📚 Newton's Method | 牛顿迭代法考点精讲
Newton's Method, also known as the Newton-Raphson method, is a powerful numerical technique for finding successively better approximations to the roots (or zeros) of a real-valued function. In IB and CIE Mathematics, this topic frequently appears in calculus and numerical methods sections, requiring students to understand both its iterative formula and its geometric interpretation. This revision guide breaks down all key points to help you master Newton's Method for your exams.
牛顿法，又称牛顿–拉夫逊方法，是一种求实值函数根（零点）的逐步逼近数值方法。在IB和CIE数学课程中，该主题常出现在微积分与数值方法部分，要求学生既掌握迭代公式，又理解其几何意义。本考点精讲为你梳理所有要点，助你全面攻克牛顿法。
1. Introduction to Newton's Method | 牛顿法简介
Newton' , provided the initial guess x₀ is sufficiently close to α and certain conditions are met.
牛顿法通过构造近似序列 {xₙ} 来逼近方程 f(x) = 0 的根 α，要求初始猜测 x₀ 足够靠近 α 并满足一定条件。
The method uses the tangent line at each approximation to predict where the function crosses the x-axis, making it both intuitive and efficient in many cases.
该方法利用每步近似值处的切线来预测函数与 x 轴的交点，在许多情况下既直观又高效。
2. Derivation of the Formula | 公式推导
Starting from a current estimate xₙ, we approximate f(x) by its tangent line at x = xₙ:
从当前估值 xₙ 出发，我们用 f(x) 在 x = xₙ 处的切线来近似函数：
f(x) ≈ f(xₙ) + f '(xₙ)(x – xₙ)
Setting the linear approximation to zero to find the next estimate xₙ₊₁ gives:
令线性近似值为零以求下一个估值 xₙ₊₁，可得：
0 = f(xₙ) + f '(xₙ)(xₙ₊₁ – xₙ)
Solving for xₙ₊₁ yields the Newton-Raphson formula.
解出 xₙ₊₁ 即得牛顿迭代公式。
xₙ₊₁ = xₙ – f(xₙ) / f '(xₙ)
3. The Iterative Formula | 迭代公式
The core iterative step for Newton's Method is:
牛顿法的核心迭代步骤为：
xₙ₊₁ = xₙ – f(xₙ) / f '(xₙ)
This formula requires computing both f(xₙ) and its derivative f '(xₙ) at each step. For many functions, this can be done analytically, but exam questions often provide the derivative or ask you to derive it first.
此公式每步都需计算 f(xₙ) 及其导数 f '(xₙ)。对多数函数可解析求导，但考试题常会给出导数，或要求你先自行求导。
4. Graphical Interpretation | 几何解释
f(x) at the point (xₙ, f(xₙ)). The tangent line crosses the x-axis at a point usually closer to the true root than xₙ, hence the method 'follows the tangent' downhill or uphill toward the root.
从几何上看，xₙ₊₁ 是曲线 y = f(x) 在点 (xₙ, f(xₙ)) 处的切线与 x 轴交点的横坐标。该交点通常比 xₙ 更接近真根，因此该方法沿着切线方向“滑向”实根。
This visualisation helps understand why a poor initial guess far from the root or near a stationary point can cause the method to diverge or oscillate.
这一几何直观有助于理解为何远离真根或靠近驻点的不良初始猜测会导致发散或振荡。
5. Conditions for Convergence | 收敛条件
For Newton's Method to converge to a root α, the following conditions should hold:
为使牛顿法收敛到根 α，应满足如下条件：
f(x) must be differentiable in a neighbourhood of α, with f '(α) ≠ 0 (simple root).
f(x) 在 α 附近必须可微，且 f '(α) ≠ 0（即 α 为单根）。
The initial guess x₀ must be sufficiently close to α.
初始猜测 x₀ 必须足够靠近 α。
f ''(x) should be continuous near α to ensure quadratic convergence.
f ''(x) 在 α 附近应连续以保证二次收敛。
Avoid intervals where f '(x) is very small; otherwise the step size f(xₙ)/f '(xₙ) becomes huge.
避开 f '(x) 极小的区间，否则步长 f(xₙ)/f '(xₙ) 会过大。
6. Rate of Convergence | 收敛速度
When an approximation is close to a simple root, Newton's Method typically exhibits quadratic convergence. This means the number of correct digits roughly doubles with each iteration:
当近似值靠近单根时，牛顿法通常具有二次收敛速度。这意味着每迭代一次，正确数字的位数大约翻倍：
|xₙ₊₁ – α| ≤ C |xₙ – α|²
For a multiple root (where f ' , which is much slower. IB and CIE may ask you to identify the order of convergence or to compare methods.
对于重根（f '(α) = 0），收敛降为线性，速度慢很多。IB 和 CIE 考试可能会要求你识别收敛阶或比较不同数值方法。
7. When Newton's Method Fails | 牛顿法失败的情形
Common failure modes include:
常见的失败情形有：
The derivative f '(xₙ) at some iteration becomes zero or very small, leading to a near-vertical tangent and a huge jump.
某次迭代的导数 f '(xₙ) 为零或极小，切线近乎垂直导致跳跃过大。
The sequence enters a cycle, oscillating between two values without approaching the root.
序列陷入循环，在两个值之间振荡而不逼近根。
Poor initial guess leads the sequence away from the intended root or toward a different root.
初始猜测不当使序列远离目标根或趋向另一根。
The function is not differentiable at some points (e.g., cusp), causing the method to break down.
