CALCULUS
from a Historical Perspective
Paul Yiu
Department of MathematicsFlorida 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 10/13Thursdays 8/27 9/3 9/10 9/17 9/24 10/1 10/8 10/15Tuesdays 10/20 10/27 11/3 11/10 11/17 11/24 12/1Thursdays 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
5 An infinite continued fraction 117
6 Ancient Chinese calculations ofπ 121
7 Existence ofπ 125
8 Archimedes’ calculation ofπ 129
9 Ptolemy’s calculations of chord lengths 131
10 The basic limit limθ→0sin θ
θ= 1 135
11 The Arithmetic - Geometric Means Inequality 139
12 Principle of nested intervals 143
13 The numbere 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
19 Normals of a parabola 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 ofy = xn 311
29 Pascal: Summation of powers of numbers in arithmetic pro-gression 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 + 122k + 1
32k + 142k + 1
52k + · · · 425
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 51545.0.1 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 triangle in terms of its sidesa, b, c and semiperimeters = a+b+c
2:
Let the sides of the triangle be7, 8, and9. . . . The result [ofmultiplying s, s − a, s − b, s − c] is 720. Take the squareroot of this and it will be the area of the triangle. Since720has not a rational square root, we shall make a close approxi-mation to the root in this manner. Since the square nearest to720 is 729, having a root27, divide27 into 720; the result is262
3; add27; the result is532
3. Take half of this; the result is
2612+ 1
3(= 265
6). Therefore the square root of720 will be very
nearly2656. For 265
6multiplied by itself gives720 1
36; so that
the difference is136
. If we wish to make the difference lessthan 1
36, instead of729 we shall take the number now found,
720 136
, and by the same method we shall find an approxima-tion differing bymuch less than 1
36.
Heron,Metrica, i.8.
Let a be a given positive integer. Consider the sequence(an) definedrecursively by
an =1
2
(
an−1 +a
an−1
)
, (1.1)
102 Extraction of square roots
with an arbitrarypositiveinitial valuea1. The sequence(an) convergesto the square root ofa regardless of the positive initial value.
Example 1.1.Approximations of√
2 with initial valuesa = 1:
n an
1 1/1 12 3/2 1.53 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 valuea1 = 2:
n an
1 2 22 7/4 1.753 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(i) with any arbitrary positivea1, all subsequentan are greater than
√a, 1
(ii) (an) is a decreasing sequence:
an
an−1=
1
2
(
1 +a
a2n−1
)
<1
2(1 + 1) = 1.
Since a decreasing sequence bounded from below converges, we writeℓ = limn→∞ xn. From (1.1), we have
ℓ =1
2
(
ℓ +a
ℓ
)
.
Solving this equation, we haveℓ =√
a.
1This follows from the arithmetic-geometric means inequality.
103
a2n − a =
1
4
(
an−1 +a
an−1
)2
− a
=1
4
(
an−1 −a
an−1
)2
=(a2
n−1 − a)2
4a2n−1
<(a2
n−1 − a)2
4a.
Therefore, beginning witha1, by iteratingn steps, we obtaina2, . . . ,an+1, with
a2n+1 − a <
1
4a· (a2
n − a)2
<1
4a· 1
(4a)2(a2
n−1 − a)4
...
<1
4a· 1
(4a)2· · · 1
(4a)2n−1 (a21 − a)2n
=1
(4a)2n−1(a2
1 − a)2n
This shows that the convergence is very fast. For example, for a = 2,if we begin witha1 = 2 and executen steps, then we have a rationalapproximationan+1 satisfying
a2n+1 − 2 <
1
22n−3.
To guarantee an accuracy up to, for example,10 decimal digits be-yond the decimal point, we need to make the error< 1
1010 . This requires22n−3 > 1010, and it is enough to iterate5 times.
104 Extraction of square roots
Reorganization
It is more convenient to calculate the numerators and denominators sep-arately. If we writean = xn
yn, then
xn
yn
= an =1
2
(xn−1
yn−1
+ayn−1
xn−1
)
=x2
n−1 + ay2n−1
2xn−1yn−1
.
This leads to two sequences of integers(xn) and(yn) defined recursively
xn = x2n−1 + ay2
n−1,
yn = 2xn−1yn−1,
with arbitrary initial valuesx1 andy1.
Exercise
(1) Find a rational approximationr of√
5 satisfying|r2 − 3| < 11030 .
(2) Construct two integer sequences(xn) and(yn) such that the(
xn
yn
)
converges to the square root of23.
(3) Let a be a positive integer. Consider a sequence(an) definedrecursively by
an =1
2
(
an−1 −a
an−1
)
.
(a) Show that the sequence does not converge regardless of the initialvalue.(b) Show that for every given positive integerℓ, an initial value can bechosen so that the sequence is periodic with periodℓ. [Hint: rewrite therecursive relation by puttingan−1 =
√a
tan θ].
(4) Start with two positive numbersa0 anda0, and iterate accordingto the rule
an =√
an−1 +√
an−2.
Does the sequence(an) converge? If so, what is the limit? (No proof isrequired).
Irrationality
Why is√
2 an irrational number? Here are two simple proofs.
105
(1) If√
2 is rational, we write√
2 = pq
in lowest terms, then a squareof sidep has the same area as two squares of sideq, and for this, thep × p square is smallest possible.
p
q
q
Yet this diagram shows that there is asmaller square which also hasthe 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 asmallestintegerq for whichq√
2is an integer. Now, the number(q
√2−q)
√2 is asmaller 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 positive integern,√
n is either irrationalor integral.2 If
√n = p
qin lowest terms, then bothq
√n = p andp
√n =
qn are integers. Since there are integersa andb satisfyingap + bq = 1,we have
√n = a · pn + b · qn, an integer.
More generally, for given positive integersn andk, the numberk√
n isirrational unlessn is thek-power of an integer. This fact follows easilyfrom the fundamental theorem of arithmetic: every positiveinteger> 1is uniquely the product of powers of distinct prime numbers.3
The ladder of Theon
Consider an integera > 1 which is not the square of an integer. Letb bethe integer closest to
√a (so that|b −√
a| < 12).
2M. Levin, The theorem that√
n is either irrational or integral,Math. Gazette, , 60 (1976) 138. Levin
has subsequently (ibid., 295) given an even shorter proof: Since the reduced fractionp2
q= p
√n = qn is
an integer,q must be equal to1, and√
n is an integer.3See the supplements to Chapter 1 for a more elementary proof.
106 Extraction of square roots
Now, for each positive integern, (b +√
a)n = xn + yn
√a for pos-
itive integersxn andyn. The sequences(xn) and(yn) can be generatedrecursively by
xn = bxn−1 + ayn−1,
yn = xn−1 + byn−1,
with x1 = b, y1 = 1.(1) Since|b −√
a| < 12, |xn − yn
√a| < 1
2n .(2) Sincexn andyn are positive integers,yn > bn and
∣∣∣∣
xn
yn
−√
a
∣∣∣∣<
1
2nyn
<1
(2b)n.
This shows thatlimn→∞xn
yn=
√a.
Chapter 2
Convergence of sequences
A sequence(an) is said to converge to a numbera if ultimately the termsof the sequence are as close toa as we like. This means that for any givenε > 0, there exists an integerN such that
|an − a| < ε whenevern > N.
We say thata is the limit of the sequence and write
limn→∞
an = a.
Example 2.1. (a) The sequence(
1n
)converges to0: limn→∞
1n
= 0.While this is intuitively clear, here is a proof. Givenε > 0, letN = ⌈1
ε⌉.
Forn > N , we haven > ⌈1ε⌉ ≥ 1
ε. This means that
∣∣ 1n− 0∣∣ = 1
n< ε.
Proposition 2.1. If limn→∞ an = A andlimn→∞ bn = B, then(a) limn→∞(an ± bn) = A ± B,(b) limn→∞ anbn = AB,(c) limn→∞
an
bn= A
B, providedbn andB are nonzero.
Proof. A convergent sequence is bounded
Example 2.2. Let f(x) be a polynomial inx. If (an) converges toa,then(f(an)) converges tof(a).
A sequence(an) is(i) bounded below if it has a lower boundA, i.e., an ≥ A for everypositive integern,
108 Convergence of sequences
(ii) bounded above if it has an upper boundB, i.e., an ≤ B for everypositive integern.
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 positivenumber in absolute value. This means that given anyM > 0, there existsan integerN > 0 such that
|an| > M whenevern > N.
Theorem 2.2.(a)An (ultimately) increasing sequence bounded above isconvergent. Its limit is theleastupper bound of the sequence.(b) A decreasing sequence bounded below is convergent. Its limit is thegreatest lower boundof the sequence.
Example 2.3. (a) Let a be a given real number. The sequence(an) isunbounded when|a| > 1.
Proof. Without loss of generality assumea > 1, and writea = 1 + b forb > 0. Note thatan = (1 + b)n > 1 + nb. Therefore, givenM > 0, letN be the least integer such that1 + nb ≥ M , i.e., n ≥ ⌈M−1
b⌉. Then,
n > M implies |an| > 1 + nb ≥ M , showing that the sequence isunbounded.
(b) The sequence(an) converges to0 if |a| < 1.
Proof. Again, it is enough to assumea > 0. Write a = 1 − b for0 < b < 1. Note that1 − b < 1
1+band
an = (1 − b)n <1
(1 + b)n<
1
nb.
Givenε > 0, let N = ⌈ 1bε⌉. Then, forn > N , n > 1
bεand 1
nb< ε. This
meansan < 1nb
< ε, andlimn→∞ an = 0.
