Tài liệu chi tiết về nhị thức Newton rất bổ ích cho các học sinh giỏi toán

1. KHOA A. VO – HOAI T. NGUYEN

2. NEWTON’S BINOMIAL FORMULA
Khoa Anh Vo - Hoai Thanh Nguyen
February 3, 2012

3. 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

4. PREFACE
This book is intended as our first English thematic for students who study in high school
or people who want to research into the history of mathematics. In detail, this talks about
the journey of John Wallis (1616 - 1703) from the Alhazen’s formulas (965 - 1040), and
the continuation of Issac Newton’s idea (1643 - 1727). Then we give some mathematical
problems in the educational programs. Therefore, we desire to provide more knownledges
for 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 collaborators
for all their helps. These include :
Name High School/ University
Thien Huu Vo Truong HCMC University of Science
Truong Nhat Thanh Mai HCMC University of Science
Quang Dang Nguyen HCMC University of Science
Minh Nhat Vu To HCMC International University
Phong Tran HCMC University of Pedagogy
Tuan Thanh Nguyen HCMC University of Economics and Law
Trang 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 Binomial
Formula” strange - looking.
The readers can find and download “Newton’s Binomial Formula” at :
http://www.forum.mathscope.org/
or
http://anhkhoavo1210.wordpress.com/
3

5. Contents
PREFACE 3
WHAT IS THE NEWTON’S BINOMIAL FORMULA? 5
1 THE INTRODUCTION 6
1.1 A FORMULA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.2 THREE PROOFS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.1 INDUCTION PROOF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2.2 COMBINATORIAL PROOF . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.2.3 DERIVATION USING CALCULUS . . . . . . . . . . . . . . . . . . . . 10
2 THE SENSITIVITY 12
2.1 JOHN WALLIS (1616 - 1703) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2 ISAAC NEWTON (1643 - 1727) . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 A JOURNEY OF JOHN WALLIS . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.4 THE CONTINUATION OF ISSAC NEWTON’S IDEA . . . . . . . . . . . . . . 25
4

6. WHAT IS THE NEWTON’S BINOMIAL
FORMULA?
5

7. 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 expansion
of a binomial n power(s) that is become a polynomial n + 1 terms.
n
n
(x + a) = Cn an−k xk
k
k=0
In order to make sense of the theorem we need to agree on some conventions. First, we
define the binomial coefficients
k n n!
Cn = =
k (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 cases
gives either an = an or xn = xn . If all of n = 0, x = 0, and a = 0 then we get the result
00 = 00 , which is not particularly meaningful, but as long as we agree on what we mean
by 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 complex
number with magnitude modulus of z less than 1, we have
∞
r
(1 + z)r = zk
k=0
k
Remark. A general formula for m (ai )’s term(s)
m n
n!
ai = an1 an2 ...anm
m
n1 !n2 !...nm ! 1 2
i=1
where n1 + n2 + ... + nm = n.
6

8. 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 small
topic 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 basis
for Pascal’s triangle.
According to Pascal’s triangle, we can order the binomial coefficients corresponding to n
power(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.
k k−1 k
Cm + Cm = Cm+1
or
m m m+1
+ =
k k−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 m m+1
+ =
k k−1 k
Proof. This is a direct calculation in which we add fractions and simplify.
7

9. 1 THE INTRODUCTION
m m m! m!
+ = +
k k−1 (m − k)!k! (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
0
(x + a)0 = x0 a0 = 1
0
Now (x + a)0 = 1 and
0
0 0
a0−k xk = a0 x0 = 1
k=0
k 0
Here we are using the conventions that
0
=1
0
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
1
1 1 1
(x + a)1 = a1−k xk = a1 x0 + a0 x1
k=0
k 0 1
This is equivalent to
1! 1!
x+a= a+ x=a+x
1!0! 0!1!
which is true. Thus we have the base cases for our induction. For the induction step we
assume that
8

