crackpot pi letter

Hippocrates of Chios, Greece (471 BC 411 BC)

the Founding Father of Mathematics. He squared

lunes, circle and semi circle (for details Pi of the

Circle, Last Chapter at www.rsjreddy.webnode.com)

Dear Students,

May God Bless You

You are very fortunate to study for your graduation in one of the top

most best Universities in the world. And at the same time you must be brilliant

too to enter here and study.

With this in my mind, I thought of informing the latest developments on

the value and nature of . Further, I believe, a genius may be there in one of

you who can see my papers, read them, no no study them, understand, think

intuitively taking a challenge that what a Zoology teacher, could do it and why

not I being a mathematician.

1. Are we to believe 3.1415926 borrowed number from the polygon

inscribed / circumscribed in and about circle and attributed to circle

believing the notion limitation principle purely based on logic is final.

2. When we call a number an approximation, that implies an exact value is

there. What ? Let us search ! you have answer in my work.

3. There are many proposals in the mathematics literature for squaring of

circle. If that is so, the mathematicians who have been searching must

have believed that squaring of circle is possible. Circle Squarer must

be then a misnomer.

4. Is C.L.F. Lindemanns 1882, proof based on questionable Eulers

formula ei

+1 = 0 right in calling as a transcendental number ? He is

right in calling 3.1415926 and not .

5. If that is so the number must be an algebraic number.

6. If that is so, further, as we are sure 3.1415926 is accepted as

transcendental number then 3.1415926 is not a number at all.

Dear Students, think over

Author

An advise: If you strongly believe that (1) 3.1415926 is the value of . (2) is a transcendental number and (3) squaring of circle is impossible, you need not look at

this work and waste your precious time.

Sir, Namasthe

In 1972, I wanted to calculate area and circumference of a circle with the line-segment radius

alone, and without pi 22/7. After 26 years of struggle two formulas with radius alone were

found in 1998. These formulas have given (14-root2)/4= 3.14644660942....as pi value. Next

16 years - 24 hour daily struggle- helped me to confirm that value as pi. 4 papers are

published recently. I want to share them with you. Kindly read and oblige the request of this

fragile old man of 68...author

IOSR Journal of Mathematics (IOSR-JM)

e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 48-49

www.iosrjournals.org

www.iosrjournals.org 48 | Page

New Method of Computing value (Siva Method)

RD Sarva Jagannada Reddy

I. Introduction

equal to 3.1415926 is an approximation. It has ruled the world for 2240 years. There is a necessity to find out the exact value in the place of this approximate value. The following method givesthe total area of

the square, and also the total area of the inscribed circle. derived from this area is thus exact.

II. Construction procedure Draw a circle with center 0 and radius a/2. Diameter is a. Draw 4 equidistant tangents on the

circle. They intersect at A, B, C and D resulting in ABCD square. The side of the square is also equal to

diameter a. Draw two diagonals. E, F, G and H are the mid points of four sides. Join EG, FH, EF, FG, GH and HE. Draw four arcs with radius a/2 and with centres A, B, C and D. Now the circle square composite

system is divided into 32 segments and number them 1 to 32. 1 to 16 are of one dimension called S1 segments

and 17 to 32 are of different dimension called S2 segments.

III. Calculations: ABCD = Square; Side = a, EFGH = Circle, diameter = a, radius = a/2

Area of the S1 segment =26 2

128a

; Area of the S2 segment = 22 2

128a

;

Area of the square = 16 S1 + 16S2 = 2 2 26 2 2 216 16

128 128a a a

Area of the inscribed circle = 16S1 + 8S2 = 2 2 26 2 2 2 14 216 8

128 128 16a a a

General formula for the area of the circle

2 2214 2

4 4 16

d aa

; where a= d = side = diameter

14 2

4

IV. How two formulae for S1 and S2 segments are derived ? 16 S1 + 16 S2 = a

2 = area of the Square Eq. (1)

New Method of Computing value (Siva Method)


16 S1 + 8 S2 =

2

4

a= area of the Circle Eq. (2)

..

(1) (2) 8S2 =

2 2 22 4

4 4

a a aa

= S2 =

2 244

32 32

a a

(2)x 2 32 S1 + 16 S2 =

22

4

a Eq. (3)

16 S1 + 16 S2 = a2

Eq. (1)

(3) (1) 16S1 =

22

2

aa

= S1 =

2 222

32 32

a a

V. Both the values appear correct when involved in the two formulae a) Official value = 3.1415926

b) Proposed value = 3.1464466 = 14 2

4

Hence, another approach is followed here to decide real value.

