$c^2=a^2+b^2$

$c$

$r$

$a$

$x$

$b$

$y$

$y^2$

\begin{align} y^2 &= r^2-x^2\\ f(x)&=\sqrt{r^2-x^2}\\ r \to 1\\ f(x)&=\sqrt{1-x^2}\\ \text{Daraus folgt:}&\\ V_{\text{Halbeinheitskreis}}&=\int_{-1}^1f(dx) =\tfrac\pi2\\ &=\int_{-1}^1\sqrt{1-x^2}dx\\ &=2\cdot\int_0^1\sqrt{1-x^2}dx\\ &=2\cdot\left[\frac{\arcsin(x)}{2}+\frac{x\cdot\sqrt{1-{x}^{2}}}{2}\right]_0^1\\ &=2\cdot\frac{\arcsin{(1)}+1\cdot\sqrt{1-1^2}}2+\frac{\arcsin{(0)}+0\cdot\sqrt{1-0^2}}2\\ &=2\cdot\frac{\arcsin{(1)}+\sqrt{1-1^2}}2+\frac{\arcsin{(0)}}2\\ &=2\cdot\frac{\arcsin{(1)}+\sqrt{1-1}}2\\ &=\frac{2\cdot\arcsin{(1)}}2\\ &=\arcsin{(1)}\\ \pi&=2\cdot\arcsin{(1)} \end{align}

$\arcsin{(1)}$

\begin{align} \pi&=2\cdot\sum_{k=0}^\infty\frac{(2k)!}{4^k\cdot(k!)^2\cdot(2k+1)}\cdot x^{2k+1}\\ x\to1&\\ &=2\cdot\sum_{k=0}^\infty\frac{(2k)!}{4^k\cdot(k!)^2\cdot(2k+1)}\cdot 1^{2k+1}\\ \pi&=2\cdot\sum_{k=0}^\infty\frac{(2k)!}{4^k\cdot(k!)^2\cdot(2k+1)} \end{align}

\begin{align} \lim_{n \to 0}&\left(\pi=4\cdot\int_0^{1}f_{Viertelkreis}(x)dx\right) \end{align}

$0$

$1$

\begin{align} \lim_{n \to 0}&\left(\pi=4\cdot\sum_{k=0}^{\tfrac1n}n\cdot\sqrt{1-(kn)^2}\right) \end{align}

\begin{align} \pi(n)&=4n\cdot\sum_{k=0}^{\tfrac1n}\sqrt{1-(kn)^2} \end{align}

$0<n\leq1$

$n=0.001$

$\pi=3.143555467$

$\pi$

$\pi=2\arcsin{(1)}$

$\pi=3\arcsin{(1)}$

$\pi=6\arcsin{(0.5)}$

\begin{align} \pi&=2\cdot\sum_{k=0}^\infty\frac{(2k)!}{4^k\cdot(k!)^2\cdot(2k+1)}\cdot x^{2k+1}\\ x\to0.5&\\ \pi&=6\cdot\sum_{k=0}^\infty\frac{(2k)!}{4^k\cdot(k!)^2\cdot(2k+1)}\cdot 0.5^{2k+1}\end{align}

$3.5419\cdot10^{-64}$

$\pi=3,141592654$

$\pi$

\begin{align} \frac\pi4&=\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}\\ \pi_{Leibniz}&=4\cdot\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \end{align}

$5\cdot10^{-3}$

$\pi=3.151493401$

\begin{align} \frac{1}{\pi} &= \frac{2\sqrt{2}}{9801} \cdot\sum_{n=0}^{\infty} \frac{(4 n)! \cdot (1103+26390 n)}{(n!)^{4} \cdot 396^{4 n}}\\ \frac1\pi&=\frac{2\sqrt{2}\cdot\sum_{n=0}^{\infty} \frac{(4 n)! \cdot (1103+26390 n)}{(n!)^{4} \cdot 396^{4 n}}}{9801}\\ \pi_{Ramanujan}&=\frac{9801}{2\sqrt{2}\cdot\sum_{n=0}^{\infty} \frac{(4 n)! \cdot (1103+26390 n)}{(n!)^{4} \cdot 396^{4 n}}} \end{align}

\begin{align} \pi_{BBP_1} &= \sum_{k = 0}^{\infty} \frac{1}{16^k}\left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6}\right)\\ \pi_{BBP_1}(n) &= \sum_{k = 0}^{n} \frac{1}{16^k}\left( \frac{4}{8k + 1} - \frac{2}{8k + 4} - \frac{1}{8k + 5} - \frac{1}{8k + 6}\right) \end{align}

$1.471\cdot10^{-124}+\varepsilon_{k=101+}$

$\pi_{BBP_1}(99)=3.141592654$

\begin{align} \pi_{BBP_2} &= \frac12\sum_{k=0}^{\infty}\frac1{16^k}\left(\frac8{8k+2} + \frac4{8k+3} + \frac4{8k+4} - \frac1{8k+7}\right)\\ \pi_{BBP_2}(n) &= \frac12\sum_{k=0}^{n}\frac1{16^k}\left(\frac8{8k+2} + \frac4{8k+3} + \frac4{8k+4} - \frac1{8k+7}\right) \end{align}

$\pi_{BBP_2}(99)=3.141592654$

$k=99$

$1.1699\cdot10^{-121}$

\begin{align} \pi_{BBP_3} &= \frac14\sum_{k=0}^{\infty}\frac1{16^k}\left(\frac8{8k+1} + \frac8{8k+2} + \frac4{8k+3} - \frac2{8k+5} - \frac2{8k+6} - \frac1{8k+7}\right)\\ \pi_{BBP_3}(n) &= \frac14\sum_{k=0}^{n}\frac1{16^k}\left(\frac8{8k+1} + \frac8{8k+2} + \frac4{8k+3} - \frac2{8k+5} - \frac2{8k+6} - \frac1{8k+7}\right) \end{align}

$\pi_{BBP_3}(99)=3.141592654$

$k=99$

$1.173\cdot10^{-121}$

\begin{align} \pi_{BBP_4} &= \;\;\sum_{k=0}^{\infty}\frac{(-1)^k}{4^k}\left(\frac2{4k+1} + \frac2{4k+2} + \frac1{4k+3}\right)\\ \pi_{BBP_4}(n) &= \;\;\sum_{k=0}^{n}\frac{(-1)^k}{4^k}\left(\frac2{4k+1} + \frac2{4k+2} + \frac1{4k+3}\right) \end{align}

$\pi_{BBP_4}(99)=3.141592654$

$k=99$

$-3.1287\cdot10^{-62}$

\begin{align} \frac{1}{\pi} &= 12\cdot\sum^\infty_{k=0} \frac{(-1)^k\cdot(6k)!\cdot(163\cdot 3344418k + 13591409)}{(3k)!(k!)^3\cdot640320^{3k + \frac32}}\\ \pi_{Chudnovsky} &= \frac1{12}\cdot\sum^\infty_{k=0} \frac{(-1)^k\cdot(6k)!\cdot(545140134k + 13591409)}{(3k)!\cdot(k!)^3\cdot640320^{3k + \frac32}} \end{align}