CALCULUS from a Historical Perspective

Paul Yiu

Department of Mathematics Florida Atlantic University

Fall 2009

Chapters 1–49

Tuesday, December 2

Tuesdays 8/25 9/1 9/8 9/15 9/22 9/29 10/6 10/13 Thursdays 8/27 9/3 9/10 9/17 9/24 10/1 10/8 10/15 Tuesdays 10/20 10/27 11/3 11/10 11/17 11/24 12/1 Thursdays 10/22 10/29 11/5 11/12 11/19 *** 12/3 Contents

1 Extraction of square roots 101

2 Convergence of sequences 107

3 Square roots by decimal digits 113

4 a + a + + √a + 115 ··· ··· q 5 An inﬁnitep continued fraction 117

6 Ancient Chinese calculations of π 121

7 Existence of π 125

8 Archimedes’ calculation of π 129

9 Ptolemy’s calculations of chord lengths 131

sin θ 10 The basic limit limθ 0 =1 135 → θ 11 The Arithmetic - Geometric Means Inequality 139

12 Principle of nested intervals 143

13 The number e 147

14 Some basic limits 151

15 The parabola 201 iv CONTENTS

16 Archimedes’ quadrature of the parabola 205

17 Quadrature of the spiral 209

18 Apollonius’ extremum problems on conics 213

19Normalsofaparabola 215

20 Envelope of normal to a conic 217

21 The cissoid 221

22 Conchoids 225

23 The quadratrix 231

24 The lemniscate 233

25 Volumes in Euclid’s Element 301

26 Archimedes’ calculation of the volume of a sphere 305

27 Cavalieri’s principle 309

28 Fermat: Area under the graph of y = xn 311

29 Pascal: Summation of powers of numbers in arithmetic progression 313

30 Pascal: On the sines of a quadrant of a circle 315

31 Newton: The fundamental theorem of calculus 317

32 Newton: The binomial theorem 319

33 Newton’s method of approximate solution of an equation 321

34 Newton’s reversion of series 323

35 Newton: The series for sine and cosine 325

36 Wallis’ product 327 CONTENTS v

37 Newton: Universal gravitation 329

38 Logarithms 401

39 Euler’s introduction to e and the exponential functions 405

40 Euler: Natural logarithms 409

41 Euler’s formula eiv = cos v + i sin v 411

42 Summation of powers of integers 417

43 Series expansions for the tangent and secant functions 421

44 Euler’s ﬁrst calculation of 1+ 1 + 1 + 1 + 1 + 425 22k 32k 42k 52k ··· 45 Euler: Triangulation of convex polygon 433

46 Inﬁnite Series with positive terms 501

47 The harmonic series 507

48 The alternating harmonic series 511

49 Conditionally convergent series 515 45.0.1 2 ...... div Chapter 1

Extraction of square roots

Heron

Heron’s numerical example illustrating the use of his formula ∆= s(s a)(s b)(s c) − − − for the area of a trianglep in terms of its sides a, b, c and semiperimeter a+b+c s = 2 : Let the sides of the triangle be 7, 8, and 9. ...The result [of multiplying s, s a, s b, s c] is 720. Take the square root of this and it− will be− the area− of the triangle. Since 720 has not a rational square root, we shall make a close approximation to the root in this manner. Since the square nearest to 720 is 729, having a root 27, divide 27 into 720; the result is 2 2 26 3 ; add 27; the result is 53 3 . Take half of this; the result is 1 1 5 26 2 + 3 (= 26 6 ). Therefore the square root of 720 will be very 5 5 1 nearly 26 6 . For 26 6 multiplied by itself gives 720 36 ; so that 1 the difference is 36 . If we wish to make the difference less 1 than 36 , instead of 729 we shall take the number now found, 1 720 36 , and by the same method we shall ﬁnd an approxima- 1 tion differing by much less than 36 . Heron, Metrica, i.8.

Let a be a given positive integer. Consider the sequence (an) deﬁned recursively by 1 a an = an 1 + , (1.1) 2 − an 1 − 102 Extraction of square roots

with an arbitrary positive initial value a1. The sequence (an) converges to the square root of a regardless of the positive initial value.

Example 1.1. Approximations of √2 with initial values a =1:

n an 1 1/1 1 2 3/2 1.5 3 17/12 1.4166666666666 4 577/408 1.4142156862745 ··· 5 665857/470832 1.4142135623746 ··· 6 886731088897/627013566048 1.4142135623730 ··· ···

Example 1.2. Approximations of √3 with initial value a1 =2:

n an 1 2 2 2 7/4 1.75 3 97/56 1.73214285714286 4 18817/10864 1.73205081001473 ··· 5 708158977/408855776 1.73205080756888 ··· 6 1002978273411373057/579069776145402304 1.73205080756888 ··· ···

Convergence

Why does the sequence (an) converge to √a? Note that 1 (i) with any arbitrary positive a1, all subsequent an are greater than √a, (ii) (an) is a decreasing sequence:

an 1 a 1 = 1+ 2 < (1+1)=1. an 1 2 an 1 2 − −

Since a decreasing sequence bounded from below converges, we write ℓ = limn xn. From (1.1), we have →∞ 1 a ℓ = ℓ + . 2 ℓ Solving this equation, we have ℓ = √a.

1This follows from the arithmetic-geometric means inequality. 103

2 2 1 a an a = an 1 + a − 4 − an 1 − − 1 a 2 = an 1 4 − − an 1 − 2 2 (an 1 a) = − 2− 4an 1 − 2 2 (an 1 a) < − − . 4a

Therefore, beginning with a1, by iterating n steps, we obtain a2,..., an+1, with

1 a2 a< (a2 a)2 n+1 − 4a · n − 1 1 2 4 < (an 1 a) 4a · (4a)2 − − . .

1 1 1 2 2n < − (a a) 4a · (4a)2 ··· (4a)2n 1 1 −

1 2 2n = 2n 1 (a1 a) (4a) − −

This shows that the convergence is very fast. For example, for a =2, if we begin with a1 = 2 and execute n steps, then we have a rational approximation an+1 satisfying

2 1 an+1 2 < 2n 3 . − 2 −

To guarantee an accuracy up to, for example, 10 decimal digits be- 1 yond the decimal point, we need to make the error < 1010 . This requires 2n 3 10 2 − > 10 , and it is enough to iterate 5 times. 104 Extraction of square roots

Reorganization It is more convenient to calculate the numerators and denominators sep- arately. If we write a = xn , then n yn

2 2 xn 1 xn 1 ayn 1 xn 1 + ayn 1 = an = − + − = − − . yn 2 yn 1 xn 1 2xn 1yn 1 − − − −

This leads to two sequences of integers (xn) and (yn) deﬁned recursively

2 2 xn = xn 1 + ayn 1, − − yn = 2xn 1yn 1, − − with arbitrary initial values x1 and y1.

Exercise (1) Find a rational approximation r of √5 satisfying r2 3 < 1 . | − | 1030 (2) Construct two integer sequences (x ) and (y ) such that the xn n n yn 2 converges to the square root of 3 . (3) Let a be a positive integer. Consider a sequence (an) deﬁned recursively by 1 a an = an 1 . 2 − − an 1 − (a) Show that the sequence does not converge regardless of the initial value. (b) Show that for every given positive integer ℓ, an initial value can be chosen so that the sequence is periodic with period ℓ. [Hint: rewrite the √a recursive relation by putting an 1 = ]. − tan θ (4) Start with two positive numbers a0 and a0, and iterate according to the rule an = √an 1 + √an 2. − − Does the sequence (an) converge? If so, what is the limit? (No proof is required).

Irrationality

Why is √2 an irrational number? Here are two simple proofs. 105

√ √ p (1) If 2 is rational, we write 2= q in lowest terms, then a square of side p has the same area as two squares of side q, and for this, the p p square is smallest possible. ×

p Yet this diagram shows that there is a smaller square which also has the same area as two squares. This contradicts the minimality of p. (Ex- ercise: what are the dimensions of the smaller squares involved?) (2) If √2 is rational, then there is a smallest integer q for which q√2 is an integer. Now, the number (q√2 q)√2 is a smaller integer multiple of √2, which is also an integer. This− contradicts the minimality of q.

Irrationality of √k n Here is a simple proof that for a positiveinteger n, √n is either irrational 2 p or integral. If √n = q in lowest terms, then both q√n = p and p√n = qn are integers. Since there are integers a and b satisfying ap + bq =1, we have √n = a pn + b qn, an integer. More generally,· for given· positive integers n and k, the number √k n is irrational unless n is the k-power of an integer. This fact follows easily from the fundamental theorem of arithmetic: every positive integer > 1 is uniquely the product of powers of distinct prime numbers. 3

The ladder of Theon

Consider an integer a> 1 which is not the square of an integer. Let b be the integer closest to √a (so that b √a < 1 ). | − | 2 2M. Levin, The theorem that √n is either irrational or integral, Math. Gazette, , 60 (1976) 138. Levin p2 has subsequently (ibid., 295) given an even shorter proof: Since the reduced fraction q = p√n = qn is an integer, q must be equal to 1, and √n is an integer. 3See the supplements to Chapter 1 for a more elementary proof. 106 Extraction of square roots

n Now, for each positive integer n, (b + √a) = xn + yn√a for positive integers xn and yn. The sequences (xn) and (yn) can be generated recursively by

xn = bxn 1 + ayn 1, − − yn = xn 1 + byn 1, − − with x1 = b, y1 =1. 1 1 (1) Since b √a < 2 , xn yn√a < 2n . | − | | − | n (2) Since xn and yn are positive integers, yn > b and x 1 1 n √a < < . y − 2ny (2b)n n n

xn This shows that lim n = √a. →∞ yn Chapter 2

Convergence of sequences

A sequence (an) is said to converge to a number a if ultimately the terms of the sequence are as close to a as we like. This means that for any given ε> 0, there exists an integer N such that

a a <ε whenever n>N. | n − | We say that a is the limit of the sequence and write

lim an = a. n →∞

1 1 Example 2.1. (a) The sequence converges to 0: limn = 0. n →∞ n While this is intuitively clear, here is a proof. Given ε> 0, let N = 1 . ⌈ ε ⌉ For n > N, we have n> 1 1 . This means that 1 0 = 1 <ε. ⌈ ε ⌉ ≥ ε n − n Proposition 2.1. If limn an = A and limn bn = B, then →∞ →∞ (a) limn (an bn)= A B, →∞ ± ± (b) limn anbn = AB, →∞ an A (c) limn = , provided bn and B are nonzero. →∞ bn B Proof. A convergent sequence is bounded

