Newton-Raphson Method | 牛顿法求根考点精讲

📚 Newton-Raphson Method | 牛顿法求根考点精讲

The Newton-Raphson method is a powerful numerical technique used to find approximate solutions to equations of the form f(x) = 0. In the CCEA GCSE Mathematics specification, this method appears within the algebra and numerical methods topics, often assessed in the non-calculator and calculator papers. Mastering the iterative formula and recognising when the method succeeds or fails are essential skills for top marks. This guide breaks down every key concept, provides worked examples, and highlights exam-focused tips tailored to the CCEA style.

牛顿-拉夫森法（牛顿法）是一种强大的数值技术，用于寻找方程 f(x)=0 的近似解。在 CCEA GCSE 数学考试大纲中，该方法属于代数与数值方法模块，常出现在可使用计算器的试题中。掌握迭代公式、判断何时方法收敛或发散是获取高分的关键。本指南将逐一拆解核心概念，提供详细示例，并针对 CCEA 考试风格给出实用技巧。

1. What is the Newton-Raphson Method? | 什么是牛顿法？

The Newton-Raphson method, often simply called Newton’s method, is an iterative procedure that starts with an initial guess x₀ and improves it step by step using the gradient (derivative) of the function. The aim is to find a root α such that f(α) = 0. Each new approximation xₙ₊₁ is generated by following the tangent line at the current point down to the x-axis. This method usually converges very quickly when the starting value is close enough to the true root.

牛顿-拉夫森法通常简称为牛顿法，是一种迭代过程：从一个初始猜测值 x₀ 出发，利用函数的梯度（导数）逐步改进近似值，目标是找到满足 f(α)=0 的根 α。每一个新近似值 xₙ₊₁ 都通过从当前点沿切线方向延伸至 x 轴得到。当初值足够靠近真实根时，该方法通常收敛得非常快。

2. The Iterative Formula | 迭代公式

The core of the method is the iteration formula:

xₙ₊₁ = xₙ − f(xₙ) / f ‘(xₙ)

You must be able to apply this formula for a given function f(x). For example, if f(x) = x² − 5, then f ‘(x) = 2x. Substituting gives:

xₙ₊₁ = xₙ − (xₙ² − 5) / (2xₙ) = (xₙ + 5/xₙ) / 2

In the exam, you may be asked to derive this specific form or simply to use it repeatedly. Always write down the derivative clearly before substituting, as marks are often awarded for correct differentiation.

该方法的核心是迭代公式：xₙ₊₁ = xₙ − f(xₙ) / f ‘(xₙ)。你必须能够对给定的函数 f(x) 应用此公式。例如，若 f(x) = x² − 5，则 f ‘(x) = 2x，代入可得：xₙ₊₁ = xₙ − (xₙ² − 5) / (2xₙ) = (xₙ + 5/xₙ) / 2。考试中可能要求你推导这一具体形式，或简单地重复使用它进行计算。代入前务必清晰地写出导数，因为正确求导往往能获得步骤分。

3. Choosing an Initial Guess | 选择初始猜测值

A good starting value x₀ dramatically affects whether the method works. You should pick a value visibly near where the graph of y = f(x) crosses the x-axis. A sketch graph, even a rough one, helps avoid divergence. If two roots exist, make sure your x₀ lies closer to the desired root than to any other critical point.

一个好的初始值 x₀ 对方法是否有效影响极大。应当选择一个在 y = f(x) 图像与 x 轴交点附近的值。即使只是粗略画图，也能帮助避免发散。如果存在两个根，要确保 x₀ 比你想要的那个根更靠近目标根，而不是更靠近其他临界点。

Common advice: choose x₀ where f(x₀) and f ”(x₀) have the same sign, which helps guarantee monotonic convergence. For CCEA, however, you are rarely required to check the second derivative; instead, focus on using a sensible integer or decimal near the suspected root.

