Newton

KHOA A. VO – HOAI T. NGUYEN

NEWTON’S BINOMIAL FORMULA

Khoa Anh Vo - Hoai Thanh Nguyen

February 3, 2012

Khoa Anh Vo - Hoai Thanh Nguyen

Vietnam National University

Ho Chi Minh City (HCMC) University of Science

Faculty of Mathematics and Computer Science

227 Nguyen Van Cu Street, District 5, Ho Chi Minh City

Vietnam

2

PREFACE

This book is intended as our first English thematic for students who study in high schoolor people who want to research into the history of mathematics. In detail, this talks aboutthe journey of John Wallis (1616 - 1703) from the Alhazen’s formulas (965 - 1040), andthe continuation of Issac Newton’s idea (1643 - 1727). Then we give some mathematicalproblems in the educational programs. Therefore, we desire to provide more knownledgesfor the positive vision that pure mathemtics bring it.

This book is also a gift which we award to our forum MathScope.Org on New Year 2012- the Year of Dragon. So we and collaborators send all nice greetings to the readers.

Acknowledgement. We (i.e. Khoa Anh Vo - Hoai Thanh Nguyen) thank the collaboratorsfor all their helps. These include :

Name High School/ University

Thien Huu Vo Truong HCMC University of ScienceTruong Nhat Thanh Mai HCMC University of Science

Quang Dang Nguyen HCMC University of ScienceMinh Nhat Vu To HCMC International University

Phong Tran HCMC University of PedagogyTuan Thanh Nguyen HCMC University of Economics and LawTrang Hien Nguyen Phan Boi Chau High School for The Gifted

Huyen Thanh Thi Nguyen Luong The Vinh High School for The Gifted

Especially, that is the approval of Dr. David Dennis (4249 Cedar Drive, San Bernardino,USA) for our translation of his documents. Furthermore, this makes “Newton’s BinomialFormula” strange - looking.

The readers can find and download “Newton’s Binomial Formula” at :

http://www.forum.mathscope.org/

or

http://anhkhoavo1210.wordpress.com/

3

Contents

PREFACE 3

WHAT IS THE NEWTON’S BINOMIAL FORMULA? 5

1 THE INTRODUCTION 61.1 A FORMULA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 THREE PROOFS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.1 INDUCTION PROOF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.2 COMBINATORIAL PROOF . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.3 DERIVATION USING CALCULUS . . . . . . . . . . . . . . . . . . . . 10

2 THE SENSITIVITY 122.1 JOHN WALLIS (1616 - 1703) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.2 ISAAC NEWTON (1643 - 1727) . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3 A JOURNEY OF JOHN WALLIS . . . . . . . . . . . . . . . . . . . . . . . . . . 142.4 THE CONTINUATION OF ISSAC NEWTON’S IDEA . . . . . . . . . . . . . . 25

4

WHAT IS THE NEWTON’S BINOMIALFORMULA?

5

1 THE INTRODUCTION

1.1 A FORMULA

As a result, Newton’s Binomial Formula was proved by two scientists : Isaac Newton (1643- 1727) and James Gregory (1638-1675). This is really a formula which uses for expansionof a binomial n power(s) that is become a polynomial n+ 1 terms.

(x+ a)n =n∑

k=0

Ckna

n−kxk

In order to make sense of the theorem we need to agree on some conventions. First, wedefine the binomial coefficients

Ckn =

(n

k

)=

n!

(n− k)!k!

using the convention that 0! = 1 to cover the cases where either n, n− k or k is 0.We will also stipulate that x0 = 1 and a0 = 1. These are questionable if x = 0 or a = 0, so

those should be dealt with as separate cases. Interpretation of the formula in those casesgives either an = an or xn = xn. If all of n = 0, x = 0, and a = 0 then we get the result00 = 00, which is not particularly meaningful, but as long as we agree on what we meanby 00 we are forced to accept the result.

In the generality case, a formula said that : Let r be a real number and z be a complexnumber with magnitude modulus of z less than 1, we have

(1 + z)r =

∞∑k=0

(r

k

)zk

Remark. A general formula for m (ai)’s term(s)(m∑i=1

ai

)n

=∑ n!

n1!n2!...nm!an11 a

n22 ...a

nmm

where n1 + n2 + ...+ nm = n.

6

1 THE INTRODUCTION

1.2 THREE PROOFS

The binomial formula can be thought of as a solution for the problem of finding an ex-pression for (x + a)n from one for (x + a)n−1 or as a way to find the coefficients of (x + a)n

directly. In this section, we have three mathematical proofs which are taken from a smalltopic Aesthetic Analysis of Proofs of the Binomial Theorem of Lawrence Neff Stout,Department of Mathematics and Computer Science, Illinois Wesleyan University.

1.2.1 INDUCTION PROOF

Many textbooks in algebra give the binomial formula as an exercise in the use of mathe-matical induction. The key calculation is in the following lemma, which forms the basisfor Pascal’s triangle.

