ad a): the coefficients of $d(x)$ in the described version are rational numbers, here the first few $d(x)=x + 1/2 x^2 - 1/4 x^3 + 1/4 x^4 - 5/16 x^5 + 27/64 x^6 - 9/16 x^7 + O(x^8)$. They diverge later with more-than-geometric-rate, but there is not yet a concise description of the general term available.2012-10-08
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    @sheldonison Not sure I understand your comment. In particular, what troubles me is that you seem to believe you are able to compute at will values of d. How would one do that, even approximately, is not clear to me because of the zero-radius problem which I started with and which Gottfried acknowledged.2012-10-08
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    @GottfriedHelms Surely we can all agree on your a). In view of your b) (you really mean radius **zero**, right?), I wonder what the phrase "computing $x_1=d(x_0)$" refers to. How does one compute d(x) for any given (nonzero) x when all one has are the functions d_n, possibly for large values of n, but with the sequence (d_n(x))_n **divergent** for every nonzero x (if I understand you correctly)?2012-10-08
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    @sheldonison And please use the @ thing to notify your comments.2012-10-08
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    @did: divergence can sometimes be handled to arrive at reasonable/valid values anyway - an impressive example is Euler's series $f(x) = 1! - 2!x + 3!x^2 - \ldots + \ldots $ which has zero-radius of convergence, but can be summed meaningfully anyway (see "Borel-summation"). For my computations I use either an adaptable version of the Noerlund-summation and/or reduce x to smaller values using the functional relations of different x by (the exact) integer iterations of the g(x)-function towards zero. I stop that reduction if I have 20 or 30 digits approximated but that can be improved.2012-10-08
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    @GottfriedHelms Which Noerlund summation? There are plenty of them... And at this point, one suddenly begins to wonder whether your results (and sheldon's) could be method-of-resummation dependent (and even details-of-said-method dependent...). But maybe you have reasons to believe the values you produce are *intrinsic*, after all (which is more interesting than *reasonable/valid*). Thanks for the explanations anyway.2012-10-08
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    @did The half iterate Taylor series is well defined for all real values except the fixed point itself. This is because there are two different superfunctions, one fast growing function for real values > fixed point that we are interested in, and another for real values < fixed point. Both together leads to the zero radius of convergence if you generate the taylor series at the fixed point itself. The superfunction for real values greater than the repelling fixed point is entire, and approaches fixedpoint-1/x as x gets more and more negative, although the parabolic case is complicated.2012-10-08
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    @did: I use a summation-method based on triangular matrices; the basic method is the matrix-implementation of the Euler-summation essentially using (powers of) the binomial-matrix. For series with hypergeometric growth-rate I use an experimental (not-yet-proven) modification of that matrix. I have a few alternative summations, and if results are suspicious I crosscheck them. Still I know and (try not to forget to) state, that the results are approximations / asymptotics so far. Finally: this all is still too short and a clever new approach is also needed to overcome the systematic problem.2012-10-08
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    A link to a possibly related problem: Consider the recursive definition $ \gamma(x)=x \gamma (x-1) + 1/e $. It differs from the recursive definition of the $\Gamma $ - function by a constant term, a bit similarly as with functions $x^2$ and $x^2+c$. Here, $\gamma(x) $ is the "incomplete gamma" and simply one of two parts of the integral-representation of the $\Gamma $ itself, defined by the bound of the integral. What I'm tinkering with is whether one could analyze the relation between our two superfunctions here in the light of that between the $\Gamma$ and the $\gamma $ (but no idea so far).2012-10-12
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    Another similar problem, is Peter Walker's [1991 paper](http://eretrandre.org/rb/files/Walker1990_90.pdf) on iterating logarithms of the superfunction(exp(z)-1). s(z) iterations of exp(z)-1 is the parabolic case, s(z)*e+e is also base eta, basechange function from tetration forum. Walker generates sexp(z) base e from $\log^{n} s(z+n)$. Walker showed that this alternative sexp function was infinitely differentiable, but on the tetration forum we showed that it is most likely nowhere analytic, so its interesting that the superfunction of x^2+0.25 generated from 2^2^z turns out to be analytic.2012-10-12
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    Im sorry Sheldon but I disagree : quote : As real z increases, if θ converges to a 1 -cyclic function, as opposed to a constant, then there are counter examples like the one I gave, If θ behaves like sin(x) + 10000 then you do not get counterexamples. If it starts to behave like sin(x) + 10000 in the limit for large x , you can have at most a finite amount of crossovers. One could argue that sin(x) + 10000 is not a possibility and such , but the basic statement is wrong and wrong statements lead to mistakes. I fear you have made this wrong interpretation in the past as well , no offense.2012-10-12
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    @mick, Perhaps a picture would help. $\alpha^{-1}(z+k)/2^{2^z}$ k is chosen to get approximately 50% duty cycle. [superfunction(z+k)/2^2^z](http://sheltx.com/share_stuff/plot_superf_22z.jpg)2012-10-12
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    @mick, Sounds like you're talking about the general case, then we also need to require that the two functions are increasing, and that theta is continuous. It comes down to $f(z+k+\theta(z))/f(z)$, where f is an increasing superfunction, and k is a constant equal to the average value of $\theta(z)$. Hope that helps.2012-10-12
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    Im not sure what you are trying to tell me. I stick with what I said and I wonder if you understand me ... hint : I quoted your last part of your OP.2012-10-13
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    @mick, I think the intended "counter example" would be $\theta=\sin(2\pi x)/2\pi+10000$, because theta's limiting behavior is 1-cyclic, but this is not a contradictory case, if we set k=-10000, to line up the functions appropriately. Consider the two infinitely intersecting functions, $f(x-10000)$ vs $f(x+\theta(x))$. Perhaps mick is complaining about my imprecise language, or lack of detail in the original post on how to line up the two functions so that they intersect an infinite number of times...2012-10-14
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    @SheldonL, on the question of complex fixpoints, there is no change in the main procedures, either in using Schroders or Abel's equation. The bad news is that one still needs to find, for example, $\alpha^{-1}\left( \frac{1}{2} + \alpha(z) \right),$ which now becomes a worse search if $z \notin \mathbb R.$ It can be done on computer, we just cannot attempt complex Newton's method because we have no closed form for $\alpha.$ Similar for Schroder's, given as an easier limit.2012-10-15
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    @WillJagy, I'm interested in real valued $\alpha$ and $\alpha^{1}$ for $x^2+c$ where c>0.25. This would be analogous to Kneser's solution for tetration for $\exp(z)$, where $\exp(z)$ has complex fixed points, but the desired tetration solution is real valued.2012-10-15
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    @SheldonL, it would appear Kneser's paper is not long. I have not seen it.2012-10-15
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    @SheldonL: I do not think we need to intersect an infinite number of times for $c$ large enough and the half-iterates not too ' wobbly '. I think that is equivalent with a theta 'not acting as you think'.2012-10-17
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    consider $g(x)=x^2+c$, with c positive, arbitrarily large, with an abel function, $\alpha(x)$. Consider an alternative abel function defined exactly as: $\alpha^{-1}_\infty(x)=\lim_{n\to\infty} g^{-1 o n} (2^{2^x})$. Then $\alpha(x)$ is defined and real valued as long as x>=0. Also, by definition, $\alpha_\infty(g(x)) = \alpha_\infty(x)+1$, so it is an alternative abel function. To get $\theta$ compare the two abel functions for c. $\theta(x)= \alpha(\alpha^{-1}_\infty(x))-x$. Either $\theta(x)$ is a constant, or it is 1-cyclic and 2^2^x and $\alpha^{-1}(x)$ intersect infinitely often.2012-10-18
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    Im I the only one who does not understand or believe Sheldon ? :) Maybe a picture will help me.2012-10-18
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    Hmm maybe this is a lot like the base change formula. After reading again Shel said converges too , not IS. Maybe I will understand some day.2012-10-18

10 Answers 10

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This may be helpful.

Let $$ f(x) = \frac{-1 + \sqrt{1 + 4 x}}{2}, \; \; x > 0 $$ We use a technique of Ecalle to solve for the Fatou coordinate $\alpha$ that solves $$ \alpha(f(x)) = \alpha(x) + 1. $$ For any $x > 0,$ let $x_0 = x, \; x_1 = f(x), \; x_2 = f(f(x)), \; x_{n+1} = f(x_n).$ Then we get the exact $$ \alpha(x) = \lim_{n \rightarrow \infty} \frac{1}{x_n} - \log x_n + \frac{x_n}{2} - \frac{x_n^2}{3} + \frac{13 x_n^3}{36} - \frac{113 x_n^4}{ 240} + \frac{1187 x_n^5}{ 1800} - \frac{877 x_n^6}{ 945} - n. $$ The point is that this expression converges far more rapidly than one would expect, and we may stop at a fairly small $n.$ It is fast enough that we may reasonably expect to solve numerically for $\alpha^{-1}(x).$

We have $$ f^{-1}(x) = x + x^2. $$ Note $$ \alpha(x) = \alpha(f^{-1}(x)) + 1, $$ $$ \alpha(x) - 1 = \alpha(f^{-1}(x)) , $$ $$ \alpha^{-1} \left( \alpha(x) - 1 \right) = f^{-1}(x). $$ It follows that if we define $$ g(x) = \alpha^{-1} \left( \alpha(x) - \frac{1}{2} \right), $$ we get the miraculous $$ g(g(x)) = \alpha^{-1} \left( \alpha(x) - 1 \right) = f^{-1}(x) = x + x^2. $$

I put quite a number of relevant pdfs at BAKER. The host computer for this was down for about a year but has recently been repaired.

EDIT, TUESDAY:

Note that $\alpha$ is actually holomorphic in an open sector that does not include the origin, such as real part positive. That is the punchline here, $\alpha$ cannot be extended around the origin as single-valued holomorphic. So, since we are finding a power series around $0,$ not only are there a $1/z$ term, which would not be so bad, but there is also a $\log z$ term. So the $\ldots -n$ business is crucial.

I give a complete worked example at my question https://mathoverflow.net/questions/45608/formal-power-series-convergence as my answer https://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765

The Ecalle technique is described in English in a book, see K_C_G PDF or go to BAKER and click on K_C_G_book_excerpts.pdf The Julia equation is Theorem 8.5.1 on page 346 of KCG. It would be no problem to produce, say, 50 terms of $\alpha(x)$ with some other computer algebra system that allows longer power series and enough programming that the finding of the correct coefficients, which i did one at a time, can be automated. No matter what, you always get the $\alpha = \mbox{stuff} - n$ when $f \leq x.$

As I said in comment, the way to improve this is to take a few dozen terms in the expansion of $\alpha(x)$ so as to get the desired decimal precision with a more reasonable number of evaluations of $f(x).$ So here is a brief version of the GP-PARI session that produced $\alpha(x):$

