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2010 年 6 月 23 日星期三

Finding the value of pi(1)

分類: Trigonometric functi...

The articles in "trigonometric function" are written for the readers who have learnt calculus,is rarely high level than the articles in "calculus", if you haven't learnt yet,click "calculus" la!

It is known that

$y=\int_{0}^{x}\frac {dt} {\sqrt{1-t^2}}=\sin^{-1}{x}$

Putting $n=-\frac{1} {2},x=-t^2$ into the binomail theorem

$(1+x)^n=\sum_{k=0}^{n}{n\choose r}x^{n-r}$ ,

we have

$\frac{1}{\sqrt{1-t^2}}$

$=(1-t^2)^{-\frac{1}{2}}$

$=1+\frac{1} {2}t^2+\frac{1\times 3} {2\times 4}t^4+\frac{1\times 3\times 5} {2\times 4\times 6} t^6+...$

Thus,it is clear that if we remove the integral notation from expression

$y=\int_{0}^{x}\frac {dt} {\sqrt{1-t^2}}=\sin^{-1}{x}$ ,

then the relationship between x and y is as following(with y repressented by the infinite series):

$y=\sin^{-1}{x}=\int_{0}^{x}\frac {dt} {\sqrt{1-t^2}}$

$=\int_0^x\left(1+\frac{1} {2}t^2+\frac{1\times 3} {2\times 4}t^4+\frac{1\times 3\times 5} {2\times 4\times 6} t^6+...\right)dt$

$=x+\frac{1} {2}\frac{x^3}{3}+\frac{1\times 3} {2\times 4}\frac{x^5}{5}+\frac{1\times 3\times 5} {2\times 4\times 6} \frac{x^7}{7}+...$ .

Putting $x=1,\sin^{-1}{x}=\frac{\pi}{2}$

$\frac{\pi}{2}=1+\frac{1} {2}\times \frac{1}{3}+\frac{1\times 3} {2\times 4}\times \frac{1}{5}+\frac{1\times 3\times 5} {2\times 4\times 6}\times \frac{1}{7}+...$ .

發表時間：2010-06-24 08:24 PM [ 訪客留言(0) ] [ 編輯日誌 ] [ 分享至FACEBOOK ]

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