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
ii
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 . . .
Chapter 15
The parabola
Let F be a given point andℓ a given straight line on a plane. The locusof a variable pointP at equal distances fromF and ℓ is a parabola,with focusF and directrixℓ. If we set up a coordinate system so thatF = (a, 0) andℓ is the linex = −a, then the parabola has equation
y2 = 4ax.
Each point on the parabola has coordinates(at2, 2at) for somet. Weshall simply call this “the pointt of the parabola”.
directrix
O F = (a, 0)−a
P(at2 , 2at)
L
We collect some useful information on the parabola.(1) The line joining the pointsP (t1) andP (t2) has equation
2x − (t1 + t2)y + 2at1t2 = 0.
This line passes through the focusF if and only if t1t2 = −1.
202 The parabola
(2) Lettingt1, t2 → t, we obtain the equation of the tangent att:
x − ty + at2 = 0.
It has slope1t
and intersects the axis of the parabola at(−at2, 0).(3) The normal att is the line perpendicular to the tangent att. It has
equationtx + y = at3 + 2at.
(4) Construction of tangent and normal att: the circle, centerF ,passing throught, intersects the axis of the parabola at two points. Oneof these is on the tangent, and the other on the normal.
O F
t
(5) The area of the triangle with verticest1, t2, t3 on the parabola is
1
2
∣∣∣∣∣∣
at21 2at1 1at22 2at2 1at23 2at3 1
∣∣∣∣∣∣
= −a2(t1 − t2)(t2 − t3)(t3 − t1).
This area is positive or negative according as the trianglet1t2t3 is tra-versed counterclockwise or clockwise.
203
Exercise
(1) Use integral calculus to find the area bounded by the parabola andthe line joining the pointst1 andt2 on the parabola.
O
t2
t1
y2 = 4ax
(2) The circle through the three pointst1, t2, t3 on the parabolay2 =4ax intersects it at a fourth point. What is this point?
204 The parabola
Chapter 16
Archimedes’ quadrature of theparabola
Archimedes solved the problem of quadrature of the parabolawithoutusing integral calculus. He expressed the answer in the following form:
Theorem 16.1(Archimedes). Given two pointsA andB on a parabola,let C be the point on the parabola the tangent at which is parallel to theline AB. Then the area of the parabola segmentAB is 4
3of the area of
the triangleABC.
O
B
A
y2 = 4ax
C
Archimedes called the pointC the vertex of the parabolic segment
206 Archimedes’ quadrature of the parabola
AB, and established (Quadrature of the Parabola, Proposition 18) thatif A andB are the pointst1 andt2 on the parabolay2 = 4ax, thenC isthe pointt1+t2
2.
Proof. The slope of the lineAB is 2t1+t2
. Since the tangent att has slope1t, it is parallel toAB if t = t1+t2
2.
This is the point on the parabola through which the parallel to the axisbisects the chordt1t2.
In Propositions 19–21, Archimedes showed that ifD1 andD2 are thevertices of the parabolic segmentsAC andCB, then
∆ACD1 = ∆CBD2 =1
8· ∆ABC.
O
B
A
y2 = 4ax
C
D1
D2
Proof. (1) ∆ABC = −a2(t1 − t2)(t2 − t1+t2
2
) (t1+t2
2− t1
)= a2
4· (t2 −
t1)3.(2) Replacingt2 by t1+t2
2, we obtain∆ACD1 = 1
8·∆ABC; similarly
with ∆CBD2, by replacingt1 by t1+t22
.
We paraphrase this with a change of terminology. Instead of callingC the vertex of the parabolic segment, we call it the midpoint of theparabolic arcAB. The above result can be restated as follows. LetC
207
be the midpoint of the parabolic arcAB. By introducing the midpointsof the parabolic arcsAC andCB, we obtain a polygon whose area is14
extra of the triangleABC. Repeated introduction of the midpoints ofparabolic arcs increases the area of the polygon by1
4of the preceding
increment. Therefore, the area of the parabolic segmentAB is the sumof the geometric progression, beginning with∆ABC, and with commonratio 1
4. This is
(
1 +1
4+
(1
4
)2
+ · · ·)
∆ABC =4
3· ∆ABC.
Remark.Prior to this, Archimedes had solved the problem in a differentway, using geometry and mechanics, detailed in his workThe Methods,which was only known in the 19-th century.
208 Archimedes’ quadrature of the parabola
Chapter 17
Quadrature of the spiral
A radius of a circle, centerO, is rotating as a constant rate. At the sametime, a pointP is moving on the radius at a constant rate, beginning withO and finishing at the end of the radius when it completes one revolution.The locus ofP is a spiral. Archimedes computes the area swept out bythe radius in one complete revolution. This is the problem ofquadratureof the spiral, with solution in his bookOn Spirals.
O A
Theorem 17.1(On Spirals, Proposition 24). The area swept out by thefirst turn of the spiral is one third of the area of the circle.
210 Quadrature of the spiral
Proof. Divide the circle inton equal parts and consider the positions ofP on the corresponding radii. The area in question is divided into nsectors. A typical one has area between the areas of two circular sectors.Specifically fork = 1, 2, . . . , n, thek-th sector has area between twocircular sector of angle1
nof the circumference, and radiik−1
n· R and
kn· R. Its areaAk satisfies
1
n
(k − 1
n
)2
<Ak
Area of circle<
1
n
(k
n
)2
.
Therefore,
12 + 22 + · · · + (n − 1)2
n3<
Area of spiralArea of circle
<12 + 22 + · · · + n2
n3.
Archimedes made use of a formula equivalent to
12 + 22 + 32 + · · ·+ n2 =1
6n(n + 1)(2n + 1),
which he established as Proposition 10.1 The above double inequalitybecomes
(n − 1)(2n − 1)
6n2<
Area of spiralArea of circle
<(n + 1)(2n + 1)
6n2.
Whenn is large, the limits of the two ends are each equal to13. There-
fore, the area of the spiral is one third of the area of the circle.
The use of the spiral for angle trisection
A
O
P
Q1
Q2
P1
P2
1Also, Lemma to Proposition 2 ofOn Conoids and spheroids.
211
To trisect an angle, place it in positionAOP with O at the vertexof the spiral,A along the initial radius, andP on the spiral. Dividethe segmentOP into three equal parts atP1 andP2. With O as center,construct circular arc throughP1, P2 to intersect the spiral atQ1 andQ2.The linesOQ1 andOQ2 are the trisectors of angleAOP .
212 Quadrature of the spiral
Chapter 18
Apollonius’ extremum problemson conics
Book V of Apollonius’ Conicstreats minimum and maximum lines forconics, in particular the parabola.
Let B(b, 0) be a point on the axis of the parabolay2 = 4ax. To finda pointP on the parabola such thatBP is minimum. LetP be the point(at2, 2at) on the parabola. The square distance
BP 2 = (at2 − b)2 + (2at)2
= a2t4 − 2a(b − a)t2 + b2
= (at2 − (b − a))2 + a(a − 2b)
is a quadratic int2.
2a
O F
P
BQ
(1) If b ≤ a, this is minimum whent = 0, i.e., P is the vertex of theparabola, and the minimumBP = b.
214 Apollonius’ extremum problems on conics
(2) If b > a, this is minimum whenb − a = at2, i.e., if BP is mini-mum andQ is the orthogonal projection ofP on the axis, thenBQ = 2a.
This is Proposition V.8 ofConics. In V.27, Apollonius shows thatBP is perpendicular to the tangent atP . In other words,BP is a normalto the parabola.
Apollonius also considers the same problem for ellipses andhyper-bolas. In the case of the ellipsex
2
a2 + y2
b2= 1, let B be the point(u, 0).
ForP (a cos t, b sin t) on the ellipse,
BP 2 = (a cos t − u)2 + (b sin t)2
= · · ·
= c2(
cos t − au
c2
)2
+b2(c2 − u2)
c2,
wherec2 = a2 − b2. The foci F andF ′ of the ellipse are the points
(±c, 0). Let Q andQ′ be the points with coordinates(
± c2
a, 0)
.
(1) If B is betweenQ andQ′, then the minimum pointP (a cos t, b sin t)has parametert given bycos t = u
OQ. In this case, the lineBP is per-
pendicular to the tangent atP .(2) If B is outside the segmentQQ′, then the minimum point is one
of the verticesA, A′ of the ellipse.