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    ? taylor( (-1 + sqrt(1 + 4 * x))/2  , x  )     %1 = x - x^2 + 2*x^3 - 5*x^4 + 14*x^5 - 42*x^6 + 132*x^7 - 429*x^8 + 1430*x^9 - 4862*x^10 + 16796*x^11 - 58786*x^12 + 208012*x^13 - 742900*x^14 + 2674440*x^15 + O(x^16)       f = x - x^2 + 2*x^3 - 5*x^4 + 14*x^5 - 42*x^6 + 132*x^7 - 429*x^8 + 1430*x^9 - 4862*x^10 + 16796*x^11 - 58786*x^12 + 208012*x^13 - 742900*x^14 + 2674440*x^15        ? fp = deriv(f)      %3 = 40116600*x^14 - 10400600*x^13 + 2704156*x^12 - 705432*x^11 + 184756*x^10 - 48620*x^9 + 12870*x^8 - 3432*x^7 + 924*x^6 - 252*x^5 + 70*x^4 - 20*x^3 + 6*x^2 - 2*x + 1       L = - f^2 + a * f^3       R = - x^2 + a * x^3      compare = L - fp * R       19129277941464384000*a*x^45 - 15941064951220320000*a*x^44 +  8891571783902889600*a*x^43 - 4151151429711140800*a*x^42 +  1752764158206050880*a*x^41 - 694541260905326880*a*x^40 +  263750697873178528*a*x^39 - 97281246609064752*a*x^38 + 35183136631942128*a*x^37  - 12571609170862072*a*x^36 + 4469001402841488*a*x^35 - 1592851713897816*a*x^34 +  575848308018344*a*x^33 - 216669955210116*a*x^32 + 96991182256584*a*x^31 +  (-37103739145436*a - 7152629313600)*x^30 + (13153650384828*a +  3973682952000)*x^29 + (-4464728141142*a - 1664531636560)*x^28 + (1475471500748*a  + 623503489280)*x^27 + (-479514623058*a - 220453019424)*x^26 + (154294360974*a +  75418138224)*x^25 + (-49409606805*a - 25316190900)*x^24 + (15816469500*a +  8416811520)*x^23 + (-5083280370*a - 2792115360)*x^22 + (1648523850*a +  930705120)*x^21 + (-543121425*a - 314317080)*x^20 + (183751830*a +  108854400)*x^19 + (-65202585*a - 39539760)*x^18 + (-14453775*a + 15967980)*x^17  + (3380195*a + 30421755)*x^16 + (-772616*a - 7726160)*x^15 + (170544*a +  1961256)*x^14 + (-35530*a - 497420)*x^13 + (6630*a + 125970)*x^12 + (-936*a -  31824)*x^11 + 8008*x^10 + (77*a - 2002)*x^9 + (-45*a + 495)*x^8 + (20*a -  120)*x^7 + (-8*a + 28)*x^6 + (3*a - 6)*x^5 + (-a + 1)*x^4       Therefore a = 1  !!!       ?      L = - f^2 +  f^3 + a * f^4      R = - x^2 +  x^3 + a * x^4       compare = L - fp * R       ....+ (1078*a + 8008)*x^10 + (-320*a - 1925)*x^9 + (95*a + 450)*x^8 + (-28*a - 100)*x^7 + (8*a + 20)*x^6 + (-2*a - 3)*x^5       This time a = -3/2  !       L = - f^2 +  f^3  - 3 * f^4 / 2  + c * f^5       R = - x^2 +  x^3 - 3 * x^4 / 2  + c * x^5         compare = L - fp * R     ...+ (2716*c - 27300)*x^11 + (-749*c + 6391)*x^10 + (205*c - 1445)*x^9 + (-55*c + 615/2)*x^8 + (14*c - 58)*x^7 + (-3*c + 8)*x^6        So c = 8/3 .       The printouts began to get too long, so I said no using semicolons, and requested coefficients one at a time..      L = - f^2 +  f^3  - 3 * f^4 / 2  + 8 * f^5 / 3 + a * f^6;       R = - x^2 +  x^3 - 3 * x^4 / 2  + 8 * x^5 / 3  + a * x^6;          compare = L - fp * R;      ? polcoeff(compare,5)     %22 = 0     ?      ?  polcoeff(compare,6)     %23 = 0     ?      ?  polcoeff(compare,7)     %24 = -4*a - 62/3      So this a = -31/6        I ran out of energy about here:       L = - f^2 +  f^3  - 3 * f^4 / 2  + 8 * f^5 / 3 - 31 * f^6 / 6 + 157 * f^7 / 15 - 649 * f^8 / 30 + 9427 * f^9 / 210 + b * f^10 ;         R = - x^2 +  x^3 - 3 * x^4 / 2  + 8 * x^5 / 3  - 31 * x^6 / 6 + 157 * x^7 / 15 - 649 * x^8 / 30 + 9427 * x^9 / 210  + b * x^10;         compare = L - fp * R;      ?      ?  polcoeff(compare, 10 )     %56 = 0     ?      ?      ?  polcoeff(compare, 11 )      %57 = -8*b - 77692/105     ?      ?        L = - f^2 +  f^3  - 3 * f^4 / 2  + 8 * f^5 / 3 - 31 * f^6 / 6 + 157 * f^7 / 15 - 649 * f^8 / 30 + 9427 * f^9 / 210 - 19423 * f^10 / 210 ;         R = - x^2 +  x^3 - 3 * x^4 / 2  + 8 * x^5 / 3  - 31 * x^6 / 6 + 157 * x^7 / 15 - 649 * x^8 / 30 + 9427 * x^9 / 210 - 19423 * x^10 / 210;         compare = L - fp * R;      ?  polcoeff(compare, 10 )     %61 = 0     ?      ?  polcoeff(compare, 11 )      %62 = 0     ?      ?  polcoeff(compare, 12)      %63 = 59184/35     ?       So R = 1 / alpha' solves the Julia equation   R(f(x)) = f'(x) R(x).      Reciprocal is alpha'      ? S =   taylor( 1 / R, x)     %65 = -x^-2 - x^-1 + 1/2 - 2/3*x + 13/12*x^2 - 113/60*x^3 + 1187/360*x^4 - 1754/315*x^5 + 14569/1680*x^6 + 532963/3024*x^7 + 1819157/151200*x^8 - 70379/4725*x^9 + 10093847/129600*x^10 - 222131137/907200*x^11 + 8110731527/12700800*x^12 - 8882574457/5953500*x^13 + 24791394983/7776000*x^14 - 113022877691/18144000*x^15 + O(x^16)       The bad news is that Pari refuses to integrate 1/x,  even when I took out that term it put it all on a common denominator,  so i integrated one term at a time to get  alpha = integral(S)  and i had to type in the terms myself, especially the log(x)      ? alpha = 1 / x - log(x) + x / 2 - x^2 / 3 + 13 * x^3 / 36 - 113 * x^4 / 240 + 1187 * x^5 / 1800 - 877 * x^6 / 945 + 14569 * x^7 / 11760 + 532963 * x^8 / 24192  

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    Very good! If I only could understand french to read Ecalle...2012-10-09
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    Will, I really appreciate your answer! I used your $\alpha$ abel function for x^2+x to generate identical results to what I posted for the half iterate of 800,000,000, by starting with 799,999,999.5 and iterating f(x) until x was very close to zero. It seemed like it required 100,000 iterations of f to get x close enough to the fixed point of zero so the $\alpha^{-1}(\alpha(x)-1/2)$ result had 13 decimal digits of precision, which matched the half iterate of $\approx$ 3898248180100, that I posted earlier.2012-10-09
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    @GottfriedHelms and sheldonison, I give a complete description of the method at http://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765 as a worked example. The cure for needing 100,000 iterations is to have, say, two dozen terms in the expansion for $\alpha$2012-10-09
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    @WillJagy I've wanted to understand the theory behind parabolic iterations for a long time; I bought Milner's book, and Gamelin's book. As you can see, I fudged it well enough to get accurate results, using polynomial interpolation. Thanks! So, that covers $x^2+x$. And the solution for $x^2+x-0.25$ is trivial, since that's analogous to iterations of $x^2$. Do you know of any links for $f(x)=x^2+x+c$, where c>0, which has complex fixed points? Also, the original question concerned half iterates of $x^2$ sometimes being small than half iterates of $x^2+0.25$. Thanks again, - Sheldon2012-10-09
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    @sheldonison, KCG is available as a new paperback http://www.cambridge.org/us/knowledge/isbn/item1120001/?site_locale=en_US for 120 or used (hardback) for about 85 with shipping, http://www.abebooks.com/servlet/SearchResults?an=kuczma&bi=0&bx=off&ds=30&recentlyadded=all&sortby=17&sts=t&tn=iterative+functional+equations&x=75&y=15 I do not know what book by Gamelin you mean.2012-10-09
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    @sheldonison, Meanwhile the history book by D. S. Alexander is very good.2012-10-09
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    Will, I did a procedure to obtain the coefficients automatically in an answer-box.2012-10-10
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    Actually Im only intrested in the first limit of Will's answer. How did you find it Will ?2012-10-12
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    @mick, I give a pretty complete tutorial at http://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765 The Julia equation is Theorem 8.5.1 on page 346 of KCG. Excerpts from that book and many other pdfs at http://zakuski.utsa.edu/~jagy/other.html Meanwhile, iteration of $x + x^2$ does not converge, we need to switch to the inverse function $(-1 + \sqrt{1+4x})/2$ to get $x_n \rightarrow 0.$2012-10-12
  • 0
    But how do you know the radius of the Julia ?2012-10-13
  • 0
    @mick, I will make a guess as to what you are asking. The solution to the Julia equation is $1/\alpha',$ this is defined in an open sector leaving (but not including) the origin, possibly with curved edges. In this case the sector would include the positive real axis, as the basic condition is $x_n \rightarrow 0.$ Otherwise, my best advice is that you read KCG and Alexander and the links I give. If you wish to email me, that is fine as well, click on my name to see how to find my email addresses.2012-10-13
  • 0
    @mick, it has been some two years, so I had to look a bit. existence and uniqueness of $\alpha$ up to an additive constant, on a sector, is the Leau-Fatou Flower Theorem and the Parabolic Linearization theorem, pages 105-107 in Milnor's book, with Julia's proof of the Flower on page 76-78 of Alexander's book, then Fatou's proof. The very specific construction relating the Abel equation to the Julia equation on the sector is due to Ecalle and may not be in any single book. But a good deal of it is at least alluded to in KCG.2012-10-13
  • 0
    I just found that the part of the finding of the Abel-function $ \lambda(f(x)) = \lambda(x) \cdot f'(x)$ can be expressed as an eigenvalue/eigenvactor problem; don't know whether this is somehow interesting here (or possibly obvious/trivial).2012-10-15
  • 0
    @GottfriedHelms, Thanks, I would not be surprised at any number of relationships. I just found the main Ecalle paper and put it at http://zakuski.utsa.edu/~jagy/other.html . I've got to admit, there seems little sign that he wrote the exact limit as I write it, but he did define $\mbox{logit}$ there. So it is a big circle of ideas. However, in the end it resembles the easier use of the Schroder equation when $|f'(z_0)| < 1.$ Neat the way the Julia equation gives us $\lambda(x)$ that is never expressed as more than a formal power series.2012-10-15
  • 0
    Considering this Ecalle limit given for the abel function , is the technique of Ecalle also capable of giving a similar limit for the superfunction ?2012-10-17
  • 0
    @mick, I don't know what Sheldon and you mean by a superfunction. As far as ordinary functions, I feel pretty confident that a real valued and analytic function on $\mathbb R$ has real analytic fractional iterates between fixpoints, but these cannot usually be extended (analytically) across fixpoints. If there are no real fixpoints, I suspect one can get real analytic fractional iterates on the entire line, but I do not currently know to go about that. I put a comment at Prof. Edgar's answer requesting clarification on that point.2012-10-17
  • 0
    The superfunction is a colloquial term for the inverse of the abel function, $\alpha^{-1}(x)$, or $f^{o x}(k)$, where k is a somewhat arbitrary value, bigger than 0, and bigger than the other real fixed point of f, if c<0.25, for $f(x)=x^2+c$.2012-10-18
  • 0
    @WillJagy, if there are complex fixed points, for c>0.25, then $\alpha^{-1}$ could be defined and real valued at the real axis, but will have a branch point for $\alpha(z<0)$. Take $f(x)=x^2+1$ as an example. $f^{-1}=0 f^0=1, f^1=2, f^2=5, f^3=11$. But what about $f^{-1}(0.9999)=0.01i$? This is a complex number, no longer real valued.... So, if $\alpha^{-1}(0)=1$, the most one could require is that it be analytic in the complex plane except for branch singularities at negative integers. There's some other uniqueness ideas we discussed in the tetration forum, that I think Trapmann proved.2012-10-18
  • 0
    Again, using $f(x)=x^2+1$. The two fixed points of f are $0.5+i\sqrt{3/4}$ and $0.5-i\sqrt{3/4}$. I would expect the unique version $\alpha^{-1}(z)$ to have $\lim_{z\to\infty}\alpha^{-1}(iz)=0.5+i\sqrt{3/4}$, and $\lim_{z\to\infty}\alpha^{-1}(-iz)=0.5-i\sqrt{3/4}$.2012-10-19
  • 0
    @SheldonL, $f(x) = x^2 + c$ for $c > 1/4$ definitely has a half iterate in an open neighborhood of the real line, using the easier Schroder equation, Milnor calls it the Koenigs Linearization Theorem (6.1) and then global Corollary (6.4 in the 1991 preprint). So far I see no adequate reason to believe the half iterate is real-valued along the real line, but perhaps there is some trick. There is one such half iterate for each fixed point, if the two versions agree on the real line that would do it.2012-10-19
  • 0
    @WillJagy Of course, the two versions from the two fixed points don't agree, but they are complex conjugates of each other. The half iterate of 1, for $f=x^2+1$ is $f^{0.5}(1)\approx 1.40634858 - 0.322908310i$. From the other fixed point $0.5-i\sqrt{3/4}$, the half iterate would be the conjugate. There is also the half iterate from $f_\infty^{0.5}(1)\approx1.41228535$, and there is the half iterate from _both_ fixed points, which would be $1.41228567$. Surprsingly, all four of these are analytic.2012-10-19
  • 0
    @SheldonL, so how do you do the version from both fixed points, how do you know it is real along the real line and analytic, and so on?2012-10-19
  • 0
    @ Sheldon : I repeat Will's question : how do you use both fixpoints ?? And also you talk about 2 fixpoints and both , thats 3 functions. Where does the 4th come from ?? Sounds intresting , but clear as mud.2012-10-19
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    @SheldonL, meanwhile, I need to go return some books, could you email me? Just click on my name for some instructions, or search with my last name at http://www.ams.org/cml/2012-10-19
  • 0
    about the half iterate from both fixed points, unfortunately, all I could do is point you to some posts in the tetration forum, and the convergence of the numerical algorithm I came up with is not proved. It might be equivalent to perturbed fatou coordinates, but I'm not able to understand the papers on that subject. What I can do is generate the numerical taylor series for the $\alpha^{-1}$ from both fixed points, to arbitrary precision. Centered at any point on the real axis. centered at $\alpha^{-1}(0)=0$, the Taylor series has a radius of convergence of 1, due to the branch singularity.2012-10-19
  • 0
    @SheldonL, thanks.2012-10-19
  • 0
    @WillJagy, Here's a short high level description of the solution from both fixed points. Time is limited today and tomorrow. The algorithm is based on Kneser's solution which requires a Riemann mapping for a function with a singularity, and it is difficult to numerically calculate. The algorithm I came up with, assuming it converges, can be shown that it would be equal to Kneser's solution, which I think I proved. I'll write a description tailored to x^2+c. The limiting behavior of $\alpha^{-1}$ as $\Re(z) \to -\infty$ and $\Im(z)>0$ converges to one fixed point, and if $\Im(z)<0$, the other.2012-10-20
  • 0
    I repeat my question : Considering this Ecalle limit given for the abel function , is the technique of Ecalle also capable of giving a similar limit for the superfunction ?2013-03-09
  • 0
    @mick, no, that is for a neighborhood of a fixed point, such that the derivative of the function at the fixed point is exactly $1.$ If the absolute value of the derivative there is not $1,$ we use the easier Schroder equation, on either the function or its inverse. With no fixed points in your set (here the real line) the only technique I know of is by Helmuth Kneser, father of Martin. He did a half iterate for $e^z$ on an open neighborhood of the real line. No real fixed points, you see.2013-03-09
  • 0
    But what prevents the existance of a similar limit for a superfunction ? Ok maybe we need a completely different method. Why cant there be a similar way to find a similar looking limit for a fixpoint with derivative > 1 ?2013-03-09
  • 0
    @mick, derivative greater than one is fine, you use Schroder (spelling?) but using the inverse function. As to your other questions, i have no idea what you are talking about.2013-03-09
  • 0
    Look , I just want to avoid using the inverse function , and have a limit that avoids taking the inverse similar to the one given here. And Schroëder does not count.2013-03-10
3

