The Newton–Raphson Method | 牛顿-拉夫森方法

📚 The Newton–Raphson Method | 牛顿-拉夫森方法

In A-Level mathematics, solving equations exactly is not always possible. The Newton–Raphson method is a powerful numerical technique for finding successively better approximations to the roots (or zeros) of a real-valued function. It uses the derivative of the function and an initial guess to converge rapidly to a solution.

在 A-Level 数学中,精确求解方程有时是不可能的。牛顿-拉夫森方法是一种强大的数值技术,用于逐步寻找实值函数根(或零点)的更好的近似值。它利用函数的导数和初始猜测值,快速收敛到解。

1. Why Numerical Methods? | 为什么需要数值方法?

Many equations, such as x = cos x or eˣ = 3x, cannot be solved algebraically to give an exact answer in terms of elementary functions. Numerical methods, including the Newton–Raphson method, allow us to find approximate solutions to a desired degree of accuracy.

许多方程,如 x = cos x 或 eˣ = 3x,无法用代数方法求得初等函数的精确解。数值方法,包括牛顿-拉夫森方法,使我们能够找到达到所需精度的近似解。

The Newton–Raphson method is particularly efficient because it frequently converges much faster than methods like bisection or linear interpolation. This makes it a favourite in both pure mathematics and applied contexts.

牛顿-拉夫森方法特别高效,因为它通常比二分法或线性插值等方法收敛得快得多。这使其在纯数学和应用情境中都备受青睐。


2. The Iterative Formula | 迭代公式

The Newton–Raphson method starts with an initial estimate x₀ for a root of the equation f(x) = 0. The iterative formula is

xₙ₊₁ = xₙ – f(xₙ) / f'(xₙ)

where f'(x) denotes the derivative of f(x).

牛顿-拉夫森方法从方程 f(x) = 0 的一个根的初始估计值 x₀ 开始。迭代公式为

xₙ₊₁ = xₙ – f(xₙ) / f'(xₙ)

其中 f'(x) 表示 f(x) 的导数。

The formula produces a sequence x₁, x₂, x₃, … that, under suitable conditions, converges to a root α.

该公式产生一个序列 x₁, x₂, x₃, …,在适当的条件下收敛到根 α。


3. Geometric Interpretation | 几何解释

Geometrically, the Newton–Raphson method approximates the root of f(x) = 0 by using tangents to the curve y = f(x). Starting at xₙ, the tangent line at (xₙ, f(xₙ)) has equation y – f(xₙ) = f'(xₙ)(x – xₙ). Setting y = 0 to find the x-intercept gives the next approximation xₙ₊₁.

从几何上看,牛顿-拉夫森方法通过使用曲线 y = f(x) 的切线来近似 f(x) = 0 的根。从 xₙ 开始,在点 (xₙ, f(xₙ)) 处的切线方程为 y – f(xₙ) = f'(xₙ)(x – xₙ)。令 y = 0 求 x 轴截距,就得到下一个近似值 xₙ₊₁。

This x-intercept is exactly xₙ – f(xₙ)/f'(xₙ). The method thus replaces the curve near the root by its tangent, and uses the intersection of the tangent with the x-axis as a better estimate.

这个 x 轴截距恰好是 xₙ – f(xₙ)/f'(xₙ)。因此该方法用曲线在根附近的切线代替曲线本身,并利用切线与 x 轴的交点作为更好的估计值。


4. Derivation Using Linear Approximation | 使用线性近似的推导

Consider the Taylor expansion of f(x) about xₙ: f(x) ≈ f(xₙ) + f'(xₙ)(x – xₙ). To find the root, we set f(x) = 0 and solve for x, assuming the linear approximation is exact. This yields 0 ≈ f(xₙ) + f'(xₙ)(x – xₙ), giving x ≈ xₙ – f(xₙ)/f'(xₙ). We take this as xₙ₊₁.

考虑 f(x) 在 xₙ 附近的泰勒展开:f(x) ≈ f(xₙ) + f'(xₙ)(x – xₙ)。为了求根,我们设 f(x) = 0 并求解 x,假设线性近似精确成立。得到 0 ≈ f(xₙ) + f'(xₙ)(x – xₙ),从而 x ≈ xₙ – f(xₙ)/f'(xₙ)。我们将此值作为 xₙ₊₁。

This derivation shows why the method works well when f'(x) is not too close to zero and the curvature is moderate – the linear term dominates the error.

这个推导表明,当 f'(x) 不太接近零且曲率适中时,该方法效果很好——线性项在误差中占主导地位。


5. Implementing the Method | 方法的实施步骤

To apply the Newton–Raphson method:

  • Rearrange the equation into the form f(x) = 0.
  • Choose an initial approximation x₀, possibly based on a graph or a sign-change interval.
  • Compute f(xₙ) and f'(xₙ).
  • Use the iterative formula to find x₁, x₂, … until the required accuracy is reached.

要应用牛顿-拉夫森方法:

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