10. 1 THE INTRODUCTION
m
m
(x + a)m = am−k xk
k=0
k
and show that the formula is true when n = m + 1.
(x + a)m+1 = (x + a)m (x + a)
m
m
= am−k xk (x + a)
k=0
k
m m
m m−k k+1 m
= a x + am−k+1 xk
k=0
k k=0
k
m
m m m m+1
= am+1 x0 + + am−k+1 xk + a0 xm+1
0 k=1
k k−1 m+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 n!
=
k (n − k)!k!
as counting the number of unordered k−subsets of an n element set. This is done by
first counting the ordered k−element strings with no repetitions : for the first element we
have n choices; for the second, n − 1; until we get to the k th which has n − k + 1 choices.
Since these choices are made in succession, we multiply to get
n!
n(n − 1)...(n − k + 1) =
(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 each
factor too use either the x or the a. The choices which result in k x’s and n − k a’s each give
n
a term of the form an−k xk . There are distinct ways to choose the k element subset
k
n
of factors from which to take the x. Thus the coefficient of an−k xk is . This tells us
k
that
n
n
(x + a) = Cn xn−k ak
k
k=0
9

11. 1 THE INTRODUCTION
1.2.3 DERIVATION USING CALCULUS
Newton’s generalization of the binomial formula gives rise to an infinite series. If we
restrict to natural number exponents, the convergence considerations are not necessary
and 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 each
of the n + 1 possible powers of x are. For concreteness let us write
n
(x + a)n = p(x) = bk xk
k=0
and show how to determine the coefficients bk .
Using the power rule and the chain rule for differentiation, we have
d
(x + a)n = n(x + a)n−1
dx
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 initial
condition p(0) = an tells us that b0 = an . We can relate later coefficients to earlier ones
using the differentiatl equation :
n
p (x) = kbk xk−1
k=1
so
n n
(x + a)p (x) = kbk xk + akbk xk−1
k=1 k=1
n−1
= ab1 + [kbk + a(k + 1)bk+1 ] xk + nbn xn
k=1
n−1
= nbk xk
k=0
Since polynomials are equal when their coefficients are equal, this tell us that
10

12. 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
n−k
bk+1 = bk
(k + 1)a
Moreover, using the face that b0 = an this gives us
b0 = an
b1 = nan−1
n(n − 1) n−2 n
b2 = a = an−2
2 2
n(n − 1)(n − 2) n−3 n
b3 = a = an−3
3.2 3
.
. .
.
. .
n(n − 1)...(n − k + 1) n−k n
bk = a = an−k
k! k
which proves the formula.
11

13. 2 THE SENSITIVITY
In this chapter, we will take from some knowledges which Dr. David Dennis’ documents
collect.
2.1 JOHN WALLIS (1616 - 1703)
“It was always my affection, even from a child, not only to learn by rote, but to know the
grounds 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 Wallis
and Joanna Chapman. He was initially educated at a local Ashford school, but moved to
James Movat’s school in Tenterden in 1625 following an outbreak of plague. Wallis was
first exposed to mathematics in 1631, at Martin Holbeach’s school in Felsted; he enjoyed
it but his study was erratic. In 1632, after decision to be a doctor, Wallis was sent in
1632 to Emmanuel College, Cambridge. While there, he kept an act on the doctrine of the
circulation of the blood; that was said to have been the first occasion in Europe on which
this theory was publicly maintained in a disputation. He received a Master’s degree in
1640, 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 of
his works. In this book, the analytic methods of Descartes and Cavalien was extended. In
addition, he also published Algebra, Opera...
12

