更改

  <math> x_{n+1}=f(x_n), \quad n=0,1,2,\ldots.</math>

  <math> x_{n+1}=f(x_n), \quad n=0,1,2,\ldots.</math>

The fixed point  is stable if the absolute value of the derivative of  at  is strictly less than 1, and unstable if it is strictly greater than 1. This is because near the point , the function  has a linear approximation with slope :

The fixed point  is stable if the absolute value of the derivative of  at  is strictly less than 1, and unstable if it is strictly greater than 1. This is because near the point , the function  has a linear approximation with slope :

当 {{Math|''a''}} 点的导数{{Math|''f''}}的绝对值严格小于1时，不动点是稳定的; 当其严格大于1时是不稳定。这是因为在该点附近，函数线性近似的斜率为:

当函数 {{Math|''f''}} 在 {{Math|''a''}} 点的导数的绝对值严格小于1时，不动点是稳定的；当在 {{Math|''a''}} 点的导数严格大于1时是不稳定的。这是因为在这个点附近，函数的斜率具有的线性近似值为:

  <math> f(x) \approx f(a)+f'(a)(x-a). </math>

  <math> f(x) \approx f(a)+f'(a)(x-a). </math>

因此

因此

  <math>x_{n+1}-x_{n} = f(x_n)-x_n \simeq f(a) + f'(a)(x_n-a)-x_n = a + f'(a)(x_n-a)-x_n = (f'(a)-1)(x_n-a) \to \frac{x_{n+1}-x_{n}}{x_n-a}=f'(a)-1</math>

  <math>x_{n+1}-x_{n} = f(x_n)-x_n \simeq f(a) + f'(a)(x_n-a)-x_n = a + f'(a)(x_n-a)-x_n = (f'(a)-1)(x_n-a) \to \frac{x_{n+1}-x_{n}}{x_n-a}=f'(a)-1</math>

==External links==

== External links==

*[http://demonstrations.wolfram.com/StableEquilibria/ Stable Equilibria] by Michael Schreiber, [[The Wolfram Demonstrations Project]].

*[http://demonstrations.wolfram.com/StableEquilibria/ Stable Equilibria] by Michael Schreiber, [[The Wolfram Demonstrations Project]].