GCSE WJEC Maths: Numerical Methods Key Points | GCSE WJEC 数学：数值方法考点精讲

📚 GCSE WJEC Maths: Numerical Methods Key Points | GCSE WJEC 数学：数值方法考点精讲

Numerical methods are essential tools in GCSE WJEC Mathematics for finding approximate solutions to equations that cannot be solved algebraically. This revision guide covers trial and improvement, iteration, accuracy checks and exam-style techniques to help you master this topic.

数值方法是 GCSE WJEC 数学考试中用于求解无法用代数方法直接求解的方程近似解的重要工具。本复习指南将涵盖试位法、迭代法、精度检验以及考试技巧，帮助你彻底掌握这一考点。

Throughout your exam, you will be expected to show working clearly, use your calculator efficiently and interpret results to the required degree of accuracy. Let’s break down every key concept.

在考试中，你需要清晰地展示计算步骤、高效使用计算器，并按要求精确度解读结果。让我们逐一击破每一个核心概念。

1. Introduction to Numerical Methods | 数值方法简介

A numerical method is a procedure that produces an approximate solution to a mathematical problem. In WJEC GCSE, you will use two main techniques: trial and improvement, and iteration. Both rely on repeated calculations that ‘home in’ on a correct answer.

数值方法是一种通过逐步逼近来获得数学问题近似解的过程。在 WJEC GCSE 考试中，你将使用两种主要方法：试位法和迭代法。两者都依赖重复计算来逐渐逼近正确答案。

Numerical methods are used when an equation cannot be factorised or solved by a simple formula, such as x³ + 2x – 5 = 0 or equations involving trigonometric terms.

当方程无法分解因式或使用简单公式求解时（例如 x³ + 2x – 5 = 0 或包含三角项的方程），就需要用到数值方法。

You must be comfortable substituting values, spotting sign changes and using your calculator’s ANS key for iterative processes.

你必须熟练掌握代入数值、观察符号变化以及使用计算器上的 ANS 键进行迭代过程。

2. Trial and Improvement Method | 试位法

Trial and improvement involves testing values of x in the equation and narrowing down the interval where the root lies. You start with two x-values that give f(x) with opposite signs, then systematically try midpoints until the required degree of accuracy is reached.

试位法是指将不同的 x 值代入方程，并逐步缩小根所在的区间。首先选取两个让 f(x) 异号的 x 值，然后系统地尝试中点值，直到达到所需的精确度。

For example, to solve x³ – x – 2 = 0 to 1 decimal place, you might test x = 1 and x = 2. Since f(1) = -2 and f(2) = 4, the root lies between 1 and 2. Next you test x = 1.5, and continue refining.

例如，精确到一位小数求解 x³ – x – 2 = 0 时，你可以测试 x = 1 和 x = 2。由于 f(1) = -2 而 f(2) = 4，根位于 1 和 2 之间。接着测试 x = 1.5，并继续细化。

In your working, you must display all trials, stating whether each result is too high or too low. The final interval width should be no more than half the required precision – e.g., for 1 decimal place, the two bounding values must differ by 0.1 or less, and you choose the midpoint rounded appropriately.

解题时，你必须列出所有测试值，并说明每次结果是偏大还是偏小。最终的区间宽度不应超过要求精度的一半——例如，对于一位小数精度，边界值之差必须小于或等于 0.1，然后适当四舍五入选择中点。

Trial x	x³ – x – 2	Comment
1	-2	Too low
2	4	Too high
1.5	-0.125	Too low
1.6	0.496	Too high
1.55	0.173…	Too high

The root correct to 1 decimal place is 1.5 because f(1.5) and f(1.6) straddle zero, and 1.55 rounds to 1.6, so final answer is 1.5.

精确到一位小数的根是 1.5，因为 f(1.5) 和 f(1.6) 跨越零点，而中点 1.55 四舍五入为 1.6，因此最终答案为 1.5。

3. What Is Iteration? | 什么是迭代法？

Iteration is a process in which you start with an initial guess and repeatedly apply an iteration formula to generate a sequence that converges to the root. The equation f(x) = 0 is first rearranged into the form x = g(x).

迭代法是指从一个初始猜测值开始，反复应用迭代公式，产生一个收敛到根值的数列。首先需要将方程 f(x) = 0 重新排列为 x = g(x) 的形式。

The iteration formula is then used as xₙ₊₁ = g(xₙ). If the sequence gets closer to a fixed number, the iteration converges; if it moves away, it diverges.