(c) The sequence((−1)n) does not converge to any number, thoughit is bounded, and the subsequence of odd numbered terms is a constantsequence, 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 toℓ, and(ii) the subsequence of even numbered terms converges to thesameℓ.Then the sequence(an) converges toℓ.
109
Proof. Givenε > 0, there are integersN1 andN2 such that(i) |an − ℓ| < ε whenevern > N1 is even,(ii) |an − ℓ| < ε whenevern > N2 is odd.Therefore,|an − ℓ| < ε whenevern > max(N1, N2), showing that(an)converges toℓ.
Proposition 2.3. Let (an), (bn), and(cn) be sequences such that(i) ultimately1 an ≤ bn ≤ cn, 2
(ii) both sequences(an) and(cn) converge toℓ.Then the sequence(bn) also converges toℓ.
Example 2.5. limn→∞1√
n2+1= 0.
Proof. 0 < 1√n2+1
< 1n
for everyn. Sincelimn→∞1n
= 0, the resultfollows from Proposition 2.3.
(b) limn→∞n2n = 0.
(c) limn→∞an
n!= 0.
Infinite series
An infinite series∞∑
n=1
an = a1 + a2 + · · ·+ an + · · ·
is convergent if the sequence ofpartial sums
sn := a1 + a2 + · · ·+ an
is convergent. The sum of the series is the limit of the sequence ofpartial sums. Otherwise, it is divergent.
For series ofpositive terms, we shall simply write∑
an < ∞ whenthe series is convergent.
(1) If the partial sums of a series ofpositive terms are bounded, thenthe series converges. Proof: The sequence of partial sums, being mono-tonic increasing and bounded above, is convergent.
(2) Comparison test for series of positive terms: Let∑
an and∑
bn
be series of positive terms.1This means that there is an integerN such that the double inequality holds whenevern > N .2One or both of the inequalities≤ may be replaced by<.
110 Convergence of sequences
(i) If an ≤ bn ultimately and∑
bn is convergent, then so is∑
an.(ii) If an ≥ bn ultimately and
∑bn is divergent, then so is
∑bn.
Theorem 2.4. A geometric series∑∞
n=0 arn converges to a1−r
provided|r| < 1.
Example 2.6.0.999 · · · = 1.
The decimal expansion of a number
Every number between0 and1 can be written uniquely in decimal form:
0.a1a2 · · ·an · · · (2.1)
where the digitsa1, . . . , an, . . . are integers between0 and9. This ex-pression is indeed an infinite series
a1
10+
a2
102+ · · ·+ an
10n+ · · · .
Consider the sequence(sn) of partial sums defined by
sn :=a1
10+
a2
102+ · · ·+ an
10n.
It is easy to see that it is an increasing sequence:
s1 ≤ s2 ≤ · · · ≤ sn ≤ · · ·
It is clearly bounded above since each
sn ≤ 9
10+
9
102+ · · ·+ 9
10n+ · · · =
910
1 − 110
= 1.
This shows that (2.1) defines a unique nonnegative number≤ 1.Conversely, given a numberα ∈ [0, 1], we determine its decimal
expansion as follows.(i) a1 is the integer satisfyinga1
10≤ α < a1+1
10.
(ii) Having defineda1, . . . , an, we putsn =∑n
j=1ak
10k and definean+1
as the integer satisfying
an+1
10n+1≤ α − sn <
an+1 + 1
10n+1.
Equivalently,an+1 = ⌊10n+1(α − sn)⌋.
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 largen, the rational number
α − 1n
is strictly betweenα andβ.(2) If α is irrational, with a decimal expansiona0 +
∑∞n=1
an
10n . Thereis a sufficiently largeN such that 1
10N < α − β. The rational number∑N
n=1an
10n is strictly betweenα andβ since
α −N∑
n=1
an
10n=
∞∑
n=N+1
an
10n≤
∞∑
n=N+1
9
10n=
1
10N< α − β.
This means thatα >∑N
n=1an
10n > β.
Exercise
(1) Prove that between any two real numbers, there is an irrational num-ber.
(2) The decimal expansion of a rational numberpq
(in lowest terms)is finite if and only if the prime divisors ofq are2 and/or5.
(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 integera is given in decimal form. To find its squareroot, divide the digits ofa into blocks of two digits beginning with theright hand side. We represent this as
a1a2 · · ·an
where eachak, k = 2, . . . , n is a 2-digit number, anda0 has either 1 or2 digits.
(1) Setb1 := a1, and letq1 be the largest integer such thatr1 :=a1 − q2
1 ≥ 0.SetQ1 := q1.
(2) Supposebk, rk andQk have been defined. Form
bk+1 := 100rk + ak+1.
Find thelargest integerqk+1 such that
rk+1 := bk+1 − (20Qk + qk+1)qk+1 ≥ 0.
SetQk+1 := 10Qk + qk+1.(3) Repeat (2) to findb1, b2, . . . ,bn.Thenb1b2 · · · 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 06 5 4 6 7 8 4 5 3 6 1 2 3 4 5 6 74
45 2 5 42 2 5
505 2 9 6 72 5 2 5
5108 4 4 2 8 44 0 8 6 4
51166 3 4 2 0 5 33 0 6 9 9 6
511726 3 5 0 5 7 6 13 0 7 0 3 5 6
5117328 4 3 5 4 0 5 2 34 0 9 3 8 6 2 4
51173365 2 6 0 1 8 9 9 4 52 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 + · · ·
Let a > 0 be a given number. What is the number√
a +
√
a + · · · +√
a + · · ·?
If this expression defines a numberℓ, it must satisfyℓ =√
a + ℓ, andso is the positive root of the quadratic equationx2 − x− a = 0, namely,ℓ = 1
2(1+
√4a + 1). We justify the existence of the numberℓ as the limit
of a convergent sequence(xn) of positive numbers defined recursivelyby
xn+1 =√
a + xn, (4.1)
regardless of the (positive) initial value. For this, it is helpful to write
x2 − x − a = (x − ℓ)(x + ℓ′) for ℓ, ℓ′ > 0.
(1) Supposexn < ℓ. Then(i) x2
n+1 − ℓ2 = a + xn − ℓ2 < a + ℓ − ℓ2 = 0, so thatxn+1 < ℓ;(ii) x2
n+1 − x2n = a + xn − x2
n = −(xn − ℓ)(xn + ℓ) > 0, so thatxn+1 > xn.Therefore, ifx1 < ℓ, then the sequence(xn) is monotonic increasing,and is bounded above. The sequence necessarily converges toℓ.
(2) Supposexn > ℓ. Then(i) x2
n+1 − ℓ2 = a + xn − ℓ2 > a + ℓ − ℓ2 = 0, so thatxn+1 > ℓ;(ii) x2
n+1−x2n = a+xn−x2
n = −(xn−ℓ)(xn+ℓ) < 0, so thatxn+1 < xn.Therefore, according asx1 > ℓ or < ℓ, the sequence(xn) is monoton-ically decreasing and bounded below, or monotonically increasing andbelow above. It necessarily converges toℓ.
116√
a +√
a + · · · +√
a + · · ·
(3) Supposex1 = ℓ. Then everyxn = ℓ.Since the sequence(xn) converges, its limitℓ satisfies, according to
(4.1),1
ℓ =√
a + ℓ.
Here are two examples fora = 5 with different initial values.
n xn xn
1 1 32 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 · · ·
Example 4.1.Fora = 1, this limit is the golden ratioϕ =√
5+12
.
Exercise
(1) Leta andb be given positive numbers. Why does the expression√
a +
√
b +
√
a +√
b + · · ·
(in whicha andb alternate indefinitely) define a real number?
(2) Identify the numbers√
1 +
√
7 +
√
1 +√
7 + · · ·
and √
7 +
√
1 +
√
7 +√
1 + · · ·.
1It is important to justify the existence of the limit before passing an recurrence relation to the limit. Hereis a counterexample. Let(xn) be defined byxn+2 = xn+1 + xn with x1 = x2 = 1. The sequencecertainly is unbounded and does not converge. The terms are the Fibonacci numbers. If we assume itconverges to a limitℓ, thenℓ = ℓ + ℓ, andℓ = 0, an impossibility.
Chapter 5
An infinite continued fraction
What is the number
a +1
a +1
a +1
.. .
for a givena > 0?If we regard this as the limitℓ of a sequence of fractions, where the
n-th term is the fraction
a +1
a +1
a +1
. .. + 1a
with n copiesa’, then ℓ = a + 1ℓ, andℓ =
√a2+4−a
2. As noted in the
preceding example, we need to justify the existence of the limit. Thesequence in question can be defined recursively as
xn+1 = a +1
xn, x1 = a.
It is possible to work out an explicit expression forxn, by writing it inthe formxn = yn
yn−1for another sequence(yn). The recurrence relation
becomesyn+1 = ayn + yn−1, y0 = 1, y1 = a.
118 An infinite continued fraction
This is a second order homogeneous linear recurrence, with character-istic equationλ2 = aλ + 1. Denote byλ1 andλ2 the two characteristicvalues. Note thatλa + λ2 = a andλ1λ2 = −1. Since their sum ispositive, we may writeλ1 = λ > 1, andλ2 = −1
λ< 0.
We haveyn = A · λn
1 + B · λn2
for appropriately chosen constantsA andB. From the initial values, wehave
A + B = 1,
Aλ1 + Bλ2 = a.
Solving these, we have
A =a − λ2
λ1 − λ2=
λ1
λ1 − λ2, B =
λ1 − a
λ1 − λ2=
−λ2
λ1 − λ2,
and
yn =λn+1
1 − λn+12
λ1 − λ2
.
It follows that
xn =λn+1
1 − λn+12
λn1 − λn
2
=λ2n+2 − (−1)n+1
λ(λ2n + 1).
