The Fascinating Number Pi

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Investigation inMathematics:

The Fascinating

Number

By Paul Asquith 10227169

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Summary

Throughout this report we have seen how various mathematicians or civilisations have discovered

approximations of. We initially started by looking at the earliest approximation of, which wasgiven by the Babylonians and has been dated to approximately 1900 BC. It was shown that the

Babylonians believed that the ratio of the perimeter of a regular inscribed hexagon to the

circumference of a circle is , which suggests the Babylonians believed the value of to be . We then went on to see second earliest approximation of was given by an ancient Egyptianscroll called the Rhind papyrus and has been dated to roughly 1650 BC. In this papyrus the

Egyptians stated that in order to find the area of a circle you must shorten the diameter of the

circle by 1/9 and then square the result, we saw that by doing this the value of suggested by theancient Egyptians was

.We then looked at the first ever theoretical calculation of

, given in 240 BC by the Sicilian born

mathematician Archimedes. By inscribing and circumscribing 96 sided polygons about a circle

Archimedes found that:

Archimedes method of inscribing and circumscribing polygons was so advanced for the time that

it was still used by mathematicians for over a thousand years in order to get better

approximations of.After Archimedes we began to look at various infinite products and series for , the first of whichwas given in 1593 by Francois Vite.

Initially we looked at how the area of a polygon inscribed in a circle changed when the number of

sides of the polygon is doubled, We then applied this to a square inscribed in the unit circle, which

proved Vites infinite product .

We then went on to look at the infinite product given in 1655 by the English mathematician John

Wallis. This infinite product was:

We saw that by evaluating the integral In we were able to prove Wallisinfinite product. This was the first time that was expressed only with rational numbers.The first infinite series we looked at was:

This was given by Gottfried Leibniz in 1674. We proved this series by looking at the area of a

quarter of the unit circle, which we calculated using infinitelysmall triangles.

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Finally we looked at the series for , which was given by Leonard Euler in 1736.

We saw that Eulers initial derivation of the series was not a fully valid proof. Euler didsubsequently proof the series for , but instead of looking at this we looked at one of theelementary proofs given after Eulers time that showed the series to be true.

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Contents

Introduction 1

Babylonians (c. 1900 BC) 2

Ancient Egyptians The Rhind Papyrus (c1650 BC) 3

Archimedes (240BC) 4

Francois Vite (1593) 11

John Wallis (1655) 14

Gottfried Leibniz (1674) 18

Leonhard Euler (1736) 23

Appendices 29

References 32

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1

Introduction

For thousands of years it has been known that the ratio of circumference to diameter of a circle is

constant[1]

. This constant is known as

and can be represented algebraically as

Or Where A is the area, C the circumference, D the diameter and r the radius of a circle.

The symbol for is the Greek letter for p and is believed to have been used because it is the firstletter inperimetron, meaning perimeter

[2]. is an irrational number, meaning it has an infinite

number of decimal places with no repeating pattern[3]

, it is also transcendental, meaning that it is

not a root of a polynomial equation with integer coefficients[4]

.

Approximations of

have dated back as far as c1900 BC where ancient Babylonian tablets have

suggested a value of as . The first theoretical calculation of was given by Archimedes in240BC, where he inscribed and circumscribed 96-sided polygons to show that:

After Archimedes, Many mathematicians including Vite, Leibniz, Newton, Euler and Gauss have

attempted to gain a greater understanding of, whether this is calculating to a greater degreeof accuracy or representing as an infinite product or series. Even in modern history, where thereis no practical use for us to know

to any more digits, it is still being calculated to increasing

degrees of accuracy and with the use of computers the current record for calculating stands at10 trillion digits [5].The aim of this report is to look at various calculations or representations of that have beendiscovered in pre-computer history, with a particular focus on the proofs of such

calculations/representations.

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2

Babylonians (c. 1900 BC)

In ancient Babylon, the Babylonians generally calculated the area of a circle by taking three times

the square of its radius; this suggests an approximation of

[6]

. However in 1936 a set of

tablets were unearthed about 200-300 miles from Babylon[7]

, these tablets have been dated toapproximately 1900-1680 BC

[6]. The tablets have suggest a value of pi as:

The tablets were focused on the ratio of areas and perimeters of regular polygons to their

respective side lengths. One particular piece of clay noted that the Babylonians believed that:

The ratio of the perimeter of a regular inscribed hexagon to the circumference of the circle is:

0;57,36

[2]

The Babylonians used a base 60, or sexagecimal, number system (Our habit of giving 360o

to a

circle, 60 seconds in a minute and 60 minutes in an hour are cultural artefacts passed down to us

from the Babylonians)[8]

.

