推导函数F(x)=f(x)/f’(x)的牛顿法公式并证明至少二次收敛

Alright, let's walk through this problem step by step — first deriving the iteration formula, then verifying its quadratic convergence to a simple root ( r ) of ( f(x) ).

Step 1: Derive the Newton Iteration Formula #

Newton's method for finding roots of a function ( G(x) ) follows the core rule:
[
x_{n+1} = x_n - \frac{G(x_n)}{G'(x_n)}
]
Here, our target function is ( F(x) = \frac{f(x)}{f'(x)} ). First, we compute ( F'(x) ) using the quotient rule:

[
F'(x) = \frac{f'(x) \cdot f'(x) - f(x) \cdot f''(x)}{\left[f'(x)\right]^2} = \frac{\left[f'(x)\right]^2 - f(x)f''(x)}{\left[f'(x)\right]^2}
]

Substitute ( F(x_n) ) and ( F'(x_n) ) into the Newton formula:
[
x_{n+1} = x_n - \frac{\frac{f(x_n)}{f'(x_n)}}{\frac{\left[f'(x_n)\right]^2 - f(x_n)f''(x_n)}{\left[f'(x_n)\right]^2}}
]

Simplify the complex fraction: the ( f'(x_n) ) in the denominator of the numerator cancels with one ( f'(x_n) ) from the denominator's numerator, leaving us with the iteration formula you already derived:
[
x_{n+1} = x_n - \frac{f(x_n)f'(x_n)}{\left[f'(x_n)\right]^2 - f(x_n)f''(x_n)}
]

Step 2: Prove Convergence is at Least Quadratic #

First, let's confirm ( f'(r) \neq 0 ): since ( r ) is a simple root of ( f(x) ), by definition ( f(r) = 0 ) and ( f'(r) \neq 0 ). Because ( f ) is three-times continuously differentiable, ( f'(x) ) is continuous near ( r ), so there's a neighborhood around ( r ) where ( f'(x) ) never vanishes. This ensures ( F(x) ) and our iteration formula are well-defined near ( r ).

To prove quadratic convergence, we analyze the iteration function ( \phi(x) = x - \frac{f(x)f'(x)}{\left[f'(x)\right]^2 - f(x)f''(x)} ). For fixed-point iterations ( x_{n+1} = \phi(x_n) ), quadratic convergence requires two key conditions: ( \phi(r) = r ) (fixed point) and ( \phi'(r) = 0 ).

1. Confirm ( \phi(r) = r ):
   Substitute ( x = r ): since ( f(r) = 0 ), the second term becomes ( \frac{0 \cdot f'(r)}{\left[f'(r)\right]^2 - 0} = 0 ). Thus ( \phi(r) = r - 0 = r ), so ( r ) is indeed a fixed point.

2. Compute ( \phi'(r) ):
   Let ( N(x) = f(x)f'(x) ) and ( D(x) = \left[f'(x)\right]^2 - f(x)f''(x) ), so ( \phi(x) = x - \frac{N(x)}{D(x)} ). Using the quotient rule for derivatives:
   [
   \phi'(x) = 1 - \frac{N'(x)D(x) - N(x)D'(x)}{\left[D(x)\right]^2}
   ]

   Calculate the necessary derivatives:

   • ( N'(x) = \left[f'(x)\right]^2 + f(x)f''(x) )
   • ( D'(x) = 2f'(x)f''(x) - \left[f'(x)f''(x) + f(x)f'''(x)\right] = f'(x)f''(x) - f(x)f'''(x) )

   Evaluate at ( x = r ) (where ( f(r) = 0 )):

   • ( N(r) = 0 ), ( D(r) = \left[f'(r)\right]^2 \neq 0 )
   • ( N'(r) = \left[f'(r)\right]^2 )
   • ( D'(r) = f'(r)f''(r) )

   Plug these into ( \phi'(r) ):
   [
   \phi'(r) = 1 - \frac{\left[f'(r)\right]^2 \cdot \left[f'(r)\right]^2 - 0}{\left[\left[f'(r)\right]^2\right]2} = 1 - \frac{\left[f'(r)\right]^{4}{\left[f'(r)\right]}4} = 1 - 1 = 0
   ]

Since ( \phi'(r) = 0 ), we use Taylor's theorem to expand ( \phi(x) ) around ( r ):
[
\phi(x) = \phi(r) + \phi'(r)(x - r) + \frac{1}{2}\phi''(\xi)(x - r)^2 = r + 0 + \frac{1}{2}\phi''(\xi)(x - r)^2
]
where ( \xi ) lies between ( x ) and ( r ). Let ( e_n = x_n - r ) (the error at step ( n )), then:
[
e_{n+1} = \phi(x_n) - r = \frac{1}{2}\phi''(\xi_n)e_n^2
]

Taking absolute values, ( |e_{n+1}| \leq C|e_n|^2 ) for some constant ( C ) (since ( \phi''(x) ) is continuous near ( r ), hence bounded). This confirms the method converges at least quadratically to ( r ).

内容的提问来源于stack exchange，提问作者james black