Newton: the Fundamental Theorem of Calculus

Chapter 31

Newton: The fundamental theorem of calculus

Relation of ﬂuxions

Isaac Newton (1665) handled the problem of tangents to a curve in the following way. Consider a Cartesian coordinate system with a point A moving along the x-axis and a point B moving along the y axis. The vertical line through A and the horizontal line through B intersect at a point P . As A and B move on the respective axes, the point P traces out a curve, which has equation f(x, y)=0. The velocities of A and B are called fluxions, and are denoted by x˙ and y˙ respectively. Newton first considered the relation between the fluxions x˙ and y˙, i j assuming f(x, y) of the form aijx y . He substituted x and y by x+xo ˙ and y +yo ˙ into f(x, y)=0 and obtained P

0= aij(x +xo ˙ )(y +yo ˙ )

i j i j 1 2 = X aijx y + aijx (jy − yo˙ + terms of o )

j i 1 2 +X aijy (ix X− xo˙ + terms of o )

i 1 j 1 + X a (ix − xo˙ + )(jy − yo˙ + ) ij ··· ··· X

i j 2 Since aijx y =0, dropping all terms involving o , we have

P i 1 j 1 aij(ix − xo˙ + jy − yo˙ )=0. X 318 Newton: The fundamental theorem of calculus

Dividing by o, Newton obtained ix˙ jy˙ + a xiyj =0, x y ij or X y˙ ja xiyj 1 = ij − . x˙ − ia xi 1yj P ij − P Fundamental Theorem of Calculus Newton considered the inverse problem of finding y in terms of x given y˙ y˙ a relation between x and the ratio of their fluxions x˙ . Specifically, x˙ = g(x).

A˙

b a x

x˙ x

Let A be the area under the curve y = f(x), between two values a and b of x. Regard this area as being swept out by the vertical segments moving from a to the right with unit ﬂuxion x˙ =1. Newton asserted that the ﬂuxion (time rate of change) of the area is A˙ = g(x) with x˙ =1, so that A˙ = g(x). x˙

This is the fundamental theorem of calculus. Chapter 32

Newton: The binomial theorem

Newton discovered the binomial theorem for an arbitrary index in the following form:

m m m m n m 2n m 3n (P +P Q) n = P n + AQ+ − BQ+ − CQ+ − DQ+ n 2n 3n 4n ··· where P + P Q signifies the quantity whose root (or even any power, of root of a power) is to be found; P signifies the first term of that quantity, m Q the remaining terms divided by the first, and n the index. Example 32.1. x2 x4 x6 5x8 √a2 + x2 = a + + + . 2a − 8a3 16c5 − 128a7 ··· x2 Here, Q = a2 . With the first term A = a, we have 1 1 x2 x2 the second term B = 2 AQ = 2 a a2 = 2a , 1 1· x·2 x2 x4 the third term C = − BQ = − = , 4 4 · 2a · a2 − 8a3 3 1 x4 x2 x6 the fourth term D = − CQ = = , 6 − 2 − 8a3 · a2 16a5 .... Here is a modern formulation of the binomial theorem: let m be a rational number,

m m m (1 + x)m =1+ x + x2 + + xk + 1 2 ··· k ··· where m m k m m = − , =1. k +1 k +1 k 0 320 Newton: The binomial theorem

It is easy to see that m m(m 1) (m k + 1) = − ··· − . k k!

Exercise Find the series expansion of 1 up to 5 terms. √1 x2 − Chapter 33

Newton’s method of approximate solution of an equation

To solve a polynomial equation n k f(x)= akx =0, k 0 X− Newton started with an approximate solution x and sought a solution of the form x = x + p: ∗ ∗ n k 0= akx k=0 Xn k = ak(x + p) ∗ k=0 Xn k k 1 = ak(x + kx − p + ) ∗ ∗ ··· k=0 Xn n k k 1 = akx + p kakx − + higher order terms of p ∗ ∗ Xk=0 Xk=0 = f(x )+ pf ′(x )+ higher order terms of p. ∗ ∗ f(x∗) Ignoring the higher terms of p, we have p ′ . Therefore, if x is ≈ − f (x∗) ∗ f(x∗) an approximate solution, then x∗ ′ is a better one. − f (x∗) 322 Newton’s method of approximate solution of an equation

Exercise (1) Find the cube root of 2 by iterating Newton’s method 3 times. (2) Show that if y3 + y + xy x3 2=0, then − − 1 x2 y =1 x + + . − 4 64 ··· Chapter 34

Newton’s reversion of series

1 Beginning with the area under the graph y = 1+x :

1 1 1 z = x x2 + x3 x4 + − 2 3 − 4 −··· Newton expressed x in terms of z in the following way. Newton used only the ﬁrst ﬁve terms of the series, and solved