According to Pascal’s triangle, we can order the binomial coefficients corresponding to npower(s).

n = 0 1

n = 1 1 1

n = 2 1 2 1

n = 3 1 3 3 1

n = 4 1 4 6 4 1

n = 5 1 5 10 10 5 1

It’s easy to observe that the pattern (4 + 6 = 10) is exactly a case of Pascal’s lemma.

Ckm + Ck−1

m = Ckm+1

or (m

k

)+

(m

k − 1

)=

(m+ 1

k

)

Of course, this lemma can be prove clearly. And the readers can prove it themself.

Lemma. For all 1 ≤ k ≤ m. Prove that(m

k

)+

(m

k − 1

)=

(m+ 1

k

)

Proof. This is a direct calculation in which we add fractions and simplify.

7

1 THE INTRODUCTION

(m

k

)+

(m

k − 1

)=

m!

(m− k)!k!+

m!

(m− k + 1)!(k − 1)!

=m!(m− k + 1)!(k − 1)! +m!(m− k)!k!

(m− k)!k!(m− k + 1)!(k − 1)!

=m!(k − 1)!(m− k)! [k + (m− k + 1)]

(m− k)!k!(m− k + 1)!(k − 1)!

=m! [k + (m− k + 1)]

k!(m− k + 1)!

=m!(m+ 1)

k!(m− k + 1)!

=(m+ 1)!

k!(m− k + 1)!

=

(m+ 1

k

)

We proceed by mathematical induction.

Proof. For the case n = 0, the formula says

(x+ a)0 =

(0

0

)x0a0 = 1

Now (x+ a)0 = 1 and

0∑k=0

(0

k

)a0−kxk =

(0

0

)a0x0 = 1

Here we are using the conventions that(0

0

)= 1

and that any number to the 0 power is 1. Given the artificiality of these assumptions,we may be happier if the base case for n = 1 is also given.

For the case n = 1 the formula says

(x+ a)1 =1∑

k=0

(1

k

)a1−kxk =

(1

0

)a1x0 +

(1

1

)a0x1

This is equivalent to

x+ a =1!

1!0!a+

1!

0!1!x = a+ x

which is true. Thus we have the base cases for our induction. For the induction step weassume that

8

1 THE INTRODUCTION

(x+ a)m =

m∑k=0

(m

k

)am−kxk

and show that the formula is true when n = m+ 1.

(x+ a)m+1 = (x+ a)m(x+ a)

=

[m∑k=0

(m

k

)am−kxk

](x+ a)

=

m∑k=0

(m

k

)am−kxk+1 +

m∑k=0

(m

k

)am−k+1xk

=

(m

0

)am+1x0 +

m∑k=1

[(m

k

)+

(m

k − 1

)]am−k+1xk +

(m+ 1

m+ 1

)a0xm+1

Completing the proof by induction.

1.2.2 COMBINATORIAL PROOF

The combinatorial proof of the binomial formula originates in Jacob Bernoulli’s Ars Con-jectandi published posthumously in 1713. It appears in many discrete mathematics texts.

Proof. We start by giving meaning to the binomial coefficient(n

k

)=

n!

(n− k)!k!

as counting the number of unordered k−subsets of an n element set. This is done byfirst counting the ordered k−element strings with no repetitions : for the first element wehave n choices; for the second, n − 1; until we get to the kth which has n − k + 1 choices.Since these choices are made in succession, we multiply to get

n(n− 1)...(n− k + 1) =n!

(n− k)!

such ordered k−tuples without repetition. Next we observe that the process of multi-plying out (x+ a)n involves adding up 2n terms each obtained by making a choice for eachfactor too use either the x or the a. The choices which result in k x’s and n− k a’s each give

a term of the form an−kxk. There are

(n

k

)distinct ways to choose the k element subset

of factors from which to take the x. Thus the coefficient of an−kxk is

(n

k

). This tells us

that

(x+ a)n =

n∑k=0

Cknx

n−kak

9

1 THE INTRODUCTION

1.2.3 DERIVATION USING CALCULUS

Newton’s generalization of the binomial formula gives rise to an infinite series. If werestrict to natural number exponents, the convergence considerations are not necessaryand a proof based on the differentiation of polynomials becomes possible.

Proof. We first note that since (x+ a) is a polynomial of degree 1, (x+ a)n will be a poly-nomial of degree n and will thus be determined once we know what the coefficients of eachof the n+ 1 possible powers of x are. For concreteness let us write

(x+ a)n = p(x) =n∑

k=0

bkxk

and show how to determine the coefficients bk.Using the power rule and the chain rule for differentiation, we have

d

dx(x+ a)n = n(x+ a)n−1

so that

(x+ a)p′(x) = np(x)

with initial condition

p(0) = (0 + a)n = an

Then we determine what the coefficients bk must be to satisfy this equation. The initialcondition p(0) = an tells us that b0 = an. We can relate later coefficients to earlier onesusing the differentiatl equation :

p′(x) =

n∑k=1

kbkxk−1

so

(x+ a)p′(x) =n∑

k=1

kbkxk +

n∑k=1

akbkxk−1

= ab1 +n−1∑k=1

[kbk + a(k + 1)bk+1]xk + nbnx

n

=

n−1∑k=0

nbkxk

Since polynomials are equal when their coefficients are equal, this tell us that

10

1 THE INTRODUCTION

ab1 = nb0

(1b1) + (a2b2) = nb1...

