I found the value of PI on my own!

This note is an extension to my previous note. In that note, I asked the community to find the mistake in my solution yielding 0=1. In that note, this mistake was the multiplication of inifinty and 0 which is basically an indeterminate form. This note is a correct version of the previous note and my formula seems to give a good approximation for π\pi.

For any nn sided regular polygon inscribed in a circle of radius rr, our objective is to calculate the perimeter of the polygon.

The figure represents the given situation. OO is the center of the circle and ABAB,BCBC are edges of a regular polygon. OLOL and OMOM are perpendiculars drawn on line segments ABAB and BCBC respectively. Since perpendiculars from center bisect the chord, thus we have,



OLB=OMB\angle OLB = \angle OMB

Therfore by RHS Criteration of congruency, ΔOLBΔOMB\Delta OLB \cong \Delta OMB.

Thus by CPCT, we have

LBO=OBM\angle LBO = \angle OBM

For any regular polygon of nn sides, each angle is given by 180(n2)n\frac{180(n-2)}{n}.

Therefore, LBM=180(n2)n\angle LBM = \frac{180(n-2)}{n}. Thus LBO=180(n2)2n\angle LBO = \frac{180(n-2)}{2n}.


BLr=cos(90(n2)n°)\frac{BL}{r} = cos(\frac{90(n-2)}{n}°)

 BL=rcos(90(n2)n°)\ BL = rcos(\frac{90(n-2)}{n}°) and thus,

BC=2BL=2rcos(90(n2)n°)BC = 2BL = 2rcos(\frac{90(n-2)}{n}°)

Since it is a regular polygon,

Perimeter = (n)(BA)=2nrcos(90(n2)n°)(n)(BA) = 2nrcos(\frac{90(n-2)}{n}°) .........................(1).........................(1)

But as nn approaches infinity the polygon tends to coincide the circle in which it is inscribed. Thus in that case the perimeter of the polygon becomes equal to the circumference of the circle in which it is inscribed.

Thus, Comparing the perimeter of the inifinite-sided polygon with the circumference of a circle we get,

ncos(90(n2)n°)=nsin(180n°)=π.........................(2)ncos(\frac{90(n-2)}{n}°)=nsin(\frac{180}{n}°) = \pi ......................... (2)

Let us take a reasonable approximation for nn which resembles infinity. Thus, let n=1018n=10^{18}

Plugging in the value of nn in the 2nd equation we get,

1018×sin(18×1017°)=π10^{18} \times sin(18 \times 10^{-17}°) = \pi which yields us 3.1415926535897932384626433832793.141592653589793238462643383279 using a scientific calculator which is actually π\pi upto 30 decimal places!

Basically, if you want a better approximation just increase the power of 10.

Note by Akshat Joshi
4 years, 6 months ago

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How did you determine a value for sin?

Scott Keadle - 3 years, 5 months ago

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This is the Archimedes's method of calculating ππ.

Atul Kumar Ashish - 4 years, 6 months ago

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I seriously did not know that Archimedes used this method. You can check my previous notes, on how I started to find out Pi. In the previous note the mistake I did was multiplying infinity and zero and wrote it as zero giving me the wrong answer. Today, I realized that instead of taking an infinite value, I can actually use a large finite value.

Akshat Joshi - 4 years, 6 months ago

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Atul Kumar Ashish - 4 years, 6 months ago

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I just found that Archimedes did not exactly use this method. Rather he had put polygons inside and outside the circle and compared the outside and inside perimeters to get a range in which Pi lies ( 3.1408 < π < 3.1429 ). And at that time, I feel that the concept of inifinity was not properly developed.

Akshat Joshi - 4 years, 6 months ago

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In fact you should read the comments of the video, which proves that this is not Archimedes' method.

Akshat Joshi - 4 years, 6 months ago

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