Numerical Methods for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学：数值方法

📚 Numerical Methods for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学：数值方法

Numerical methods appear throughout Edexcel A-Level Pure Mathematics, especially in Paper 2 and Paper 3. They are used when an equation cannot be solved exactly using algebra, or when a definite integral has no standard antiderivative. This article covers locating roots, iteration, Newton-Raphson, the trapezium rule, error analysis, and exam technique in one complete revision guide.

数值方法贯穿 Edexcel A-Level 纯数学，尤其是 Paper 2 和 Paper 3。当一个方程无法用代数精确求解，或者一个定积分没有标准原函数时，就需要使用数值方法。本文完整覆盖定位根、迭代、牛顿-拉弗森法、梯形法则、误差分析以及考试技巧。

1. Locating Roots | 定位根

A root of the equation f(x) = 0 is a value x = α such that f(α) = 0. If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then there is at least one root in the interval (a, b). This is the sign-change rule and it is the standard justification for a root lying in a given interval.

方程 f(x) = 0 的根是使 f(α) = 0 的值 x = α。如果 f 在 [a, b] 上连续，且 f(a) 与 f(b) 异号，那么在区间 (a, b) 内至少存在一个根。这就是符号变化法则，也是证明根位于给定区间内的标准依据。

Always quote the values f(a) and f(b) to justify a sign change.
State that f is continuous if required by the question.
Use a calculator with care: a tiny value close to zero still counts as a sign change only if the sign is clear.

始终列出 f(a) 和 f(b) 的值来证明符号变化。
如果题目要求，需要说明 f 连续。
使用计算器时要小心：只有符号明确时，接近零的极小值才能算作符号变化。

2. Rearranging Equations into Iteration Form | 将方程改写为迭代形式

To solve f(x) = 0 by iteration, rearrange the equation into the form x = g(x). A common trick is to isolate one x, or add x to both sides, or use a reciprocal. The choice of rearrangement affects the convergence of the iterative sequence.

要用迭代法求解 f(x) = 0，需要把方程改写成 x = g(x) 的形式。常见的技巧是分离出一个 x，或在等式两边同时加上 x，或使用倒数。重排方式的选择会影响迭代序列的收敛性。

For example, x³ – 5x + 1 = 0 can be rearranged as x = (x³ + 1) / 5, or as x = ∛(5x – 1). Both have the form x = g(x), but only one may produce a convergent sequence from a given starting value.

例如，x³ – 5x + 1 = 0 可以重排为 x = (x³ + 1) / 5，也可以重排为 x = ∛(5x – 1)。两者都具有 x = g(x) 的形式，但从给定的初值出发，可能只有一个能产生收敛序列。

3. Iterative Sequences and Convergence | 迭代序列与收敛性

Once the equation is written as x = g(x), the iterative formula is xₙ₊₁ = g(xₙ). Starting with x₀, you generate x₁, x₂, x₃, and so on. If the sequence approaches a limit α, then α satisfies α = g(α), so α is a root of the original equation.

一旦方程写成 x = g(x)，迭代公式就是 xₙ₊₁ = g(xₙ)。从 x₀ 开始，依次生成 x₁、x₂、x₃ 等。如果序列趋近于极限 α，那么 α 满足 α = g(α)，因此 α 是原方程的一个根。

xₙ₊₁ = g(xₙ), n = 0, 1, 2, …

Convergence is not guaranteed. In many exam questions, you are given x₀ and asked to calculate x₁, x₂, x₃ to a stated accuracy. Always show at least six decimal places in intermediate work and round only the final answer.

收敛并非必然。在许多考试题中，会给出 x₀，并要求你计算到指定精度的 x₁、x₂、x₃。中间过程至少保留六位小数，最后才进行舍入。

4. Cobweb and Staircase Diagrams | 蛛网图与阶梯图

A cobweb diagram or staircase diagram is a graphical way to show how an iterative sequence behaves. You draw the curve y = g(x) and the line y = x. Starting at (x₀, 0), move vertically to y = g(x), then horizontally to y = x, and repeat. The pattern is a staircase if g(x) is increasing, and a cobweb if g(x) is decreasing near the root.

蛛网图或阶梯图是展示迭代序列行为的图形方法。画出曲线 y = g(x) 和直线 y = x。从 (x₀, 0) 出发，竖直移动到 y = g(x)，再水平移动到 y = x，重复进行。如果 g(x) 在根附近递增，图形为阶梯；如果 g(x) 递减，则为蛛网。

The behaviour gives a visual check for convergence: if the steps move toward the intersection of y = g(x) and y = x, the sequence converges; if they move away, it diverges.

这种行为提供了收敛性的直观判断：如果阶梯或蛛网路径逐步靠近 y = g(x) 与 y = x 的交点，则序列收敛；如果逐步远离，则发散。

5. Newton-Raphson Method | 牛顿-拉弗森法

The Newton-Raphson method uses the tangent to the curve y = f(x) at a point xₙ to find a better approximation to the root. The iterative formula is:

牛顿-拉弗森法利用曲线 y = f(x) 在点 xₙ 处的切线来求得根的一个更好近似。迭代公式为：

xₙ₊₁ = xₙ – f(xₙ) / f'(xₙ)

It converges very quickly when x₀ is close to the root, often doubling the number of correct decimal places at each step. You must differentiate f(x) correctly before applying the formula.

