📚 GCSE CCEA Maths: Numerical Methods | 数值方法考点精讲
Numerical methods provide powerful techniques to solve equations that cannot be solved algebraically. In GCSE CCEA Mathematics, you are expected to master trial and improvement, as well as iterative methods to approximate solutions to a required degree of accuracy. This revision guide covers all key concepts with step-by-step explanations.
数值方法提供了强大的技巧，用于求解无法用代数方法解决的方程。在 GCSE CCEA 数学中，你需要掌握试位法和迭代法，以近似求解并达到所需的精度。本考点精讲涵盖所有关键概念，并配有逐步解释。
1. Introduction to Numerical Methods | 数值方法简介
Numerical methods are used when exact algebraic solutions are impossible or impractical. Instead of finding an exact answer, we approximate the solution to a specified number of decimal places or significant figures.
当精确的代数解不可能或不实用时，就会使用数值方法。我们不是找到精确答案，而是将解近似到指定的小数位数或有效数字。
The CCEA exam may require you to use trial and improvement or iteration to find roots of equations such as x³ – x = 20.
CCEA 考试可能会要求你使用试位法或迭代法求解方程如 x³ – x = 20 的根。
2. Trial and Improvement Method | 试位法
The trial and improvement method involves substituting values of x into an equation and checking whether the result is too high or too low. You systematically narrow down the interval containing the root.
试位法是将 x 值代入方程，并检查结果是否太大或太小。你系统地缩小包含根的区间。
For example, to solve x³ – x = 20, we try x = 2: 2³ – 2 = 6, too low. x = 3: 27 – 3 = 24, too high. The root lies between 2 and 3.
例如，解方程 x³ – x = 20，我们试 x = 2：2³ – 2 = 6，太小。x = 3：27 – 3 = 24，太大。根位于 2 和 3 之间。
We then try the midpoint 2.5 and continue adjusting based on whether the result is below or above 20.
接着我们试中点 2.5，并根据结果低于还是高于 20 继续调整。
3. Setting Up a Trial and Improvement Table | 建立试位法表格
A clear table helps organise trials and avoids confusion. Columns typically show the x value, the substituted expression, and the result compared to the target.
清晰的表格有助于组织试位步骤，避免混乱。各列通常包括 x 值、代入的表达式以及结果与目标值的比较。
Exam questions often provide a partially completed table. You must fill in missing entries and use the information to determine the solution to a given accuracy. Below is an example table for x³ – x = 20.
考试题目常会提供一个部分完成的表格。你必须填写缺失的项，并利用信息确定给定精度的解。以下是 x³ – x = 20 的示例表格。
| x | x³ – x | Comparison to 20 |
|---|---|---|
| 2 | 6 | Too low |
| 3 | 24 | Too high |
| 2.8 | 19.152 | Too low |
| 2.9 | 21.489 | Too high |
From the table, the solution lies between 2.8 and 2.9 because the result crosses 20 there. You would then trial 2.85 to refine the answer.
从表格中可以看到，解位于 2.8 和 2.9 之间，因为结果在这里跨越了 20。然后你可以试 2.85 来进一步精化答案。
4. Determining the Solution to a Given Accuracy | 确定给定精度的解
Once you identify two consecutive x values where the result changes sign (below to above target), the root lies between them. To find the solution to 1 decimal place, you test the midpoint of this interval and see which side gives a closer result.
一旦你确定了两个相邻的 x 值，使得结果发生符号变化（低于到高于目标值），根就位于它们之间。要找到精确到 1 位小数的解，就测试该区间的中点，看哪一边的结果更接近目标。
For accuracy of 2 decimal places, you may need to test x values with increments of 0.01 near the approximation until the rounded value no longer changes. Always double-check by substituting your final approximated x back into the original equation.
要精确到 2 位小数，你可能需要以 0.01 为增量在近似值附近测试 x 值，直到四舍五入后的值不再变化。始终要将最终近似的 x 值代回原方程进行双重检验。
5. Iteration: Rearranging Equations | 迭代法：方程重组
Iteration is an alternative numerical method that generates a sequence of approximations using a formula xₙ₊₁ = g(xₙ). First, rearrange the given equation f(x) = 0 into the form x = g(x).
迭代法是另一种数值方法，它使用公式 xₙ₊₁ = g(xₙ) 生成一系列近似值。首先，将给定的方程 f(x) = 0 重组为 x = g(x) 的形式。
For the equation x³ – x – 20 = 0, one possible rearrangement is x = ³√(x + 20). Another is x = (x + 20)^{⅓}. Both are equivalent.
对于方程 x³ – x – 20 = 0，一种可能的变形是 x = ³√(x + 20)。另一种是 x = (x + 20)^{⅓}。两者等价。
Not all rearrangements lead to a convergent sequence. CCEA questions usually give the iterative formula directly, so you do not need to invent one, but you should practise forming simple rearrangements.
并非所有重组形式都会产生收敛序列。CCEA 考题通常会直接给出迭代公式，因此你不需要自己创造，但应练习构造简单的变形。
6. Using an Iterative Formula | 使用迭代公式
Given an iterative formula, such as xₙ₊₁ = ³√(20 + xₙ), you start with an initial guess x₀ (e.g., x₀ = 2.5). Substitute x₀ into the right-hand side to obtain x₁. Then use x₁ to find x₂, and so on.
给定一个迭代公式，例如 xₙ₊₁ = ³√(20 + xₙ)，你从一个初始猜测值 x₀（如 x₀ = 2.5）开始。将 x₀ 代入右侧得到 x₁。然后用 x₁ 求 x₂，以此类推。
Record each iteration clearly, often rounding to a set number of decimal places. The sequence should converge to a fixed point, which is the solution.
清楚地记录每一次迭代，通常四舍五入到指定的小数位数。序列应收敛到不动点，即所求的解。
Example: x₀ = 2.5 → x₁ = ³√(20 + 2.5) = ³√22.5 ≈ 2.824, x₂ = ³√(20 + 2.824) ≈ 2.857, and so on. Continue until values settle to the required accuracy.
示例：x₀ = 2.5 → x₁ = ³√(20 + 2.5) = ³√22.5 ≈ 2.824, x₂ = ³√(20 + 2.824) ≈ 2.857，等等。继续迭代直到值稳定在所需精度。
xₙ₊₁ = ³√(20 + xₙ)
7. Convergence and Divergence | 收敛与发散
An iterative formula converges if the sequence approaches a limit. This happens when |g'(x)| < 1 near the root. In CCEA, you may be asked to test whether a given rearrangement works by performing a few iterations.
如果序列趋近于一个极限，则迭代公式收敛。当根的附近 |g'(x)| < 1 时发生收敛。在 CCEA 考试中，你可能会被要求通过进行几次迭代来测试给定的重组是否有效。
If the numbers grow larger or oscillate wildly, the method diverges. You can identify divergence by spotting that the values are getting further from each other.
如果数值变得越来越大或剧烈振荡，则该方法是发散的。你可以通过发现值彼此远离来识别发散。
Always use the iterative formula exactly as given to avoid divergence. If a formula causes divergence,
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