[Math]Pi(1)

數學知識忘地太快，在博客記錄一下pi的生成。

• 100 Decimal places
  • 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679
• Approximations
  • 22/7 3 decimal places (used by Egyptians around 1000BC)
  • 666/212 4 decimal places
  • 355/113 6 decimal places
  • 104348/33215 8 decimal places
• Series Expansions
  • English mathematician John Wallis in 1655.

       4 * 4 * 6 * 6 * 8 * 8 * 10 * 10 * 12 * 12 .....

  pi = 8 * -------------------------------------------------

       3 * 3 * 5 * 5 * 7 * 7 * 9 * 9 * 11 * 11 ....

  • Scottish mathematician and astronomer James Gregory in 1671

  pi = 4 * (1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + ....)

  • Swiss mathematician Leonard Euler.

  pi = sqrt(12 - (12/22) + (12/32) - (12/42) + (12/52) .... )        …… (1)

  pi = sqrt[6 * ( 1/1^2 + 1/2^2 + 1/3^2 + 1/4^2 + ...)]　　　　 …… (2)

下面則試證一下 Gregory’s Series

1. Taylor series

\begin{equation}\label{E1}
f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{ { f^{\left( n \right)}}\left( a \right)}}{{n!}}} {\left( {x - a}\right)^n}
\end{equation}

2. Maclaurin series

\begin{equation}\label{E2}
f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{ f^{\left( n \right)}\left( 0 \right) }{n!} } { x^n }
\end{equation}

3. arctan(x)一階導數

\begin{align*}
&y = f \left( x \right) = \arctan \left( x \right) \\
&x = tan \left( y \right)
\end{align*}
\begin{align*}
\Longrightarrow dx &= \sec^{2}y * dy \\
f^{ \prime }{ \left( x \right) }&= { \frac {dx}{dy} } = {\frac{1}{ x^{2}+1 } }
\end{align*}

4. 推導過程

(1).y=arctan(x)的n階導能夠用下面的方法求得：

\begin{align*}
\because &\arctan \left( x \right) = \int \nolimits_0^x \frac{1}{ 1+t^{2} } \,dt \\
&\frac{1}{1+x^{2} } = \frac{1}{2}( \frac{1}{1-ix} + \frac{1}{1+ix} ) \\
\therefore &\arctan \left( x \right) = \frac{1}{2}i \left[ \ln (1-ix) -\ln (1+ix) \right]
\end{align*}

(2).若按原始方法，得先記住分數函數的求導方式：

$$ \left( \frac { f \left( x \right) } { g \left( x \right)} \right)^{\prime} = \frac { { f^{ \prime } \left( x \right) } { g \left( x \right) } - { f \left( x \right) } { g^{ \prime } \left( x \right) } } { g^{2} \left( x \right) } $$

(3).f(x)的n階導數

\begin{align*}
& f ^{\left( 1 \right)}\left( x \right) = {\frac{1}{ x^{2}+1 } } \\
& f ^{\left( 2 \right)}\left( x \right) = {\frac{-2x}{ \left(x^{2}+1\right)^{2} } } \\
& f ^{\left( 3 \right)}\left( x \right) = {\frac{2\left( 3x^{2}-1 \right) }{ \left(x^{2}+1\right)^{3} } } \\
& f ^{\left( 4 \right)}\left( x \right) = {\frac{-24x\left(x^{2}-1\right) }{ \left(x^{2}+1\right)^{4} } } \\
& f ^{\left( 5 \right)}\left( x \right) = {\frac{24\left(5x^{4}-10x^{2}+1\right) }{ \left(x^{2}+1\right)^{5} } } \\
& ...\\
& f ^{\left( n \right)}\left( x \right) = \frac {1}{2} (-1)^{n} i \left[ (-i+x)^{-n}-(i+x)^{-n} \right] (n-1)! \\
& ...\\ 
\end{align*}

(4).f(x) Taylor Series Expansion 的係數

\begin{align*}
k_{1} &= \frac{ f ^{\left( 1 \right)}\left( 0 \right) } { 1! } = 1\\
k_{2} &= \frac{ f ^{\left( 2 \right)}\left( 0 \right) } { 2! } = 0\\
k_{3} &= \frac{ f ^{\left( 3 \right)}\left( 0 \right) } { 3! } = \frac {-1}{3}\\
k_{4} &= \frac{ f ^{\left( 4 \right)}\left( 0 \right) } { 4! } = 0\\
k_{5} &= \frac{ f ^{\left( 5 \right)}\left( 0 \right) } { 5! } = \frac {1}{5}\\
& ...\\
\end{align*}

5. get the conclusion, Maclaurin Series.

『Gregory's series』 or 『Leibniz's series』

\begin{align*} \because \arctan \left( x \right) &= \sum \limits_{n=0}^{\infty} (-1)^{n} { \frac{1}{2n+1} } x^{2n+1} \\ &= x - \frac{1}{3}x^{3} + \frac{1}{5}x^{5} - \frac{1}{7}x^{7} + ...\\ \therefore \arctan \left( 1 \right) &= 1-\frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} -\frac{1}{11}+... =\frac{ \pi }{4}\end{align*}