So in Sexagecimal numeral system

0; 57, 36 So perimeter of inscribed regular hexagon =

(circumference of circle)

[9]

=

A regular hexagon is made of six equilateral triangles. Since a

hexagon is inscribed in the circle, the length of each side of the

triangle is r and the perimeter of the hexagon is therefore 6r.

So 6r =

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3

Ancient Egyptians The Rhind Papyrus (c1650 BC)

In 1858 a Scottish lawyer and antiquarian, Alexander Henry Rhind, on one of his trips to the Nile

valley purchased a document that had been found a few years earlier in the ruins of a small

building in Thebes[10]

. This Document, subsequently named the Rhind papyrus, was 18 feet long

and 13 inches high, it contained 87 different mathematical problems with their solutions,including equations, volumes of cylinders and areas of triangles, rectangles and circles

[11]. The

Rhind papyrus is also known as the Ahmes papyrus in honour of Ahmes the scribe who copied it in

1650 BC[12]

. Ahmes begins by saying that he has copied this work from a very old scroll from a

couple of hundred years earlier, then goes on to say that he will provide a "complete and

thorough study of all things" and will reveal "the knowledge of all secrets" , which refers to the 87

problems contained in the papyrus[13]

.

Our focus is going to be on problem 50 in the papyrus. We will show that the ancient Egyptians

approximation of pi was . Unfortunately the way in which they managed to find thisapproximation was not given in the papyrus

[14].

Problem 50 of the Rhind papyrus reads:

A circular field has diameter 9 khet. What is its area?(A khet being a measurement of length

approximately equivalent to 50 metres)[15]

.

The solution given in the papyrus is:

Take away 1/9 of it, namely 1; the remainder is 8. Multiply 8 times 8; it makes 64. Therefore it

contains 64 setat[15]

. (A setat being one square khet).

What does this mean?

Ahmes has stated that to find the area of this circle we must shorten the diameter by 1/9 and

square the result.

So given a circle of diameter of 9 khet:

( ) = 88

2= 64 setat

And this, according to the papyrus, is the area of the circle of diameter 9 khet.

The General Solution

If we take the solution to this problem as a general formula, then we can acquire a formula for thearea of a circle (A) given the diameter of the circle (d).

A = We know that A So

[15]

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4

Archimedes (240BC)

Archimedes was born in the city of Syracuse on the island of Sicily in 287 BC. He was the son of an

astronomer and mathematician named Phidias[16]

. He was also on intimate terms, if not related

to king Hieron II of Syracuse[17]

. In his younger years Archimedes travelled to Egypt in order to

pursue his mathematics education in Alexandria[18]

. It is here that he studied under the followersof the renowned mathematician Euclid

[19]. But, he soon returned to Syracuse, where he

accomplished most of his work[18]

.

Although he was renowned for his contribution to mathematics, Archimedes also designed many

mechanical inventions[18]

. One particular story recounts how a perplexed King Hieron was unable

to empty rainwater from the hull of one of his ships and turned to Archimedes for assistance.

Archimedes solution was to create a machine, now known as Archimedes screw, which was a

screw like contraption contained in a hollow tube and when turned by a handle at the end it

would raise the water out of the hull of the ship. This is still used as a method of irrigation in

developing countries to this day[16]

.

Because his skill in mechanical objects was unequalled, king Hieron often required Archimedes to

improve the defences of the city[18]

. Archimedes obliged and invented multiple devices for the

citys defence. Archimedes' claw was one such invention; it was shaped like a crane arm, from

which a large metal hook was balanced. When the claw was dropped on an attacking ship, it

would lift the ship by swinging the arm upwards and then sink the ship[20]

. Another such invention

of Archimedes is known as the burning mirrors. He erected large mirrors placed at designated

locations on land and positioned them in such a way that it targeted enemy ships at sea. Once the

sunbeams reflected off the mirrors, it ignited the enemy ships within moments[21]

. Other notable

devices were catapults ingeniously constructed to be equally accurate at short or long range and

machines for discharging showers of missiles through holes made in the walls[17]

.

Despite his numerous inventions Archimedes was said to have been more dedicated to pure

theory rather than the practical applications of mathematics[18]

. An example of this is Archimedes'

use of an exhaustion method, cutting up shapes into infinitely small pieces to discover their

volumes. This method paved the way for what we now call integral calculus[22]

. Archimedes

regarded these discoveries regarding the volumes and surface areas of solids as his most

important and most beautiful achievements. Archimedes favourite was said to have been his

discovery that a sphere has a volume and surface area two-thirds that of the cylinder that

circumscribes it[23]

.