VI. Involvement of line-segments are chosen to decide real value. A line-segment equal to the value of ( - 2) in S1 segments formula and second line-segment equal to the

value of (4 - ) in S2 segments formula are searched in the above construction.

a) Official : - 2 = 3.1415926 - 2 = 1.1415926.

Proposed : - 2 = 14 22

4

= 6 2

4

The following calculation gives a line-segment for 6 2

4

and no line-segment for 1.1415926..

IM and LR two parallel lines to DC and CB; OK = OJ = Radius = 2

a; JOK = triangle

JK = Hypotenuse = 2

2

a

Third square = LKMC; KM = CM = Side = ?

KM = 2 1 2 2

2 2 2 4

IM JK aa a

; Side of first square DC = a

DC + CM = 2 2 6 2

4 4a a a

b) Official = 4 - = 4 3.1415926 = 0.8584074.

Proposed = 4 - = 14 2 2 2

44 4

No line-segment for 0.8584074 in this diagram.

MB line-segment is equal to 2 2

4

. How ?

Side of the first square CB = a

MB = CB CM = 2 2 2 24 4

a a a

VII. Conclusion: This diagram not only gives two formulae for the areas of S1& S2 segments andalso shows two line-

segments for ( - 2) and (4 - ). Line-segment is the soul of Geometry.


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. I. (Jan. 2014), PP 58-59



Jesus Method to Compute the Circumference of A Circle and

Exact Value

RD Sarava Jagannada Reddy

I. Introduction The Holy Bible has said value is equal to 3. Mathematicians were not satisfied with the value. They thought over. Pythagorean theorem came in the mean time. A regular polygon with known perimeter was

inscribed in a circle and the sides doubled successively until the inscribed polygon touches the circumference,

leaving no gap between them. Hence this method is called Exhaustion method. The value of the perimeter of

the inscribed polygon is calculated applying Pythagorean theorem and is attributed to the circumference of the

circle. This method was interpreted, first time, on scientific lines by Archimedes of Syracuse, Greece. He has

said value is less than 3 1/7. Later mathematicians have refined the Exhaustion method and found many decimals. The value is

3.1415926 and this value has been made official. From 15

th century (Madhava (1450) of South India) onwards infinite series has been used for more

decimals to compute 3.1415926 of geometrical method. Notable people are Francois Viete (1579), Van

Ceulen (1596), John Wallis (1655), William Brouncker (1658) James Gregory (1660), G.W. Leibnitz

(1658), Isaac Newton (1666), Machin (1776), Euler (1748), S. Ramanujan (1914), Chudnovsky brothers

(1989). The latest infinite series for the computation of value is that of David Bailey, Peter Borwein and Simon Plouffe (1996) and is as follows:

0

1 4 2 1 1

16 8 1 8 4 8 5 8 6ii i i i i

Using above formula Yasumasa Kanada of Tokyo University, Japan, calculated trillions of decimals

to 3.1415926.. with the help of super computer. Mathematics is an exact science. We have compromised with an approximate value. Hence, many

have tried to find exact value. This author is one among the millions. What is ? It is the ratio of circumference of a circle to its diameter. However, in Exhaustion method, perimeter of the inscribed polygon is

divided by the diameter of the outside circle. Thus 3.1415926. violates the definition of . This is about the

value of . Next, about the nature of . C.L.F. Lindemann (1882) has said is a transcendental number based

on Eulers formula 1 0ie . In Mathematical Cranks, Underwood Dudley has said s only position in mathematics is its relation to infinite services (and) that has no relation to the circle. Lindemann proclaimed the squaring of the circle impossible, but Lindemanns proof is misleading for he uses numbers (which are approximate in themselves) in his proof.

Hence, pre-infinite series days of geometrical method is approached again to find out exact value

and squaring of circle. This author has struggled for 26 years (1972 to 1998) and calculated the exact value of

in March, 1998. The following method calculates the total length of circumference and thus the exact value has been derived from it.

Procedure: Draw a square. Draw two diagonals. Inscribe a circle. Side = a,

Diagonal = 2a , Diameter is also = a = d. 1) Straighten the square. Perimeter = 4a

Perimeter Sum of the lengths of two diagonals = 4 2 2a a = esp esp = end segment of the perimeter of the square.

Jesus Method To Compute The Circumference Of A Circle And Exact Value


2) Straighten similarly the circumference of the inscribed circle

3 diameters plus some length, is equal to the length of the circumference.