1 1 1 1 x5 x4 x3 + x2 + x z =0 5 − 4 − 3 2 − for x (in terms of z). Beginning with the ﬁrst approximation z for x, he wrote x = z + p, substituted into the above equation, suppressing higher order terms of p and got

1 1 1 1 z2 + z3 z4 + z5 + p(1 z + z2 z3 + z4) −2 3 − 4 5 − − 1 3z2 1 + p2 + z +2z3 + p3 z +2z2 −2 − 2 3 − 1 1 + p4 + z + p5 =0. (34.1) −4 5 Using only the ﬁrst two terms, he obtained

1 z2 + 1 z3 1 z4 + 1 z5 1 p − 2 3 − 4 5 = z2 + ≈ − 1 z + z2 z3 + z4 2 ··· − − 324 Newton’s reversion of series

1 2 This gives the second approximation x z + 2 z . Then Newton put 1 2 ≈ p = 2 z + q into (34.1) and obtain 1 1 1 1 z3 + z4 z5 + q 1 z + z2 + =0. −6 8 − 20 − 2 ··· From this, 1 z3 + 1 z4 1 z5 1 q = − 6 8 − 20 = z3 + . − 1 z + 1 z2) 6 ··· − 2 This gives the third approximation 1 1 x z + z2 + z3 + . ≈ 2 6 ··· Continuing in the same way, Newton obtained 1 1 1 1 x = z + z2 + z3 + z4 + z5 + . 2 6 24 120 ···

Now, since z = log(1 + x) is equivalent to ez 1, this gives the expansion − 1 1 1 1 ez =1+ z + z2 + z3 + z4 + z5 + . 2 6 24 120 ··· Chapter 35

Newton: The series for sine and cosine

The series for arcsin x The series of arcsin x can be found from the quadrature of the circle. More precisely, the area under the circular arc AQ can be found in two ways: 1 1 x θ + x√1 x2 = √1 x2dx. 2 2 − − Z0 y

1 √1 x2 − θ x x O P B Note that x x 1 1 1 5 √1 x2dx = 1 x2 x4 x6 x8 + dx − − 2 − 8 − 16 − 128 ··· Z0 Z0 1 1 1 = x x3 x5 x7 + − 6 − 40 − 112 ··· 326 Newton: The series for sine and cosine

Therefore, with θ = arcsin x, we have

x arcsin x = 2 √1 x2dx x√1 x2 − − − Z0 1 1 1 = 2 x x3 x5 x7 + − 6 − 40 − 112 ··· 1 1 1 x 1 x2 x4 x6 + − − 2 − 8 − 16 ··· 1 3 5 = x + x3 + x5 + x7 + . 6 40 112 ··· Newton then obtain the series of sin θ by reversing 1 3 5 θ = x + x3 + x5 + x7 + . 6 40 112 ··· This is 1 1 sin θ = θ θ3 + θ5 + − 6 120 − ··· From this, the series of cos θ also follows. Chapter 36

Wallis’ product

Wallis has obtained an inﬁnite product expression for π by interpola- tion. 1 Precisely,

4 3 3 5 5 7 7 9 9 11 11 = × × × × × × × × × ×··· . π 2 4 4 6 6 8 8 10 10 12 × × × × × × × × × ×···

π 2 n Let In := 0 sin dx. (1) Establish the reduction formula R n 1 In = − In 2. n · −

(2) From this we see that the subsequences (I2n) and (I2n+1) are monotonic decreasing. Since every In is positive, these two subsequences are convergent. (3) We show that they indeed converge to the same limit. Since 2n+1 2n 2n 1 π sin x sin x sin − x for x 0, , we have I I ≤ ≤ ∈ 2 2n+1 ≤ 2n ≤ I2n 1, and − 2n I I = 2n+1 2n 1. 2n +1 I2n 1 ≤ I2n 1 ≤ − −

I2n Therefore, limn =1, and indeed limn I2n = limn I2n+1. →∞ I2n−1 →∞ →∞ π (4) Now, from the recurrence relation and the initial values I0 = 2

1J. F. Scott, The Work of John Wallis, Chelsea, 1981; pp.47–60. 328 Wallis’ product

and I1 =1, we have 1 3 2n 1 π I = − , 2n 2 · 4 ··· 2n · 2 2 4 2n I = . 2n+1 3 · 5 ··· 2n +1 Since I π 1 3 3 5 5 7 (2n 3) 2n 1 2n = × × × × × ×···× − − , I2n 1 2 × 2 2 4 4 6 6 2n 2 · 2n 2 − × × × × × ×···× − − we conclude that 4 3 3 5 5 7 (2n 3) (2n 1) = lim × × × × ×···× − × − . π n 2 4 4 6 6 (2n 2) (2n 2) →∞ × × × × ×···× − × −

Exercise Suppose we construct a sequence of rectangles as follows. We begin with a square of area one. We then alternate adjoining a rectangle of area one alongside or on top of the previous rectangle. Find the limiting ratio of length to height.