IGCSE CIE Mathematics: Numerical Methods Key Points | IGCSE CIE 数学：数值方法 考点精讲

📚 IGCSE CIE Mathematics: Numerical Methods Key Points | IGCSE CIE 数学：数值方法 考点精讲

When exact algebraic solutions are impossible or difficult to obtain, numerical methods provide systematic ways to find approximate roots of equations to a desired degree of accuracy. In IGCSE CIE Mathematics, you will need to apply trial and improvement, use given iterative formulas, and understand how to judge when an answer is accurate enough. This guide covers all the essential techniques, common pitfalls, and exam-style tips to help you score full marks.

当代数解难以求出或不可能求出时，数值方法提供了系统化的方式，来求出方程满足所需精度的近似根。在IGCSE CIE数学中，你需要会使用试位法、代入给定的迭代公式，并会判断答案何时达到足够的精度。本指南涵盖所有关键方法、常见错误及考试技巧，助你拿满分数。

1. What Are Numerical Methods? | 什么是数值方法？

Numerical methods are techniques to find approximate solutions to equations, typically by starting with an initial guess and refining it step by step. Instead of solving an equation like x³ + 2x − 5 = 0 algebraically, we home in on a root using repeated calculations. The core idea is iteration: produce a sequence of values that get closer and closer to the true solution.

数值方法是求方程近似解的技术，通常从一个初始猜测值开始，再逐步修正。我们不用代数方法去解像 x³ + 2x − 5 = 0 这样的方程，而是通过重复计算逼近根。核心思想是迭代：生成一列越来越接近真解的值。

2. Trial and Improvement Method | 试位法

Trial and improvement (or ‘trial and error’) is a systematic guessing strategy. Choose two x-values that give function outputs with opposite signs – this confirms a root lies between them. Keep testing midpoints until the function value is very close to zero or until the x-values agree to the required decimal places.

试位法是一种有系统的猜测策略。选取两个使函数值异号的 x 值——由此确定根在它们之间。不断测试中点，直到函数值非常接近零，或者 x 值在小数位上达到要求的一致性。

Example: Solve x³ − x − 2 = 0, correct to 1 decimal place. Try x=1: 1−1−2=−2 (negative). Try x=2: 8−2−2=4 (positive). Root is between 1 and 2. Try 1.5: 3.375−1.5−2=−0.125 (negative). Try 1.6: 4.096−1.6−2=0.496 (positive). Between 1.5 and 1.6. Try 1.55: 3.723875−1.55−2=0.173875 (positive). 1.5 gives negative, so root between 1.5 and 1.55. Check 1.52: 3.511808−1.52−2=−0.008192 (negative). 1.53: 3.581577−1.53−2=0.051577 (positive). The root lies between 1.52 and 1.53. Both give 1.5 to 1 decimal place, so answer is 1.5.

例：解 x³ − x − 2 = 0，精确到1位小数。试 x=1: 负；x=2: 正。根在1和2之间。试1.5: 负；1.6: 正。介于1.5与1.6。试1.55: 正，而1.5为负，故根在1.5和1.55之间。试1.52: 负；1.53: 正。根在1.52与1.53之间。两者都给出1.5（精确到1位小数），故答案为1.5。

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3. Bisection Method (Interval Halving) | 二分法（区间半分法）

The bisection method repeatedly halves the interval that contains the root. If f(a) and f(b) have opposite signs, find the midpoint c = (a+b)/2. If f(c) has the same sign as f(a), replace a with c; otherwise replace b with c. This guarantees the interval size halves each time, giving a very controlled convergence.

二分法反复把包含根的区间减半。若 f(a) 与 f(b) 异号，取中点 c = (a+b)/2。若 f(c) 的符号与 f(a) 相同，就用 c 替换 a；否则替换 b。这样每次区间长度减半，收敛非常稳定可控。

It is useful when a function is continuous and you can easily identify a sign change. The required number of steps to achieve a certain accuracy can be calculated: for interval width (b−a), after n steps width is (b−a)/2ⁿ. Set this less than the desired tolerance to find n.

