📚 The Newton–Raphson Method | 牛顿-拉夫森方法
The Newton– 0. Unlike algebraic factorisation, it uses an iterative process based on tangents to a curve, and it is especially useful when exact solutions are difficult or impossible to obtain.
牛顿-拉夫森方法是一种强大的数值方法，用于求方程 f(x)=0 的近似根。与代数因式分解不同，它利用曲线的切线进行迭代，特别适用于难以或无法得到精确解的情况。
1. What is the Newton–Raphson method? | 什么是牛顿-拉夫森方法？
the tangent line to the curve y=f(x) at the current point until it intersects the x-axis.
该方法从方程 f(x)=0 的一个初始估计值 x₀ 开始，然后生成序列 x₁、x₂、x₃……在合适条件下，该序列会快速收敛到某个根。每个新的近似值都是沿着曲线 y=f(x) 在当前点处的切线，找到切线与 x 轴的交点而得到的。
In Edexcel A-Level Mathematics, Newton–Raphson appears in the numerical methods topic. You are expected to use the iteration formula correctly, understand its geometric basis, apply it with a calculator, and describe situations where it may fail.
在 Edexcel A-Level 数学中，牛顿-拉夫森方法属于数值方法专题。考试要求你正确使用迭代公式，理解其几何基础，会用计算器执行迭代，并能说明其可能失效的情形。
2. The iteration formula | 迭代公式
If xₙ is an approximation to a root of f(x)=0, then the next approximation is given by:
如果 xₙ 是方程 f(x)=0 的一个近似根，那么下一个近似值由下式给出：
xₙ₊₁ = xₙ − f(xₙ) / f′(xₙ)
Here f′(xₙ) is the derivative of f evaluated at xₙ. The formula assumes that f′(xₙ) ≠ 0; division by zero would make the iteration undefined.
这里 f′(xₙ) 是函数 f 在 xₙ 处的导数。该公式要求 f′(xₙ) ≠ 0；除以零会使迭代无定义。
tion becomes xₙ₊₁ = xₙ − (xₙ²−5)/(2xₙ). Starting from x₀=2 gives rapid improvement towards √5.
例如，若 f(x)=x²−5，则 f′(x)=2x，迭代公式变为 xₙ₊₁ = xₙ − (xₙ²−5)/(2xₙ)。从 x₀=2 开始，可以快速逼近 √5。
3. Geometric interpretation | 几何解释
xₙ). Setting y=0 gives the x-axis intersection, which is exactly xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ).
在 x=xₙ 处，曲线 y=f(x) 的切线斜率为 f′(xₙ)。切线方程为 y − f(xₙ) = f′(xₙ)(x − xₙ)。令 y=0 得到切线与 x 轴的交点，正是 xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ)。
This means each step replaces the curve locally by its tangent. If the curve is smooth and the starting point is close to a root, the tangent line points almost directly at the root, so the approximation improves quickly.
这意味着每一步都用切线局部代替曲线。如果曲线光滑且起点接近根，切线几乎直接指向根，因此近似值迅速改善。
4. When to use Newton–Raphson | 何时使用牛顿-拉夫森方法
Newton– ly using algebraic methods. Examples include transcendental equations such as x = cos x, eˣ = 3x, or higher-degree polynomials without simple factors.
牛顿-拉夫森方法最适合用于无法用代数方法精确求解的方程。例子包括超越方程，如 x = cos x、eˣ = 3x，或没有简单因式的高次多项式。
In Edexcel exam questions, you are often asked to show that a root lies in an interval, then use Newton–Raphson with a given starting value. You may also need to explain why the method is appropriate or why it fails.
在 Edexcel 的考试题中，通常先要求证明某个区间内存在根，然后给出初始值使用牛顿-拉夫森方法。你也可能需要解释为什么该方法适用或为什么失效。
5. Choosing a starting value | 选择初始值
A good starting value x₀ should be close to the desired root. You can obtain it by sketching the graph, using a sign-change method, or using a graphical calculator. A poor choice can cause the sequence to converge to a different root or diverge.