14. 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 a
success.”
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 enough
at birth. When he was barely three years old Newton’s mother, Hanna, placed her first
born with his grandmother in order to remarry and raise a second family with Barnabas
Smith, a wealthy rector from nearby North Witham. Much has been made of Newton’s
posthumous birth, his prolonged separation from his mother, and his unrivaled hatred
of his stepfather. Until Hanna returned to Woolsthorpe in 1653 after the death of her
second husband, Newton was denied his mother’s attention, a possible clue to his complex
character. Newton’s childhood was anything but happy, and throughout his life he verged
on emotional collapse, occasionally falling into violent and vindictive attacks against friend
and 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 returned
to Woolsthorpe in midyear. For in those days I was in my prime of age for invention, and
minded mathematics and philosophy more than at any time since. Especially in 1666, he
observed the fall of an apple in his garden at Woolsthorpe, later recalling, ’In the same year
I began to think of gravity extending to the orb of the Moon’. In mathematics, Newton later
became involved in a dispute with Leibniz over priority in the development of infinitesimal
calculus. Most modern historians believe that Newton and Leibniz developed infinitesimal
calculus independently, although with very different notations. Moreover, he found the
generality formula of binomial and give the definition of light theory.
He published Philosophiae Naturalis Principia Mathematica in 1687 which was
the important book all over the world. In addition, he wrote Opticks.
13

15. 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 derived
his formulas by laying out a sequence of rectangles whose areas represent the terms of the
sum.
• Look at a rectangle whose length is n + 1 and width is n, we divide this rectangle into
several rectangles (see Figure 1). Thus its area must equals to
n n
n(n + 1) = i+ i
i=1 i=1
= 2 (1 + 2 + 3 + ... + n)
Hence,
1
1 + 2 + 3 + ... + n = n(n + 1)
2
Figure 1
n
• Consider a rectangle whose length is i and width is n + 1, we also divide this
i=1
n
rectangle into several rectangles (see Figure 2). Apply the above formula (i.e. i=
i=1
1 1 1
n (n + 1) = n2 + n), its area must equals to
2 2 2
14

16. 2 THE SENSITIVITY
n n n i
i (n + 1) = i2 + k
i=1 i=1 i=1 k=1
n n
1 2 1 1 2 1
n + n (n + 1) = i2 + i + i
2 2 2 2
i=1 i=1
n n
1 3 1 1 1 1 2 1
n + n2 + n = i + 2
i2 + n + n
2 2 2 2 2 2
i=1 i=1
n
1 3 3 2 1 3
n + n + n = i2
2 4 4 2
i=1
Hence,
1 1 1
12 + 22 + 32 + ... + n2 = n3 + n2 + n
3 2 6
Figure 2
• Using this similar method, we can find out the sum of the cubes or the fourth powers
n
or more if we want. Thus we continue to view a rectangle whose length is i2 and
i=1
width is n + 1, we also divide this rectangle into several rectangles (see Figure 3).
n n
1 1 1 1 1
Apply the above formulas (i.e. i = n (n + 1) = n2 + n and i2 = n3 + n2 +
2 2 2 3 2
i=1 i=1
1
n), its area must equals to
6
15

17. 2 THE SENSITIVITY
n n n i
i2 (n + 1) = i3 + k2
i=1 i=1 i=1 k=1
n n
1 3 1 2 1 1 3 1 2 1
n + n + n (n + 1) = i3 + i + i + i
3 2 6 3 2 6
i=1 i=1
n n
1 4 5 3 2 2 1 1
3 1 1 3 1 2 1 1 1 2 1
n + n + n + n = i + i3 + n + n + n + n + n
3 6 3 6 3 2 3 2 6 6 2 2
i=1 i=1
n
1 4 2 3 1 2 4
n + n + n = i3
3 3 3 3
i=1
Hence,
1 1 1
13 + 23 + 33 + ... + n3 = n4 + n3 + n2
4 2 4
Figure 3
At that time, the definition of fractional exponents is not correct, people still have a
question for its existence. For example, the concept of this is suggested in various ways in
the works of Oresme (14th century), Girard and Stevin (16th century).
The Geometry, first published in 1638, of René Descartes was the first published treatise
to use positive integer exponents written as superscripts. He saw exponents as an index for
repeated multiplication. That is to say he wrote x3 in place of xxx. Wallis adopted this use
of 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 in
contrast to Descartes’ work which always began with a geometrical construction. Descartes
always 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 proved
its existence successful.
And the Arithmetica Infinitorum contains a detailed investigation of the behavior
of sequences and ratios of sequences from which a variety of geometric results are then
concluded. We shall look at one of the most important examples. Consider the ratio of the
sum of a sequence of a fixed power to a series of constant terms all equal to the highest
value appearing in the sum. Wallis researched into ratios of the form :
0k + 1k + 2k + ... + nk
A=
nk + nk + nk + ... + nk
16