a plug

For some material on fractional iterates of $x^2+c$ see the last section of...
"Fractional Iteration of Series and Transseries" by G. A. Edgar ... LINK
To appear in Trans. Amer. Math. Soc.

  • 0
    section 6 "Julia Example," pdf pages 23-26. Do you know the OP's two main questions, (A) does $x^2 + c$ have a real analytic half iterate for $c > 1/4,$ and (B) if $c_1 < c_2,$ assuming (A), does the half iterate with the larger $c$ value exceed the other, for example as $x \rightarrow + \infty?$2012-10-16
  • 0
    I looked at the two examples on page 23. The fractional iterates of x^2 are the _same_ when generated from $\infty$ and when generated from the fixed point of x=1. Perhaps the fractional iterates of $x^2-2$ are also the _same_ when generated from $\infty$ and when generated from the fixed point of x=2; at least it appears so. The fractional iterates of $x^2+\frac{1}{4}$ are _not_ the same when generated from $\infty$ and when generated from the parabolic fixed point of x=0.5. Also, thanks for the pdf, much to learn .....2012-10-17
  • 0
    Consider $g(x)=x^2-2$, then the following limit, adapted from my post above, should be pretty easy to show. $\alpha^{-1}_\infty(z)=\lim_{n\to\infty}g^{-1 on}(2^{2^{z+n}}) = 2\cosh(2^{z+k})$. This is the same as from page 23 for c=-2. The same limit equation for c=0.25, $g(x)=x^2+\frac{1}{4}$, gives a different $\alpha^{-1}$ inverse abel funtion solution than the parabolic solution. The limit should work for other values of c as well. The limit converges as long as |2^2^(z+n)| is growing arbitrarily large in the complex plane.2012-10-18
  • 0
    Via section 6 page 23: "For other values of c no closed form is known, and it is likely that there is none"2017-08-09
2

Remark: this is not an answer but only a work-out based on Will's Pari/GP protocol

\\ Pari/GP-code \ps 64      \\ define taylor-series-extension sufficiently high f= taylor( (-1 + sqrt(1 + 4 * x))/2  , x  )  \\  should be: x - x^2 + 2*x^3 - 5*x^4 + 14*x^5 - 42*x^6 + ... fp = deriv(f)    \\ should be: 1 - 2*x + 6*x^2 - 20*x^3 + 70*x^4 - 252*x^5 + ...   listf = vectorv(24);  \\ provide the required powers of f beforehand as constants     listf[1]=f;     for(k=2,#listf,listf[k] = listf[k-1]*f ) listx = vectorv(#listf,r,x^r) \\ that list for powers of x is not really needed valpha = vectorv(#listf); \\ shall get the sought coefficents     valpha[1]=0; valpha[2]=-1  \\ known constants at the beginning  {for(j=2,#listf-1,     L = sum(k=2,j,va[k]*listf[k]) + 'a*listf[j+1];     R = sum(k=2,j,va[k]*listx[k]) + 'a*listx[j+1];     Compare = L-fp*R;     coefx = polcoeff(Compare,j+2);print(coefx);     ac=-polcoeff(coefx,0)/polcoeff(coefx,1);     valpha[j+1]=ac;   );}   

Now check this:

valpha \\ display coefficients  /* should be:   [0, -1, 1, -3/2, 8/3, -31/6, 157/15, -649/30, 9427/210, -19423/210,      6576/35, -2627/7, 853627/1155, -2007055/1386, 3682190/1287, -29646689/5148,     212029715/18018, -1077705008/45045, 3291567542/69615, -4216011601/46410,       1728974695307/9699690, -3696738921829/9699690, 12315245049166/14549535,      -8505662174957/5290740]~ */        alpha=Ser(valpha) /* comes out to be:     -x + x^2 - 3/2*x^3 + 8/3*x^4 - 31/6*x^5 + 157/15*x^6 - 649/30*x^7 +         9427/210*x^8 - 19423/210*x^9 + 6576/35*x^10 - 2627/7*x^11 + 853627/1155*x^12    + O(x^13)    */ 

However, I didn't catch it how to proceed now...


Ok, I got it now working. Only I had to do one "magic step", indicated by (**) in the comment; ( I missed one link from that coefficients by Will's above procedure to arrive at R and S).
Now as it is working, it is really miraculous... ;-)

\\ I found heuristically examining your document, that it must be result = intformal( 1/( x*alpha ) + 1/x ) \\ (**)          \\ the +1/x in the expressions allows "formal integration" for Pari/GP  coeffs_abel=Vec(result)    \\ put the result into a coefficientsvector #coeffs_abel \\ = 63 in my example \\ getting :  [1, 0, 1/2, -1/3, 13/36, -113/240] for x^-1,x^0,x^1,...      \\ your example-function f(x) myf(x,h=0)=for(k=1,h,x=(-1+sqrt(1+4*x))/2);x   \\ then the Abel-function alpha(x) as given in the beginning of your example {fAbel(x,n=0)=local(xn);  xn = myf(x,n);  \\ here n -> infty, but n~20 suffices    sum(k=-1,#coeffs_abel-2,coeffs_abel[2+k]*xn^k) - log(xn) - n } 

Now test the functions:

\\ testing:  maxn=20  \\ try some sufficient n (=maxn) for the Abel-function x0  = 0.125 x12 = myf(x0,12)     \\ x12=0.0521939337419 is 12 iterations from x0  a0=fAbel(x0  , maxn)     \\    =10.1373406515 a1=fAbel(x12 , maxn)     \\    =22.1373406515 a1-a0       \\ comes out to be =12.0000000000  \\ how to find the 0.5-iterate from x0=0.25 (with a0=Abel(x0)) x_05=solve(x=0.01,x0-0.001, (fAbel(x,maxn)-a0) -1/2) \\ comes out to be 0.118366472264 \\check a0 - fAbel(x_05,maxn)  \\ comes out to be -0.5 (a0 - fAbel(x_05,maxn)) - (-1/2)       \\ < 5e-201 using internal float precision of 200 digits 


@Will: Could you make the missing step visible in your protocol; my move in the integral-expression using $x*alpha$ was simply a heuristic.



Data of the experiment:

x_0 - the initial value
x_1 - the correct value by one integer iteration using the original formula
abel_x_05 - "half-iterate" using the Abel-mechanism
abel_x_10 - "unit-iterate" by applying "half-iteration" to the abel_x_05
should equal the original x_1
h - the "height" of iteration = 0.5, thus: "half-iterate"
a0 - the Abel-function-value of x_0
a05 - the Abel function-value of the half-iterate x_05
a05-a0-1/2 - the difference between the abel-values should be 1/2. This is the error
x_1-abel_x_10 - if the difference is zero, then the Abel-function is exact. This is the error

The table:

