Let \[ f(x)=\frac14+x-x^2. \] For \(x\in\RR\), define \(x_0=x\) and \[ x_{n+1}=f(x_n). \] If the sequence converges, write its limit as \(x_\infty\).
- For \(x=0\), show that \((x_n)\) is bounded and nondecreasing and find its limit \(\lambda\).
- Determine all \(y\in\RR\) for which \(y_\infty=\lambda\).
Solution
For an orbit \((x_n)\), set \[ u_n\coloneqq\frac12-x_n. \]
1. The recurrence for \((u_n)\) is\[ u_{n+1}=u_n^2. \]
Completing the square gives \[ f(x)=\frac12-\left(x-\frac12\right)^2. \] Hence \[ u_{n+1} =\frac12-x_{n+1} =\left(x_n-\frac12\right)^2 =u_n^2. \]
2. For the initial value \(x_0=0\),\[ x_n=\frac12-2^{-2^n}, \]so \((x_n)\) is bounded and strictly increasing and\[ \lambda=\boxed{\frac12}. \]
Here \(u_0=1/2\). Iterating step 1 gives \[ u_n=u_0^{2^n}=2^{-2^n}, \] and therefore \[ x_n=\frac12-u_n=\frac12-2^{-2^n}. \] The positive sequence \(2^{-2^n}\) is strictly decreasing to \(0\). Thus \(x_n\) is strictly increasing, is bounded above by \(1/2\), and converges to \(1/2\).
3. For an initial value \(y\in\RR\), one has \(y_\infty=\lambda\) exactly when\[ y\in\boxed{\left(-\frac12,\frac32\right)}. \]
Let \((y_n)\) be the orbit with \(y_0=y\), and set \(v_n=1/2-y_n\). By step 1, \[ v_n=v_0^{2^n}=\left(\frac12-y\right)^{2^n}. \] Since \(y_n=1/2-v_n\), the equality \(y_\infty=\lambda=1/2\) holds exactly when \(v_n\to0\). This occurs exactly when \[ \abs{v_0}=\abs{\frac12-y}<1, \] which is equivalent to \(-1/2<y<3/2\).
At either endpoint, \(\abs{v_0}=1\) and \(v_n=1\) for all \(n\geq1\), so the limit is \(-1/2\) rather than \(1/2\). If \(\abs{v_0}>1\), then \(v_n\to+\infty\), so the orbit does not converge to \(1/2\).
4. Q.E.D.
Step 2 proves part 1, and step 3 proves part 2.