Sinceλ > 1, it is clear that the limit isλ.
Example 5.1.Fora = 1, this limit is (again) the golden ratioϕ =√
5+12
.
Exercise
(1) For which values ofx1 will (xn) converge, where
xn+1 = 3 − 2
xn
?
(2) Show that regardless of initial value, the sequence(xn) definedrecursively by
xn+1 = 1 − 1
xn
119
does not converge.(3) Let λ ∈ (0, 1) be a fixed number. Givena, 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 ofπ
The ancient Chinese classics theNine Chapters of the Mathematical Artadopted 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, andexplainedthe mensuration of the circle by the method of dissection.
Consider a circle of radiusR. Denote byA its area. Inscribe in thecircle a regular polygon ofn sides, each of lengthan. For the regularn-gon, denote by(i) pn the perimeter,(ii) An the area,(iii) dn the distance from the center to a side, and(iv) cn = R − dn.
Beginning witha6 = 1, LIU Hui first computeda12, making use of
the right triangle theorem:d6 =√
R2 −(
a6
2
)2, c6 = R − d6, and
a12 =
√
c26 +
(a6
2
)2
=
√√√√
(
R −√
R2 −(a6
2
)2)2
+(a6
2
)2
.
122 Ancient Chinese calculations ofπ
d6
c6
R
a12
n an dn cn pn = n · an
6 1 0.8660254 0.1339746 612 0.517638 0.9659258 0.0340742 6.211656
More generally,
a2n =
√√√√
(
R −√
R2 −(an
2
)2)2
+(an
2
)2
. (6.1)
LIU Hui iterated the process several times and obtained
n an dn cn pn = n · an
6 1 0.8660254 0.1339746 612 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
He actually recorded the values ofa22n and extracted their square roots
to finda2n: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
a212 = 0.267949193445 (0.267949192431 · · · )
a224 = 0.068148349466 (0.068148347421 · · · )
a248 = 0.017110278813 (0.017110277252 · · · )
a296 = 0.004282154012 (0.004282153522 · · · )
The area of the polygon can be computed asAn = n · 12an ·dn. Indeed
the area of the regular2n-gon isA2n = n2· an · R. LIU Hui made use of
the following inequality to estimate the area of the circle:
A2n < A < A2n + (A2n − An).
Now thatA2n − An = n · 12· an · cn, andA12 = 6 · 1
2R2 = 3R2.
n an A2n − An A2n A2n + (A2n − An)6 1 312 0.517638 0.105828 3.105828 3.21165624 0.261052 0.026796 3.132624 3.15942048 0.130806 0.006720 3.139344 3.14606496 0.065438 0.001680 3.141024 3.142704
From these, LIU Hui concluded that the area of the circle is 3.14correct to two places of decimal.
In the fifth century, ZU Chongzi (429–500) gave the value ofπ cor-rect to 6 decimal places, as between 3.1415926 and 3.1415927. His
124 Ancient Chinese calculations ofπ
manuscriptSushuwas lost. But it is generally believed that he carried outLIU Hui’s program further to regular polygons of sides6 · 211 = 12288and6 · 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 · · · 3.1416627435768 0.0081812080 · · · 0.0000262842 · · · 3.1415838921 · · · 3.14161017631536 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 · · ·
Zu also gave thecrudeapproximation227
and thefineapproximation355113
for π.
Chapter 7
Existence ofπ
Area of circle in Euclid’s Elements
Eucl. X.I
Two unequal magnitudes being set out, if from the greater there be sub-tracted a magnitude greater than its half, and from that which is left amagnitude greater than its half, and if this process be repeated continu-ally, there will be left some magnitude which will be less than the lessermagnitude set out.
Reformulation
Let (an) be a sequence of real numbers satisfyingan+1 < 12·an for every
n. Givenε > 0, there exists an integerN such thatan < ε whenevern > 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 magnitudesubtracted each time be greater than a certain proportion ofthe magni-tude (from which it is subtracted)?
126 Existence ofπ
Eucl. XII.1
Similar polygons inscribed in circles are to one another as the squareson 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, diameterd, then for anyd1, d2, C(d1) : C(d1) = d2
1 : d22.
H
N
E
K
F
L
G
M
(i) Let C be the area of a circle. Suppose there is an inscribedn-gonof areaAn, each arc smaller than a semicircle. Bisecting the arcs, oneobtains an inscribed2n-gon, of areaA2n. Then,C −A2n < 1
2(C −An).
△EKF >1
2segment EKF
=⇒sum of shaded segments >1
2segment EKF.
(ii) Starting from an inscribed square, one can approximate, by re-peated bisection of arcs, the areaC arbitrarily closely: given any positiveǫ, there is an inscribed regularn-gon for whichC − An < ǫ.
(iii) Given two circles of diametersd1, d2, Euclid shows that ifd21 :
d22 = C(d1) : X, thenX cannot be smaller thanC(d2). Suppose to
the contrary thatX < C(d2). By (ii), we can find a regularn−gon,inscribed inC(d2) with areaA′
n satisfyingC(d2) − A′n < C(d2) − X.
Note thatA′n > X.
Now, in the circleC(d1), construct a (similar) regularn-gon of areaAn. Euclid has shown in XII.1 thatd2
1 : d22 = An : A′
n. It follows that
An : A′n = d2
1 : d22 = C(d1) : X,
andAn : C(d1) = A′n : X. But this is a contradiction sinceAn < C(d1)
andA′n > X.
127
(iv) Suppose nowd21 : d2
2 = C(d1) : X with X > C(d2). Wecan rewrite this asd2
1 : d22 = Y : C(d2) with Y < C(d1). The same
reasoning in (iii) shows a contradiction. From this,D21 : d2
2 = C(d1) :C(d2).
Exercise
(3) A circle is approximated by a regular octagon obtained bycutting outcorners from its circumscribed square. What is the approximate value ofπ?
128 Existence ofπ
Chapter 8
Archimedes’ calculation ofπ
Archimedes, in hisMeasurement of a Circle, gave the following inter-esting bounds forπ:
31
7> π > 3
10
71.
To obtain the upper bound, Archimedes began with a regular hexagoncircumscribing the circle, doubling the number of sides anddeterminedthe length of a side of a circumscribed regular polygon of6·2n+1 sides interms of one of6 ·2n sides. He did the same thing for inscribed polygonsto obtain the lower bound.
CO
A
A′
B
X
D
Consider a tangentAC to a circle, centerO such that∠AOC is 12k
ofthe circle,k = 6 · 2n, so thatAC is one half of the length of a side ofcircumscribed regular polygon ofk sides.
If the bisector of angleAOC intersectsAC at A′, thenA′C is onehalf of a side of a circumscribed regular polygon of2k sides.
If OA intersects the circle atB, andX the projection ofB on OC,thenBC is a side of an inscribed regular polygon of2k sides, andBX
130 Archimedes’ calculation ofπ
is one half of a side of an inscribed polygon ofk sides.Let an andbn be the perimeters of regular polygons ofk = 6 · 2n
sides, respectively circumscribed and inscribed in the circle. We have
AC =an
2k, A′C =
an+1
4k, BC =
bn+1
2k, BX =
bn
2k.
(1) SinceOA′ bisects angleAOC,
A′C : AC = OC : OA + OC = OB : OA + OB = BX : AC + BX.
This givesan+1
4kan
2k
=bn
2kan
2k+ bn
2k
.
From this, we have
an+1 =2anbn
an + bn.
(2) SinceBC2 = CX · CD, we haveCX = BC2
CD. Also, from the
similarity of trianglesOA′C andBCX, we have
A′C
OC=
CX
BX=
BC2
CD · BX,
andan+1
4k= A′C =
OC · BC2
CD · BX=
BC2
2 · BX=
b2n+1
4k · bn
.
Therefore,b2n+1 = an+1bn.
If we take the diameter of the circle to be1, thena0 = 2√
3 andb0 = 3. The sequences(an) and(bn) defined recursively by
an+1 =2anbn
an + bn
, bn+1 =√
an+1bn
both converge to the circumference of the circle, namely,π.
n 6 · 2n an bn
0 6 3.464101615 31 12 3.21539 3.1058285412 24 3.15966 3.1326286133 48 3.14609 3.1393502034 96 3.14271 3.1410319515 192 3.14187 3.141452472
This was the basis of practically all calculations ofπ before the 16thcentury. Correct to the first34 decimal places,
π = 3.1415926535897932384626433832795028 · · · · · ·
Chapter 9
Ptolemy’s calculations of chordlengths
Ptolemy Theorem
In a cyclic quadrilateral, the sum of the products of two pairs of oppositesides is equal to the product of the diagonals.
O
C D
A
B
E
Proof. Choose a pointE on the diagonalAC such that∠ADE = ∠BDC.Then trianglesADE andBDC are similar, and
AD
AE=
BD
BC⇒ AD · BC = AE · BD.
Also, trianglesABD andECD are similar, and
AB
BD=
EC
CD⇒ AB · CD = EC · BD.
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 into360 equal arcs, andthe diameter into120 parts, and expressed fractions of these parts in thesexagesimal system.
x 180◦ 60◦ 90◦ 120◦
crd(x) 120p 60p 84p51′10′′ 103p55′23′′
Clearly, by the Pythagorean theorem,
crd(x)2 + crd(180 − x)2 = 4.
A
B
D
C
O
xy
C
D
A
B
O
x2
x2
Applying Ptolemy’s theorem, one has
crd(x ± y) =1
2[crd(x)crd(180 − y) ± crd(180 − x)crd(y)] .
In particular,
crd(2x) = crd(x)crd(180 − x).
