Newton-Raphson Method for Root Finding | 牛顿-拉夫逊法求根详解

📚 Newton-Raphson Method for Root Finding | 牛顿-拉夫逊法求根详解

The Newton-Raphson method, also known simply as Newton’s method, is one of the most powerful and widely used numerical techniques for finding roots of a function. In A-Level Mathematics, it is a core topic in numerical methods, and it offers a fast, iterative approach to approximating solutions to equations that cannot be solved algebraically.

牛顿-拉夫逊法（Newton-Raphson method），通常简称为牛顿法，是求解函数零点最强大且应用最广泛的数值方法之一。在A-Level数学中，它是数值方法章节的核心内容，提供了一种快速、迭代的方式来逼近无法用代数方法求解的方程。

1. The Idea Behind Newton-Raphson | 牛顿-拉夫逊法的基本思想

The method is based on a simple geometric idea: to find a root of the equation f(x) = 0, we start with an initial guess x₀, then draw the tangent line to the curve y = f(x) at that point. The x-intercept of this tangent line gives us a new, usually better, approximation x₁. We repeat this process until the value converges to the root.

该方法的几何原理十分直观：要求解方程 f(x) = 0 的根，我们先取一个初始猜测值 x₀，然后在曲线 y = f(x) 上该点处作切线。切线的 x 轴截距就给出了一个新的、通常更精确的近似值 x₁。重复这个过程，直到数值收敛到方程的根。

The derivation of the iterative formula uses the point-slope form of a tangent line. At x = x₀, the tangent has slope f'(x₀), and passes through the point (x₀, f(x₀)). Setting the tangent line equation equal to zero at the intercept leads directly to the recurrence relation.

迭代公式的推导基于切线的点斜式方程。在 x = x₀ 处，切线的斜率为 f'(x₀)，并且经过点 (x₀, f(x₀))。令切线方程在截距处等于零，即可直接得到递推关系。

xₙ₊₁ = xₙ − f(xₙ) / f'(xₙ), provided f'(xₙ) ≠ 0

Here, n = 0, 1, 2, … represents the iteration number, and each new value xₙ₊₁ is computed from the previous value xₙ. The condition f'(xₙ) ≠ 0 is crucial, as the tangent line must not be horizontal for the intercept to exist.

其中 n = 0, 1, 2, … 表示迭代次数，每个新值 xₙ₊₁ 由前一个值 xₙ 计算得出。条件 f'(xₙ) ≠ 0 至关重要，因为切线不能是水平的，否则截距不存在。

2. Derivation of the Formula | 公式的推导

To derive the Newton-Raphson formula rigorously, consider the tangent line to the curve y = f(x) at x = x₀. The equation of this tangent line is given by the point-slope form:

为了严格推导牛顿-拉夫逊公式，考虑曲线 y = f(x) 在 x = x₀ 处的切线。该切线的方程由点斜式给出：

y − f(x₀) = f'(x₀)(x − x₀)

We want the x-value where this tangent crosses the x-axis, i.e., where y = 0. Substituting y = 0 into the tangent equation gives:

我们要求这条切线与 x 轴交点处的 x 值，即 y = 0 的地方。将 y = 0 代入切线方程：

0 − f(x₀) = f'(x₀)(x₁ − x₀)

Rearranging for x₁, we obtain x₁ = x₀ − f(x₀)/f'(x₀). Generalising this result by replacing x₀ with xₙ and x₁ with xₙ₊₁ yields the general iteration formula. This is the recurrence relation used for all subsequent approximations.

整理得到 x₁ = x₀ − f(x₀)/f'(x₀)。将 x₀ 推广为 xₙ，x₁ 推广为 xₙ₊₁，即可得到通用迭代公式。这就是用于后续所有近似计算的核心递推关系。

3. Worked Example – Finding a Root Step by Step | 实例演示——逐步求根

Let us apply the Newton-Raphson method to find a root of f(x) = x³ − 2x − 5 = 0. This classic example appears frequently in A-Level textbooks. We first compute the derivative: f'(x) = 3x² − 2. Taking an initial guess of x₀ = 2, we can begin the iterative process.