For two years the machines developed for Syracuses defence had kept the Romans, led by

General Marcellus, at bay. Nevertheless, during the siege of Syracuse in 212 BC, Marcellus forcemanaged to take the city

[16]. Apparently Archimedes was not concerned with the siege taking

place as his attentions were focused on making mathematical diagrams in the sand at his home.

Then, even though Marcellus had given strict orders that the mathematician not be harmed, a

roman soldier broke into Archimedes house and spoiled his diagram, and then when Archimedes

voiced his displeasure, the soldier promptly slew him. At this time Archimedes was 75 years old[18]

. Marcellus was greatly distressed upon hearing the news of Archimedes' death, and ordered

that he be buried with honours. Archimedes' tombstone was, as he had wished, engraved with an

image of a sphere within a cylinder[16]

.

Our focus will be on Archimedes approximation of pi which he gives in On the Measurement of

the Circle. He used a method of fitting polygons inscribed and circumscribed about a given circle

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5

and knew that the circumference of the circle will lie between the two perimeters of the polygons[24]

.

He initially uses a hexagon circumscribing the circle then uses it to calculate the perimeters of a

12, 24, 48 and finally 96 sided polygon; he then repeats using inscribed polygons[24]

. By this

method Archimedes managed to find that:

We will now look at the Method Archimedes used to get this approximation of pi. Please note

that throughout this proof Archimedes used approximations to various square roots and left no

indication to how he got these approximations[24]

. When one of these approximations appear it

will be indicated with ***. Through the proof Archimedes also used Proposition 3 of book VI of

Euclids Elements, this will be denoted as Euclid VI. 3 and can be found inAppendix A1.

Part 1 Fitting Circumscribed Polygon

Let AB be the diameter of any circle, O its centre, AC the tangent at A and let the angle AOC beone third of a right angle.

Then OA : AC = : 1 265 : 153 (*** ) (see Appendix A2) OC : AC = 2 : 1 = 306 : 153 (see Appendix A2)Draw OD bisecting the angle AOC and meeting AC at point D

Now OC : OA = CD : AD (Euclid VI. 3)

*This Means

(CD+AD = AC)

(multiply by OA and divide by AC)Therefore OC + OA : AC = OA : AD

**So by using & we can calculate OA : AD

Hence OA : AD 571 : 153We Know OD

2= OA

2+ AD

2(Pythagoras theorem)

[25]

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6

So that OD : AD : 153 (*** )Secondly, let OE bisect the angle AOD Meeting AD at point E

Now OD : OA = DE : AE (see Euclid VI. 3)

OD + OA : AD = OA : AE (By using similar steps to *)

Now by using we can calculate OA : AE OA : AE (by similar steps to **)We know

OE2

= OA2

+ AE2

(Pythagoras theorem)

So that

OE : AE

(***

Thirdly, let OF Bisect angle AOE and meet the line AE At point F

Now OE : OA = EF : AF

OE + OA : AE = OA : AF (By using similar steps to *)

Now by using we can calculate OA : AF OA : AF (by similar steps to **)We Know

OF2 = OA2 + AF2 (Pythagoras theorem) So that OF : AF (*** Fourthly, let OG bisect the angle AOF, meeting AF at point G

Now OF : OA = FG : GA (see Euclid VI. 3)

OF + OA : AF = OA : AG (By using similar steps to *)

Now by using we can calculate OA : AGOA : AG (by similar steps to **)

What Have we Done Up Until This Point?

The angle AOC (one third of a right angle) has

been bisected four times, the fourth time by line

OG, from this information we can deduce that:

AOG (a right angle) (a right angle)Now create angle AOH on the other side of OA,let this angle be equal to the angle AOG.Let AG

produced meet OH at H (see diagram right) then:

[24]

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7

GOH (a right angle) (a right angle)So GH is one side of a regular polygon of 96 sides (1/24 of a right angle is 1/96 of a complete circle). This

96-gon circumscribes the given circle because we have constructed the polygon so that centre of every side

touches the circle (in the case of side GH, the centre, point A, touches the circle).

Now Recall OA : AG , AB = 2OA & GH = 2AGThis means

So

We know that

because the perimeter of the 96-gon islarger than the circumference of the given circle (since the 96-gon circumscribes the circle).

Hence Part 2 fitting inscribed polygon

Let AB be the diameter of a circle and let AC, meeting the circle at C make the angle CAB which is equal to

one third of a right angle. Now Join Points B and C.

Then AC : BC = : 1 1351 : 780 (*** ) (see Appendix A2)Let AD bisect CAB, meet BC at point d and meet the circle at point D. Join points B and DThe angles at points C and D are both right angles because the angle inscribed on a semi-circle is always a

right angle (see Appendix A3).

So (because the line AD bisects BAC leaving theses two equal angles) (angle at C is a right angle) (opposite angles of two intersecting lines are always equal) (angle at D is a right angle)It has just been shown that the triangles ADB, ACd and BDd are similar, because we have just shown that all

of their angles must be equal.