Let us say circumference = x.

Circumference 3 diameters = x 3a = esc esc = end segment of the circumference of the circle.

3) When the side of the square is equal to a, the radius of the inscribed circle is equal to a/2. So, the radius is 1/8

th of the perimeter of the square.

4) The above relation also exists between the end segment of the circumference of the circle and the end segment of the perimeter of the square.

Thus as radius 2

a

of the inscribed circle is to the perimeter of the square (4a), i.e., 1/8th

of it,

so also, is the end segment of the circumference of the circle, to the end segment of the perimeter

of the square.

So, the end segment of the circumference = 8

end segment of the perimeter of the square

4 2 23

8 8

esp a aesc x a

14 2

4

a ax

5) Circumference of the circle = d = a (where a = d = diameter)

14 2

4

a aa

14 2

4

II. Conclusion

value, derived from the Jesus proof is algebraic, being a root of 2 56 97 0x x but also that it

differs from the usually accepted value in the third decimal place, being 3.146..


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 1 Ver. IV. (Feb. 2014), PP 14-15



Durga Method of Squaring A Circle

RD Sarva Jagannadha Reddy

Abstract: Squaring of circle is an unsolved problem with the official value 3.1415926 with the new value

1/4 (14- 2 ) it is done in this paper.

Keywords: Exact Pi value = 1/4 (14- 2 ), Squaring of circle, Hippocrates squaring of lunes.

I. Introduction Squaring a circle is defined as constructing a square having an area equal to that of a given circle. It is

also called as quadrature of the circle.

This concept has been there from the days of Rhind Papyrus (1800 B.C) written by a scribe named

Ahmes. Hippocrates of Chios (450 B.C) has squared lunes, full circle and semicircle along with lunes. He

fore saw the algebraic nature of the value. value 3.1415926 has failed to find a place for it in the squaring of lunes. Though the World of Mathematics has accepted his squaring of lunes, they became silent for why

3.1415926 is a misfit in his constructions. Further, there is a false opinion that Hippocrates could not square a circle. However, Hippocrates did square a full circle and a semicircle along with a lune. In both the cases

squaring a lune, squaring a circle along with a lune the new value, 14 2

4

has explained perfectly well the

constructions of Hippocrates. Thus the propositions of Hippocrates which remained theoretical all these 2400

years, have become practical constructions with the discovery of 14 2

4

. It is clear therefore, we have

misunderstood Hippocrates because, we believed 3.1415926 as the value of . I therefore apologize to Hippocrates on behalf of mathematics community for the past mistake done by us. And to atone the

academic sin committed by us, I bow my head and dedicate the explained parts (for details: Pi of the

Circle, last chapter: Latest work, Pages from 273 to 281) to Hippocrates, in www.rsjreddy.webnode.com

James Gregory (1660) has said squaring of circle is impossible. His view has been confirmed by

C.L.F. Lindemann (1882) based on Eulers formula ei+1 = 0. Von K. Weiertrass (1815-1897) and David Hilbert (1893) have supported the proof of Lindemann by their proofs.

S. Ramanujan (1913) has squared a circle upto some decimals of 3.1415926 Prof. Underwood Dudley doesnt accept Lindemanns proof because this is based on numbers which are approximate in themselves.

Now, the exact value is discovered. It is 1 14 24

. It is an algebraic number. The following is

the procedure how to square a circle.

II. Procedure We have to obtain a side of the square

whose value is 1

2 ; when

14 2

4

, then

1 1 14 2 14 2

2 2 4 4

CD = a, OK = OF = radius = 2

a, FK =

2

2

a, JK =

FG = GC,

GC = 1

JG KF2

= 2a 1

a2 2

=

Durga Method Of Squaring A Circle


2 2a

4

; GB = BC GC = 2 2

a a4

= 2 2

a4

;

Bisect GB. GH = HB = 2 2

a8

. Bisect HB. 2 2

a16

= HI = BI

CI = BC BI = a 2 2

a16

= 14 2

a16

= 4

; Area of the circle =

214 2 a16

CB = diameter = a;

Draw a semicircle on CB, with radius a

2 and center O; CO = OB =

1

2 where a = 1

Draw a perpendicular line on CB at I, which meets semicircle at Y. Apply altitude theorem to obtain IY length.

IY = 14 2 2 2 26 12 2

CI IB16 16 16

Connect YC which is the side of the square CYUT whose area is equal to that of the inscribed circle in the

square ABCD.