18. 2 THE SENSITIVITY
For each fixed integer value of k, Wallis investigated the behavior of these ratios as n
increases. When k = 1, he calculates :
0+1+2 1
=
2+2+2 2
0+1+2+3 1
=
3+3+3+3 2
0+1+2+3+4 1
=
4+4+4+4+4 2
... = ...
This can be seen from the well known Alhazen’s fomulas. We have
n
in (n + 1)
2 1
A = i=1 = =
n (n + 1) n (n + 1) 2
1
Then Wallis called the characteristic ratio of the index k = 1.
2
When k = 2, Wallis continued to compute the following ratios :
02 + 1 2 1 1
= +
12 + 1 2 3 6
02 + 12 + 22 1 1
= +
22 + 22 + 22 3 12
0 2 + 12 + 2 2 + 3 2 1 1
= +
32 + 32 + 32 + 32 3 18
02 + 12 + 22 + 3 2 + 4 2 1 1
= +
42 + 42 + 42 + 42 + 42 3 24
0 2 + 12 + 22 + 32 + 4 2 + 5 2 1 1
= +
52 + 5 2 + 5 2 + 5 2 + 5 2 + 5 2 3 30
... = ...
1 1
Wallis claimed that the right hand side is always equals + and this can be checked
3 6n
by Alhazen’s fomulas.
n
i2 1
n (n + 1) (2n + 1) 1 1
A = 2 i=1 = 6 2 (n + 1)
= +
n (n + 1) n 3 6n
1 1 1 1
As n increases this ratio approaches (we can see lim + = now), so Wallis
3 n→∞ 3 6n 3
1
then defined the characteristic ratio of the index k = 2 as equal to . In a similar way,
3
1 1
Wallos computed the characteristic ratio of k = 3 as , and k = 4 as and so forth.
4 5
1
Thus he made the general claim that the characteristic ratio of the index k is for all
k+1
positive integers.
17

19. 2 THE SENSITIVITY
Next, Wallis show that these characteristic ratios yielded most of the familiar ratios
of area and volume known from geometry. It means he showed that his arithmetic was
consistent with the accepted truths of geometry.
He assumed that an area is a sum of an infinite number of parallel line segments, and
that a volume is a sum of an infinite number of parallel areas, his basis assumptions were
taken from Cavalieri’s Geometria Indivisibilibus Continuorum which was published
in 1635. Wallis first considered the area under the curve y = xk (see Figure 4). He wanted
to 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 of
the sequence with index k. In specific, the terms in the numerator are the lengths of the
line segments that make up the shaded area while the terms in the denominator are the
lengths of the line segments that make up the rectangle (hence constant). He imagined
the increment or scale as very small while the number of the terms is very large.
1
And then this characteristic ratio of holds for all parabolas, not just y = x2 . In detail,
3
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 :
ˆ 1 1
2 x3 5
S1 = 5x dx = 5 =
0 3 0 3
An area of a rectangle which encloses this curve (S2 ) equals 5, so that we have
1
S1 = S2
3
This example means that characteristic ratio depends only on the exponent and not on
the coefficient, or we can say that characteristic ratio is not linear.
18

