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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) –  Gottfried Helms Oct 8 '12 at 2:34
    
Hey Gottfried, thanks for your comments –  Sheldon L Oct 8 '12 at 13:09
    
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 ... –  Sheldon L Oct 8 '12 at 13:15
1  
Sorry to interrupt this conversation but when $f:x\mapsto x^2+\frac14$, how do you DEFINE the function $h$? –  Did Oct 8 '12 at 14:58
1  
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? –  Did Oct 8 '12 at 15:45

9 Answers 9

up vote 5 down vote accepted

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 http://mathoverflow.net/questions/45608/formal-power-series-convergence as my answer http://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 $\alpha(x):$

=======

    ? 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... –  Gottfried Helms Oct 9 '12 at 7:01
    
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. –  Sheldon L Oct 9 '12 at 12:34
    
@GottfriedHelms and sheldonison, I give a complete description of the method at mathoverflow.net/questions/45608/… as a worked example. The cure for needing 100,000 iterations is to have, say, two dozen terms in the expansion for $\alpha$ –  Will Jagy Oct 9 '12 at 18:56
    
@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, - Sheldon –  Sheldon L Oct 9 '12 at 21:33
1  
Will, I did a procedure to obtain the coefficients automatically in an answer-box. –  Gottfried Helms Oct 10 '12 at 13:55

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

share|improve this answer
    
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 mathoverflow.net/questions/45608/… I admit it is a good deal of work. –  Will Jagy Oct 10 '12 at 18:41
    
@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. –  Gottfried Helms Oct 10 '12 at 18:52
    
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 $x + x^2.$ –  Will Jagy Oct 10 '12 at 19:03
    
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. –  Will Jagy Oct 10 '12 at 19:54
    
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. –  Will Jagy Oct 10 '12 at 20:35

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.

share|improve this answer
    
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 $x \rightarrow + \infty?$ –  Will Jagy Oct 16 '12 at 20:46
    
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 ..... –  Sheldon L Oct 17 '12 at 22:42
    
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. –  Sheldon L Oct 18 '12 at 3:22

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
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0.2356194490192398   52.30753131765482    0.2345247921385231    0.2334453638559151    0.2334453638559106    4.549e-15
0.2268928027592682   56.4986387155308    0.225915626691418    0.2249510543438699    0.2249510543438702    -2.537e-16
0.2181661564992965   61.2064903053345    0.2172977079441771    0.2164396139381293    0.216439613938108    2.128e-14
0.2094395102393248   66.51913636418844    0.2086713809927613    0.2079116908177978    0.2079116908177645    3.331e-14
0.2007128639793532   72.54419017165753    0.2000369899148535    0.1993679344171978    0.1993679344172024    -4.546e-15
0.1919862177193815   79.41428325980672    0.1913948778158753    0.1908089953765355    0.19080899537655    -1.447e-14
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 <iostream>
#include <stdlib.h>
#include <fstream>
#include <sstream>
#include <list>
#include <set>
#include <math.h>
#include <iomanip>
#include <string>
#include <algorithm>
#include <iterator>
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

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

share|improve this answer

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


share|improve this answer
    
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. –  Sheldon L Oct 11 '12 at 13:36
    
@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. –  Gottfried Helms Oct 11 '12 at 14:00
    
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. –  Sheldon L Oct 11 '12 at 14:16

** 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
share|improve this answer
    
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()"). –  Gottfried Helms Oct 13 '12 at 5:05
    
@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. –  Sheldon L Oct 13 '12 at 6:34

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 <iostream>
#include <stdlib.h>
#include <fstream>
#include <sstream>
#include <list>
#include <set>
#include <math.h>
#include <iomanip>
#include <string>
#include <algorithm>
#include <iterator>
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

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

share|improve this answer
    
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"). –  Gottfried Helms Oct 11 '12 at 6:27

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
share|improve this answer
    
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. –  Will Jagy Oct 11 '12 at 15:47
    
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. –  Will Jagy Oct 11 '12 at 16:20

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)) 

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

share|improve this answer
    
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) –  Gottfried Helms Oct 13 '12 at 5:19
    
@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 ams.org/cml –  Will Jagy Oct 13 '12 at 14:43
    
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. –  Sheldon L Oct 15 '12 at 1:46
    
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. –  Sheldon L Oct 15 '12 at 1:58
    
@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. –  Will Jagy Oct 15 '12 at 2:15

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