迭代公式随后用作 xₙ₊₁ = g(xₙ)。如果数列逐渐接近某个固定数，则迭代收敛；如果离得越来越远，则发散。

For the equation x² – 3x – 2 = 0, one possible rearrangement is x = √(3x + 2). This gives the iteration formula xₙ₊₁ = √(3xₙ + 2).

对于方程 x² – 3x – 2 = 0，一种可能的重排形式是 x = √(3x + 2)，得出迭代公式 xₙ₊₁ = √(3xₙ + 2)。

In WJEC exams you may be given the rearrangement or asked to form one yourself. Always check that the rearrangement is valid and avoids division by zero.

在 WJEC 考试中，你可能会得到重排后的式子，也可能需要自己构建。务必检查重排是否合理，并避免出现除以零的情况。

4. Setting Up an Iteration Equation | 建立迭代方程

To obtain an iteration formula, isolate x in the equation f(x) = 0. For example, starting from x³ + 2x – 5 = 0, you could write x³ = 5 – 2x, so x = ³√(5 – 2x). The iteration becomes xₙ₊₁ = ³√(5 – 2xₙ).

要得到迭代公式，需在 f(x) = 0 中将 x 独立出来。例如，从 x³ + 2x – 5 = 0 出发，可写成 x³ = 5 – 2x，因此 x = ³√(5 – 2x)，迭代公式变为 xₙ₊₁ = ³√(5 – 2xₙ)。

Alternatively, you might rearrange to x = (5 – x³)/2, yielding xₙ₊₁ = (5 – xₙ³)/2. Both are correct rearrangements, but they may have different convergence properties.

或者，你也可以重排成 x = (5 – x³)/2，得到 xₙ₊₁ = (5 – xₙ³)/2。两种重排都是正确的，但它们的收敛特性可能不同。

Always test your rearrangement by substituting a value to see if the original equation balances. In the exam, you are usually instructed which form to use.

务必通过代入数值检验重排结果是否能让原方程成立。考试中通常会指定使用哪一种重排形式。

5. Performing Iterations Using a Calculator | 用计算器执行迭代

The most efficient way to perform iterations is by using the ANS key on your calculator. Enter the starting value x₀ and press ‘=’. Then type the iteration formula, using ANS to represent xₙ. Press ‘=’ repeatedly to generate the sequence.

执行迭代最有效的方式是使用计算器上的 ANS 键。输入初始值 x₀ 并按 ‘=’，然后输入迭代公式，用 ANS 表示 xₙ。反复按 ‘=’ 即可生成数列。

For example, with xₙ₊₁ = √(3ANS + 2) and x₀ = 3, the display will show 3, then √(3×3+2) = √11 ≈ 3.3166, then √(3×3.3166+2) ≈ 3.4641, and so on.

例如，使用 xₙ₊₁ = √(3ANS + 2) 且 x₀ = 3，屏幕将依次显示 3、√(3×3+2) = √11 ≈ 3.3166、√(3×3.3166+2) ≈ 3.4641，依此类推。

Record the values to the required number of decimal places, usually 4 or 5, to observe convergence. Stop when the values stabilise to the required accuracy, e.g., when consecutive terms round to the same 2 decimal places.

按要求的位数（通常四到五位小数）记录数值以观察收敛情况。当数值稳定到所需精度时即可停止，例如当相邻两项四舍五入到相同两位小数时。

Always show at least the first three iterates clearly in your answer booklet, even if the calculator does many more.

即使在计算器上进行了更多次迭代，也务必在答题纸上清晰展示至少前三次迭代结果。

6. Checking Accuracy and Error Bounds | 精度检查与误差界限

For trial and improvement, the root is correct to a given number of decimal places if the upper and lower bounds round to the same value. For 1 decimal place, you must check that the interval lies entirely within the range where rounding gives that digit.

在试位法中，若上界和下界四舍五入后得到相同值，则所求根在指定位小数下是正确的。对于一位小数，必须检查区间是否完全落在舍入得该位数的范围之内。

For iteration, accuracy is confirmed when the difference between successive terms is less than the required tolerance, e.g., |xₙ₊₁ – xₙ| < 0.0005 for 3 decimal places. Some questions ask you to find a root to 2 decimal places and you may need to show that both x₅ and x₆ round to the same value.

对于迭代法，当相邻两项的差值小于所需容差时即可确认精度，例如，对于三位小数，要求 |xₙ₊₁ – xₙ| < 0.0005。有些问题要求将根值精确到两位小数，你可能需要证明 x₅ 与 x₆ 四舍五入后结果一致。

In WJEC mark schemes, you must state clearly that the root has been found to the stated accuracy, often by writing the final bounds or iterates.

