二项式估计 / Binomial Estimation
在工程和科学中，经常需要为复杂函数找到简单的近似值。
In engineering and science, it is often useful to find simple approximations for complicated functions.
如果 \(x\) 的值小于1，那么 \(x^n\) 随着 \(n\) 的增大而变小。
If the value of \(x\) is less than 1, then \(x^n\) gets smaller as \(n\) gets larger.
如果 \(x\) 很小，有时可以忽略 \(x\) 的高次幂来近似函数或估计值。
If \(x\) is small you can sometimes ignore large powers of \(x\) to approximate a function or estimate a value.
例题9 / Example 9
a) 求 \((1-\frac{x}{4})^{10}\) 按 \(x\) 的升幂展开的前四项。
a) Find the first four terms of the binomial expansion, in ascending powers of \(x\), of \((1-\frac{x}{4})^{10}\).
b) 使用你的展开式估计 \(0.975^{10}\) 的值，答案保留4位小数。
b) Use your expansion to estimate the value of \(0.975^{10}\), giving your answer to 4 decimal places.
解答 / Solution
a) \((1-\frac{x}{4})^{10}\)
a) \((1-\frac{x}{4})^{10}\)
使用二项式展开：
Using binomial expansion:
\((1-\frac{x}{4})^{10} = 1^{10} + \binom{10}{1} \cdot 1^9 \cdot (-\frac{x}{4}) + \binom{10}{2} \cdot 1^8 \cdot (-\frac{x}{4})^2 + \binom{10}{3} \cdot 1^7 \cdot (-\frac{x}{4})^3 + \ldots\)
\((1-\frac{x}{4})^{10} = 1^{10} + \binom{10}{1} \cdot 1^9 \cdot (-\frac{x}{4}) + \binom{10}{2} \cdot 1^8 \cdot (-\frac{x}{4})^2 + \binom{10}{3} \cdot 1^7 \cdot (-\frac{x}{4})^3 + \ldots\)
\(= 1 + 10(-\frac{x}{4}) + 45(\frac{x^2}{16}) + 120(-\frac{x^3}{64}) + \ldots\)
\(= 1 + 10(-\frac{x}{4}) + 45(\frac{x^2}{16}) + 120(-\frac{x^3}{64}) + \ldots\)
\(= 1 - 2.5x + 2.8125x^2 - 1.875x^3 + \ldots\)
\(= 1 - 2.5x + 2.8125x^2 - 1.875x^3 + \ldots\)
b) 估计 \(0.975^{10}\)
b) Estimate \(0.975^{10}\)
注意到 \(0.975 = 1 - 0.025\)，所以 \(0.975^{10} = (1-0.025)^{10}\)
Note that \(0.975 = 1 - 0.025\), so \(0.975^{10} = (1-0.025)^{10}\)
设 \(x = 0.1\)，则 \(\frac{x}{4} = 0.025\)，所以 \(x = 0.1\)
Let \(x = 0.1\), then \(\frac{x}{4} = 0.025\), so \(x = 0.1\)
代入展开式：
Substituting into the expansion:
\(0.975^{10} \approx 1 - 2.5(0.1) + 2.8125(0.1)^2 - 1.875(0.1)^3\)
\(0.975^{10} \approx 1 - 2.5(0.1) + 2.8125(0.1)^2 - 1.875(0.1)^3\)
\(= 1 - 0.25 + 0.028125 - 0.001875\)
\(= 1 - 0.25 + 0.028125 - 0.001875\)
\(= 0.77625\)
\(= 0.77625\)
保留4位小数：\(0.7763\)
To 4 decimal places: \(0.7763\)
忽略高次项的应用 / Applications of Ignoring Higher Order Terms
如果 \(x\) 很小，使得 \(x^3\) 及更高次项可以忽略，证明：
If \(x\) is so small that terms of \(x^3\) and higher can be ignored, show that:
证明过程 / Proof Process
步骤1： 展开 \((1-3x)^5\)
Step 1: Expand \((1-3x)^5\)
\((1-3x)^5 = 1 + 5(-3x) + 10(-3x)^2 + \ldots\)
\((1-3x)^5 = 1 + 5(-3x) + 10(-3x)^2 + \ldots\)
\(= 1 - 15x + 90x^2 + \ldots\)
\(= 1 - 15x + 90x^2 + \ldots\)
步骤2： 乘以 \((2+x)\)
Step 2: Multiply by \((2+x)\)
\((2+x)(1-3x)^5 = (2+x)(1-15x+90x^2+\ldots)\)
\((2+x)(1-3x)^5 = (2+x)(1-15x+90x^2+\ldots)\)
\(= 2(1-15x+90x^2) + x(1-15x+90x^2)\)
\(= 2(1-15x+90x^2) + x(1-15x+90x^2)\)
\(= 2 - 30x + 180x^2 + x - 15x^2 + 90x^3\)
\(= 2 - 30x + 180x^2 + x - 15x^2 + 90x^3\)
\(= 2 - 29x + 165x^2 + 90x^3\)
\(= 2 - 29x + 165x^2 + 90x^3\)
忽略 \(x^3\) 项：\(\approx 2 - 29x + 165x^2\)
Ignoring \(x^3\) term: \(\approx 2 - 29x + 165x^2\)