...

(kbk) + [a(k + 1)bk+1] = nbk

nbn = nbn

Thus for k = 1, n− 1, we get

bk+1 =n− k

(k + 1)abk

Moreover, using the face that b0 = an this gives us

b0 = an

b1 = nan−1

b2 =n(n− 1)

2an−2 =

(n

2

)an−2

b3 =n(n− 1)(n− 2)

3.2an−3 =

(n

3

)an−3

......

bk =n(n− 1)...(n− k + 1)

k!an−k =

(n

k

)an−k

which proves the formula.

11

2 THE SENSITIVITY

In this chapter, we will take from some knowledges which Dr. David Dennis’ documentscollect.

2.1 JOHN WALLIS (1616 - 1703)

“It was always my affection, even from a child, not only to learn by rote, but to know thegrounds or reasons of what I learnt; to inform my judgement as well as to furnish my

memory.”

John Wallis was an English mathematician who is given partial credit for the develop-ment of infinitesimal calculus and was credited with introducing the symbol∞ for infinity.

He was born in 1616, Kent, England, the third of five children of Reverend John Wallisand Joanna Chapman. He was initially educated at a local Ashford school, but moved toJames Movat’s school in Tenterden in 1625 following an outbreak of plague. Wallis wasfirst exposed to mathematics in 1631, at Martin Holbeach’s school in Felsted; he enjoyedit but his study was erratic. In 1632, after decision to be a doctor, Wallis was sent in1632 to Emmanuel College, Cambridge. While there, he kept an act on the doctrine of thecirculation of the blood; that was said to have been the first occasion in Europe on whichthis theory was publicly maintained in a disputation. He received a Master’s degree in1640, afterwards entering the priesthood. Wallis was elected to a fellowship at Queens’College, Cambridge in 1644, which he however had to resign following his marriage.

Wallis made significant contributions to trigonometry, calculus, geometry, and the anal-ysis of infinite series. Especially, Arithfumetica Infinitorum was the most important ofhis works. In this book, the analytic methods of Descartes and Cavalien was extended. Inaddition, he also published Algebra, Opera...

12

2 THE SENSITIVITY

2.2 ISAAC NEWTON (1643 - 1727)

“If you ask a good skating how to be successful, he will say to you that fall, get up is asuccess.”

Isaac Newton was an English physicist, mathematician, astronomer, natural philoso-pher, alchemist, and theologian.

He was born in 1643, Lincolnshire, England. The fatherless infant was small enoughat birth. When he was barely three years old Newton’s mother, Hanna, placed her firstborn with his grandmother in order to remarry and raise a second family with BarnabasSmith, a wealthy rector from nearby North Witham. Much has been made of Newton’sposthumous birth, his prolonged separation from his mother, and his unrivaled hatredof his stepfather. Until Hanna returned to Woolsthorpe in 1653 after the death of hersecond husband, Newton was denied his mother’s attention, a possible clue to his complexcharacter. Newton’s childhood was anything but happy, and throughout his life he vergedon emotional collapse, occasionally falling into violent and vindictive attacks against friendand foe alike.

In 1665 Newton took his bachelor’s degree at Cambridge without honors or distinction.Since the university was closed for the next two years because of plague, Newton returnedto Woolsthorpe in midyear. For in those days I was in my prime of age for invention, andminded mathematics and philosophy more than at any time since. Especially in 1666, heobserved the fall of an apple in his garden at Woolsthorpe, later recalling, ’In the same yearI began to think of gravity extending to the orb of the Moon’. In mathematics, Newton laterbecame involved in a dispute with Leibniz over priority in the development of infinitesimalcalculus. Most modern historians believe that Newton and Leibniz developed infinitesimalcalculus independently, although with very different notations. Moreover, he found thegenerality formula of binomial and give the definition of light theory.

He published Philosophiae Naturalis Principia Mathematica in 1687 which wasthe important book all over the world. In addition, he wrote Opticks.

13

2 THE SENSITIVITY

2.3 A JOURNEY OF JOHN WALLIS

Beginning of Alhazen’s Summation Formulas, Ahazen (965 - 1040) - the Iraqi mathemati-cian who stated some formulas which affected the later results of Wallis. Ahazen derivedhis formulas by laying out a sequence of rectangles whose areas represent the terms of thesum.