当 x₀ 接近根时，该方法收敛非常快，通常每一步正确小数位数会翻倍。在应用公式之前，必须先正确求出 f'(x)。

For example, to solve x² – 2 = 0, let f(x) = x² – 2 and f'(x) = 2x. Then the formula becomes xₙ₊₁ = xₙ – (xₙ² – 2) / (2xₙ). Starting from x₀ = 1.5 gives approximations to √2.

例如，求解 x² – 2 = 0，令 f(x) = x² – 2，f'(x) = 2x。则公式变为 xₙ₊₁ = xₙ – (xₙ² – 2) / (2xₙ)。从 x₀ = 1.5 出发可得到 √2 的近似值。

6. Failure Cases of Newton-Raphson | 牛顿-拉弗森法的失败情形

Newton-Raphson can fail if f'(xₙ) = 0, because the tangent is horizontal and never meets the x-axis. It can also cycle, or converge to the wrong root if the starting point is poorly chosen. A graph should be drawn to check suitability when a question asks about failure.

如果 f'(xₙ) = 0，牛顿-拉弗森法会失败，因为切线水平，永远不会与 x 轴相交。它也可能出现循环，或者当初值选择不当时收敛到错误的根。当题目涉及失败问题时，应画图检查适用性。

For instance, f(x) = x³ – x has f'(x) = 0 at x = ±1/√3, so using those as starting values would cause division by zero. Near these points, the method may be unstable.

例如，f(x) = x³ – x 在 x = ±1/√3 处有 f'(x) = 0，因此以这些值作为初值会导致除以零。在这些点附近，该法可能不稳定。

7. The Trapezium Rule | 梯形法则

The trapezium rule estimates the area under a curve by dividing the interval [a, b] into n equal strips of width h = (b – a) / n, and approximating each strip by a trapezium. The formula is:

梯形法则通过将区间 [a, b] 分成 n 个等宽为 h = (b – a) / n 的小段，并用梯形近似每一小段来估计曲线下的面积。公式为：

∫ₐᵇ f(x) dx ≈ h/2 [y₀ + 2(y₁ + y₂ + … + yₙ₋₁) + yₙ]

Here yᵢ = f(xᵢ), and xᵢ = a + ih. The first and last ordinates are multiplied by 1; all the inner ordinates are multiplied by 2.

其中 yᵢ = f(xᵢ)，xᵢ = a + ih。第一个和最后一个纵坐标乘以 1；所有内部纵坐标乘以 2。

If the curve is concave upward, the trapezium rule overestimates the integral; if concave downward, it underestimates. This is a common exam question requiring a sketch or a second derivative sign.

如果曲线是凹向上，梯形法则会高估积分；如果是凹向下，则会低估。这是常见考题，需要画草图或使用二阶导数的符号判断。

8. Accuracy and Error Bounds | 精度与误差界

Numerical answers are rarely exact. When a question says “give your answer to 3 decimal places”, you must work with more decimal places during calculation and round only at the end. For iteration, the root is usually confirmed when two successive approximations agree to the required number of decimal places.

数值答案很少精确。当题目要求“答案保留 3 位小数”时，必须在计算过程中保留更多小数位，最后才舍入。对迭代而言，通常当两个连续近似值在所需小数位下一致时，就确认了根。

The error in the trapezium rule decreases as n increases. More strips mean smaller h, but also more terms. Exam questions often ask whether an estimate is an overestimate or underestimate, rather than for a formal error bound.

梯形法则的误差随 n 增大而减小。更多的分段意味着更小的 h，但也意味着更多的项。考试题通常问估算是高估还是低估，而不会要求正式误差界。

9. Worked Example: Iteration | 例题：迭代

Question: The equation x³ – 3x – 5 = 0 has a root near x = 2. Show that x = (3x + 5)^(1/3) is a possible iterative formula, then use x₀ = 2 to find the root to 3 decimal places.

例题：方程 x³ – 3x – 5 = 0 在 x = 2 附近有一个根。证明 x = (3x + 5)^(1/3) 是一个可行的迭代公式，然后从 x₀ = 2 出发求根，精确到 3 位小数。

Starting from x³ – 3x – 5 = 0, rearrange x³ = 3x + 5, so x = ∛(3x + 5). Applying the formula with x₀ = 2:

由 x³ – 3x – 5 = 0，移项得 x³ = 3x + 5，所以 x = ∛(3x + 5)。应用公式，取 x₀ = 2：

x₁ = ∛(3(2) + 5) = ∛11 ≈ 2.22398

x₂ = ∛(3(2.22398) + 5) = ∛11.67194 ≈ 2.26837

x₃ = ∛(3(2.26837) + 5) = ∛11.80511 ≈ 2.27651

x₄ ≈ 2.27789, x₅ ≈ 2.27812, x₆ ≈ 2.27816

To 3 decimal places, the root is x ≈ 2.278. The iterations x₅ and x₆ agree when rounded to 3 decimal places, confirming the answer.