当函数连续且能轻松找到符号改变时，二分法很有用。要达到某一精度所需的步数可以计算：若初始区间宽度为 (b−a)，n 步后宽度为 (b−a)/2ⁿ。令其小于所需容差即可求出 n。

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4. Working with Given Iterative Formulas | 使用给定的迭代公式

CIE often provides an iteration formula of the form xₙ₊₁ = g(xₙ) and asks you to find an approximate root. You start with a given x₀ (or choose a reasonable value), substitute into the right side to get x₁, then x₂, and so on, until consecutive x‑values agree to the required number of decimal places.

CIE 常给出形如 xₙ₊₁ = g(xₙ) 的迭代公式，让你求近似根。你从给定的 x₀（或选一个合理的值）开始，代入右侧求出 x₁，再求 x₂，依此类推，直到连续两个 x 值在所求小数位上一致。

Example: Use the iteration xₙ₊₁ = √(5 − xₙ) with x₀ = 2 to find a root to 2 decimal places. x₁ = √(5−2) = √3 ≈ 1.73205. x₂ = √(5−1.73205) = √3.26795 ≈ 1.80773. x₃ = √(5−1.80773) = √3.19227 ≈ 1.78670. x₄ = √(5−1.78670) = √3.21330 ≈ 1.79258. Continue until stable to 2 d.p. (eventually about 1.79).

例：用迭代公式 xₙ₊₁ = √(5 − xₙ)，设 x₀ = 2，求根精确到 2 位小数。x₁ = √(5−2) = √3 ≈ 1.73205；x₂ ≈ 1.80773；x₃ ≈ 1.78670；x₄ ≈ 1.79258；继续直到两位小数稳定（最终约1.79）。

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5. Rearranging Equations into Iteration Form | 把方程变形为迭代形式

If a question asks you to show that x = h(x) is a suitable rearrangement of f(x) = 0, you need to check that the root of f(x)=0 is also a fixed point of the iteration and that the iteration converges near the root. Often, you simply rearrange algebraically: e.g. from x² − 3x + 1 = 0, one form is x = (x²+1)/3. Another is x = √(3x−1). You must be able to derive these.

如果题目要求你说明 x = h(x) 是 f(x)=0 的一个适当的变形，你需要验证 f(x)=0 的根也是该迭代的不动点，并且迭代在根附近收敛。通常就是把方程作代数变形：比如由 x² − 3x + 1 = 0 可得 x = (x²+1)/3，也可得 x = √(3x−1)。你必须能自己得出这些形式。

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6. Testing Accuracy and Stopping Criteria | 检验精度与停止条件

For trial and improvement or bisection, the stopping condition is usually when the two bounding x-values round to the same number at the required level of accuracy (e.g. both 1.5 to 1 d.p.). For iterative formulas, you stop when successive iterates agree to the required decimal places. The examination expects you to explicitly state this.

对试位法或二分法，停止条件通常是所夹的两个 x 值在所需精度下四舍五入后相同（如都精确到 1 位小数得 1.5）。对迭代公式，当相继两个迭代值在所求小数位上一致时停止。考试中要求你明确表述这一点。

Never stop just because the function value is small, unless the question says so. Always check the actual x-values. Also, be careful with rounding: carry out calculations to at least one more decimal place than required, to avoid premature or incorrect rounding.

绝不能仅因为函数值很小就停止，除非题目明确要求。一定要检查 x 值本身。此外，小心舍入：计算时至少保留比要求多一位的小数，以免过早舍入或产生舍入误差。

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7. Using Graphs to Estimate Roots | 利用图像估计根

A graph can provide a good starting interval. Plot f(x) or use a given graph to see where the curve crosses the x-axis. You can then read off approximate x-values to initiate a trial and improvement or iteration process. In some exam questions, you will be given a graph and asked to complete a table of values to further refine the root.

图像可以提供一个良好的初始区间。画出 f(x) 的图像，或利用给出的图像观察曲线与 x 轴的交点。然后读取近似的 x 值，启动试位法或迭代过程。有些考题会给出图像，让你补全数值表以进一步细化根。

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8. Common Mistakes and How to Avoid Them | 常见错误与如何避免

Mistake 1: Using a rounded value instead of the full display in intermediate steps. This can propagate error and lead to a different final answer. Always use the calculator’s memory or keep maximum precision until the end.

错误1：在中间步骤使用舍入后的值，而非计算器显示的全部数值。这会将误差传递下去，导致最终答案不同。应始终使用计算器存储的精确值，或到最后才舍入。

Mistake 2: Stopping the trial and improvement process too early. Just because one trial gives a value near zero does not mean the x-value is accurate to the required decimals. Verify that the two neighbouring x-values round to the same target decimal.