• Look at a rectangle whose length is n+1 and width is n, we divide this rectangle intoseveral rectangles (see Figure 1). Thus its area must equals to

n(n+ 1) =n∑

i=1

i+n∑

i=1

i

= 2 (1 + 2 + 3 + ...+ n)

Hence,

1 + 2 + 3 + ...+ n =1

2n(n+ 1)

Figure 1

• Consider a rectangle whose length isn∑

i=1

i and width is n + 1, we also divide this

rectangle into several rectangles (see Figure 2). Apply the above formula (i.e.n∑

i=1

i =

1

2n (n+ 1) =

1

2n2 +

1

2n), its area must equals to

14

2 THE SENSITIVITY(n∑

i=1

i

)(n+ 1) =

n∑i=1

i2 +n∑

i=1

i∑k=1

k

(1

2n2 +

1

2n

)(n+ 1) =

n∑i=1

i2 +n∑

i=1

(1

2i2 +

1

2i

)1

2n3 + n2 +

1

2n =

n∑i=1

i2 +1

2

n∑i=1

i2 +1

2

(1

2n2 +

1

2n

)1

2n3 +

3

4n2 +

1

4n =

3

2

n∑i=1

i2

Hence,

12 + 22 + 32 + ...+ n2 =1

3n3 +

1

2n2 +

1

6n

Figure 2

• Using this similar method, we can find out the sum of the cubes or the fourth powers

or more if we want. Thus we continue to view a rectangle whose length isn∑

i=1

i2 and

width is n + 1, we also divide this rectangle into several rectangles (see Figure 3).

Apply the above formulas (i.e.n∑

i=1

i =1

2n (n+ 1) =

1

2n2 +

1

2n and

n∑i=1

i2 =1

3n3 +

1

2n2 +

1

6n), its area must equals to

15

2 THE SENSITIVITY(n∑

i=1

i2

)(n+ 1) =

n∑i=1

i3 +n∑

i=1

i∑k=1

k2

(1

3n3 +

1

2n2 +

1

6n

)(n+ 1) =

n∑i=1

i3 +n∑

i=1

(1

3i3 +

1

2i2 +

1

6i

)1

3n4 +

5

6n3 +

2

3n2 +

1

6n =

n∑i=1

i3 +1

3

n∑i=1

i3 +1

2

(1

3n3 +

1

2n2 +

1

6n

)+

1

6

(1

2n2 +

1

2n

)1

3n4 +

2

3n3 +

1

3n2 =

4

3

n∑i=1

i3

Hence,

13 + 23 + 33 + ...+ n3 =1

4n4 +

1

2n3 +

1

4n2

Figure 3

At that time, the definition of fractional exponents is not correct, people still have aquestion for its existence. For example, the concept of this is suggested in various ways inthe works of Oresme (14th century), Girard and Stevin (16th century).

The Geometry, first published in 1638, of René Descartes was the first published treatiseto use positive integer exponents written as superscripts. He saw exponents as an index forrepeated multiplication. That is to say he wrote x3 in place of xxx. Wallis adopted this useof an index and tried to extend it, and tested its validity across multiple representations.

Wallis took from Fermat the idea of using an equation to generate a curve, which was incontrast to Descartes’ work which always began with a geometrical construction. Descartesalways constructed a curve geometrically first, and then analyzed it by finding its equa-tion. Wallis mixed these ideas, so he defined what is the fractional exponents and provedits existence successful.

And the Arithmetica Infinitorum contains a detailed investigation of the behaviorof sequences and ratios of sequences from which a variety of geometric results are thenconcluded. We shall look at one of the most important examples. Consider the ratio of thesum of a sequence of a fixed power to a series of constant terms all equal to the highestvalue appearing in the sum. Wallis researched into ratios of the form :

A =0k + 1k + 2k + ...+ nk

nk + nk + nk + ...+ nk

16

2 THE SENSITIVITY

For each fixed integer value of k, Wallis investigated the behavior of these ratios as nincreases. When k = 1, he calculates :

0 + 1 + 2

2 + 2 + 2=

1

20 + 1 + 2 + 3

3 + 3 + 3 + 3=

1

20 + 1 + 2 + 3 + 4

4 + 4 + 4 + 4 + 4=

1

2... = ...

This can be seen from the well known Alhazen’s fomulas. We have

A =

n∑i=1

i

n (n+ 1)=

n (n+ 1)

2n (n+ 1)

=1

2

Then Wallis called1

2the characteristic ratio of the index k = 1.

When k = 2, Wallis continued to compute the following ratios :

02 + 12

12 + 12=

1

3+

1

6

02 + 12 + 22

22 + 22 + 22=

1

3+

1

12

02 + 12 + 22 + 32

32 + 32 + 32 + 32=

1

3+

1

18

02 + 12 + 22 + 32 + 42

42 + 42 + 42 + 42 + 42=

1

3+

1

24

02 + 12 + 22 + 32 + 42 + 52

52 + 52 + 52 + 52 + 52 + 52=

1

3+

1

30... = ...

Wallis claimed that the right hand side is always equals1

3+

1

6nand this can be checked

by Alhazen’s fomulas.

A =

n∑i=1

i2

n2 (n+ 1)=

1

6n (n+ 1) (2n+ 1)

n2 (n+ 1)=

1

3+

1

6n

As n increases this ratio approaches1

3(we can see lim

n→∞

(1

3+

1

6n

)=

1

3now), so Wallis

then defined the characteristic ratio of the index k = 2 as equal to1

3. In a similar way,

Wallos computed the characteristic ratio of k = 3 as1

4, and k = 4 as

1

5and so forth.