O FF ′ QQ′ B
AA′
P
t
Chapter 19
Normals of a parabola
Given a pointA with coordinates(x, y), what is the point on the parabolaP : y2 = 4ax closest toA? If P is the pointt on the parabola, the squaredistance ofAP is given by
F (t) = (at2 − x)2 + (2at − y)2.
Consider the derivatives ofF as a function oft:
F ′(t) = 2at(at2 − x) + 2a(2at − y)
= 2a(at3 + 2at − tx − y),
F ′′(t) = 2a(3at2 + 2a − x).
If F (t) is minimum, thenF ′(t) = 0. This condition is exactly the sameas the equation of the normal att. Thus,if AP is minimum, thenP lieson the normal throughA.
In general, through a point(x, y), there are3 normal lines. The cubicequation
at3 + (2a − x)t − y = 0
has a double root if and only if(
2a − x
3a
)3
+( y
2a
)2
= 0,
or27ay2 = 4(x − 2a)3.
This is a semicubical parabolaQ parametrized by
x = 2a + 3as2, y = 2as3.
216 Normals of a parabola
The pointQ(s) lies on the normals ofP at P (−s) andP (2s). SinceF ′′(−s) = 0 andF ′′(2s) > 0, P (−s) is a point of inflection andP (2s)a minimum forF (t).
O F 2a
Q(s)
P (−s)
P (2s)
Q(−2s)
Proposition 19.1.The normal atP (t) to P
(i) intersectsP again atP (t′), wheret′ = −t − 2t,
(ii) is tangent toQ at Q(−t), and intersects it atQ(− t
2
).
Exercise
(1) Find the normal toy2 = 4ax which passes throughP (t).(2) Find the minimum normal chords of the parabolay2 = 4ax.(3) Calculate the length of the semicubical parabolaQ between the
pointsQ(0) andQ(s).
Chapter 20
Envelope of normal to a conic
What we have found in§19 for parabolas extend to parametric curves.Consider a curveC with parametrization
x = f(t), y = g(t).
At the pointt on the curve, the tangents and the normals are the lines
g′(t)(x − f(t)) − f ′(t)(y − g(t)) = 0,
f ′(t)(x − f(t)) + g′(t)(y − g(t)) = 0.
We consider the problem of minimizing the distance of a pointon thecurve from a given pointP (x, y). The square distance is given by
F (t) = (x − f(t))2 + (y − g(t))2.
dF (t)
dt= −2(f ′(t)(x − f(t)) + g′(t)(y − g(t)).
This means thatt is a critical point ofF if and only if the normal attcontainsP .
The envelope of the normal
Consider the normal ofC at the pointt, with equation given by
f ′(t)(x − f(t)) + g′(t)(y − g(t)) = 0.
218 Envelope of normal to a conic
At a neighboring pointt′, the normal is given by the same equation witht replaced byt′. If we write t′ = t + ε for a smallε, and replacef(t′) byf(t) + f ′(t)ε, f ′′(t) by f ′(t) + f ′′(t)ε, and similarly forg(t) andg′(t),then the normal at this neighboring point is the line
f ′(t)(x − f(t)) + g′(t)(y − g(t))
+ ε(f ′′(t)(x −′ f(t)) + g′′(t)(y − g′(t))
= 0.
Therefore, solving the system of equations
f ′(t)(x − f(t)) + g′(t)(y − g(t)) = 0,
f ′′(t)(x − f ′(t)) + g′′(t)(y − g′(t)) = 0,
we obtain the intersection of two neighoring normals, namely,{
x = f(t) − g′(t) · f ′(t)2+g′(t)2
f ′(t)g′′(t)−f ′′(t)g′(t),
y = g(t) + f ′(t) · f ′(t)2+g′(t)2
f ′(t)g′′(t)−f ′′(t)g′(t).
(20.1)
This is called thecenter of curvature at P (t). With this point ascenter, the circle throughP is tangent to the curveC. The radius of thiscircle is
ρ =
∣∣∣∣∣
(f ′(t)2 + g′(t)2)32
f ′(t)g′′(t) − f ′′(t)g′(t)
∣∣∣∣∣.
Thecurvature atP (t) is κ := 1ρ.
Example 20.1.The normals to a circle all pass through the center of thecircle. This center is the center of curvature for every point on the circle.The circle has constant curvature.
219
Example 20.2.For the parabola(x, y) = (at2, 2at), the center of curva-ture ofP (t) is the pointQ(−t).
O F 2a
P (t)
Q(−t)
We may also take (20.1) as defining theenvelopeof the normal asa parametric curveC∗. This is also called theevolute of C. Thus, thetangents ofC∗ are the normals toC, generalizing Proposition 19.1.
Example 20.3.Consider the ellipsex2
a2 + y2
b2= 1 with parametrization
x = a cos t, y = b sin t,
wherea > b (so that the foci are on thex-axis). The evolute is theparametric curve
x =a2 − b2
a· cos3 t, y = −a2 − b2
b· sin3 t.
Eliminatingt, we have
(ax)23 + (by)
23 = c
43 .
This curve is called anastroid.
220 Envelope of normal to a conic
O
P
Q
Example 20.4. For the hyperbolax2
a2 − y2
b2= 1 with parametrization
(x, y) = (a sec t, b tan t), the evolute is the parametric curve
x =a2 + b2
a· sec3 t, y = −a2 + b2
b· tan3 t.
O
P
Q
Chapter 21
The cissoid
The cissoid of Diocles was invented to solve the problem of two meansproportions: Given two quantitiesa andb, to find x andy such thata,x, y, b are in geometric progression. Ifb = 2a, this is the problem ofduplication of the cube.
Given a diameterAB of a circle(O), for every pointP on the circle,let P ′ be the reflection ofP in the diameter perpendicular toAB, andQthe intersection of the lineAP and the perpendicular toAB throughP ′.The locus ofQ is the cissoid.
BAO
PP ′
Q
X
222 The cissoid
Consider(O) as the unit circle,A andB the points(−1, 0) and(1, 0)respectively. If angle∠PAB = t, thenP has coordinates(cos 2t, sin 2t),andQ is the point
(−1, 0) +1 − cos 2t
1 + cos 2t· (1 + cos 2t, sin 2t) = (− cos 2t, 2 sin2 t tan t).
Proposition 21.1.Let the lineP ′Q intersect the diameterAB at X. Thefour quantitiesXB, XP ′, AX, andXQ are in geometric progression.
Proof. The four quantities
XB = 1 + cos 2t = 2 cos2 t,
XP ′ = sin 2t = 2 sin t cos t,
AX = 1 − cos 2t = 2 sin2 t,
XQ = 2 sin2 t tan t
form a geometric progression with common ratiotan t.
Application to the duplication of the cube
Let CD be the diameter perpendicular toAB, andM the midpoint ofOC. JoinB, M to intersect the cissoid atQ. Let X be the intersectionof the diameterAB with the perpendicular fromQ. ThenXB : XQ =OB : OM = 2 : 1, andXP ′ andAX are the two mean proportionsbetween them. In particular,AX = 3
√2 · OM .
B
M
AO
P ′
Q
C
D
X
223
MacLaurin’s trisectrix
Given two curvesC1, C2, and a poleA, for every line throughA inter-secting the curves atP1 andP2, let Q be the point on the line such thatAQ = AP1 −AP2. The locus ofQ is the cissoid with respect to the twogiven curves.
Consider a circleO(A) and the perpendicular bisector of the radiusOA. The cissord with respect to these two curves and with poleA isMacLaurin’s trisectrix.
ABO
Q
P1
P2
To trisect an angle, place it in positionBOQ with Q on the trisectrix.JoinA, Q to intersect the circle atP1. Then the lineOP1 a trisector ofthe angle.
Proof. Note that trianglesOAQ andOP1P2 are congruent. Therefore,OQ = OP2 = AP2 = QP1, and
∠BOP1 = 2∠OP1Q = 2∠QOP1.
This shows thatOP1 trisects angleBOQ.
224 The cissoid
Exercise
(1) Show that the trisectrix has polar equationr = a cos t − a4 cos t
.(2) Calculate the area enclosed by the loop of the trisectrix.
Newton’s construction of the cissoid.