Also,crd(x
2
)
=√
2 − crd(180 − x).
Ptolemy then made use of these to determinecrd(1◦).
1. Calculations with a regular pentagon gavecrd(72◦) = 70p32′3′′.
2. crd(12◦) = crd(72◦ − 60◦) = · · · = 12p32′36′′.
133
3. crd(6◦) = crd(
12· 12◦
)= · · · ; thencrd(3◦), and
crd
(
11
2
◦)
= 1p34′15′′,
crd
(3
4
◦)
= 47′8′′.
4. To computecrd(1◦), Ptolemy made use of an interpolation basedon an inequality
crd(x)
crd(y)<
x
y
for arcsx > y smaller than a quadrant of a circle.(i) crd(1◦) : crd
(34
◦)< 1 : 3
4,
(ii) crd(11
2
◦): crd(1◦) < 11
2: 1.
Therefore,
4
3crd
(3
4
◦)
> crd(1◦) >2
3crd
(
11
2
◦)
.
The two ends are respectively1p2′5023
′′and1p2′50′′. Sincecrd(1◦)
is both less and greater than a length which differs from1p2′50′′
insignificantly, Ptolemy took this forcrd(1◦).
From this Ptolemy deduced that crd(
12
◦) is very nearly31′25′′. Mak-ing use of this and the above relations, he was able to complete thisTable of Chords for arcs subtending angles increasing from1
2
◦to 180◦
by increments of12
◦.
Modern reformulation
If we take diameter of the circle to be2, and writesin x = 12crd(2x),
cos x = 12crd(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 sinx cos x;
cos 2x = cos2 x − sin2 x = 2 cos2 x − 1 = 1 − 2 sin2 x;
sin2 x
2=
1 − cos x
2;
cos2 x
2=
1 + cos x
2.
Chapter 10
The basic limit limθ→0sin θθ = 1
Theorem 10.1(Ptolemy). For circular arcsx > y smaller than a quad-rant of a circle,
crd(x)
crd(y)<
x
y
O
C
D
A
B
E F
G
H
Proof. ([?, pp.ii 281–282]). With reference to the diagram above,crd(AB) <crd(BC), we prove that
crd(CB)
crd(BA)<
arcCB
arcBA. (10.1)
Bisect angleABC to intersectAC atE and the circumference of the
136 The basic limit limθ→0sin θ
θ= 1
circle atD. The arcsAD andDC are equal, so are the chordsAD andDC. SinceCB : BA = CE; EA, we haveCE > EA.
ConstructDF perpendicular toAC. Note thatAD > DE > DF .Therefore, the circle, centerD, radiusFE intersectsDA at G, and theextension ofDF at a pointH. Now,
FE : EA = ∆FED : ∆AED
< sectorHED : sectorGED
< ∠FDE : ∠EDA.
From this,
FA : AE < ∠FDA : ∠ADE, componendo
CA : AE < ∠CDA : ∠ADE,
CE : EA < ∠CDE : ∠EDA; separando
CB : BA < ∠CDB : ∠BDA sinceCB : BA = CE : EA
< arc CB: arcBA.
This establishes (10.1) above.
Theorem 10.2.limθ→0sin θ
θ= 1.
Proof. Given a small arc of lengthθ on a unit circle of circumferenceC,choose an integern large enough so that the sectorθ is between1
nand
12n
of the circle:C
2n< θ <
C
n.
By Ptolemy’s theorem,
s2n
C2n
>sin θ
θ>
sn
Cn
2n · s2n
C>
sin θ
θ>
n · sn
C.
As θ → 0, n → ∞. Since the two ends of the double inequality have thesame limit1 asn → ∞, we conclude thatlimθ→0
sin θθ
= 1.
Corollary 10.3. limθ→01−cos θ
θ= 0.
Proof. This follows from1 − cos θ = 2 sin2 θ2.
1 − cos θ
θ=
θ
2·(
sin θ2
θ2
)2
.
137
As θ → 0, the two factors have limits0 and1 respectively. Hence thelimit the product is0.
Corollary 10.4. limθ→01−cos θ
θ2 = 12.
Differentiation of sine and cosine
d(sin x)
dx= lim
h→0
sin(x + h) − sin x
h
= limh→0
sin x cos h + cos x sin h − sin x
h
= limh→0
(
cos x · sin h
h− sin x · 1 − cos h
h
)
= cos x · limh→0
sin h
h− sin x · lim
h→0
1 − cos h
h= cos x · 1 − sin x · 0= cos x.
Exercise
Prove thatd(cos x)dx
= − sin x.
138 The basic limit limθ→0sin θ
θ= 1
Chapter 11
The Arithmetic - GeometricMeans Inequality
A.M. ≥ G.M. ≥ H.M.
Lemma 11.1.For positive numbersa ≤ b,
a ≤ 2ab
a + b≤
√ab ≤ a + b
2≤ b;
equality holds if and only ifa = b.
a b
a
b
OP
G
H
A
Proof. In the diagram above,
OA =a + b
2, PG =
√ab, PH =
2ab
a + b.
Clearly,a ≤ PH ≤ PG ≤ OA ≤ b. Two of these are equal if and onlythe corresponding pointsA, G, H coincide. This happens only whena = b.
140 The Arithmetic - Geometric Means Inequality
Forn given positive numbersa1, a2, . . . ,an, we define(i) thearithmetic mean A(a1, a2, . . . , an) := 1
n(a1 + a2 + · · ·+ an),
(ii) the geometric meanG(a1, a2, . . . , an) := n√
a1 · a2 · · ·an, and(iii) the harmonic meanH(a1, a2, . . . , an) := n
1a1
+ 1a2
+···+ 1an
.
Note that
1
H(a1, a2, · · · , an)= A
(1
a1
,1
a2
, · · · ,1
an
)
.
Theorem 11.2(AGI). For positive numbersa1, a2, . . . ,an,
A(a1, a2, . . . , an) ≥ G(a1, a2, . . . , an).
Equality holds if and only ifa1 = a2 = · · · = an.
Corollary 11.3. For positive numbersa1, a2, . . . ,an,
A(a1, a2, . . . , an) ≥ G(a1, a2, . . . , an) ≥ H(a1, a2, . . . , an).
Any two of these means are equal if and only ifa1 = a2 = · · · = an.
Lemma 11.1 is the arithmetic-geometric means inequality for 2 posi-tive 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 statementsP(n) for n = 1, 2, . . . :
P(n) : a1+a2+···+an
n≥ n
√a1a2 · · ·an for a1, . . . , an ≥ 0.
(1) P(2) is valid by Lemma 11.1.(2) Assuming the validity ofP(2k), we proceed to validateP(2k+1).
This means that for nonnegative numbers
a1, a2, . . . , a2k , a2k+1, a2k+2, . . . , a2k+1 ,
we have to show that
a1 + · · ·a2k + a2k+1 + · · ·+ a2k+1
2k+1≥ 2k+1√a1 · · ·a2ka2k+1 · · ·a2k+1 .
1See the supplements for other proofs.
141
Now,
a1 + · · ·a2k + a2k+1 + · · ·+ a2k+1
2k+1
=1
2
(a1 + · · ·a2k
2k+
a2k+1 + · · · + a2k+1
2k
)
≥ 1
2
(2k√a1 · · ·a2k + 2k√a2k+1 · · ·a2k+1
)by assumption ofP(2k)
≥(
2k√a1 · · ·a2k · 2k√a2k+1 · · ·a2k+1
) 12 by P(2)
= 2k+1√a1 · · ·a2ka2k+1 · · ·a2k+1 .
Therefore, we have establishedP(2k) for all integersk ≥ 1.(3) AssumeP(n). We have to validateP(n − 1). This means that
given n − 1 nonnegative numbersa1, . . . , an−1 with arithmetic meanAn−1 and geometric meanGn−1, we have to show thatAn−1 ≥ Gn−1.
In order to applyP(n), we need an extra numberan. For this, we takean = An−1. Note that the arithmetic mean is
An =(n − 1)An−1 + An−1
n= An−1,
and the geometric mean is
Gn = n
√
Gn−1n−1An−1.
SinceAn ≥ Gn, we haveAnn−1 ≥ Gn−1
n−1An−1. From this,An−1n−1 ≥ Gn−1
n−1,andAn ≥ Gn. Equality holds if and only ifa1 = a2 = · · · = an−1 =An−1.
With these, we claim thatP(n) is valid for every integern ≥ 1. Infact, givenn, there is an integerk such that2k−1 < n ≤ 2k. Note thatP(n) is valid by (2) above. Ifn = 2k, thenP(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 functionf defined on a closed interval[a, b] is concaveiff(
x+y2
)≥ f(x)+f(y)
2for x, y ∈ [a, b].
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
x1, x2, . . . , xn ∈ [a, b],
f
(x1 + x2 + · · ·+ xn
n
)
≥ f(x1) + f(x2) + · · · + f(xn)
n.
2The base of logarithm may be taken to be anya > 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 everyym.
If the difference|xn − yn| can be made arbitrarily small, then the twosequences converge to a common limit.
The conditions (i), (ii), (iii) can be reformulated as
x1 ≤ x2 ≤ · · ·xn ≤ · · · ≤ yn ≤ · · · y2 ≤ y1,
or as a sequence ofnested intervals
[x1, y1] ⊇ [x2, y2] ⊇ · · · ⊇ [xn, yn] ⊇ · · ·
Leibniz test for convergence of alternating series
Theorem 12.1.If (an) is adecreasingsequence of positive real numbersconverging to0, then thealternating series
∑∞n=0(−1)nan converges.