我们来用牛顿-拉夫逊法求 f(x) = x³ − 2x − 5 = 0 的一个根。这个经典例题在A-Level教材中经常出现。先计算导数：f'(x) = 3x² − 2。取初始猜测值 x₀ = 2，然后开始迭代。

Step 1: Evaluate f(2) = 8 − 4 − 5 = −1, and f'(2) = 12 − 2 = 10. Applying the formula gives x₁ = 2 − (−1)/10 = 2.1.

第一步：计算 f(2) = 8 − 4 − 5 = −1，f'(2) = 12 − 2 = 10。代入公式得 x₁ = 2 − (−1)/10 = 2.1。

Step 2: Evaluate f(2.1) = 9.261 − 4.2 − 5 = 0.061, and f'(2.1) = 13.23 − 2 = 11.23. Then x₂ = 2.1 − 0.061/11.23 ≈ 2.094568.

第二步：计算 f(2.1) = 9.261 − 4.2 − 5 = 0.061，f'(2.1) = 13.23 − 2 = 11.23。于是 x₂ = 2.1 − 0.061/11.23 ≈ 2.094568。

Step 3: Applying the formula again gives x₃ ≈ 2.094551. The values are converging quickly. After just three iterations, the root has stabilised to five decimal places, illustrating the quadratic convergence of this method.

第三步：再次代入公式得到 x₃ ≈ 2.094551。数值快速收敛。仅经过三次迭代，根已稳定到小数点后五位，体现了该方法的二次收敛特性。

Iteration n	xₙ	f(xₙ)
0	2.000000	−1.000000
1	2.100000	0.061000
2	2.094568	0.000185
3	2.094551	≈ 0.000000

4. Choosing the Initial Value | 如何选择初始值

The choice of the initial approximation x₀ is critical to the success of the Newton-Raphson method. A good initial guess should be sufficiently close to the true root. In A-Level exams, you are often asked to show that a root lies within a given interval, typically using the change-of-sign theorem, before applying the method.

初始近似值 x₀ 的选择对牛顿-拉夫逊法的成败至关重要。一个好的初始猜测值应当足够接近真实的根。在A-Level考试中，通常会要求你先用变号定理证明某个区间内存在根，然后再使用该方法。

To locate an interval containing a root, evaluate f(a) and f(b). If f(a) and f(b) have opposite signs, and f is continuous on [a, b], then by the Intermediate Value Theorem there exists at least one root in (a, b). The midpoint or an endpoint can then serve as a reasonable starting value.

要定位包含根的区间，先计算 f(a) 和 f(b)。如果 f(a) 与 f(b) 异号，且 f 在 [a, b] 上连续，那么根据介值定理，(a, b) 内至少存在一个根。此时可用中点或端点作为合理的初始值。

For our example f(x) = x³ − 2x − 5, we observe f(2) = −1 < 0 and f(3) = 16 > 0. Since the signs are opposite, the root lies between 2 and 3. Choosing x₀ = 2 is a safe and natural starting point, though other values within the interval would also work.

对于本例 f(x) = x³ − 2x − 5，我们观察到 f(2) = −1 < 0 而 f(3) = 16 > 0。由于符号相反，根位于 2 和 3 之间。选择 x₀ = 2 是安全且自然的起点，区间内的其他值通常也同样适用。

5. When Does Newton-Raphson Fail? | 牛顿-拉夫逊法何时失效？

The Newton-Raphson method is not infallible. There are several situations in which the method fails to converge to a root. The most common failure occurs when f'(xₙ) = 0 at some iteration. When the tangent line is horizontal, it never intersects the x-axis, and the formula becomes undefined due to division by zero.

牛顿-拉夫逊法并非万无一失。有几种情况会导致该方法无法收敛到一个根。最常见的失败情况是某次迭代时 f'(xₙ) = 0。当切线水平时，它永远不会与 x 轴相交，公式因除以零而无法定义。

Another failure mode is oscillation. For certain functions, the method may cycle between two values without ever converging to a root. This often happens when the initial guess is chosen poorly, or when the function has a local maximum or minimum near the root.

另一种失败模式是振荡。对于某些函数，该方法可能在两个值之间来回循环，永远无法收敛到一个根。这通常发生在初始猜测值选取不当，或者函数在根附近存在局部极大值或极小值的情况下。

Divergence is also possible: even when the tangent line intersects the x-axis, the new point may be further away from the root than the previous one. As the iterations continue, the values may move farther and farther from the true solution. This highlights the importance of choosing a sensible starting value.