在 WJEC 评分标准中，你必须明确说明根已经达到指定精度，通常需要写出最终边界值或迭代值。

7. Graphical Interpretation of Iteration | 迭代的图形解释

An iterative sequence can be visualised by drawing the graphs of y = x and y = g(x) on the same axes. The x-coordinate of the intersection point represents the root of the original equation.

可以在同一坐标系中画出 y = x 和 y = g(x) 的图像来直观理解迭代数列。两线交点的 x 坐标即代表原方程的根。

Starting from x₀, draw a vertical line to y = g(x) to get x₁, then a horizontal line to y = x, and repeat. This creates a cobweb or staircase pattern converging towards the intersection.

从 x₀ 出发，作垂直线交 y = g(x) 得到 x₁，然后作水平线交 y = x，重复此过程。这会形成向交点收敛的蛛网图或阶梯图。

This interpretation is not heavily examined in WJEC GCSE but can appear in questions that ask you to explain why an iteration converges or diverges.

这部分图形解释在 WJEC GCSE 考试中出现不多，但可能会问及为何迭代会收敛或发散的问题。

8. Convergence and Divergence | 收敛与发散

An iteration converges if successive values get closer to the root. This typically happens when the gradient of g(x) near the root is between -1 and 1. If |g'(root)| < 1, the iteration is locally convergent.

若连续值逐渐靠近根，则迭代收敛。这通常发生在根附近 g(x) 的梯度介于 -1 到 1 之间时。若 |g'(根)| < 1，迭代局部收敛。

A divergent iteration produces values that move away from the root, often spiralling out. This can happen if you choose the wrong rearrangement or a poor starting value.

发散迭代产生的值会远离根，往往呈螺旋状扩大。如果你选择了错误的重排方式或不当的起始值，就可能发生这种情况。

In your exam, you will not be asked to calculate derivatives, but you may be asked to test whether an iteration formula works by using a given starting value and observing the output.

考试中你不会被要求计算导数，但可能会让你尝试一个给定的迭代公式，使用给定起始值并观察输出结果，以判断其是否有效。

9. Common Mistakes and How to Avoid Them | 常见错误及其避免方法

One common error is rearranging the equation incorrectly. Always double-check by substituting a number into your rearranged equation to see if the original equation is satisfied.

一个常见错误是方程重排错误。务必通过代入数值到重排后的等式检验是否满足原方程，进行双重检查。

Another mistake is forgetting to record all trials or iterates. Examiners require clear evidence of the iterative process; missing steps can lose marks.

另一个错误是忘记记录所有的试验值或迭代值。考官要求清晰的迭代过程证据，遗漏步骤可能导致失分。

Also, some students stop iteration too early. Ensure that the required decimal places have truly stabilised, and always state which iterate you are using as the final answer.

此外，有些学生过早停止迭代。请确保指定小数位数已真正稳定，并始终注明你将哪个迭代值作为最终答案。

Using the wrong starting value can lead to divergence. If told to use a specific x₀, stick to it. If not, choose a value near the expected root based on a quick trial.

使用错误的起始值会导致发散。如果题目指定了特定的 x₀，就严格按照要求使用。如果没有指定，可基于快速试验选择一个接近预期根的值。

10. Worked Example: Trial and Improvement | 实例解析：试位法

Problem: The equation x³ + 3x – 5 = 0 has a root between 1 and 2. Find this root correct to 1 decimal place using trial and improvement.

问题：方程 x³ + 3x – 5 = 0 在 1 和 2 之间有一个根。使用试位法求出该根，精确到一位小数。

Step-by-step: f(1) = 1+3-5 = -1 (negative), f(2) = 8+6-5 = 9 (positive). Try x = 1.5: f(1.5) = 3.375 + 4.5 – 5 = 2.875 (positive) → too high, so root is between 1 and 1.5.

逐步解答：f(1) = 1+3-5 = -1（负），f(2) = 8+6-5 = 9（正）。尝试 x = 1.5：f(1.5) = 3.375+4.5-5 = 2.875（正）→ 偏大，故根在 1 与 1.5 之间。

Try x = 1.3: f(1.3) = 2.197+3.9-5 = 1.097 (positive) → still too high. Try x = 1.1: f(1.1) = 1.331+3.3-5 = -0.369 (negative). Root between 1.1 and 1.3.

