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 .
0 件のコメント:
コメントを投稿