# Beautiful Formulas for pi=3.14… The number $\pi$ is defined a s the ratio of a circle’s circumference $C$ to its diameter $d$:
$\pi=\frac{C}{d}.$

$\pi$ in decimal starts with 3.14… and never end.

I will show you several beautiful formulas for $\pi$. Add to solve later

## Art Museum of formulas for $\pi$

### Beautiful formula of $\pi$ (continued fraction).

$\pi$ is an irrational number. This means that $\pi$ can not be written as a ratio of two integers:$\pi \neq \frac{n}{m}$ for any integers $n, m$.

However, $\pi$ can be written as an infinite series of nested fractions, known as continued fraction.
There are several known continued fractions that are equal to $\pi$.
\begin{align*}
\pi&= 3 + \cfrac{1}{7
+ \cfrac{1}{15
+ \cfrac{1}{1 + \cfrac{1}{292 + \cdots}}}}
\\[20pt] \pi&= \cfrac{4}{1
+ \cfrac{1^2}{2
+ \cfrac{3^2}{2 + \cfrac{5^2}{2
+\cfrac{7^2}{2 + \cdots}}}}}
\\[20pt] \frac{4}{\pi}&= 1+\cfrac{1}{3
+ \cfrac{2^2}{5
+ \cfrac{3^2}{7 + \cfrac{4^2}{9
+\cfrac{5^2}{11 + \cdots}}}}}
\end{align*}

It is mysterious that $\pi$ in decimal shows no pattern but the expressions of $\pi$ in continued fractions have simple patterns.

### Beautiful formula of $\pi$.

\begin{align*}
\frac{\pi}{4}&=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots= \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} &&\text{Leibniz formula for $\pi$}\\[12pt] \frac{2}{\pi}&=\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{2+\sqrt{2}}}{2}\cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdots &&\text{Franciscus Vieta}\\[12pt] \frac{\pi}{2}&=\frac{2}{1}\cdot \frac{2}{3}\cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5}\cdot \frac{6}{7}\cdot \frac{8}{7} \cdot \frac{8}{9}\cdots &&\text{John Wallis}\\[12pt] \frac{\pi^2}{6}&=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}\cdots && \substack{\text{Leonhard Euler}\\ \text{ the value of Riemann zeta function $\zeta(2)$}}\\[12pt] \sqrt{\pi}&=\int_{-\infty}^{\infty}e^{-x^2}\, dx &&\text{Gaussian integral}\\[12pt] \frac{\pi}{4}&=4\arctan{\frac{1}{5}}-\arctan{\frac{1}{239}} && \text{John Machin}\\[12pt] e^{2\pi i}&=-1 && \text{Euler’s formula}
\end{align*}

### 1 Response

1. 07/13/2017

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