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Assume that $f(x)$ and $g(x)$ are analytic and that $f$ has a fixed point, without loss of generality $f(0)=0$ and therefore $f(f(0))=g(0)=0$. The Taylors series of $g^n(x)$ can be constructed. using Bell polynomials of iterated functions and then the iterate can be evaluated at $1/2$.

Taylors series of iterated functions

Let $$f(x)=g^{1/2}(x)$$ then \begin{eqnarray*} g^n(x) &=& g_1^n x+ \frac{1}{2} g_2(\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2}) x^2 \\ &+& \frac{1}{6}[g_3(\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3}) +3g_2^2 (\sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5}) ] x^3 + \cdots \end{eqnarray*}

Proof.

Let $h(x)=g^{n-1}(x)$, then

\begin{eqnarray*} Dg(h(z))&=&g'(h(x))h'(x)\\ &=&g'(g^{n-1}(x))Dg^{n-1}(x)\\ &=&\prod^{n-1}_{k_1=0}g'(g^{n-k_1-1}(x))\\ \end{eqnarray*} \begin{eqnarray} Dg^n(0)&=&g'(0)^n \\ &=&g_1^n = g_1^{1/2} \end{eqnarray} For the second derivative \begin{eqnarray} D^2g^n(0)&=&g_2g_1^{2n-2}+g_1 D^2g^{n-1}(0) \\ &=&g_1^0g_2 g_1^{2n-2} \\ &&+g_1^1g_2 g_1^{2n-4} \\ &&+\cdots \\ &&+g_1^{n-2}g_2 g_1^2 \\ &&+g_1^{n-1}g_2 g_1^0 \\ &=&g_2\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2} \end{eqnarray} For the third derivative \begin{eqnarray} D^3g^n(0)&=&g_3g_1^{3n-3}+3 g_2^2\sum_{k_1=0}^{n-1}g_1^{3n-k_1-5} +g_1 D^3g^{n-1}(0) \\ &=&g_3\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3} +3g_2^2 \sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5} \end{eqnarray}

Note that the index $k_1$ from the second derivative is renamed $k_2$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

$\blacksquare$

Algebraic expansion shows that $$g^m(g^n(x))=g^{m+n}(x)$$.

Classification of Fixed Points

Note that $g^n(x)$ was not evaluated at $n=1/2$. What at first may appear to be a defect is a strength as it reveals the impact of the classification of fixed points. If $g'(0)$ equals a root of unity, then the geometrical progressions evaluate very differently than the standard method of evaluation. This formula works regardless of whether dealing with hyperbolic fixed points, parabolically neutral fixed points and rationally and irrationally neutral fixed points.

See Bell polynomials of iterated functions and What are efficient ways to compute the derivatives of iterated functions? for more details.

Assume that $f(x)$ and $g(x)$ are analytic and that $f$ has a fixed point, without loss of generality $f(0)=0$ and therefore $f(f(0))=g(0)=0$. The Taylors series of $g^n(x)$ can be constructed. using Bell polynomials of iterated functions and then the iterate can be evaluated at $1/2$.

Taylors series of iterated functions

Let $$f(x)=g^{1/2}(x)$$ then \begin{eqnarray*} g^n(x) &=& g_1^n x+ \frac{1}{2} g_2(\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2}) x^2 \\ &+& \frac{1}{6}[g_3(\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3}) +3g_2^2 (\sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5}) ] x^3 + \cdots \end{eqnarray*}

Proof.

Let $h(x)=g^{n-1}(x)$, then

\begin{eqnarray*} Dg(h(z))&=&g'(h(x))h'(x)\\ &=&g'(g^{n-1}(x))Dg^{n-1}(x)\\ &=&\prod^{n-1}_{k_1=0}g'(g^{n-k_1-1}(x))\\ \end{eqnarray*} \begin{eqnarray} Dg^n(0)&=&g'(0)^n \\ &=&g_1^n = g_1^{1/2} \end{eqnarray} For the second derivative \begin{eqnarray} D^2g^n(0)&=&g_2g_1^{2n-2}+g_1 D^2g^{n-1}(0) \\ &=&g_1^0g_2 g_1^{2n-2} \\ &&+g_1^1g_2 g_1^{2n-4} \\ &&+\cdots \\ &&+g_1^{n-2}g_2 g_1^2 \\ &&+g_1^{n-1}g_2 g_1^0 \\ &=&g_2\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2} \end{eqnarray} For the third derivative \begin{eqnarray} D^3g^n(0)&=&g_3g_1^{3n-3}+3 g_2^2\sum_{k_1=0}^{n-1}g_1^{3n-k_1-5} +g_1 D^3g^{n-1}(0) \\ &=&g_3\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3} +3g_2^2 \sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5} \end{eqnarray}

Note that the index $k_1$ from the second derivative is renamed $k_2$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

$\blacksquare$

Algebraic expansion shows that $$g^m(g^n(x))=g^{m+n}(x)$$.