Let ABCD be a given square, each side of lengtha. Move an identicalsquarePQRS so that one vertexP slides on the sideAD and the sideQR always passes through the fixed pointB. The locus of the midpointof PQ is the cissoid.
Chapter 22
Conchoids
The conchoids of Nicomedes (2nd century BC) was invented to solvethe problem of neuxus, and applied to the trisection of an angle. Givena lineℓ, a pointA, and a lengthb, the chord with axisℓ and poleA forlengthb is the locus of a pointQ such that when the lineAQ intersectℓatP , the length ofPQ is equal tob.
A
P
Q
Q′
Let ℓ be thex-axis, A = (0,−a) for a > 0. Given a nonzero realnumberb, we consider the pointQ on the line joiningA to P = (t, 0)such thatPQ = b.
(1) The coordinates ofQ are
(x, y) = (t, 0) +b√
a2 + t2· (t, a).
226 Conchoids
(2) If we write t = a tan θ, this becomes
(x, y) = (a tan θ + b sin θ, b cos θ).
(i) |b| < a.
a
θ
A
O P
Q
Q′
(ii) |b| = a.
A
O
P
Q
Q′
(iii) |b| > a shown above.
Exercise
If b > a, the conchoid with length−b has a loop. What is the areaenclosed by the loop?
A
227
This is how the conchoid can be used for the trisection of an angle.Given an acute angleOAB to trisect (assumingOB perpendicular toOA), with A as pole and the lineOB as axis, construct the conchoidwith length2 · AB. Construct the parallel toOA throughB, to intersectthe conchoid atQ. Then the lineAQ trisects angleAOB.
A
O B
C
Q
P
M
Proof. JoinA andQ to intersect the axis atP . Let M be the midpointof PQ. SincePQ = 2 · AB and ∠PBQ is a right angle, we haveMQ = MB = AB, and
∠MAB = ∠AMB = ∠MBQ + ∠MQB = 2∠MQB = 2∠MAO.
This shows thatAM trisects angleOAB.
More general conchoids
Let C be a given curve, andA a point,k a fixed length. IfP is a varyingpoint onC and on the lineAP we mark a pointQ such that the directedlengthPQ = k (positive if PQ is along the same direction ofAP ), thelocus ofQ is called theconchoidCA,k.
228 Conchoids
Conchoids on a circle
The conchoids on a circle are easily described in polar coordinates:
r = a cos θ − b.
(i) |b| < a (limacon):
AB
Q
Q′
P
(ii) |b| = a (cardioid):
AB
Q
Q′
P
229
(iii) |b| > a:
AB
Q
Q′
P
Application to trisection of an angle
Etienne Pascal (1588-1640)1 made use of the conchoid witha = 2b totrisect an angle.
AB
Q
P
O
Exercise
(1) Consider the unit circlex2 + y2 = 1. Show that The envelope of theline joining (cos t, sin t) to (cos 2t, sin 2t) is a cardioid.
1Father of Blaise Pascal (1623–1662).
230 Conchoids
(2) Show that for a fixedk, the locus of the point dividing the segmentjoining (cos t, sin t) to (cos 2t, sin 2t) in the ratiok : 1−k is a conchoidon a circle.
The witch of Agnesi
Let AB be a diameter with a circle. Given a pointP on the circle, joinA andP to meet the tangent atB at a pointR. The parallel throughPand the perpendicular throughQ to AB intersect atQ. The locus ofQ isthewitch of Agnesi. 2 It has parametric equations
x = a cot t, y = a sin2 t.
A
R
QP
B
t
t
t
a cot t
a
a sin t
??
Exercise
(1) Locate the points of inflection of the witch of Agnesi.(2) Calculate the area under the witch of Agnesi (and above the x-
axis).
a
2The infamous name of this curve resulted from a mistake in translation. See [?].
Chapter 23
The quadratrix
The quadratrix of Hippias was invented to solve the problemsof angletrisection and quadrature of the circle. This latter means finding thelength a quadrant of a circle in terms of the radius. Considerthe quadrantof the circleOAC inside a squareOABC. Suppose a radius is rotatinguniformly from positionOA toOC. At the same time, a segment parallelto AB is moving uniformly from positionAB to OC. The locus ofthe intersection of the rotating radius and the moving segment is thequadratrix.
O
A
X Y
B
C
P
Q
G
Suppose each side of the square has lengtha.
(x, y) = (2a
π· t cot t,
2a
π· t) for 0 < t ≤ π
2.
232 The quadratrix
Application to angle trisection
Suppose an acute angle, given as the arcCOQ, is to be trisected. Letthe radiusOQ intersect the quadratrix atP . ThroughP construct thesegmentXY parallel toOC, with X onOA andY onCB respectively.Trisect the segmentCY atY1 andY2. The parallels toOC through thesepoints intersect the quadratrix at two pointsP1 andP2 such thatOP1 andOP2 trisect the given angleCOP .
O
A
X Y
B
C
P
Q
G
Y1
Y2
X1
X2
P1
P2
Quadrature of the circle
Let G be the limiting position on the radiusOC of the intersectionP ofthe rotating radius and moving segment generating the quadratrix. Thelength of the quadrantAQC is equal toAB2
OG.
Chapter 24
The lemniscate
Given two pointsA andB at a distance2a apart, the lemniscate is thelocus of pointP such thatAP · BP = a2.
a a AB O
P
If (x, y) is a point on the locus, then
((x − a)2 + y2)((x + a)2 + y2) = (a2)2.
Rewriting this as
(x2 + y2 + a2 − 2ax)(x2 + y2 + a2 + 2ax) = a4.
In terms of polar coordinates, this is
(ρ2 + a2 − 2aρ cos θ)(ρ2 + a2 + 2aρ cos θ) = a4.
234 The lemniscate
Cancelling a common terma4 and then a common factorρ2, we have
ρ2 + 2a2 − 4a2 cos2 θ = 0,
orρ2 = 2a2(2 cos2 θ − 1) = 2a2 cos 2θ.
Remark.The rectangular equation is
(x2 + y2)2 − 2a2(x2 − y2) = 0.
Exercise
(1) Calculate the area enclosed by the lemniscate.
Perimeter of the lemniscate
To compute the perimeter of the lemniscate we recall the formula for arclength of polar curveρ = ρ(θ):
∫ β
α
√
ρ2 +
(dρ
dθ
)2
dθ.
For the lemnicateρ2 = 2a2 cos 2θ, we have2ρρ′ = −4a2 sin 2θ, ρ′ =−2a2 sin 2θ
ρand
ρ2 + ρ′2 = ρ2 +4a4 sin2 2θ
2a2 cos 2θ= 2a2 cos 2θ +
2a2 sin2 θ
cos 2θ=
2a2
cos 2θ.
The perimeter of the lemniscate is given by the integral
4
∫ π4
0
√2a · dθ√
cos 2θ= 4
√2a
∫ π4
0
dθ√cos 2θ
Exercise
(2) Show that the perimeter of the lemniscate is also given by
4√
2a
∫ 1
0
dt√1 − t4
.
cci
Supplement 16: Quadrature of parabola according toArchimedes’ Method
Proposition 0.1 (Method, Proposition 1). Let ABC be a segment of aparabola bounded by the straight lineAC and the parabolaABC, andlet D be the middple point ofAC. Draw the straight lineDBE parallelto the axis of the parabola and joinAB, BC. Then shall the segmentABC be 4
3of the triangleABC.
A
C
F
K
H
B
D
E
M
N
P
O
FromA drawAKF parallel toDE, and let the tangent to the parabolaat C meetDBE in E andAKF in F . ProduceCB to meetAF in K,and again produceCK to H, makingKH equal toCK.
ConsiderCH as the bar of a balance,K being its middle point.Let MO be any straight line parallel toED, and let it meetCF , CK,
AC in M , N , O and the curve inP .Now, sinceCE is a tangent to the parabola andCD the semi-ordinate,
EB = BD;
“for this is proved in the Elements [of Conics].”1
SinceFA, MO are parallel toED, it follows that
FK = KA, MN = NO.
1Heath’s footnote: i.e. the works on conics by Aristaeus and Euclid.
ccii The lemniscate
Now, by the property of the parabola, “proved in a lemma,”
MO : OP = CA : AO
= CK : KN= HK : KN.
