Solution

Let

x^{*} (λ) = 1 - 1/ λ

and center the map at this parameter-dependent fixed point:

G (X, μ) = F (x^{*} (2 + μ) + X, 2 + μ) - x^{*} (2 + μ) .

X

, with the coefficients expanded at

(x, λ) = (1/ 2, 2)

, gives

G (X, μ) = - X + αμ X + β X^{2} + γ X^{3} + O (μ^{2})

under the period-doubling scaling

X = O (μ)

. The coefficients are

α = \frac{d}{d λ} F_{x} (x^{*} (λ), λ)_{λ = 2} = - 2,

β = \frac{1}{2} F_{xx} (\frac{1}{2}, 2) = - 32, γ = \frac{1}{6} F_{xxx} (\frac{1}{2}, 2) = - 2.

Here we used

F_{x} (x^{*} (λ), λ) = 3 - 2 λ

F_{xx} = - 6 λ x

, and

F_{xxx} = - 6 λ

Composing the local map with itself and retaining the terms of order

μ X

and

X^{3}

gives

G (G (X, μ), μ) - X = - 2 αμ X - 2 (γ + β^{2}) X^{3} + o (μ X + X^{3}) .

Besides the fixed solution

X = 0

, the two leading solutions therefore satisfy

X^{2} = - \frac{αμ}{γ + β ^{2}} .

Thus the period-two points born at

λ = 2

are

x_{\pm} = x^{*} \pm \frac{- αμ}{γ + β ^{2}}

to leading order. With the coefficients above, the radicand is positive for

μ > 0

Solved by gpt-5.6-sol high.

 Ancestors (11)