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NEWTON’S BINOMIAL FORMULA Khoa Anh Vo - Hoai Thanh Nguyen February 3, 2012
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Khoa Anh Vo - Hoai Thanh NguyenVietnam National UniversityHo Chi Minh City (HCMC) University of ScienceFaculty of Mathematics and Computer Science227 Nguyen Van Cu Street, District 5, Ho Chi Minh CityVietnam 2
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PREFACEThis 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 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 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
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1 THE INTRODUCTION1.1 A FORMULAAs 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. 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, wedeﬁne 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, sothose 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 ∞ r (1 + z)r = zk k=0 kRemark. 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
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1 THE INTRODUCTION1.2 THREE PROOFSThe 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)ndirectly. 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 PROOFMany 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. 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 kProof. This is a direct calculation in which we add fractions and simplify. 7
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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 weassume that 8
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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+1Completing the proof by induction.1.2.2 COMBINATORIAL PROOFThe 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 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 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 eachfactor too use either the x or the a. The choices which result in k x’s and n − k a’s each give na term of the form an−k xk . There are distinct ways to choose the k element subset k nof factors from which to take the x. Thus the coefﬁcient of an−k xk is . This tells us kthat n n (x + a) = Cn xn−k ak k k=0 9
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1 THE INTRODUCTION1.2.3 DERIVATION USING CALCULUSNewton’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 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 initialcondition p(0) = an tells us that b0 = an . We can relate later coefﬁcients to earlier onesusing 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
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1 THE INTRODUCTION ab1 = nb0 (1b1 ) + (a2b2 ) = nb1 . . . . . . (kbk ) + [a(k + 1)bk+1 ] = nbk nbn = nbnThus for k = 1, n − 1, we get n−k bk+1 = bk (k + 1)aMoreover, 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! kwhich proves the formula. 11
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2 THE SENSITIVITYIn 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 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 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
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2 THE SENSITIVITY2.2 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 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
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2 THE SENSITIVITY2.3 A JOURNEY OF JOHN WALLISBeginning 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 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
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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=1Hence, 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
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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 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 : 0k + 1k + 2k + ... + nk A= nk + nk + nk + ... + nk 16
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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 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 6nby 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 1then deﬁned the characteristic ratio of the index k = 2 as equal to . In a similar way, 3 1 1Wallos computed the characteristic ratio of k = 3 as , and k = 4 as and so forth. 4 5 1Thus he made the general claim that the characteristic ratio of the index k is for all k+1positive integers. 17
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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. 1 And then this characteristic ratio of holds for all parabolas, not just y = x2 . In detail, 3we 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 iscalculated : 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 onthe coefﬁcient, or we can say that characteristic ratio is not linear. 18
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2 THE SENSITIVITY Figure 5 1 In addition, that above ratio also shows that the volume of a pyramid is of the box that 3surrounds 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 1y = xk had a ratio of to the rectangle that encloses it. However, Wallis went on to k+1 √ 1assert that if we deﬁne the index of x as , the claim remains true. Since the area under √ 2the curve y = x is the complement of the area under y = x2 (i.e. the unshaded area 2 1in Figure 4), it must have a characteristic ratio of = . The same can be seen for 3 1 +1 2 √ 3 1y = x, whose characteristic ratio must be = 3 ... 4 1 +1 3 It was this coordination of two separate representations that gave Wallis the conﬁdence √ pto claim that the appropriate index of y = q xp must be , and that its characteristic ratio q 1must 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 √ 3way 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 thatmotivated 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 , andconsidered it in the ﬁrst quadrant. He wanted to determine the ratio of its area to the r bysquare that contains it. π Of course Wallis knew that this ratio is, from various geometric constructions going 4back 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 family √ √ pof curves deﬁned by the equations y = ( q r − q x) . Its graphs is showed in the unit square(r = 1) (see Figure 6). 19
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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 thepth 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 1have 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 pand q are integers. This table records the ratio of the rectangle to the shaded area for each √ √ pof the curves y = ( q r − q x) . (see Table 1) 20