Example 2.2. Let f(x) be a polynomial in x. If (an) converges to a, then (f(an)) converges to f(a). A sequence (an) is (i) bounded below if it has a lower bound A, i.e., an A for every positive integer n, ≥ 108 Convergence of sequences

(ii) bounded above if it has an upper bound B, i.e., an B for every positive integer n. ≤ The sequence is bounded if it has a lower bound and an upper bound. Otherwise, the sequence is unbounded, i.e., it exceeds any given positive number in absolute value. This means that given any M > 0, there exists an integer N > 0 such that a > M whenever n>N. | n|

Theorem 2.2. (a) An (ultimately) increasing sequence bounded above is convergent. Its limit is the least upper bound of the sequence. (b) A decreasing sequence bounded below is convergent. Its limit is the greatest lower bound of the sequence. Example 2.3. (a) Let a be a given real number. The sequence (an) is unbounded when a > 1. | | Proof. Without loss of generality assume a> 1, and write a =1+ b for b > 0. Note that an = (1+ b)n > 1+ nb. Therefore, given M > 0, let M 1 N be the least integer such that 1+ nb M, i.e., n b− . Then, n > M implies an > 1 + nb M,≥ showing that≥ the ⌈ sequence⌉ is unbounded. | | ≥ (b) The sequence (an) converges to 0 if a < 1. | | Proof. Again, it is enough to assume a > 0. Write a = 1 b for 0

1 1 1 Given ε> 0, let N = bε . Then, for n > N, n> bε and nb <ε. This n 1 ⌈ ⌉ n means a < <ε, and limn a =0. nb →∞ (c) The sequence (( 1)n) does not converge to any number, though it is bounded, and the subsequence− of odd numbered terms is a constant sequence, as is that of even numbered terms.

Example 2.4. Let (an) be a sequence such that (i) the subsequence of odd numbered terms converges to ℓ, and (ii) the subsequence of even numbered terms converges to the same ℓ. Then the sequence (an) converges to ℓ. 109

Proof. Given ε> 0, there are integers N1 and N2 such that (i) an ℓ <ε whenever n > N1 is even, (ii)| a − ℓ| <ε whenever n > N is odd. | n − | 2 Therefore, an ℓ <ε whenever n> max(N1, N2), showing that (an) converges to| ℓ.− |

Proposition 2.3. Let (an), (bn), and (cn) be sequences such that (i) ultimately 1 a b c , 2 n ≤ n ≤ n (ii) both sequences (an) and (cn) converge to ℓ. Then the sequence (bn) also converges to ℓ. 1 Example 2.5. limn 2 =0. →∞ √n +1 1 1 1 Proof. 0 < 2 < for every n. Since limn = 0, the result √n +1 n →∞ n follows from Proposition 2.3.

n (b) limn 2n =0. →∞ an (c) limn =0. →∞ n!

Inﬁnite series

An inﬁnite series

∞ a = a + a + + a + n 1 2 ··· n ··· n=1 X is convergent if the sequence of partial sums

s := a + a + + a n 1 2 ··· n is convergent. The sum of the series is the limit of the sequence of partial sums. Otherwise, it is divergent. For series of positive terms, we shall simply write an < when the series is convergent. ∞ (1) If the partial sums of a series of positive terms areP bounded, then the series converges. Proof: The sequence of partial sums, being monotonic increasing and bounded above, is convergent. (2) Comparison test for series of positive terms: Let an and bn be series of positive terms. P P 1This means that there is an integer N such that the double inequality holds whenever n > N. 2One or both of the inequalities may be replaced by <. ≤ 110 Convergence of sequences

(i) If a b ultimately and b is convergent, then so is a . n ≤ n n n (ii) If an bn ultimately and bn is divergent, then so is bn. ≥ P P n a ∞ Theorem 2.4. A geometric seriesP n=0 ar converges toP1 r provided r < 1. − | | P Example 2.6. 0.999 =1. ··· The decimal expansion of a number Every number between 0 and 1 can be written uniquely in decimal form:

0.a a a (2.1) 1 2 ··· n ··· where the digits a1, ..., an, . . . are integers between 0 and 9. This expression is indeed an inﬁnite series a a a 1 + 2 + + n + . 10 102 ··· 10n ···

Consider the sequence (sn) of partial sums deﬁned by a a a s := 1 + 2 + + n . n 10 102 ··· 10n It is easy to see that it is an increasing sequence:

s s s 1 ≤ 2 ≤···≤ n ≤··· It is clearly bounded above since each

9 9 9 9 s + + + + = 10 =1. n ≤ 10 102 ··· 10n ··· 1 1 − 10 This shows that (2.1) defines a unique nonnegative number 1. Conversely, given a number α [0, 1], we determine≤ its decimal expansion as follows. ∈ a1 a1+1 (i) a1 is the integer satisfying 10 α< 10 . ≤ n ak (ii) Having defined a1, ..., an, we put sn = j=1 10k and define an+1 as the integer satisfying P a a +1 n+1 α s < n+1 . 10n+1 ≤ − n 10n+1 Equivalently, a = 10n+1(α s ) . n+1 ⌊ − n ⌋ 111

Theorem 2.5. Between any two real numbers, there is a rational number. Proof. Let α > β be two given real numbers. (1) If α is rational, then with a sufficiently large n, the rational number 1 α n is strictly between α and β. − an ∞ (2) If α is irrational, with a decimal expansion a0 + n=1 10n . There 1 is a sufficiently large N such that 10N < α β. The rational number N an − P n=1 10n is strictly between α and β since P N a ∞ a ∞ 9 1 α n = n = < α β. − 10n 10n ≤ 10n 10N − n=1 n=N+1 n=N+1 X X X N an This means that α> n=1 10n > β. P Exercise (1) Prove that between any two real numbers, there is an irrational number. p (2) The decimal expansion of a rational number q (in lowest terms) is finite if and only if the prime divisors of q are 2 and/or 5. (3) Why is the decimal expansion of a rational number periodic? 112 Convergence of sequences Chapter 3

Square roots by decimal digits

Suppose a positive integer a is given in decimal form. To ﬁnd its square root, divide the digits of a into blocks of two digits beginning with the right hand side. We represent this as

a a a 1 2 ··· n

where each ak, k = 2,...,n is a 2-digit number, and a0 has either 1 or 2 digits. (1) Set b1 := a1, and let q1 be the largest integer such that r1 := a q2 0. 1 − 1 ≥ Set Q1 := q1. (2) Suppose bk, rk and Qk have been deﬁned. Form

bk+1 := 100rk + ak+1.

Find the largest integer qk+1 such that r := b (20Q + q )q 0. k+1 k+1 − k k+1 k+1 ≥

Set Qk+1 := 10Qk + qk+1. (3) Repeat (2) to ﬁnd b1, b2,..., bn. Then b1b2 bn = √a . The calculation··· may⌊ be continued⌋ beyond the decimal points by adding pairs of zeros. 114 Square roots by decimal digits

2 5 5 8 6 6 8 5 0 6 5 4 6 7 8 4 5 3 6 1 2 3 4 5 6 7 4 45 2 5 4 2 2 5 505 2 9 6 7 2 5 2 5 5108 4 4 2 8 4 4 0 8 6 4 51166 3 4 2 0 5 3 3 0 6 9 9 6 511726 3 5 0 5 7 6 1 3 0 7 0 3 5 6 5117328 4 3 5 4 0 5 2 3 4 0 9 3 8 6 2 4 51173365 2 6 0 1 8 9 9 4 5 2 5 5 8 6 6 8 2 5 511733700 4 3 2 3 1 2 0 6 7

4 3 2 3 1 2 0 6 7 Chapter 4

a + a + + √a + · · · · · · q p Let a> 0 be a given number. What is the number

a + a + + √a + ? ··· ··· r q If this expression defines a number ℓ, it must satisfy ℓ = √a + ℓ, and so is the positive root of the quadratic equation x2 x a =0, namely, 1 √ − − ℓ = 2 (1+ 4a + 1). We justify the existence of the number ℓ as the limit of a convergent sequence (xn) of positive numbers defined recursively by xn+1 = √a + xn, (4.1) regardless of the (positive) initial value. For this, it is helpful to write 2 x x a =(x ℓ)(x + ℓ′) for ℓ,ℓ′ > 0. − − − (1) Suppose xn <ℓ. Then 2 2 2 2 (i) xn+1 ℓ = a + xn ℓ < a + ℓ ℓ =0, so that xn+1 <ℓ; (ii) x2 − x2 = a +−x x2 = −(x ℓ)(x + ℓ) > 0, so that n+1 − n n − n − n − n xn+1 > xn. Therefore, if x1 < ℓ, then the sequence (xn) is monotonic increasing, and is bounded above. The sequence necessarily converges to ℓ. (2) Suppose xn >ℓ. Then 2 2 2 2 (i) xn+1 ℓ = a + xn ℓ > a + ℓ ℓ =0, so that xn+1 >ℓ; (ii) x2 −x2 = a+x −x2 = (x −ℓ)(x +ℓ) < 0, so that x < x . n+1 − n n − n − n − n n+1 n Therefore, according as x1 > ℓ or < ℓ, the sequence (xn) is monotonically decreasing and bounded below, or monotonically increasing and below above. It necessarily converges to ℓ. 116 a + a + + √a + · · · · · · q p (3) Suppose x1 = ℓ. Then every xn = ℓ. Since the sequence (xn) converges, its limit ℓ satisfies, according to (4.1), 1 ℓ = √a + ℓ. Here are two examples for a =5 with different initial values.

n xn xn 1 1 3 2 2.449489743 2.828427125 3 2.729375339 ··· 2.797932652 ··· 4 2.780175415 ··· 2.79247787 ··· 5 2.789296581 ··· 2.791501007··· 6 2.790931131 ··· 2.79132603 ··· 7 2.791223949 ··· 2.791294687··· 8 2.791276401 ··· 2.791289073 ··· 9 2.791285797 ··· 2.791288067 ··· 10 2.79128748 ··· 2.791287887 ··· 11 2.791287782··· 2.791287855 ··· 12 2.791287836 ··· 2.791287849 ··· 13 2.791287845 ··· 2.791287848 ··· 14 2.791287847 ··· 2.791287848 ··· 15 2.791287847 ··· 2.791287847 ··· ··· ··· √5+1 Example 4.1. For a =1, this limit is the golden ratio ϕ = 2 .

Exercise (1) Let a and b be given positive numbers. Why does the expression

a + b + a + √b + s ··· r q (in which a and b alternate indeﬁnitely) deﬁne a real number?