Apply Pythagorean theorem to get CY from the triangle CIY.

Side of the square CY =

22

2 2 14 2 26 12 2 14 2CI IY16 16 4

Area of the square CYUT =

2

14 2 14 2

4 16

= area of the inscribed circle in the square ABCD.

III. Conclusion

14 2

4

is the exact value of circle. Hence, squaring of circle is done now. The misnomer Circle

squarer will sink into oblivion. Hippocrates will now gets his deserving throne of greatness though delayed unfortunately for 2400 years.


e-ISSN: 2278-3008, p-ISSN:2319-7676. Volume 10, Issue 2 Ver. II (Mar-Apr. 2014), PP 09-12



Supporting Evidences To the Exact Value from the Works Of Hippocrates Of Chios, Alfred S. Posamentier And Ingmar

Lehmann

R.D. Sarva Jagannadha Reddy 19-9-73/D3, Sri Jayalakshmi Colony, S.T.V. Nagar, Tirupati 517 501, A.P., India

Abstract: Till very recently we believed 3.1415926 was the final value of .And no body thought exact value would be seen in future. One drawback with 3.1415926is, that it is not derived from any line-segment of the circle. In fact, 3.1415926 is derived from the line-segment of the inscribed/ circumscribed polygon in and about circle, respectively. Surprisingly, when any line-segment of the circle is involved two things

happened: they are 1. Exact value is derived and 2 that exact value differs from 3.1415926 from its 3rd decimal onwards, being 3.1464466 Two geometrical constructions of Hippocrates of Chios, Greece (450 B.C.) and Prof. Alfred S. Posamentier of New York, USA, and Prof. Ingmar Lehmann of Berlin, Germany, are

the supporting evidences of the new value. They are detailed below. Keywords: value, lune, triangle, area of curved regions

I. Introduction In the days of Hippocrates, value 3 of the Holy Bible was followed in mathematical calculations.

He did not evince interest in knowing the correct value of . He wrote a book on Geometry. This was the first book on Geometry. This book became later, a guiding subject for Euclids Elements. He is very famous for his squaring of lunes. Prof. Alfred S. Posamentier and Prof. Ingmar Lehmann wrote a very fine

collaborative book on . They have chosen two regions and have proved both the regions, though appear very different in their shapes, still both of them are same in their areas. These areas are represented by a formula

2 12

r

. The symbol r is radius. , here must be, the universally accepted 3.1415926

Every subject in Science is based on one important point. It would be its soul. In Geometry, the soul is

a line-segment. The study of right relationship between two or more line-segments help us to find out areas,

circumference of a circle, perimeters of a triangle, polygon etc. For example, we have side in the square, base,

altitudein the triangle. The same concept is extended here, to show its inevitable importance in the study of

two regions of Professors of USA and Germany. The lengths of the concerned line-segments have been arrived

at and associated with 2 12

r

. 3.1415926 does not agree with the value of line-segments of two regions.

However, the new value 3.1464466 = 14 2

4

has agreed in to-to with the line-segments of the two regions

of the Professors. This author does believe this argument involving interpretation of 2 12

r

with the line-

segments, is acceptable to these great professors and the mathematics community. It is only a humble

submission to the World of Mathematics. Judgment is yours. If this argument in associating line-segment with

the formula looks specious or superficial, this author may beexcused.

II. Procedure The two methods are as follows:

1. Hippocrates' Method of Squaring Lunes And Computation of The Exact Value

Archimedes's procedure for finding approximate numerical values of (without, of course, referring

to as a number), by establishing narrower and narrower limits between which the value must lie, turned out to be the only practicable way of squaring the circle. But the Greeks also tried to square the circle exactly, that is

they tried to find a method, employing only straight edge and compasses, by which one might construct a square

equivalent to the given circle. All such attempts failed, though Hippocrates of Chios did succeed in squaring

lunes.

Hippocrates begins by noting that the areas of similar segments of circles are proportional to the

squares of the chords which subtend them

Supporting Evidences To The Exact Value From The Works Of Hippocrates Of Chios, Alfred S.


Consider a semi-circle ACB with diameter AB. Let us inscribe in this semi-circle an isosceles triangle

ACB, and then draw the circular arc AMB which touches the lines CA and CB at A and B respectively. The

segments ANC, CPB and AMB are similar. Their areas are therefore proportional to the squares of AC, CB and

AB respectively, and from Pythagoras's theorem the greater segment is equivalent to the sum of the other two.