Classification of Fixed Points

Note that $g^n(x)$ was not evaluated at $n=1/2$. What at first may appear to be a defect is a strength as it reveals the impact of the classification of fixed points. If $g'(0)$ equals a root of unity, then the geometrical progressions evaluate very differently than the standard method of evaluation. This formula works regardless of whether dealing with hyperbolic fixed points, parabolically neutral fixed points and rationally and irrationally neutral fixed points.

See Bell polynomials of iterated functions and What are efficient ways to compute the derivatives of iterated functions? for more details.

Assume that $f(x)$ and $g(x)$ are analytic and that $f$ has a fixed point, without loss of generality $f(0)=0$ and therefore $f(f(0))=g(0)=0$. The Taylors series of $g^n(x)$ can be constructed. using Bell polynomials of iterated functions and then the iterate can be evaluated at $1/2$.

Taylors series of iterated functions

Let $$f(x)=g^{1/2}(x)$$ then \begin{eqnarray*} g^n(x) &=& g_1^n x+ \frac{1}{2} g_2(\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2}) x^2 \\ &+& \frac{1}{6}[g_3(\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3}) +3g_2^2 (\sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5}) ] x^3 + \cdots \end{eqnarray*}

Proof.

Let $h(x)=g^{n-1}(x)$, then

\begin{eqnarray*} Dg(h(z))&=&g'(h(x))h'(x)\\ &=&g'(g^{n-1}(x))Dg^{n-1}(x)\\ &=&\prod^{n-1}_{k_1=0}g'(g^{n-k_1-1}(x))\\ \end{eqnarray*} \begin{eqnarray} Dg^n(0)&=&g'(0)^n \\ &=&g_1^n = g_1^{1/2} \end{eqnarray} For the second derivative \begin{eqnarray} D^2g^n(0)&=&g_2g_1^{2n-2}+g_1 D^2g^{n-1}(0) \\ &=&g_1^0g_2 g_1^{2n-2} \\ &&+g_1^1g_2 g_1^{2n-4} \\ &&+\cdots \\ &&+g_1^{n-2}g_2 g_1^2 \\ &&+g_1^{n-1}g_2 g_1^0 \\ &=&g_2\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2} \end{eqnarray} For the third derivative \begin{eqnarray} D^3g^n(0)&=&g_3g_1^{3n-3}+3 g_2^2\sum_{k_1=0}^{n-1}g_1^{3n-k_1-5} +g_1 D^3g^{n-1}(0) \\ &=&g_3\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3} +3g_2^2 \sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5} \end{eqnarray}

Note that the index $k_1$ from the second derivative is renamed $k_2$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

$\blacksquare$

Classification of Fixed Points

Note that $g^n(x)$ was not evaluated at $n=1/2$. What at first may appear to be a defect is a strength as it reveals the impact of the classification of fixed points. If $g'(0)$ equals a root of unity, then the geometrical progressions evaluate very differently than the standard method of evaluation. This formula works regardless of whether dealing with hyperbolic fixed points, parabolically neutral fixed points and rationally and irrationally neutral fixed points.

See Bell polynomials of iterated functions and What are efficient ways to compute the derivatives of iterated functions? for more details.

Assume that $f(x)$ and $g(x)$ are analytic and that $f$ has a fixed point, without loss of generality $f(0)=0$ and therefore $f(f(0))=g(0)=0$. The Taylors series of $g^n(x)$ can be constructed. using Bell polynomials of iterated functions and then the iterate can be evaluated at $1/2$.

Taylors series of iterated functions

Let $$f(x)=g^{1/2}(x)$$ then \begin{eqnarray*} g^n(x) &=& g_1^n x+ \frac{1}{2} g_2(\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2}) x^2 \\ &+& \frac{1}{6}[g_3(\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3}) +3g_2^2 (\sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5}) ] x^3 + \cdots \end{eqnarray*}

Proof.

Let $h(x)=g^{n-1}(x)$, then

\begin{eqnarray*} Dg(h(z))&=&g'(h(x))h'(x)\\ &=&g'(g^{n-1}(x))Dg^{n-1}(x)\\ &=&\prod^{n-1}_{k_1=0}g'(g^{n-k_1-1}(x))\\ \end{eqnarray*} \begin{eqnarray} Dg^n(0)&=&g'(0)^n \\ &=&g_1^n = g_1^{1/2} \end{eqnarray} For the second derivative \begin{eqnarray} D^2g^n(0)&=&g_2g_1^{2n-2}+g_1 D^2g^{n-1}(0) \\ &=&g_1^0g_2 g_1^{2n-2} \\ &&+g_1^1g_2 g_1^{2n-4} \\ &&+\cdots \\ &&+g_1^{n-2}g_2 g_1^2 \\ &&+g_1^{n-1}g_2 g_1^0 \\ &=&g_2\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2} \end{eqnarray} For the third derivative \begin{eqnarray} D^3g^n(0)&=&g_3g_1^{3n-3}+3 g_2^2\sum_{k_1=0}^{n-1}g_1^{3n-k_1-5} +g_1 D^3g^{n-1}(0) \\ &=&g_3\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3} +3g_2^2 \sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5} \end{eqnarray}

Note that the index $k_1$ from the second derivative is renamed $k_2$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

$\blacksquare$

Algebraic expansion shows that $$g^m(g^n(x))=g^{m+n}(x)$$.