Thus he made the general claim that the characteristic ratio of the index k is1

k + 1for all

positive integers.

17

2 THE SENSITIVITY

Next, Wallis show that these characteristic ratios yielded most of the familiar ratiosof area and volume known from geometry. It means he showed that his arithmetic wasconsistent with the accepted truths of geometry.

He assumed that an area is a sum of an infinite number of parallel line segments, andthat a volume is a sum of an infinite number of parallel areas, his basis assumptions weretaken from Cavalieri’s Geometria Indivisibilibus Continuorum which was publishedin 1635. Wallis first considered the area under the curve y = xk (see Figure 4). He wantedto compute the ratio of the shaded area to the area of the rectangle which encloses it.

Figure 4

Wallis claimed that this geometric problem is an example of the characteristic ratio ofthe sequence with index k. In specific, the terms in the numerator are the lengths of theline segments that make up the shaded area while the terms in the denominator are thelengths of the line segments that make up the rectangle (hence constant). He imaginedthe increment or scale as very small while the number of the terms is very large.

And then this characteristic ratio of1

3holds for all parabolas, not just y = x2. In detail,

we can use Riemann’s integral for proving that results.Consider a set [0; 1] and a curve y = 5x2, we have an area under this curve (S1) which is

calculated :

S1 =

ˆ 1

05x2dx = 5

[x3

3

]10

=5

3

An area of a rectangle which encloses this curve (S2) equals 5, so that we have

S1 =1

3S2

This example means that characteristic ratio depends only on the exponent and not onthe coefficient, or we can say that characteristic ratio is not linear.

18

2 THE SENSITIVITY

Figure 5

In addition, that above ratio also shows that the volume of a pyramid is1

3of the box that

surrounds it (see Figure 5). Hence Wallis saw this as another example of his computationof the characteristic ratio for the index k = 2.

These geometric results were not knew. Fermat, Roberval, Cavalieri and Pascal had allpreviously made this claim that when k is a positive integer; the area under the curve

y = xk had a ratio of1

k + 1to the rectangle that encloses it. However, Wallis went on to

assert that if we define the index of√x as

1

2, the claim remains true. Since the area under

the curve y =√x is the complement of the area under y = x2 (i.e. the unshaded area

in Figure 4), it must have a characteristic ratio of2

3=

11

2+ 1

. The same can be seen for

y = 3√x, whose characteristic ratio must be

3

4=

11

3+ 1

...

It was this coordination of two separate representations that gave Wallis the confidenceto claim that the appropriate index of y = q

√xp must be

p

q, and that its characteristic ratio

must be1

p

q+ 1

. Wallis continued to assert that this claim remained true even when the

index is irrational because he gave√

3 as an example. But in many cases, Wallis had noway to directly verify the characteristic ratio of an index, for example : y =

3√x2.

How can we determine the characteristic ratio of the circle? This is the question thatmotivated Wallis to study a particular family of curves from which he could interpolatethe value for circle. He wrote the equation of the circle of radius r, as y =

√r2 − x2, and

considered it in the first quadrant. He wanted to determine the ratio of its area to the r bysquare that contains it.

Of course Wallis knew that this ratio isπ

4, from various geometric constructions going

back to Archimedes, but he wanted to test his theory of index, characteristic ratio andinterpolation by arriving at this result in a new way. Therefore, he considered the familyof curves defined by the equations y = ( q

√r − q√x)

p. Its graphs is showed in the unit square(r = 1) (see Figure 6).

19

2 THE SENSITIVITY

Figure 6

If p and q are both integers, he knew that by expanding the binomial ( q√r − q√x)

p to thepth power and using his rule for characteristic ratios he could determine the ratio for thesecurves. For example, when p = q = 2, then : y = (

√r −√x)

2= r− 2

√r√x+ x, and so must

have a characteristic ratio of 1− 2.2

3+

1

2=

1

6.

Figure 7

Next, Wallis, Pascal and others made a table of these ratios after computing it when p

and q are integers. This table records the ratio of the rectangle to the shaded area for eachof the curves y = ( q

√r − q√x)

p. (see Table 1)

20

2 THE SENSITIVITY

Table 1

At this point, Wallis temporarily abandoned both the geometric and algebraic represen-tations and began to work solely in the table representation. The question then became,how does one interpolate the missing values in this table? Firstly, he worked on the rowswith integer values of q. We can summarize his works :

• When q = 0, we see the constant value of 1.

• When q = 1, we see an arithmetic progression whose common difference is1

2.

• In the row q = 2, we have the triangular numbers which are the sums of the integersin the row q = 1. Thus we can use the formula for the sum of consecutive integers,s2 + s

2, where s = p + 1. Putting the intermediate values into this formula allows us

to complete the row q = 2. For example, letting s =3

2in this formula yields

15

8, which

becomes the entry where p =1

2.

• The numbers in the row q = 3 are the pyramidal numbers each of which is the sumof integers in the row q = 2. Hence the appropriate formula is found by summing theformula from the row q = 2. Applying Alhazen’s formulas and then collecting terms,

we gain1

2

s∑i=1

i2 + i =s3 + 3s2 + 2s

6, where s = p + 1. For example, letting s =

3

2, the

formula yields105

48, which becomes the table entry where p =

1

2.

