2. 원주율의 간략한 역사 - A brief history of pi

Archimedes (c.287 – c. 212 BC) $$ \begin{align} \frac{223}{71} < \pi < \frac{22}{7} \end{align} $$ $$ \begin{align} \pi \sim 3.14 \end{align} $$

 

Ptolemy (c.100 – c.170 AD) $$ \begin{align} \pi \sim 3.1416 \end{align} $$

 

Zu Chongzhi (430 - 501 AD) $$ \begin{align} \pi \sim \frac{355}{113} \quad (=3.1415929\cdots) \end{align} $$

 

Francois Viete (1540 - 1615) $$ \begin{align} \frac{2}{\pi} = \frac{ \sqrt{2} }{2} \cdot \frac{ \sqrt{ 2 + \sqrt{2} } }{2} \cdot \frac{ \sqrt{ 2 + \sqrt{ 2 + \sqrt{2} } } }{2} \cdot \cdots \end{align} $$

 

John Wallis (1616 - 1703) $$ \begin{align} \frac{\pi}{2} = \left( \frac{1 \cdot 3}{2 \cdot 2} \right) \left( \frac{3 \cdot 5}{4 \cdot 4} \right) \left( \frac{5 \cdot 7}{6 \cdot 6} \right) \cdots \end{align} $$

 

Isaac Newton (1642 - 1727) $$ \begin{align} \frac{ \pi }{ 12 } + \frac{ \sqrt{3} }{ 8 } &= \int_0^{\frac12} \sqrt{ 1 - x^2 } \, \textrm{d}x \\ \\ \frac{ \pi }{ 12 } + \frac{ \sqrt{3} }{ 8 } &= \frac12 - \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{4m+4} m!(m+1)! (2m+3) } \\ \\ \Rightarrow \qquad \pi &\sim 3.14161 \end{align} $$

 

Proof. We consider $$ \begin{align} \frac1{ \sqrt{ 1 - x^2 } } &= \left( 1-x^2 \right)^{ - \frac12 } \\ &= 1 - \frac12 \left( -x^2 \right) + \frac{ \left( - \frac12 \right) \left( - \frac32 \right) }{ 2! } \left( -x^2 \right)^2 + \frac{ \left( - \frac12 \right) \left( - \frac32 \right) \left( - \frac52 \right) }{ 3! } \left( -x^2 \right)^3 + \cdots \\ &= 1 + \frac12 x^2 + \frac{ 1 \cdot 3 }{ 2! \cdot 2^2 } \left( x^2 \right)^2 + \frac{ 1 \cdot 3 \cdot 5 }{ 3! \cdot 2^3 } \left( x^2 \right)^3 + \cdots \\ &\qquad + \frac{ 1 \cdot 3 \cdot 5 \cdots (2m-1) }{ m! \cdot 2^m } \left( x^2 \right)^m + \cdots \end{align} $$

 

We find $$ \begin{align} \frac{ 1 \cdot 3 \cdot 5 \cdots (2m-1) }{ m! \cdot 2^m } = \frac{ (2m)! }{ m! \cdot 2^m \cdot (2m)(2m-2)(2m-4) \cdots 2 } = \frac{ (2m)! }{ 2^{2m} (m!)^2 } \end{align} $$ which gives $$ \begin{align} \Rightarrow \qquad \frac1{ \sqrt{ 1 - x^2 } } = \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } x^{2m} \end{align} $$

 

Now, for $\int \sqrt{ 1 - x^2 } \, \textrm{d}x$, we note: $$ \begin{align} && \frac{ \textrm{d} }{ \textrm{d}x } \sqrt{ 1 - x^2 } &= - \frac{ x }{ \sqrt{ 1 - x^2 } } = - x \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } x^{2m} \\ \\ &\Rightarrow & \int_0^x \left( \frac{ \textrm{d} }{ \textrm{d}t } \sqrt{ 1 - x^2 } \right) \, \textrm{d}t&= \sqrt{ 1 - x^2 } - 1 = - \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } \int_0^x t^{2m+1} \, \textrm{d}t \\ \\ &\Rightarrow & \sqrt{ 1 - x^2 } &= 1 - \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{ x^{2m+2} }{ 2m+2 } \\ \\ &\Rightarrow & \int_0^{\frac12} \sqrt{ 1 - x^2 } \, \textrm{d}x &= \left[ x - \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{ x^{2m+3} }{ (2m+2)(2m+3) } \right]_0^{\frac12} \\ &&&= \frac12 - \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{ \left( \frac12 \right)^{2m+3} }{ (2m+2)(2m+3) } \\ &&&= \frac12 - \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{4m+4} m!(m+1)! (2m+3) } \\ \end{align} $$