函数在某些点不可微（如尖点），使方法失效。
8. Application: Finding Square Roots | 应用：求平方根
One classic application is approximating √a. Let f(x) = x² – a. Then f '(x) = 2x, and the iteration becomes:
经典应用之一为近似 √a。设 f(x) = x² – a，则 f '(x) = 2x，迭代公式化为：
xₙ₊₁ = ½ ( xₙ + a / xₙ )
Starting from a reasonable x₀ > 0, this produces rapid convergence to √a. For example, to estimate √2 using x₀ = 1: x₁ = 1.5, x₂ ≈ 1.4167, x₃ ≈ 1.4142157 – already very accurate.
只要从合理的 x₀ > 0 出发，就能快速收敛到 √a。例如以 x₀ = 1 估计 √2：x₁ = 1.5，x₂ ≈ 1.4167，x₃ ≈ 1.4142157，已极为精确。
9. Application: Solving Polynomial Equations | 应用：解多项式方程
Newton' oots of higher-degree polynomials where algebraic solutions are unavailable. For f(x) = x³ – 2x – 5, a typical exam problem might ask for the root near 2, applying the iteration:
牛顿法广泛用于求解无法用代数解的高次多项式实根。对于 f(x) = x³ – 2x – 5，典型考题可能要求逼近 x = 2 附近的根，应用迭代：
xₙ₊₁ = xₙ – (xₙ³ – 2xₙ – 5) / (3xₙ² – 2)
Students are often asked to perform two or three iterations and give the result to a specified accuracy. Always show the full substitution to earn method marks.
学生常被要求进行两到三次迭代，并按指定精度给出结果。务必展示完整代入过程以获取方法分。
10. Exam Tips for IB and CIE | IB与CIE考试技巧
Always write down the formula xₙ₊₁ = xₙ – f(xₙ) / f '(xₙ) before substituting. This demonstrates understanding and guards against errors.
代入前务必先写出公式 xₙ₊₁ = xₙ – f(xₙ) / f '(xₙ)，这既表明理解也能减少错误。
Show the derivative calculation clearly. In CIE, you may be given f '(x) already; in IB, you might need to differentiate.
清晰展示导数计算。CIE 中有时会直接给出 f '(x)；IB 可能需自己求导。
Use sufficient decimal places during working (at least 5 or 6) and round only the final answer to required accuracy.
计算过程中保留足够小数位数（至少5或6位），只对最终答案按要求精度四舍五入。
State the stopping criterion: either a fixed number of iterations, or when |xₙ₊₁ – xₙ| < ε.
说明停止准则：或为固定迭代次数，或当 |xₙ₊₁ – xₙ| < ε 时停止。
If asked about convergence, check that f '(α) ≠ 0 and that x₀ is near α. Mention the risk of divergence near turning points.
若问及收敛性，检查 f '(α) ≠ 0 且 x₀ 靠近 α；提及在驻点附近可能发散。
11. Worked Example | 典型例题
Find the root of f(x) = eˣ – 4x near x = 0.5, correct to 4 decimal places.
求 f(x) = eˣ – 4x 在 x = 0.5 附近的根，精确至 4 位小数。
Derivative: f '(x) = eˣ – 4.
导数：f '(x) = eˣ – 4。
Using x₀ = 0.5:
使用 x₀ = 0.5：
x₁ = 0.5 – (e⁰·⁵ – 2) / (e⁰·⁵ – 4)
Compute e⁰·⁵ ≈ 1.64872: f(0.5) ≈ 1.64872 – 2 = –0.35128; f '(0.5) ≈ 1.64872 – 4 = –2.35128. So x₁ ≈ 0.5 – (–0.35128)/(–2.35128) ≈ 0.5 – 0.14940 = 0.35060.
计算 e⁰·⁵ ≈ 1.64872：f(0.5) ≈ –0.35128；f '(0.5) ≈ –2.35128。故 x₁ ≈ 0.35060。
Next iteration: x₂ = 0.35060 – (e⁰·³⁵⁰⁶⁰ – 4×0.35060) / (e⁰·³⁵⁰⁶⁰ – 4). e⁰·³⁵⁰⁶⁰ ≈ 1.41997; f ≈ 1.41997 – 1.40240 = 0.01757; f ' ≈ –2.58003. Then x₂ ≈ 0.35060 – 0.01757/(–2.58003) ≈ 0.35060 + 0.00681 = 0.35741. Repeat until stable. Final root ≈ 0.3574 (4 d.p.).
下一次迭代：x₂ ≈ 0.35741。重复直至稳定。最终根 ≈ 0.3574（4位小数）。
12. Common Mistakes | 常见错误
| Mistake | Why it happens |
|---|---|
| Forgetting the derivative or differentiating incorrectly | Rushing into iteration without checking f '(x) |
| Writing xₙ₊₁ = f(xₙ)/f '(xₙ) without the minus sign | Misremembering the formula |
| Using a poor initial guess leading to divergence | Not sketching the function or checking local behaviour |
| Rounding intermediate values too early | Desire to save time, but this spoils accuracy |
| Stopping after one iteration on accuracy questions | Not checking the required stopping condition |
中英对照：
| 常见错误 | 原因 |
|---|---|
| 忘记求导或求导错误 | 急于迭代而未检查 f '(x) |
| 写成 xₙ₊₁ = f(xₙ)/f '(xₙ)，漏掉负号 | 公式记忆不清 |
| 初始猜测不当导致发散 | 未画草图或未分析局部特性 |
| 过早舍入中间值 | 为省时间，但破坏精度 |
| 在精度要求题中只迭代一次就停止 | 未检查所需的收敛条件 |
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