• In a similar fashion, we sum the previous cubic formula to obtain a formula forthe row q = 4. Using the Alhazen’s formulas and then collecting terms, we gets4 + 6s3 + 11s2 + 6s

24, where s = p+ 1.

Therefore, Wallis built the following table by Ahazen’s formulas and his formulas for in-terpolation. Since the table is symmetrical this also allows us to fill in the correspondingcolumns when p is an integer. (see Table 2)

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2 THE SENSITIVITY

Table 2

Wallis began to turn his attention to the row q =1

2. Each of the entries that now appear

there is calculated by using each of the successive interpolation formulas. And he see thateach of these formulas has a higher algebraic degree.

What pattern exists in the formation of these numbers which will allow us to interpolatebetween them to find the missing entries? Remember that the first missing entry is the

q = p =1

2(i.e. the ratio of the square to the area of the quarter circle), we can find the

characteristic ratio of this as equals to4

π.

Hence Wallis first tried to fill in this row with arithmetic averages. The average of 1 and3

2is

5

4, so this is not equal to

4

π.

Wallis now observed that each of the numerators in these fractions is the product ofconsecutive odd integers, while each of the denominators is the product of consecutive even

intergers; this means15

8=

3.5

2.4;105

48=

3.5.7

2.4.6;945

384=

3.5.7.9

2.4.6.8. Hence to move two entries to

the right in this row one multiplies byn

n− 1.

For the entries that appear so far, n is always odd; so Wallis assumed that to get fromone missing entry to the next one, he should still multiply by

n

n− 1, but this time n would

have to be the intermediate even number. Denoting the first missing entry by Ω whose

coefficient is 1, then the next missing entry should be 1.4

4− 1=

4

3, so its result is

4

3Ω. In a

similar fashion, we get the one after should be4.6

3.5Ω =

8

5Ω, and the one after that should

be4.6.8

3.5.7Ω =

64

35Ω and so forth.

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2 THE SENSITIVITY

Then the q =1

2row becomes :

So the column p =1

2can now be filled in by symmetry. The row q =

3

2has a similar

pattern (i.e. products of consecutive odd over consecutive enven number) but there theentries two spaces to the right are always multiply by

n

n− 3(note that we go on n = 6).

We can also double check, as Wallis did, that this law of formation agrees with usual lawfor the formation of binomials (i.e. each entry is the sum the entries two up, and two tothe left). The full table now show on Table 3 :

Table 3

Wallis saw that he could evaluate π which is put in Ω. He had to find a way to calculateΩ using his priciple of interpolation so that he could check his value against the one knownfrom geometry.

Thus returning once again to the row q =1

2where moving two spaces to the right from

the nth entry multiplies that entry byn

n− 1, Wallis noted that as n increases the fraction

n

n− 1gets closer and closer to 1

(i.e. lim

n→∞

n

n− 1= 1

). Hence the number two spaces to

the right must change very little as we go further out the sequence. This is true of thecalculated fractions as well as the multiples of Ω.

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2 THE SENSITIVITY

Wallis argued that since the whole sequence is increasing steadily, that consecutiveterms must also be getting close to 1 as we proceed. Hence he built these terms to be-come

4.6.8.10...

3.5.7.9...Ω ≈ 3.5.7.9...

2.4.6.8...

Ω ≈ 3.3.5.5.7.7.9.9...

2.4.4.6.6.8.8.10...

Because of4

π= Ω, hence

π = 2.2.2.4.4.6.6.8.8...

1.3.3.5.5.7.7.9...

The empirical methods of Wallis led the young Isaac Newton to his first profound math-ematical creation; the expansion of functions in binomial series. Thus Wallis’ method ofinterpolation became for Newton the basis of his notion of continuity. So Newton wantedto generalize the methods of Wallis...

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2 THE SENSITIVITY

2.4 THE CONTINUATION OF ISSAC NEWTON’S IDEA

In 1661, the nineteen-year-old Isaac Newton read the Arithmetica Infinitorum and wasmuch impressed. In 1664 and 1665 he made a series of annotations from Wallis whichextended the concepts of interpolation and extrapolation. It was here that Newton firstdeveloped his binomial expansions for negative and fractional exponents.

Newton made a series of extensions of the ideas in Wallis. He extended the tables ofareas to the left to include negative powers and found new patterns upon which to baseinterpolations. Perhaps his significant deviation from Wallis was that Newton abandonedthe use of ratios of areas and instead sought direct expressions which would calculate thearea under a portion of a curve from the value of the abscissa. Using what he knew fromWallis he could write down area expressions for the integer powers. Referring back toFigure 4, we have :

Area under xn

Area of containing rectangle=

1

n+ 1

Area of containing rectangle = x.xn = xn+1

Hence

Area under xn =xn+1

n+ 1

Next, he considered the positive and negative integer powers of 1 + x, that is

... , y =1

1 + x, y = 1 , y = 1 + x , y = (1 + x)2 , y = (1 + x)3 , y = (1 + x)4 , ...