尝试 x = 1.3：f(1.3) = 2.197+3.9-5 = 1.097（正）→ 仍偏大。尝试 x = 1.1：f(1.1) = 1.331+3.3-5 = -0.369（负）。根在 1.1 与 1.3 之间。

Try x = 1.2: f(1.2) = 1.728+3.6-5 = 0.328 (positive). Root between 1.1 and 1.2. Examine midpoint 1.15: f(1.15) = 1.520875+3.45-5 = -0.029125 (negative). So root between 1.15 and 1.2. To 1 d.p., 1.15 rounds up to 1.2 and 1.2 stays 1.2, so the root is 1.2 (1 d.p.).

尝试 x = 1.2：f(1.2) = 1.728+3.6-5 = 0.328（正）。根在 1.1 与 1.2 之间。检验中点 1.15：f(1.15) = 1.520875+3.45-5 = -0.029125（负）。因此根在 1.15 与 1.2 之间。精确到一位小数，1.15 舍入为 1.2，1.2 仍为 1.2，因此根为 1.2（一位小数）。

11. Worked Example: Iteration | 实例解析：迭代法

Problem: Show that the equation x² – 4x – 1 = 0 can be rearranged into x = 4 + 1/x. Use the iteration xₙ₊₁ = 4 + 1/xₙ with x₀ = 4 to find a root correct to 2 decimal places.

问题：证明方程 x² – 4x – 1 = 0 可以重排为 x = 4 + 1/x。使用迭代公式 xₙ₊₁ = 4 + 1/xₙ 且 x₀ = 4，求出一个精确到两位小数的根。

Rearrangement proof: x² – 4x – 1 = 0 ⇒ x² – 4x = 1 ⇒ x(x – 4) = 1 ⇒ x – 4 = 1/x ⇒ x = 4 + 1/x. This is valid as long as x ≠ 0.

重排证明：x² – 4x – 1 = 0 ⇒ x² – 4x = 1 ⇒ x(x – 4) = 1 ⇒ x – 4 = 1/x ⇒ x = 4 + 1/x。只要 x ≠ 0 就成立。

Iteration: x₀ = 4. x₁ = 4 + 1/4 = 4.25. x₂ = 4 + 1/4.25 = 4.235294… x₃ = 4 + 1/4.235294 = 4.236111… x₄ = 4 + 1/4.236111 = 4.236065… x₅ = 4 + 1/4.236065 = 4.236068…

迭代过程：x₀ = 4。x₁ = 4 + 1/4 = 4.25。x₂ = 4 + 1/4.25 = 4.235294… x₃ = 4 + 1/4.235294 = 4.236111… x₄ = 4 + 1/4.236111 = 4.236065… x₅ = 4 + 1/4.236065 = 4.236068…

After x₄ and x₅, the values round to 4.24 when rounded to two decimal places. The difference is very small, so the root is 4.24 (2 d.p.).

从 x₄ 和 x₅ 之后，四舍五入到两位小数均为 4.24。差值极小，因此根为 4.24（两位小数）。

12. WJEC Exam Tips Summary | WJEC 考试技巧总结

Always write down your first guess and every iteration or trial value clearly. Use a table if it helps organise trial and improvement.

务必清晰写下你的首个猜测值以及每一个迭代值或试验值。如果进行试位法，使用表格有助于整理数据。

For iteration questions, copy the given formula accurately and show the sequence at least up to the point where two successive answers round to the same required accuracy.

对于迭代题，准确抄下所给公式，并展示数列直到两个连续答案四舍五入到相同指定精度为止。

Check convergence by comparing successive terms. If the values start to repeat or oscillate, you may have encountered a divergence or a cycle – review your rearrangement or starting value.

通过比较连续项来检查收敛性。若数值开始重复或振荡，可能遇到了发散或循环——此时需复查重排方式或起始值。

State your final answer clearly, e.g., ‘The root is 2.7 correct to 1 decimal place’, and always include the unit or degree of accuracy as required by the question.

清晰地陈述最终答案，例如“该根为 2.7，精确到一位小数”，并始终根据题目要求包含单位或精确度说明。

Lastly, manage your time: numerical methods questions are often worth several marks and require careful bookkeeping – rushing leads to simple arithmetic errors.

最后，合理分配时间：数值方法题目通常占分较多且需要细心记录——匆忙作答会导致简单的算术错误。

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GCSE WJEC Maths: Numerical Methods Key Points | GCSE WJEC 数学：数值方法 考点精讲