Question

2(Xn+1)² - 4Xn+1 + Xn = 0 X1 = 1/2 lim x-->inf (Xn)^a^n = L L is non zero and finite, find a and L

Answer 1

a=0, L=1

Answer 2

The recurrence relation is $2(X_{n+1})^2 - 4X_{n+1} + X_n = 0$. Solving for $X_{n+1}$ using the quadratic formula: $X_{n+1} = 1 \pm \frac{1}{2}\sqrt{4 - 2X_n}$. Given $X_1 = 1/2$. The branch $X_{n+1} = 1 + \frac{1}{2}\sqrt{4 - 2X_n}$ leads to $X_2 \approx 1.866$, and for $n \ge 2$, $X_n$ appears to increase and may exceed 2, leading to complex numbers. The branch $X_{n+1} = 1 - \frac{1}{2}\sqrt{4 - 2X_n}$ with $X_1 = 1/2$ yields $X_2 = 1 - \frac{\sqrt{3}}{2} \approx 0.134$. This sequence is decreasing and bounded below by 0, converging to $X=0$. We need $\lim_{n \to \infty} (X_n)^{a^n} = L$, with $L \neq 0$ and finite. If $a=0$, then $a^n=0$ for $n \ge 1$. Since $X_n \to 0$ and $X_n > 0$, $(X_n)^0 = 1$. Thus $L=1$. This gives $(a, L) = (0, 1)$. If $a \neq 0$, let $Y_n = (X_n)^{a^n}$, so $\ln(Y_n) = a^n \ln(X_n)$. As $X_n \to 0$, $\ln(X_n) \to -\infty$. If $a > 1$, $a^n \to \infty$, so $\ln(Y_n) \to -\infty$, $Y_n \to 0$, contradicting $L \neq 0$. If $0 < |a| < 1$, $a^n \to 0$. Using $X_n \sim C(1/4)^n$ for large $n$, $\ln(Y_n) = a^n (\ln C - n \ln 4) \to 0$. Thus $Y_n \to e^0 = 1$, so $L=1$. This gives $(a, L) = (a, 1)$ for $a \in (-1, 0) \cup (0, 1)$. If $a=-1$, $a^n$ alternates between $1$ and $-1$. The limit may not exist or be finite. For uniqueness, the conventional choice is $a=0$.