As we know that, mathematicians heretofore proposed/suggested many mathematical expressions, applying which one can calculate the value of the ratio of circumference to the corresponding diameter of any circle of any size, popularly known as ‘π’. But, neither any one among them is based on theoretical knowledge nor any of these expressions is mathematically derived one. Also all these expressions failed completely to explain why the value of the said ratio of the circumference to the corresponding diameter is independent of the size of the circle. Fact is that all these expressions are nothing more than proposals/propositions or empirical formula without any theoretical base, whatsoever. And, in this type of background where lack of knowledge makes it dark and hazy entirely, the formulas presented here, being mathematically derived, are just enough/sufficient to end this endless darkness of ignorance with its glow of complete knowledge in this regard. The formulas put forward here is new and based on derivation.

INTRODUCTION

The paper is an extract of the original works (i.e. books) namely “ITIBRITTA’’ and “APECKHAK BISTRITI EBONG” very fundamental books on mathematical analysis and principles. In this selected part I am going to introduce the algebraic expression which actually correspond to/stands for the ratio of circumference to the corresponding diameter of any circle of any size what is popularly known as ‘π’.

And the derived formula to evaluate ‘π’ is

$\pi =a.({2}^{n+1})\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2+...{n}^{thterm}+\sqrt{a}}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$
Where

i) ‘a’ = { 2,3 } ( i.e. a set whose elements are 2 and 3 ) that is the value of ‘a’ can be either 2 or otherwise 3.

ii) ‘n’ can have any positive integral value including 0 and can be any large value without any limitation.

And this general form of the expression, which resemblance and actually represent the ratio of circumference to its diameter of any circle i. e. ‘π’ when the value of ‘n’ is sufficiently large, the formula is named as ‘JANAKEE NATH’s formula for ‘π’ in loving memory of author’s grandfather Janakee Nath De.

Now if we take the value of ‘ a ‘ as 2 in this general ‘ Janakee Nath’s Formula of ‘π’ we get the first particular formula, namely ‘MEGHA’S FORMULA‘ of π. And hence Megha’s formula of ‘π’ ultimately takes the form as

$\pi =2.({2}^{n+1})\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2+...{n}^{thterm}+\sqrt{2}}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$
And if we take the value of ‘a’ as 3 in the general ‘ Janakee Nath’s formula of ‘π’ we get the second particular formula, namely ‘ANKITA’S FORMULA of ‘π’. And hence Ankita’s formula of ‘π’ ultimately takes the form as

$\pi =3.({2}^{n+1})\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2+...{n}^{thterm}+\sqrt{3}}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$
And in calculating the value ‘π’ by using these newly invented formulas, greater value of ‘n’ will yield more accurate value of ‘π’. And when the value of ‘n’ is required infinity then the value of ‘π’ so calculated will be correct up to infinite place after decimal point. That is we will get exact value of ‘π’ when the value of ‘n’ is required infinity. Now let us check the value of π as calculated by using the two newly invented formulas, namely ‘Megha’s Formula of π’ and‘Ankita’s Formula of π’ , for different values of ‘n’. Now starts with the ‘Megha’s Formula of ‘π’.

*Part A*

Megha’s formula states that

$\pi =2.({2}^{n+1})\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2+...{n}^{thterm}+\sqrt{2}}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$

**1) For n = 0 :**

$\mathrm{\pi}=[2.\{{2}^{0+1}\left\}\right].(2\u2013\sqrt{2}{)}^{1/2},\mathrm{\pi}=(4).(0.7653668647301795...)$
**= 3.0614674589207181…**

**2) For n = 1 :**

$\mathrm{\pi}=[2.\{{2}^{1+1}\left\}\right].\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$
= (8) . (0.39018064403225653 ……)

**= 3.1214451522580522855……**

**3) For n = 2 :**

$\mathrm{\pi}=[2.\{{2}^{2+1}\left\}\right].\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$
= (16) . (0.19603428065912120….)

**= 3.136548490545939263814….**

**4) For n = 3 :**

$\mathrm{\pi}=[2.\{{2}^{3+1}\left\}\right].\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$

= (32) . (0.098135348654836……)

**= 3.14033115695475291231…….**

**4) Likewise for n = 12 :**

$\mathrm{\pi}=[2.\{{2}^{12+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}}}}}}}}}{\}}^{1/2}$

= (16384) . (0.0001917475981919546917410….)

**= 3.141592648776985669485107969 …….**

**5) In the same way for n = 20** :

$\mathrm{\pi}=[2.\{{2}^{20+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{2+\sqrt{2+....+\sqrt{{20}^{th}2+\sqrt{2}}}}}{\}}^{1/2}$
= (4194304) . (0.000191747598191954…..)