20. 2 THE SENSITIVITY
Figure 5
1
In addition, that above ratio also shows that the volume of a pyramid is of the box that
3
surrounds it (see Figure 5). Hence Wallis saw this as another example of his computation
of the characteristic ratio for the index k = 2.
These geometric results were not knew. Fermat, Roberval, Cavalieri and Pascal had all
previously made this claim that when k is a positive integer; the area under the curve
1
y = xk had a ratio of to the rectangle that encloses it. However, Wallis went on to
k+1
√ 1
assert that if we define the index of x as , the claim remains true. Since the area under
√ 2
the curve y = x is the complement of the area under y = x2 (i.e. the unshaded area
2 1
in Figure 4), it must have a characteristic ratio of = . The same can be seen for
3 1
+1
2
√ 3 1
y = x, whose characteristic ratio must be =
3
...
4 1
+1
3
It was this coordination of two separate representations that gave Wallis the confidence
√ p
to claim that the appropriate index of y = q xp must be , and that its characteristic ratio
q
1
must be p . Wallis continued to assert that this claim remained true even when the
+1
q √
index is irrational because he gave 3 as an example. But in many cases, Wallis had no
√
3
way to directly verify the characteristic ratio of an index, for example : y = x2 .
How can we determine the characteristic ratio of the circle? This is the question that
motivated Wallis to study a particular family of curves from which he could interpolate
√
the 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 by
square that contains it.
π
Of course Wallis knew that this ratio is, from various geometric constructions going
4
back to Archimedes, but he wanted to test his theory of index, characteristic ratio and
interpolation by arriving at this result in a new way. Therefore, he considered the family
√ √ p
of curves defined by the equations y = ( q r − q x) . Its graphs is showed in the unit square
(r = 1) (see Figure 6).
19

21. 2 THE SENSITIVITY
Figure 6
√ √ p
If p and q are both integers, he knew that by expanding the binomial ( q r − q x) to the
pth power and using his rule for characteristic ratios he could determine the ratio for these
√ √ 2 √ √
curves. For example, when p = q = 2, then : y = ( r − x) = r − 2 r x + x, and so must
2 1 1
have a characteristic ratio of 1 − 2. + = .
3 2 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 each
√ √ p
of the curves y = ( q r − q x) . (see Table 1)
20

22. 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 rows
with integer values of q. We can summarize his works :
• When q = 0, we see the constant value of 1.
1
• When q = 1, we see an arithmetic progression whose common difference is .
2
• In the row q = 2, we have the triangular numbers which are the sums of the integers
in the row q = 1. Thus we can use the formula for the sum of consecutive integers,
s2 + s
, where s = p + 1. Putting the intermediate values into this formula allows us
2
3 15
to complete the row q = 2. For example, letting s = in this formula yields , which
2 8
1
becomes the entry where p = .
2
• The numbers in the row q = 3 are the pyramidal numbers each of which is the sum
of integers in the row q = 2. Hence the appropriate formula is found by summing the
formula from the row q = 2. Applying Alhazen’s formulas and then collecting terms,
s
1 s3 + 3s2 + 2s 3
we gain i2 + i = , where s = p + 1. For example, letting s = , the
2 6 2
i=1
105 1
formula yields , which becomes the table entry where p = .
48 2
• In a similar fashion, we sum the previous cubic formula to obtain a formula for
the row q = 4. Using the Alhazen’s formulas and then collecting terms, we get
s4 + 6s3 + 11s2 + 6s
, where s = p + 1.
24
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 corresponding
columns when p is an integer. (see Table 2)
21