    x_0                  x_1             abel_x_05        abel_x_1       h     a_0             a_05            a05-a0-1/2            x_1-abel_x_1   0.0100000000000  0.00990195135928  0.00995073533545  0.00990195135928  1/2  104.610137209  105.110137209   1.11696228987E-201  -2.85779229102E-97   0.0200000000000   0.0196152422707   0.0198057704819   0.0196152422707  1/2  53.9218924877  54.4218924877   3.97098709435E-202  -6.15809353856E-82   0.0300000000000   0.0291502622129   0.0295691127718   0.0291502622129  1/2  36.8546006147  37.3546006147   4.97268862342E-202  -4.06098551075E-74   0.0400000000000   0.0385164807135   0.0392444803983   0.0385164807135  1/2  28.2383644612  28.7383644612  -3.54446782891E-200  -3.59148072904E-69   0.0500000000000   0.0477225575052   0.0488353314257   0.0477225575052  1/2  23.0199413289  23.5199413289  -1.92438083583E-202  -1.07323790193E-65   0.0600000000000   0.0567764362830   0.0583448891277   0.0567764362830  1/2  19.5089497541  20.0089497541   3.82913315022E-200  -4.30261434261E-63   0.0700000000000   0.0656854249492   0.0677761642099   0.0656854249492  1/2  16.9784545543  17.4784545543   2.30176349353E-200  -4.68144861850E-61   0.0800000000000   0.0744562646538   0.0771319743721   0.0744562646538  1/2  15.0637628558  15.5637628558    -1.959630265E-200  -2.06820942631E-59   0.0900000000000   0.0830951894845   0.0864149615923   0.0830951894845  1/2  13.5615925326  14.0615925326              0.E-202  -4.75931307811E-58    0.100000000000   0.0916079783100   0.0956276074506   0.0916079783100  1/2  12.3495715644  12.8495715644     2.612840354E-200  -6.71587352419E-57    0.110000000000    0.100000000000    0.104772246757    0.100000000000  1/2  11.3495715644  11.8495715644              0.E-202  -6.49893010190E-56    0.120000000000    0.108276253030    0.113851079713    0.108276253030  1/2  10.5093372632  11.0093372632    -1.469722699E-200  -4.66951632156E-55    0.130000000000    0.116441400297    0.122866182786    0.116441400297  1/2  9.79257475074  10.2925747507     -6.53210088E-201  -2.64025433320E-54    0.140000000000    0.124499799840    0.131819518477    0.124499799840  1/2  9.17327627451  9.67327627451    -1.143117654E-200  -1.22717201784E-53    0.150000000000    0.132455532034    0.140712944100    0.132455532034  1/2  8.63230833801  9.13230833801     3.266050442E-201  -4.84797799860E-53    0.160000000000    0.140312423743    0.149548219701    0.140312423743  1/2  8.15527503721  8.65527503721      9.79815132E-201  -1.67081681025E-52    0.170000000000    0.148074069841    0.158327015221    0.148074069841  1/2  7.73113278533  8.23113278533      8.16512610E-201  -5.12835185230E-52    0.180000000000    0.155743852430    0.167050916985    0.155743852430  1/2  7.35126498055  7.85126498055      9.79815132E-201  -1.42532084917E-51    0.190000000000    0.163324958071    0.175721433593    0.163324958071  1/2  7.00884764373  7.50884764373    -1.633025221E-201  -3.63574721484E-51    0.200000000000    0.170820393250    0.184340001282    0.170820393250  1/2  6.69840449769  7.19840449769     1.469722699E-200  -8.60676914865E-51    0.210000000000    0.178232998313    0.192907988820    0.178232998313  1/2  6.41548854806  6.91548854806    -1.633025221E-201  -1.90833380748E-50    0.220000000000    0.185565460040    0.201426701971    0.185565460040  1/2  6.15645005622  6.65645005622      9.79815132E-201  -3.99383229632E-50    0.230000000000    0.192820323028    0.209897387587    0.192820323028  1/2  5.91826470908  6.41826470908     1.633025221E-201  -7.94107754605E-50    0.240000000000    0.200000000000    0.218321237354    0.200000000000  1/2  5.69840449769  6.19840449769     4.899075662E-201  -1.50846028308E-49    0.250000000000    0.207106781187    0.226699391244    0.207106781187  1/2  5.49473939600  5.99473939600      6.53210088E-201  -2.75054364650E-49    0.260000000000    0.214142842854    0.235032940678    0.214142842854  1/2  5.30546158398  5.80546158398    -1.143117654E-200  -4.83408189236E-49    0.270000000000    0.221110255093    0.243322931449    0.221110255093  1/2  5.12902639712  5.62902639712    -4.899075662E-201  -8.21796258865E-49    0.280000000000    0.228010988928    0.251570366421    0.228010988928  1/2  4.96410584104  5.46410584104     2.612840354E-200  -1.35554771981E-48    0.290000000000    0.234846922835    0.259776208015    0.234846922835  1/2  4.80955165341  5.30955165341     1.633025221E-201  -2.17542886532E-48    0.300000000000    0.241619848710    0.267941380520    0.241619848710  1/2  4.66436569742  5.16436569742    -1.143117654E-200  -3.40480227973E-48    0.310000000000    0.248331477355    0.276066772226    0.248331477355  1/2  4.52767604024  5.02767604024    -1.143117654E-200  -5.20802211274E-48    0.320000000000    0.254983443527    0.284153237414    0.254983443527  1/2  4.39871747998  4.89871747998      6.53210088E-201  -7.80012141557E-48    0.330000000000    0.261577310586    0.292201598193    0.261577310586  1/2  4.27681558319  4.77681558319     -9.79815132E-201  -1.14578285473E-47    0.340000000000    0.268114574787    0.300212646221    0.268114574787  1/2  4.16137351452  4.66137351452    -1.143117654E-200  -1.65319303514E-47    0.350000000000    0.274596669241    0.308187144298    0.274596669241  1/2  4.05186110361  4.55186110361     2.449537831E-200  -2.34609807510E-47    0.360000000000    0.281024967591    0.316125827860    0.281024967591  1/2  3.94780571723  4.44780571723     3.266050442E-201  -3.27863351540E-47    0.370000000000    0.287400787401    0.324029406368    0.287400787401  1/2  3.84878459717  4.34878459717              0.E-202  -4.51684740363E-47    0.380000000000    0.293725393319    0.331898564609    0.293725393319  1/2  3.75441839607  4.25441839607     1.633025221E-201  -6.14045635954E-47    0.390000000000    0.300000000000    0.339733963915    0.300000000000  1/2  3.66436569742  4.16436569742    -1.633025221E-201  -8.24471876728E-47    0.400000000000    0.306225774830    0.347536243297    0.306225774830  1/2  3.57831834906  4.07831834906      6.53210088E-201  -1.09424173357E-46    0.410000000000    0.312403840464    0.355306020520    0.312403840464  1/2  3.49599747214  3.99599747214     -9.79815132E-201  -1.43659422864E-46    0.420000000000    0.318535277187    0.363043893101    0.318535277187  1/2  3.41715003390  3.91715003390     -6.53210088E-201  -1.86694656416E-46    0.430000000000    0.324621125124    0.370750439252    0.324621125124  1/2  3.34154589332  3.84154589332     4.899075662E-201  -2.40311964896E-46    0.440000000000    0.330662386292    0.378426218767    0.330662386292  1/2  3.26897524487  3.76897524487    -3.266050442E-201  -3.06557066840E-46    0.450000000000    0.336660026534    0.386071773851    0.336660026534  1/2  3.19924639910  3.69924639910     1.633025221E-201  -3.87763161862E-46    0.460000000000    0.342614977318    0.393687629910    0.342614977318  1/2  3.13218384914  3.63218384914     1.633025221E-201  -4.86575271568E-46    0.470000000000    0.348528137424    0.401274296286    0.348528137424  1/2  3.06762658100  3.56762658100     -9.79815132E-201  -6.05974959408E-46    0.480000000000    0.354400374532    0.408832266957    0.354400374532  1/2  3.00542659239  3.50542659239     1.143117654E-200  -7.49305322423E-46    0.490000000000    0.360232526704    0.416362021194    0.360232526704  1/2  2.94544759052  3.44544759052     -9.79815132E-201  -9.20296150448E-46    0.500000000000    0.366025403784    0.423864024184    0.366025403784  1/2  2.88756384413  3.38756384413              0.E-202  -1.12308915176E-45    0.510000000000    0.371779788708    0.431338727620    0.371779788708  1/2  2.83165916874  3.33165916874     -8.16512610E-201  -1.36226314832E-45    0.520000000000    0.377496438739    0.438786570254    0.377496438739  1/2  2.77762602736  3.27762602736    -1.143117654E-200  -1.64285914844E-45    0.530000000000    0.383176086633    0.446207978426    0.383176086633  1/2  2.72536473159  3.22536473159     -8.16512610E-201  -1.97040520998E-45    0.540000000000    0.388819441732    0.453603366565    0.388819441732  1/2  2.67478273021  3.17478273021              0.E-202  -2.35094101264E-45    0.550000000000    0.394427191000    0.460973137658    0.394427191000  1/2  2.62579397425  3.12579397425     3.266050442E-201  -2.79104206351E-45    0.560000000000    0.400000000000    0.468317683702    0.400000000000  1/2  2.57831834906  3.07831834906    -1.633025221E-201  -3.29784346620E-45    0.570000000000    0.405538513814    0.475637386133    0.405538513814  1/2  2.53228116531  3.03228116531     -6.53210088E-201  -3.87906318943E-45    0.580000000000    0.411043357914    0.482932616224    0.411043357914  1/2  2.48761270178  2.98761270178    -1.633025221E-201  -4.54302477715E-45    0.590000000000    0.416515138991    0.490203735478    0.416515138991  1/2  2.44424779394  2.94424779394     3.266050442E-201  -5.29867944782E-45    0.600000000000    0.421954445729    0.497451095989    0.421954445729  1/2  2.40212546307  2.90212546307      6.53210088E-201  -6.15562753640E-45    0.610000000000    0.427361849550    0.504675040790    0.427361849550  1/2  2.36118858117  2.86118858117      6.53210088E-201  -7.12413923790E-45    0.620000000000    0.432737905309    0.511875904189    0.432737905309  1/2  2.32138356786  2.82138356786    -1.143117654E-200  -8.21517461673E-45    0.630000000000    0.438083151965    0.519054012082    0.438083151965  1/2  2.28266011564  2.78266011564      6.53210088E-201  -9.44040285135E-45    0.640000000000    0.443398113206    0.526209682255    0.443398113206  1/2  2.24497094044  2.74497094044    -4.899075662E-201  -1.08122206882E-44    0.650000000000    0.448683298051    0.533343224672    0.448683298051  1/2  2.20827155486  2.70827155486    -1.633025221E-201  -1.23437700836E-44    0.660000000000    0.453939201417    0.540454941749    0.453939201417  1/2  2.17252006161  2.67252006161    -1.633025221E-201  -1.40489550174E-44    0.670000000000    0.459166304663    0.547545128614    0.459166304663  1/2  2.13767696515  2.63767696515      8.16512610E-201  -1.59424574642E-44    0.680000000000    0.464365076099    0.554614073360    0.464365076099  1/2  2.10370499971  2.60370499971      9.79815132E-201  -1.80397525144E-44    0.690000000000    0.469535971483    0.561662057284    0.469535971483  1/2  2.07056897183  2.57056897183    -4.899075662E-201  -2.03571226387E-44    0.700000000000    0.474679434481    0.568689355110    0.474679434481  1/2  2.03823561638  2.53823561638    -1.633025221E-201  -2.29116710935E-44    0.710000000000    0.479795897113    0.575696235217    0.479795897113  1/2  2.00667346430  2.50667346430      9.79815132E-201  -2.57213344685E-44    0.720000000000    0.484885780180    0.582682959838    0.484885780180  1/2  1.97585272133  2.47585272133     -6.53210088E-201  -2.88048943794E-44    0.730000000000    0.489949493661    0.589649785270    0.489949493661  1/2  1.94574515637  2.44574515637    -3.266050442E-201  -3.21819883116E-44    0.740000000000    0.494987437107    0.596596962058    0.494987437107  1/2  1.91632399887  2.41632399887     1.633025221E-201  -3.58731196224E-44    0.750000000000    0.500000000000    0.603524735182    0.500000000000  1/2  1.88756384413  2.38756384413      9.79815132E-201  -3.98996667126E-44    0.760000000000    0.504987562112    0.610433344234    0.504987562112  1/2  1.85944056601  2.35944056601    -1.469722699E-200  -4.42838913794E-44    0.770000000000    0.509950493836    0.617323023586    0.509950493836  1/2  1.83193123628  2.33193123628     3.266050442E-201  -4.90489463626E-44    0.780000000000    0.514889156509    0.624194002553    0.514889156509  1/2  1.80501405007  2.30501405007              0.E-202  -5.42188821009E-44    0.790000000000    0.519803902719    0.631046505547    0.519803902719  1/2  1.77866825684  2.27866825684     3.266050442E-201  -5.98186527137E-44    0.800000000000    0.524695076596    0.637880752227    0.524695076596  1/2  1.75287409642  2.25287409642      8.16512610E-201  -6.58741212246E-44    0.810000000000    0.529563014099    0.644696957644    0.529563014099  1/2  1.72761273971  2.22761273971      8.16512610E-201  -7.24120640468E-44    0.820000000000    0.534408043279    0.651495332378    0.534408043279  1/2  1.70286623365  2.20286623365     1.796327743E-200  -7.94601747474E-44    0.830000000000    0.539230484541    0.658276082669    0.539230484541  1/2  1.67861744997  2.17861744997     3.266050442E-201  -8.70470671114E-44    0.840000000000    0.544030650891    0.665039410547    0.544030650891  1/2  1.65485003771  2.15485003771    -4.899075662E-201  -9.52022775248E-44    0.850000000000    0.548808848170    0.671785513954    0.548808848170  1/2  1.63154837883  2.13154837883     3.266050442E-201  -1.03956266698E-43    0.860000000000    0.553565375285    0.678514586862    0.553565375285  1/2  1.60869754695  2.10869754695      6.53210088E-201  -1.13340420751E-43    0.870000000000    0.558300524426    0.685226819385    0.558300524426  1/2  1.58628326890  2.08628326890              0.E-202  -1.23387051676E-43    0.880000000000    0.563014581273    0.691922397891    0.563014581273  1/2  1.56429188873  2.06429188873     -6.53210088E-201  -1.34129397209E-43    0.890000000000    0.567707825203    0.698601505104    0.567707825203  1/2  1.54271033417  2.04271033417    -1.633025221E-201  -1.45601620124E-43    0.900000000000    0.572380529476    0.705264320212    0.572380529476  1/2  1.52152608528  2.02152608528      9.79815132E-201  -1.57838806969E-43    0.910000000000    0.577032961427    0.711911018956    0.577032961427  1/2  1.50072714504  2.00072714504    -1.633025221E-201  -1.70876966271E-43    0.920000000000    0.581665382639    0.718541773732    0.581665382639  1/2  1.48030201191  1.98030201191     -8.16512610E-201  -1.84753026232E-43    0.930000000000    0.586278049120    0.725156753679    0.586278049120  1/2  1.46023965409  1.96023965409    -1.633025221E-201  -1.99504831930E-43    0.940000000000    0.590871211464    0.731756124764    0.590871211464  1/2  1.44052948530  1.94052948530     -6.53210088E-201  -2.15171142049E-43    0.950000000000    0.595445115010    0.738340049873    0.595445115010  1/2  1.42116134220  1.92116134220      9.79815132E-201  -2.31791625155E-43    0.960000000000    0.600000000000    0.744908688889    0.600000000000  1/2  1.40212546307  1.90212546307    -3.266050442E-201  -2.49406855556E-43    0.970000000000    0.604536101719    0.751462198770    0.604536101719  1/2  1.38341246783  1.88341246783    -1.633025221E-201  -2.68058308730E-43    0.980000000000    0.609053650641    0.758000733628    0.609053650641  1/2  1.36501333924  1.86501333924    -1.143117654E-200  -2.87788356377E-43    0.990000000000    0.613552872566    0.764524444801    0.613552872566  1/2  1.34691940522  1.84691940522      8.16512610E-201  -3.08640261096E-43     1.00000000000    0.618033988750    0.771033480925    0.618033988750  1/2  1.32912232216  1.82912232216     -8.16512610E-201  -3.30658170700E-43 




[update]: Another protocol, as requested by Will Jagy is at my website (to save space here) at go.helms-net.de