Take a straight lineTG equal toOP , and place it with its center ofgravity atH, so thatTH = HG; then, sinceN is the center of gravityof the straight lineMO, and
MO : TG = HK : KN,
it follows thatTG atH andMO atN will be in equilibrium aboutK.[On the Equilibrium of Planes, I. 6,7]
Similarly, for all other straight lines parallel toDE and meeting thearc of the parabola,
(1) the portion intercepted betweenFC, AC with its middle point onKC and
(2) a length equal to the intercept between the curve andAC placedwith its center of gravity atH
will be in equilibrium aboutK.ThereforeK is the center of gravity of the whole system consisting(1) of all the straight lines asMO intercepted betweenFC, AC and
placed as they actually are in the figure, and(2) of all the straight lines placed atH equal to the straight lines as
PO intecepted between the curve andAC.And, since the triangleCFA is made up of all the parallel lines like
MO, andthe segmentCBA is made up of all the straight lines likePO within
the curve,it follows that the triangle, placed where it is in the figure,is in equi-
librium aboutK with the segmentCBA placed with its center of gravityatH.
Divide KC atW so thatCK = 3KW ;then W is the center of gravity of the traignleACF , “for this is
proved in the books on equilibrium”. [On the Equilibrium of Planes, I.15].
Therefore,△ACF : segmentABC = HK : KW = 3 : 1.Therefore, segmentABC = 1
3△ACF .
But,△ACF = 4△BAC.
cciii
Therefore, segmentABC = 43△ABC.
Now the fact here stated is not actually demonstrated by the argumentused; but that argument has given a sort of indication that the conclusionis true. Seeing then that the theorem is not demonstrated, but at the sametime suspecting that the conclusion is true, we shall have recourse to thegeometrical demonstration which I myself discovered and have alreadypublished.
cciv The lemniscate
Supplement 17: Archimedes’ summation of12+22+· · ·+n2
Lemma to Proposition 2 ofOn Conoids and Spheroids
If A1, A2, A3, . . . , An be n lines forming an ascending arithmeticalprogression in which the common difference is equal to the lest termA1,then
(n + 1)A2n + A1(A1 + A2 + · · ·+ An) = 3(A2
1 + A22 + · · ·+ A2
n).
Let the linesAn, An−1, An−2, . . . ,A1 be placed in a row from left toright. ProduceAn−1, An−2, . . . , A1 until they are each equal toAn, sothat the parts produced are respectively equal toA1, A2, . . . ,An−1.
AnAn−1
A1
An−2
A2
A2
An−2
A1
An−1
· · ·
Taking each line successively, we have
2A2n = 2A2
n,
(A1 + An−1)2 = A2
1 + A2n−1 + 2 · A1 · An−1,
(A2 + An−2)2 = A2
2 + A2n−2 + 2 · A2 · An−2,
...
(An−1 + A1)2 = A2
n−1 + A21 + 2 · An−1 · A1.
ccv
And, by addition,
(n + 1)A2n = 2(A2
1 + A22 + · · ·+ A2
n)
+ 2A1 · An−1 + 2A2 · An−2 + · · ·+ 2An−1 · A1.
Therefore, in order to obtain the required result, we have toprove that
2(A1 · An−1 + A2 · An−2 + · · ·+ An−1 · A1)
+ A1(A1 + A2 + · · · + An)
= A21 + A2
2 + · · ·+ A2n.
Now,
2A2 · An−2 = A1 · 4An−2 becauseA2 = 2A1,
2A3 · An−3 = A1 · 6An−2 becauseA3 = 3A1,
...
2An−1 · A1 = A1 · 2(n − 1)A1.
It follows that
2(A1 · An−1 + A2 · An−2 + · · · + An−1 · A1)
+ A1(A1 + A2 + · · ·+ An)
= A1(An + 3An−1 + 5An−2 + · · ·+ (2n − 1)A1).
And this last expression can be proved to be equal to
A21 + A2
2 + · · ·+ A2n.
For
A2n = A1(n · An)
= A1(An + (n − 1)An)
= A1(An + 2(An−1 + An−2 + · · ·+ A1)),
because
(n − 1)An = An−1 + A1
+ An−2 + A2
+ · · ·+ A1 + An−1.
ccvi The lemniscate
Similarly,
A2n−1 = A1(An−1 + 2(An−2 + An−3 + · · ·+ A1)),
...
A22 = A1(A2 + 2A1),
A21 = A1 · A1.
Whence, by addition,
A21 + A2
2 + · · ·+ A2n
= A1(An + 3An−1 + 5An−2 + · · · + (2n − 1)A1).
and the result follows.
ccvii
Supplement 20A: Generation of the ellipse and hyperbola
Consider a fixed pointF (c, 0) and a variable pointP on the circlex2 +y2 = a2. We find the envelope of the perpendicular toFP at P . If Phas coordinates(a cos θ, a sin θ), then this line has equationF (θ) = 0,where
F (θ) : = (x − a cos θ)(c − a cos θ) + (−a sin θ)(y − a sin θ)
= (c − a cos θ)x − a sin θ · y + a(a − c cos θ)
= −a(x + c) cos θ − ay · sin θ + (a2 + cx).
To find the envelope of the line, we eliminateθ from F (θ) = 0 andF ′(θ) = 0.
a(x + c) cos θ + ay · sin θ = (a2 + cx),
−a(x + c) sin θ + ay · cos θ = 0.
Note that the sum of the squares of the expressions on the lefthandside is independent ofθ:
a2(x + c)2 + a2y2 = (a2 + cx)2.
Collecting terms, we have
(a2 − c2)x2 + a2y2 = a2(a2 − c2),
orx2
a2+
y2
a2 − c2= 1.
Therefore, this is an ellipse ifc2 < a2 and a hyperbola ifc2 > a2. ThepointF is a focus.
If c = a, thenF is on the given circle and all such perpendicularspass through the antipodal point ofF . The envelope degenerate into thissingle point.
ccviii The lemniscate
Supplement 20B: More examples of envelopes
Example 0.1. Consider the lines whose intercepts in the first quadranthas a fixed lengtha. If this line makes an anglet with the (negative)xaxis, it has equation x
a cos t+ y
a sin t= 1, or
f(x, y, t) := x sin t + y cos t − a sin t cos t = 0.
Here,ft(x, y, t) = x cos t−y sin t−a(cos2 t−sin2 t). Solvingf(x, y, t) =0 andft(x, y, t) = 0 simultaneously forx andy, we have
x = a cos3 t, y = a sin3 t.
The envelope is the astroidx23 + y
23 = a
23 .
Figure 1: The astroid
Example 0.2. Consider the family of lines whose intercepts on thex-andy-axes have a fixed total lengtha. If thex-intercept has lengtht ≤ a,then the line has equationx
t+ y
a−t= 1, or
f(x, y, t) := (a − t)x + ty − t(a − t) = 0.
Here,ft(x, y, t) = −x + y − (a − 2t). Solvingf = 0 andft = 0, wehave
x =t2
a, y =
(a − t)2
a.
This is the parabola√
x +√
y =√
a. Its focus is the point(
a2, a
2
), and
its directrix is the linex + y = 0. The distances from(
t2
a, (a−t)2
a
)
to the
focus and the directrix are both2t2−2at+a2√
2a.
ccix
F
Figure 2: The parabola
Example 0.3. Consider the circle with center(at2, 2at) on the parabolay2 = 4ax and passing through the origin. It is given by
f(x, y, t) = x2 + y2 − 2at2x − 4aty.
Here,ft(x, y, t) = −4a(tx + y). Solvingf = 0 andft = 0, we have
(x, y) =
(−2at2
1 + t2,
2at3
1 + t2
)
.
ccx The lemniscate
Exercise
(1) Find the cartesian equation of the envelope.(2) Consider the circle through the origin and with center
(at, a
t
)on
the rectangular hyperbolaxy = a2.(i) Find the equation of the circle.(ii) Find the envelope of the circle as a parametrized curve.(iii) Find the Cartesian equation of the envelope.
Example 0.4. Consider a pointP (t) = (a cos t, b sin t) on the ellipsex2
a2 + y2
b2= 1. The equation of the circle with centerP (t) and passing
through the origin is
x2 + y2 − 2a cos t · x − 2b sin t · y = 0
To find the envelope, we eliminatet from
ax cos t + by sin t =1
2(x2 + y2),
−ax sin t + by cos t =0
by noting that the sum of the squares of the expressions on theleft handside is independent oft:
a2x2 + b2y2 =1
4(x2 + y2)2.
