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Binomial Estimation | 二项式估计

2026-08-29 23:30 · 本文部分内容由AI辅助创作

📚 Binomial Estimation | 二项式估计

Binomial estimation is a powerful technique in A-Level Mathematics that uses the binomial expansion to approximate numerical values. When the power is not a positive integer, the binomial series provides an infinite polynomial that can be truncated to give a very accurate approximation for suitable values of x.

二项式估算是 A-Level 数学中一项至关重要的技巧,它利用二项展开式来近似计算数值。当幂次不是正整数时,二项级数构成一个无穷多项式;在合适的 x 取值下,截取前几项即可得到非常精确的近似值。


1. The Binomial Theorem for Positive Integers | 正整数指数的二项式定理

For a positive integer n, the binomial expansion is exact and finite:

对于正整数 n,二项展开是精确且有限的:

(a + b)ⁿ = aⁿ + ⁿC₁ aⁿ⁻¹b + ⁿC₂ aⁿ⁻²b² + … + bⁿ

eries. For estimation, however, we usually need the case where a = 1, written as (1 + x)ⁿ.

这个常见形式是通用二项级数的基础。然而在估算中,我们通常需要 a = 1 的情形,即 (1 + x)ⁿ。


2. Extending to Rational and Negative Powers | 推广到有理数和负数指数

For any rational number n, provided |x| < 1, the expansion becomes an infinite series:

对于任意有理数 n,只要满足 |x| < 1,展开式就变成一个无穷级数:

(1 + x)ⁿ = 1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + …

This expansion is valid when n is negative or fractional, and it is the core of binomial estimation.

当 n 为负数或分数时,该展开式依然有效,这正是二项式估算的核心。


3. The Convergent Condition | 收敛条件

The condition |x| < 1 guarantees that the infinite series converges. For small values of x, the terms decrease rapidly, so keeping only the first few terms yields a good approximation.

条件 |x| < 1 保证了无穷级数收敛。当 x 很小时,各项递减很快,因此只保留前几项就能得到良好的近似。


4. Rearranging Expressions | 表达式的变形

Many expressions are not given as (1 + x)ⁿ directly. To apply binomial estimation, rewrite them in the form aⁿ(1 + b/a)ⁿ, ensuring |b/a| < 1.

许多表达式并不是直接以 (1 + x)ⁿ 的形式给出。为了应用二项式估算,我们需要将其改写成 aⁿ(1 + b/a)ⁿ,并确保 |b/a| < 1。

For example, √17 = √16 + 1 = 4(1 + 1/16)1/2. Here b/a = 1/16, which is much less than 1.

例如,√17 = √16 + 1 = 4(1 + 1/16)1/2。这里 b/a = 1/16,远小于 1。


5. Estimation Steps | 估算步骤

Follow these systematic steps when performing binomial estimation:

进行二项式估算时,请遵循以下系统步骤:


6. Worked Example 1 | 例题一

Expand (1 + 2x)1/2 up to the term in x², and hence estimate √1.02.

将 (1 + 2x)1/2 展开到 x² 项,并据此估算 √1.02。

(1 + 2x)1/2 = 1 + (1/2)(2x) + (1/2)(−1/2)/2! (2x)² + …

Simplifying gives:

化简得到:

1 + x − ½x² + …

To estimate √1.02, write √1.02 = (1 + 0.02)1/2. Set 2x = 0.02, so x = 0.01. Then:

为了估算 √1.02,我们写 √1.02 = (1 + 0.02)1/2。令 2x = 0.02,即 x = 0.01,则:

√1.02 ≈ 1 + 0.01 − ½(0.01)² = 1.00995

This is accurate to at least five decimal places.

该结果至少精确到小数点后五位。


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7. Worked Example 2 | 例题二

Use the binomial expansion to estimate √101 correct to four decimal places.

利用二项展开式估算 √101,精确到四位小数。

First, rewrite √101 as:

首先,将 √101 改写为:

√101 = 10√1 + 1/100 = 10(1 + 0.01)1/2

Using the expansion (1 + x)1/2 with x = 0.01:

使用 (1 + x)1/2 的展开式,令 x = 0.01:

(1 + 0.01)1/2 ≈ 1 + 0.01/2 − (0.01)²/8 + (0.01)³/16

≈ 1 + 0.005 − 0.0000125 + 0.0000000625 = 1.0049875625

Multiplying by 10 gives √101 ≈ 10.049875625, which rounds to 10.0499.

乘以 10 得到 √101 ≈ 10.049875625,四舍五入为 10.0499。


8. Accuracy and Error | 精度与误差

The error from truncating a binomial series is approximately the magnitude of the first omitted term. For an alternating series, this bound is particularly reliable.

截断二项级数所产生的误差大约等于被省略的第一项的大小。对于交错级数,这个误差界尤为可靠。

For example, in the expansion of (1 + x)1/2 .01, this term is about 10⁻⁶, so the approximation is accurate to around 6 decimal places.

例如,在 (1 + x)1/2 的展开式中,被省略的第一项通常是 x³ 量级。若 x = 0.01,该项约为 10⁻⁶,因此近似值精度大约到 6 位小数。


9. Common Mistakes | 常见错误


10. Exam Tips | 考试技巧

In Edexcel A-Level exams, always state the range of validity when asked for a binomial expansion. Choose a convenient value of x that is small and satisfies the convergence condition.

在 Edexcel A-Level 考试中,当被要求进行二项展开时,务必注明有效范围。选择方便且满足收敛条件的小 x 值。


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