  • 0
    I made a small change in the code portion of my answer... Looking at your fAbel, you do have the log() but i do not see the all-important subtraction of $n$ itself. The bootstrapping aspect is that $$ \alpha(x_0) = \alpha(x_n) - n, $$ and with a finite approximation to $\alpha$ we can get a correct limit as $n \rightarrow \infty.$ Do you think you could work up a display for small $x,$ as in the numerical portion at the end of http://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765 I admit it is a good deal of work.2012-10-10
  • 0
    @Will: concerning the *n* - well, it cancels, if I compute the difference between two Abel-values ***with the same n***- so I didn't see an effect. (I only was wondering, when I noticed, that the individual Abel-values increased so much when I increased *n* for precision. ;-) I'll correct the formula for this... thanks for the hint.2012-10-10
  • 0
    Glad this is getting somewhere. Note that $\alpha$ must always be calculated, given some $h(x), h(0) = 0, h'(0) = 1,$ with the choice of either $h(x), h^{-1}(x)$ that is slightly below $x.$ So, for my $\sin x$ answer I used $\sin x,$ but here, for $x + x^2,$ one must use the inverse. Hence the $-1/2$ in the definition of $g(x)$ in my answer above, where $\sin x$ used $+1/2.$ Well, if you can stand it, please produce the final numerical outcome for, say, $0 < x \leq 2$ by intervals of $1/10$ for the half iterate of $x + x^2.$2012-10-10
  • 0
    Alright, I think you have displayed the calculation of $\alpha.$ In order to have $\alpha^{-1}(x)$ available I had the computer solve $\alpha(t) = x$ numerically, by bisection as I recall, in C++. With that I got another C++ function that gives the $g(x)$ in my answer, displayed that, displayed $g(g(x))$ and $x + x^2$ and the error $x + x^2 - g(g(x))$ in parallel columns.2012-10-10
  • 0
    I put a new answer with the C++ program and output for the problem $f(f(x)) = \sin x,$ with $0 < x \leq \frac{\pi}{2}.$ A fair amount of work was involved.2012-10-10
  • 0
    @WillJagy, Gottfried, Thanks both of you for your hard work. I haven't had time to recreate the closed for coefficients for the abel function (fatou coordinate), for x^2+x. Its interesting that the function is centered at the singularity, and has zero radius of convergence, with all derivatives defined, and is useful!. I ordered "Iterative Functional Equations", based on your recommendation. Also, I think the $\theta(z)$ from my original post is analytic, and is analytic, well defined, for $\Im(z)<\pi/(2\log(2))$, and has a fractal boundary. - Sheldon2012-10-10
  • 0
    That is the limit as integer n goes to infinity, $\theta(z+n)$ is well defined, and converges to an analytic 1-cyclic function, I'll post some pictures maybe later. $$\theta(z)= \lim_{n\to\infty}\text{abel}_{x^2+0.25} (2^{2^{z+n}})-z-n$$2012-10-10
  • 0
    @sheldonison, glad you are getting the book. Ecalle's method is in there in full, you just need to dig a bit. I posted the half iterate of $\sin x$ as a separate answer. Note that there is nothing special about $1/2,$ the same program with $1/3$ in the same location would solve $f(f(f(x))) = \sin x.$ Meanwhile, $\alpha$ as i have it should be holomorphic anywhere the limit must exist, meaning anywhere $x_n, \arg x_n \rightarrow 0.$ For $x + x^2,$ this should include all of $\mathbb C$ except $0$ and the negative real axis. Principal square root branch!2012-10-10
  • 0
    @Will: I've prepared a protocol for the half-iterates of the original function f(x)=x^2 + 0.25. See http://go.helms-net.de/math/divers/mse/MSE_20121010.htm2012-10-11
  • 0
    @GottfriedHelms, thanks, I posted the output and C++ program I had in mind as yet another answer. The accuracy is not superb, just about everything needs fine adjustment.2012-10-11
2

Gottfried, here are the output and the C++ program for the half iterate of $\sin x.$ You should be able to copy these and paste to text files, print out for closer scrutiny. If you go through the C+ program you will find a number of choices I had to make, bounds I had to put in. The short version is that a computer does not really do mathematics. Most such bounds would need to change for the $x + x^2$ problem.

=========================

jagy@phobeusjunior:~$  g++ -o abel_sine   abel_sine.cc -lm        jagy@phobeusjunior:~$  jagy@phobeusjunior:~$      jagy@phobeusjunior:~$ ./abel_sine          x               alpha(x)              f(x)                f(f(x))                sin x               f(f(x))- sin x  1.570796326794897   2.089622719673273    1.140179476167262    1.000000000000167    1    1.67e-13 1.562069680534925   2.089797249258235    1.140115090046273    0.9999619230634524    0.9999619230641713    -7.188e-13 1.553343034274953   2.09032097448571    1.139921975900568    0.999847695158399    0.9998476951563913    2.008e-12 1.544616388014982   2.091194304923151    1.139600266203484    0.9996573249780338    0.9996573249755573    2.477e-12 1.53588974175501   2.0924179237329    1.139150181135067    0.9993908270177291    0.9993908270190958    -1.367e-12 1.527163095495039   2.093992788553488    1.138572027671961    0.9990482215816853    0.9990482215818578    -1.725e-13 1.518436449235067   2.095920132741632    1.137866198271987    0.9986295347537874    0.9986295347545739    -7.866e-13 1.509709802975096   2.098201466844743    1.137033169308497    0.9981347984222052    0.998134798421867    3.382e-13 1.500983156715124   2.10083858053253    1.136073499125411    0.9975640502629188    0.9975640502598243    3.095e-12 1.492256510455153   2.103833544989774    1.134987825712907    0.9969173337335647    0.9969173337331281    4.367e-13 1.483529864195181   2.107188715362888    1.133776864276473    0.9961946980874663    0.9961946980917457    -4.279e-12 1.47480321793521   2.110906733837137    1.132441404386233    0.9953961983660398    0.9953961983671789    -1.139e-12 1.466076571675238   2.114990533073489    1.130982306919422    0.9945218953721769    0.9945218953682734    3.903e-12 1.457349925415266   2.119443339917354    1.129400500817922    0.9935718556769257    0.9935718556765877    3.381e-13 1.448623279155295   2.124268679484612    1.127696979720126    0.9925461516392783    0.9925461516413222    -2.044e-12 1.439896632895323   2.129470379858582    1.125872798278496    0.991444861375245    0.9914448613738106    1.434e-12 1.431169986635352   2.135052576998492    1.123929068488904    0.9902680687417381    0.9902680687415705    1.676e-13 1.42244334037538   2.141019720127247    1.121866955896414    0.9890158633592981    0.989015863361917    -2.619e-12 1.413716694115409   2.147376577526611    1.119687675701728    0.9876883405944573    0.987688340595138    -6.807e-13 1.404990047855437   2.154128243013393    1.117392488792027    0.9862856015385387    0.9862856015372317    1.307e-12 1.396263401595466   2.161280142477607    1.114982697899367    0.9848077530109615    0.9848077530122084    -1.247e-12 1.387536755335494   2.168838041301966    1.112459643576912    0.9832549075641427    0.9832549075639549    1.877e-13 1.378810109075522   2.176808052031916    1.109824700383979    0.9816271834461333    0.9816271834476643    -1.531e-12 1.370083462815551   2.185196642699624    1.107079272988684    0.9799247046204426    0.97992470462083    -3.875e-13 1.361356816555579   2.194010645601362    1.104224792442635    0.978147600735532    0.9781476007338061    1.726e-12 1.352630170295608   2.203257266737447    1.101262712496418    0.9762960071208225    0.9762960071199339    8.887e-13 1.343903524035636   2.212944095790644    1.09819450604305    0.9743700647819381    0.9743700647852358    -3.298e-12 1.335176877775665   2.223079116682825    1.095021661692928    0.9723699203987797    0.9723699203976772    1.102e-12 1.326450231515693   2.233670718878459    1.091745680449532    0.970295726276226    0.9702957262759971    2.288e-13 1.317723585255722   2.244727709254562    1.088368072577014    0.9681476403767416    0.9681476403781085    -1.367e-12 1.30899693899575   2.256259324701092    1.084890354600779    0.9659258262894636    0.9659258262890691    3.945e-13 1.300270292735779   2.268275245578629    1.081314046433883    0.96363045320776    0.9636304532086238    -8.638e-13 1.291543646475807   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0.1832595714594099   87.29438125450602    0.1827453868744002    0.1822355254921534    0.1822355254921526    7.663e-16 0.1745329251994382   96.39172246024259    0.1740888583870049    0.1736481776669286    0.1736481776669355    -6.867e-15 0.1658062789394665   106.9695114145022    0.1654256328125449    0.1650476058606701    0.1650476058606828    -1.272e-14 0.1570796326794949   119.3660806949922    0.1567560498155935    0.156434465040227    0.156434465040236    -9.006e-15 0.1483529864195232   134.0221665724577    0.1480804483095538    0.1478094111296127    0.1478094111296158    -3.05e-15 0.1396263401595516   151.5204761717553    0.1393991664991375    0.1391731009600686    0.1391731009600706    -1.952e-15 0.1308996938995799   172.6443090412062    0.1307125419223942    0.1305261922200602    0.1305261922200567    3.51e-15 0.1221730476396082   198.4664891916352    0.1220209114922434    0.121869343405156    0.1218693434051526    3.391e-15 0.1134464013796366   230.4879361506713    0.1133246115375998    0.1132032137679124    0.1132032137679119    5.441e-16 0.104719755119665   270.8602612491742    0.1046239778440629    0.1045284632676639    0.1045284632676586    5.296e-15 0.09599310885969331   322.7560914893005    0.0959193456942896    0.09584575252023451    0.09584575252022914    5.365e-15 0.08726646259972166   391.0107887508433    0.0872110499079478    0.08715574274766681    0.08715574274766334    3.473e-15 0.07853981633975002   483.2891665285358    0.0784994248814307    0.07845909572784977    0.07845909572785012    -3.432e-16 0.06981317007977837   612.3322384664045    0.06978480462721946    0.06975647374413034    0.06975647374413048    -1.372e-16 0.06108652381980673   800.5982996267156    0.06106752281302281    0.06104853953486369    0.06104853953486206    1.641e-15 0.05235987755983508   1090.729452077929    0.0523479128006657    0.05233595624295204    0.05233595624294902    3.018e-15 0.04363323129986343   1571.988867808221    0.04362630768477163    0.04361938736533912    0.04361938736534119    -2.074e-15 0.03490658503989179   2458.078758458438    0.03490304033128337    0.03489949670250996    0.03489949670250617    3.79e-15 0.02617993877992014   4372.703870691502    0.02617844341578261    0.02617694830787946    0.02617694830787835    1.107e-15 0.0174532925199485   9843.561200591173    0.01745284946174855    0.01745240643728656    0.01745240643728871    -2.155e-15          x               alpha(x)              f(x)                f(f(x))                sin x               f(f(x))- sin x  jagy@phobeusjunior:~$      jagy@phobeusjunior:~$  

==========================

#include  #include  #include  #include  #include  #include  #include  #include  #include  #include  #include  using namespace std;     //    lines after double slashes are comments  //   also on a line with a command, anything after // is  commentary  //  on a Unix or Linux computer,  compile using line    //        g++ -o abel_sine   abel_sine.cc -lm   //  then run the program  with  //  ./abel_sine       double abel(double x) {     double eps = 0.000000001;    eps = eps / 100000.0;   double f = x ;   double g = 1.0, g_old = 100.0, diff = 1.0 ;  for( int n = 0; n <= 100000  && diff >= eps ; ++n)  {      g =  3.0 / (f * f)  +  6.0 * log(f) / 5.0 +  79.0 * f * f/ 1050.0 + 29.0 * f * f * f * f /2625.0   - n;      diff = fabs(g - g_old);  //  cout.precision(16);  //  cout << n << "  " << x  << "  "  << f  << "  " << g <<  "   " << diff << endl ;    f = sin ( 1.0 * f);     g_old = g;  }   return g; } // abel   double inverse_abel(double x) {   int count = 0;   double eps = 0.000000001;      eps = eps / 100000.0;   double middle, left, right;   if( x < 2.089607) return 0.0;   else   {      left = 0.001;     right = 2.0 * atan(1.0) ;     middle = ( left + right) / 2.0;      double left_val = abel(left) , right_val = abel(right), middle_val = abel(middle);     while ( right - left > eps)     {       if (middle_val < x )       {         right = middle;         middle = ( left + right) / 2.0;          right_val = abel(right);         middle_val = abel(middle);       }       else       {         left = middle;         middle = ( left + right) / 2.0;          left_val = abel(left);         middle_val = abel(middle);       }       count++;    //   cout << count;     //  cout.precision(16);     //  cout << "   " << x << "  " << middle << endl;      } // while not accurate   }  // else in range    return  middle; } // inverse_abel  double half_iterate(double x) {   return inverse_abel( 1/2.0 + abel(x)  ); }    int main() {    double my_pi = 4.0 * atan(1.0) ;    double phlegm = 5.0;   // cout << my_pi / 2.0 << "   " <<  abel( my_pi / 2.0) << endl;  //  cout << my_pi / 2.0 << "   " <<   half_iterate( my_pi / 2.0) << endl;      cout <<  "         x               alpha(x)              f(x)                f(f(x))                sin x               f(f(x))- sin x " << endl;      for( double x =  my_pi / 2; x >= 0.01 ; x -= my_pi / 360.0)   { //     cerr << x << endl;  cout.precision(16);     cout << x << "   " <<  abel( x) << "    "  << half_iterate( x) << "    "  << half_iterate(half_iterate( x)) << "    " <<  sin(1.0 * x)   ;   cout.precision(4);  cout << "    "  << half_iterate(half_iterate( x)) -  sin(1.0 * x)   << endl; // cout <<  inverse_abel(abel(x))  -  x   << endl;   }  cout <<  "         x               alpha(x)              f(x)                f(f(x))                sin x               f(f(x))- sin x " << endl;     return 0 ; }    //  end of main     //        g++ -o abel_sine   abel_sine.cc -lm     //   x                 alpha(x)              f(x)                 f(f(x))            f(f(x))- sin x //1.570796326794897   2.089622719673273    1.140179476167262    1.000000000000167    1.67e-13  //1.562069680534925   2.089797249258235    1.140115090046273    0.9999619230634524    -7.188e-13  //1.553343034274953   2.090320974485711    1.139921975900568    0.999847695158399    2.008e-12  //1.544616388014982   2.091194304923151    1.139600266203484    0.9996573249780338    2.477e-12  //1.53588974175501   2.0924179237329    1.139150181135067    0.9993908270177291    -1.367e-12  //1.527163095495039   2.093992788553489    1.138572027671961    0.9990482215816853    -1.725e-13  //1.518436449235067   2.095920132741632    1.137866198271987    0.9986295347537874    -7.866e-13  //1.509709802975096   2.098201466844743    1.137033169308497    0.9981347984222052    3.382e-13  //1.500983156715124   2.10083858053253    1.136073499125411    0.9975640502629188    3.095e-12  //1.492256510455153   2.103833544989774    1.134987825712907    0.9969173337335647    4.367e-13 