常见建议：选择使 f(x₀) 与 f ”(x₀) 同号的 x₀，这有助于保证单调收敛。但在 CCEA 考试中，很少要求检查二阶导数；重点在于使用接近猜测根的一个合理整数或小数。

4. Step-by-Step Process | 逐步求解过程

Follow this sequence in any Newton-Raphson question:

在任何牛顿法问题中请遵循以下步骤：

Identify f(x) and compute f ‘(x).

确定 f(x) 并计算导数 f ‘(x)。
Write down the iterative formula xₙ₊₁ = xₙ − f(xₙ) / f ‘(xₙ).

写出迭代公式。
Substitute the given starting value x₀ to find x₁.

代入给定的初始值 x₀ 求得 x₁。
Repeat the substitution using x₁ to find x₂, and so on.

用 x₁ 重复代入求出 x₂，依此类推。
Round each value to the required degree of accuracy (e.g. 4 decimal places).

将每个值四舍五入到要求的精度（如 4 位小数）。
Stop when successive approximations agree to the specified accuracy, or when you have performed the requested number of iterations.

当相邻近似值在指定精度下一致，或完成题目要求次数的迭代后停止。

5. Example 1: Solving a Quadratic Equation | 示例1：解二次方程

Find a root of f(x) = x² − 7 using the Newton-Raphson method, starting with x₀ = 2.5. Work to 4 decimal places.

用牛顿法求 f(x) = x² − 7 的一个根，初值 x₀ = 2.5，结果保留 4 位小数。

f ‘(x) = 2x, so the iteration is xₙ₊₁ = xₙ − (xₙ² − 7) / (2xₙ) = (xₙ + 7/xₙ) / 2.

f ‘(x) = 2x，故迭代公式为 xₙ₊₁ = (xₙ + 7/xₙ) / 2。

x₀ = 2.5 → x₁ = (2.5 + 7/2.5) / 2 = (2.5 + 2.8) / 2 = 2.65.

x₂ = (2.65 + 7/2.65) / 2 ≈ (2.65 + 2.6415) / 2 = 2.64575 ≈ 2.6458.

x₃ = (2.6458 + 7/2.6458) / 2 ≈ (2.6458 + 2.64575) / 2 = 2.645775 ≈ 2.6458.

Since x₂ and x₃ agree to 4 decimal places, the root is approximately 2.6458. The exact value is √7 ≈ 2.6458, confirming rapid convergence.

x₂ 与 x₃ 在 4 位小数下一致，因此根约为 2.6458。准确值为 √7 ≈ 2.6458，证实方法收敛迅速。

6. Example 2: Solving a Cubic Equation | 示例2：解三次方程

Given f(x) = x³ − 2x − 5, use x₀ = 2 to find the root correct to 3 decimal places.

已知 f(x) = x³ − 2x − 5，取 x₀ = 2，求根精确到 3 位小数。

f ‘(x) = 3x² − 2. Iteration: xₙ₊₁ = xₙ − (xₙ³ − 2xₙ − 5) / (3xₙ² − 2).

f ‘(x) = 3x² − 2。迭代公式：xₙ₊₁ = xₙ − (xₙ³ − 2xₙ − 5) / (3xₙ² − 2)。

x₀ = 2 → f(2)=8−4−5=−1, f ‘(2)=12−2=10. x₁ = 2 − (−1)/10 = 2.1.

x₁ = 2.1 → f(2.1)=9.261−4.2−5=0.061, f ‘(2.1)=13.23−2=11.23. x₂ = 2.1 − 0.061/11.23 ≈ 2.1 − 0.00543 = 2.09457.

x₂ ≈ 2.0946 → f(2.0946)≈9.1898−4.1892−5=−0.0006, f ‘(2.0946)≈13.162−2=11.162. x₃ ≈ 2.0946 − (−0.0006)/11.162 ≈ 2.09465.

x₂ and x₃ both round to 2.095, so the root is 2.095 to 3 d.p.

x₂ 与 x₃ 均四舍五入为 2.095，因此根精确到 3 位小数为 2.095。

7. Convergence and Divergence | 收敛与发散

The method converges when successive approximations get closer to the true root. However, it can diverge if x₀ is poorly chosen, for instance near a turning point where f ‘(x) is close to zero. In such cases, the tangent line may shoot far away, leading to wild oscillations or values that leave the domain.

