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2019/02/28

a formula for π 4 ( arctan )

The inverse function of \tan x is called \arctan x .
That is, if \tan \theta =x , then \arctan x=\theta .

We have already gotten
\int_{-\infty}^{\infty} \frac{1}{1+x^2}dx =\pi . 

By some calculations. we will get
\int_0^1 \frac{1}{1+x^2}dx = \frac{\pi}{4} . 

In additions,
\arctan y =\int_0^y \frac{1}{1+x^2}dx

Thus,
\arctan y =\int_0^y (1-x^2+x^4-x^6+\cdots (-1)^nx^{2n})dx +R_n(y) 
\arctan y =y-\frac{y^3}{3}+\frac{y^5}{5}-\frac{y^7}{7}+\cdots+(-1)^n\frac{y^{2n+1}}{2n+1}+R_n(y)  
As R_n(y)\rightarrow 0 ,
  \arctan y = y-\frac{y^3}{3}+\frac{y^5}{5}-\cdots 

In this place, we put y=1 .
\arctan 1 =\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots
Therefore,
\pi=4\left( \lim_{n\rightarrow \infty}\sum_{k=0}^n(-1)^k\frac{1}{2k+1}   \right)

We got the series expression of \pi .











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