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2010 年 6 月 24 日星期四

Finding the value of pi(2)

分類: Trigonometric functi...

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

At first,don't mind any grammatical mistakes occured as the following sin la !Ha!

In the article "Finding the value of pi(1)",some concepts are clearly known for finding the value of π by the method of calculus.In this article,there is another way to find the value of π using trigonometric function.

Something is known that

$\int_0^x \frac{dt}{1+t^2}=\tan^{-1}{x}$ ,

do you know why it is happening?

To prove it,the method of substitution is needed to be used.

When we put $t = tan x ,$

$ie.$

∵ $dt=\sec^2{x}dx=\left(1+\tan^2{x}\right)dx$

∴ $\int_0^x \frac{dt}{1+t^2}=\int_0^x \frac{1+\tan^2{x}}{1+\tan^2{x}}dx$ $=\left[ x \right]_0^x$

$=\left[ \tan^{-1}{t} \right]_0^x$ $= tan - 1 x$

$Do you know how to prove the previous equation$ $\int_0^x \frac{dt}{\sqrt{1-t^2}}=\sin^{-1}{x}$ ,now?Putting t=sinx only!

In fact,we can rationalize $\frac{1}{1+t^2}$ by using a geometric series $a+ar+ar^2+...+ar^{n-1}=\frac{a(1-r^2)}{1-r}$ ,

if $n \to \infty$ ,the series will be changed as $a+ar+ar^2+...=\frac{a}{1-r}$ ,have you learnt it?

Putting a=1,r=t² into the series,

we have

$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...$ (|t|<1).

ie:

$\int_0^x \frac{dt}{1+t^2}=\int_0^x 1-t^2+t^4-t^6+...dt=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+...$

∴ $\tan^{-1}{x}=x-\frac{1}{3}x^3+\frac{1}{5}x^5-\frac{1}{7}x^7+...$

Putting x=1, $\tan^{-1}{x}=\frac{\pi}{4}$

∴ $\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$

$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...=\sum_{r=0}^{\infty}\frac{(-1)^r}{2r+1}$

It is also proved that $\tan^{-1}{x}=\sum_{r=0}^{\infty}\frac{(-1)^r}{2r+1}x^{2r+1}$

發表時間：2010-06-24 09:16 PM [ 訪客留言(4) ] [ 編輯日誌 ] [ 分享至FACEBOOK ]

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