当逐次近似值越来越接近真实根时，方法收敛。但如果 x₀ 选择不当，例如靠近导数为零的驻点，切线可能猛烈偏离，导致大幅振荡或跑出定义域。

CCEA questions sometimes ask you to explain why the method fails for a particular starting value. A typical answer: “Because f ‘(x₀) is very small, the correction term f(x₀)/f ‘(x₀) becomes very large, causing the next approximation to be far from the root.” Also, if the function is not defined for some values (e.g. square root of a negative), the iterations might break down.

CCEA 试题有时会要求解释为何某特定初始值导致方法失败。典型答案为：“因为 f ‘(x₀) 非常小，修正项 f(x₀)/f ‘(x₀) 变得很大，导致下一个近似值远离真实的根。”此外，若函数在某些点上无定义（如负数开平方根），迭代可能会中断。

8. Using a Calculator Efficiently | 高效使用计算器

Most CCEA Newton-Raphson questions permit a scientific calculator. Use the ANS key to store the latest x-value, which greatly speeds up repeated substitution. For example, after computing x₁, store it in memory or leave it as Ans. Then type the iteration expression entirely in terms of Ans. Each press of “=” generates the next iteration. Always write down the intermediate values to the required accuracy; do not round prematurely during the calculation.

大多数 CCEA 牛顿法题目允许使用科学计算器。利用 ANS 键存储最新的 x 值，可以极大加快重复代入的速度。例如，求出 x₁ 后将其存入记忆或保留为 Ans，然后用完全由 Ans 表达的迭代式输入。每按一次“=”就生成下一次迭代。务必按规定精度记录中间值，计算过程中不要过早四舍五入。

For the cubic example above, you could enter: Ans − (Ans³ − 2Ans − 5)/(3Ans² − 2) and then repeatedly press “=” after inputting 2 as the first Ans. This reduces keying errors and saves time.

就上述三次方程例子，你可以输入：Ans − (Ans³ − 2Ans − 5)/(3Ans² − 2)，并在首次将 2 作为 Ans 输入后反复按“=”。这能减少按键错误并节省时间。

9. Common Mistakes to Avoid | 常见错误

Forgetting the derivative: Using the wrong f ‘(x) is the single most frequent error. Double-check differentiation, especially for terms like 1/x or √x.

忘记求导：用错 f ‘(x) 是最常见的错误。务必复查导数，尤其是处理 1/x 或 √x 这类项时。
Incorrect substitution: Mixing up xₙ and xₙ₊₁. Always keep the iteration direction clear: old value on the right, new value on the left.

代入错误：混淆 xₙ 和 xₙ₊₁。始终保持迭代方向清晰：右侧为旧值，左侧为新值。
Premature rounding: Rounding intermediate answers too early can cause the final answer to drift outside the tolerance. Keep at least one more guard digit than required, and round only the final result.

过早四舍五入：太早将中间结果四舍五入可能导致最终答案超出允许误差。多保留至少一位保护数字，仅在最终结果处四舍五入。
Failing to stop: After sufficient iterations, the answer may stabilise. Continuing further wastes time and may introduce rounding errors. Stop when two consecutive approximations agree to the required precision.

不知何时停止：经过足够迭代后答案可能稳定。继续迭代浪费时间并可能引入四舍五入误差。当连续两次近似值达到所需精度时即可停止。

10. Exam Tips for CCEA GCSE | CCEA GCSE 考试技巧

In CCEA papers, Newton-Raphson questions typically carry 5–8 marks. They often provide the iterative formula, so you do not need to derive it from scratch. However, you may be asked to show that it reduces to a given form, which simply involves algebraic simplification. Marks are allocated for correct substitution, accurate calculation to the stated accuracy, and the final statement of the root.