Newton drew the following graph of several members of this family of curves (see Figure8), he found the coefficients in simple binomial.

25

2 THE SENSITIVITY

Figure 8

According to Figure 8, we get CK = CD = 1, letting DE = x, then E (1 + x, 0). This

shows that EB =1

1 + x, EF = 1 , EG = 1 + x and EH = (1 + x)2. He then wrote down a

series of expressions which calculate areas under the curves over the segment DE as :

SAFED = x , SAGED = x+x2

2, SAHED = x+

2x2

2+x3

3

So it is easy to verify these results by Riemann’s integral. For example, we look atSAHED :

SAHED =

ˆ x+1

1t2dt =

[t3

3

]x+1

1

=(x+ 1)3 − 1

3= x+

2x2

2+x3

3

Although the higher power curves did not appear in the graph, Newton went on to writedown more area expressions for curves in this family. He obtainedd the following areaexpressions by first expanding and then finding the area term by term.

• Third power : x+3x2

2+

3x3

3+x4

4

• Fourth power : x+4x2

2+

6x3

3+

4x4

4+x5

5

• Fifth power : x+5x2

2+

10x3

3+

10x4

4+

5x5

5+x6

6

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2 THE SENSITIVITY

At this point, Newton wanted to find a pattern which would which would allow him toextend his calculations to include the areas under the negative powers of 1 +x. He noticedthat the denominators form an arithmetic sequence while the numerators follow the bino-mial patterns. He then made the following table of the area expressions for (1 + x)p (seeTable 4). And the question becomes : how can one fill in the missing entries?

Table 4

This binomial table is different from Wallis’ table in that the rows are all nudged succes-sively to the right so that the diagonals of the Wallis’ table become the columns of Newton’stable. The binomial pattern of formation is now such that each entry is the sum of the en-try to the left of it and the one above that one. So that we can find that the ? must be equalto −1.

Hence Newton filled in the table of coefficients for the area expressions under the curves(1 + x)p. (see Table 5)

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2 THE SENSITIVITY

Table 5

According to Table 5, we can evaluate the area under the hyperbola y =1

1 + xwhat we

now call the natural logarithm of 1 + x, and Newton considered it as SABED.

SABED = x− x2

2+x3

3− x4

4+x5

5− x6

6+x7

7...

(=∞∑n=0

(−1)nxn+1

n+ 1,−1 < x < 1

)

Next, Newton returned to the table of characteristic ratios made by Wallis (see Table 3).As discussed previously, Newton abandoned Wallis’ use of area ratios and set out to makea table of coefficients for a sequence of explicit expressions for calculating areas. He used

the same set of curves whose characteristic ratios Wallis had tabulated in the row q =1

2inside a unit square. Hence he saw the areas under the following sequence of curves (seeFigure 9).

... , y = 1 , y =√

1− x2 , y = 1− x2 , y =(1− x2

)√1− x2 , y =

(1− x2

)2 , ...

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2 THE SENSITIVITY

Figure 9

He let AD = DC = 1 and DE = x then E (x, 0); EF,EB,EG,EH,EI,EN are then theordinates of his series of curves respectively.

For the integer powers of 1− x2, Newton could write down the areas in Figure 9 as :

SAFED = x , SAGED = x− 1

2x3 , SAIED = x− 2

3x3 +

1

5x5

In a similar fashion, he continued this sequence of area expressions for(1− x2

)p asfollows :

• p = 3 : x− 3

3x3 +

3

5x5 − 1

7x7

• p = 4 : x− 4

3x3 +

6

5x5 − 4

7x7 +

1

9x9

• p = 5 : x− 5

3x3 +

10

5x5 − 10

7x7 +

5

9x9 − 1

11x11

According to the pattern of Table 5 (i.e. each entry is the sum of the entry to the left of itand the one above that one), he made a table of these results including an extension intothe negative powers. (see Table 6)

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2 THE SENSITIVITY

Table 6

However, this method is not true when p is not integer. He first noted that integerbinomial tables obey the following additive pattern of formation (see Table 7). This patternis formed by starting with a constant sequence (a, a, a, ...) and an arbitrary left column(a, b, c, d, ...); and then forming each entry as the sum of the one to the left and the oneabove that. Newton saw that old method is not appropriate for the integers. In specific,the entries in the top row must all be 1 in all the interpolated tables (i.e. a = 1) and theincrement fo the second row must be 1. So this is unreasonable because the distance ofcoefficients of p = 0 and p = −1 are 1.

Table 7

This forced that Newton must found other methods which filled in that table. To getaround this difficulty, he rewrote this pattern so as to unlink the rows of Table 7. That isto say, he preserved the pattern within each individual row but he changed the names ofthe variables so that each variable appeared in only one row.

As you move down the rows each new row can be described using successively one morevariable. Changing the names of variables so that each row is independent of the others,the pattern now becomes Table 8.