错误2：过早终止试位法。某次试值使函数值接近零，并不代表 x 值在所需小数位上准确。要验证相邻两个 x 值四舍五入后一致。

Mistake 3: Not checking sign change. If f(a) and f(b) do not have opposite signs, the root might be outside that interval, or the function could touch the axis without crossing, which often indicates a multiple root – but typically at IGCSE a clear sign change is required.

错误3：未检查符号变化。若 f(a) 和 f(b) 同号，根可能不在该区间内，或者函数与轴相切却没有穿过——这通常对应着重根，但就IGCSE而言，通常要求有明显的符号改变。

9. Worked Example: Trial and Improvement with Table | 例题：用表格进行试位法

The equation x³ + x = 10 has a solution between 2 and 3. Find the solution correct to 1 decimal place.

方程 x³ + x = 10 在 2 与 3 之间有一个解。求该解，精确到 1 位小数。

x	x³ + x	Comment
2	10	Equal to 10 – root at 2? Check. Actually 2³+2=10 exactly. This is exact solution, but example continues: suppose equation was x³+x=11 for illustration. Let’s adjust: x³+x=11, root near 2.

Let us use a proper example: solve x³ + 2x = 20, root between 2 and 3, answer to 1 d.p.

x	x³ + 2x	Comparison to 20
2.0	12	too low
3.0	33	too high
2.5	20.625	too high
2.3	16.767	too low
2.4	19.024	too low
2.45	20.5586	too high
2.42	20.0143	slightly high
2.41	19.8239	too low

From the table, root lies between 2.41 and 2.42. Both round to 2.4 to 1 decimal place. Hence the solution is x = 2.4 (1 d.p.).

从表中可见，根在2.41和2.42之间。两者都精确到1位小数得2.4。故解为 x = 2.4（1 d.p.）。

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10. Iteration and Convergence Conditions | 迭代与收敛条件

Not every rearrangement leads to a convergent sequence. For xₙ₊₁ = g(xₙ) to converge to a root near r, we need |g'(r)| < 1 (at IGCSE this is not always tested, but you may be told 'the iteration converges'. You should still follow the process). If the sequence diverges (values getting larger, or oscillating wildly), you may need to try a different starting value or a different rearrangement.

并非每种变形都能得到收敛序列。要使迭代 xₙ₊₁ = g(xₙ) 收敛到 r 附近的根，需要 |g'(r)| < 1（在IGCSE中不总要求，但可能会提示“该迭代收敛”）。如果序列发散（值越来越大，或剧烈振荡），你可能需要换一个初值或换一种变形。

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11. Using Calculator Efficiently | 高效使用计算器

Use the ANS key to speed up iteration. Enter the initial value, then type the formula using ANS. Pressing equals repeatedly performs the iteration without retyping. For trial and improvement, store intermediate values in memory and use the table feature if available. This saves time and reduces transcription errors.

使用 ANS 键可以加速迭代。输入初值，然后用 ANS 键入公式。反复按等号即可连续迭代，无需重新输入。对于试位法，可将中间值存入记忆，若可用表格功能则更佳。这能节约时间并减少抄写错误。

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12. Summary and Exam Tips | 总结与考试提示

• Always show clear substitution steps, not just the final result. Write down each trial or iterate.

  始终展示清晰的代入步骤，不要只写最终结果。写下每一次试值或迭代值。

• State explicitly the reason for choosing the final answer: e.g. ‘root lies between 2.41 and 2.42, both round to 2.4 to 1 d.p.’

  明确陈述选择最终答案的理由，例如：“根介于 2.41 与 2.42 之间，两者四舍五入到 1 位小数均为 2.4。”

• Keep full accuracy during calculation, round only for the final answer. Use at least one extra decimal place in working.

  计算过程中保留全部精度，只在最后答案舍入。过程中至少多保留一位小数。

• If an iteration formula is given, use it exactly as shown. Do not attempt to simplify it.

  如果给出迭代公式，严格按照原式使用。不要试图化简。

• Check that your final answer makes sense: substitute back into the original equation to see if it is close to zero.

  检查最终答案是否合理：代回原方程，看看是否接近零。

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