关于特定迭代序列收敛性及收敛速率的形式化证明咨询

Hey there! Let's break down how to formally prove the convergence rate of this sequence and link it to second-order (quadratic) convergence step by step.

1. Recap the Given Setup #

First, let's restate all key definitions to align our understanding:

• Starting point: $x_0 = 1$
• Target function: $f(x) = x^5$ (with derivatives $f'(x) = 5x^4$, $f''(x) = 20x^3$)
• Auxiliary function: $F(x) = \frac{f(x)}{f'(x)} = \frac{x}{5}$ (valid for $x \neq 0$)
• Iteration rule: Newton's method applied to $F(x)$, written as $x_{n+1} = x_n - \frac{F(x_n)}{F'(x_n)}$. You also noted this can be rewritten as $x_{n+1} = x_n - \frac{f(x_n)f''(x_n)}{\left[f'(x_n)\right]^2}$
• Observed sequence: $x_0=1, x_1=\frac{4}{5}, x_2=\frac{3}{5}, x_3=\frac{2}{5}, x_4=\frac{1}{5}, x_5=0$ (converges to 0 in exactly 5 steps)

2. Formal Convergence Rate Proof #

First, let's recall standard convergence rate definitions for a sequence ${x_n}$ converging to limit $\alpha$:

• Linear convergence: There exists a constant $0 < C < 1$ such that for large enough $n$, $|x_{n+1} - \alpha| \leq C |x_n - \alpha|$
• Quadratic (second-order) convergence: There exists a constant $C > 0$ such that for large enough $n$, $|x_{n+1} - \alpha| \leq C |x_n - \alpha|^2$
• Finite termination: The sequence reaches $\alpha$ exactly after a fixed number of iterations (this is stronger than any fixed-order convergence)

For your specific sequence:

• The limit $\alpha = 0$
• For each $n$ from 0 to 4, $|x_n - 0| = \frac{5-n}{5}$, and $|x_{n+1} - 0| = |x_n - 0| - \frac{1}{5}$

We can formalize the finite termination with induction:

1. Base case: $n=0$, $x_0=1$, $x_1=1 - \frac{1}{5} = \frac{4}{5}$ (matches your result)
2. Inductive step: Assume $x_k = \frac{5 - k}{5}$ for some $k \leq 4$. Then $x_{k+1} = x_k - \frac{1}{5} = \frac{5 - k}{5} - \frac{1}{5} = \frac{5 - (k+1)}{5}$
3. Conclusion: For $k=5$, $x_5 = \frac{5-5}{5} = 0$, so the sequence reaches the limit exactly at step 5.

3. Connecting to Second-Order Convergence #

The iteration rule you provided ($x_{n+1} = x_n - \frac{f(x_n)f''(x_n)}{\left[f'(x_n)\right]^2}$) is a variant of a root-finding method designed to handle multiple roots—a scenario where standard Newton's method only converges linearly.

To link this to second-order convergence, consider a general function $f(x)$ with a root $\alpha$ of multiplicity $m$ (meaning $f(\alpha)=f'(\alpha)=...=f^{(m-1)}(\alpha)=0$, but $f^{(m)}(\alpha) \neq 0$):

• Write $f(x) = (x - \alpha)^m g(x)$, where $g(\alpha) \neq 0$
• Compute derivatives and simplify the iteration term:
  $$
  \frac{f(x)f''(x)}{\left[f'(x)\right]^2} = \frac{m-1}{m} + O(x-\alpha)
  $$
• Let $e_n = x_n - \alpha$ (error at step $n$). The iteration becomes:
  $$
  e_{n+1} = e_n - \left(\frac{m-1}{m} + O(e_n)\right) = \frac{e_n}{m} + O(e_n^2)
  $$

While this looks linear at first glance, the $O(e_n^2)$ term hints at the method's connection to second-order convergence. For non-monomial functions with multiple roots, variants of this iteration (like Schröder's method) restore quadratic convergence by eliminating the linear error term.

4. Why Your Sequence Terminates in 5 Steps #

Since $f(x)=x^5$ is a degree-5 monomial, the modified iteration simplifies to a linear subtraction of $\frac{1}{5}$ each step. Starting from $x_0=1$, this means we hit 0 exactly after 5 iterations. This is a special case of finite termination, which can occur when root-finding methods are applied to polynomials with simple structures like monomials.

内容的提问来源于stack exchange，提问作者Andi