**Pi! What is π? How to prove its value is approximately 3.1415?**

"Pi" is one of the fascinating numbers in mathematics history. I have seen some poems in Sanskrit whose character sequence gives the value of phi. It was great Indian heritage. Here I am providing my own proof.

There are more than 50 solutions found to prove phi value. They range from simple geometrical based proof to advanced Fourier series based proofs. Given below is one way, which I thought during my engineering (

There are more than 50 solutions found to prove phi value. They range from simple geometrical based proof to advanced Fourier series based proofs. Given below is one way, which I thought during my engineering (

*of-course that day the class was boring*). Note that phi is irrational number.*After our proof we can conclude it*.Note that when we increase the diameter of circle its Area and Perimeter will also increase. It is obvious by means of visual observation. What interests to a mathematician is, how much is the increase ratio? Is it constant?

How can we devise an algorithm or method that can be understood by school kid (may be 5

^{th}standard)?We trace phi value in two parts. PART 1 deals with calculating area of circle with infinitesimal pieces of radii. PART 2 deals with calculating area of circle using Pythagoras Theorem.

**PART - I**

The base for our proof is Pythagoras Theorem. The sum of areas of squares formed by adjacent sides to the right angle is equal to the square formed by opposite side.

Most of us know that Area of circle is A = πr

^{2}, and Perimeter P = 2πr. Or we can write them in the following way.**Π = 2πr/2r = 2πr/D = P/D where D is diameter of circle. [Note it as equation 1]**

It means, the ratio of perimeter of circle to its diameter is constant and it is π.

*I know it is poor derivation, after all we are trying for value of π*. How much is its value? Can we make use of area of circle to find π value? Yes.How can we calculate area of circle? Have a look at the following diagram.

You might have guessed. Just divide the circle into 4 parts as in Figure 1, and join them. We will not get perfect rectangle share. Try cutting the circle into further smaller pieces, and join them.

Now, cut the circle into multiple pieces of infinitesimal width and length will obviously be its radius. If we join all these pieces, we will get the rectangle as shown in the picture.

From the diagram we can calculate the circle area as

**A = Pr/2**.**[Note this as equation 2]****PART - II**

We need bit patience. Consider upper right part of the circle in the Figure 1. We can find area of this piece using Pythagoras Theorem.

- The area of rectangle triangle OAB can be easily found which is
**½ r * r**. - We left with area of the half convex formed by chord AB and perimeter of the circle. How can we calculate it? As usual, recurs.
- Draw a perpendicular OC on to AB from the origin O. Join the lines AC and BC (those in red color). By symmetry, area of the part ADC and BDC are same.
- In-order to find area of ADC and BDC, we need length of DC. We can easily calculate the lengths OD and DC. Also note that OD + OC = radius, and AD = DB = AB/2 by symmetry. We know AB from Pythagoras which is r
^{2}+ r^{2}. From the figure we know that OD^{2}+ BD^{2}= r^{2}(Using trigonometry also we can find OD, try it). - After this step we can calculate OD and hence DC which is
**r * (SQRT(2)-1)/SQRT(2)**. AD and DB also known [**r/SQRT(2)**]. We need to repeat the above approach for pieces, ADC over the arc of circle and BDC over the arc of circle. After this step we will left with**DC =****r * (SQRT(2)-1)/SQRT(2)**,**AD = r/SQRT(2)**, which gives us area of**∆ADC = ∆BDC = r**.^{2}*(SQRT(2) - 1)/4 - Again draw perpendiculars on to AC and CB.
- At this stage the process repeats from step 3.

We will stop after sufficient accuracy.

*By now, we can understand why phi is irrational, never ending and non repeating fraction (it includes lots of irrational numbers in calculation).*The area of the upper right part AOB of circle can be integrated over these small areas found by brute force method.

There will be one triangle ∆AOB that forms first term in the series is T1 =

**½ r * r**.There will be two triangles (∆ADC and ∆BDC) that will form the second term (T2 =

**r**) in the series.^{2}/sqrt(8)There will be four triangles that will form the third term (T3) in the series.

There will be eight triangles that will form the third term (T4) in the series.

So, area of the upper right part of circle is given by

**Area of Upper Right = T1 + 2 x T2 + 4 x T3 + 8 * T4 + …**

The above series leaves only ¼ of the area of actual circle. The area of circle can be found by multiplying it with 4.

**Area of circle = 4 x [Area of Upper Right]**

**= 4 x [T1 + 2 x T2 + 4 x T3 + 8 * T4 + …]**

**= 4 x [**

**½ r**

^{2}+ 2 x**r**+ 4 x T3 + 8 * T4 + …^{2}*(SQRT(2) - 1)/4**]**

**= 2 x**

**r**

^{2}

**+ 2 x**

**r**

^{2}

***(SQRT(2) - 1)**

**+ 16 x T3 + 32 * T4 + … [Note this as equation 3]**

From equations 2 and 3, we can conclude these two areas must be same. Therefore

**Pr/2**

**=**

**2 x**

**r**

^{2}

**+ 2 x**

**r**

^{2}

***(SQRT(2) - 1)**

**+ 16 x T3 + 32 * T4**

**+ …**

**Pr/2r**

^{2}**=**

**2 x**

**r**

^{2}

**+ 2 x**

**r**

^{2}

***(SQRT(2) - 1)**

**+ 16 x T3 + 32 * T4**

**+ …**

**P/2r**

**=**

**2 x**

**r**

^{2}

**+ 2 x**

**r**

^{2}

***(SQRT(2) - 1)**

**+ 16 x T3 + 32 * T4**

**+ …**

**P/D**

**=**

**2**

**+ 2 x**

**(SQRT(2) - 1)**

**+ 16 x T3 + 32 * T4**

**+ …**

**Π**= 2

**+ 0.8284 + 16 x T3 + 32 * T4 + … --->**

**from equation 1**

Which leaves a fraction about

**3.1415…**when we include few more terms. Note that the actual value is never ending and non-repeating number.*Lengthy but interesting.*