精确到 3 位小数，根为 x ≈ 2.278。x₅ 和 x₆ 在舍入到 3 位小数时一致，从而确认了答案。

10. Worked Example: Newton-Raphson | 例题：牛顿-拉弗森法

Question: Use Newton-Raphson with x₀ = 2 to find the root of f(x) = eˣ – 4x in the interval [0, 1] to 4 decimal places. You may use the fact that f'(x) = eˣ – 4.

例题：用牛顿-拉弗森法，从 x₀ = 2 出发，求 f(x) = eˣ – 4x 在区间 [0, 1] 内精确到 4 位小数的根。已知 f'(x) = eˣ – 4。

The formula is xₙ₊₁ = xₙ – (eˣⁿ – 4xₙ) / (eˣⁿ – 4). Starting from x₀ = 0.5:

公式为 xₙ₊₁ = xₙ – (eˣⁿ – 4xₙ) / (eˣⁿ – 4)。从 x₀ = 0.5 出发：

x₁ = 0.5 – (e^0.5 – 2) / (e^0.5 – 4) ≈ 0.35999

x₂ = 0.35999 – (e^0.35999 – 1.43996) / (e^0.35999 – 4) ≈ 0.35740

x₃ ≈ 0.35740, x₄ ≈ 0.35740

To 4 decimal places, the root is x ≈ 0.3574. The rapid convergence is typical of Newton-Raphson when the starting value is close.

精确到 4 位小数，根为 x ≈ 0.3574。当初值接近时，牛顿-拉弗森法典型的快速收敛特性在此得到体现。

11. Worked Example: Trapezium Rule | 例题：梯形法则

Question: Use the trapezium rule with four strips to estimate ∫₀² 1/(1 + x²) dx, giving your answer to 4 decimal places.

例题：用四个分段下的梯形法则估计 ∫₀² 1/(1 + x²) dx，答案保留 4 位小数。

Here a = 0, b = 2, n = 4, so h = (2 – 0) / 4 = 0.5. The x-values are 0, 0.5, 1, 1.5, 2. The corresponding y-values are:

此处 a = 0，b = 2，n = 4，所以 h = (2 – 0) / 4 = 0.5。x 值为 0、0.5、1、1.5、2。对应的 y 值为：

x	0	0.5	1	1.5	2
y = 1/(1+x²)	1	0.8	0.5	0.30769	0.2

Using the trapezium rule:

使用梯形法则：

∫₀² 1/(1 + x²) dx ≈ 0.5/2 [1 + 2(0.8 + 0.5 + 0.30769) + 0.2]

= 0.25 [1 + 2(1.60769) + 0.2]

= 0.25 [1 + 3.21538 + 0.2] = 0.25 × 4.41538 ≈ 1.10385

So the estimate is approximately 1.1039 (4 d.p.). The exact value is arctan 2 ≈ 1.1071, so the trapezium rule underestimates slightly, which is consistent with the curve’s concavity on part of the interval.

因此估计值约为 1.1039（保留 4 位小数）。精确值为 arctan 2 ≈ 1.1071，因此梯形法则略为低估，这与该曲线在区间部分位置的凹凸性一致。

12. Exam Technique and Revision Checklist | 考试技巧与复习清单

For Edexcel A-Level Maths, numerical methods questions often appear as structured exam items. You must be able to justify a root’s location using sign change, apply an iterative formula accurately, explain when Newton-Raphson fails, and use the trapezium rule with a table of ordinates.

对 Edexcel A-Level 数学而言，数值方法题常以结构化考题出现。你必须能够用符号变化证明根的位置，准确应用迭代公式，解释牛顿-拉弗森法何时失效，并使用纵坐标表应用梯形法则。

Always quote f(a) and f(b) with signs for a root-location question.
Show as many decimal places as possible in working, then round at the end.
Draw a sketch to decide whether a trapezium rule estimate is an overestimate or underestimate.
Check that f'(xₙ) is not zero before using Newton-Raphson.
Write the iterative formula clearly and use subscript notation correctly.

对于定位根的题目，始终列出 f(a) 和 f(b) 及其符号。
计算过程尽可能多保留小数位，最后再舍入。
画草图来判断梯形法则结果是高估还是低估。
使用牛顿-拉弗森法前，检查 f'(xₙ) 不为零。
清楚写出迭代公式，正确使用下标记号。

Mastering these techniques will give you confidence in the numerical methods section of Edexcel A-Level Mathematics. Practise past paper questions under timed conditions, and always check your calculator is in radian mode when trigonometric functions are involved.

掌握这些技巧将使你在 Edexcel A-Level 数学的数值方法部分充满信心。在限时条件下练习历年真题，并在涉及三角函数时始终检查计算器是否处于弧度模式。

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Numerical Methods for Edexcel A-Level Pure Maths | Edexcel A-Level 纯数学：数值方法