Therefore AD : BD = BD : Dd = AC : Cd

By using Euclid VI. 3 on triangle ABC

Cd : Bd = AC : AB AB : Bd = AC : Cd (multiply each ratio by AB and divide by Cd)So AD : BD = AC : Cd = AB : Bd

= (AB + AC) : (Bd + Cd)

= (AB + AC) : BC

We Know

AB : BC = 2:1 = 1560 : 780AC : BC : 1 1351 : 780 (see Appendix A2)

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8

So AD : BD 1560 + 1351 : 780 2911 : 780 (by similar steps to **)Then AB

2= AD

2+ BD

2(by Pythagoras Theorem)

So that AB : BD (*** Secondly let AE bisect angle BAD, meet BD at point e and meet the circle at point E. Join points

B and E.

We then use the same approach as in the last part i.e. by showing that triangles AEB, ADe & BEe are similar,

then using similar steps to the last part.

By doing this we see that by using we can calculate AE : BEAE : BE = AB + AD : BD

(by similar steps to **)

So AE : BE 1823 : 240We know AB

2= AE

2+ BE

2 (by Pythagoras Theorem)

Therefore AB : BE (*** )Thirdly, Let AF bisect BAE meeting the circle at Point F

We will prove this in the same way as the previous two parts.

So by Using we can calculate AF : BFAF : BF = AB + AE : BE (by similar steps to **)So AF : BF 1007 : 66We know AB

2= AF

2+ BF

2(by Pythagoras Theorem)

Therefore AB : BF (*** )Fourthly, Let angle BAF be bisected by line AG meeting the circle at G

Again this will be proved in the same way as the previous three parts parts

So by using we can calculate AG : BGAG : BG = AB + AF : BF (by similar steps to **)

So AG : BG

We know AB2 = AG2 + BG2 (by Pythagoras Theorem)

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9

Therefore AB : BG (*** )Meaning that BG : AB What have we done Up Until This

Point?

The angle BAG is the fourth bisection of

angle BAC, which is one third of a right

angle.

So BAG (a right angle)

(a right angle)

Therefore BOG (a right angle)(See Appendix 4)

Therefore BG is one side of a regular 96-gon that is inscribed in the circle (1/24 of a right angle is 1/96 of a

complete circle), it is inscribed in the circle because we have constructed it in such a way that each vertex

of the polygon touches the circle.

So we have

(using result from

)

We know (because polygon is

inscribed in the circle)

Hence So we have seen that the ratio of the circumference to the diameter of a circle, or , is:

[24, 26 & 27]Archimedes approximation of pi is the first ever theoretical calculation of pi and it is a testament

to his method that a more accurate approximation for pi was not obtained for about 400 years.

There was no reason that Archimedes could not have continued his iterations and fitted polygons

with a larger number of sides to get a more accurate approximation of pi. In fact numerous

people did exactly that, including:

Ptolemy (150 AD) 4 places 360-gon

[24]

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10

Francois Viete (1593) 9 places - 393,216-gon Romanus (1593) 17 places - -gon Van Ceulen (1600) 35 places - -gon [28]

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11

Francois Vite (1593)

Franois Vite (or Vieta, as he is often known by his Latinized name) was born in 1540 in

Fontenay-le-Comte, France[29]

. He attended the University of Poitiers and graduated with a law

degree in 1560, only to abandon the profession four years later[30]

in order to pursue a career in

science and mathematics and in 1571 he published his first mathematical work calledmathematical laws applied to triangles

[31].

Vite lived a very political life. In 1573, Charles IX appointed him to the government of Brittany. In

1580, Henry III appointed him royal privy counsellor[32]

.But, in 1584 Vite was banished from the

court for his protestant faith, he then continued to spend the next five years devoting himself to

mathematical pursuits. Then in 1588, when Henry III was forced to flee to Paris, Vite was made a

member of the kings parliament at Tours. In 1589, upon the murder of Henry III, Vite entered

the service of Henry IV[31]

. During this service he was tasked with decoding a difficult Spanish

cipher, which had a 500-character code. When Vite deciphered the code the King of Spain,

Convinced that it was unbreakable, accused the French of using black magic[33]

, but in this case it

was mathematics rather than sorcery[31]

.

Aside from his political work, Vite continued his personal pursuit of mathematics[33]

and in 1591

he published Introduction to the Analytic Art[31]

, which is considered to be his most important

work. Vite's major contribution in this book was his innovative treatment of algebraic equations[34]

. He introduced the use of letters as variables to denote both the known and unknown

quantities[33]

. This was the beginning of a new type of algebra, expressed in terms of abstract

formulas and general rules, instead of the geometric visualizations[34]

, Allowing the painless

rearrangement of equations. Vites pioneering symbol system completely altered the field of

mathematics, rightfully earning him the title as the Father of Modern Algebra[33]

.