(2) Identify the numbers

1+ 7+ 1+ √7+ s ··· r q and

7+ 1+ 7+ √1+ . s ··· r q 1It is important to justify the existence of the limit before passing an recurrence relation to the limit. Here is a counterexample. Let (xn) be deﬁned by xn+2 = xn+1 + xn with x1 = x2 = 1. The sequence certainly is unbounded and does not converge. The terms are the Fibonacci numbers. If we assume it converges to a limit ℓ, then ℓ = ℓ + ℓ, and ℓ = 0, an impossibility. Chapter 5

An inﬁnite continued fraction

What is the number 1 a + 1 a + 1 a + . .. for a given a> 0? If we regard this as the limit ℓ of a sequence of fractions, where the n-th term is the fraction 1 a + 1 a + 1 a + . .. 1 + a

1 √a2+4 a with n copies a’, then ℓ = a + ℓ , and ℓ = 2 − . As noted in the preceding example, we need to justify the existence of the limit. The sequence in question can be deﬁned recursively as 1 xn+1 = a + , x1 = a. xn

It is possible to work out an explicit expression for xn, by writing it in the form x = yn for another sequence (y ). The recurrence relation n yn−1 n becomes

yn+1 = ayn + yn 1, y0 =1, y1 = a. − 118 An inﬁnite continued fraction

This is a second order homogeneous linear recurrence, with character- 2 istic equation λ = aλ +1. Denote by λ1 and λ2 the two characteristic values. Note that λa + λ2 = a and λ1λ2 = 1. Since their sum is −1 positive, we may write λ1 = λ> 1, and λ2 = −λ < 0. We have y = A λn + B λn n · 1 · 2 for appropriately chosen constants A and B. From the initial values, we have

A + B = 1,

Aλ1 + Bλ2 = a. Solving these, we have a λ λ λ a λ A = − 2 = 1 , B = 1 − = − 2 , λ λ λ λ λ λ λ λ 1 − 2 1 − 2 1 − 2 1 − 2 and λn+1 λn+1 y = 1 − 2 . n λ λ 1 − 2 It follows that λn+1 λn+1 λ2n+2 ( 1)n+1 x = 1 − 2 = − − . n λn λn λ(λ2n + 1) 1 − 2 Since λ> 1, it is clear that the limit is λ.

√5+1 Example 5.1. For a =1, this limit is (again) the golden ratio ϕ = 2 .

Exercise

(1) For which values of x1 will (xn) converge, where 2 xn+1 =3 ? − xn

(2) Show that regardless of initial value, the sequence (xn) deﬁned recursively by 1 xn+1 =1 − xn 119

does not converge. (3) Let λ (0, 1) be a ﬁxed number. Given a, b, deﬁne a sequence ∈ (xn) recursively by

x1 = a, x2 = b, xn = λxn 1 + (1 λ)xn 2. − − −

Show that the sequence (xn) converges and ﬁnd its limit. 120 An inﬁnite continued fraction Chapter 6

Ancient Chinese calculations of π

The ancient Chinese classics the Nine Chapters of the Mathematical Art adopted the formula

Area of circle = product of half-circumference and half-diameter with the rule “diameter one, circumference 3”. The third century com- mentator LIU Hui pointed out the inadequacy of this rule, and explained the mensuration of the circle by the method of dissection. Consider a circle of radius R. Denote by A its area. Inscribe in the circle a regular polygon of n sides, each of length an. For the regular n-gon, denote by (i) pn the perimeter, (ii) An the area, (iii) dn the distance from the center to a side, and (iv) c = R d . n − n Beginning with a6 = 1, LIU Hui ﬁrst computed a12, making use of 2 the right triangle theorem: d = R2 a6 , c = R d , and 6 − 2 6 − 6 q 2 a 2 a 2 a 2 a = c2 + 6 = R R2 6 + 6 . 12 6 v r 2 u − r − 2 ! 2 u t 122 Ancient Chinese calculations of π

a12 c6

d 6 R

n an dn cn pn = n an 61 0.8660254 0.1339746 6 · 12 0.517638 0.9659258 0.0340742 6.211656

More generally,

2 a 2 a 2 a = R R2 n + n . (6.1) 2n v u − r − 2 ! 2 u t LIU Hui iterated the process several times and obtained

n an dn cn pn = n an 61 0.8660254 0.1339746 6 · 12 0.517638 0.9659258 0.0340742 6.211656(a) 24 0.261052 0.9914448 0.0085552(b) 6.265248(c) 48 0.130806 0.9978589 0.0021411 6.278688(d) 96 0.065438 6.282048(e)

1 2 He actually recorded the values of a2n and extracted their square roots to ﬁnd a2n:

1The rounding off of the 6th digit after the decimal point is not correct. To 10 places, these are (a) 6.2116570824; (b) 0.0085551386; (c) 6.2652572265; (d) 6.2787004060; (e) 6.2820639017. 123

2 2431 a12 = 0.267949193445 (0.26794919 ) 2 7421 ··· a24 = 0.068148349466 (0.06814834 ) 2 7252 ··· a48 = 0.017110278813 (0.01711027 ) a2 = 0.004282154012 (0.004282153522 ··· ) 96 ··· 1 The area of the polygon can be computed as An = n 2 an dn. Indeed n · · the area of the regular 2n-gon is A2n = 2 an R. LIU Hui made use of the following inequality to estimate the area· of· the circle: A

Now that A A = n 1 a c , and A =6 1 R2 =3R2. 2n − n · 2 · n · n 12 · 2

n an A2n An A2n A2n +(A2n An) 61− 3 − 12 0.517638 0.105828 3.105828 3.211656 24 0.261052 0.026796 3.132624 3.159420 48 0.130806 0.006720 3.139344 3.146064 96 0.065438 0.001680 3.141024 3.142704

From these, LIU Hui concluded that the area of the circle is 3.14 correct to two places of decimal. In the ﬁfth century, ZU Chongzi (429–500) gave the value of π correct to 6 decimal places, as between 3.1415926 and 3.1415927. His 124 Ancient Chinese calculations of π manuscript Sushu was lost. But it is generally believed that he carried out LIU Hui’s program further to regular polygons of sides 6 211 = 12288 and 6 212 = 25576: · ·

n an A2n An A2n A2n + (A2n An) − − 192 0.0327234632 0.0004205213 3.1414524722 3.1418729936 384 0.0163622792 ··· 0.0001051356 ··· 3.1415576079 ··· 3.1416627435 ··· 768 0.0081812080 ··· 0.0000262842 ··· 3.1415838921 ··· 3.1416101763 1536 0.0040906125 ··· 0.0000065710 ··· 3.1415904632 ··· 3.1415970343 3072 0.0020453073 ··· 0.0000016427 ··· 3.1415921059 ··· 3.1415937487 ··· 6144 0.0010226538 ··· 0.0000004106 ··· 3.1415925166 ··· 3.1415929273 ··· 12288 0.0005113269 ··· 0.0000001026 ··· 3.1415926193 ··· 3.1415927220 ··· ··· ··· ··· ··· 22 Zu also gave the crude approximation 7 and the ﬁne approximation 355 113 for π. Chapter 7

Existence of π

Area of circle in Euclid’s Elements

Eucl. X.I

Two unequal magnitudes being set out, if from the greater there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continu- ally, there will be left some magnitude which will be less than the lesser magnitude set out.

Reformulation

Let (a ) be a sequence of real numbers satisfying a < 1 a for every n n+1 2 · n n. Given ε > 0, there exists an integer N such that an < ε whenever n > N.

Exercise

(1) In Eucl. X.1, if each occurrence of the word “half” is replaced by “one third”, is the proposition still valid? (2) Is the proposition still valid without specifying that the magnitude subtracted each time be greater than a certain proportion of the magnitude (from which it is subtracted)? 126 Existence of π

Eucl. XII.1 Similar polygons inscribed in circles are to one another as the squares on the diameters.

Eucl. XII.2 Circles are to one another as the squares on their diameters. Proof. If C(d) denotes the area of a circle, diameter d, then for any 2 2 d1,d2, C(d1) : C(d1)= d1 : d2. E

K N

F H

M L

G (i) Let C be the area of a circle. Suppose there is an inscribed n-gon of area An, each arc smaller than a semicircle. Bisecting the arcs, one obtains an inscribed 2n-gon, of area A . Then, C A < 1 (C A ). 2n − 2n 2 − n 1 EKF > segment EKF △ 2 1 = sum of shaded segments > segment EKF. ⇒ 2 (ii) Starting from an inscribed square, one can approximate, by repeated bisection of arcs, the area C arbitrarily closely: given any positive ǫ, there is an inscribed regular n-gon for which C An < ǫ. − 2 (iii) Given two circles of diameters d1,d2, Euclid shows that if d1 : 2 d2 = C(d1) : X, then X cannot be smaller than C(d2). Suppose to the contrary that X X. Now, in the circle C(d1), construct a (similar) regular n-gon of area 2 2 An. Euclid has shown in XII.1 that d1 : d2 = An : An′ . It follows that 2 2 An : An′ = d1 : d2 = C(d1) : X, and An : C(d1)= An′ : X. But this is a contradiction since An X. 127

2 2 (iv) Suppose now d1 : d2 = C(d1) : X with X >C(d2). We 2 2 can rewrite this as d1 : d2 = Y : C(d2) with Y

Exercise (3) A circle is approximated by a regular octagon obtained by cutting out corners from its circumscribed square. What is the approximate value of π? 128 Existence of π Chapter 8

Archimedes’ calculation of π

Archimedes, in his Measurement of a Circle, gave the following interesting bounds for π: 1 10 3 >π> 3 . 7 71 To obtain the upper bound, Archimedes began with a regular hexagon circumscribing the circle, doubling the number of sides and determined the length of a side of a circumscribed regular polygon of 6 2n+1 sides in terms of one of 6 2n sides. He did the same thing for inscribed· polygons to obtain the lower· bound.