23. 2 THE SENSITIVITY
Table 2
1
Wallis began to turn his attention to the row q = . Each of the entries that now appear
2
there is calculated by using each of the successive interpolation formulas. And he see that
each of these formulas has a higher algebraic degree.
What pattern exists in the formation of these numbers which will allow us to interpolate
between them to find the missing entries? Remember that the first missing entry is the
1
q = p = (i.e. the ratio of the square to the area of the quarter circle), we can find the
2
4
characteristic ratio of this as equals to .
π
Hence Wallis first tried to fill in this row with arithmetic averages. The average of 1 and
3 5 4
is , so this is not equal to .
2 4 π
Wallis now observed that each of the numerators in these fractions is the product of
consecutive odd integers, while each of the denominators is the product of consecutive even
15 3.5 105 3.5.7 945 3.5.7.9
intergers; this means = ; = ; = . Hence to move two entries to
8 2.4 48 2.4.6 384 2.4.6.8
n
the right in this row one multiplies by .
n−1
For the entries that appear so far, n is always odd; so Wallis assumed that to get from
n
one missing entry to the next one, he should still multiply by , but this time n would
n−1
have to be the intermediate even number. Denoting the first missing entry by Ω whose
4 4 4
coefficient is 1, then the next missing entry should be 1. = , so its result is Ω. In a
4−1 3 3
4.6 8
similar fashion, we get the one after should be Ω = Ω, and the one after that should
3.5 5
4.6.8 64
be Ω = Ω and so forth.
3.5.7 35
22

24. 2 THE SENSITIVITY
1
Then the q = row becomes :
2
1 3
So the column p = can now be filled in by symmetry. The row q = has a similar
2 2
pattern (i.e. products of consecutive odd over consecutive enven number) but there the
n
entries two spaces to the right are always multiply by (note that we go on n = 6).
n−3
We can also double check, as Wallis did, that this law of formation agrees with usual law
for the formation of binomials (i.e. each entry is the sum the entries two up, and two to
the 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 calculate
Ω using his priciple of interpolation so that he could check his value against the one known
from geometry.
1
Thus returning once again to the row q = where moving two spaces to the right from
2
n
the nth entry multiplies that entry by , Wallis noted that as n increases the fraction
n−1
n n
gets closer and closer to 1 i.e. lim = 1 . Hence the number two spaces to
n−1 n→∞ n − 1
the right must change very little as we go further out the sequence. This is true of the
calculated fractions as well as the multiples of Ω.
23

25. 2 THE SENSITIVITY
Wallis argued that since the whole sequence is increasing steadily, that consecutive
terms 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...
3.3.5.5.7.7.9.9...
Ω ≈
2.4.4.6.6.8.8.10...
4
Because of = Ω, hence
π
2.2.4.4.6.6.8.8...
π = 2.
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 of
interpolation became for Newton the basis of his notion of continuity. So Newton wanted
to generalize the methods of Wallis...
24

26. 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 was
much impressed. In 1664 and 1665 he made a series of annotations from Wallis which
extended the concepts of interpolation and extrapolation. It was here that Newton first
developed his binomial expansions for negative and fractional exponents.
Newton made a series of extensions of the ideas in Wallis. He extended the tables of
areas to the left to include negative powers and found new patterns upon which to base
interpolations. Perhaps his significant deviation from Wallis was that Newton abandoned
the use of ratios of areas and instead sought direct expressions which would calculate the
area under a portion of a curve from the value of the abscissa. Using what he knew from
Wallis he could write down area expressions for the integer powers. Referring back to
Figure 4, we have :
Area under xn

1
=


Area of containing rectangle n+1
Area of containing rectangle = x.xn = xn+1

Hence
xn+1
Area under xn =
n+1
Next, he considered the positive and negative integer powers of 1 + x, that is
1
... , y = , y = 1 , y = 1 + x , y = (1 + x)2 , y = (1 + x)3 , y = (1 + x)4 , ...
1+x
Newton drew the following graph of several members of this family of curves (see Figure
8), he found the coefficients in simple binomial.
25

27. 2 THE SENSITIVITY
Figure 8
According to Figure 8, we get CK = CD = 1, letting DE = x, then E (1 + x, 0). This
1
shows that EB = , EF = 1 , EG = 1 + x and EH = (1 + x)2 . He then wrote down a
1+x
series of expressions which calculate areas under the curves over the segment DE as :
x2 2x2 x3
SAF ED = x , SAGED = x + , SAHED = x + +
2 2 3
So it is easy to verify these results by Riemann’s integral. For example, we look at
SAHED :
ˆ x+1
x+1
t3 (x + 1)3 − 1 2x2 x3
SAHED = t2 dt = = =x+ +
1 3 1 3 2 3
Although the higher power curves did not appear in the graph, Newton went on to write
down more area expressions for curves in this family. He obtainedd the following area
expressions by first expanding and then finding the area term by term.
3x2 3x3 x4
• Third power : x + + +
2 3 4
4x2 6x3 4x4 x5
• Fourth power : x + + + +
2 3 4 5
5x2 10x3 10x4 5x5 x6
• Fifth power : x + + + + +
2 3 4 5 6
26