Vite also wrote books on trigonometry and geometry, which contained geometrical solutions todoubling a cube and trisecting an angle. As well as this in 1593 he calculated pi to 10 places using

a polygon of 393216 sides. He also represented pi as an infinite product using only square roots

and twos[32]

. This infinite product will be what we focus on now.

Proof

Let 2

be the angle at the centre of a regular polygon that has N sides and is inscribed in a circle

of radius r, therefore

is the angle at the centre of the inscribed regular polygon with 2N sides.

We can see that:

OF = OA = r AB = 2AF = 2(OA ) = 2r

Let AN denote the area of the N-sided polygon

Then AN = = = N

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12

If the number of sides of a polygon is doubled, then the angle at the centre of the polygon is

halved, hence:

A2N = 2Nr2 = Nr2 (by the double angle formula)

We can see that:

AN = A2N Similarly

A4N = 4Nr2 = 2Nr2 (by double angle formula)

So A2N = A4N By using similar steps to above we see *

AN = A2N A2N = A4N A4N = A8N

So let us use this information. We will begin by looking at a Square inscribed in the circle of radius

one.

Calculate Area of inscribed Square (A4)

A4 = (HI)2

= = 2So for the inscribed square,

use * with: N = 4, A4 = 2 & = 2 = A4 = A8

By reiterating we get

2 = A16 = A32 =

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13

As the number of sides of the inscribed polygon tends to infinity the area of the polygon will

tend towards the area of the circumscribed circle.

i.e. when k tends to infinity, tends towards A (where A is the area of the circle)So 2 = A The circumscribed circle is of radius one, so A = = Therefore 2 =

= We know that

Let U0 = U1 = = = U2 = = =

Un = So

= U0 U1 U2 U3 where = so U0 = = (See Appendix B)Using this we get

Or alternatively

= U0 U1 U2 U3 Un = = = & U0 = = And using this we get

[35 & 36]

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14

John Wallis (1655)

John Wallis was born on November 23rd

1616, in Ashford, England. Wallis was said to have been

the greatest English mathematician of his time[37]

and also said to have been the most influential

mathematician prior to Sir Isaac Newton [38].

Wallis attended the famous school of Martin Holbeach at Felsted, Essex, where he learnt Latin,

Greek and Hebrew[39]

, but not mathematics because it was not considered an important subject

in the best schools of the time[40]

. Wallis was first introduced to mathematics when he was fifteen

years old during his Christmas holidays where he saw a book of arithmetic in his brothers hand

and was curious at the odd signs and symbols, so he borrowed the book and in a fortnight, with

his brothers help, had mastered the subject[41]

. In 1632 Wallis went to Emmanuel College in

Cambridge, where he studied ethics, metaphysics, geography, astronomy and medicine[37]

, but

nobody in Cambridge at this time could direct his mathematical studies[40]

. Wallis completed his

bachelors degree in 1637 and went on to obtain his masters degree in 1640. Shortly after

obtaining his masters degree, Wallis was ordained and served as a chaplain over the next fewyears. Wallis still exhibited great mathematical skill and his career took a turn when, during the

civil war, he managed to decipher an encoded royalist message in only two hours, making him

popular with the parliamentarians[37]

.

In 1645, Wallis began to meet regularly with a group of scientists interested in medicine,

geometry, astronomy and mechanics. This group later became the Royal Society[37]

. It was

through these meetings that he encountered William Oughtreds Clavis Mathematicae (the Key to

Mathematics), after reading it his love of mathematics, which had never found the opportunity to

flourish, now came pouring out[40]

and he began to start his own investigations[37]

. Wallis then

wrote a book titled Treatise of Angular Sections and discovered methods of solving equations of

degree four [40]. Wallis mathematical talent was rewarded when he was made Savilian professorof geometry at Oxford in 1649

[39], a position he held for 50 years until his death

[40]. He was also

made keeper of the Oxford University archives in 1657[37]

.Wallis then took up the systematic

study of all the major mathematical literature available to him in the libraries at Oxford[39]

,

allowing him to further his mathematical work.

One of Wallis main works was his treatise, Tract on Conic Sections, it was here that he introduced

the sign for infinity and used to represent, for example, the height of an infinitely smalltriangle. In 1655 Wallis published the book titledArithmetica infinitorum (The Arithmetic of

Infinitesimals), it was this book along with his work on conic sections that Wallis fame as a

mathematician is based[39]

. It was in this book that, by a sequence of interpolations, he produced

the following infinite product for . This was the first time that pi had appeared as the limit of a sequence of rational numbers

[42]and

is quite a beautiful result.