Exercise
Find the polar equation of the curve.
ccxi
Supplement 21: Application to the duplication of the cube (correc-tion)
Let CD be the diameter perpendicular toAB, and M the midpointof OC. Join B, M to intersect the cissoid atQ. Let X be the in-tersection of the diameterAB with the perpendicular fromQ. ThenXB : XQ = OB : OM = 2 : 1, andXP ′ andAX are the two meanproportions between them. Therefore,XP ′
XB· OB and AX
AQ· OM are two
mean proportions betweenOB andOM . The latter is 3√
2 · OM .
B
M
AO
P ′
Q
C
D
X
ccxii The lemniscate
Supplement 22A: The Philo line
Given a pointP in an angleXCY , the line throughP which cuts a min-imum segmentAB between the sides of the angle is called the Philo linefor P with respect to the angle. This line is characterized by the follow-ing equivalent properties:(1) If Q the orthogonal projection ofC on AB, thenP andQ are iso-tomic with respect toAB, i.e., AP = QB.(2) The perpendiculars toCA at A, CB at B, andAB at P are concur-rent.
P
CX
Y
Q
A
B
Z
O
Exercise
(1) and (2) are equivalent.
ccxiii
Let p andq be the lengths of the perpendiculars fromP to CX andCY respectively, the length of the segmentAB is
c =p
sin A+
q
sin B. (1)
Here, the anglesA andB depends on the position of the line, but
A + B + C = π.
Differentiating (1), we have
dc = −p cos A
sin2 A· dA − q cos B
sin2 B· dB.
For minimumc, we requiredc = 0. Noting thatdA + dB = 0, we have
p cos A
sin2 A=
q cos B
sin2 B.
SinceAP = psin A
andBP = qsin B
, this condition is the same as
AP cotA = PB cot B,
which is the same as the concurrency of three perpendicularsin (2)above. SinceZ andC are antipodal points on the circle, with orthogonalprojectionsP andQ onAB, it follows thatAP = QB.
Exercise
SupposeCX andCY are perpendicular to each other. Show thattan3 A =pq
for the Philo lineAB.
ccxiv The lemniscate
Construction
The Philo line cannot be constructed using ruler and compass. Here is aconstruction using a conic section.
P
C X
Y
Q
A
B
O
(1) Construct the hyperbola passing throughP with two given linesas asymptotes.
(2) Construct the circle withCP as diameter to intersect the hyper-bola again atQ.
(3) The linePQ is the Philo line; it intersects the given lines atA andB such thatAP = QB.
Remark.Related to the Philo line problem is the one for which triangleABC has minimum perimeter.2 The solution has a very easy descrip-tion: it is the tangent atP to the circle throughP which is tangent to thetwo given lines. It is well known that this circle can be constructed byruler and compass. Here is a simple construction: letQ be the reflectionof P in the bisector of the given angle. JoinPQ to intersect one of thelines, say,CX, atM . On the half lineMX, construct the pointN suchthat MN2 = MP · MQ. The circle throughP , Q, N is the requiredone.
2The problem for minimum area is trivial.
ccxv
Supplement 22B: Durer’s conchoid
Consider a long ruler with a segment of length2a and midpointT .Durer’s conchoid is the locus of the endpoints of the segments whenTmoves on thex-axis with coordinates(t, 0) and the ruler passes througha point(0, b − t) on they-axis.
These endpoints are the intersections of the
(x − t)2 + y2 = a2,x
t+
y
b − t= 1.
The solutions of this system are quite simple:
x = t +at√
2t2 − 2bt + b2, y =
−a(b − t)√2t2 − 2bt + b2
,
and
x = t − at√2t2 − 2bt + b2
, y =a(b − t)√
2t2 − 2bt + b2,
If we put t = b2(1 + tan θ), then
y =a(b − t)√
2t2 − 2bt + b2=
12ab(1 − tan θ)√
12b sec θ
=a√2(cos θ − sin θ).
at√2t2 − 2bt + t2
=12ab(1 + tan θ)√
12b sec θ
=a√2(cos θ + sin θ).
x =b
2(1 + tan θ) +
a√2(cos θ + sin θ),
y = − a√2(cos θ − sin θ).
Allowing negative values fora, Durer’s conchoid is simply parametrizedby
ccxvi The lemniscate
x =b
2(1 + tan θ) − a√
2(cos θ + sin θ) =
(b√
2 cos θ− a
)
cos(
θ − π
4
)
,
y =a√2(cos θ − sin θ) = a cos
(
θ +π
4
)
.
Figure 3:a = 2b
Figure 4:a = b
ccxvii
Figure 5:a =23b
ccxviii The lemniscate
Supplement 24A: Rectangular linkages
Given two robot arms of lengthb hinged at fociA =(−a
2, 0)
andB =
(a2, 0), connect the free ends with a third arm of lengtha. This
configuration is a rectangular linkage. LetM be the midpoint of themiddle robot arm. We determine the locus ofM as P varies on thecircle (B).
b
b
P
Q
M
OA
BK
Let (ρ, θ) be the polar coordinates ofM (relative toO and the lineAB). Note thatAQBP is a symmetric trapezoid. If the diagonalsPQand AB intersect atK, then KO = KM . Therefore,∠MKB =2∠MOB = 2θ. In triangleBKP , we have
KB = OB − OK =a
2− ρ
2 cos θ,
KP = MP + KM =a
2+
ρ
2 cos θ.
Applying the law of cosines, we have
b2 =(a
2− ρ
2 cos θ
)2
+(a
2+
ρ
2 cos θ
)2
− 2(a
2− ρ
2 cos θ
)(a
2+
ρ
2 cos θ
)
cos 2θ
=1
2
(
a2 +ρ2
cos2 θ
)
− 1
2
(
a2 − ρ2
cos2 θ
)
cos 2θ
=1
2a2(1 − cos 2θ) +
1
2· ρ2
cos2 θ(1 + cos 2θ)
= a2 sin2 θ + ρ2.
Therefore, the polar equation of the middle pointM of the middlerobot arm is
ρ2 = b2 − a2 sin2 θ,
ccxix
b
b
M
O
Q
M
O
defined forθ satisfying|sin θ| ≤ ba. If b =
√2a, this is the lemniscate
ρ2 = 2a2 cos 2θ.Here is a more general result. Since the vectorQP = a(cos 2θ, sin 2θ),
we can easily determine, for a fixedt ∈ [0, 1], the locus of the pointPt
dividing PQ in the ratio1 − t : t.
Pt = M +1
2(2t − 1)QP
=√
b2 − a2 sin2 θ(cos θ, sin θ) +1
2(2t − 1) (a cos 2θ, a sin 2θ)
=(a
2(1 − 2t), 0
)
+ (√
b2 − a2 sin2 θ + a(2t − 1) cos θ)(cos θ, sin θ).
If we shift the pole to the point(
a2(1 − 2t), 0
), we have the polar
curveρ =
√
b2 − a2 sin2 θ + a(2t − 1) cos θ
defined forθ satisfying|sin θ| ≤ ba.
Figure 6:t =12
ccxx The lemniscate
Figure 7:t =45
Figure 8:t =13
Figure 9:t =35
Figure 10:t =25
Figure 11:t =110
ccxxi
Supplement 24B: The lemniscate integral
The integral for the perimeter of the lemniscate is an example of an el-liptic integral.
Theorem 0.2(Gauss). Leta ≤ b be given positive real numbers.∫ ∞
−∞
dx√
(x2 + a2)(x2 + b2)=
π
agM(a, b).
Proof. We first show that the integral is invariant ifa andb are replacedby
√ab and a+b
2respectively.
Consider the substitutionx = 12
(
y − aby
)
. As y ranges from0 to ∞,
x ranges from−∞ to ∞. Let a1 =√
ab andb1 = a+b2
. Note that
x2 + a21 = x2 + ab =
1
4
(
y − ab
y
)2
+ ab =1
4
(
y +ab
y
)2
=(y2 + ab)2
4y2,
x2 + b21 = x2 +
(a + b
2
)2
=1
4
((
y − ab
y
)2
+
(a + b
2
)2)
=(y2 + a2)(y2 + b2)
4y2,
dx =1
2
(
1 +ab
y2
)
dy =(y2 + ab)dy
2y2.