28. 2 THE SENSITIVITY
At this point, Newton wanted to find a pattern which would which would allow him to
extend his calculations to include the areas under the negative powers of 1 + x. He noticed
that 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 (see
Table 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’s
table. 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 equal
to −1.
Hence Newton filled in the table of coefficients for the area expressions under the curves
(1 + x)p . (see Table 5)
27

29. 2 THE SENSITIVITY
Table 5
1
According to Table 5, we can evaluate the area under the hyperbola y = what we
1+x
now call the natural logarithm of 1 + x, and Newton considered it as SABED .
∞
x2 x3 x4 x5 x6 x7 xn+1
SABED = x − + − + − + ... = (−1)n , −1 < x < 1
2 3 4 5 6 7 n+1
n=0
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 make
a table of coefficients for a sequence of explicit expressions for calculating areas. He used
1
the same set of curves whose characteristic ratios Wallis had tabulated in the row q =
2
inside a unit square. Hence he saw the areas under the following sequence of curves (see
Figure 9).
√ √ 2
... , y = 1 , y = 1 − x2 , y = 1 − x2 , y = 1 − x2 1 − x2 , y = 1 − x2 , ...
28

30. 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 the
ordinates of his series of curves respectively.
For the integer powers of 1 − x2 , Newton could write down the areas in Figure 9 as :
1 2 1
SAF ED = x , SAGED = x − x3 , SAIED = x − x3 + x5
2 3 5
p
In a similar fashion, he continued this sequence of area expressions for 1 − x2 as
follows :
3 3 1
• p = 3 : x − x3 + x5 − x7
3 5 7
4 6 4 1
• p = 4 : x − x3 + x5 − x7 + x9
3 5 7 9
5 10 10 5 1
• p = 5 : x − x3 + x5 − x7 + x9 − x11
3 5 7 9 11
According to the pattern of Table 5 (i.e. each entry is the sum of the entry to the left of it
and the one above that one), he made a table of these results including an extension into
the negative powers. (see Table 6)
29

31. 2 THE SENSITIVITY
Table 6
However, this method is not true when p is not integer. He first noted that integer
binomial tables obey the following additive pattern of formation (see Table 7). This pattern
is 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 one
above 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 the
increment fo the second row must be 1. So this is unreasonable because the distance of
coefficients of p = 0 and p = −1 are 1.
Table 7
This forced that Newton must found other methods which filled in that table. To get
around this difficulty, he rewrote this pattern so as to unlink the rows of Table 7. That is
to say, he preserved the pattern within each individual row but he changed the names of
the 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 more
variable. Changing the names of variables so that each row is independent of the others,
the pattern now becomes Table 8.
30

32. 2 THE SENSITIVITY
Table 8
Using this table, if any entry in the first row is known the whole row is known. If any
two entries in the second row are known then one can solve for b and c and fill in the entire
row. If any three entries in the third row are known one can solve for d, e and f and fill in
the entire row. Thus with a sufficient number of known values in a given row one could
solve 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

 1
f =
4
1



e =−
8
We can now complete the entire row using these values, but it should be noted here that
although we used three equations to find d, e and f there are actually an infinite number
of equations involving these three variables.
And Newton’s method is satisfied because the values he found agree with Wallis and
with the additive pattern of table formation. We can see Table 9 :
31