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15

Let us look at a proof of Wallis formula.

For n=0, 1, 2, 3. We define In as

In So I0

I1 Now suppose that n Note that: In

Using integration by parts

u u (see appendix C)v v In

Now we know Therefore So now we have

In

(

)

In

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16

Hence for any integer n we have

*

Now, let be even i.e. p= 1, 2, 3..So by * we have

Recall

So And hence

Now let be an odd number i.e. p=1, 2, 3..So by * we have

Recall So [43 & 44]Using & we can see

**Now lets look closer at for We know that for

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17

Therefore , *** So In is a decreasing function of n.

Using *** we can more specifically see that Using *** we can also see

Using * we can see

Putting this information together we can see

By looking at the limit as Hence So by the squeeze rule we have got

So by **

We can write this as [42, 43 & 44]

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18

Gottfried Leibniz (1674)

Gottfried Wilhelm Leibniz was born in 1646 in Germany[45]

. When he was seven Leibniz entered

the Nicolai school in his hometown of Leipzig. Even though he was taught Latin at school, he

managed to teach himself a far more advanced Latin than the school had taught and also some

Greek all by the age of 12[46]

.

In 1661, at the age of fourteen, Leibniz entered the University of Leipzig, where he studied

philosophy, mathematics and law[47]

, He graduated with a bachelors degree in 1663[45]

. The

following summer Leibniz studied at the University of Jena, it was here that he met Erhard Weigel,

who was professor of mathematics at the university. Through Weigel, Leibniz began to

understand the importance of the method of mathematical proof. After the summer, Leibniz was

went back to the University of Leipzig, where he was awarded his masters degree in philosophy

for a dissertation that combined aspects of philosophy, law and mathematical ideas that he had

learnt in the summer from Weigel[46]

. Leibniz spent the next three years devoted to his legal

studies[48]

and in 1667, after failing to get his doctorate in law at Leipzig, He moved to the

University at Altdorf, where this time he succeeded in getting his law doctorate[45]

.

Leibnizs first big steps in his law career came after he met Baron Johann Christian von Boineburg.

By 1667 Leibniz had moved to Frankfurt and was employed by Boineburg and took up residence

at the courts of Mainz[46]

. It was here he was charged with improving the Roman civil law code,

which he saw as part of his lifelong aim to collate all human knowledge[45]

. In 1672 Leibniz went

to Paris, on behalf of Boineburg, on political business. Whilst in Paris Leibniz made contact with

mathematicians and philosophers. He ended up studying mathematics and physics under

Christian Huygens and it was under Huygens that Leibniz began to get acquainted with summing

series[46]

. In 1673 Leibniz went on another political visit, this time to London. Here he promoted

his ideas about calculus to the secretary of the Royal Society and was, shortly after, elected a

member of the Royal Society

[47]

. In 1674 Leibniz began to study the geometry of infinitesimals

[46]

.But it wasnt until 1675, when Leibniz was back in Paris, that he made his most important

contribution to mathematics. Leibniz managed to derive most of the calculus in the form that we

still use to this day[49]

. This included the first use of the notation. In 1676 Leibniz alsodiscovered the incredibly useful result that [46]. Because of this, Leibniz is known,along with Newton, as the co-discoverer of calculus.

We will be focusing on Leibnizs infinite series for , which he found using the idea ofinfinitesimal triangles to find:

Let us have a look at Leibnizs proof of this infinite series.

Let us begin by setting up our proof

-Let OAT be a quarter of a unit circle, where O is

the origin, A= (1, 0) and T= (1, 1)

-Join O & T by a straight line

-Let P & Q be points on the arc OT

-Draw straight lines OP & OQ

-Let ds be the arc length PQ

So, as Q tends towards P, ds becomes very small

and will tend towards the straight line segment PQ.

This means OPQ can be considered a triangle. It

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also means that P can be considered a tangent to the quarter circle.

-Extend the straight line PQ through the

horizontal axis, meeting the y-axis at

point S.-From point R on this newly made line,

draw a perpendicular line to O.

-Create the right angled triangle PQU,

where ds is the hypotenuse of the

triangle.

-let dx be the horizontal distance

between P & Q, i.e. PU.

We have now set up the Leibnizs proof.

Let us look at triangles UPQ & OSR

For this part it may help to imagine that we join points S &

P, where ds remains on line RQ (see diagram right)

RQ is a straight line so

So

So we have & Since two angles in each triangle are the same, the third

angles must be the same.

Therefore triangles UPQ & OSR are similar.