′ A

C D X O ∠ 1 Consider a tangent AC to a circle, center O such that AOC is 2k of the circle, k = 6 2n, so that AC is one half of the length of a side of circumscribed regular· polygon of k sides. If the bisector of angle AOC intersects AC at A′, then A′C is one half of a side of a circumscribed regular polygon of 2k sides. If OA intersects the circle at B, and X the projection of B on OC, then BC is a side of an inscribed regular polygon of 2k sides, and BX 130 Archimedes’ calculation of π is one half of a side of an inscribed polygon of k sides. n Let an and bn be the perimeters of regular polygons of k = 6 2 sides, respectively circumscribed and inscribed in the circle. We have·

an an+1 bn+1 bn AC = , A′C = , BC = , BX = . 2k 4k 2k 2k

(1) Since OA′ bisects angle AOC,

A′C : AC = OC : OA + OC = OB : OA + OB = BX : AC + BX. This gives an+1 bn 4k = 2k . an an bn 2k 2k + 2k From this, we have 2anbn an+1 = . an + bn

2 (2) Since BC2 = CX CD, we have CX = BC . Also, from the · CD similarity of triangles OA′C and BCX, we have A C CX BC2 ′ = = , OC BX CD BX and · 2 2 2 an+1 OC BC BC bn+1 = A′C = · = = . 4k CD BX 2 BX 4k bn 2 · · · Therefore, bn+1 = an+1bn. If we take the diameter of the circle to be 1, then a0 = 2√3 and b0 =3. The sequences (an) and (bn) deﬁned recursively by

2anbn an+1 = , bn+1 = an+1bn an + bn both converge to the circumference of the circle,p namely, π.

n n 6 2 an bn · 0 6 3.464101615 3 1 12 3.21539 3.105828541 2 24 3.15966 3.132628613 3 48 3.14609 3.139350203 4 96 3.14271 3.141031951 5 192 3.14187 3.141452472 This was the basis of practically all calculations of π before the 16th century. Correct to the ﬁrst 34 decimal places, π =3.1415926535897932384626433832795028 ······ Chapter 9

Ptolemy’s calculations of chord lengths

Ptolemy Theorem In a cyclic quadrilateral, the sum of the products of two pairs of opposite sides is equal to the product of the diagonals.

B A

E O

C D

Proof. Choose a point E on the diagonal AC such that ∠ADE = ∠BDC. Then triangles ADE and BDC are similar, and AD BD = AD BC = AE BD. AE BC ⇒ · · Also, triangles ABD and ECD are similar, and AB EC = AB CD = EC BD. BD CD ⇒ · · Therefore, AD BC + AB CD =(AE + EC) BD = AC BD. · · · · 132 Ptolemy’s calculations of chord lengths

Calculation of chord lengths Ptolemy divided the circumference of the circle into 360 equal arcs, and the diameter into 120 parts, and expressed fractions of these parts in the sexagesimal system.

x 180◦ 60◦ 90◦ 120◦ p p p p crd(x) 120 60 84 51′10′′ 103 55′23′′

Clearly, by the Pythagorean theorem,

crd(x)2 + crd(180 x)2 =4. −

B D

A C

x y x 2 O O x 2

D A

Applying Ptolemy’s theorem, one has 1 crd(x y)= [crd(x)crd(180 y) crd(180 x)crd(y)] . ± 2 − ± −

In particular,

crd(2x) = crd(x)crd(180 x). − Also, x crd = 2 crd(180 x). 2 − − p Ptolemy then made use of these to determine crd(1◦).

p 1. Calculations with a regular pentagon gave crd(72◦)=70 32′3′′. p 2. crd(12◦) = crd(72◦ 60◦)= = 12 32′36′′. − ··· 133

1 3. crd(6◦)= crd 12◦ = ; then crd(3◦), and 2 · ··· 1 ◦ p crd 1 = 1 34′15′′, 2 3 ◦ crd = 47′8′′. 4

4. To compute crd(1◦), Ptolemy made use of an interpolation based on an inequality crd(x) x < crd(y) y

for arcs x>y smaller than a quadrant of a circle. 3 ◦ 3 (i) crd(1◦) : crd 4 < 1 : 4 , (ii) crd 1 1 ◦ : crd(1 ) < 1 1 :1. 2 ◦ 2 Therefore,

4 3 ◦ 2 1 ◦ crd > crd(1◦) > crd 1 . 3 4 3 2

p 2 ′′ p The two ends are respectively 1 2′50 3 and 1 2′50′′. Since crd(1◦) p is both less and greater than a length which differs from 1 2′50′′ insigniﬁcantly, Ptolemy took this for crd(1◦).

1 ◦ From this Ptolemy deduced that crd 2 is very nearly 31′25′′. Mak- ing use of this and the above relations, he was able to complete this 1 ◦ Table of Chords for arcs subtending angles increasing from 2 to 180◦ 1 ◦ by increments of 2 .

Modern reformulation

1 If we take diameter of the circle to be 2, and write sin x = 2 crd(2x), 1 cos x = 2 crd(180◦ 2x) = sin(90◦ x), Ptolemy’s relations become our modern basic trigonometric− identities:− 134 Ptolemy’s calculations of chord lengths

sin(x y) = sin x cos y cos x sin y, ± ± cos(x y) = cos x cos y sin x sin y; ± ∓ sin 2x = 2 sin x cos x; cos2x = cos2 x sin2 x = 2 cos2 x 1=1 2 sin2 x; − − − x 1 cos x sin2 = − ; 2 2 x 1 + cos x cos2 = . 2 2 Chapter 10 lim sin θ =1 The basic limit θ 0 θ →

Theorem 10.1 (Ptolemy). For circular arcs x>y smaller than a quadrant of a circle, crd(x) x < crd(y) y

H A C E F

Proof. ([?, pp.ii 281–282]). With reference to the diagram above, crd(AB) < crd(BC), we prove that crd(CB) arc CB < . (10.1) crd(BA) arc BA Bisect angle ABC to intersect AC at E and the circumference of the lim sin θ =1 136 The basic limit θ→0 θ circle at D. The arcs AD and DC are equal, so are the chords AD and DC. Since CB : BA = CE; EA, we have CE >EA. Construct DF perpendicular to AC. Note that AD > DE > DF . Therefore, the circle, center D, radius F E intersects DA at G, and the extension of DF at a point H. Now, F E : EA = ∆FED : ∆AED < sector HED : sector GED < ∠F DE : ∠EDA. From this, F A : AE < ∠F DA : ∠ADE, componendo CA : AE < ∠CDA : ∠ADE, CE : EA < ∠CDE : ∠EDA; separando CB : BA < ∠CDB : ∠BDA since CB : BA = CE : EA < arc CB : arc BA. This establishes (10.1) above.

sin θ Theorem 10.2. limθ 0 =1. → θ Proof. Given a small arc of length θ on a unit circle of circumference C, 1 choose an integer n large enough so that the sector θ is between n and 1 2n of the circle: C C <θ< . 2n n By Ptolemy’s theorem,

s2n sin θ sn C > > C 2n θ n 2n s sin θ n s · 2n > > · n . C θ C As θ 0, n . Since the two ends of the double inequality have the → →∞ sin θ same limit 1 as n , we conclude that limθ 0 =1. →∞ → θ 1 cos θ Corollary 10.3. limθ 0 − =0. → θ Proof. This follows from 1 cos θ = 2 sin2 θ . − 2 2 θ 1 cos θ θ sin 2 − = θ . θ 2 · 2 ! 137

As θ 0, the two factors have limits 0 and 1 respectively. Hence the limit the→ product is 0.

1 cos θ 1 Corollary 10.4. limθ 0 − 2 = . → θ 2

Differentiation of sine and cosine

d(sin x) sin(x + h) sin x = lim − h 0 dx → h sin x cos h + cos x sin h sin x = lim − h 0 h → sin h 1 cos h = lim cos x sin x − h 0 · h − · h → sin h 1 cos h = cos x lim sin x lim − h 0 h 0 · → h − · → h = cos x 1 sin x 0 · − · = cos x.

Exercise Prove that d(cos x) = sin x. dx − lim sin θ =1 138 The basic limit θ→0 θ Chapter 11

The Arithmetic - Geometric Means Inequality

A.M. G.M. H.M. ≥ ≥ Lemma 11.1. For positive numbers a b, ≤ 2ab a + b a √ab b; ≤ a + b ≤ ≤ 2 ≤ equality holds if and only if a = b.

A G

a P O b Proof. In the diagram above, a + b 2ab OA = , PG = √ab, PH = . 2 a + b Clearly, a PH PG OA b. Two of these are equal if and only the corresponding≤ ≤ points ≤A, G, ≤H coincide. This happens only when a = b. 140 The Arithmetic - Geometric Means Inequality

For n given positive numbers a1, a2,..., an, we deﬁne 1 (i) the arithmetic mean A(a1, a2,...,an) := n (a1 + a2 + + an), n ··· (ii) the geometric mean G(a1, a2,...,an) := √a1 a2 an, and (iii) the harmonic mean · n ··· . H(a1, a2,...,an) := 1 + 1 + + 1 a1 a2 ··· an Note that 1 1 1 1 = A , , , . H(a , a , , a ) a a ··· a 1 2 ··· n 1 2 n

Theorem 11.2 (AGI). For positive numbers a1, a2,..., an, A(a , a ,...,a ) G(a , a ,...,a ). 1 2 n ≥ 1 2 n Equality holds if and only if a = a = = a . 1 2 ··· n Corollary 11.3. For positive numbers a1, a2,..., an, A(a , a ,...,a ) G(a , a ,...,a ) H(a , a ,...,a ). 1 2 n ≥ 1 2 n ≥ 1 2 n Any two of these means are equal if and only if a = a = = a . 1 2 ··· n Lemma 11.1 is the arithmetic-geometric means inequality for 2 positive numbers. We present several proofs of Theorem 11.2. We give the proof of AGI by A. L. Cauchy (1789–1857). 1

Cauchy’s proof Consider the statements P(n) for n =1, 2,... :

a1+a2+ +an P(n) : ··· √n a a a for a ,...,a 0. n ≥ 1 2 ··· n 1 n ≥

(1) P(2) is valid by Lemma 11.1. (2) Assuming the validity of P(2k), we proceed to validate P(2k+1). This means that for nonnegative numbers

a1, a2,...,a2k , a2k+1, a2k+2, ...,a2k+1 , we have to show that

a + a k + a k + + a k+1 1 2 2 +1 2 2k+1 ··· ··· √a1 a k a k a k+1 . 2k+1 ≥ ··· 2 2 +1 ··· 2 1See the supplements for other proofs. 141

Now,

a + a k + a k + + a k+1 1 ··· 2 2 +1 ··· 2 2k+1 1 a + a k a k + + a k+1 = 1 ··· 2 + 2 +1 ··· 2 2 2k 2k 1 k 2 2k k √a1 a k + √a k a k+1 by assumption of P(2 ) ≥ 2 ··· 2 2 +1 ··· 2 1 k 2 2k 2 √a a k √a k a k+1 by P(2) ≥ 1 ··· 2 · 2 +1 ··· 2 2k+1 = √a1 a k a k a k+1 . ··· 2 2 +1 ··· 2 Therefore, we have established P(2k) for all integers k 1. (3) Assume P(n). We have to validate P(n 1).≥ This means that − given n 1 nonnegative numbers a1, ..., an 1 with arithmetic mean − − An 1 and geometric mean Gn 1, we have to show that An 1 Gn 1. − − − ≥ − In order to apply P(n), we need an extra number an. For this, we take an = An 1. Note that the arithmetic mean is −

