= Solution
Write the local <Taylor series> at the <critical point> as
$$
g(x)=1+c x^d+O(x^{d+2}),\qquad c<0,
$$
where $d$ is even, and put $a=g(1)$. The fixed-point equation first gives $g(g(0))=g(1)=a$, consistently with normalization. Expanding the inner function at zero and the outer one at one,
$$
g(ax)=1+c a^d x^d+O(x^{d+2}),
$$
$$
a^{-1}g(g(ax))=1+c a^{d-1}g'(1)x^d+O(x^{d+2}).
$$
The quadratic remainder in the outer expansion is of order $x^{2d}$, at least $x^{d+2}$ because $d\geq2$. Comparing the nonzero $x^d$ coefficients proves the exact identity
$$
\boxed{a^{d-1}g'(1)=1.}
$$
To obtain the requested first-order relation, use the <leading polynomial approximation to a renormalization fixed point>: retain only $g(x)\approx1+c x^d$. Then $a\approx1+c$, while $g'(1)\approx dc$, giving the truncated equation
$$
\boxed{d(a-1)a^{d-1}=1\quad\text{in the first-order polynomial approximation}.}
$$
The equality belongs to this truncation; higher analytic coefficients generally change $g'(1)$ and $a$, so it is not an additional exact equation for the full <fixed point>.
For $d=2$, solve $2a(a-1)=1$. The root compatible with $-1<a<0$ is $a=(1-\sqrt3)/2$; the other root is inadmissible. Thus
$$
\boxed{g(x)\approx1-\frac{1+\sqrt3}{2}x^2,\qquad
\alpha\approx-\frac1a=1+\sqrt3\simeq2.73205.}
$$
This is a coarse leading approximation, not the more accurate universal value $2.5029\ldots$ obtained after retaining higher coefficients.
Back to article page