在 CCEA 试卷中，牛顿法题目通常占 5–8 分。题目常常直接给出迭代公式，因此无需从零推导。但可能要求你展示其化简为给定形式，这仅涉及代数简化。分值分配在正确代入、按规定精度准确计算以及最终写出根的步骤上。

When asked to “show that the equation can be rearranged into the form…”, carefully do the algebra step by step, showing each manipulation clearly. Additionally, if the question asks for the root correct to a certain number of decimal places, you must explicitly check the next digit to justify rounding. Writing “2.6458 to 4 d.p.” is safer than simply “2.6458”, because it demonstrates you know the precision required.

当题目要求“证明方程可变形为……形式”时，请逐步仔细进行代数运算，清晰地展示每一步变换。此外，若题目要求根精确到某位小数，你必须通过检查下一位数字来证明四舍五入的正确性。写出“2.6458（保留4位小数）”比仅仅写“2.6458”更稳妥，因为这表明你清楚所要求的精度。

Also, be ready to interpret diagrammatic representations. Some questions show a graph with a tangent drawn from x₀, and ask you to read off x₁ from the x-intercept. This tests your understanding of the geometry behind the method. The x-intercept of the tangent is exactly the next approximation.

此外，要准备好解读图形表示。有些题目会展示一幅图象，并画出从 x₀ 出发的切线，要求你从 x 轴截距读取 x₁。这考察你对方法背后几何意义的理解。切线的 x 轴截距正是下一个近似值。

11. Relationship to Other Numerical Methods | 与其他数值方法的关系

Newton-Raphson is not the only numerical method in the CCEA specification. You should also be familiar with trial and improvement (decimal search) and interval bisection. Newton’s method is generally faster but requires calculus knowledge. In questions that ask you to compare methods, comment on speed, reliability, and the need for a derivative. For example: “Newton’s method converges more rapidly than decimal search, but it may fail if f ‘(x) is zero near the starting point, whereas interval bisection always works if a sign change is present.”

牛顿法并非 CCEA 大纲中唯一的数值方法。你还应熟悉试位法（十进制搜索）和区间二分法。牛顿法通常更快，但需要微积分知识。在要求比较方法的题目中，请评论速度、可靠性和求导需求。例如：“牛顿法比十进制搜索收敛更快，但若 f ‘(x) 在初始点附近为零则可能失败，而区间二分法只要存在符号变化就总能成功。”

Understanding these contrasts deepens your grasp of when to apply each technique, which is exactly the kind of reasoning high-achieving candidates demonstrate.

理解这些对比能加深你对何时应用每种技术的掌握，而这正是高分考生所能展现的推理能力。

12. Practice and Further Study | 练习与拓展

To master the Newton-Raphson method, work through past CCEA paper questions, paying attention to the mark schemes. Practice deriving the iterative formula for functions such as f(x) = eˣ − 4x or f(x) = x − cos x, even if such functions are beyond the core GCSE syllabus, because extended problems can use simple trigonometric or exponential forms. Always check that your final answer satisfies the original equation approximately.

要掌握牛顿法，请反复练习 CCEA 历年真题，并留意评分标准。练习为诸如 f(x) = eˣ − 4x 或 f(x) = x − cos x 等函数推导迭代公式，即便这些函数超出了 GCSE 的核心范围，因为进阶题目可能引入简单的三角或指数形式。务必检查最终答案是否近似满足原方程。

Remember, the method is named after Isaac Newton and Joseph Raphson, and its elegance lies in transforming a difficult root-finding problem into a simple repetitive calculation. With careful computation and a good initial guess, you can conquer any equation that comes your way in the examination hall.

请记住，该方法以艾萨克·牛顿和约瑟夫·拉夫森命名，其精妙之处在于将困难求根问题转化为简单的重复计算。只要细心计算并选择合理的初始猜测，你就能在考场上攻克任何遇到的方程。

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Newton-Raphson Method | 牛顿法求根考点精讲