Since the triangles are similar we can see that

By looking at - OR - we can see that it is the height of triangle OPQ

- ds - we can see it is the base of triangle OPQ

(see diagram right).

So the area of OPQ Or

(where

is the area of the small triangle OPQ)

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So to find the area of the segment that is bounded by the straight line OT and the arc OT we will

take the sum of all the small triangles, OPQ, as the horizontal value of P varies from 0 to 1.

This means

Let So Now let us use integration by parts

Note:

Let

So

* + Hence Now we want to get x in terms of y so that we can carry out this integration

-let the horizontal value of P equal x

-make the straight line PA-make the straight line SA

-let Consider the Triangles SAO and SAP

Now (Recall that P can be considered the tangent of the

quarter circle as Q tends towards P)

Now (because both are the radius of the quarter unit circle)We know that both triangles share the side ASSo SAO & SAP are both right angled triangles that have two sides of equal length. This means SAO

and SAP are congruent to each other.

This means We know Therefore

Now that we know this angle we can see that

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&

, so

So we have We will now use the trigonometric identity With

So Now let us look at

Now we know that So we have just shown that

We have now got x in terms of y, so we can finally return back to our integral

Now let us rewrite this using with So

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* +

So we have found an expression for the area bound by the

line OT and arc OT

Now if we find the area of the remaining triangle and add

this to C it will give us the area of a quarter of a unit circle.

Find area of triangle OAT

AOAT Area of a quarter of a unit circle

So putting this information together we get

[50, 51, 52 & 53]

It is worth noting that this result can also be obtained by using the series for arctan that was

discovered by Gregory in 1671.

For

| |

By using this with x=1 we obtain the series for. [54]

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Leonhard Euler (1736)

Leonhard Euler was born in Basel, Switzerland, in 1707. Eulers father educated him in

mathematics when he was young; having himself being interested in the subject and even

attended lectures given by Jacob Bernoulli[55]

. At age 7 Euler began school[56]

, but this school was

a rather poor one and Euler was taught no mathematics from the school. However Eulers interestin mathematics had already been sparked by his fathers early tutelage, so he read mathematics

books on his own and took private lessons[57]

.

In 1720, at the young age of 14, Euler entered the University of Basel in order to obtain a general

education before attempting more advanced studies[57]

. It was at the university that he came

under the guidance of Johann Bernoulli[55]

, who seeing Eulers potential, gave Euler private tuition[57]

. In 1722, he received his Bachelor of Arts degree. A year later he completed his masters

degree in philosophy. Euler then began to study to become a minister, as this was his fathers

wish, but this was short lived as Eulers strength clearly lied in mathematics[55]

, so Euler with

support from his tutor, Johann Bernoulli, who was his fathers friend, persuaded his father to

allow Euler to pursue mathematics instead

[58]

. In 1726 Euler completed his mathematics studiesat the University of Basel

[57].

After his studies, because of a lack of opportunities available in Switzerland, Euler accepted an

offer to work at the St Petersburg Academy of Science, in Russia[55]

. Here he was appointed to

the mathematical-physical division of the Academy[57]

. Then In 1731, he managed to advance and

became professor of physics and in 1733 became professor of mathematics. In 1740 Euler

accepted an offer to work in Berlin at the Society of the Sciences[55]

. Whilst here he undertook

duties such as: supervising the observatory, publishing various calendars and geographical maps[57]

and also undertook practical problems such as correcting the level of the Finow Canal. Euler

stayed a total of 25 years in Berlin, during which time he produced more than 380 works in areas

such as calculus of variations, ballistics analysis, calculation of planetary orbits, motion of themoon and differential calculus[55]

.

In 1766 Euler returned to St Petersburg. Soon after his return to Russia he was left completely

blind after an illness. But, incredibly, even after his blindness, at the age of 59, Euler with the help

of his assistants was still able to produce roughly half of his total works[57]

. In 1783 Euler died of a

brain haemorrhage[55]

. But even after his death, so vast were Eulers works that the St Petersburg

Academy of Sciences continued to publish Eulers unpublished work for 50 more years[57]

.

As previously mentioned, the mathematical works of Euler were vast, publishing over 800 papers

in his life, making him the one of the most prolific mathematical writers of all time. Throughout

these works Euler made many important mathematical contributions. He introduced the use of

many mathematical symbols includingf(x) for a functionfofx, ifor the square root of -1[59]

, forsummations and the modern notation for the trigonometric functions [60]. Euler also introducedthe symbol for the base of natural logarithms (e); he then showed that e is irrational

[59],

discovered a power series expansion for e and defined the exponential function for complex

numbers[60]

as: When we get a special case that is known as Eulers identity

Which, to many, is considered the most beautiful equation in all mathematics because it features

the five most important numbers: e, i, , 1 & 0 [60].