(n 1)An 1 + An 1 An = − − − = An 1, n − and the geometric mean is

n n 1 Gn = Gn−1An 1. − − n q n 1 n 1 n 1 Since An Gn, we have An 1 Gn−1An 1. From this, An−1 Gn−1, ≥ − ≥ − − − ≥ − and An Gn. Equality holds if and only if a1 = a2 = = an 1 = ≥ ··· − An 1. −With these, we claim that P(n) is valid for every integer n 1. In k 1 k ≥ fact, given n, there is an integer k such that 2 − < n 2 . Note that P(n) is valid by (2) above. If n = 2k, then P(n) is valid.≤ Otherwise, applying (3) 2k n times, we have − P(2k) P(2k 1) P(n + 1) P(n). ⇒ − ⇒···⇒ ⇒

Exercise We say that a function f deﬁned on a closed interval [a, b] is concave if f x+y f(x)+f(y) for x, y [a, b]. 2 ≥ 2 ∈ 142 The Arithmetic - Geometric Means Inequality

(1) Show that the logarithm function is concave on (0, ). 2 (2) Show that the sine function is concave on [0, π]. ∞ (3) Let f be a concave function on [a, b]. Show that for arbitrary x , x ,...,x [a, b], 1 2 n ∈ x + x + + x f(x )+ f(x )+ + f(x ) f 1 2 ··· n 1 2 ··· n . n ≥ n

2The base of logarithm may be taken to be any a> 1. Chapter 12

Principle of nested intervals

Let (xn) and (yn) be two sequences of numbers satisfying (i) (xn) is increasing (non-decreasing), (ii) (yn) is decreasing (non-increasing), and (iii) every xn is smaller than every ym. If the difference xn yn can be made arbitrarily small, then the two sequences converge| to a− common| limit. The conditions (i), (ii), (iii) can be reformulated as x x x y y y , 1 ≤ 2 ≤··· n ≤···≤ n ≤··· 2 ≤ 1 or as a sequence of nested intervals [x ,y ] [x ,y ] [x ,y ] 1 1 ⊇ 2 2 ⊇···⊇ n n ⊇···

Leibniz test for convergence of alternating series

Theorem 12.1. If (an) isa decreasing sequence of positive real numbers n converging to 0, then the alternating series ∞ ( 1) a converges. n=0 − n Proof. Since the sequence (an) is decreasing,P the partial sums sn = n ( 1)ka satisfy k=0 − k P s1

These give a sequence of nested intervals

[s , s ] [s , s ] [s , s ] . 1 0 ⊃ 3 2 ⊃···⊃ 2n+1 2n ⊃···

Note that s2n s2n+1 = a2n+1. Since (an) converges to 0, then the nested intervals− deﬁne a unique real number s, which is the sum of the alternating series. Thus, for example, the alternating harmonic series 1 1 1 1 1 1 + + + − 2 3 − 4 5 − 6 −··· 1 converges since the sequence n decreases to 0. Remarks. (1) We shall show later that this series converge to log2. (2) Similarly, the alternating series 1 1 1 ( 1)n 1 + + + − + − 3 5 − 7 ··· 2n +1 ··· π also converges. The sum is 4 .

Arithmetic-harmonic mean

Consider the sequences (xn) and (yn) deﬁned by

2xnyn xn + yn xn+1 = , yn+1 = , xn + yn 2

with initial values x1 = a, y1 = b. Note that xn+1 and yn+1 are respectively the harmonic and arithmetic means of xn and yn. If we begin with two positive x1 y1, this recurrence deﬁnes a sequence of nested intervals ≤ [x ,y ] [x ,y ] [x ,y ] . 1 1 ⊇ 2 2 ⊇···⊇ n n ⊇···

Since

2 xn + yn 2xnyn (yn xn) yn xn yn xn yn+1 xn+1 = = − = − − − 2 − xn + yn 2(xn + yn) yn + xn · 2 y x n − n , (12.1) ≤ 2 145

it is clear that limn (yn xn)=0. Therefore, the nested intervals →∞ − deﬁne a unique real number as the common limit of (xn) and (yn). Here are two examples. (a) With x1 =1, y1 =5; limn xn = limn yn = √5. →∞ →∞

n xn yn 1 1 5 2 1.6666666666666666666666666666 3 3 2.1428571428571428571428571428 ··· 2.3333333333333333333333333333 4 2.2340425531914893617021276595 ··· 2.2380952380952380952380952381 ··· 5 2.2360670593565926597190756683 ··· 2.2360688956433637284701114488 ··· 6 2.2360679774996011987237537947 ··· 2.2360679774999781940945935585 ··· 7 2.2360679774997896964091736607 ··· 2.2360679774997896964091736766 ··· 8 2.2360679774997896964091736687 ··· 2.2360679774997896964091736687 ··· ··· ···

(b) With x1 =2, x2 =5; limn xn = limn yn = √10. →∞ →∞

n xn yn 1 2 5 2 2.857142857142857142857142857142 3.5 3 3.146067415730337078651685393258 ··· 3.178571428571428571428571428571 4 3.162235898737389759533024554279 ··· 3.162319422150882825040128410915 ··· 5 3.162277659892622371735257166789 ··· 3.162277660444136292286576482597 ··· 6 3.162277660168379331986870264172 ··· 3.162277660168379332010916824693 ··· 7 3.162277660168379331998893544432 ··· 3.162277660168379331998893544432 ··· ··· ···

Exercise Find this common limit in terms of a and b. Justify your answer.

The arithmetic-geometric mean

Given two positive numbers a < b, let x1 = a, y1 = b, and x + y x = √x y , y = n n . n+1 n n n+1 2 Then 2 2 xn + yn (√yn √xn) (yn xn) yn+1 xn+1 = √xnyn = − = − 2 − 2 − 2 2(√yn + √xn) (y x )2 y x y x y x < n − n = n − n n − n < n − n . (12.2) 2(yn + xn) yn + xn · 2 2

Therefore, the sequence of nested intervals [xn,yn] deﬁnes a unique real number. This is called the arithmetico-geometric mean (agM) of a and b. 1 1C. F. Gauss (1777–1854) discovered the agM when he was 15 years old, and later its connections with the elliptic integrals. 146 Principle of nested intervals

Example 12.1. agM(1, 2)

n xn yn 1 1.0000000000000000000 2.0000000000000000000 2 1.4142135623730950488 ··· 1.5000000000000000000 ··· 3 1.4564753151219702609 ··· 1.4571067811865475244 ··· 4 1.4567910139395549462 ··· 1.4567910481542588926 ··· 5 1.4567910310469068190 ··· 1.4567910310469069194 ··· ··· ···

Example 12.2. agM(1, √2)

n xn yn 1 1.0000000000000000000 1.4142135623730950488 2 1.1892071150027210667 ··· 1.2071067811865475244 ··· 3 1.1981235214931201226 ··· 1.1981569480946342956 ··· 4 1.1981402346773072058 ··· 1.1981402347938772091 ··· 5 1.1981402347355922074 ··· 1.1981402347355922074 ··· ··· ··· Remark. The estimates (12.1) and (12.2) can be improved to explain the fast convergence in both cases. For arbitrary initial values a < b, the difference yn xn is ultimately less than 2a. We may as well assume b a< 2a. Now− we have − 2 (yn xn) 1 2 yn+1 xn+1 < − < (yn xn) . − 2(yn + xn) 4a · − Iterating, we have

2 1 1 2 yn+1 xn+1 < (yn 1 xn 1) − 4a · 4a · − − − < ··· n (b a)2 < − 2n 1 (4a) − n b a 2 = 4a − . · 2a Chapter 13

The number e

Nested intervals deﬁning e

1 n 1 n+1 Let an := 1+ n and bn := 1+ n . (1) Consider the n +1 numbers 1 1 1 1+ , 1+ , ..., 1+ , 1. n n n n

n(1+ 1 )+1 | n{z }1 Their arithmetic mean is n+1 = 1+ n+1 , and their product is 1 n 1 n+1 1 n 1+ n . By the AGI, 1+ n+1 > 1+ n , showing that the sequence (an) is increasing. (2) A similar reasoning applied to the n +1 numbers 1 1 1 1 , 1 , ..., 1 , 1 − n − n − n n

shows that the sequence| (bn) is decreasing.{z } (3) Since the increasing sequence (an) is bounded above by b1, it is convergent (to a ﬁnite number). Now, since b a = 1 a , we have n − n n · n limn (bn an)=0. →∞ − 1 n 1 n+1 Therefore, the nested intervals 1+ n , 1+ n deﬁne a unique number : e h i 1 n 1 n+1 e = lim 1+ = lim 1+ . n n n n →∞ →∞ 148 The number e

1 n 1 n+1 n (1 + n ) (1 + n ) 1 2. 4. 2 2.25 3.375 3 2.37037037 3.160493827 4 2.44140625 3.051757813 5 2.48832 2.985984 6 2.521626372 2.941897434 7 2.546499697 2.910285368 8 2.565784514 2.886507578 9 2.581174792 2.867971991 10 2.59374246 2.853116706 20 2.653297705 2.78596259 30 2.674318776 2.763462735 40 2.685063838 2.752190434 50 2.691588029 2.74541979 60 2.695970139 2.740902975 70 2.699116371 2.737675176 80 2.701484941 2.735253503 90 2.703332461 2.733369488 100 2.704813829 2.731861968 1000 2.716923932 2.719640856 10000 2.718145927 2.718417741 The convergence of the sequences is quite slow.