Proof. Since the sequence(an) is decreasing, the partial sumssn =∑n
k=0(−1)kak satisfy
s1 < s0,
s1 < s3 < s2 < s0,
s1 < s3 < s5 < s4 < s2 < s0,
· · · .
144 Principle of nested intervals
These give a sequence of nested intervals
[s1, s0] ⊃ [s3, s2] ⊃ · · · ⊃ [s2n+1, s2n] ⊃ · · · .
Note thats2n − s2n+1 = a2n+1. Since(an) converges to0, then thenested intervals define a unique real numbers, which is the sum of thealternating series.
Thus, for example, the alternating harmonic series
1 − 1
2+
1
3− 1
4+
1
5− 1
6+ − · · ·
converges since the sequence(
1n
)decreases to0.
Remarks.(1) We shall show later that this series converge tolog 2.(2) Similarly, the alternating series
1 − 1
3+
1
5− 1
7+ · · ·+ (−1)n
2n + 1+ · · ·
also converges. The sum isπ4.
Arithmetic-harmonic mean
Consider the sequences(xn) and(yn) defined by
xn+1 =2xnyn
xn + yn
, yn+1 =xn + yn
2,
with initial valuesx1 = a, y1 = b. Note thatxn+1 andyn+1 are re-spectively the harmonic and arithmetic means ofxn andyn. If we beginwith two positivex1 ≤ y1, this recurrence defines a sequence of nestedintervals
[x1, y1] ⊇ [x2, y2] ⊇ · · · ⊇ [xn, yn] ⊇ · · · .
Since
yn+1 − xn+1 =xn + yn
2− 2xnyn
xn + yn=
(yn − xn)2
2(xn + yn)=
yn − xn
yn + xn· yn − xn
2
≤ yn − xn
2, (12.1)
145
it is clear thatlimn→∞(yn − xn) = 0. Therefore, the nested intervalsdefine a unique real number as the common limit of(xn) and(yn).
Here are two examples.(a) Withx1 = 1, y1 = 5; limn→∞ xn = limn→∞ yn =
√5.
n xn yn
1 1 52 1.6666666666666666666666666666 · · · 33 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 52 2.857142857142857142857142857142 · · · 3.53 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 ofa andb. Justify your answer.
The arithmetic-geometric mean
Given two positive numbersa < b, let x1 = a, y1 = b, and
xn+1 =√
xnyn, yn+1 =xn + yn
2.
Then
yn+1 − xn+1 =xn + yn
2−√
xnyn =(√
yn −√xn)2
2=
(yn − xn)2
2(√
yn +√
xn)2
<(yn − xn)2
2(yn + xn)=
yn − xn
yn + xn· yn − xn
2<
yn − xn
2. (12.2)
Therefore, the sequence of nested intervals[xn, yn] defines a uniquereal number. This is called the arithmetico-geometric mean(agM) of aandb. 1
1C. F. Gauss (1777–1854) discovered the agM when he was15 years old, and later its connections withthe 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 thefast convergence in both cases. For arbitrary initial values a < b, thedifferenceyn − xn is ultimately less than2a. We may as well assumeb − a < 2a. Now we have
yn+1 − xn+1 <(yn − xn)2
2(yn + xn)<
1
4a· (yn − xn)2.
Iterating, we have
yn+1 − xn+1 <1
4a·(
1
4a· (yn−1 − xn−1)
2
)2
< · · ·
<(b − a)2n
(4a)2n−1
= 4a ·(
b − a
2a
)2n
.
Chapter 13
The number e
Nested intervals defininge
Let an :=(1 + 1
n
)nandbn :=
(1 + 1
n
)n+1.
(1) Consider then + 1 numbers
1 +1
n, 1 +
1
n, . . . , 1 +
1
n︸ ︷︷ ︸
n
, 1.
Their arithmetic mean isn(1+ 1
n)+1
n+1= 1 + 1
n+1, and their product is
(1 + 1
n
)n. By the AGI,
(1 + 1
n+1
)n+1>(1 + 1
n
)n, showing that the
sequence(an) is increasing.(2) A similar reasoning applied to then + 1 numbers
1 − 1
n, 1 − 1
n, . . . , 1 − 1
n︸ ︷︷ ︸
n
, 1
shows that the sequence(bn) is decreasing.(3) Since the increasing sequence(an) is bounded above byb1, it is
convergent (to a finite number). Now, sincebn − an = 1n· an, we have
limn→∞(bn − an) = 0.
Therefore, the nested intervals[(
1 + 1n
)n,(1 + 1
n
)n+1]
define a unique
numbere:
e = limn→∞
(
1 +1
n
)n
= limn→∞
(
1 +1
n
)n+1
.
148 The number e
n (1 + 1n
)n (1 + 1n
)n+1
1 2. 4.2 2.25 3.3753 2.37037037 3.1604938274 2.44140625 3.0517578135 2.48832 2.9859846 2.521626372 2.9418974347 2.546499697 2.9102853688 2.565784514 2.8865075789 2.581174792 2.867971991
10 2.59374246 2.85311670620 2.653297705 2.7859625930 2.674318776 2.76346273540 2.685063838 2.75219043450 2.691588029 2.7454197960 2.695970139 2.74090297570 2.699116371 2.73767517680 2.701484941 2.73525350390 2.703332461 2.733369488
100 2.704813829 2.7318619681000 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 defininge, with much fasterconvergence.
(
1 +1
n
)n
= 1 +n∑
k=1
(
n
k
)(1
n
)k
= 1 +n∑
k=1
n(n − 1) · · · (n − k + 1)
k!·
1
nk
= 1 +n∑
k=1
n(n − 1) · · · (n − k + 1)
nk·
1
k!
= 1 + 1 +
n∑
k=2
(
1 −1
n
)(
1 −2
n
)
· · ·
(
1 −k − 1
n
)
·1
k!
< 1 +
n∑
k=1
1
k!
< 1 +n∑
k=1
1
2k−1= 1 +
1 − ( 12)n
1 −12
= 3 −1
2n−1< 3.
Let sn :=∑n
k=01k!
. From the above calculation, it is clear that(1 + 1
n
)n< sn. Also, note that
149
(
1 −1
n
)(
1 −2
n
)
· · ·
(
1 −k − 1
n
)
> 1 −1 + 2 + · · · + (k − 1)
n= 1 −
k(k − 1)
2n.
This gives
(
1 +1
n
)n
>2 +n∑
k=2
(
1 − k(k − 1)
2n
)1
k!
=
n∑
k=0
1
k!− 1
2n
n−2∑
k=0
1
k!>
n∑
k=0
1
k!− 3
2n.
Therefore,
sn − 3
2n<
(
1 +1
n
)n
< sn.
Since(sn) is an increasing sequence bounded above (by3), it convergesto a number, say,e′. From the above double inequality, we havee′−0 ≤e ≤ e′, showing thate′ = e. Therefore, we have established followingthe important theorem.
Theorem 13.1.e =∑∞
n=01n!
.
Note thate −∑n
k=01k!
=∑∞
k=n+11k!
<∑∞
k=n+11
2k−1 = 12k . This
gives a much faster calculation ofe:
n 1 + 1 + 12!
+ · · · + 1n!
1 2.00000000000000000002 2.50000000000000000003 2.66666666666666666674 2.70833333333333333335 2.71666666666666666676 2.71805555555555555567 2.71825396825396825408 2.71827876984126984139 2.7182815255731922399
10 2.718281801146384479711 2.718281826198492865212 2.718281828286168563913 2.718281828446759002314 2.718281828458229747915 2.718281828458994464316 2.718281828459042259117 2.718281828459045070518 2.718281828459045226719 2.718281828459045234920 2.7182818284590452353
150 The number e
Irrationality of e
Theorem 13.2(Euler). The numbere is irrational.
Proof. Supposee = ab
= 1 +∑b
k=11k!
+∑∞
k=b+11k!
. Multiplying by b!and rearranging terms, we have aninteger
a(b − 1)! − b!
(
1 +b∑
k=1
1
k!
)
=
∞∑
k=b+1
b!
k!
=1
b + 1+
1
(b + 1)(b + 2)+
1
(b + 1)(b + 2)(b + 3)+ · · ·
<1
b + 1+
1
(b + 1)2+
1
(b + 1)3+ · · ·
<1
b< 1,
which is a contradiction.
Chapter 14
Some basic limits
Theorem 14.1.limn→∞ n√
a = 1 for a > 0.
Proof. Without loss of generality we may assumea > 1. Applying theAGI to then numbers1, . . . , 1
︸ ︷︷ ︸
n−1
anda, we have
1 < n√
a <n − 1 + a
n= 1 +
a − 1
n.
Sincelimn→∞(1 + a−1
n
)= 1, we conclude thatlimn→∞ n
√a = 1.
Theorem 14.2.limn→∞ n√
n = 1.
Proof. Applying the AGI to then numbers1, . . . , 1︸ ︷︷ ︸
n−1
and√
n, we have
1 <n
√√n <
n − 1 +√
n
n,
1 < 2n√
n < 1 +1√n
.
Squaring, we obtain
1 < n√
n <
(
1 +1√n
)2
.
Sincelimn→∞
(
1 + 1√n
)2
= 1, we conclude thatlimn→∞ n√
n = 1.
Theorem 14.3.limn→∞n√
n!n
= 1e.
152 Some basic limits
Proof. Fork = 1, 2, . . . , n, we have
(k + 1)k
kk< e <
(k + 1)k+1
kk+1.
Multiplying then double inequalities together, we have
n∏
k=1
(k + 1)k
kk< en <
n∏
k=1
(k + 1)k+1
kk+1;
(n + 1)n
n!< en <
(n + 1)n+1
n!;
n + 1
e<
n√
n! <(n + 1)1+ 1
n
e;
(n + 1
n
)
· 1
e<
n√
n!
n<
n + 1
n· (n + 1)
1n
e.