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24

Euler also showed that there are an infinite number of primes, proved the binomial theorem is

valid for any rational exponent[59]

and also stated the theorem that an algebraic polynomial of

degree n has n roots, known as the fundamental theorem of algebra[55]

. Euler also popularised

the use of the standard notation for pi (

)

[61].

As well as popularising the symbol for pi, Euler also gave an infinite series involving pi, this was:

Let us now look at how Euler derived this series for

.Look at the Taylor series expansion for

Dividing each side by x we get

For So

when and This means the roots of are the non-zero roots of , i.e. Now we know the roots of

we may now factorise it

By expanding each group of two brackets we get

Then by multiplying out the brackets we get

Note: the one comes from multiplying all the ones in each bracket. The terms come frommultiplying the coefficient in each bracket by all the ones in the remaining brackets.So we know

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Equating the coefficients we get

[62, 63 & 64]This is how Euler initially derived the series for

. But this is not a proof because Euler has usedthe property of a polynomial; that if you know the roots of a polynomial you can factorise it. He

has then assumed that this property holds true for infinite series. Even without full justification,

Euler was able to verify it numerically and was confident to announce it to the mathematical

community[65]

.

So we have looked at Eulers original derivation of the series, now let us have a look at a fullproof of it.Full proof of the series for

Let where and let be an odd integerBy De Moivres Theorem

()

By the Binomial Theorem

() where () () () ( ) ()

() () () () () () So we now know

() () () () () ()

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26

Equate the imaginary parts

() () ()

() () () *Now let n = 2m+1, where m is a positive integer (n is still a positive odd integer)Let where r = 1, 2m (note )This means So

(because )Hence So by using * with we get

() () ()

()

()

()

Since we can divide both sides by () () () Subbing in n = 2m+1

( ) ( ) ( ) Let

( ) ( ) ( ) () Now because is one-one for we can see that (where r=1, 2, 3, , m)are the distinct roots of the m

thdegree polynomial equation:

( ) ( ) ( ) ()

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Looking at Vites Formula

For any polynomial of degree n

For

Where the polynomial has n roots . Then the sums of the roots ofthe polynomial are [66]

Applying Vites formula to our mth

degree polynomial we get

( ) ( ) ( ) () We know has m distinct roots

i.e.

So by Vites formula

( )( ) ( )( )

Now looking at

We can see that Using this equality we can see that

By looking at the inequalities & (for

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28

Now by using this inequality for each value of r = 1, 2m then adding themtogether we get:

Multiplying through by

Now as m tends to infinity

So by the squeeze rule

[65 & 67]

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Appendix A

A1: Proposition 3 of Book VI of Euclid's Elements (Euclid VI. 3)

If an angle of a triangle be bisected and the straight line cutting the

angle cut the base also, the segments of the base will have the

same ratio as the remaining sides of the triangle; and, if thesegments of the base have the same ratio as the remaining sides of

the triangle, the straight line joined from the vertex to the point of

section will bisect the angle of the triangle

I.e. If AD bisects BAC, then BD : CD = BA : AC[24]A2: 30-60-90 triangle proof

Consider the Equilateral triangle of side length 2X, and let AD bisect the angle at A.

Therefore

DAC =

DAB = 30

BD = CD because the triangles ACD & ABD share the side AD and thesides AB and AC are of equal length (because they are the sides of an

equilateral triangle). We also know that DAC =DAB = 30, all thisinformation tells us that the triangles are congruent, hence BD = CD = X

So we have BA = 2X & BD = X, then by Pythagoras theorem:

Therefore BD : AD : AB = 1 : : 2

A3: Thales TheoremThales' theorem states that if A, B and C are points on a circle where the

line AB is a diameter of the circle, then the angle ACB is a right angle[28]

Proof

Let M be the centre of the circle, where A , B and C are points on the

circle.

Then AM = BM = CM, therefore the triangles AMC and BMC are isosceles.

If

BMC = then

MCB = 90

and

CMA=180 .

Therefore ACM= and ACB = MCB + ACM = 90 [68 & 69]A4: Inscribed Angles Conjecture IAn inscribed angle is an angle formed by two chords in a circle which have a

common endpoint. The other two endpoints define what we call an intercepted

arc on the circle. A central angle is any angle whose vertex is located at the

centre of a circle.

In a circle, the measure of an inscribed angle is half the measure of the central

angle with the same intercepted arc

I.e. [70]

A

BC

2X

2X2X

XX D

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30

Appendix B

cos

12

= AC2

+ BC2

12

= 2AC2

(AC = BC because it is an isosceles

triangle)

1 = AC = AC =BCSo cos

=

=

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31

Appendix C

Differentiation ofby the chain ruleLet

and

So

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