近似计算 π 的值

isee

源自知乎提问

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生活里要求$\pi$ 的近似值不算高，可取
\begin{align*}\pi\approx \sqrt2+\sqrt3\approx 1.414+1.732=3.146.\end{align*}

上式的来源：在单位圆中内接正八边形的面积为
\begin{align*}S_8=8\cdot \left(\frac12\cdot 1^2\cdot \sin\frac\pi4\right)=2\sqrt2.\end{align*}
其外切正六边形的面积为
\begin{align*}S_6=6\cdot \left(\frac12\cdot \Big(\frac2{\sqrt3}\Big)^2\cdot \sin\frac\pi3\right)=2\sqrt3.\end{align*}
取这两面积的平均就约等于单位圆的面积 $S=\pi\cdot 1^2=\pi$ ，即有
\begin{align*}\frac{S_8+S_6}2\approx3.146\approx S=\pi.\end{align*}

hejoseph

用这个近似值还不如用祖冲之提到的约率和密率呢
\begin{align*}
\pi&\approx 3.141592654\\
\frac{22}{7}&\approx 3.142857143\\
\frac{355}{113}&\approx 3.141592920
\end{align*}
这两个数简单得多了。

hbghlyj

密率355/113是π的一个渐近分数（参见连分数），是分母小于16604的所有既约分数中最接近π的一个（参见最佳逼近）。它的小数点后六位皆与π相同，与其仅有0.000009%的差距，即小于1/3748629。更加精确的分数近似值，则是52163/16604，也仍然只有小数点后六位数字相同。而要达到7位数字相同，则要86953/27678才得以实现。若要8位數字相同，則要102928/32763才成。

hbghlyj

The accuracy of Milü to the true value of π can be explained using the continued fraction expansion of π, the first few terms of which are [3; 7, 15, 1, 292, 1, 1, ...]. A property of continued fractions is that truncating the expansion of a given number at any point will give the "best rational approximation" to the number. To obtain Milü, truncate the continued fraction expansion of π immediately before the term 292; that is, π is approximated by the finite continued fraction [3; 7, 15, 1], which is equivalent to Milü. Since 292 is an unusually large term in a continued fraction expansion (corresponding to the next truncation introducing only a very small term, 1/292, to the overall fraction), this convergent will be especially close to the true value of π:\[ \pi =3+{\cfrac {1}{7+{\cfrac {1}{15+{\cfrac {1}{1+{\color {magenta}{\cfrac {1}{292+\cdots }}}}}}}}}\quad \approx \quad 3+{\cfrac {1}{7+{\cfrac {1}{15+{\cfrac {1}{1+{\color {magenta}0}}}}}}}={\frac {355}{113}}\][3]

isee

源自知乎提问区

题主核心意思是： \begin{gather*}
\pi\approx8\arctan\frac5{12}.
\end{gather*}
这是可以证明的.

由正切倍角公式 \begin{gather*}
\tan\left(2\arctan\frac5{12}\right)=\frac{2\cdot \frac5{12}}{1-\left(\frac5{12}\right)^2}=\frac{120}{119},
\end{gather*} 于是 \begin{gather*}
\tan\left(2\arctan\frac5{12}-\frac{\pi}4\right)=\frac{\frac{120}{119}-1}{1+\frac{120}{119}\cdot 1}=\frac{120-119}{119+120}=\frac1{239},
\end{gather*} 因此知 \begin{gather*}
2\arctan\frac5{12}-\frac{\pi}4=\arctan\frac1{239},\\[1ex]
\iff \pi=8\arctan\frac5{12}-4\arctan\frac1{239},
\end{gather*} 而 $\frac1{239}=0.\dot 004184\dot1$ ，非常接近零，且当 $x\to 0$ 时， $\arctan x\sim x$ ，从而知 \[\arctan\frac1{239}\approx 0.00418,\] 进一步有 \begin{align*}
\pi&\approx 8\arctan\frac5{12}-4\times 0.00418\\[1ex]
&\approx 8\arctan\frac5{12}.
\end{align*}

题主为什么又提第一组勾股数呢？


顺着这个思想，且联想到马青公式\begin{gather*}
\pi=16\arctan\frac15-4 \arctan\frac1{239},
\end{gather*}

哦，原来是 \[\arctan\frac5{12}=2\arctan\frac15.\] 题主最后那个239，就都溯到源了——马青公式.

hbghlyj

Proof that $22/7 > π$
$$0<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1+x^{2}}}\,dx={\frac {22}{7}}-\pi .$$
Proof that $355/113 >π$
$$0<\int _{0}^{1}{\frac {x^{8}\left(1-x\right)^{8}\left(25+816x^{2}\right)}{3164\left(1+x^{2}\right)}}\,dx={\frac {355}{113}}-\pi .$$

kuing

hbghlyj 发表于 2024-5-13 17:42
Proof that $22/7 > π$
$$0<\int _{0}^{1}{\frac {x^{4}\left(1-x\right)^{4}}{1+x^{2}}}\,dx={\frac {22} ...

之前 86鱼 给我发过一个视频破解这种积分玩法嘀：
bilibili.com/video/BV1nb4y1u72V/