发散同样可能发生：即使切线与 x 轴相交，新点也可能比前一个点离根更远。随着迭代的继续，数值可能离真实解越来越远。这凸显了选择合理初始值的重要性。

6. Geometric Interpretation of Divergence | 发散的几何解读

A tangent drawn at an unsuitable point may point away from the root. For example, if f'(x) is very small in magnitude, the denominator in the formula is small, producing a very large step. This can throw the next estimate far away from the region where the root lies.

在不合适的点处作切线，可能会指向远离根的方向。例如，如果 f'(x) 的绝对值很小，公式中的分母就很小，从而会产生一个非常大的步长。这可能会使下一个估计值远远偏离根所在的区域。

Consider a function with a stationary point near the root. If x₀ is close to the stationary point, then f'(x₀) ≈ 0, and the first iteration x₁ may be extremely large. Mathematically, this means the tangent is nearly horizontal, so its intersection with the x-axis is far away.

考虑一个在根附近存在驻点的函数。如果 x₀ 接近驻点，那么 f'(x₀) ≈ 0，第一次迭代产生的 x₁ 可能非常巨大。从数学上讲，这意味着切线接近水平，因此它与 x 轴的交点非常遥远。

In such cases, you should abandon the current initial value and choose a different one. A good strategy is to sketch the graph or evaluate the function at several points to ensure that the initial guess lies on a “well-behaved” part of the curve where f'(x) is reasonably large.

在这种情况下，应放弃当前的初始值并另选一个。一个好的策略是先画出函数草图，或在多个点上计算函数值，确保初始猜测落在曲线”性状良好”的区域，即 f'(x) 较大的部分。

7. Stopping Criteria and Accuracy | 停止条件与精度控制

In practice, we cannot iterate forever. We need a stopping criterion to determine when to halt the process. A common criterion in A-Level exams is to continue iterating until two successive approximations agree to a specified number of decimal places. For example, if consecutive estimates xₙ and xₙ₊₁ are both 2.09455 when rounded to 5 decimal places, we can stop and report this as the approximation of the root.

在实际操作中，我们无法永远迭代下去。需要一个停止条件来决定何时终止过程。A-Level考试中常用的标准是：持续迭代，直到相邻两次的近似值在指定位数的小数上一致。例如，如果连续两个估计值 xₙ 和 xₙ₊₁ 四舍五入到小数点后5位都为 2.09455，则可以停止并将此作为根的近似值。

Absolute error: Stop when |xₙ₊₁ − xₙ| < ε, where ε is a pre-set tolerance such as 0.0001.
绝对误差：当 |xₙ₊₁ − xₙ| < ε 时停止，其中 ε 是预设的容差，如 0.0001。
Function value: Stop when |f(xₙ₊₁)| is sufficiently small, since the function value at the root must be zero.
函数值：当 |f(xₙ₊₁)| 足够小时停止，因为根处的函数值应为零。
Fixed iterations: Some questions specify a fixed number of iterations, e.g., “perform 3 iterations.”
固定迭代次数：有些题目指定固定的迭代次数，例如”执行3次迭代”。

When reporting your answer, use the degree of accuracy requested by the question. If asked to give the root to 4 decimal places, ensure that both the final value and the function’s sign check support the accuracy of your approximation.

在报告答案时，务必使用题目要求的精度。如果要求将根精确到小数点后4位，请确保最终值以及函数值的符号检验都支持你的近似精度。

8. Comparison with Other Root-Finding Methods | 与其他求根方法的比较

A-Level Mathematics also introduces other numerical methods for solving equations, namely the change-of-sign method (bisection) and the fixed-point iteration (x = g(x)). Each method has its own advantages and disadvantages.

A-Level数学中还介绍了其他数值解方程的方法，即变号法（二分法）和不动点迭代法（x = g(x)）。每种方法各有优劣。

The bisection method is simple and guaranteed to converge as long as the initial interval contains a sign change. However, it is relatively slow, requiring many iterations to achieve high accuracy. Each iteration only halves the interval, so it may take ten or more steps to gain just three decimal places of accuracy.