The result follows fromlimn→∞n√
n + 1 = 1.
ci
Supplement 1
Irrationality of k√
n
More generally, for given positive integersn andk, the numberk√
n isirrational unlessn is thek-power of an integer. This fact follows fromthe fundamental theorem of arithmetic. Here is an elementary prooffollowing [?, ?].
Suppose to the contrary thatk√
n is rational. Then, for eachpositive integeri, k
√ni is rational. Therefore, there is an inte-
gerQ such that the numbersQ · k√
ni, i = 1, 2, . . . , k − 1, areall integers. LetP be thesmallestof these. Sincen is not ak-power, there is an integerm such thatmk < n < (m + 1)k.Let c = P · k
√n−Pm. It is a positive integer less thanP . But
for i = 1, 2, . . . , k − 2,
c · k√
ni = P · k√
ni+1 − P · m · k√
ni
is an integer. And fori = k − 1,
c · k√
nk−1 = Pn − m · P · k√
nk−1
is also an integer. This contradicts the minimality ofP .
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 twosequences defined recursively by
an+1 = an + bn,
bn+1 = an + an+1.
Beginning witha1 = 1, b1 = 1, we obtain
n an bn
1 1 12 2 33 5 74 12 175 29 41...
......
Thenlimn→∞bn
an=
√2.
cii Some basic limits
Supplement 2
Example2.5 (b)limn→∞n2n = 0.
Proof. If n ≥ 4, then2n = (1 + 1)n >(
n2
)> n. Therefore,
0 <n
2n<
n12n(n − 1)
=2
n − 1.
Sincelimn→∞2
n−1= 0, we conclude thatn
2n = 0.
(c) limn→∞an
n!= 0.
Proof. We may assumea > 0. Let N = ⌊a⌋, andA = an
N !. Then for
n > N ,
0 <an
n!=
aN
N !· a
N + 1· a
N + 2· · · a
n< A ·
(a
N + 1
)n−N
.
Sincea < N + 1, limn→∞ A ·(
aN+1
)n−N= 0, we conclude that
limn→∞an
n!= 0 as well.
Example 0.1.The following sequences are convergent because they arebounded, monotone sequences:
(a)((
1 − 12
) (1 − 1
4
) (1 − 1
8
)· · ·(1 − 1
2n
));
(b)((
1 + 12
) (1 + 1
4
) (1 + 1
8
)· · ·(1 + 1
2n
));
(c)((
1 − 12
) (1 − 1
4
) (1 − 1
16
)· · ·(1 − 1
22n
));
(d)((
1 + 12
) (1 + 1
4
) (1 + 1
16
)· · ·(1 + 1
22n
)).
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
an =
(
1 +1
2
)(
1 +1
4
)(
1 +1
16
)
· · ·(
1 +1
22n
)
.
Note that1
2an =
(
1 − 1
2
)(
1 +1
2
)(
1 +1
4
)(
1 +1
16
)
· · ·(
1 +1
22n
)
=
(
1 − 1
4
)(
1 +1
4
)(
1 +1
16
)
· · ·(
1 +1
22n
)
...
= 1 − 1
22n+1 .
ciii
From this,limn→∞ an = 2.
Harmonic representation of a real number
Every number between0 and1 can be uniquely written in the form
β =b2
2!+
b3
3!+ · · ·+ bn
n!+ · · ·
for nonnegative integersbn < n, n = 2, 3, . . . .We call this the harmonic representation of the number.(i) Describe precisely the construction ofbn.(ii) Show thatβ is rational if and only if there exists an integerN
such thatbn = 0 for n > N .(iii) Find the harmonic representation of1
7.
(iv) What is the number
∞∑
n=2
n − 1
n!=
1
2!+
2
3!+ · · ·+ n − 1
n!+ · · ·?
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-rowarrangement on a calculating board, each column for one digit:
(i) The second row is labeledDividend. One begins by entering inthis row the number (dividend) whose square root is to be extracted.
(ii) The square root is to appear in the first row, labeledQuotient. 1
(iii) The third row, labeledDivisor, is auxiliary in the computation.(iv) The bottom row is labeledUnit. 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 theNine Chapters of MathematicalArt on the extraction of the square root of3, 972, 150, 625.
1 Quotient3 9 7 2 1 5 0 6 2 5 Dividend
Divisor1 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 theleft, two places in the bottom row for each move of one place inthe toprow, until the unit in the bottom is in the leftmost, below a digit of thequotient. The location of the unit in the top row is the leftmost digit ofthe square root.
The computation begins with finding the largest number whosesquaredoes not exceed the number with unit indicated in the bottom row. In thiscase, the largest square not exceeding39 is 62 = 36. Here we replacethe unit in the top row by6, and subtract36 from the second row.
6 Quotient3 7 2 1 5 0 6 2 5 Dividend
Divisor1 Unit
Multiply the quotient by 2, enter it in the third row (asdivisor), shift-ing one place to the right. At the same time, shift the unit (rod) in thebottom rowtwo places to the right.
1The extraction of square root is seen as a modification of the process of divsion, hence the word quotientfor the square root.
cv
6 Quotient3 7 2 1 5 0 6 2 5 Dividend1 2 Divisor
1 Unit
For the next digit in the quotient, find thelargestnumber which, whenmultiplied to the numberwith the same units digit in the third row, givesa product not exceeding the number in the second row. In the presentcase, this number is3. Subtract the product3 × 123 = 369 from thesecond row.
6 3 Quotient3 1 5 0 6 2 5 Dividend
1 2 3 Divisor1 Unit
Double the number in the first row, enter it in the third row,2 shiftingone place to the right. Shift the unit rod in the bottom row twoplaces tothe right.
6 3 Quotient3 1 5 0 6 2 5 Dividend
1 2 6 Divisor1 Unit
Repeat the same procedure till the unit rod appears at the end. Thenumber 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 twoplaces to the right.
6 3 0 Quotient3 1 5 0 6 2 5 Dividend
1 2 6 0 Divisor1 Unit
2This has the same effect asdoublingthe unit digits of the divisor in the third row.
cvi Some basic limits
6 3 0 Quotient3 1 5 0 6 2 5 Dividend1 2 6 0 Divisor
1 Unit
6 3 0 2 Quotient3 1 5 0 6 2 5 Dividend1 2 6 0 2 Divisor
1 Unit
6 3 0 2 Quotient6 3 0 2 2 5 Dividend
1 2 6 0 2 Divisor1 Unit
6 3 0 2 5 Quotient6 3 0 2 2 5 Dividend1 2 6 0 4 5 Divisor
1 Unit
The next number should be 5. Since5×126045 = 630225, this leavesthe second row blank, and the calculation terminates, giving 63025 forthe square root.
6 3 0 2 5 QuotientDividend
1 2 6 0 4 5 Divisor1 Unit
If the dividend appearing in the last step is a nonzero numberr, thesquare root in question is not “exact”. In this case, the ancient Chineseadopted one of the following options.
(i) Round off the square root with the fractionr2q+1
or r2q
, q being thequotient in the first row. In other words,
√
q2 + r ≈ q +r
2q + 1or q +
r
2q.
cvii
(ii) Continue the calculation “beyond the decimal point” bytreatinga 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)71824, (iv)564, 7521
4.
cviii Some basic limits
Supplement 4
The infinite radical√
2 +
√
2 + · · ·+√
2 + · · · = 2,
as the limit of the sequence(xn) defined recursively by
xn+1 =√
2 + xn, x1 = 2.
In this case, if we putxn = 2 cos tn, then
xn+1 =√
2(1 + cos tn) = 2 costn2
.
Sincex1 =√
2 = 2 cos π4, t1 = π
4, and more generally,tn = π
2n+1 .Therefore, we have an alternative expression
xn = 2 cosπ
2n+1.
From this it is also clear thatlimn→∞ xn = 2.Now, the sequence(4n(2 − xn)) converges to an interesting number.
Note that
4n(2 − xn) = 4n(
2 − 2 cosπ
2n+1
)
= 4n+1 sin2 π
2n+2
=π2
4·(
sin π2n+1
π2n+1
)2
.
Sincelimθ→0sin θ
θ= 1, we have
limn→∞
4n(2 − xn) =π2
4.
Equivalently,
limn→∞
2n ·
√√√√
2 −
√
2 +
√
2 +
√
2 + · · · +√
2
︸ ︷︷ ︸
(n+1)−fold square root
=π
2.
cix
Supplement 6
The equation (6.1) can be simplified:
a2n =
√
2 −√
4 − a2n,
where we have assumed for convenienceR = 1.(i) Beginning witha6 = 1, we have
a6·2k =
√
2 −√
2 +
√
2 + · · · +√
3 (k + 1) − fold square root
Therefore,
limn→∞
2k
√
2 −√
2 +
√
2 + · · ·+√
3︸ ︷︷ ︸
(k+1)−fold square root
=π
3.
(ii) Beginning witha4 = 1, we have
a4·2k =
√
2 −√
2 +
√
2 + · · · +√
2 (k + 1) − fold square root
Therefore,
limn→∞
2k
√
2 −√
2 +
√
2 + · · ·+√
2︸ ︷︷ ︸
(k+1)−fold square root
=π
2.
cx Some basic limits
Supplement 7
Exercise (2). The answer is no. Here is an example. Ifa2 = 34· a1,
a3 = 89· a2, a4 = 15
16· a3, . . . ,an+1 =
(1 − 1
n2
)an, then
an+1 =
(n + 1
n· n − 1
n
)(n
n − 1· n − 2
n − 1
)
· · ·(
4
3· 2
3
)(3
2· 1
2
)
=
(n + 1
n· n
n − 1· · · 4
3· 3
2
)(n − 1
n· n − 2
n − 1· · · 2
3· 1
2
)
=n + 1
2· 1
n
=n + 2
2n.