It follows that∫ x=∞
x=−∞
dx√
(x2 + a21)(x
2 + b21)
=
∫ y=∞
y=0
2dy√
(y2 + a2)(y2 + b2)
=
∫ x=∞
x=−∞
dx√
(x2 + a2)(x2 + b2).
If we continue replacing by geometric and arithmetic means,we obtaina sequence of nested intervals
[a, b] ⊃ [a1, b1] ⊃ [a2, b2] ⊃ · · · ⊃ [an, bn] ⊃ · · ·converging to agM(a, b). Moreover,
∫ ∞
−∞
dx√
(x2 + a2n)(x2 + b2
n)=
∫ ∞
−∞
dx√
(x2 + a2)(x2 + b2)
ccxxii The lemniscate
for everyn. Since∫ ∞
−∞
dx
x2 + b2n
≤∫ ∞
−∞
dx√
(x2 + a2n)(x2 + b2
n)≤∫ ∞
−∞
dx
x2 + a2n
,
we haveπ
bn≤∫ ∞
−∞
2dy√
(x2 + a2)(x2 + b2)≤ π
an
for everyn. Since(an) and(bn) both converge to agM(a, b), we have
π
agM(a, b)≤∫ ∞
−∞
dx√
(x2 + a2)(x2 + b2)≤ π
agM(a, b)
and ∫ ∞
−∞
dx√
(x2 + a2)(x2 + b2)=
π
agM(a, b).
Theorem 0.3.The perimeter of the lemniscateρ2 = 2a2 cos 2θ is 4√
2πaagM(1,
√2)
.
Proof. The integral for the perimeter of the lemniscate, namely,∫ 1
0dt√1−t4
,can be transformed into one of the form considered in the theorem above,by putting
t =1√
x2 + 1.
Note thatx ranges from0 to ∞ oppositelyast ranges from0 to 1, and
dt =−xdx
(x2 + 1)32
,
1 − t4 = 1 − 1
(x2 + 1)2=
x2(x2 + 2)
(x2 + 1)2.
Therefore, the perimeter of the lemniscate is
4√
2a
∫ t=1
t=0
dt√1 − t4
= 4√
2a
∫ x=∞
x=0
dx√
(x2 + 1)(x2 + 2)=
2√
2πa
agM(1,√
2).
ccxxiii
Supplement 24C: The Cassini curves
The Cassini curves are defined as the loci of points whose distancesfrom two fixed points have constant products. Given two points F± :=(±a, 0), andλ > 0, let P (x, y) be a point with polar coordinates(ρ, θ)such thatPF+ · PF− = λ2a2. we have
(ρ2 + a2 − 2aρ cos θ)(ρ2 + a2 + 2aρ cos θ) = λ4 · a4,
which can be rewritten as
ρ4 − 2a2ρ2 cos 2θ + (1 − λ4)a4 = 0.
From this we make some observations.
1. If λ = 1, this becomes
ρ2 = 2a2 cos 2θ,
the lemniscate.
Figure 12: The lemniscate
2. If λ > 1, there is only one positive value forρ2 for everyθ:
ρ2 = a2 cos 2θ + a2√
λ4 − sin2 2θ.
3. If λ < 1, write λ2 = sin α for an acute angleα:
ρ2 = a2 cos 2θ ± a2√
sin2 α − sin2 2θ
ccxxiv The lemniscate
has real solutions forθ ∈ [−α, α] ∪ [π − α, π + α]. For eachinterior pointθ in these intervals, there are two positive values ofρ. This means that the curve consists of two disjoint simple closedcurves.
Figure 13: Cassini ovals
Consider a circley2 + (z − a)2 = b2 (with a > b) in theyz-plane. Ifthe circle is rotated about they-axis, it generates the surface of a torus.The equation of the surface is obtained by replacingz by
√x2 + z2:
y2 + (√
x2 + z2 − a)2 = b2,
or(x2 + y2 + z2 + a2 − b2)2 − 4a2(x2 + z2) = 0.
The horizontal sections of the surface, cut out by planesz = k, arecalled cassini curves:
C : (x2 + y2 + k2 + a2 − b2)2 − 4a2(x2 + k2) = 0.
Proposition 0.4. A cassini curve is the locus of points whose distancesfrom two fixed points have a constant product.
ccxxv
The polar equation:
ρ2 = cos 2θ ±√
λ4 − sin2 2θ
if the foci are at distances2 and the product= λ2.
Chapter 25
Volumes in Euclid’s Element
Definitions (Euclid XI.12). A pyramid is a solid figure, contained byplanes, which is constructed from one plane to one point.
Notation:(V, ABC . . .K).(XI.13). A prism is a solid figure contained by planes two of which,namely those which are opposite, are equal, similar, and parallel, whilethe rest are parallelograms. A right triangular prism is a prism whose twobases are congruent right trianglesABC andA′B′C ′ in which the linesAA′, BB′, CC ′ are perpendicular to the planes ofABC andA′B′C ′.
Notation:(ABC, A′B′C ′).
Proposition 25.1(Euclid XII.3). Any pyramid which has a triangularbase is divided into two pyramids equal and similar to one another, sim-ilar to the whole and having triangular bases, and into two equal prisms;and the two prisms are greater than the half of the whole pyramid.
Heath’s outline.We will denote a pyramid with vertexD and baseABCby D(ABC) or D−ABC and the triangular prism with trianglesGCF ,HLK for bases by(GCF, HLK). The following are the steps of theproof.
I. To prove pyramidH(AEG) equal and similar to pyramidD(HKL).Since sides of△DAB are bisected atH, E, K,
HE//DB, and HK//AB.
HenceHK = EB = EA,
andHE = KB = DK.
302 Volumes in Euclid’s Element
A B
D
H K
C
G
F
E
L
A B
D
H K
C
G
F
E
L
A B
D
H K
C
G
F
E
L
Therefore (1)△HAE and△DHK are equal and similar.Similarly (2)△HAG and△DHL are equal and similar.Again, LH, HK are respectively parallel toGA, AE in a different
plane; therefore∠GAE = ∠LHK.And LH, HK are respectively equal toGA, AE.Therefore (3)△GAE and△LHK are equal and similar.Similarly (4)△HGE and△DLK are equal and similar.Therefore1 the pyramidsH(AEG) andD(HKL) are equal and sim-
ilar.II. To prove the pyramidD(HKL) similar to the pyramidD(ABC). 2
(1)△DHK and△DAB are equiangular and therefore similar.Similarly (2)△DLH and△DCA are similar, as also(3)△DLK and△DCB.Again, BA, AC are respectively parallel toKH, HL in a different
plane; therefore∠BAC = ∠KHL. And BA : AC = KH : HL.Therefore (4)△BAC and△KHL are similar.Consequently the pyramidD(ABC) is similar to the pyramidD(HKL),
and therefore also to the pyramidH(AEG).III. To prove prism(GCF, HLK) equal to prism(HGE, KFB).The prisms may be regarded as having the same height (the distance
between the planesHKL, ABC) and having for bases(1)△CGF and(2) the parallelogramEBFG, which is the double of△CGF .Therefore, by XI.39,3 the prisms are equal.IV. To prove the prisms greater than the small pyramids. Prism (HGE, KFB)
1Eucl. XI, Definition 10: Equal and similar solid figures are those contained by similar planes equal inmultitude and in magnitude.
2Eucl. XI, Definition 9: Similar solid figures are those contained bysimilar planes equal in multitude.3If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and
if the parallelogram be double of the triangle, the prisms will be equal.
303
is clearly greater than pyramidK(EFB) and therefore greater than pyra-mid H(AEG).
Therefore each of the prisms is greater than each of the smallpyra-mids; and the sum of the two prisms is greater than the sum of the twosmall pyramids, which, with the two prisms, make up the wholepyra-mid.
Proposition 25.2(Euclid XII.4). If there be two pyramids of the sameheight which have triangular bases, and each of them be divided intotwo pyramids equal to one another and similar to the whole, and intotwo equal prisms, then, as the base of the one pyramid is to thebaseof the other pyramid, so will all the prisms in one pyramid be to all theprisms, being equal in multitude, in the other pyramid.