**= 3.141592653589719800**

From the above demonstrations it is now evident and clear that for greater and greater value of ‘n’ one can calculate, by using MEGHA’s FORMULA, more and more accurate value of ‘π’.

Now starts with the ‘Ankita’s Formula of ‘π’

*Part B*

Ankita’s formula states that

$\pi =3.({2}^{n+1})\left\{\begin{array}{l}\\ \end{array}\right.2\u2013\sqrt{2+\sqrt{2+\sqrt{2+...{n}^{thterm}+\sqrt{3}}}}{\left.\begin{array}{r}\\ \end{array}\right\}}^{1/2}$

**1) For n = 0 :**

$\mathrm{\pi}=[3.\{2{}^{0+1}\left\}\right].(2\u2013\sqrt{3}{)}^{1/2},\mathrm{\pi}=(6\left)\right(0.51763809020504152469...)$

= 3.105828541230249148186….

**2) For n = 1 :**

$\mathrm{\pi}=[3.\{{2}^{1+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{3}}{\}}^{1/2}$

= (12) . (0.2610523844401031830968……)

**= 3.1326286132812381971617…..**

**3) For n = 2 :**

$\mathrm{\pi}=[3.\{{2}^{2+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{2+\sqrt{3}}}{\}}^{1/2}$

= (24) . (0.130806258460286133630….)

**= 3.13935020304686720713514…**

**4) For n = 3 :**

$\mathrm{\pi}=[3.\{{2}^{3+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}{\}}^{1/2}$

= (48) . (0.06543816554355228412731985….)

**= 3.1410319508905096381113529…**

**4) Likewise for n = 12 :**

$\mathrm{\pi}=[3.\{{2}^{12+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}}}}}}}}}{\}}^{1/2}$

= (24576) . (0.00012783173223676626186947….)

**= 3.14159265145076765170425364……**

**5) In the same way for n = 20** :

$\mathrm{\pi}=[3.\{{2}^{20+1}\left\}\right].\{2\u2013\sqrt{2+\sqrt{2+\sqrt{2+....+\sqrt{{20}^{th}2+\sqrt{3}}}}}{\}}^{1/2}$

= (6291456) . (0.0000004993427043898519833…..)

**= 3.1415926535897605995224…..**

From the above demonstrations it is now evident and clear that for greater and even greater values of n one can calculate, by using ANKITA’s FORMULA, more and even more accurate value of ‘π’.

DISCUSSION & FINDINGS :

Now the reader may be well aware and also it will be worth full to mention at this juncture that many people had already suggested many expressions to calculate the value of ‘π’. Now the author presenting a few of them here for easy and ready reference for the reader.

1) Karl Theodor Wilhelm Weierstrass in 1841 AD suggested :

$\pi \mathit{}\mathit{=}\mathit{}{\mathit{\int}}_{\mathit{\u2013}\mathit{1}}^{\mathit{+}\mathit{1}}\frac{\mathit{d}\mathit{x}}{\sqrt{\mathit{1}\mathit{\u2013}{\mathit{x}}^{\mathit{2}}}}$

2) Continued fractions :

a) b)

$\mathrm{\pi}=\frac{4}{1+{\displaystyle \frac{{1}^{2}}{2+{\displaystyle \frac{{3}^{2}}{2+{\displaystyle \frac{{5}^{2}}{2+{\displaystyle \frac{{7}^{2}}{2+{\displaystyle \frac{{9}^{2}}{2+{\displaystyle \frac{....}{}}}}}}}}}}}}}\mathrm{\pi}=3+\frac{{1}^{2}}{6+{\displaystyle \frac{{3}^{2}}{6+{\displaystyle \frac{{5}^{2}}{6+{\displaystyle \frac{{7}^{2}}{6+{\displaystyle \frac{{9}^{2}}{6+{\displaystyle \frac{{11}^{2}}{6+{\displaystyle \frac{....}{}}}}}}}}}}}}}$

3) Francois Viête in 1593 AD suggested :

$\frac{\mathrm{\pi}}{2}=\frac{\sqrt{2}}{2}.\frac{\sqrt{2+\sqrt{2}}}{2}.\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}.....$

4) John Wallis in 1655 AD suggested that

$\frac{2}{\mathrm{\pi}}=\frac{2}{1}.\frac{2}{3}.\frac{4}{3}.\frac{4}{5}.\frac{6}{5}.\frac{6}{7}.\frac{8}{7}.\frac{8}{9}......$

5) Srinivasa Ramanujam in 1914 AD suggested that (sato series)