 

Gottfried Wilhelm Leibniz (1646 - 1716), James Gregory (1638 - 1675) $$ \begin{align} \frac{\pi}{4} = 1 - \frac13 + \frac15 - \frac17 + \frac19 - \cdots \end{align} $$

 

Proof. We consider: $$ \begin{align} \arctan x = \int_0^x \frac1{ 1 + t^2 } \, \textrm{d}t \\ \end{align} $$ By binomial expansion, $$ \begin{align} \left( 1 + t^2 \right)^{-1} &= 1 - t^2 + t^4 - t^6 + \cdots \\ &= \sum_{m=0}^\infty (-1)^m t^{2m} \end{align} $$ Then, $$ \begin{align} \Rightarrow \qquad \arctan x &= \int_0^x \sum_{m=0}^\infty (-1)^m t^{2m} \, \textrm{d}t \\ &= \sum_{m=0}^\infty (-1)^m \frac{x^{2m+1}}{2m+1} \end{align} $$ This Maclaurin series is also known as Gregory's series. The Leibniz's $\pi$ formula is obtained when $x=1$, i.e. $$ \begin{align} \Rightarrow \qquad \frac{\pi}{4} = \arctan 1 = \sum_{m=0}^\infty \frac{ (-1)^m }{2m+1} = 1 - \frac13 + \frac15 - \frac17 + \cdots \qquad \square \end{align} $$ Note: This is also related to the Dirichlet beta function.

 

Karl Weierstrass (1815 - 1897) $$ \begin{align} \pi = \int_{-1}^1 \frac1{ \sqrt{ 1 - x^2 } } \, \textrm{d}x = \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{ 2 }{ 2m + 1 } \end{align} $$

 

Proof. We note that: $$ \begin{align} \frac1{ \sqrt{ 1 - x^2 } } &= \left( 1-x^2 \right)^{ - \frac12 } \\ &= 1 - \frac12 \left( -x^2 \right) + \frac{ \left( - \frac12 \right) \left( - \frac32 \right) }{ 2! } \left( -x^2 \right)^2 + \frac{ \left( - \frac12 \right) \left( - \frac32 \right) \left( - \frac52 \right) }{ 3! } \left( -x^2 \right)^3 + \cdots \\ &= 1 + \frac12 x^2 + \frac{ 1 \cdot 3 }{ 2! \cdot 2^2 } \left( x^2 \right)^2 + \frac{ 1 \cdot 3 \cdot 5 }{ 3! \cdot 2^3 } \left( x^2 \right)^3 + \cdots \\ &\qquad + \frac{ 1 \cdot 3 \cdot 5 \cdots (2m-1) }{ m! \cdot 2^m } \left( x^2 \right)^m + \cdots \end{align} $$

 

We find $$ \begin{align} \frac{ 1 \cdot 3 \cdot 5 \cdots (2m-1) }{ m! \cdot 2^m } = \frac{ (2m)! }{ m! \cdot 2^m \cdot (2m)(2m-2)(2m-4) \cdots 2 } = \frac{ (2m)! }{ 2^{2m} (m!)^2 } \end{align} $$ which gives $$ \begin{align} \Rightarrow \qquad \frac1{ \sqrt{ 1 - x^2 } } = \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } x^{2m} \end{align} $$ Finally, $$ \begin{align} I&=\int_{-1}^1 \frac1{ \sqrt{ 1 - x^2 } } \, \textrm{d}x = \arcsin 1 - \arcsin (-1) = \pi \\ &= \int_{-1}^1 \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } x^{2m} \, \textrm{d}x = \sum_{m=0}^\infty \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{ 2 }{ 2m + 1 } \qquad \square \end{align} $$

 

Srinivasa Ramanujan (1887 - 1920) $$ \begin{align} \frac1{\pi} = \frac{ 2 \sqrt{2} }{ 99^2 } \sum_{k=0}^\infty \frac{ (4k)! }{ (k!)^4 } \frac{ 26390 k + 1103 }{ 396^{4k} } \end{align} $$