A
B
C
G
P
M
N
KO
LD
E
F
H
S
T
U
QV
R
Proposition 25.3 (Euclid XII.5(6)). Pyramids which are of the sameheight and have triangular (polygonal) bases are to anotheras the bases.
Proposition 25.4(Euclid XII.7). Any prism which has a triangular baseis divided into three pyramids equal to one another which have triangu-lar bases.
Porism
From this it is manifest that any pyramid is a third part of theprismwhich has the same base with it and equal height.
(i) C(ABD) = (C, DEB) = (D, EBC) by XII.5.(ii) (D, EBC) = (D, ECF ).Thus, (C, ABD) = (D, EBC) = (D, ECF ). Since these three
pyramids together make up the prism, each of them is one thirdof theprism.
304 Volumes in Euclid’s Element
B A
C
E D
F
B A
C
E D
F
B A
C
E D
F
Figure 25.1:
Proposition 25.5(Euclid XII.8). Similar pyramids which have triangu-lar bases are in the triplicate ratio of their correspondingsides.
Proposition 25.6(Euclid XII.9). In equal pyramids which have trian-gular bases the bases are reciprocally proportional to the heights; andthose pyramids in which the bases are reciprocally proportional to theheights are equal.
Proposition 25.7(Euclid XII.10). Any cone is a third part of the cylinderwhich has the same base with it and equal height.
Proposition 25.8(Euclid XII.11). Cones and cylinders which are of thesame height are to one another as their bases.
Proposition 25.9 (Euclid XII.12). Similar cones and cylinders are toone another in the triplicate ratio of the diameters in theirbases.
Proposition 25.10(Euclid XII.13). If a cylinder be cut by a plane whichis parallel to its opposite planes, then as the cylinder is tothe cylinder,so will the axis be to the axis.
Proposition 25.11(Euclid XII.14). Cones and cylinders which are onequal bases are to one another as their heights.
Proposition 25.12(Euclid XII.15). In equal cones and cylinders thebases are reciprocally proportional to the heights; and those cones andcylinders in which the bases are reciprocally proportionalto the heightsare equal.
Proposition 25.13(Euclid XII.18). Spheres are to one another in thetriplicate ratio of their respective diameters.
Chapter 26
Archimedes’ calculation of thevolume of a sphere
Proposition 26.1(Method, Proposition 2). (1) Any sphere is (in respectof solid content) four times the cone with base equal to a great circle ofthe sphere and height equal to its radius.
(2) The cylinder with base equal to a great circle of the sphere andheight equal to the diameter is11
2times the sphere.
G
N
F
L
M
E
B DK
W C Y
V A X
O P
S
R
Q
H
(1) LetABCD be a great circle of a sphere, andAC, BD diametersat right angles to one another.
306 Archimedes’ calculation of the volume of a sphere
Let a circle be drawn aboutBD as diameter and in a plane perpen-dicular toAC, and on this circle as base let a cone be described withA as vertex. Let the surface of this cone be produced and then cut by aplane throughC parallel to its base; the section will be a circle onEFas diameter. On this circle as base let a cylinder be erected with heightand axisAC, and produceCA to H, makingAH equal toCA.
Let CH be regarded as the bar of a balance,A being its middle point.Draw any straight lineMN in the plane of the circleABCD and
parallel toBD. Let MN meet the circle inO, P , the diameterAC in S,and the straight linesAE, AF in Q, R respectively. JoinAO.
ThroughMN draw a plane at right angles toAC; this plane will cutthe cylinder in a circle with diameterMN , the sphere in a circle withdiameterOP , and the cone in a circle with diameterQR.
Now, sinceMS = AC, andQS = AS,
MS · SQ = CA · AS = AO2 = OS2 + SQ2.
And, sinceHA = AC,
HA : AS =CA : AS
=MS : SQ
=MS2 : MS · SQ
=MS2 : (OS2 + SQ2)
=MN2 : (OP 2 + QR2)
=(circle, diameter MN) : (circle, diameter OP + circle, diameter QR).
That is,HA : AS = (circle in cylinder) : (circle in sphere + circle incone).
Therefore the circle in the cylinder, placed where it is, is in equilib-rium, aboutA, with the circle in the sphere together with the circle inthe cone, if both the latter circles are placed with their centers of gravityatH.
Similarly for the three corresponding sections made by a plane per-pendicular toAC and passing through any other straight line in the par-allelogramLF parallel toEF .
If we deal in the same way with all the sets of three circles in whichplanes perpendicular toAC cut the cylinder, the sphere and the cone,and which make up those solids respectively, it follows thatthe cylinder,
307
in the place where it is, will be in equilibrium about A with the sphereand the cone together, when both are placed with their centers of gravityatH.
Therefore, sinceK is the centre of gravity of the cylinder,HA : AK = (cylinder): (sphere + coneAEF ).
But HA = 2AK; therefore cylinder = 2 (sphere + coneAEF ). Nowcylinder = 3 (coneAEF ); 1 therefore coneAEF = 2 (sphere). But, sinceEF = 2BD, coneAEF = 8 (coneABD); therefore sphere = 4 (coneABD).
(2) ThroughB, D drawV BW , XDY parallel toAC; and imagine acylinder which hasAC for axis and the circles onV X, WY as diametersfor bases.
Then cylinderV Y = 2 (cylinderV D) = 6 (coneABD) = 32
(sphere),from above.
From this theorem, to the effect that a sphere is four times asgreatas the cone with a great circle of the sphere as base and with heightequal to the radius of the sphere, I conceived the notion thatthe surfaceof any sphere is four times as great as a great circle in it; for, judgingfrom the fact that any circle is equal to a triangle with base equal to thecircumference and height equal to the radius of the circle, Iapprehendedthat in like manner, any sphere is equal to a cone with base equal to thesurface of the sphere and height equal to the radius.
1Eucl. XII. 10.
308 Archimedes’ calculation of the volume of a sphere
Proposition 26.2(Method, Proposition 7). Any segment of a sphere hasto the cone with the same base and height the ratio which the sum of theradius of the sphere and the height of the complementary segment has tothe height of the complementary segment.
G
N
F
L
M
E
B DK
W C Y
V A X
O P
S
R
Q
H
D BFE G
segmentBAD : coneABD =1
2AC + GC : GC.
Chapter 27
Cavalieri’s principle
If two solids between two horizontal planes have equal sectional areas atevery level, then they have equal volumes.
Consider two solids of the same height:(i) a hemisphere of radiusa (with a horizontal base),(ii) a cylinder of base radiusa and heighta, with the right circular coneon the top face and apex the center of the base removed.
x xa
x
a a
The bases of the two solids are on the same horizontal plane. At eachlevelx above the base,(i) the section of the hemisphere is a circle of radius
√a2 − x2,
(ii) the section of the other solid is an annulus of outer radiusa and innerradiusx.
These two sections clearly have equal areas. Therefore, by Cavalieri’sprinciple, the two solids have equal volumes. The volume of the cylinderwith cone removed is
πa2 · a − 1
3πa2 · a =
2
3πa3.
Therefore, the volume of a sphere of radiusa is 43πa3.
310 Cavalieri’s principle
Chapter 28
Fermat: Area under the graphof y = xn
To find the area under the graph ofy = xn, above thex-axis, and be-tween the two vertical linesx = a andx = b, Fermat introducedN − 1mean proportions betweena andb, namely,
a < ar < ar2 < · · ·arN−1 < arN = b,
for r = N
√ba. The region is divided intoN small regions. The widths of
the intervals form a geometric progression of common ratior. The val-ues of at the points of division form a geometric progressionof commonratiorn.
Therefore if we consider the rectangles with the left boundaries assides, these form a geometric progression ofN terms with common ratiorn+1 and first terman+1(r − 1). This has sum
an+1(r − 1) · (rn+1)N − 1
rn+1 − 1
= an+1(r − 1) · (rN)n+1 − 1
rn+1 − 1
= an+1((rN)n+1 − 1) · r − 1
rn+1 − 1
= ((arN)n+1 − an+1) · 1
1 + r + r2 + · · ·+ rn
= (bn+1 − an+1) · 1
1 + r + r2 + · · ·+ rn.