A faster calculation of e Here is another sequence of nested intervals deﬁning e, with much faster convergence.

n 1 n n 1 k 1+ = 1+ n k n k=1 ! Xn n(n − 1) · · · (n − k + 1) 1 = 1+ · k! nk k=1 Xn n(n − 1) · · · (n − k + 1) 1 = 1+ · nk k! k=1 X n 1 2 k − 1 1 = 1+1+ 1 − 1 − · · · 1 − · n n n k! k=2 n X 1 < 1+ k! k=1 Xn 1 1 − ( 1 )n 1 < 1+ =1+ 2 = 3 − < 3. 2k−1 − 1 2n−1 k 1 2 X=1

n 1 Let sn := k=0 k! . From the above calculation, it is clear that 1+ 1 n

~~1 2 k − 1 1+2+ · · · +(k − 1) k(k − 1) 1 − 1 − · · · 1 − > 1 − = 1 − . n n n n 2n ~~

~~This gives~~

~~1 n n k(k 1) 1 1+ >2+ 1 − n − 2n k! k=2 n X n 2 n 1 1 − 1 1 3 = > . k! − 2n k! k! − 2n Xk=0 Xk=0 Xk=0 Therefore, 3 1 n s < 1+~~

1 ∞ Theorem 13.1. e = n=0 n! . n 1 1 1 1 ∞ ∞ Note that e kP=0 k! = k=n+1 k! < k=n+1 2k−1 = 2k . This gives a much faster− calculation of e: P P P n 1+1+ 1 + + 1 2! ··· n! 1 2.0000000000000000000 2 2.5000000000000000000 3 2.6666666666666666667 4 2.7083333333333333333 5 2.7166666666666666667 6 2.7180555555555555556 7 2.7182539682539682540 8 2.7182787698412698413 9 2.7182815255731922399 10 2.7182818011463844797 11 2.7182818261984928652 12 2.7182818282861685639 13 2.7182818284467590023 14 2.7182818284582297479 15 2.7182818284589944643 16 2.7182818284590422591 17 2.7182818284590450705 18 2.7182818284590452267 19 2.7182818284590452349 20 2.7182818284590452353 150 The number e

~~Irrationality of e Theorem 13.2 (Euler). The number e is irrational.~~

a b 1 1 ∞ Proof. Suppose e = b =1+ k=1 k! + k=b+1 k! . Multiplying by b! and rearranging terms, we have an integer P P b 1 a(b 1)! b! 1+ − − k!! Xk=1 ∞ b! = k! kX=b+1 1 1 1 = + + + b +1 (b + 1)(b + 2) (b + 1)(b + 2)(b + 3) ··· 1 1 1 < + + + b +1 (b + 1)2 (b + 1)3 ··· 1 < < 1, b which is a contradiction. Chapter 14

~~Some basic limits~~

~~n Theorem 14.1. limn √a =1 for a> 0. →∞ Proof. Without loss of generality we may assume a > 1. Applying the AGI to the n numbers 1, ..., 1 and a, we have n 1 −~~

~~|n {z n} 1+ a a 1 1 < √a< − =1+ − . n n~~

~~a 1 n Since limn 1+ − =1, we conclude that limn √a =1. →∞ n →∞ n Theorem 14.2.limn √n =1. →∞ Proof. Applying the AGI to the n numbers 1, ..., 1 and √n, we have n 1 −~~

~~n n 1+| √{zn } 1 < √n< − , n q 1 1 < 2√n n< 1+ . √n Squaring, we obtain~~

~~1 2 1 < √n n< 1+ . √n 2 1 n Since limn 1+ =1, we conclude that limn √n =1. →∞ √n →∞ n √n! 1 Theorem 14.3. limn = . →∞ n e 152 Some basic limits~~

~~Proof. For k =1, 2,...,n, we have (k + 1)k (k + 1)k+1~~

n (k + 1)k n (k + 1)k+1 < en < ; kk kk+1 kY=1 Yk=1 (n + 1)n (n + 1)n+1 < en < ; n! n! 1+ 1 n +1 n (n + 1) n < √n! < ; e e n 1 n +1 1 √n! n +1 (n + 1) n < < . n · e n n · e n The result follows from limn √n +1=1. →∞ ci

~~Supplement 1~~

Irrationality of √k n More generally, for given positive integers n and k, the number √k n is irrational unless n is the k-power of an integer. This fact follows from the fundamental theorem of arithmetic. Here is an elementary proof following [?, ?]. Suppose to the contrary that √k n is rational. Then, for each positive integer i, √k ni is rational. Therefore, there is an integer Q such that the numbers Q √k ni, i =1, 2,...,k 1, are all integers. Let P be the smallest· of these. Since n−is not a k-power, there is an integer m such that mk

~~The ladder of Theon The original ladder of Theon was about computing √2 by addition only, and was formulated in a slightly different way. Let (an) and (bn) be two sequences deﬁned recursively by~~

~~an+1 = an + bn,~~

~~bn+1 = an + an+1.~~

~~Beginning with a1 =1, b1 =1, we obtain~~

~~n an bn 1 1 1 2 2 3 3 5 7 4 12 17 5 29 41 ......~~

~~bn Then limn = √2. →∞ an cii Some basic limits~~

~~Supplement 2~~

n Example 2.5 (b) limn n =0. →∞ 2 Proof. If n 4, then 2n = (1+1)n > n > n. Therefore, ≥ 2 n n 2 0 < < = . 2n 1 n(n 1) n 1 2 − − 2 n Since limn n 1 =0, we conclude that 2n =0. →∞ − an (c) limn =0. →∞ n! an Proof. We may assume a > 0. Let N = a , and A = N! . Then for n > N, ⌊ ⌋ n N an aN a a a a − 0 < = < A . n! N! · N +1 · N +2 ··· n · N +1 n N a − Since a < N + 1, limn A N+1 = 0, we conclude that an →∞ · limn =0 as well. →∞ n! Example 0.1. The following sequences are convergent because they are bounded, monotone sequences: 1 1 1 1 (a) 1 2 1 4 1 8 1 2n ; − 1 − 1 − 1 ··· − 1 (b) 1+ 2 1+ 4 1+ 8 1+ 2n ; 1 1 1 ··· 1 (c) 1 2 1 4 1 16 1 22n ; − 1 − 1 − 1 ··· − 1 (d) 1+ 1+ 1+ 1+ n . 2 4 16 22 Calculating their limits is a different··· question, often much more dif- ﬁcult. I do not know the answers to (a), (b), (c). But (d) is quite easy. Let 1 1 1 1 a = 1+ 1+ 1+ 1+ . n 2 4 16 ··· 22n Note that 1 1 1 1 1 1 a = 1 1+ 1+ 1+ 1+ 2 n − 2 2 4 16 ··· 22n 1 1 1 1 = 1 1+ 1+ 1+ − 4 4 16 ··· 22n . . 1 = 1 . − 22n+1 ciii

~~From this, limn an =2. →∞~~

~~Harmonic representation of a real number Every number between 0 and 1 can be uniquely written in the form b b b β = 2 + 3 + + n + 2! 3! ··· n! ···~~

for nonnegative integers bn < n, n =2, 3,... . We call this the harmonic representation of the number. (i) Describe precisely the construction of bn. (ii) Show that β is rational if and only if there exists an integer N such that bn =0 for n > N. 1 (iii) Find the harmonic representation of 7 . (iv) What is the number

~~∞ n 1 1 2 n 1 − = + + + − + ? n! 2! 3! ··· n! ··· n=2 X civ Some basic limits~~

~~Supplement 3: The anicent Chinese calculating board~~

In ancient China, the extraction of square roots is carried out in a 4-row arrangement on a calculating board, each column for one digit: (i) The second row is labeled Dividend. One begins by entering in this row the number (dividend) whose square root is to be extracted. (ii) The square root is to appear in the ﬁrst row, labeled Quotient. 1 (iii) The third row, labeled Divisor, is auxiliary in the computation. (iv) The bottom row is labeled Unit. It controls the auxiliary multi- plications in the process. The extraction of a square root begins with a set-up for the unit. We illustrate this with Problem IV.16 of the Nine Chapters of Mathematical Art on the extraction of the square root of 3, 972, 150, 625. 1 Quotient 3972150625 Dividend Divisor 1 Unit

After the dividend is in place in the second row, a counting rod (rep- resenting a unit) is placed in each of top (Quotient) and bottom (Unit) rows, at the rightmost position. They are simultaneously moved to the left, two places in the bottom row for each move of one place in the top row, until the unit in the bottom is in the leftmost, below a digit of the quotient. The location of the unit in the top row is the leftmost digit of the square root. The computation begins with ﬁnding the largest number whose square does not exceed the number with unit indicated in the bottom row. In this case, the largest square not exceeding 39 is 62 = 36. Here we replace the unit in the top row by 6, and subtract 36 from the second row.

6 Quotient 372150625 Dividend Divisor 1 Unit Multiply the quotient by 2, enter it in the third row (as divisor), shifting one place to the right. At the same time, shift the unit (rod) in the bottom row two places to the right.

~~1The extraction of square root is seen as a modiﬁcation of the process of divsion, hence the word quotient for the square root. cv~~

~~6 Quotient 372150625 Dividend 1 2 Divisor 1 Unit~~

For the next digit in the quotient, ﬁnd the largest number which, when multiplied to the number with the same units digit in the third row, gives a product not exceeding the number in the second row. In the present case, this number is 3. Subtract the product 3 123 = 369 from the second row. ×

~~6 3 Quotient 3150625 Dividend 1 2 3 Divisor 1 Unit~~

~~Double the number in the ﬁrst row, enter it in the third row, 2 shifting one place to the right. Shift the unit rod in the bottom row two places to the right.~~

~~6 3 Quotient 3150625 Dividend 1 2 6 Divisor 1 Unit~~

Repeat the same procedure till the unit rod appears at the end. The number appearing in the ﬁrst row would be the square root. In the present example, the next number in the ﬁrst row is 0. Thus, one simply shifts the divisor one place to the right, and the unit rod two places to the right.

~~6 3 0 Quotient 3150625 Dividend 1 2 6 0 Divisor 1 Unit~~

~~2This has the same effect as doubling the unit digits of the divisor in the third row. cvi Some basic limits~~

~~6 3 0 Quotient 3150625 Dividend 1 2 6 0 Divisor 1 Unit~~

~~6 3 0 2 Quotient 3150625 Dividend 1260 2 Divisor 1 Unit~~

~~6 3 0 2 Quotient 630225 Dividend 1260 2 Divisor 1 Unit~~

~~6302 5 Quotient 630225 Dividend 12604 5 Divisor 1 Unit~~

~~The next number should be 5. Since 5 126045 = 630225, this leaves the second row blank, and the calculation× terminates, giving 63025 for the square root.~~

~~63025 Quotient Dividend 126045 Divisor 1 Unit~~

If the dividend appearing in the last step is a nonzero number r, the square root in question is not “exact”. In this case, the ancient Chinese adopted one of the following options. r r (i) Round off the square root with the fraction 2q+1 or 2q , q being the quotient in the ﬁrst row. In other words, r r q2 + r q + or q + . ≈ 2q +1 2q p cvii

~~(ii) Continue the calculation “beyond the decimal point” by treating a unit as 10 subunits, 100 sub-subunits, 1000 sub-sub-subunits etc.~~

~~Exercise Find the square roots of the following numbers (i) 55225, (ii) 25281, (iii) 1 71824, (iv) 564, 752 4. cviii Some basic limits~~

~~Supplement 4~~

~~The inﬁnite radical~~

~~2+ 2+ + √2+ =2, ··· ··· r q as the limit of the sequence (xn) deﬁned recursively by~~