Therefore the lune ACBMA is equivalent to the triangle ACB. It can therefore be squared.

The Circular arc AMB which touches the lines CA and CB at A and B

respectively can be drawn by taking E as the centre and radius equal to EA or EB.

AB = diameter, d. DE = DC = radius, d/2; F = mid point of AC

N = mid point of arc AC

NF = 2

2 2

d d; DM =

2

2

d d; MC =

2

2

d d

With the guidance of the formulae of earlier methods of the author where a

Circle is inscribed with the Square, the formulae for the areas of ANC, CPB, ACM

and BCM are devised.

1. Area of ANC = Area of CPB =

2d 2 12 1

32 2 2

2. Area of AMB = Areas of ANC + CPB (Hippocrates)

3. Area of ACM = Area of BCM =

2

2d

16

2 18

2

4. Area of ACB triangle = 1 d

d2 2

5. According to Hippocrates the area of the lune ACBMA is equivalent to the area of the triangle ACB

Lune ACBMA = triangle ACB

(ANC + ACM + BCM + CPB)

i.e.

2

2

2d

d 2 1 1 d164 1 2 d32 2 22 2 2 1

82

ANC + CPB ACM+BCM ACB

From the above equation it is clear that the devised formulae for the areas of different segments is

exactly correct.

6. Area of AMB = Areas of ANC + CPB

7. Area of the semicircle = 2d

8

= Areas of ANC + CPB + ACM + BCM + AMB

8. 2

8 Area of thesemicircle

d

=

2

2 2

2

2d

8 d 2 1 d 2 1164 1 2 4 132 32d 2 2 2 22 1

82

=

14 2

4

2. Alfred S. Posamentiers similarity of the two areas and decimal similarity between an area and its line-segments

Prof. A.S. Posamentier has established that areas of A and B regions are



equal. His formula is 2 1

2r

for the above regions. This author is grateful to the professor of New York for

the reason through his idea this author tries to show that his new value equal to 1 14 24

is exactly right.

1. Arc = BCA; O = Centre; OB = OA = OC = Radius = r

2. Semicircles : BFO = AFO; E and D = Centres; OD=DA = BE = OE= radius= 2

r

OF = 2

2

r; FC = OC OF =

2

2

rr =

2 2

2

r r

3. Petal = OKFH; EK = 2

r; ED =

2

2

r; EJ =

2

4

r; JK = EK EJ =

2

2 4

r r =

2 2

4

r r;

JK = JH, HK = JH + JK = 2 2

2

r r

4. So, FC of region A = HK of region B = 2 2

2

r r

5. BFAC = OKFH i.e. areas of A and B regions are equal (A.S. Posamentier and I. Lehmann).

(By Courtesy: From their book )

Formula for A and B is 2

2 1 22 2

rr

Here r = radius = 1

From March 1998, there are two values. The official value is 3.1415926 and the new value is

14 2

4

= 3.1464466 and which value is exact and true ?

Let us substitute both the values in 2

22

r , then

Official value = 2

3.1415926 22

r =

2

1.1415926...2

r

(It is universally accepted that 3.1415926 is approximate at its last decimalplace however astronomical it is in its magnitude.)

New value = 2

3.1464466 22

r =

2

1.1464466...2

r

6. FC = HK (HJ + JK) line segments = 2 2

2

r r

7. Half of HC and HK are same 2 2 1

2 2 2 2

FC HK r r

=

2 2

4

r r = 0.1464466..

8. Area of A/B region equal to 1.1464466 is similar in decimal value of half of FC/HK line segment i.e. 0.1464466

9. Formulae a2, 4a of square and ab of triangle are based on side of the square and altitude, base of triangle, respectively. In this construction, FC and HK are the line segments of A and B regions,

respectively.



As the value 0.1464466 which is half of FC or HK is in agreement with the area value of A/B region

equal to 1.1464466 in decimal part, it is argued that new value equal to 1 14 24

= 3.1464466

is exactly correct.

The decimals 0.1415926 of the official value 3.1415926 does not tally beyond 3rd decimal with the half

the lengths of HK and FC, whose value is 0.1464466, thus, the official value is partially right. Whereas, FC

& HK are incompatible with the areas of A & B calculated using official value. Then, which is real, Sirs?

III. Conclusion 3.1415926 agrees partially (upto two decimals only) with the line-segments of curved geometrical

constructions. When these line-segments agree totally and play a significant role in these constructions a

different value, exact value 14 2

4

= 3.1464466 invariably appears. Hence,

14 2

4

is the true valueof

.