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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. 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. 24Therefore, 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) 21
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2 THE SENSITIVITY Table 2 1 Wallis began to turn his attention to the row q = . Each of the entries that now appear 2there 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 1q = p = (i.e. the ratio of the square to the area of the quarter circle), we can ﬁnd the 2 4characteristic ratio of this as equals to . π Hence Wallis ﬁrst tried to ﬁll in this row with arithmetic averages. The average of 1 and3 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 ofconsecutive odd integers, while each of the denominators is the product of consecutive even 15 3.5 105 3.5.7 945 3.5.7.9intergers; this means = ; = ; = . Hence to move two entries to 8 2.4 48 2.4.6 384 2.4.6.8 nthe 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 none missing entry to the next one, he should still multiply by , but this time n would n−1have to be the intermediate even number. Denoting the ﬁrst missing entry by Ω whose 4 4 4coefﬁcient is 1, then the next missing entry should be 1. = , so its result is Ω. In a 4−1 3 3 4.6 8similar fashion, we get the one after should be Ω = Ω, and the one after that should 3.5 5 4.6.8 64be Ω = Ω and so forth. 3.5.7 35 22
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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 2pattern (i.e. products of consecutive odd over consecutive enven number) but there the nentries 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 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. 1 Thus returning once again to the row q = where moving two spaces to the right from 2 nthe 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 ton−1 n→∞ n − 1the 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. 23
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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... 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 ofinterpolation became for Newton the basis of his notion of continuity. So Newton wantedto generalize the methods of Wallis... 24
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2 THE SENSITIVITY2.4 THE CONTINUATION OF ISSAC NEWTON’S IDEAIn 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 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 Figure8), he found the coefﬁcients in simple binomial. 25
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2 THE SENSITIVITY Figure 8 According to Figure 8, we get CK = CD = 1, letting DE = x, then E (1 + x, 0). This 1shows that EB = , EF = 1 , EG = 1 + x and EH = (1 + x)2 . He then wrote down a 1+xseries 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 atSAHED : 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 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. 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
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2 THE SENSITIVITYAt 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) 27
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2 THE SENSITIVITY Table 5 1 According to Table 5, we can evaluate the area under the hyperbola y = what we 1+xnow 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 makea table of coefﬁcients for a sequence of explicit expressions for calculating areas. He used 1the same set of curves whose characteristic ratios Wallis had tabulated in the row q = 2inside a unit square. Hence he saw the areas under the following sequence of curves (seeFigure 9). √ √ 2 ... , y = 1 , y = 1 − x2 , y = 1 − x2 , y = 1 − x2 1 − x2 , y = 1 − x2 , ... 28
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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 : 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 asfollows : 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 11According 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) 29
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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
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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 1 f = 4 1 e =− 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 : 31
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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 1which will validate his method in a geometric representation. So the column p = gives √ 2an 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. 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 anew caculation of π : π 1 1 1 5 7 =1− − − − − − ... 4 6 40 112 1152 2816 32
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2 THE SENSITIVITY Figure 10 Moreover, Newton also pointed out that this series allowed him to compute arcsin x. 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 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 thecolumns 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 1arithmetic sequences. Since the circle was so important to him, he studied the p = col- 2umn ﬁrst. Factoring the numbers in these fractions, he found that they could be producedby continued multiplication as : 33
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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 2as : 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 1the binomial table at intervals of (see Table 10). Using the patterns from Table 8 and 3solving 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 1the 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 33 (ignoring the ﬁrst term). And the same thing happens where p = ; but by increments 2 2of 2. At this point, Newton wrote down an explicit formula for the binomial numbers in an xarbitrary 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
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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 xn+1useful for root extractions. He simply had to replace the area terms with the original n+1terms xn from which they came. The coefﬁcients in the tables remain the same. So we havethe 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 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
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