~~xn+1 = √2+ xn, x1 =2.~~

In this case, if we put xn = 2 cos tn, then t x = 2(1 + cos t ) = 2 cos n . n+1 n 2 √ pπ π π Since x1 = 2 = 2 cos 4 , t1 = 4 , and more generally, tn = 2n+1 . Therefore, we have an alternative expression π x = 2 cos . n 2n+1

From this it is also clear that limn xn =2. n →∞ Now, the sequence (4 (2 xn)) converges to an interesting number. Note that − π 4n(2 x )= 4n 2 2 cos − n − 2n+1 π = 4n+1 sin2 2n+2 2 π 2 π sin 2n+1 = π . 4 · n 2 +1 sin θ Since limθ 0 =1, we have → θ 2 n π lim 4 (2 xn)= . n →∞ − 4 Equivalently,

~~π lim 2n v2 2+ 2+ 2+ + √2 = . n · u − s ··· 2 →∞ u r q t (n+1) fold square root − | {z } cix~~

~~Supplement 6~~

~~The equation (6.1) can be simpliﬁed:~~

~~a = 2 4 a2 , 2n − − n q where we have assumed for conveniencep R =1. (i) Beginning with a6 =1, we have~~

~~a6 2k = 2 2+ 2+ + √3 (k + 1) fold square root · s − ··· − r q Therefore,~~

~~π lim 2k 2 2+ 2+ + √3 = . n s − ··· 3 →∞ r q (k+1) fold square root − | {z } (ii) Beginning with a4 =1, we have~~

~~a4 2k = 2 2+ 2+ + √2 (k + 1) fold square root · s − ··· − r q Therefore,~~

~~π lim 2k 2 2+ 2+ + √2 = . n s − ··· 2 →∞ r q (k+1) fold square root − | {z } cx Some basic limits~~

~~Supplement 7~~

Exercise (2). The answer is no. Here is an example. If a = 3 a , 2 4 · 1 a = 8 a , a = 15 a ,..., a = 1 1 a , then 3 9 · 2 4 16 · 3 n+1 − n2 n n +1 n 1 n n 2 4 2 3 1 a = − − n+1 n · n n 1 · n 1 ··· 3 · 3 2 · 2 − − n +1 n 4 3 n 1 n 2 2 1 = − − n · n 1 ··· 3 · 2 n · n 1 ··· 3 · 2 − − n +1 1 = 2 · n n +2 = . 2n 1 From this, limn an = . →∞ 2 cxi

~~Supplement 8: Archimedes’ circle area formula~~

~~From Archimedes’ Measurement of a Circle:~~

~~Proposition 1 The area of any circle is equal to a right-angled triangle in which one of the sides about the right angle is equal to the radius, and the other to the circumference of the circle.~~

~~T G H~~

~~D A~~

~~E O~~

~~K B C~~

Proof. Let ABCD be the given circle, K the triangle described. Then, if the circle is not equal to K, it must be either greater or less. (1) If possible, let the circle be greater than K. Inscribe a square ABCD, bisect the arcs AB, BC, CD, DA, then bisect (if necessary) the halves, and so on, until the sides of the inscribed polygon whose angular points are the points of division subtend segments whose sum is less than the excess of the area of the circle over K. Thus, the area of the polygon is greater than K. Let AE be any side of it, and ON the perpendicular on AE from the center O. Then ON is less than the radius of the circle and therefore less than one of the sides about the right angle in K. Also the perimeter of the polygon is less than the circumference of the circle, i.e. less than the other side about the right angle in K. Therefore, the area of the polygon is less than K; which is inconsis- tent with the hypothesis. Thus the area of the circle is not greater than K. (2) If possible, let the circle be less than K. Circumscribe a square, and let two adjacent sides, touching the circle in E, H, meet in T . Bisect the arcs between adjacent points of contact cxii Some basic limits and draw the tangents at the points of bisection. Let A be the middle point of the arc EH, and F AG the tangent at A. Then the angle T AG is a right angle. Therefore, TG>GA = GH. It follows that the triangle FTG is greater than half the area TEAH. Similarly, if the arc AH be bisected and the tangent at the point of bisection be drawn, it will cut off from the area GAH more than one- half. Thus, by continuing the process, we shall ultimately arrive at a circumscribed polygon such that the spaces intercepted between it and the circle are together less than the excess of K over the area of the circle. Thus, the area of the polygon will be less than K. Now, since the perpendicular from O on any side of the polygon is equal to the radius of the circle, while the perimeter of the polygon is greater than the circumference of the circle, it follows that the area of the polygon is greater than the triangle K; which is impossible. Therefore the area of the circle is not less than K. Since then the area of the circle is neither greater nor less than K, it is equal to it.

An ‘explanation’ of the rational approximation for √3 adopted by Archimedes Basic inequalities: b b a > √a2 b > a . ± 2a ± ± 2a +1 (i) Start with 2 1 > √3 > 2 1 or 7 > √3 > 5 , or 21 > √27 > 5. − 4 − 3 4 3 4 (ii) Apply the basic inequalities to √27 = √52 +2: 2 2 26 19 5+ > √27 > 5+ , > √3 > . 10 11 15 11 (iii) Note that 3 152 = 675 = 676 1=262 1. · − − Exercise Use the basic inequalities to deduce that 1351 265 > √3 > . 780 153 cxiii

~~Supplement 10: Vieta’s inﬁnite product for π~~

Beginning with the trigonometric identity θ θ sin θ = 2 sin cos , 2 2 x x x we replace by θ successively by x, 2 , x2 ,..., 2n−1 , multiply the results together, simplifying, we obtain sin x n x n x = cos k . 2 sin 2n 2 kY=1 For a ﬁxed x> 0, x x sin n lim 2n sin = lim x 2 = x. n n n x →∞ 2 →∞ · 2n Therefore, we have an inﬁnite product formula:

~~∞ x sin x cos = . 2n x n=1 Y π π In particular, with x = 2 , let cn := cos 2n+1 . These values can be computed iteratively by~~

~~2cn+1 = √2+2cn. These are~~

~~√2 c = , 1 2 2+ √2 c2 = , p 2 2+ 2+ √2 c3 = , q p2 . . From these, we have Vieta’s formula 3~~

~~2 √2 2+ √2 2+ 2+ √2 = . π 2 p 2 ! q p2  ···~~

~~3Franc¸ois Vieta (1540–1603). This is historically the ﬁrst exact expression forπ. cxiv Some basic limits cxv~~

~~Supplement 11A: Proofs of the AGI~~

~~Second proof~~

Denote by A and G the arithmetic and geometric means of a1, a2, ..., an. (1) If a = a = = a , then clearly A = G. 1 2 ··· n (2) If a1, a2,..., an are not all equal, we arrange them in descending order: a a a . 1 ≥ 2 ≥···≥ n In this sequence, a1 >G and an

~~a1an a′ = G, a′ = a for i =2,...,n 1, a′ = . 1 i i − n G~~

Let A′ and G′ be their arithmetic and geometric means respectively. Clearly, G′ = G. On the other hand, 1 A′ = (a1′ + an′ + a2′ + an′ 1) n ··· − 1 a1an = (G + + a2 + an 1). n G ··· −

~~a1an Note that G + G < a1 + an, because~~

~~G2 + a a G(a + a )=(G a )(G a ) < 0. 1 n − 1 n − 1 − n~~

~~This gives A′ < A. Sorting the numbers a1′ , a2′ ,..., an′ in descending order, we obtain a new sequence~~

b b b 1 ≥ 2 ≥···≥ n in which (i) the geometric mean remains the same, (ii) the arithmetic mean is diminished, and (iii) the number of terms equal to G is increased by 1. In view of (iii), in a ﬁnite number of steps (not more than n 1), the sequence would be replaced by one in which each term is equal− to G. For such a sequence, the arithemtic mean is equal to G. It follows that A>G if the given numbers are not all equal to begin with. cxvi Some basic limits

~~Third Proof~~

~~Given positive numbers a1, ..., an, ..., denote by An and Gn respectively the arithmetic and geometric means of a1,..., an. For a given n, consider the function~~

~~1 n+1 (nAn + x) fn(x)= n+1 Gn x n · Using the quotient rule, we computep the derivative (exercise):~~

~~n(x An) n fn′ (x)= − n+2 . 2 n+1 (n + 1) Gn x n+1 ·~~

From this it is clear that fn(x) attains its minimum at x = An, and the minimum value is n A n+1 f (A )= n . n n G n With this preparation, we prove the AGI by induction. (i) A2 G2; equality holds if a1 = a2 (Lemma 11.1 above). (ii) Assume≥ A G . Then for a ,...,a , we have n ≥ n 1 n+1 n A A n+1 n+1 = f(a ) f(A )= n 1, G n+1 ≥ n G ≥ n+1 n showing A G . Moreover, A = G if and only if n+1 ≥ n+1 n+1 n+1 f(an+1)=1. This means f(an+1) = f(An)=1, and An = Gn. From this it follows that a1 = = an = an+1. Therefore, by the principle··· of mathematical induction, the arithmetic and geometric means inequality holds for any number of positive quan- tities.

~~Exercise Give a proof of the AGI by considering the maximum of the function~~

~~n f(x)= n √a1a2 an 1x x · ··· − −~~

~~for x> 0 and a1 > 0, a2 > 0,..., an 1 > 0. − cxvii~~

Fourth proof We make use of the concavity of the function of y = log x. 4 Consider the tangent line at the point (1, 0). This is the line y = x 1, and is entirely above the graph of y = log x. − y = x − 1

~~y = log x~~

~~x1 x2 G G~~

~~1 x3 ···xn−1 xn G G G~~

~~Let A and G be respectively the arithmetic and geometric means of n positive numbers x1, x2,..., xn.~~

n n nA − nG x x x x · · · x Gn = k − 1 ≥ log k = log 1 2 n = log = 0. G G G Gn Gn k k X=1 X=1 Therefore, A G, and equality holds if and only if x = x = = ≥ 1 2 ··· xn(= G). Remark. This “proof with few words” can be found in CMJ, 31 (2000) x1 x2 xn 68, where G , G ,..., G are erroneously all depicted on the right hand side of 1.

