A Simple Estimation

특정한 조건을 만족시키는 함수의 iteration에 대한 간단한 estimation입니다.

Assumptions. Suppose for \( a \neq 0 \), there is \( m > 0 \) such that \[ f(x) = x \left( 1 - (a + o(1)) x^{m} \right) \] near \( x = 0 \). Also, suppose the sequence \( (x_k) \) given by the recurrence relation \[ x_{k+1} = f(x_k) \] is positive and vanishes as \( k \to \infty \).

Calculation 1. By simple calculation, we have \[ \frac{1}{x_{k+1}^{m}} - \frac{1}{x_{k}^{m}} = ma + o(1).\] Summing, we have \[ \frac{1}{x_{k}^{m}} = mak + o(k), \] hence we have \[ x_{k}^{m} = \frac{1}{mak} \left( 1 + o (1) \right). \]

Calculation 2. Now suppose \[ f(x) = x \left(1 - a x^{m} + (b + o(1)) x^{2m} \right) \] instead. Then \[ \frac{1}{x_{k+1}^{m}} - \frac{1}{x_{k}^{m}} = ma + \frac{m (m+1) a^2 - 2 m b}{2} x_{k}^{m} + o \left( x_{k}^{m} \right).\] We know that \[ x_{k}^{m} = \frac{1}{ma(k+1)} + o \left( \frac{1}{k+1} \right), \] so plugging this back yields \[ \frac{1}{x_{k+1}^{m}} - \frac{1}{x_{k}^{m}} = ma + \frac{(m+1) a^2 - 2 b}{2a} \frac{1}{k+1} + o \left( \frac{1}{k+1} \right) .\] Summing, we have \[ \frac{1}{x_{k}^{m}} = mak + \frac{(m+1) a^2 - 2 b}{2a} \log k + o\left( \log k \right),\] hence we have \[ x_{k}^{m} = \frac{1}{mak} \left( 1 - \frac{(m+1) a^2 - 2 b}{2ma^2} \frac{\log k}{k} + o\left( \frac{\log k}{k} \right) \right). \]

Calculation 3. Finally, suppose \[ f(x) = x \left(1 - a x^{m} + b x^{2m} + O \left( x^{3m} \right) \right). \] Then \[ \frac{1}{x_{k+1}^{m}} - \frac{1}{x_{k}^{m}} = ma + \frac{m (m+1) a^2 - 2 m b}{2} x_{k}^{m} + O \left( x_{k}^{2m} \right),\] and by the estimation obtained by Calculation 2 shows that \[ \frac{1}{x_{k+1}^{m}} - \frac{1}{x_{k}^{m}} = ma + \frac{(m+1) a^2 - 2 b}{2a} \frac{1}{k+1} + O \left( \frac{\log k}{(k+1)^2} \right).\] Summing up yields \[ \frac{1}{x_{k}^{m}} = mak + \frac{(m+1) a^2 - 2 b}{2a} \log k + O\left( 1 \right),\] hence we obtain\[ x_{k}^{m} = \frac{1}{mak} \left( 1 - \frac{(m+1) a^2 - 2 b}{2ma^2} \frac{\log k}{k} + O \left( \frac{1}{k} \right) \right). \] In particluar, taking \( \frac{1}{m} \)-th power to both sides, we have \[ x_{k} = \left( \frac{1}{mak} \right)^{\frac{1}{m}} \left( 1 - \frac{(m+1) a^2 - 2 b}{2m^2a^2} \frac{\log k}{k} + O \left( \frac{1}{k} \right) \right). \]