📚 A-Level Maths: Iterative Methods for Approximating Roots | A-Level 数学：迭代法求方程近似根
Many equations cannot be solved exactly using simple algebra. Iterative methods give us a practical way to approximate roots to any required degree of accuracy.
许多方程无法通过简单代数精确求解。迭代法为我们提供了一种实用方法，可以按所需精度近似求出方程的根。
1. What Is an Iterative Method? | 什么是迭代法？
An iterative method is a process in which we start with an initial guess and repeatedly apply a rule to produce a sequence of values. Each new value is called an iteration, and the sequence is designed to get closer and closer to the true root.
迭代法是这样一种过程：我们先给出一个初始猜测值，然后反复应用同一规则产生一列数值。每一个新值称为一次迭代，这一数列被设计为越来越接近真实根。
We write the sequence as x₀, x₁, x₂, x₃, ... where x₀ is the starting value and xₙ₊₁ is calculated from xₙ.
我们将数列记为 x₀, x₁, x₂, x₃, ...，其中 x₀ 是初始值，xₙ₊₁ 由 xₙ 计算得出。
2. Locating Roots: The Change of Sign | 找根：变号法
Before iterating, we must first confirm that a root exists in a given interval. If f(x) is continuous and f(a) and f(b) have opposite signs, then there is at least one root between a and b.
在迭代之前，我们必须先确认某个区间内是否存在根。如果 f(x) 连续，且 f(a) 与 f(b) 异号，那么在 a 与 b 之间至少存在一个根。
For example, for f(x) = x³ - 2x - 5:
例如，对于 f(x) = x³ - 2x - 5：
f(2) = 8 - 4 - 5 = -1 < 0
f(3) = 27 - 6 - 5 = 16 > 0
Since the sign changes between x = 2 and x = 3, there is a root in the interval (2, 3).
因为 x = 2 和 x = 3 之间符号发生改变，所以在区间 (2, 3) 内存在一个根。
3. Fixed-Point Iteration | 不动点迭代
arrange the equation f(x) = 0 into the form x = g(x). Then we apply the iteration formula:
为了使用不动点迭代，我们将方程 f(x) = 0 改写为 x = g(x) 的形式。然后我们应用迭代公式：
xₙ₊₁ = g(xₙ)
nverges, it approaches a number α such that α = g(α). This number α is a root of f(x) = 0.
如果数列收敛，它会趋近于一个满足 α = g(α) 的数 α。这个 α 就是 f(x) = 0 的一个根。
The value α is called a fixed point of g(x), because applying g leaves it unchanged.
α 被称为 g(x) 的不动点，因为对 α 应用 g 后其值不变。
4. Rearranging Equations into x = g(x) | 将方程改写为 x = g(x)
The same equation can often be rearranged in several different ways. Some rearrangements lead to convergent iterations, while others do not.
同一个方程通常可以有多种不同的改写方式。有些改写能使迭代收敛，有些则不能。
For x³ - 2x - 5 = 0, one possible rearrangement is:
对于 x³ - 2x - 5 = 0，一种可能的改写是：
x = ∛(2x + 5)
So the iteration formula becomes xₙ₊₁ = ∛(2xₙ + 5).
因此迭代公式变为 xₙ₊₁ = ∛(2xₙ + 5)。
e x = (x³ - 5)/2, but this often fails to converge. Choosing a suitable form is essential.
另一种改写是 x = (x³ - 5)/2，但它常常无法收敛。选择合适的形式至关重要。
5. Convergence Condition | 收敛条件
fixed-point iteration xₙ₊₁ = g(xₙ) to converge to a root α, a sufficient condition is that |g'(α)| < 1 near the root.
对于不动点迭代 xₙ₊₁ = g(xₙ)，若要收敛到根 α，一个充分条件是根附近 |g'(α)| < 1。
If |g'(α)| > 1, the errors become larger with each step and the iteration diverges. If |g'(α)| = 1, convergence is uncertain and may be very slow.
如果 |g'(α)| > 1，每一步误差会越来越大，迭代发散。如果 |g'(α)| = 1，收敛性不确定，且可能非常缓慢。
For g(x) = ∛(2x + 5), the derivative is:
对于 g(x) = ∛(2x + 5)，其导数为：
g'(x) = 2 / (3(2x + 5)2/3)
Near x ≈ 2.09, this value is approximately 0.15, so the iteration should converge.
在 x ≈ 2.09 附近，该值约为 0.15，因此迭代应当收敛。
6. Worked Example: x³ - 2x - 5 = 0 | 例题：x³ - 2x - 5 = 0
Let us use the iteration xₙ₊₁ = ∛(2xₙ + 5) with x₀ = 2.
我们使用迭代公式 xₙ₊₁ = ∛(2xₙ + 5)，取 x₀ = 2。
x₁ = ∛(2(2) + 5) = ∛9 ≈ 2.0801
x₂ = ∛(2(2.0801) + 5) = ∛9.1602 ≈ 2.0924
x₃ = ∛(2(2.0924) + 5) = ∛9.1848 ≈ 2.0940
x₄ = ∛(2(2.0940) + 5) = ∛9.1880 ≈ 2.0943
lues are settling near 2.0946. To verify, f(2.0946) ≈ 0.0005, which is very close to zero.