二分法简单，只要初始区间包含变号就保证收敛。然而它的速度相对较慢，需要大量迭代才能达到较高精度。每次迭代只是将区间减半，因此获得三位小数精度可能需要十步以上。

Newton-Raphson, by contrast, converges very rapidly, typically doubling the number of accurate digits with each iteration. However, it requires the derivative f'(x) to be computed, and it is not guaranteed to converge from every starting point. As mentioned earlier, a poor initial guess can lead to divergence or oscillation.

相比之下，牛顿-拉夫逊法收敛非常快，通常每迭代一次精确位数就会翻倍。然而它需要计算导数 f'(x)，而且并非从任何起始点都保证收敛。如前所述，糟糕的初始猜测可能导致发散或振荡。

9. Exam Tips and Common Pitfalls | 考试技巧与常见误区

Many students lose marks on Newton-Raphson problems due to avoidable errors. Here are the most common pitfalls and how to avoid them. First, always check whether the derivative is correct before starting the iteration, as an incorrect f'(x) will produce incorrect approximations.

许多学生在牛顿-拉夫逊法题目上丢分都是因为一些可以避免的错误。以下是最常见的误区及规避方法。首先，在开始迭代之前务必检查导数是否正确，因为错误的 f'(x) 会产生错误的近似值。

Second, do not truncate intermediate values. Keep all values in your calculator’s memory and only round at the final stage, if at all. Truncating intermediate results can introduce significant errors in the final approximation.

其次，不要对中间值进行截断。将所有数值保留在计算器内存中，仅在最终阶段（如果需要）进行四舍五入。截断中间结果会给最终近似值引入显著的误差。

Third, be mindful of the notation. In the formula xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ), the subscript n + 1 must be written carefully. Confusing xₙ and xₙ₊₁ in your working is a common source of careless mistakes.

第三，注意符号的书写。在公式 xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ) 中，下标 n+1 必须仔细书写。在演算过程中混淆 xₙ 和 xₙ₊₁ 是粗心错误的常见来源。

Finally, if the question asks for the root to 5 decimal places, you must provide evidence by showing two consecutive iterations give the same value to 5 decimal places. Simply quoting the final value without this justification may not earn full marks.

最后，如果题目要求将根精确到小数点后5位，你必须通过展示连续两次迭代在小数点后5位上得到相同的值来提供证据。仅给出最终值而没有这一论证，可能无法获得满分。

10. Worked Exam-Style Question | 考试风格例题精解

Let us attempt a typical A-Level question: Given that f(x) = eˣ − 3x, show that the equation f(x) = 0 has a root in the interval [0, 1], and use the Newton-Raphson method with x₀ = 0.5 to find the root correct to 4 decimal places.

让我们来解答一道典型的A-Level题目：已知 f(x) = eˣ − 3x，证明方程 f(x) = 0 在区间 [0, 1] 内有一个根，并使用牛顿-拉夫逊法以 x₀ = 0.5 为初始值求根，精确到小数点后4位。

Solution | 解答： First, evaluate f(0) = e⁰ − 0 = 1 > 0 and f(1) = e − 3 ≈ −0.2817 < 0. Since the signs of f(0) and f(1) are opposite, and f is continuous on [0, 1], the Intermediate Value Theorem guarantees a root in this interval.

解答： 首先计算 f(0) = e⁰ − 0 = 1 > 0 和 f(1) = e − 3 ≈ −0.2817 < 0。由于 f(0) 与 f(1) 异号，且 f 在 [0, 1] 上连续，由介值定理保证该区间内存在一个根。

Next, compute the derivative: f'(x) = eˣ − 3. Starting with x₀ = 0.5:

接着计算导数：f'(x) = eˣ − 3。以 x₀ = 0.5 开始：

x₁ = 0.5 − (e⁰·⁵ − 1.5)/(e⁰·⁵ − 3) = 0.5 − (1.6487 − 1.5)/(1.6487 − 3) = 0.5 − 0.1487/(−1.3513) ≈ 0.6100.

x₁ = 0.5 − (e⁰·⁵ − 1.5)/(e⁰·⁵ − 3) = 0.5 − (1.6487 − 1.5)/(1.6487 − 3) = 0.5 − 0.1487/(−1.3513) ≈ 0.6100。

x₂ = 0.6100 − (e⁰·⁶¹ − 1.8300)/(e⁰·⁶¹ − 3) ≈ 0.6190.