~~Exercise~~

(1) Let a1, a2,..., an be positive real numbers. Show that 1 2 3 n s + + + + , 3 n s a1 √a2 √a3 ··· √an ≥ √a1a2 an ··· 4This has been shown in Exercise 1 above. Therefore, Exercise 3 furnishes a proof following the line of the First proof. cxviii Some basic limits

~~where s =1+2+3+ + n. ··· 2 n2 (2) Find the positive integer n for which n + 125 is least possible. What if 125 is replaced by 52? and by 53?~~

~~Remark. Let a1 a2 an be positive numbers with arithmetic ≤ ≤···≤ 2 mean A and geometric mean G. Let S = 1 i~~

~~Fifth proof~~

log x We make use of the fact that the function F (x) = x has maximum 1 F (e) = e . Given positive numbers a1,..., an, with arithmetic mean A ake and geometric mean G, let xk = for k = 1, 2,...,n. For each of a e G G log k these, F (x )= G < 1 , and k ake e a log a +1 log G< k . k − G Combining these inequalities, we have

~~n n a (log a +1 log G) k , k − ≤ G Xk=1 Xk=1 nA log(a a )+ n n log G . 1 ··· n − ≤ G n nA Since log(a1 an) = log G = n log G, this becomes n G and A G. ··· ≤ ≥ Sixth proof~~

~~x n This proof makes use of the increasing property of the sequence 1+ n for every positive number x. More precisely, we shall make use of n 1 x n x − 1+ 1+ , n ≥ n 1 − cxix~~

~~for x 0. Here, equality holds if and only if x =0. ≥ Given n positive numbers a1,..., an, assumed in ascending order, so the each number ak is no more than the arithmetic mean of ak, ak+1,..., an.~~

a + a + + a n na (n 1)a + a + + a n 1 2 ··· n = 1 − − 1 2 ··· n na na 1 1 (n 1)a + a + + a n = 1+ − − 1 2 ··· n na1 n 1 (n 1)a + a + + a − 1+ − − 1 2 ··· n ≥ (n 1)a1 n−1 a + + a − = 2 ··· n . (n 1)a − 1 Simplifying and continuing, we have

n n 1 a + a + + a a + + a − 1 2 ··· n a 2 ··· n n ≥ 1 · n 1 − n 2 a + + a − a a 3 ··· n ≥ 1 2 · n 2 . − . 2 an 1 + an a1a2 an 2 − ≥ ··· − · 2 a a a . ≥ 1 2 ··· n Equality holds if and only if for each k, ak is the arithmetic mean of a , a ,..., a . This is the case when a = a = = a . k k+1 n 1 2 ··· n Seventh proof Note that

n 2 n 2 n 3 x nx + n 1=(x 1) (x − +2x − + +(n 1)) 0 − − − ··· − ≥ for x > 0. Now given positive numbers a , a , ..., a , let x = An . 1 2 n An−1 We have A n A n n n +(n 1) 0, An 1 − · An 1 − ≥ − − cxx Some basic limits

~~or n n 1 n 1 An An−1(nAn (n 1)An 1)= An−1 an. ≥ − − − − − · Continuing, we have~~

~~n n 1 n 2 n An An−1 an An−2 an 1an A1 a2 an = a1a2 an = Gn. ≥ − · ≥ − · − ≥···≥ · ··· ··· Therefore, A G . See [?]. n ≥ n cxxi~~

~~Supplement 11B: Some useful inequalities~~

~~Cauchy-Schwarz inequality~~

~~Given real numbers a1, a2,..., an and b1, b2,..., bn, (a b + a b + + a b )2 (a2 + + a2 )(b2 + + b2 ). 1 1 2 2 ··· n n ≤ 1 ··· n 1 ··· n~~

~~Equality holds if and only if (a1, a2,...,an) and (b1, b2,...,bn) are linearly dependent. Proof. The quadratic equation~~

~~(a t b )2 +(a t b )2 + +(a t b )2 =0 1 − 1 2 − 2 ··· n − n has at most one real roots. Therefore, its discriminant~~

~~(a b + a b + + a b )2 (a2 + + a2 )(b2 + + b2 ) 0. 1 1 2 2 ··· n n − 1 ··· n 1 ··· n ≤~~

~~Equality holds if and only if there is a real number t satisfying akt = bk for k =1, 2,...,n.~~

~~Bernoulli’s inequality If n> 1 is an integer,~~

(1 + x)n > 1+ nx for x 1. ≥ − For positive x this is clear by expansion using the binomial theorem. To extend the range of x for the validity of the inequality, let us assume that the inequality is true for a positive integer n. Then

(1 + x)n+1 = (1+ x)n (1 + x) > (1 + nx)(1 + x) > 1+(n + 1)x, · provided that 1+ x is positive. Since the inequality (1 + x)2 > 1+2x is clearly true (for all values of x), by induction, Bernoulli’s inequality is true for n 2 and x> 1. ≥ −

~~Bernoulli’s inequality indeed holds for x 2. This is clear for n =1. We shall assume n 2. ≥ − ≥ cxxii Some basic limits~~

Let x [ 2, 1). Then (i) < 1 ∈ 1+− x<− 0 implies 1 (1 + x)n < 1; (ii) −2 ≤x< 1 implies 1 −2n ≤1+ nx < 1 n. For− n≤ 2, we− have − ≤ − ≥ 1 2n 1+ nx < 1 n 1 (1 + x)n < 1. − ≤ − ≤ − ≤ This shows that Bernoulli’s inequality can be improved to x 2. ≥ − cxxiii

~~Supplement 11C: Maxima and minima without calculus~~

Example 0.2. Find the dimensions of the rectangle of maximum area x2 y2 that can be inscribed in the ellipse a2 + b2 =1. Solution. If the corner of the rectangle in the ﬁrst quadrant has coordi- nates (x, y), its area is 4xy. Now,

1 x2 y2 x2 y2 xy + = 2 a2 b2 ≥ a2 · b2 ab r 2 2 gives xy 1 ab, with equality when x = y = 1 , i.e., x = a and ≤ 2 a2 b2 2 √2 y = b . The dimensions of the maximum rectangle are √2a √2b. √2 × Example 0.3. Find the maximum volume of a circular cylinder inscribed in a right circular cone of base radius r and height h. Solution. Suppose the cylinder has radius x and height y. It has volume 2 x y x x y V = πx y. Note that r + h = 1. Rewriting this as 2r + 2r + h = 1, we see that the product x 2 y = 1 x2y is maximum when x = 2r · h 4r2h · 2r y = 1 , i.e., x = 2r and y = h . The maximum volume of the cylinder is h 3 3 3 V = π 2r 2 h = 4πr2h . max 3 · 3 27 Example 0.4. What is the length of the shortest line in the ﬁrst quadrant x2 y2 that is tangent to the ellipse a2 + b2 =1. Solution. If the point of tangency is (p, q), the equation of the tangent px qy a2 is a2 + b2 = 1. This intersects the x- and y- axes at the points p , 0 b2 and 0, q . The squared length of the tangent in the ﬁrst quadrant is a4 b4 a4 b4 p2 q2 a2 p b2 q 2 + = + + + =(a + b)2. p2 q2 p2 q2 a2 b2 ≥ p · a q · b The minimum length of the tangent is therefore a + b. Equality holds a2 b2 p q p2 q2 when p , q and a , b are linearly dependent, i.e., a3 = b3 : p2 q2 p2 q2 2 2 2 + 2 1 a = b = a b = . a b a + b a + b

~~2 a3 2 b3 a b Therefore, p = a+b and q = a+b . The pointof tangencyis a a+b , b a+b . p q cxxiv Some basic limits~~

b Example 0.5. To cut equal squares of dimension x from the four corners of a rectangle of length a and breadth b so that the box obtained by folding along the creases has a greatest capacity. The capacity of the box is V = x(a 2x)(b 2x). We ﬁnd appropriate numbers λ and µ and maximize − − 2(λ + µ)x λ(a 2x) µ(b 2x). · − · − These three factors have a constant sum λa + µb. Their product is maximum when 2(λ + µ)x = λ(a 2x)= µ(b 2x), − − or λ + µ λ µ λ + µ 1 = 1 = 1 = 1 1 . 2x a 2x b 2x a 2x + b 2x − − − − In other words, 1 1 1 = + . 2x a 2x b 2x − − This reduces to the quadratic equation 12x2 4(a + b)x + ab =0, with a root − a + b √a2 ab + b2 1 x = − − < min(a, b). 6 2 · This maximizes the product 2(λ + µ)x λ(a 2x) µ(b 2x); hence also V = x(a 2x)(b 2x). · − · − − − a2 b2 Example 0.6. Find the maximum and minimum of y = x + a x . As- sume a>b> 0. − Solution. Rewrite the given relation as a quadratic equation in x: y x2 (a2 b2 + ay)x + a3 =0. · − − Since this has real roots, we must have (a2 b2 + ay)2 4a3y 0, or − − ≥ (ay (a b)2)(ay (a + b)2) 0. − − − ≥ cxxv

~~(a b)2 (a+b)2 − From this, we have y a (maximum) or y a (minimum). ≤a2 a2 ≥ Correspondingly, x = a b and a+b . − Exercise~~

~~(a x)(x b) (a b)2 2ab − − − Show that x2 has maximum 4ab when x = a+b . Example 0.7. Find the maximum volume of a right circular cone inscribed in a sphere of radius r. cxxvi Some basic limits~~

~~1 n 1 n+1 Supplement 13: On the numbers 1 + n and 1 + n Theorem 0.4 (Euler). If x~~

~~1 n 1 n+1 (x, y)= 1+ , 1+ n n ! for some positive integer n.~~

y y x m Proof. Rewrite the equation as y = x and put x =1+ n for relatively prime integers m and n. m 1+ n m+n x = n x, m · n m+n x = n . n 1 (m+n) m (m+n)n m Therefore, x = n = n . n m n Since m and n do not have common divisors, this is a rational number if and only if both (m+n)n and nn are m-th powers. Since gcd(m, n)= 1, this means that m + n and n are both m-th powers. Write n = am and m + n = bm for integers a < b. Now, bm am = m is possible only when m =1. Therefore, − (1 + n)n 1 n x = = 1+ , nn n and 1 1 n+1 y = 1+ x = 1+ . n n

Exercise Determine ﬁve pairs of positive integers p > q for which there exist geometric progression in which the q-th term is p, the p-th term is q, and the (p + q)-term is an integer. p 1 q 1 p+q 1 Solution. Suppose ar − = q, ar − = p, and ar − = b, where b is an integer. Then, 1 1 b q b p r = = , q p 1

and q p p b q b = . b b Therefore, there is an integer nfor which p (n + 1)n+1 q (n + 1)n n(n + 1)n = , = = . b nn+1 b nn nn+1 It follows that p = k(n + 1)n+1, q = kn(n + 1)n and b = knn+1 for some integer k. The common ratio is

~~1 n k(n+1)n r = . n +1 With k =1, we have the following 5 pairs of p > q (along with b):~~

~~p 4 27 256 3125 46656 . . . q 2 18 192 2500 38800 . . . b 1 8 81 1024 15625 . . .~~