Acknowledgements This author is greatly indebted to Hippocrates of Chios, Prof. Alfred S. Posamentier, and Prof.

Ingmar Lehmann for using their ingenious and intuitive geometrical constructions as a supportive evidence of

the new value of .

Reference [1]. T. Dantzig (1955), The Bequest of the Greeks, George Allen & Unwin Ltd., London. [2]. P. Dedron and J. Itard (1973). Mathematics and Mathematicians, Vol.2, translated from French by J.V. Field, The Open

University Press, England.

[3]. Alfred S. Posamentier&Ingmar Lehmann (2004). A Biography of the Worlds Most Mysterious Number. Prometheus Books, New York, Pages 178 to 181.

[4]. RD Sarva Jagannadha Reddy (2014), Pi of the Circle, a Canto on-line edition, in the free website: www.rsjreddy.webnode.com

times of area of the circle is equal to area of the triangle (Arthanaareeswara method)

Square ABCD

Side AB = 1

Diagonal = AC = 2

Take a paper and construct a square whose side is 1 (=10 cm) and diagonal 2 . Fold the paper

along the diagonal AC. Then bring the two points of A and C of the folded triangle touching each

other in the form of a ring, such that AC becomes the length of the circumference of the circle

whose value is 2 . Now the folded paper finally looks like a paper crown.

Let us find out the area of the circle

Circumference = 2 = d; d = 2

; Area =

2

4

d=

2 2 1 2

4 4

Area of the triangle = 1

2; x Area of circle =

2 1

4 2

Second method

This time let us bring A and D or D and C close together, touching just in such a way they form a

ring (= circle)

Side = AD = 1 (=10 cm); Circumference = 1 = d;

d = 1

;

2 1 1 1 1

4 4 4

d

; 2 x Area of circle =

1 12

4 2

is a simple and an algebraic number.

Half body of Lord Shivaa and the remaining half body of His consort goddess

Parvathi is Arthanaareeswara

D 1 C

A 1 B

1 1 2

R.D. SARVA JAGANNADHA REDDY HOME PAGE

1. Mother : Dhanalakshmi

2. Father : Venkata Reddy

3. Date of Birth : 13.04.1946

4. Native Place : Bandarupalli Village

Yerpedu Mandal,

Chittoor District, AP, India.

5. Phone No. : 0877-2244370

6. E-mail : [email protected]

7. Present Address (Temporary) : 19-9-73/D3,

Sri Jayalakshmi Colony,

S.T.V. Nagar,

Tirupati 517 501, INDIA

8. Education : B.Sc., Zoology (Major),

Botany, Chemistry (minors) 1963-66

M.Sc., Zoology 1966-68

at S.V. University College, Tirupati.

9. Books : 1. Origin of Matter

2. Origin of the Universe

3. Organic Bloom (on Animal Evolution)

4. Pi of the Circle

Telugu Books

5. Sarvam Pavithram

6. Pavana Prapancham

7. Mahabhagavatham Maanavaavirbhavam

8. Abhinandana

9. Mattipella

10. Janthu Pravarthana (Animal behavior) for B.Sc.,

11. Kachhapi

10. Wife : Late Savithri

Children : Shyam Sundar Reddy, Gowri Devi, Sarada

11. Profession : Lecturer in Zoology, Retired on 30.06.2003

12. Donation : As a mark of Gurudakshina to my Alma Mater Sri

Venkateswara University, Tirupati a granite stone-

sphere of 6 feet diameter and another granite stone-

sphere of 3 feet diameter to Govt. Junior College,

Piler, Chittoor district (where, this author got the idea

in 1972, while working as Junior Lecturer in Zoology,

one can get formulae for the computation of area and

circumference of a circle without using constant

22/7 as in r2 and 2 r) have been humbly donated.

Stone at S.V.U. Mathematics Department, Tirupati, A.P., India.

Stone at Govt. Jr. College, Piler, Chittoor District, A.P., India.

Donated 11 Feet high Sivalingam to S.V. Higher Secondary School, Prakasam

Road, Tirupati, and consecrated at TTDs, S.V. Dhyanaramam, Opposite to

Regional Science Centre, Alipiri, Tirupati.

Donated 6 Feet high Sivalingam to Beriveedhi Elementary School, Tirupati and

consecrated at Rayalacheruvu Katta, 15 km away from Tirupati.

4TH PUBLISHED PAPER.pdfB010220912LET 07-03-14

crackpot pi letter

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