Finding the value of pi(2) - Mathematics Blog

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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 [ 編輯日誌 ] [ 分享至FACEBOOK ]

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PMSplayer 於 2010-06-30 12:21 AM 發表：

咁您咪係度玩下您ge不等式lo, 您玩得起就唔怕悶ga la...
好似我舉果個例咁...黃金比(Golden radio)同歐拉恆等式(Euler Identity)
唔識果D就當然覺得悶, 因為佢地本來無興趣睇, 可以唔洗理...
但讀過呢兩條呢兩條公式ge人又覺得呢兩條數無新意, 平平無奇~

E+將兩條式combine...雖然係好普通ge代入,但呢D咁簡單ge方法正正就係令其他數學愛好者, 甚至數學家眼前一亮ge idea
數學係美妙, 但都要人玩得過癮先得嘛...玩黎玩去都係人地玩過ge數論, 真係悶蛋la...何況您E+打果D仲係人地ge proof...
但如果您可以玩不等式玩到又簡單, 人地又一時三刻估您唔到或者從來無諗過ge...人地就自然覺得好玩ga la, 甚至同您一齊玩tim...
例如：Let Max[g(x)] = f(x), then h(x) ≥ g(x)
↑如果您係計算過程運用類似呢D數論係計算過程...點都好過日日都係數學練習丫~

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PMSplayer 於 2010-06-29 07:02 PM 發表：

唔係丫, 我係話您有無一D好玩ge數學想法...好過您打D連練習都有ge數學...
直接D講, 去圖書館借本數學練習好過係您到睇la~

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網主回覆

又係既,我講成篇太理論性既文章都幾死板-.-

但我都諗唔到有咩好玩既數學想法

對我黎講inequality係幾好玩,不過其他人會覺得好悶

Posted at 2010-06-29 08:31 PM [ 編輯 ] [ 刪除 ]

PMSplayer 於 2010-06-26 12:31 AM 發表：

I am so sorry.=_="
I mean "Do you have your own idea in Maths?"...
For example:
Can you combine golden ratio, mathematical constant, the circumference-to-diameter ratio in one function like this?
→Let a : b = φ : 1 where φ is golden ratio
∵ a + b / a = a / b
φ + 1 = φ^2
∴φ^2 - φ -1 = 0
e^(ix) = cos x + i sin x (Euler's complex number formula)
[where e is mathematical constant]
Sub x = π where π is the circumference-to-diameter ratio,
e^[i(π)] = cos (π) + i sin (π)
e^(iπ) = (-1) + i (0)
= -1 [Euler Identity]
Sub e^(iπ) = -1 into φ^2 - φ -1 = 0
φ^2 - φ + [e^(iπ)] = 0
φ^2 = φ -(e^iπ)
φ = 1 - {[(e^(iπ)]/ φ} [Functional equation]

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You mean the interplay between different theories?

Posted at 2010-06-28 12:36 PM [ 編輯 ] [ 刪除 ]

PMSplayer 於 2010-06-25 12:30 AM 發表：

Do you have your own idea about Mathematics?

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網主回覆

Mathematics,not only possesses truth,but supreme beauty.

Many people who enjoy maths(especially mathematicians) could understand this kind of aesthetics,including myself,of course!

Mathematics is elegant,charming,we can describe this subject as art,golden ratio,topology,chaotic theory...are best reflected it.

Oh,how can I explain that beauty?I am sorry,the answer cannot be explained easily,if you enjoy solving mathematics problems,you will know!

To me,this is an absorbing subject,solving the difficult problems correctly is one way for me to feel proud.

Posted at 2010-06-26 12:05 AM [ 編輯 ] [ 刪除 ]

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