==========================

2

Arguably an off-subject remark:

If only you relented to allow c < 0, there is the celebrated ("chaotic " logistic map) closed form example (p302) of Ernst Schroeder himself (1870); namely, for
$$ h(x)= x^2-2, $$ it follows directly that for $$ y=\frac{x\pm \sqrt{x^2-4}}{2} $$ that is $$ x=y+y^{-1}, $$ one has $$ h(x)=y^2+y^{-2}\equiv h_1(x). $$ Whence, subscripting the iteration index, $$ h_n(x)= y^{2^n}+ y^{-2^n}. $$

This, then, specifies the whole iteration group: so your functional square root is just $$ h_{\sqrt2} (x)=y^{\sqrt 2} +y^{-\sqrt 2}. $$

Pardon if the point has been made, explicitly, or implicitly, in the outstanding answers above. If not, it might well offer guidance or continuation ideas.

More formally, in E.S.'s language of conjugacy, $\psi(x)=\frac{x\pm \sqrt{x^2-4}}{2}$, $~f(y)=y^2$, $~f_n(y)=y^{2^n}$; so that $h(x)= \psi^{-1} \circ f \circ \psi (x)$, and $$h_n= \psi^{-1} \circ f_n \circ \psi ~.$$

A conjugacy iteration approximation method is available in our 2011 paper: Approximate solutions of Functional equations. Apologies if this late lark answer is only proffering coals to Newcastle, but, in my experience, this is the canonical gambit of chaos discussions--naturally, domains and ranges are chosen suitably for the answer to make sense.

  • 2
    very nice. I didn't know about that closed form solution. I'll have to read your paper; which will take some time.2017-06-01
1

Remark: Shel, possibly I misunderstood something in your post and this pictures here may be completely crap. I expected diff/theta-function-curve crossing the x-axis, but see only the wobbling around a certain y-value. So if this is all wrong, please let me know and I'll improve or delete this post


An image for the theta-function in your (Sheldon's) original post. I understand the z-parameter in the theta-function as "height"-parameter, when some number $x_0$ is iterated $h$ - (or $z$ -) times to the number $x_h$ .
here is how I implemented the diff-function:

{shtheta(h,x0=1)= local(a,xh,h1,l2=log(2));   xh = iterateByAbelfunction(x0,h);   h1 = log(log(xh)/l2)/l2;  \\ h1 should give the height-difference in terms of           \\ the other function $x^2$   return(h1-h);} 

Your example of wobbling was at $x_0=800000000$ - here I begin at $x_0=60$ and show the iterates in steps of 1/10 up to $x_6$ which crosses your 800000000 at height of about $2.3239$ . This is the blue curve in the first plot. The magenta curve is the equivalent, but begins at $x_0=70$ and it should be a left-shift of the blue curve by some small $h$ (just to improve the visualization of the problem):

picture1

The next picture is the detail of bigger "heights" (from $x_1 \approx 3600 $ on) and the magenta-curve shifted to match at the last point at $h=6$ to make the fine sinusoidal form visible.

picture2


[Added]: Hmm, I think now I understand the question and what's going on better now after some more consideration. And I leave the pictures so far, because they are still informative even if not directly to the point.

My hypothese for now: the "wobbling" which leads to the change of sign in your theta-function is caused by differences or better by a different behave of the functions when derivatives with respect to the height-parameter are considered. Without exact inspection I assume, that the derivatives of all orders of the $x^2$-function with respect to the iteration-height-parameter are always positive but that of the Abel-iteration may be mixed so that the change of the function-value is not "completely smooth".

I hope I could made this comprehensible so far, perhaps I can do better later ...


[added2]: I took a closer look at your theta-function and searched for change-signs earlier than your $x_0 = 8e8 $. I found some, for instance $x_0 = 2000 $ and the first 20 iterates in steps of 1/10. Then I scanned 16 areas beginning at $x_0 = 10^{k/2} $ and iterating from $ x_0 $ 20 times by height of 1/10. Each of the latter trajectories make a line in the following plot, also the lines are normalized such that their amplitude is between $ \pm 1 $. Only that lines are drawn which contain at least one sign-change.

sign-changes


  • 0
    Hey Gottfried; your pictures match the results I've generated. The theta function converges to a 1-cyclic function, and not to a constant. That is why the half iterates of one function are sometimes bigger and sometimes smaller than the half iterates of the other functions, as the two superfunctions $\alpha_{x^2}^{-1}$, and $\alpha_{x^2+0.25}^{-1}$ criss cross each other an infinite number of times, if lined up at larger real values. Thanks for all the work again.2012-10-11
  • 0
    @Shel: have you data/analytics of the derivatives with respect to the height parameter? I've tried to do this numerically but don't arrive at high derivatives due to the binary-search-implementation of the inverse Abel-function. I got negative derivatives when order gt 7 but am very unsure about the reliability of that toy-computations.2012-10-11
  • 0
    yes, I can generate accurate numerical values for the 1-cyclic function that $\theta(z)$ converges to, as well as for any of its derivatives, although I switched to generating $\theta_2(z)= \alpha_{x^2+0.25}(\alpha^{-1}_{x^2}) = \alpha_{x^2+0.25}(2^{2^z})$ I can generate $\theta_2(z)$ in the complex plane as well, as long as $\Im(z)<\pi/(2\log(2))$, which I believe is its analytic boundary.2012-10-11
1

** ADDITIONAL UPDATES, answer **

Again, thanks, Gottfried, and Will, for your updates and answers, and for the important theoretical background proving the existence of the parabolic solution. For this section of updates, I use $\alpha(z)$ as the abel function of $x^2+0.25$, so that
$\alpha(z)=\alpha(z^2+0.25)-1$
$\theta(z)=\lim_{(n \to \infty)} {\alpha(s(z+n))-z-n}$
$s(z)=2^{2^z}$, where 2^2^z is the superfunction for $x^2$, $s(z)=s^2(z-1)$

The reason for the switch, is that 2^2^z is well defined in the complex plane, making it easier to identify the analytic boundary of $\theta(z)$ in the complex plane. Earlier, I was generating a slightly different $\theta(z)$ from the composition of the abel function of $z^2$ with the superfunction of $z^2+0.25$. The key is that 2^2^z is periodic in the complex plane with period=$2\pi i/\log(2)$. In addition, as z increases, the absolute value of 2^2^z grows without bounds in the neighborhood of the real axis if $|\Im(z)|<0.5\pi/\log(2)$. The $\theta(z)$ function only converges to a 1-cyclic function if 2^2^z is growing in magnitude. To understand this, consider the function $f(z)=\sqrt{z^2-0.25}-z$. If the magnitude of z is large enough, than f(z) is an arbitrarily small function.

To help understand the definition of $\theta(z)$, consider one other function as an "alternative" abel/superfunction function of x^2+0.25. Define $g(z)=\sqrt{x-0.25}$, $g^{-1}(z)=z^2+0.25$, and consider the following "alternative" abel function for x^2+0.25.
$\alpha_{alt}(z)=\lim_{(n \to \infty)} \log_2(\log_2(g^{-1 o n}(z)))-n$
$\alpha_{alt}^{-1}(z)=\lim_{(n \to \infty)} {g^{o n}(2^{2^{z+n}})}$
This $\alpha_{alt}^{-1}(z)$ alternative inverse abel for (x^2+0.25) is not as well behaved as 2^2^z in the complex plane, but it is defined if $\Im(z)<\pi/2\log(2)$. In addition, this alternative function corresponds to generating $\alpha(z)$ from the super attracting fixed point at infinity, instead of the fixed point of 0.5. Because is is generated from the fixed point at infinity, half iterates for real numbers>1 generated with this alternative abel function, are always bigger than the half iterates of $x^2$!
$\alpha_{alt}^{-1}(\alpha_{alt}(x)+0.5)>x^{\sqrt{2}}$, for real(x)>1

Also, $\theta(z)=\alpha(\alpha_{alt}^{-1}(z))$, which is easy to show. Hopefully, this is not too confusing, as my time this morning is limited, and I want to post some plots of $\theta(z)$, at the real axis, and in the complex plane.

$\theta$ at the real axis. Here, I arbitrarily set $\theta(n)=0$, for large enough integers. Note that 2^2^9, is a really big number, 10^154, so theta(z) has converged. theta real axis

$\theta$ at $\Im(z)=1$, note the magnitude of $\theta(z)$ is much larger here.
theta imag i=1

And, here is the analytic limit of theta, $\Im(z)=0.5\pi/\log(2)$, showing the fractal behavior since 2^2^z is no longer increasing, but instead, |2^2^z|=1. theta imag i=1

Finally, here is a plot of the ratio of the ratio of the two superfunctions; the inverse abel function for $z^2+0.25$, and 2^2^z, lined up to approximately 50% duty cycle as z increases.
superfunction(z)/2^2^z

Given that $\theta(z)$ is defined in the complex plane, as opposed to just at the real axis, it is fairly straightforward to generate the derivatives of $\theta(z)$. Results are posted below. A Fourier series is also an appropriate representation, and I also generated coefficients for that representation of $\theta(z)$. - Sheldon

Taylor series coefficients for $\theta(x)$, centered at integer values for large enough x. The results were calculated to around 50 decimal digits accuracy, with 32 decimal digits printed.

      a0=   0.0, my method can't calculate a unique value       a1=   0.00000028810398845902074305989277221548       a2=   0.00000089435733793739252528458588523408       a3=  -0.0000018956451499697646411943344197949       a4=  -0.0000029423289610212918024529854670052       a5=   0.0000037418289741301058019029496736133       a6=   0.0000038720089580678152184208095047739       a7=  -0.0000035170858822412227427467114593047       a8=  -0.0000027298239627872774249635651034448       a9=   0.0000019282555213557966088957347187964       a10=  0.0000011977279049053074677056810297193       a11= -0.00000069174982884319540796335734464650       a12= -0.00000035856584190142105144853204971408       a13=  0.00000017476443978345605933412052585987       a14=  0.000000078082360773990075295730752702844       a15= -0.000000032632929388465892972414066625782       a16= -0.000000013044250611865030061306500138718       a17=  0.0000000046077520459399449519271392494947       a18=  0.0000000017862361640653315104084905464566       a19= -0.00000000047244000496147095322980527968311       a20= -2.2853200883620998276187332225037 E-10       a21=  2.2006495266220113739934318292157 E-11       a22=  3.4223321847536372255896822143837 E-11       a23=  5.3079836937900696515727605371273 E-12       a24= -6.6588645528547823638468825694018 E-12       a25= -2.2182507837443852330724254905386 E-12       a26=  1.4315559424375709748743291952982 E-12       a27=  5.5535195868969985670819915243637 E-13       a28= -2.9118965305410181912463403433221 E-13       a29= -1.1519357480572864323459930397988 E-13       a30=  5.3266530517343176431264459132840 E-14       a31=  2.1564860886672909324064036274316 E-14       a32= -8.7692103463831850128049893554813 E-15       a33= -3.8711319443724945736122721505205 E-15       a34=  1.3300279324610041843065021692979 E-15       a35=  7.0546171262916079313373944244875 E-16       a36= -1.9335235895886778013423987940850 E-16       a37= -1.3461650784510169419141972232730 E-16       a38=  2.8225495528332258646098244330668 E-17       a39=  2.6504795309806104325089128767594 E-17 
  • 0
    Hi Shel- thanks for the protocol. Well, what I was trying was to compute the derivatives of the function $f(x,h)$ ( where $f(x,1)=x^2+0.25$, and *h* is the iteration height ) with respect to varying *h* (I always understand this is meant as being the *superfunction* for $x^2 + 0.25 $ when *x* is some default value ), with respect to the iteration-height *h* . I could only go to an estimate up to the 5'th or 6'th derivative (numerically) because of the binary-search in the inverse abel-function, which seems to be not very efficient in the obvious implementation in Üari/GP ("solve()").2012-10-13
  • 0
    @GottfriedHelms Since that inverse abel function is difficult, you might take the abel function of 2^2^z. theta(z)=abel(2^2^z)-z is what I used for this most recent post. We are interested in the limiting behavior as theta converges to a 1-cyclic periodic function. I calculated the derivatives with the cauchy integral sampling, which works well for analytic functions, as long as the nearest singularity is not too close, else more and more samples required. 2.266i is the singularity; I sampled at the real axis with radius=1. A little tricky to calculate abel(2^2^z) in the complex plane.2012-10-13
0

