Fractals/Iterations in the complex plane/Fatou coordinate for f(z)=z+z^2)

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Will Jagy[edit]

http://math.stackexchange.com/questions/208996/half-iterate-of-x2c?

" This may be helpful.

Let


We use a technique of Ecalle to solve for the Fatou coordinate that solves


For any


let


Then we get the exact


The point is that this expression converges far more rapidly than one would expect, and we may stop at a fairly small .

It is fast enough that we may reasonably expect to solve numerically for .

We have

Note




It follows that if we define


we get the miraculous


...

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

The Julia equation is Theorem 8.5.1 on page 346 of KCG.

It would be no problem to produce, say, 50 terms of 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


when


As I said in comment, the way to improve this is to take a few dozen terms in the expansion of so as to get the desired decimal precision with a more reasonable number of evaluations of :

=======

        ? 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

======

"

Jonathan Lubin[edit]

From http://math.stackexchange.com/questions/911818/how-to-obtain-fx-if-it-is-known-that-ffx-x2x?

"Here’s a technique for finding the first few terms of a formal power series representing the fractional iterate of a given function like

.

I repeat that this is a formal solution to the problem, and leaves unaddressed all considerations of convergence of the series answer.

I’m going to find the first six terms of

the “half-th” iterate of , out to the -term.

Let’s write down the iterates of , starting with the zero-th.

where the congruences are modulo all terms of degree and more.

Now look at the coefficients

  • of the -term: always .
  • Of the -term? In , it’s .
  • The coefficient of in is , as one can see by inspection.

Now, a moment’s thought (well, maybe several moments’) tells you that , the coefficient of in , is a polynomial in of degree .

And a familiar technique of finite differences shows you that

I won’t go into the details of that technique. The upshot is that, modulo terms of degree and higher, you have

.

Now, you just plug in in this formula to get your desired series.

And I’ll leave it to you to go one degree higher, using the iterates I’ve given you."