$\frac{1}{\mathrm{\pi}}=\frac{2\sqrt{2}}{9801}\sum _{k=0}^{\infty}\frac{\left(4k\right)!(1103\u201326390k)}{(k!{)}^{4}({396}^{4k})}$

6) David Volfovich Chudnovsky and Gregory Volfovich Chudnovsky in 1985 AD suggested that (sato series):

$\frac{1}{\mathrm{\pi}}=\frac{12}{(640230{)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\sum _{k=0}^{\infty}\frac{\left(6k\right)!(13591409+5451401k)}{\left(3k\right)!(k!{)}^{3}(\u2013640320{)}^{3k}}$

7) Simon Plouffe in 2006 suggested that :

${\mathrm{\pi}}^{k}=\sum _{\mathrm{n}=1}^{\infty}\frac{1}{{\mathrm{n}}^{\mathrm{k}}}\{\frac{\mathrm{a}}{\mathrm{qn}\u20131}+\frac{\mathrm{b}}{\mathrm{qn}\u20131}+\frac{\mathrm{c}}{\mathrm{qn}\u20131}\}$

Where 𝑞 = 𝑒𝜋 ( Gelfond’s constant )

K = an odd number

a, b, c = certain rational numbers

But none of the suggested expressions is able to explain why and how the

expression, as suggested by different persons, represent the ratio of the circumference to the diameter of a given circle, neither those inventors of those expressions were able to explain that the said ratio of the circumference to the diameter of a circle is independent of the size of the circle that is why, the said ratio of circumference to the diameter i. e. ‘π’ does not depend on the size of the concerned circle. Actually, all these suggested/proposed expressions/Propositions by different persons are empirical in nature and not based on theoretical knowledge whatsoever neither any of them is mathematically derived one. Especially and particularly what Karl Theodor Wilhelm Weierstrass had suggested in 1841 AD is the most ridiculous one, perhaps, among the lot.

$\mathrm{\pi}={\int}_{\u20131}^{+1}\frac{\mathrm{dx}}{\sqrt{1\u2013{\mathrm{x}}^{2}}}=[{\mathrm{Sin}}^{\u20131}\mathrm{x}{]}_{\u20131}^{+1}=[{\mathrm{Sin}}^{\u20131}\left(1\right)\u2013{\mathrm{Sin}}^{\u20131}(\u20131)]$
$=\left[2Si{n}^{\u20131}\right(1\left)\right]=2.90degree=2.\frac{\mathrm{\pi}}{2}radian=\mathrm{\pi}\mathrm{radian}$

1) Now that whatever may be the value of x (only condition to be maintained here is

(|x| ≤ 1), {\( sin^{-1} x\) }will and shall represent some angle, that means \(sin^{-1} x \) will never be a unit less pure number but will represent a physical quantity i. e. angle and hence will bear a suitable and appropriate unit of angle. But the left-hand side of the equation being the ratio of two same physical quantity that is length, is a unit less pure number.

2) To evaluate \(sin^{-1} x \), particularly in radian unit, we are bound to use the relation \(\frac{perimeter\, of\, a\, circle}{diameter\, of\, that\, circle} \) = \(\frac{p}{d}\) . [ for evaluation of the term \(sin^{-1} x \) in degree unit or gradian unit we can do it without using the relation \(\pi = \frac{p}{d}\) .] And hence by integrating \( \frac{1}{ \sqrt{1-x^{2} } } \) with respect to x from (-1) to (+1) we will get \(2sin^{-1} (1) \), but to get \(2sin^{-1} (1) = \pi\) there is no other way but to use the relation \(\pi = \frac{p}{d}\) . And thus Karl Theodor Wilhelm Weierstrass established \(\pi = \frac{p}{d}\) by using the same relation \(\pi = \frac{p}{d}\) , how idiotic indeed his approach was, in this regard !!

And in this situation where there is no theoretically and mathematically derived formula to evaluate ‘π’ that is the ratio of circumference of any circle of any size to its diameter till date, my formulas are completely different from the other empirical formulas or propositions proposed by different person at different places, some of them have presented above.

Because the formulas presented here are not any empirical formula nor mere any proposition. On the contrary author has invented these formula on the base of theoretical knowledge and derived them mathematically. And hence

1) ‘Janakee Nath’s formula of π’

2) ‘Megha’s formula of π’

3) ‘Ankita’s formula of π‘ are neither empirical formula nor mere propositions only.

And being derived ones, these formulae are of immense importance. For complete derivation of these formulae one shall go through the books namely ‘ITIBRITTA’ and ‘APECKHAK BISTRITI EBONG‘ authored by me (in Bengali language).

CONCLUSSION: No, mathematically derived formula of π was available till date. And with the invention of this formula of π, the search comes to an end.