30

2 THE SENSITIVITY

Table 8

Using this table, if any entry in the first row is known the whole row is known. If anytwo entries in the second row are known then one can solve for b and c and fill in the entirerow. If any three entries in the third row are known one can solve for d, e and f and fill inthe entire row. Thus with a sufficient number of known values in a given row one couldsolve a system of linear equations for all the variables in that row.

For example, we can consider the third row, using the known values where 0, ?, 0, ?, 1.We have a set of linear equations as follows :

d = 0

f + 2e+ d = 0

6f + 4e+ d = 1

We obtain d = 0

f =1

4

e = −1

8

We can now complete the entire row using these values, but it should be noted here thatalthough we used three equations to find d, e and f there are actually an infinite numberof equations involving these three variables.

And Newton’s method is satisfied because the values he found agree with Wallis andwith the additive pattern of table formation. We can see Table 9 :

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2 THE SENSITIVITY

Table 9

With the completion of this table, Newton will also obtain a new way to calculate π

which will validate his method in a geometric representation. So the column p =1

2gives

an infinite series which calculates the area under any portion of a circle y =√

1− x2. Thatis to say, that SABED we can see it in Figure 9.

SABED = x− 1

2.x3

3− 1

8.x5

5− 3

48.x7

7− 15

384.x9

9− 105

3840.x11

11− ...

Letting x = 1 in this series, we calculate the area of one quarter of the circle and yield anew caculation of π :

π

4= 1− 1

6− 1

40− 1

112− 5

1152− 7

2816− ...

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2 THE SENSITIVITY

Figure 10

Moreover, Newton also pointed out that this series allowed him to compute arcsinx. Byadding a line from D to B in Figure 9 (see Figure 10), and subtracting the area of DBEfrom ABED, it obtains the area of the circular sector ABD. So we getarcsinx = EBD

EB‖AD

Hence

arcsinx = BDA

Since the circular radius is 1, twice the area of sector ABD equals BDA (i.e. 2SABD =

BDA). According to the area of ABED, we only calculate the area of DBE.

SBDE =1

2DE.EB =

1

2x√

1− x2

Thus we can evaluate arcsinx.Satisfied with his interpolation methods Newton began searching for a pattern in the

columns of his table which would allow him to continue each series without having torepeat his tedious interpolation procedure row by row. Note that some of the fractionsin Table 9 are not reduced. In earlier tabulations Newton reduced the fractions but hesoon became aware that this would only obscure any possible patterns in their formations.Following the example set by Wallis, he sought a pattern of continued multiplication of

arithmetic sequences. Since the circle was so important to him, he studied the p =1

2col-

umn first. Factoring the numbers in these fractions, he found that they could be producedby continued multiplication as :

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2 THE SENSITIVITY

1

1.1

2.−1

4.−3

6.−5

8.−7

10.−9

12.−11

14...

Similarly, the entries in the p =3

2column can be produced by continued multiplication

as :

1

1.3

2.1

4.−1

6.−3

8.−5

10.−7

12.−9

14...

In order to further investigate these patterns, Newton carried out an interpolation of

the binomial table at intervals of1

3(see Table 10). Using the patterns from Table 8 and

solving the systems of equations for the variables in each row, he produced the followinginterpolated Table 10.

Table 10

For a pattern of repeated multiplication of arithmetic sequences that would generate

the columns of this table, Newton discerned the following pattern for the column p =1

3.

1

1.1

3.−2

6.−5

9.−8

12.−11

15.−14

18.−17

21...

Note that the sequence of numerators and denominators both change by increments of

3 (ignoring the first term). And the same thing happens where p =1

2;3

2but by increments

of 2.At this point, Newton wrote down an explicit formula for the binomial numbers in an

arbitrary column p =x

y.

1

1.x

y.x− y

2y.x− 2y

3y.x− 3y

4y.x− 4y

5y.x− 5y

6y...

Or this is equivalent to

p

1.p− 1

2.p− 2

3.p− 3

4.p− 4

5.p− 5

6...

34

2 THE SENSITIVITY

His original interpolations were designed to calculate areas under families of curves butNewton soon saw that by changing the terms to which the coefficients were applied, hecould see these numbers to calculate the points on the curve as well. This was particularly

useful for root extractions. He simply had to replace the area termsxn+1

n+ 1with the original

terms xn from which they came. The coefficients in the tables remain the same. So we havethe examples :

• For p =1

2in Table 9, we can calculate :

√1− x2 = 1− 1

2x2 − 1

8x4 − 1

16x6 − 1

128x8 − ...

• Or p =1

3in Table 10, we get :

3√

1− x2 = 1− 1

3x2 − 1

9x4 − 5

81x6 − 10

243x8 − ...

3√

1 + x = 1 +1

3x− 1

9x2 +

5

81x3 − 10

243x4...

These series appear in the form in the letters to Oldenburg in which Newton explainedhis binomial series at the request of Liebniz in 1676.

Hence, using the methods of Newton, we can represent the binomial expansion for allreal numbers; this is very important in our life. In detail, this helps us investigate thegraph or the approximately values of more functions, so this is also a basis of all afterseries. Newton is such a great mathematician worthy of the naming for this binomialformula.

35