Classification of Fixed Points

Note that $g^n(x)$ was not evaluated at $n=1/2$. What at first may appear to be a defect is a strength as it reveals the impact of the classification of fixed points. If $g'(0)$ equals a root of unity, then the geometrical progressions evaluate very differently than the standard method of evaluation. This formula works regardless of whether dealing with hyperbolic fixed points, parabolically neutral fixed points and rationally and irrationally neutral fixed points.

See Bell polynomials of iterated functions and What are efficient ways to compute the derivatives of iterated functions? for more details.

Assume that $f(x)$ and $g(x)$ are analytic and that $f$ has a fixed point, without loss of generality $f(0)=0$ and therefore $f(f(0))=g(0)=0$. The Taylors series of $g^n(x)$ can be constructed. using Bell polynomials of iterated functions and then the iterate can be evaluated at $1/2$.

Let $$f(x)=g^{1/2}(x)$$ then \begin{eqnarray} Dg^n(0)&=&g'(0)^n \\ &=&g_1^n = g_1^{1/2} \end{eqnarray} For the second derivative \begin{eqnarray} D^2g^n(0)&=&g_2g_1^{2n-2}+g_1 D^2g^{n-1}(0) \\ &=&g_1^0g_2 g_1^{2n-2} \\ &&+g_1^1g_2 g_1^{2n-4} \\ &&+\cdots \\ &&+g_1^{n-2}g_2 g_1^2 \\ &&+g_1^{n-1}g_2 g_1^0 \\ &=&g_2\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2} \end{eqnarray}

See Bell polynomials of iterated functions and What are efficient ways to compute the derivatives of iterated functions? for more details.

Assume that $f(x)$ and $g(x)$ are analytic and that $f$ has a fixed point, without loss of generality $f(0)=0$ and therefore $f(f(0))=g(0)=0$. The Taylors series of $g^n(x)$ can be constructed. using Bell polynomials of iterated functions and then the iterate can be evaluated at $1/2$.

Taylors series of iterated functions

Let $$f(x)=g^{1/2}(x)$$ then \begin{eqnarray*} g^n(x) &=& g_1^n x+ \frac{1}{2} g_2(\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2}) x^2 \\ &+& \frac{1}{6}[g_3(\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3}) +3g_2^2 (\sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5}) ] x^3 + \cdots \end{eqnarray*}

Proof.

Let $h(x)=g^{n-1}(x)$, then

\begin{eqnarray*} Dg(h(z))&=&g'(h(x))h'(x)\\ &=&g'(g^{n-1}(x))Dg^{n-1}(x)\\ &=&\prod^{n-1}_{k_1=0}g'(g^{n-k_1-1}(x))\\ \end{eqnarray*} \begin{eqnarray} Dg^n(0)&=&g'(0)^n \\ &=&g_1^n = g_1^{1/2} \end{eqnarray} For the second derivative \begin{eqnarray} D^2g^n(0)&=&g_2g_1^{2n-2}+g_1 D^2g^{n-1}(0) \\ &=&g_1^0g_2 g_1^{2n-2} \\ &&+g_1^1g_2 g_1^{2n-4} \\ &&+\cdots \\ &&+g_1^{n-2}g_2 g_1^2 \\ &&+g_1^{n-1}g_2 g_1^0 \\ &=&g_2\sum_{k_1=0}^{n-1}g_1^{2n-k_1-2} \end{eqnarray} For the third derivative \begin{eqnarray} D^3g^n(0)&=&g_3g_1^{3n-3}+3 g_2^2\sum_{k_1=0}^{n-1}g_1^{3n-k_1-5} +g_1 D^3g^{n-1}(0) \\ &=&g_3\sum_{k_1=0}^{n-1}g_1^{3n-2k_1-3} +3g_2^2 \sum_{k_1=0}^{n-1} \sum_{k_2=0}^{n-k_1-2} g_1^{3n-2k_1-k_2-5} \end{eqnarray}

Note that the index $k_1$ from the second derivative is renamed $k_2$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

$\blacksquare$

Classification of Fixed Points

Note that $g^n(x)$ was not evaluated at $n=1/2$. What at first may appear to be a defect is a strength as it reveals the impact of the classification of fixed points. If $g'(0)$ equals a root of unity, then the geometrical progressions evaluate very differently than the standard method of evaluation. This formula works regardless of whether dealing with hyperbolic fixed points, parabolically neutral fixed points and rationally and irrationally neutral fixed points.

See Bell polynomials of iterated functions and What are efficient ways to compute the derivatives of iterated functions? for more details.