From this,limn→∞ an = 12.
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 oneof the sides about the right angle is equal to the radius, and the other tothe circumference of the circle.
O
HT
F
E
B C
DA
G
K
Proof. Let ABCD be the given circle,K the triangle described.Then, if the circle is not equal toK, it must be either greater or less.(1) If possible, let the circle be greater thanK.Inscribe a squareABCD, bisect the arcsAB, BC, CD, DA, then
bisect (if necessary) the halves, and so on, until the sides of the inscribedpolygon whose angular points are the points of division subtend seg-ments whose sum is less than the excess of the area of the circle overK.
Thus, the area of the polygon is greater thanK.Let AE be any side of it, andON the perpendicular onAE from the
centerO.ThenON is less than the radius of the circle and therefore less than
one of the sides about the right angle inK. Also the perimeter of thepolygon is less than the circumference of the circle,i.e. less than theother side about the right angle inK.
Therefore, the area of the polygon is less thanK; which is inconsis-tent with the hypothesis.
Thus the area of the circle is not greater thanK.(2) If possible, let the circle be less thanK.Circumscribe a square, and let two adjacent sides, touchingthe circle
in E, H, meet inT . Bisect the arcs between adjacent points of contact
cxii Some basic limits
and draw the tangents at the points of bisection. LetA be the middlepoint of the arcEH, andFAG the tangent atA.
Then the angleTAG is a right angle.Therefore,TG > GA = GH.It follows that the triangleFTG is greater than half the areaTEAH.Similarly, if the arcAH be bisected and the tangent at the point of
bisection be drawn, it will cut off from the areaGAH more than one-half.
Thus, by continuing the process, we shall ultimately arriveat a cir-cumscribed polygon such that the spaces intercepted between it and thecircle are together less than the excess ofK over the area of the circle.
Thus, the area of the polygon will be less thanK.Now, since the perpendicular fromO on any side of the polygon is
equal to the radius of the circle, while the perimeter of the polygon isgreater than the circumference of the circle, it follows that the area ofthe polygon is greater than the triangleK; which is impossible.
Therefore the area of the circle is not less thanK.Since then the area of the circle is neither greater nor less thanK, it
is equal to it.
An ‘explanation’ of the rational approximation for√
3 adopted byArchimedes
Basic inequalities:
a ± b
2a>
√a2 ± b > a ± b
2a + 1.
(i) Start with2− 14
>√
3 > 2− 13
or 74
>√
3 > 53, or 21
4>
√27 > 5.
(ii) Apply the basic inequalities to√
27 =√
52 + 2:
5 +2
10>
√27 > 5 +
2
11,
26
15>
√3 >
19
11.
(iii) Note that3 · 152 = 675 = 676 − 1 = 262 − 1.
Exercise
Use the basic inequalities to deduce that
1351
780>
√3 >
265
153.
cxiii
Supplement 10: Vieta’s infinite product for π
Beginning with the trigonometric identity
sin θ = 2 sinθ
2cos
θ
2,
we replace byθ successively byx, x2, x
x2 , . . . , x2n−1 , multiply the results
together, simplifying, we obtain
sin x
2n sin x2n
=
n∏
k=1
cosx
2k.
For a fixedx > 0,
limn→∞
2n sinx
2n= lim
n→∞x · sin x
2n
x2n
= x.
Therefore, we have an infinite product formula:∞∏
n=1
cosx
2n=
sin x
x.
In particular, withx = π2, let cn := cos π
2n+1 . These values can becomputed iteratively by
2cn+1 =√
2 + 2cn.
These are
c1 =
√2
2,
c2 =
√
2 +√
2
2,
c3 =
√
2 +√
2 +√
2
2,
...
From these, we have Vieta’s formula3
2
π=
√2
2
(√
2 +√
2
2
)
√
2 +√
2 +√
2
2
· · · .
3Francois Vieta (1540–1603). This is historically the firstexact expression forπ.
cxiv Some basic limits
cxv
Supplement 11A: Proofs of the AGI
Second proof
Denote byA andG the arithmetic and geometric means ofa1, a2, . . . ,an.
(1) If a1 = a2 = · · · = an, then clearlyA = G.(2) If a1, a2, . . . ,an are not all equal, we arrange them in descending
order:a1 ≥ a2 ≥ · · · ≥ an.
In this sequence,a1 > G andan < G.(3) Consider the new set of numbers
a′1 = G, a′
i = ai for i = 2, . . . , n − 1, a′n =
a1an
G.
Let A′ and G′ be their arithmetic and geometric means respectively.Clearly,G′ = G. On the other hand,
A′ =1
n(a′
1 + a′n + a′
2 + · · ·a′n−1)
=1
n(G +
a1an
G+ a2 + · · ·an−1).
Note thatG + a1an
G< a1 + an, because
G2 + a1an − G(a1 + an) = (G − a1)(G − an) < 0.
This givesA′ < A. Sorting the numbersa′1, a′
2, . . . ,a′n in descending
order, we obtain a new sequence
b1 ≥ b2 ≥ · · · ≥ bn
in which(i) the geometric mean remains the same,(ii) the arithmetic mean is diminished, and(iii) the number of terms equal toG is increased by1.
In view of (iii), in a finite number of steps (not more thann − 1), thesequence would be replaced by one in which each term is equal to G.For such a sequence, the arithemtic mean is equal toG. It follows thatA > G if the given numbers are not all equal to begin with.
cxvi Some basic limits
Third Proof
Given positive numbersa1, . . . , an, . . . , denote byAn andGn respec-tively the arithmetic and geometric means ofa1, . . . ,an. For a givenn,consider the function
fn(x) =1
n+1(nAn + x)
n+1√
Gnn · x
Using the quotient rule, we compute the derivative (exercise):
f ′n(x) =
n(x − An)
(n + 1)2Gn
n+1n · xn+2
n+1
.
From this it is clear thatfn(x) attains its minimum atx = An, and theminimum value is
fn(An) =
(An
Gn
) nn+1
.
With this preparation, we prove the AGI by induction.(i) A2 ≥ G2; equality holds ifa1 = a2 (Lemma 11.1 above).(ii) AssumeAn ≥ Gn. Then fora1, . . . , an+1, we have
An+1
Gn+1
= f(an+1) ≥ f(An) =
(An
Gn
) nn+1
≥ 1,
showingAn+1 ≥ Gn+1. Moreover,An+1 = Gn+1 if and only iff(an+1) = 1. This meansf(an+1) = f(An) = 1, andAn = Gn.From this it follows thata1 = · · · = an = an+1.
Therefore, by the principle of mathematical induction, thearithmeticand 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
f(x) = n · n√
a1a2 · · ·an−1x − x
for x > 0 anda1 > 0, a2 > 0, . . . ,an−1 > 0.
cxvii
Fourth proof
We make use of the concavity of the function ofy = log x. 4 Considerthe tangent line at the point(1, 0). This is the liney = x − 1, and isentirely above the graph ofy = log x.
y = x − 1
y = log x
x3G
xn−1G
xnG
· · ·
x1G
x2G
1
Let A andG be respectively the arithmetic and geometric means ofnpositive numbersx1, x2, . . . ,xn.
nA − nG
G=
n∑
k=1
(xk
G− 1)
≥
n∑
k=1
logxk
G= log
x1x2 · · ·xn
Gn= log
Gn
Gn= 0.
Therefore,A ≥ G, and equality holds if and only ifx1 = x2 = · · · =xn(= G).
Remark.This “proof with few words” can be found inCMJ, 31 (2000)68, wherex1
G, x2
G, . . . , xn
Gare erroneously all depicted on the right hand
side of1.
Exercise
(1) Leta1, a2, . . . ,an be positive real numbers. Show that
1
a1+
2√a2
+3
3√
a3+ · · · + n
n√
an≥ s
s√
a1a2 · · ·an,
4This has been shown in Exercise 1 above. Therefore, Exercise3 furnishes a proof following the line ofthe First proof.
cxviii Some basic limits
wheres = 1 + 2 + 3 + · · ·+ n.(2) Find the positive integern for which 2
n+ n2
125is least possible.
What if 125 is replaced by52? and by53?
Remark.Let a1 ≤ a2 ≤ · · · ≤ an be positive numbers with arithmeticmeanA and geometric meanG. Let S =
∑
1≤i<j≤n(aj − ai)2.
S
2n2 · an≤ A − G ≤ S
2n2 · a1,
a31 · S
2n2 · a4n
≤ G − H ≤ a3n · S
2n2 · a41
.
See [?, ?].
Fifth proof
We make use of the fact that the functionF (x) = log xx
has maximumF (e) = 1
e. Given positive numbersa1, . . . , an, with arithmetic meanA
and geometric meanG, let xk = akeG
for k = 1, 2, . . . , n. For each of
these,F (xk) =G log
ake
G
ake< 1
e, and
log ak + 1 − log G <ak
G.
Combining these inequalities, we have
n∑
k=1
(log ak + 1 − log G) ≤n∑
k=1
ak
G,
log(a1 · · ·an) + n − n log G ≤ nA
G.
Since log(a1 · · ·an) = log Gn = n log G, this becomesn ≤ nAG
andA ≥ G.
