# [SOLVED] How to calculate PI

• Nov 17th 2006, 06:07 AM
Magician
[SOLVED] How to calculate PI
Hi Every bdy

please how to PI that equal 3.14
• Nov 17th 2006, 06:10 AM
Quick
Quote:

Originally Posted by Magician
Hi Every bdy

please how to PI that equal 3.14

Are you asking how to find $\displaystyle \pi$?
• Nov 17th 2006, 06:55 AM
CaptainBlack
Quote:

Originally Posted by Magician
Hi Every bdy

please how to PI that equal 3.14

RonL
• Nov 17th 2006, 07:47 AM
ThePerfectHacker
CaptainBlank post is not ideal because it is speaking how to find a fraction.
---

The way mathematicians did it is by an infinite series of an integral.
If you are familar with some Calculus, the curve $\displaystyle \sqrt{r^2-x^2}$ is a circle with radius $\displaystyle r$. So the area below it is,
$\displaystyle \int_{-r}^r \sqrt{r^2-x^2}dx=\pi r^2$
Setting $\displaystyle r=2$ and some manipulation,
$\displaystyle \int_0^2 \sqrt{4-x^2}=\pi$
You can calculate the integral on the top by means of an approximation and you have a value of $\displaystyle \pi$.

Another was is to use an infinite series,
$\displaystyle \frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$
But this series discovered by Leiniz converges too slowly, that means it does approach the value of $\displaystyle \pi$ but way too slow.

A quicker one I can imagine was proven by Euler, the Basel problem,
$\displaystyle 1+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6}$

But one of the most quickest converging series came from Ramanjuan, it is not as elegant looking as the ones above but it is far more efficient. I think each term add about 10 decimal places. So maybe this is the one that is programed into a computer to find decimal places.
• Nov 17th 2006, 07:53 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
CaptainBlank post is not ideal because it is speaking how to find a fraction.
---

The way mathematicians did it is by an infinite series of an integral.
If you are familar with some Calculus, the curve $\displaystyle \sqrt{r^2-x^2}$ is a circle with radius $\displaystyle r$. So the area below it is,
$\displaystyle \int_{-r}^r \sqrt{r^2-x^2}dx=\pi r^2$

Except that it give the area of a semi-circle, you mean something like:

$\displaystyle \int_{0}^r \sqrt{r^2-x^2}dx=\pi r^2/4$

but this is of no use if you don't know calculus.

Code:

This is EULER, Version 2.3 RL-06.
Type help(Return) for help.
Enter command: (20971520 Bytes free.)
Processing configuration file.
Done.
>dx=0.01
0.01
>x=dx/2:dx:1;
>
>y=sqrt(1^2-x^2);
>
>Pi=sum(y)*dx*4
3.14194
>

RonL
• Dec 1st 2006, 11:36 AM
WeeG
you can also calculate PI using the Monte Carlo simulation method and the Buffon niddle problem.
• Dec 1st 2006, 12:40 PM
TriKri
here is another thread discussing different methods of calculating $\displaystyle \pi$.