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π的萊布尼茨公式

維基百科,自由的百科全書(由于維基被禁,搬到此)

在數學領域,π的萊布尼茨公式說明

   π 4 ​ = 1   −   1 3   +   1 5   −   1 7   +   1 9   −   ⋯    {\displaystyle \;{\frac {\pi }{4}}\!=1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,{\frac {1}{7}}\,+\,{\frac {1}{9}}\,-\,\cdots \;} 4π​=1−31​+51​−71​+91​−⋯

右邊的展式是一個無窮級數,被稱為萊布尼茨級數,這個級數收斂到 π 4 {\displaystyle {\frac {\pi }{4}}} 4π​。它通常也被稱為格雷戈裡-萊布尼茨級數,用以紀念萊布尼茨同時代的天文學家兼數學家詹姆斯·格雷戈裡。使用求和符号可記作:

   π 4 = ∑ n = 0 ∞   ( − 1 ) n 2 n + 1 {\displaystyle \;{\frac {\pi }{4}}=\sum _{n=0}^{\infty }\,{\frac {(-1)^{n}}{2n+1}}} 4π​=n=0∑∞​2n+1(−1)n​

目錄

1 證明

1.1 初等證明

2 參考文獻

3 外部連結

證明

考慮下面的幾何數列:

1 &ThinSpace; − &ThinSpace; x 2 &ThinSpace; + &ThinSpace; x 4 &ThinSpace; − &ThinSpace; x 6 &ThinSpace; + &ThinSpace; x 8 &ThinSpace; − &ThinSpace; ⋯ &ThickSpace; = &ThickSpace; 1 1 + x 2 , ∣ x ∣ &lt; 1. &NegativeThinSpace; {\displaystyle 1\,-\,x^{2}\,+\,x^{4}\,-\,x^{6}\,+\,x^{8}\,-\,\cdots \;=\;{\frac {1}{1+x^{2}}},\qquad |x|&lt;1.\!} 1−x2+x4−x6+x8−⋯=1+x21​,∣x∣<1.

對等式兩邊積分可得到反正切的幂級數:

x &ThinSpace; − &ThinSpace; x 3 3 &ThinSpace; + &ThinSpace; x 5 5 &ThinSpace; − &ThinSpace; x 7 7 &ThinSpace; + &ThinSpace; x 9 9 &ThinSpace; − &ThinSpace; ⋯ &ThickSpace; = &ThickSpace; tan ⁡ − 1 x , ∣ x ∣ &lt; 1. &NegativeThinSpace; {\displaystyle x\,-\,{\frac {x^{3}}{3}}\,+\,{\frac {x^{5}}{5}}\,-\,{\frac {x^{7}}{7}}\,+\,{\frac {x^{9}}{9}}\,-\,\cdots \;=\;\tan ^{-1}x,\qquad |x|&lt;1.\!} x−3x3​+5x5​−7x7​+9x9​−⋯=tan−1x,∣x∣<1.

将 x = 1 x = 1 x=1 代入,便得萊布尼茲公式(1的反正切是 π ⁄ 4 π ⁄ 4 π⁄4)。這種推理産生的一個問題是1不在幂級數的收斂半徑以内。是以,需要額外論證當 x = 1 x = 1 x=1時級數收斂到 tan ⁡ − 1 ( 1 ) \tan^{−1}(1) tan−1(1)。一種方法是利用交替級數判别法,然後使用阿貝爾定理證明級數收斂到 tan ⁡ − 1 ( 1 ) \tan^{−1}(1) tan−1(1)。然而,也可以用一個完全初等的證明。

初等證明

考慮如下分解

1 1 + x 2 &ThickSpace; = &ThickSpace; 1 &ThinSpace; − &ThinSpace; x 2 &ThinSpace; + &ThinSpace; x 4 &ThinSpace; − &ThinSpace; ⋯ &ThinSpace; + &ThinSpace; ( − 1 ) n x 2 n &ThickSpace; + &ThickSpace; ( − 1 ) n + 1 &ThinSpace; x 2 n + 2 1 + x 2 . &NegativeThinSpace; {\displaystyle {\frac {1}{1+x^{2}}}\;=\;1\,-\,x^{2}\,+\,x^{4}\,-\,\cdots \,+\,(-1)^{n}x^{2n}\;+\;{\frac {(-1)^{n+1}\,x^{2n+2}}{1+x^{2}}}.\!} 1+x21​=1−x2+x4−⋯+(−1)nx2n+1+x2(−1)n+1x2n+2​.