The $x + x^2$ problem with some output and the C++ code

============================

jagy@phobeusjunior:~$      jagy@phobeusjunior:~$ date Wed Oct 10 19:41:20 PDT 2012 jagy@phobeusjunior:~$      jagy@phobeusjunior:~$ g++ -o abel_any_function   abel_any_function.cc -lm  jagy@phobeusjunior:~$      jagy@phobeusjunior:~$ ./abel_any_function          x               alpha(x)              f(x)                f(f(x))               x + x^2         f(f(x))- (x+x^2)  4                     -0.3590448941269863      7.95040053721441     19.99999999998791    20                    -1.209e-11 3.9                   -0.3377026486408653      7.688520999700604    19.10999999997955    19.11                 -2.045e-11 3.8                   -0.3155523104802599      7.42911723768848     18.24000000000991    18.24                  9.912e-12 3.7                   -0.2925389073553167      7.172223398023817    17.39000000111184    17.39                  1.112e-09 3.6                   -0.2686020190344757      6.917874938470664    16.55999999997078    16.56                 -2.922e-11 3.5                   -0.2436750604472868      6.666108711998744    15.75000000000014    15.75                  1.386e-13 3.399999999999999     -0.2176844459454219      6.416963058840906    14.96000000005435    14.96                  5.435e-11 3.299999999999999     -0.1905486107379348      6.170477907091147    14.19000000001195    14.19                  1.195e-11 3.199999999999999     -0.1621768597700163      5.926694883235451    13.44000000000269    13.43999999999999      2.696e-12 3.099999999999999     -0.1324680068816612      5.685657433446336    12.7099999999996     12.70999999999999     -3.948e-13 2.999999999999999     -0.1013087576522246      5.447410957290053    12.00000000001379    11.99999999999999       1.38e-11 2.899999999999999     -0.0685717768874564      5.212002955613553    11.3100000000063     11.30999999999999      6.304e-12 2.799999999999999     -0.03411336553416398     4.979483148557       10.64000000404555    10.63999999999999      4.046e-09 2.699999999999999      0.002229349696170169    4.749903886596367     9.989999999999235    9.989999999999993     -7.58e-13 2.599999999999999      0.04064183933210896     4.523319895799798     9.359999999998582    9.359999999999992     -1.41e-12 2.499999999999999      0.08133637080178369     4.299788963018781     8.750000000001108    8.749999999999993     1.116e-12 2.399999999999999      0.1245572014856252      4.07937196231825      8.159999999995687    8.159999999999991    -4.305e-12 2.299999999999998      0.1705870546326995      3.862133188725139     7.590000000005771    7.589999999999992      5.78e-12 2.199999999999998      0.2197552692584715      3.648140684327197     7.040000000005112    7.039999999999991      5.12e-12 2.099999999999998      0.2724481590443035      3.43746660924917      6.510000000004961    6.509999999999991      4.97e-12 1.999999999999998      0.3291223221651317      3.230187665721837     6.000000000002764    5.999999999999991     2.773e-12 1.899999999999998      0.3903219461201083      3.026385585337975     5.509999999994411    5.509999999999991     -5.58e-12 1.799999999999998      0.4567016008063509      2.826147692182384     5.040000000000555    5.039999999999991     5.641e-13 1.699999999999998      0.5290566937088022      2.629567557541728     4.589999999999334    4.589999999999991    -6.573e-13 1.599999999999998      0.6083648146507388      2.436745766287358     4.159999999997735    4.159999999999991    -2.256e-12 1.499999999999998      0.6958428672560889      2.247790820483765     3.750000000000785    3.749999999999991     7.936e-13 1.399999999999998      0.7930276008088424      2.062820213392563     3.359999999998415    3.359999999999991    -1.576e-12 1.299999999999998      0.9018917081211759      1.881961717357213     2.989999999999875    2.989999999999991    -1.168e-13 1.199999999999998      1.025015540937898       1.705354943301408     2.639999999997569    2.639999999999992    -2.422e-12 1.099999999999997      1.165848685499068       1.533153249914835     2.309999999999998    2.309999999999992     6.519e-15 0.9999999999999974     1.329122322165132       1.365526109632838     1.999999999999481    1.999999999999992     -5.11e-13 0.8999999999999975     1.521526085277808       1.202662081575776     1.709999999999741    1.709999999999993    -2.523e-13 0.7999999999999975     1.752874096417759       1.044772606288851     1.440000000000155    1.439999999999994     1.619e-13 0.6999999999999975     2.038235616381106       0.8920969377256853    1.189999999999717    1.189999999999994    -2.774e-13 0.5999999999999975     2.402125463067824       0.7449086888890672    0.9599999999997391   0.9599999999999945   -2.555e-13 0.4999999999999976     2.887563844128977       0.6035247351814985    0.7499999999997691   0.7499999999999951    -2.26e-13 0.3999999999999976     3.578318349061931       0.4683176837021201    0.5599999999999967   0.5599999999999956    1.079e-15 0.2999999999999976     4.664365697417611       0.3397339639146319    0.390000000000061    0.3899999999999962    6.486e-14 0.1999999999999976     6.698404497689143       0.2183212373542653    0.2399999999999996   0.2399999999999966    2.968e-15 0.09999999999999759   12.34957156441279        0.1047722467573414    0.1100000000000074   0.1099999999999971    1.029e-14          x               alpha(x)              f(x)                f(f(x))                 x + x^2       f(f(x))- (x+x^2)    jagy@phobeusjunior:~$  

=============================

#include  #include  #include  #include  #include  #include  #include  #include  #include  #include  #include  using namespace std;  // file named abel_any_function.cc   //    lines after double slashes are comments  //   also on a line with a command, anything after // is  commentary  //  on a Unix or Linux computer,  compile using line    //        g++ -o abel_any_function   abel_any_function.cc -lm   //  then run the program  with  //  ./abel_any_function      double any_function(double x) {    return    ( -1.0 + sqrt( 1.0 + 4 * x )  ) / 2.0; }    double abel(double x) {     double eps = 0.000000001;    eps = eps / 100000.0;   double f = x ;   double g = 1.0, g_old = 1000.0, diff = 1.0 ;  for( int n = 0; n <= 100000  && diff >= eps ; ++n)  {   //   g =  3.0 / (f * f)  +  6.0 * log(f) / 5.0 +  79.0 * f * f/ 1050.0 + 29.0 * f * f * f * f /2625.0   - n;  double f2 = f * f; double f3 = f * f2; double f4 = f * f3; double f5 = f * f4; double f6 = f * f5; double f7 = f * f6; double f8 = f * f7; double f9 = f * f8; double f10 = f * f9;       g =   1.0 / f - log(1.0 * f) + f / 2.0 - f2 / 3.0 + 13.0 * f3 / 36.0 - 113.0 * f4 / 240.0 + 1187.0 * f5 / 1800.0 - 877.0 * f6 / 945.0 + 14569.0 * f7 / 11760.0 + 532963.0 * f8 / 24192.0  + 1819157.0 * f9 / 1360800.0  - 70379.0 * f10 / 47250.0    - n ;      diff = fabs(g - g_old);  //  cout.precision(16);  //  cout << n << "  " << x  << "  "  << f  << "  " << g <<  "   " << diff << endl ;    f = any_function ( 1.0 * f);     g_old = g;  }   return g; } // abel     //        g++ -o abel_any_function   abel_any_function.cc -lm    double inverse_abel(double x) {   int count = 0;   double eps = 0.000000001;      eps = eps / 100000.0;   double middle, left, right;   if( x < -10.0) return 0.0;   else   {      left = 0.01;     right = 110.0 ;     middle = ( left + right) / 2.0;      double left_val = abel(left) , right_val = abel(right), middle_val = abel(middle);     while ( right - left > eps)     {       if (middle_val < x )       {         right = middle;         middle = ( left + right) / 2.0;          right_val = abel(right);         middle_val = abel(middle);       }       else       {         left = middle;         middle = ( left + right) / 2.0;          left_val = abel(left);         middle_val = abel(middle);       }       count++;    //   cout << count;     //  cout.precision(16);     //  cout << "   " << x << "  " << middle << endl;      } // while not accurate   }  // else in range    return  middle; } // inverse_abel  double half_iterate(double x) {   return inverse_abel( -1/2.0 + abel(x)  ); }    int main() {    double my_pi = 4.0 * atan(1.0) ;    double phlegm = 5.0;   // cout << my_pi / 2.0 << "   " <<  abel( my_pi / 2.0) << endl;  //  cout << my_pi / 2.0 << "   " <<   half_iterate( my_pi / 2.0) << endl;      cout <<  "         x               alpha(x)              f(x)                f(f(x))                g( x)               f(f(x))- g( x) " << endl;      for( double x =   4.0; x >= 0.005 ; x -= 0.1)   { //     cerr << x << endl;  cout.precision(16);     cout << x << "   " <<  abel( x) << "    "  << half_iterate( x) << "    "  << half_iterate(half_iterate( x)) << "    " <<  x + x * x   ;    cout.precision(4);  cout << "    "  << half_iterate(half_iterate( x)) -  x - x * x   << endl; // cout <<  inverse_abel(abel(x))  -  x   << endl;   }  cout <<  "         x               alpha(x)              f(x)                f(f(x))                g( x)               f(f(x))- g( x) " << endl;     return 0 ; }    //  end of main    //        g++ -o abel_any_function   abel_any_function.cc -lm  

===========================

It was easy to run increasing $x$ and ask how high $n$ needed to be in evaluating the limit with $x_n$ for $\alpha.$ Even for very large $x,$ the number was generally from 55-75, with gusts into the 100's.

==================

x               alpha  (x)            1   count  54  alpha  1.329122322165132 10   count  58  alpha  -0.9983537653455241 100   count  68  alpha  -1.968971136021889 1000   count  66  alpha  -2.552472649197334 10000   count  63  alpha  -2.967413457600026 100000   count  61  alpha  -3.289334360104431 1000000   count  63  alpha  -3.552368279442522 10000000   count  71  alpha  -3.774760625378939 100000000   count  60  alpha  -3.967405626083742 1000000000   count  63  alpha  -4.137330604148414 10000000000   count  71  alpha  -4.28933373355295 100000000000   count  61  alpha  -4.426837311274541 1000000000000   count  61  alpha  -4.552368227229554 10000000000000   count  320  alpha  -4.667845444457483 100000000000000   count  72  alpha  -4.774760620903448 1000000000000000   count  67  alpha  -4.874296255743502 1e+16   count  65  alpha  -4.967405625692374 1e+17   count  71  alpha  -5.05486844668566 1e+18   count  65  alpha  -5.137330604113803 1e+19   count  64  alpha  -5.215333128877907 1e+20   count  61  alpha  -5.289333733549392 1e+21   count  66  alpha  -5.359723089041725 1e+22   count  62  alpha  -5.426837311274541 9.999999999999999e+22   count  65  alpha  -5.49096766961696 1e+24   count  62  alpha  -5.552368227229554 9.999999999999999e+24   count  68  alpha  -5.611261920378335 9.999999999999999e+25   count  321  alpha  -5.667845444457483 9.999999999999999e+26   count  65  alpha  -5.722293217232147 1e+28   count  73  alpha  -5.774760620903448 9.999999999999999e+28   count  69  alpha  -5.825386674958394 9.999999999999999e+29   count  68  alpha  -5.874296255743502 

============

  • 0
    I've encountered a difference between the coefficients as I computed them and that one, which you use in your "abel"-function for *g*. See the table in the new answer-box ("g-coefficients").2012-10-11
0

remark: this is the comment "g-coefficients" to Will's answer containing the c-program for the Abel-function.