Sixth proof
This proof makes use of the increasing property of the sequence((
1 + xn
)n)
for every positive numberx. More precisely, we shall make use of
(
1 +x
n
)n
≥(
1 +x
n − 1
)n−1
,
cxix
for x ≥ 0. Here, equality holds if and only ifx = 0.Givenn positive numbersa1, . . . ,an, assumed in ascending order, so
the each numberak is no more than the arithmetic mean ofak, ak+1, . . . ,an.
(a1 + a2 + · · · + an
na1
)n
=
(na1 − (n − 1)a1 + a2 + · · ·+ an
na1
)n
=
(
1 +−(n − 1)a1 + a2 + · · · + an
na1
)n
≥(
1 +−(n − 1)a1 + a2 + · · ·+ an
(n − 1)a1
)n−1
=
(a2 + · · · + an
(n − 1)a1
)n−1
.
Simplifying and continuing, we have(
a1 + a2 + · · ·+ an
n
)n
≥ a1 ·(
a2 + · · · + an
n − 1
)n−1
≥ a1a2 ·(
a3 + · · ·+ an
n − 2
)n−2
...
≥ a1a2 · · ·an−2 ·(
an−1 + an
2
)2
≥ a1a2 · · ·an.
Equality holds if and only if for eachk, ak is the arithmetic mean ofak, ak+1, . . . ,an. This is the case whena1 = a2 = · · · = an.
Seventh proof
Note that
xn − nx + n − 1 = (x − 1)2(xn−2 + 2xn−3 + · · · + (n − 1)) ≥ 0
for x > 0. Now given positive numbersa1, a2, . . . , an, let x = An
An−1.
We have (An
An−1
)n
− n · An
An−1
+ (n − 1) ≥ 0,
cxx Some basic limits
orAn
n ≥ An−1n−1(nAn − (n − 1)An−1) = An−1
n−1 · an.
Continuing, we have
Ann ≥ An−1
n−1·an ≥ An−2n−2·an−1an ≥ · · · ≥ A1·a2 · · ·an = a1a2 · · ·an = Gn
n.
Therefore,An ≥ Gn. See [?].
cxxi
Supplement 11B: Some useful inequalities
Cauchy-Schwarz inequality
Given real numbersa1, a2, . . . ,an andb1, b2, . . . ,bn,
(a1b1 + a2b2 + · · ·+ anbn)2 ≤ (a21 + · · ·+ a2
n)(b21 + · · · + b2
n).
Equality holds if and only if(a1, a2, . . . , an) and (b1, b2, . . . , bn) arelinearly dependent.
Proof. The quadratic equation
(a1t − b1)2 + (a2t − b2)
2 + · · ·+ (ant − bn)2 = 0
has at most one real roots. Therefore, its discriminant
(a1b1 + a2b2 + · · · + anbn)2 − (a21 + · · · + a2
n)(b21 + · · ·+ b2
n) ≤ 0.
Equality holds if and only if there is a real numbert satisfyingakt = bk
for k = 1, 2, . . . , n.
Bernoulli’s inequality
If n > 1 is an integer,
(1 + x)n > 1 + nx for x ≥ −1.
For positivex this is clear by expansion using the binomial theorem.To extend the range ofx for the validity of the inequality, let us as-
sume that the inequality is true for a positive integern. Then
(1 + x)n+1 = (1 + x)n · (1 + x) > (1 + nx)(1 + x) > 1 + (n + 1)x,
providedthat1 + x is positive. Since the inequality(1 + x)2 > 1 + 2xis clearly true (for all values ofx), by induction, Bernoulli’s inequalityis true forn ≥ 2 andx > −1.
Bernoulli’s inequality indeed holds forx ≥ −2. This is clear forn = 1.We shall assumen ≥ 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 implies1 − 2n ≤ 1 + nx < 1 − n.
Forn ≥ 2, we have
1 − 2n ≤ 1 + nx < 1 − n ≤ −1 ≤ (1 + x)n < 1.
This shows that Bernoulli’s inequality can be improved tox ≥ −2.
cxxiii
Supplement 11C: Maxima and minima without calculus
Example 0.2. Find the dimensions of the rectangle of maximum areathat can be inscribed in the ellipsex2
a2 + y2
b2= 1.
Solution. If the corner of the rectangle in the first quadrant has coordi-nates(x, y), its area is4xy. Now,
1
2
(x2
a2+
y2
b2
)
≥√
x2
a2· y2
b2=
xy
ab
givesxy ≤ 12ab, with equality whenx2
a2 = y2
b2= 1
2, i.e., x = a√
2and
y = b√2. The dimensions of the maximum rectangle are
√2a ×
√2b.
Example 0.3.Find the maximum volume of a circular cylinder inscribedin a right circular cone of base radiusr and heighth.Solution. Suppose the cylinder has radiusx and heighty. It has volumeV = πx2y. Note thatx
r+ y
h= 1. Rewriting this asx
2r+ x
2r+ y
h= 1,
we see that the product(
x2r
)2 · yh
= 14r2h
· x2y is maximum whenx2r
=yh
= 13, i.e., x = 2r
3andy = h
3. The maximum volume of the cylinder is
Vmax = π(
2r3
)2 · h3
= 4πr2h27
.
Example 0.4.What is the length of the shortest line in the first quadrantthat is tangent to the ellipsex
2
a2 + y2
b2= 1.
Solution. If the point of tangency is(p, q), the equation of the tangent
is pxa2 + qy
b2= 1. This intersects thex- andy- axes at the points
(a2
p, 0)
and(
0, b2
q
)
. The squared length of the tangent in the first quadrant is
a4
p2+
b4
q2=
(a4
p2+
b4
q2
)(p2
a2+
q2
b2
)
≥(
a2
p· p
a+
b2
q· q
b
)2
= (a + b)2.
The minimum length of the tangent is thereforea + b. Equality holds
when(
a2
p, b2
q
)
and(
pa, q
b
)are linearly dependent,i.e., p2
a3 = q2
b3:
p2
a2
a=
q2
b2
b=
p2
a2 + q2
b2
a + b=
1
a + b.
Therefore,p2 = a3
a+bandq2 = b3
a+b. The point of tangency is
(
a√
aa+b
, b√
ba+b
)
.
cxxiv Some basic limits
a
b
x
Example 0.5.To cut equal squares of dimensionx from the four cornersof a rectangle of lengtha and breadthb so that the box obtained byfolding along the creases has a greatest capacity.
The capacity of the box isV = x(a−2x)(b−2x). We find appropriatenumbersλ andµ and maximize
2(λ + µ)x · λ(a − 2x) · µ(b − 2x).
These three factors have a constant sumλa + µb. Their product ismaximum when
2(λ + µ)x = λ(a − 2x) = µ(b − 2x),
orλ + µ
12x
=λ1
a−2x
=µ1
b−2x
=λ + µ
1a−2x
+ 1b−2x
.
In other words,1
2x=
1
a − 2x+
1
b − 2x.
This reduces to the quadratic equation12x2 − 4(a + b)x + ab = 0, witha root
x =a + b −
√a2 − ab + b2
6<
1
2· min(a, b).
This maximizes the product2(λ + µ)x · λ(a − 2x) · µ(b − 2x); hencealsoV = x(a − 2x)(b − 2x).
Example 0.6. Find the maximum and minimum ofy = a2
x+ b2
a−x. As-
sumea > b > 0.Solution. Rewrite the given relation as a quadratic equation inx:
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
From this, we havey ≤ (a−b)2
a(maximum) ory ≥ (a+b)2
a(minimum).
Correspondingly,x = a2
a−band a2
a+b.
Exercise
Show that(a−x)(x−b)x2 has maximum(a−b)2
4abwhenx = 2ab
a+b.
Example 0.7. Find the maximum volume of a right circular cone in-scribed in a sphere of radiusr.
cxxvi Some basic limits
Supplement 13: On the numbers(1 + 1
n
)nand
(1 + 1
n
)n+1
Theorem 0.4(Euler). If x < y are rational numbers satisfying the equa-tion xy = yx, then
(x, y) =
((
1 +1
n
)n
,
(
1 +1
n
)n+1)
for some positive integern.
Proof. Rewrite the equation asy = xy
x and putyx
= 1 + mn
for relativelyprime integersm andn.
x1+ mn = m+n
n· x,
xmn = m+n
n.
Therefore,x = (m+n)nm
nnm
=(
(m+n)n
nn
) 1m
.
Sincem andn do not have common divisors, this is a rational numberif and only if both(m+n)n andnn arem-th powers. Sincegcd(m, n) =1, this means thatm+n andn are bothm-th powers. Writen = am andm + n = bm for integersa < b. Now, bm − am = m is possible onlywhenm = 1. Therefore,
x =(1 + n)n
nn=
(
1 +1
n
)n
,
and
y =
(
1 +1
n
)
x =
(
1 +1
n
)n+1
.
Exercise
Determine five pairs of positive integersp > q for which there existgeometric progression in which theq-th term isp, thep-th term isq, andthe(p + q)-term is an integer.Solution. Supposearp−1 = q, arq−1 = p, andarp+q−1 = b, whereb isan integer. Then,
r =
(b
q
) 1q
=
(b
p
) 1p
,
1
and(p
b
) q
b
=(q
b
)p
b
.
Therefore, there is an integern for which
p
b=
(n + 1)n+1
nn+1,
q
b=
(n + 1)n
nn=
n(n + 1)n
nn+1.
It follows thatp = k(n + 1)n+1, q = kn(n + 1)n andb = knn+1 forsome integerk. The common ratio is
r =
(n
n + 1
) 1k(n+1)n
.
With k = 1, we have the following5 pairs ofp > q (along withb):
p 4 27 256 3125 46656 . . .
q 2 18 192 2500 38800 . . .
b 1 8 81 1024 15625 . . .