一个好的初始值 x₀ 应该接近目标根。你可以通过画图、使用变号法或图形计算器来获得。初始值选择不当可能导致序列收敛到另一个根或发散。
For Edexcel papers, the starting value is normally given. However, if you need to choose one, first locate the root using a sign change between two integers, for example f(1)<0 and f(2)>0 suggests a root near 1.5.
在 Edexcel 试卷中，初始值通常已经给出。但如果需要自己选择，可先用两个整数之间的符号变化确定根的位置，例如 f(1)<0 且 f(2)>0 说明根在 1.5 附近。
6. Worked example | 实例解答
Let f(x) = x³ − 3x + 1. Use the Newton– method with x₀ = 0 to find the root near x = 0.3, giving your answer to 3 decimal places.
设 f(x) = x³ − 3x + 1。使用牛顿-拉夫森方法，取 x₀ = 0，求 x = 0.3 附近的根，答案精确到 3 位小数。
We first find f′(x) = 3x² − 3. The iteration formula is:
先求导数 f′(x) = 3x² − 3。迭代公式为：
xₙ₊₁ = xₙ − (xₙ³ − 3xₙ + 1) / (3xₙ² − 3)
Starting at x₀ = 0: f(0)=1, f′(0)=−3, so x₁ = 0 − 1/(−3) = 0.333333...
从 x₀ = 0 开始：f(0)=1，f′(0)=−3，因此 x₁ = 0 − 1/(−3) = 0.333333……
≈ 0.037037, f′(0.333333) ≈ −2.666667, so x₂ ≈ 0.333333 − 0.037037/(−2.666667) ≈ 0.347222.
下一步，x₁ = 0.333333：f(0.333333) ≈ 0.037037，f′(0.333333) ≈ −2.666667，因此 x₂ ≈ 0.333333 − 0.037037/(−2.666667) ≈ 0.347222。
≈ 0.347296, which is 0.347 to 3 decimal places. The method converged in only three steps.
继续迭代得到 x₃ ≈ 0.347296，即精确到 3 位小数为 0.347。该方法仅三步就收敛了。
7. Convergence and divergence | 收敛与发散
Convergence occurs when the successive approximations approach a fixed value. Newton–Raphson converges quadratically near a simple root, meaning the number of correct decimal places roughly doubles with each step once you are close enough.
当逐次近似值趋向某个固定值时，我们称之为收敛。牛顿-拉夫森方法在单根附近具有二次收敛性，即一旦足够接近，正确的小数位数每步大约翻倍。
Divergence can occur if the starting value is too far from a root, if f′(x) is close to zero near the starting point, or if the function has a discontinuity or a horizontal tangent between x₀ and the root. The sequence may oscillate, move away, or cycle without settling.
如果初始值离根太远、f′(x) 在起点附近接近零，或者函数在 x₀ 与根之间存在间断或水平切线，就可能发散。序列可能振荡、远离或循环而不稳定。
8. Failure cases | 失效情况
se the tangent is horizontal or nearly horizontal, it does not meet the x-axis near the root, and the next value may be far away.
考试中最常见的失效情况是 f′(x₀) = 0 或非常小。由于切线水平或接近水平，它不会在根附近与 x 轴相交，下一个近似值可能远离目标根。
undefined. Similarly, a starting value on one side of a maximum may send the tangent intersection to the other side, causing oscillation.
另一种失效发生在驻点或拐点附近。例如，若 f(x)=x² 且从 x₀=0 开始，则 f′(0)=0，公式无定义。类似地，位于极大值一侧的初始值可能使切线与 x 轴的交点跳到另一侧，导致振荡。
Exam questions may ask you to explain why the Newton–Raphson method fails for a particular x₀. The answer should mention the derivative being zero or close to zero, or the tangent being horizontal, rather than simply saying 'it does not work'.
考试题
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