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Historical Background of Infinite Series

The Problem: Area of a Circle

Imagine being a 17th-century mathematician facing a simple-looking question: How can we compute the area of a circle exactly?

The formula is A=πr2A = \pi r^2, but what is π\pi precisely? People knew π3.14\pi \approx 3.14 and needed more digits.

Limits of Classical Geometry

Archimedes estimated π\pi by inscribed and circumscribed polygons (hexagon, dodecagon, etc.), but this was laborious and only modestly accurate.

The Geometric-Series Formula

Mathematicians noticed

11x=1+x+x2+x3+x4+\frac{1}{1 - x} = 1 + x + x^2 + x^3 + x^4 + \cdots

When x=12x = \frac{1}{2},

1112=1+12+14+18+=2,\frac{1}{1 - \frac{1}{2}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 2, an infinite addition giving a finite result.

If you want Euler’s proof:

Euler's proof idea
  1. Let S=1+x+x2+x3+S = 1 + x + x^2 + x^3 + \cdots
  2. Multiply by xx: xS=x+x2+x3+xS = x + x^2 + x^3 + \cdots
  3. Subtract: SxS=1S - xS = 1
  4. Factor: (1x)S=1(1 - x)S = 1
  5. Solve: S=11xS = \frac{1}{1 - x}

Converges when x<1|x| < 1. The algebra is simple; its insight was profound for later analysis.

Birth of Infinite Series

This showed a finite number can be expressed as an infinite sum—opening the door to infinite series.

Computing π\pi

Leibniz later found

π4=113+1517+19,\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \cdots, derived from the power series of arctanx\arctan x at x=1x=1—simple operations to approximate π\pi, though slowly convergent.

Leibniz discovery sketch
  1. Use arctanx=xx33+x55\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots
  2. Substitute x=1x = 1 to get arctan1=π4\arctan 1 = \frac{\pi}{4}
  3. Hence π4=113+1517+\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

Historically the first infinite-series formula for π\pi.

练习题

练习 1

Compute the first four-term sum of 1+12+14+181 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}.

参考答案

Sum: 1+0.5+0.25+0.125=1.8751 + 0.5 + 0.25 + 0.125 = 1.875.

练习 2

Use the Leibniz series to approximate π\pi with the first three terms.

参考答案

π4113+15=0.8667\frac{\pi}{4} \approx 1 - \frac{1}{3} + \frac{1}{5} = 0.8667
π4×0.86673.4668\pi \approx 4 \times 0.8667 \approx 3.4668.


总结

本文出现的符号

符号类型读音/说明在本文中的含义
π\pi希腊字母Pi(派)圆周率,约等于 3.14159
\sum希腊字母Sigma(西格玛)求和符号,表示级数
\infty数学符号无穷大表示无穷级数,项数无限
xx数学符号变量级数中的变量

中英对照

中文术语英文术语音标说明
圆周率pi/paɪ/圆的周长与直径的比值,记作 π\pi
几何级数geometric series/dʒiːəˈmetrɪk ˈsɪəriːz/形如 n=0arn\sum_{n=0}^{\infty} ar^n 的级数
等比级数geometric progression/dʒiːəˈmetrɪk prəˈɡreʃən/几何级数的另一种称呼
收敛convergence/kənˈvɜːdʒəns/级数部分和序列有有限极限
无穷级数infinite series/ˈɪnfɪnɪt ˈsɪəriːz/项数无限的级数
幂级数power series/ˈpaʊə ˈsɪəriːz/形如 n=0anxn\sum_{n=0}^{\infty} a_n x^n 的级数
反正切函数arctangent function/ɑːkˈtændʒənt ˈfʌŋkʃən/反正切函数,记作 arctanx\arctan x

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