4.5 Binomial Estimation - 练习题

Binomial Estimation - Exercises

导航菜单
主页
知识点总结
练习题
教材内容
彩色打印版

基础练习 / Basic Exercises

练习1 / Exercise 1

a) 求 \((1-\frac{x}{10})^6\) 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 \((1-\frac{x}{10})^6\).
b) 通过代入适当的 \(x\) 值，求 \((0.99)^6\) 的近似值。
b) By substituting an appropriate value for \(x\), find an approximate value for \((0.99)^6\).

答案 / Answer

a) \((1-\frac{x}{10})^6 = 1 - \frac{6x}{10} + \frac{15x^2}{100} - \frac{20x^3}{1000} + \ldots\)

\(= 1 - 0.6x + 0.15x^2 - 0.02x^3 + \ldots\)

b) 设 \(x = 0.1\)，则 \(\frac{x}{10} = 0.01\)

\((0.99)^6 = (1-0.01)^6 \approx 1 - 0.6(0.1) = 0.94\)

练习2 / Exercise 2

a) 写出 \((2+\frac{x}{5})^{10}\) 的二项式展开的前四项。
a) Write down the first four terms of the binomial expansion of \((2+\frac{x}{5})^{10}\).
b) 通过代入适当的 \(x\) 值，求 \((2.1)^{10}\) 的近似值。
b) By substituting an appropriate value for \(x\), find an approximate value for \((2.1)^{10}\).

答案 / Answer

a) \((2+\frac{x}{5})^{10} = 2^{10} + 10 \cdot 2^9 \cdot \frac{x}{5} + 45 \cdot 2^8 \cdot \frac{x^2}{25} + 120 \cdot 2^7 \cdot \frac{x^3}{125} + \ldots\)

\(= 1024 + 1024x + 230.4x^2 + 30.72x^3 + \ldots\)

b) 设 \(x = 0.5\)，则 \(\frac{x}{5} = 0.1\)

\((2.1)^{10} = (2+0.1)^{10} \approx 1024 + 1024(0.5) = 1536\)

进阶练习 / Advanced Exercises

练习3 / Exercise 3

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:

\((2+x)(1-3x)^5 \approx 2 - 29x + 165x^2\)

答案 / Answer

\((1-3x)^5 = 1 + 5(-3x) + 10(-3x)^2 + \ldots = 1 - 15x + 90x^2 + \ldots\)

\((2+x)(1-3x)^5 = (2+x)(1-15x+90x^2+\ldots)\)

\(= 2(1-15x+90x^2) + x(1-15x+90x^2)\)

\(= 2 - 30x + 180x^2 + x - 15x^2 + 90x^3\)

\(= 2 - 29x + 165x^2 + 90x^3\)

忽略 \(x^3\) 项：\(\approx 2 - 29x + 165x^2\)

练习4 / Exercise 4

如果 \(x\) 很小，使得 \(x^3\) 及更高次项可以忽略，且
If \(x\) is so small that terms of \(x^3\) and higher can be ignored, and

\((2-x)(3+x)^4 \approx a + bx + cx^2\)

求常数 \(a\)、\(b\) 和 \(c\) 的值。
find the values of the constants \(a\), \(b\) and \(c\).

答案 / Answer

\((3+x)^4 = 3^4 + 4 \cdot 3^3 \cdot x + 6 \cdot 3^2 \cdot x^2 + \ldots = 81 + 108x + 54x^2 + \ldots\)

\((2-x)(3+x)^4 = (2-x)(81+108x+54x^2+\ldots)\)

\(= 2(81+108x+54x^2) - x(81+108x+54x^2)\)

\(= 162 + 216x + 108x^2 - 81x - 108x^2 - 54x^3\)

\(= 162 + 135x + 0x^2 - 54x^3\)

忽略 \(x^3\) 项：\(\approx 162 + 135x + 0x^2\)

所以：\(a = 162\)，\(b = 135\)，\(c = 0\)

练习5 / Exercise 5

a) 写出 \((1+2x)^8\) 展开式的前四项。
a) Write down the first four terms in the expansion of \((1+2x)^8\).
b) 通过代入适当的 \(x\) 值（应说明），求 \((1.02)^8\) 的近似值。
b) By substituting an appropriate value of \(x\) (which should be stated), find an approximate value of \((1.02)^8\).

答案 / Answer

a) \((1+2x)^8 = 1 + 8(2x) + 28(2x)^2 + 56(2x)^3 + \ldots\)

\(= 1 + 16x + 112x^2 + 448x^3 + \ldots\)

b) 设 \(x = 0.01\)，则 \(2x = 0.02\)

\((1.02)^8 = (1+0.02)^8 \approx 1 + 16(0.01) = 1.16\)

使用更高精度：\(\approx 1 + 16(0.01) + 112(0.01)^2 = 1.16 + 0.0112 = 1.1712\)

挑战练习 / Challenge Exercises

练习6 / Exercise 6

设 \(f(x) = (1-5x)^{30}\)
Let \(f(x) = (1-5x)^{30}\)
a) 求 \(f(x)\) 的二项式展开中按 \(x\) 的升幂排列的前四项。
a) Find the first four terms, in ascending powers of \(x\), in the binomial expansion of \(f(x)\).
b) 使用你的答案估计 \((0.995)^{30}\) 的值，答案保留6位小数。
b) Use your answer to part a to estimate the value of \((0.995)^{30}\), giving your answer to 6 decimal places.
c) 使用计算器计算 \((0.995)^{30}\) 并计算你在b部分的答案的百分比误差。
c) Use your calculator to evaluate \((0.995)^{30}\) and calculate the percentage error in your answer to part b.

答案 / Answer

a) \((1-5x)^{30} = 1 + 30(-5x) + 435(-5x)^2 + 4060(-5x)^3 + \ldots\)

\(= 1 - 150x + 10875x^2 - 507500x^3 + \ldots\)

b) 设 \(x = 0.001\)，则 \(5x = 0.005\)

\((0.995)^{30} = (1-0.005)^{30} \approx 1 - 150(0.001) + 10875(0.001)^2 - 507500(0.001)^3\)

\(= 1 - 0.15 + 0.010875 - 0.0005075 = 0.8603675\)

保留6位小数：\(0.860368\)

c) 计算器值：\((0.995)^{30} \approx 0.860708\)

百分比误差：\(\frac{|0.860708 - 0.860368|}{0.860708} \times 100\% \approx 0.04\%\)

练习7 / Exercise 7

a) 求 \((3-\frac{x}{5})^{10}\) 的二项式展开中按 \(x\) 的升幂排列的前三项。
a) Find the first 3 terms, in ascending powers of \(x\), of the binomial expansion of \((3-\frac{x}{5})^{10}\).
b) 解释你将如何使用你的展开式来估计 \((2.98)^{10}\) 的值。
b) Explain how you would use your expansion to give an estimate for the value of \((2.98)^{10}\).

答案 / Answer

a) \((3-\frac{x}{5})^{10} = 3^{10} + 10 \cdot 3^9 \cdot (-\frac{x}{5}) + 45 \cdot 3^8 \cdot (-\frac{x}{5})^2 + \ldots\)

\(= 59049 - 39366x + 13122x^2 + \ldots\)

b) 设 \(x = 0.1\)，则 \(\frac{x}{5} = 0.02\)

\((2.98)^{10} = (3-0.02)^{10} \approx 59049 - 39366(0.1) = 59049 - 3936.6 = 55112.4\)

使用更高精度：\(\approx 59049 - 39366(0.1) + 13122(0.1)^2 = 55112.4 + 131.22 = 55243.62\)

教材全解系统

何老师 用心做国际教育