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

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

" 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

$$ g(g(x)) =   \alpha^{-1} \left( \alpha(x) - 1 \right)  = f^{-1}(x) = x + x^2  $$

...

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] http://zakuski.utsa.edu/~jagy/K_C_G_book_excerpts.pdf
 * [BAKER] http://zakuski.utsa.edu/~jagy/other.html

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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=Jonathan Lubin= 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

$$f(x)=x+x^2$$.

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

$$f^{\circ1/2}(x)$$

the “half-th” iterate of $$f$$, out to the $$x^5$$-term.

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

$$ \begin{align} f^{\circ0}(x)&=x\\ f^{\circ1}=f&=x&+x^2\\ f^{\circ2}&=x&+2x^2&+2x^3&+x^4\\ f^{\circ3}&\equiv x&+3x^3&+6x^3& + 9x^4& + 10x^5& + 8x^6\\ f^{\circ4}&\equiv x &+ 4x^2& + 12x^3& + 30x^4& + 64x^5& + 118x^6\\ f^{\circ5}&\equiv x& + 5x^2& + 20x^3& + 70x^4& + 220x^5& + 630x^6\\ f^{\circ6}&\equiv x& + 6x^2& + 30x^3& + 135x^4& + 560x^5& + 2170x^6\\ f^{\circ7}&\equiv x& + 7x^2& + 42x^3& + 231x^4& + 1190x^5& + 5810x^6\,, \end{align} $$

where the congruences are modulo all terms of degree $$7$$ and more.

Now look at the coefficients
 * of the $$x$$-term: always $$1$$.
 * Of the $$x^2$$-term? In $$f^{\circ n}$$, it’s $$C_2(n)=n$$.
 * The coefficient of $$x^3$$ in $$f^{\circ n}$$ is $$C_3(n)=n(n-1)=n^2-n$$, as one can see by inspection.

Now, a moment’s thought (well, maybe several moments’) tells you that $$C_j(n)$$, the coefficient of $$x^j$$ in $$f^{\circ n}$$, is a polynomial in $$n$$ of degree $$j-1$$.

And a familiar technique of finite differences shows you that

$$ \begin{align} C_4(n)&=\frac{2n^3-5n^2+3n}2\\ C_5(n)&=\frac{3n^4-13n^3+18n^2-8n}3\,, \end{align} $$

I won’t go into the details of that technique. The upshot is that, modulo terms of degree $$6$$ and higher, you have

$$f^{\circ n}(x)\equiv x+nx^2+(n^2-n)x^3+\frac12(2n^3-5n^2+3n)x^4+\frac13(3n^4-13n^3+18n^2-8n)x^5$$.

Now, you just plug in $$n=\frac12$$ 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."