这些值逐渐稳定在 2.0946 附近。验证可得 f(2.0946) ≈ 0.0005，已非常接近零。
7. Using a Calculator or Spreadsheet | 使用计算器或电子表格
In an exam, you will often be asked to record iterations in a table. A spreadsheet or graphing calculator can perform repeated substitutions quickly.
在考试中，你常需要将迭代过程记录在表格中。电子表格或图形计算器可以快速完成重复代入。
| n | xₙ | xₙ₊₁ = ∛(2xₙ + 5) |
| 0 | 2.0000 | 2.0801 |
| 1 | 2.0801 | 2.0924 |
| 2 | 2.0924 | 2.0940 |
| 3 | 2.0940 | 2.0943 |
| 4 | 2.0943 | 2.0946 |
Always keep a consistent number of decimal places in your table and write down the values you enter into your calculator.
在表格中始终保持一致的小数位数，并写下你输入计算器的数值。
8. Accuracy and Stopping Rules | 精度与停止规则
When asked to give a root correct to a certain number of decimal places, you must continue until two successive approximations agree to that accuracy.
当题目要求将根精确到指定位数的小数时，你必须继续迭代，直到连续两次近似值在该精度下一致。
For example, to find a root correct to 3 decimal places, keep iterating until xₙ and xₙ₊₁ both round to the same three-decimal value.
例如，要求根精确到 3 位小数时，应持续迭代，直到 xₙ 与 xₙ₊₁ 四舍五入后得到相同的三位小数。
If xₙ = 2.0946 and xₙ₊₁ = 2.0946, then root ≈ 2.095 (3 d.p.)
Always check your final answer by substituting it back into the original equation.
务必通过将最终答案代回原方程来检验。
9. Staircase and Cobweb Diagrams | 阶梯图与蛛网图
Graphical representations help us understand why an iteration converges or diverges.
图形表示有助于我们理解迭代为何收敛或发散。
the x-axis, move vertically to the curve, then horizontally to the line y = x, and repeat.
在 y = g(x) 与直线 y = x 的图上，从 x 轴上的 x₀ 开始，竖直移动到曲线，再水平移动到直线 y = x，然后重复。
If the steps form a shrinking square spiral around the intersection, the iteration converges. This is called a cobweb diagram.
如果这些路径在交点周围形成逐渐缩小的方形螺旋，则迭代收敛。这称为蛛网图。
If the graph is shallow near the intersection, the steps look like a staircase moving toward the fixed point.
如果曲线在交点附近较平缓，这些路径看起来像一段段楼梯，逐步移向不动点。
10. When Iteration Fails | 迭代失败的情形
If you choose a rearrangement with |g'(α)| > 1, the sequence may move away from the root even if x₀ is very close to it.
如果你选择的改写形式满足 |g'(α)| > 1，那么即使 x₀ 非常接近根，数列也可能远离根。
For example, using xₙ₊₁ = (xₙ³ - 5)/2 for the same equation gives:
例如，对同一方程使用 xₙ₊₁ = (xₙ³ - 5)/2 会得到：
x₀ = 2 → x₁ = 1.5 → x₂ = -0.8125 → x₃ = -5.268...
The values quickly move away from the root. This shows that the choice of rearrangement is not just a matter of algebra; it determines whether the iteration works at all.
数值迅速远离根。这说明改写形式的选择不仅是代数问题，它直接决定迭代是否有效。
11. Comparison with Newton-Raphson | 与牛顿-拉弗森法的比较
Newton's method is a powerful iterative technique using the formula:
牛顿法是一种强大的迭代技术，其公式为：
xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)
It converges much faster than simple fixed-point iteration when it works, but it requires the derivative f'(x) and a good starting value.
在有效时，它比简单不动点迭代收敛快得多，但需要计算导数 f'(x)，且需要良好的初始值。
Fixed-point iteration is simpler to set up and does not require differentiation, which is why it appears in many A-Level papers as a standard numerical method.
不动点迭代设置更简单，且不需要求导，因此它经常作为标准数值方法出现在许多 A-Level 试卷中。
12. Exam Tips and Summary | 考试技巧与总结
Always show your iteration formula clearly and state your starting value x₀.
务必清晰写出迭代公式，并说明初始值 x₀。
Check for a sign change before starting.
开始前先检查符号变化。
Rearrange the equation into x = g(x) and check |g'(α)| < 1 if possible.
将方程改写为 x = g(x)，并尽可能检查 |g'(α)| < 1。
Keep four or five decimal places in your working to avoid rounding errors.
计算中保留四位或五位小数，以避免舍入误差。
Stop only when two successive values agree to the required accuracy.
只有当连续两次取值达到所需精度时才停止。
Substitute your final answer back into f(x) to confirm it is close to zero.
将最终答案代回 f(x)，确认其接近零。
An iterative method gives an approximation, not an exact answer. The key is to choose a convergent formula and record your working systematically.
迭代法给出的是近似值，而非精确答案。关键在于选择收敛的公式，并系统地记录你的计算过程。
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