x₂ = 0.6100 − (e⁰·⁶¹ − 1.8300)/(e⁰·⁶¹ − 3) ≈ 0.6190。

x₃ = 0.6190 − (e⁰·⁶¹⁹ − 1.8570)/(e⁰·⁶¹⁹ − 3) ≈ 0.6191.

x₃ = 0.6190 − (e⁰·⁶¹⁹ − 1.8570)/(e⁰·⁶¹⁹ − 3) ≈ 0.6191。

Since x₂ and x₃ both round to 0.6191, the root is 0.6191 correct to 4 decimal places. A final check with f(0.6191) ≈ 0 confirms the result.

由于 x₂ 和 x₃ 四舍五入后均为 0.6191，因此根为 0.6191（精确到小数点后4位）。最后用 f(0.6191) ≈ 0 验证结果无误。

11. Advanced Considerations: Order of Convergence | 进阶思考：收敛阶

The Newton-Raphson method is said to have quadratic convergence, meaning that if the error at step n is eₙ, then the error at step n+1 is roughly proportional to eₙ². This mathematical property explains why the number of correct digits roughly doubles after each iteration, making the method extremely efficient when it does converge.

牛顿-拉夫逊法具有二次收敛的性质，即如果第 n 步的误差为 eₙ，那么第 n+1 步的误差大致与 eₙ² 成正比。这一数学特性解释了为什么每次迭代后正确数字的个数大约翻倍，使得该方法在收敛时极为高效。

However, this rapid convergence relies on an important assumption: the root must be a simple root, meaning f'(r) ≠ 0 at the root r, and the function must be sufficiently smooth. If f'(r) = 0, the root is called a repeated root, and Newton’s method loses its quadratic convergence, degrading to linear convergence — much slower.

然而，这种快速收敛依赖一个重要假设：根必须是单根，即根 r 处满足 f'(r) ≠ 0，并且函数需要足够光滑。如果 f'(r) = 0，则该根被称为重根，此时牛顿法的二次收敛性会退化为线性收敛——速度大大降低。

This is because when the curve touches the x-axis at a repeated root, the tangent at any nearby point approximates the curve poorly relative to the flatness at the root. Special adaptations are required to restore quadratic convergence in such cases, but these are beyond the standard A-Level syllabus.

这是因为当曲线在重根处与 x 轴相切时，附近任意点处的切线对曲线的近似效果都会因根处的平坦性而变差。要在此类情况下恢复二次收敛，需要特殊的修正技巧，但这超出了A-Level标准大纲的范围。

12. Summary and Key Takeaways | 总结与核心要点

The Newton-Raphson method is an essential numerical technique in A-Level Mathematics. Mastering it requires understanding the underlying tangent-line geometry, memorising the iteration formula, and practising the computational steps fluently. The method’s fast convergence makes it the preferred choice in many numerical problems.

牛顿-拉夫逊法是A-Level数学中一项重要的数值技术。掌握它需要理解背后的切线几何原理、牢记迭代公式并熟练地进行计算步骤。该方法收敛速度快，使其成为许多数值问题中的首选方法。

Formula: xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ), provided f'(xₙ) ≠ 0.
公式：xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ)，前提是 f'(xₙ) ≠ 0。
Starting value: Choose x₀ near the root, often using the sign-change rule to locate an interval first.
初始值：选择靠近根的 x₀，通常先用变号规则确定含根区间。
Failure modes: Division by zero, oscillation, and divergence can all occur with poor guesses or pathological functions.
失效模式：猜测不佳或函数异常时，可能出现除零、振荡和发散。
Stopping: Iterate until consecutive approximations agree to the required accuracy.
停止条件：持续迭代直到连续近似值达到所需精度。

With consistent practice and attention to detail, Newton-Raphson questions can become one of the most reliable marks on your exam paper. Always remember to show your iterations clearly and justify your final answer to the requested degree of accuracy.

通过持续练习和对细节的重视，牛顿-拉夫逊法题目可以成为你考卷上最稳拿分的题型之一。务必清晰地展示迭代过程，并按照题目要求证明最终答案达到指定精度。

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Newton-Raphson Method for Root Finding | 牛顿-拉夫逊法求根详解