33. 2 THE SENSITIVITY
Table 9
With the completion of this table, Newton will also obtain a new way to calculate π
1
which will validate his method in a geometric representation. So the column p = gives
√ 2
an infinite series which calculates the area under any portion of a circle y = 1 − x2 . That
is to say, that SABED we can see it in Figure 9.
1 x3 1 x5 3 x7 15 x9 105 x11
SABED = x − . − . − . − . − . − ...
2 3 8 5 48 7 384 9 3840 11
Letting x = 1 in this series, we calculate the area of one quarter of the circle and yield a
new caculation of π :
π 1 1 1 5 7
=1− − − − − − ...
4 6 40 112 1152 2816
32

34. 2 THE SENSITIVITY
Figure 10
Moreover, Newton also pointed out that this series allowed him to compute arcsin x. By
adding a line from D to B in Figure 9 (see Figure 10), and subtracting the area of DBE
from ABED, it obtains the area of the circular sector ABD. So we get

arcsin x = EBD
EB AD
Hence
arcsin x = 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.
1 1 √
SBDE = DE.EB = x 1 − x2
2 2
Thus we can evaluate arcsin x.
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 to
repeat his tedious interpolation procedure row by row. Note that some of the fractions
in Table 9 are not reduced. In earlier tabulations Newton reduced the fractions but he
soon 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
1
arithmetic sequences. Since the circle was so important to him, he studied the p = col-
2
umn first. Factoring the numbers in these fractions, he found that they could be produced
by continued multiplication as :
33

35. 2 THE SENSITIVITY
1 1 −1 −3 −5 −7 −9 −11
. . . . . . . ...
1 2 4 6 8 10 12 14
3
Similarly, the entries in the p = column can be produced by continued multiplication
2
as :
1 3 1 −1 −3 −5 −7 −9
. . . . . . . ...
1 2 4 6 8 10 12 14
In order to further investigate these patterns, Newton carried out an interpolation of
1
the binomial table at intervals of (see Table 10). Using the patterns from Table 8 and
3
solving the systems of equations for the variables in each row, he produced the following
interpolated Table 10.
Table 10
For a pattern of repeated multiplication of arithmetic sequences that would generate
1
the columns of this table, Newton discerned the following pattern for the column p = .
3
1 1 −2 −5 −8 −11 −14 −17
. . . . . . . ...
1 3 6 9 12 15 18 21
Note that the sequence of numerators and denominators both change by increments of
1 3
3 (ignoring the first term). And the same thing happens where p = ; but by increments
2 2
of 2.
At this point, Newton wrote down an explicit formula for the binomial numbers in an
x
arbitrary column p = .
y
1 x x − y x − 2y x − 3y x − 4y x − 5y
. . . . . . ...
1 y 2y 3y 4y 5y 6y
Or this is equivalent to
p p−1 p−2 p−3 p−4 p−5
. . . . . ...
1 2 3 4 5 6
34

36. 2 THE SENSITIVITY
His original interpolations were designed to calculate areas under families of curves but
Newton soon saw that by changing the terms to which the coefficients were applied, he
could see these numbers to calculate the points on the curve as well. This was particularly
xn+1
useful for root extractions. He simply had to replace the area terms with the original
n+1
terms xn from which they came. The coefficients in the tables remain the same. So we have
the examples :
1
• For p = in Table 9, we can calculate :
2
√ 1 1 1 1 8
1 − x2 = 1 − x2 − x4 − x6 − x − ...
2 8 16 128
1
• Or p = in Table 10, we get :
3
√ 1 1 5 10 8
3
1 − x2 = 1 − x2 − x4 − x6 − x − ...
3 9 81 243
√ 1 1 5 10 4
3
1 + x = 1 + x − x2 + x3 − x ...
3 9 81 243
These series appear in the form in the letters to Oldenburg in which Newton explained
his binomial series at the request of Liebniz in 1676.
Hence, using the methods of Newton, we can represent the binomial expansion for all
real numbers; this is very important in our life. In detail, this helps us investigate the
graph or the approximately values of more functions, so this is also a basis of all after
series. Newton is such a great mathematician worthy of the naming for this binomial
formula.
35