對于 ∣ x ∣ &lt; 1 |x| &lt; 1 ∣x∣<1,右側的分式是餘下的幾何級數的和。然而,上面的方程并沒有包含無窮級數,并且對任何實數 x x x 成立。上式兩端從0到1積分可得:

π 4 &ThickSpace; = &ThickSpace; 1 &ThinSpace; − &ThinSpace; 1 3 &ThinSpace; + &ThinSpace; 1 5 &ThinSpace; − &ThinSpace; ⋯ &ThinSpace; + ( − 1 ) n 2 n + 1 &ThickSpace; + &ThickSpace; ( − 1 ) n + 1 &NegativeThinSpace;&NegativeThinSpace; ∫ 0 1 x 2 n + 2 1 + x 2 &ThinSpace; d x . &NegativeThinSpace; {\displaystyle {\frac {\pi }{4}}\;=\;1\,-\,{\frac {1}{3}}\,+\,{\frac {1}{5}}\,-\,\cdots \,+{\frac {(-1)^{n}}{2n+1}}\;+\;(-1)^{n+1}\!\!\int _{0}^{1}{\frac {x^{2n+2}}{1+x^{2}}}\,dx.\!} 4π​=1−31​+51​−⋯+2n+1(−1)n​+(−1)n+1∫01​1+x2x2n+2​dx.

當 n → ∞ &NegativeThinSpace; {\displaystyle n\rightarrow \infty \!} n→∞ 時,除積分項以外的項收斂到萊布尼茨級數。同時,積分項收斂到 0:

0 ≤ ∫ 0 1 x 2 n + 2 1 + x 2 &ThinSpace; d x ≤ ∫ 0 1 x 2 n + 2 &ThinSpace; d x &ThickSpace; = &ThickSpace; 1 2 n + 3 &ThickSpace; → &ThickSpace; 0 &NegativeThinSpace; &ThinSpace;&ThinSpace; 當 n → ∞ &NegativeThinSpace; {\displaystyle 0\leq \int _{0}^{1}{\frac {x^{2n+2}}{1+x^{2}}}\,dx\leq \int _{0}^{1}x^{2n+2}\,dx\;=\;{\frac {1}{2n+3}}\;\rightarrow \;0\!}\,\, 當 {\displaystyle n\rightarrow \infty \!} 0≤∫01​1+x2x2n+2​dx≤∫01​x2n+2dx=2n+31​→0當n→∞

這便證明了萊布尼茨公式。

參考文獻

Jonathan Borwein, David Bailey & Roland Girgensohn, Experimentation in Mathematics - Computational Paths to Discovery, A K Peters 2003, ISBN 1-56881-136-5, pages 28–30.

外部連結

Implementation of the Leibniz formula for TI Basic

Leibniz Formula in C, x86 FPU Assembly, x86-64 SSE3 Assembly, and DEC Alpha Assembly

補充

π 4 = ∫ 0 1 1 1 + x 2 &ThinSpace; d x . \frac {\pi}{4}=\int _{0}^{1}{\frac {1}{1+x^{2}}}\,dx. 4π​=∫01​1+x21​dx.

π 2 = ∫ 0 ∞ 1 1 + x 2 &ThinSpace; d x . \frac {\pi}{2}=\int _{0}^{\infty}{\frac {1}{1+x^{2}}}\,dx. 2π​=∫0∞​1+x21​dx.

π = ∫ − ∞ ∞ 1 1 + x 2 &ThinSpace; d x . \pi=\int _{-\infty}^{\infty}{\frac {1}{1+x^{2}}}\,dx. π=∫−∞∞​1+x21​dx.

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