Newton 迭代
N e w t o n Newton Newton法基本格式
设给定 f ( x ) = 0 f(x)=0 f(x)=0的求根问题，我们构造以下迭代格式：
A
x k + 1 = x k − f ( x k ) f ′ ( x k ) x_{k+1}=x_k-\frac{f(x_k)}{f^{\prime}(x_k)} xk+1=xk−f′(xk)f(xk)
迭代函数为
φ ( x ) = x − f ( x ) f ′ ( x ) \varphi (x)=x-\frac{f(x)}{f^{\prime }(x)} φ(x)=x−f′(x)f(x)
其中容易得到收敛阶
lim k → ∞ x k + 1 − x ∗ ( x k − 1 − x ∗ ) 2 = f ′ ′ ( x ∗ ) 2 f ′ ( x ∗ ) \lim_{k\rightarrow \infty}\frac{x^{k+1}-x^*}{(x_{k-1}-x^*)^2}=\frac {f^{\prime\prime}(x^*)}{2f^{\prime}(x^*)} k→∞lim(xk−1−x∗)2xk+1−x∗=2f′(x∗)f′′(x∗)
N e w t o n Newton Newton求根公式收敛的充分条件
函数 f ( x ) f(x) f(x)在区间 [ a , b ] [a,b] [a,b]上二阶光滑，且在区间 [ a , b ] [a,b] [a,b]上单调， ∣ f ′ ′ ( x ) ∣ ≤ M |f^{\prime \prime}(x)|\leq M ∣f′′(x)∣≤M，对于选取的初解 x 0 ∈ [ a , b ] x_0 \in [a,b] x0∈[a,b]满足：
∣ x 0 − x ∗ ∣ ≤ 2 ∣ f ′ ( x ∗ ) ∣ M |x_0-x^*|\leq \frac{2|f^{\prime}(x^*)|}{M} ∣x0−x∗∣≤M2∣f′(x∗)∣
其中 x ∗ x^* x∗为 f ( x ) = 0 f(x)=0 f(x)=0的根，则牛顿迭代收敛且至少为二阶。
其它相近迭代格式
简化牛顿法
x k + 1 = x k − f ( x k ) f ′ ( x 0 ) x_{k+1}=x_k-\frac{f(x_k)}{f^{\prime}(x_0)} xk+1=xk−f′(x0)f(xk)
牛顿下山法
λ \lambda λ取 1 2 n \frac{1}{2^n} 2n1:
x ‾ k + 1 = x k − f ( x k ) f ′ ( x k ) \overline{x}_{k+1}=x_k -\frac{f(x_k)}{f^{\prime}(x_k)} xk+1=xk−f′(xk)f(xk)
x k + 1 = ( 1 − λ ) x k + λ x ‾ k + 1 x_{k+1}=(1-\lambda) x_k+\lambda \overline{x}_{k+1} xk+1=(1−λ)xk+λxk+1
重根情形
x k + 1 = x k − m f ( x k ) f ′ ( x k ) x_{k+1}=x_k-m\frac{f(x_k)}{f^{\prime}(x_k)} xk+1=xk−mf′(xk)f(xk)
弦截法
使用向后均差代替导数
x k + 1 = x k − f ( x k ) f ( x k ) − f ( x k + 1 ) ( x k − x k + 1 ) x_{k+1}=x_{k}-\frac{f(x_k)}{f(x_k)-f(x_{k+1})}(x_k-x_{k+1}) xk+1=xk−f(xk)−f(xk+1)f(xk)(xk−xk+1)