@Will: Here is the table of coefficients. (Your numbers are originally given as floats(double)) The first significant coefficients-difference is at f8. I don't know which ones are correct and didn't think about the possible nonsignificance due to the n'th iteration of x towards the fixpoints.

                Helms                  Jagy                -1  log(x)             -1   log(f)                 -1    n                -1   n                      1  x^-1                1   1/f                 0   x^0                0    --               1/2   x^1              1/2   f              -1/3   x^2             -1/3   f2             13/36   x^3            13/36   f3          -113/240   x^4         -113/240   f4         1187/1800   x^5        1187/1800   f5          -877/945   x^6         -877/945   f6       14569/11760   x^7      14569/11760   f7   -----------------------------------------------------    -176017/120960   x^8     532963/24192   f8   *** here it begins to differ   1745717/1360800   x^9  1819157/1360800   f9     -88217/259875  x^10     -70379/47250  f10 
  • 0
    Gottfried, I'm awake. I did the 8,9,10 coefficients in a bit of a hurry last night, plus I am not that great with Pari. The true test is this: drop the $-n,$ call the result $\alpha(x).$ Take the derivative $\alpha'(x).$ Take the reciprocal, i called that $\lambda(x) = 1 / \alpha'(x).$ Find the power series for $h(x) = (-1 + \sqrt{1+4x})/2.$ Then $\lambda(x)$ should solve $ \lambda(h(x)) = h'(x) \cdot \lambda(x).$ Meaning, in Pari, the power series for $ \lambda(h(x)) - h'(x) \cdot \lambda(x)$ should have coefficient $0$ until $x^{11}$ or so.2012-10-11
  • 0
    Looked at it a bit, i think the real trouble is that Pari gave me power series up to O(x^16) but I cut them off early.2012-10-11
0

FRIDAY: I've been playing with the C++. Gottfried says my coefficients for $\alpha$ are correct up to $x^7,$ so I did that, with $x$ increasing by $1/10$ up to 10. The second column is $\alpha,$ the next column is $g(x) = \alpha^{-1} \left(- \frac{1}{2} + \alpha(x) \right),$ the final column is $g(g(x))$ which compares very well with $x+ x^2,$ as you can see with the integral $x.$

=============

   x               alpha(x)             g(x)             g(g(x))    0.1        12.34957156441259  0.1047722467573381  0.1099999999999914   0.2        6.698404497688887  0.2183212373542643  0.2400000000000055   0.3        4.664365697417439  0.3397339639146599  0.3900000000000158   0.4        3.578318349061967  0.4683176837021162  0.5599999999999985   0.5        2.887563844129021  0.6035247351815045   0.749999999999915   0.6        2.402125463067622  0.7449086888889984  0.9600000000000521   0.7        2.038235616380761  0.8920969377256855    1.19000000000007   0.8        1.752874096417655   1.044772606289162   1.439999999999584   0.9        1.521526085277657   1.202662081575782   1.710000000000312     1        1.329122322164689   1.365526109633105   2.000000000000555   1.1        1.165848685498792   1.533153249914976   2.309999999999806   1.2        1.025015540937656   1.705354943302132   2.640000000000098   1.3       0.9018917081212492    1.88196171735636   2.990000000000656   1.4       0.7930276008088036   2.062820213391945   3.360000000000213   1.5       0.6958428672559145   2.247790820483585   3.750000000000525   1.6       0.6083648146504035    2.43674576628842   4.159999999999291   1.7       0.5290566937086119   2.629567557542234   4.590000000000048   1.8       0.4567016008063152   2.826147692182283   5.039999999999818   1.9       0.3903219461197646   3.026385585339394   5.510000000000232     2       0.3291223221646891   3.230187665722821   5.999999999999915   2.1       0.2724481590442369   3.437466609251579   6.510000000001385   2.2       0.2197552692584598   3.648140684326019   7.039999999999509   2.3       0.1705870546325254   3.862133188724311   7.589999999999385   2.4        0.124557201485502   4.079371962317376   8.159999999999702   2.5      0.08133637080242444   4.299788963015828   8.750000000000881   2.6      0.04064183933224629    4.52331989579711   9.359999999998582   2.7     0.002229349696022672   4.749903886596474   9.989999999999988   2.8     -0.03411336553461632   4.979483194345788   10.63999999999846   2.9      -0.0685717768879244   5.212002955613551   11.30999999999516     3       -0.101308757652073   5.447410957287675    12.0000000000025   3.1      -0.1324680068818291   5.685657433445996   12.71000000000289   3.2      -0.1621768597702231   5.926694883244602   13.43999999999888   3.3      -0.1905486107381383   6.170477907090149   14.18999999999903   3.4       -0.217684445944716   6.416963058834378   14.95999999999939   3.5      -0.2436750604476325   6.666108712005801   15.75000000000014   3.6      -0.2686020190349739   6.917874938477105   16.56000000000593   3.7      -0.2925389073554089   7.172223398030425   17.38999999999782   3.8      -0.3155523104805156   7.429117237690733   18.24000000001208   3.9      -0.3377026486412144   7.688520999701667   19.11000000000036     4       -0.359044894127126   7.950400537216488   20.00000000000308   4.1       -0.379629188807631   8.214722936914129   20.90999999999864   4.2       -0.399501378150432   8.481456447790578   21.84000000000423   4.3      -0.4187034747632079   8.750570415576934   22.79000000000375   4.4      -0.4372740621842076   9.022035222145501   23.75999999999652   4.5      -0.4552486478217184   9.295822229482983     24.749999999992   4.6      -0.4726599724392348   9.571903727807236   25.75999999998796   4.7       -0.489538282369249   9.850252887366977   26.78999999998784   4.8       -0.505911569651961    10.1308437137012   27.83999999997848   4.9      -0.5218057844694055   10.41365100596458   28.91000000000528     5      -0.5372450235774485   10.69865031810888    30.0000000000268   5.1      -0.5522516978769516   10.98581792261065   31.10999999998392   5.2      -0.5668466818022829   11.27513077665873   32.23999999999783   5.3      -0.5810494468112958   11.56656649034594   33.38999999998659   5.4      -0.5948781809449553   11.86010329714074    34.5599999999896   5.5      -0.6083498961375551   12.15572002587118    35.7499999999838   5.6      -0.6214805247424423   12.45339607470347   36.96000000000471   5.7      -0.6342850065223927   12.75311138645592   38.18999999999465   5.8      -0.6467773672082882   13.05484642552751   39.44000000003192   5.9      -0.6589707895703519   13.35858215607981   40.71000000000579     6      -0.6708776778353109   13.66430002151068   41.99999999994029   6.1       -0.682509716169434   13.97198192501955   43.30999999997566   6.2      -0.6938779218702816   14.28161021134496   44.63999999997695   6.3      -0.7049926938174378   14.59316764936057   45.99000000000397   6.4      -0.7158638566799534   14.90663741576357   47.36000000003547   6.5      -0.7265007013094341   15.22200307949671   48.74999999993412   6.6      -0.7369120217049638   15.53924858705377   50.15999999998954   6.7      -0.7471061488852925   15.85835824845936   51.58999999998022   6.8      -0.7570909819765037   16.17931672414124   53.03999999999729   6.9       -0.766874016772743   16.50210901219836   54.51000000000624     7      -0.7764623720190872   16.82672043645934   55.99999999936385   7.1      -0.7858628136233136   17.15313663519212   57.51000000001224   7.2      -0.7950817769841955   17.48134355009424   59.04000000000003   7.3      -0.8041253876168806    17.8113274160783   60.58999999997775   7.4      -0.8129994802177363   18.14307475139146   62.15999999999015   7.5      -0.8217096163081531   18.47657234815847   63.74999999999534   7.6      -0.8302611005888889   18.81180726349173   65.36000000004154   7.7      -0.8386589961029491   19.14876681089618   66.98999999994771   7.8      -0.8469081383179831    19.4874385520576    68.6400000000553   7.9      -0.8550131482122515   19.82781028897663   70.30999999995979     8      -0.8629784444526734    20.1698700565776   71.99999999990217   8.1      -0.8708082547322369   20.51360611552482   73.70999999996329   8.2      -0.8785066263364341   20.85900694520567   75.44000000000185   8.3      -0.8860774360049619   21.20606123733967   77.19000000000509   8.4      -0.8935243991361352   21.55475788956775   78.96000000013811   8.5      -0.9008510783898448   21.90508599935687   80.75000000002262   8.6      -0.9080608917366347   22.25703485834448   82.55999999999338   8.7      -0.9151571199876521   22.61059394661323   84.38999999996526   8.8      -0.9221429138533149   22.96575292741191   86.23999999995436   8.9      -0.9290213005570148    23.3225016419572   88.10999999998333     9      -0.9357951900460741   23.68083010453317   90.00000000007395   9.1      -0.9424673808176133   24.04072849765743   91.91000000003487   9.2      -0.9490405653974854   24.40218716764626   93.84000000000023   9.3      -0.9555173354899633    24.7651966200477   95.78999999998206   9.4      -0.9619001868221458   25.12974751543187   97.76000000002301   9.5       -0.968191523710482   25.49583066539916    99.7500000000214   9.6       -0.974393663357775   25.86343702851241   101.7600000000278   9.7      -0.9805088399074678   26.23255770657607   103.7900000000062   9.8      -0.9865392082677702   26.60318394084462   105.8399999998493   9.9      -0.9924868477235212   26.97530710881871   107.9100000000317    10      -0.9983537653457113   27.34891872036212   109.9999999999982    x               alpha(x)             g( x)            g(g( x))  

=============

  • 0
    Ah, yes, thanks! The precision is thus something about 1e-10. Why do you work with the c++ and the "double" precision only? Would you like to get my Pari/GP-script where you can without hassle use internal precision up to 200, 400 digits precision which leads then to 60 or 70 correct digits for the results? (you can also use my email, but note that in the next week our courses begin and I'll likely have less time (and patience) in the first two or three weeks)2012-10-13
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    @GottfriedHelms, yes, I'd like to see that. I should also try to see what I can do with GMP in C++. As to choices, I know C++ pretty well. With Pari. I know how to do interactive sessions, but have never succeeded in running an actual program. I see, if short you could display the Pari stuff here, but email may be more reliable for something longer, my addresses can be found (if not visible on my profile) by searching with my last name at http://www.ams.org/cml/2012-10-13
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    I haven't received KCG, but I spent some time playing with the formal power series for $\alpha(x)$, and my results match Gottfried's 1st 10 coefficients, with the x^11 coefficient = -14763581/109771200. I generated 25 terms of the abel equation, and computations matched my earlier results to 46 decimal digits when x<=0.01. Interestingly, my algorithm for the formal power series differs from the algorithm Will Jaggy posted earlier, details next.2012-10-15
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    If we know the form of the solution is $\alpha(x)=-1/x + \log(x) + \sum_{n \to \infty}a_n x^n$, and we know that $\alpha(x+x^2)=\alpha(x)-1$ Then it can be solved term by term without integration or recipricals. First, $-1/(x+x^2) = -1/x + 1/(x+1)$, and $\log(x+x^2) = \log(x) + \log(x+1)$. Then $\alpha(x+x^2)+1-\alpha(x) = -1/(x+1) - \log(x+1) + \sum_{n \to \infty}-a_n x^n + \sum_{n \to \infty}a_n(x^2+x)^n$, which can be solved term by term, without requiring a reciprical calculation.2012-10-15
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    @SheldonL It is permitted that subtracting off the appropriate $\log x$ term from $\alpha $ gives something meromorphic at the origin, although it is known that $\alpha$ itself can only be so when the original function is a certain Mobius transformation. This is a paraphrase of Thm 8.5.3 on page 347 of KCG, which is about the solution $\lambda(x)$ of the related Julia equation.2012-10-15
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    There is the concept of a super attracting fixed point at infinity (or zero), for iterations of x^2. For x^2+c, there is a super attracting fixed point at infinity, and the $\alpha_{alt}$ for x^2+0.25 in my answer above corresponds to abel/fatou function from the super attracting fixed point at infinity, which I think makes the half iterates nicely behaved (ordered), since then $\theta(z)$ would converge to a constant. So, for c=0.25, there is both a parabolic abel solution whose $\alpha^{-1}$ is entire, and a different well defined analytic $\alpha_{alt}^{-1}$ with a fractal boundary.2012-10-16
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    How do you know there is a parabolic abel solution whose inverse abel is entire ?2012-10-18
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    @mick The parabolic solution is the one that Will Jagy gave the formal power series solution for, and its inverse abel function acts like 1/z in the left half of the complex plane. Then given that the iterating function is a finite series, (x^2+x), iterating finite functions doesn't give singularities. I'm not claiming that I could prove it though I think it's covered in Milnor's book.2012-10-18