312 Fermat: Area under the graph ofy = xn
On the other hand, if we consider the rectangles with the right bound-aries as sides, these form a geometric progression ofN terms with thesame common ratiorn+1 but first terman+1(r − 1)rn. This has sum
(bn+1 − an+1)rn
1 + r + r2 + · · · + rn.
The true area is between these two:
bn+1 − an+1
1 + r + r2 + · · ·+ rn<
∫ b
a
xndx <(bn+1 − an+1)rn
1 + r + r2 + · · · + rn.
SincelimN→∞ r = 1, we have
limN→∞
bn+1 − an+1
1 + r + r2 + · · ·+ rn= lim
N→∞
(bn+1 − an+1)rn
1 + r + r2 + · · ·+ rn=
bn+1 − an+1
n + 1.
Therefore, we conclude that∫ b
a
xndx =bn+1 − an+1
n + 1.
Exercise
Apply the same method to find the area under the graph ofy = 1x, above
thex-axis, and between the vertical linesx = a andx = b.
Chapter 29
Pascal: Summation of powers ofnumbers in arithmeticprogression
Pascal outlined a method of finding the sums of powers of integers inarithmetic progression, namely,
Σk := ak + (a + d)k + · · ·+ (a + (n − 1)d)k
for givena, d, andn. Clearly,Σ0 = n andΣ1 = n(2a+(n−1)d)2
. In general,
(a + jd + d)k+1 = (a + jd)k+1 +
(
k + 1
1
)
(a + jd)kd + · · ·+
(
k + 1
k
)
(a + jd)dk + dk+1
for j = 0, 1, . . . , n − 1. Combining the equations
(a + d)k+1 = ak+1 +(k + 1
1
)
akd + · · · +(k + 1
k
)
adk + dk+1,
(a + 2d)k+1 = (a + d)k+1 +(k + 1
1
)
(a + d)kd + · · · +(k + 1
k
)
(a + d)dk + dk+1,
...
(a + (n − 1)d)k+1 = (a + (n − 2)d)k+1 +(k + 1
1
)
(a + (n − 2)d)kd + · · · +(k + 1
k
)
(a + (n − 2)d)dk + dk+1,
(a + nd)k+1 = (a + (n − 1)d)k+1 +(k + 1
1
)
(a + (n − 1)d)kd + · · · +(k + 1
k
)
(a + (n − 1)d)dk + dk+1,
we obtain, after cancelling common terms,
(a + nd)k+1 = ak+1 +
(k + 1
1
)
Σkd + · · ·+(
k + 1
k
)
Σ1dk + n · dk+1.
314 Pascal: Summation of powers of numbers in arithmetic progression
KnowingΣ1, Σ2, . . . ,Σk−1, one can use this to determineΣk.Fora = d = 1, denoteΣk by Sk. Thus,
Sk := 1k + 2k + · · ·nk.
We have
S1 =1
2n(n + 1) =
1
2n2 +
1
2n,
S2 =1
6n(n + 1)(2n + 1) =
1
3n3 +
1
2n2 +
1
6n,
S3 =1
4n2(n + 1)2 =
1
4n4 +
1
2n3 +
1
4n2
...
Sk is a polynomial of degreek + 1 in n:
Sk =1
k + 1nk+1 +
1
2nk + lower order terms.
This is enough to give
limn→∞
1k + 2k + · · ·+ nk
nk+1=
1
k + 1,
solving the quadrature problem ofxn (for a positive integral powern):∫ a
0
xkdx =ak+1
k + 1.
Chapter 30
Pascal: On the sines of aquadrant of a circle
CA
B
D
RR
E
E
K
I
Proposition I
The sum of the sines of any arc of a quadrant is equal to the portion of the base betweenthe extreme sines, multiplied by the radius.
Preparation for the proof. Let any arc BP be divided into an infinite number of partsby the points D from which we draw the sines PO, DI etc. Let us take in the otherquadrant of the circle the segment AQ, equal to AO (which measures the distancebetween the extreme sines of the arc, BA, PO). Let AQ be divided into an infinitenumber of equal parts by the points H , at which the ordinates HL will be drawn.
Proof of Proposition I. I say that the sum of the sines DI (each of them multipliedof course by one of the equal small arcs DD) is equal to the segment AO multiplied bythe radius AB. Indeed, let us draw at all the point D the tangent DE, each of whichintersects its neighbor at the points E; if we drop the perpendiculars ER it is clear thateach sine DI multiplied by the tangent EE is equal to each distance RR multiplied
316 Pascal: On the sines of a quadrant of a circle
AI H
D L
G N
O
P
by the radius AB. Therefore, all the quadrilaterals formed by the sines DI and theirtangents EE (which are all equal to each other) are equal to tall the quadrilaterals,formed by all the portions RR with the radius AB; that is, (since one of the tangents EE
multiplies each of the sines, and since the radius AB multiplies each of the distances),the sum of the sines DI , each of them multiplied by one of the tangents EE, is equal tothe sum of the distances RR, each multiplied by AB. But each tangent EE is equal toeach one of the equal arcs DD. Therefore the sum of the sines multiplied by one of theequal small arcs is equal to the distances AO multiplied by the radius.
Note: It should not cause surprise when I say that all the distances RR are equal toAO and likewise that each tangent EE is equal to each of the small arcs DD, since itis well known that, even though this equality is not true when the number of the sinesis finite, nevertheless the equality is true when the number is infinite; because then thesum of all the equal tangents EE differs from the entire arc BD, or from the sum of allthe equal arcs DD, by less than any given quantity; similarly the sum of the RR formthe entire AC.
Pascal’s formula Translation∑
DI · EE = AB∑
RR∫ β
αr sin θd(rθ) = r2(cos α − cos β)
∑DI2
· EE = AB∑
DI · RR∫ β
αr2 sin2 θd(rθ) = r
∫ β
αr sin θd(r cos θ)
∑DI3
· EE = AB∑
DI2· RR
∫ β
αr3 sin3 θd(rθ) = r
∫ β
αr2 sin2 θd(r cos θ)
Chapter 31
Newton: The fundamentaltheorem of calculus
Relation of fluxions
Isaac Newton (1665) handled the problem of tangents to a curve in thefollowing way. Consider a Cartesian coordinate system witha pointAmoving along thex-axis and a pointB moving along they axis. Thevertical line throughA and the horizontal line throughB intersect at apointP . AsA andB move on the respective axes, the pointP traces outa curve, which has equationf(x, y) = 0. The velocities ofA andB arecalled fluxions, and are denoted byx andy respectively.
Newton first considered the relation between the fluxionsx and y,assumingf(x, y) of the form
∑aijx
iyj. He substitutedx andy byx+xoandy + yo into f(x, y) = 0 and obtained
0 =∑
aij(x + xo)(y + yo)
=∑
aijxiyj +
∑
aijxi(jyj−1yo + terms ofo2)
+∑
aijyj(ixi−1xo + terms ofo2)
+∑
aij(ixi−1xo + · · · )(jyj−1yo + · · · )
Since∑
aijxiyj = 0, dropping all terms involvingo2, we have
∑
aij(ixi−1xo + jyj−1yo) = 0.
318 Newton: The fundamental theorem of calculus
Dividing by o, Newton obtained
∑(
ix
x+
jy
y
)
aijxiyj = 0,
ory
x= −
∑jaijx
iyj−1
∑iaijxi−1yj
.
Fundamental Theorem of Calculus
Newton considered the inverse problem of findingy in terms ofx givena relation betweenx and the ratio of their fluxionsy
x. Specifically, y
x=
g(x).
x
A
xab
x
A
Let A be the area under the curvey = f(x), between two valuesaandb of x. Regard this area as being swept out by the vertical segmentsmoving froma to the right with unit fluxionx = 1. Newton asserted thatthe fluxion (time rate of change) of the area isA = g(x) with x = 1, sothat
A
x= g(x).
This is the fundamental theorem of calculus.
Chapter 32
Newton: The binomial theorem
Newton discovered the binomial theorem for an arbitrary index in thefollowing form:
(P+PQ)mn = P
mn +
m
nAQ+
m − n
2nBQ+
m − 2n
3nCQ+
m − 3n
4nDQ+· · ·
whereP + PQ signifies the quantity whose root (or even any power, ofroot of a power) is to be found;P signifies the first term of that quantity,Q the remaining terms divided by the first, andm
nthe index.
Example 32.1.
√a2 + x2 = a +
x2
2a− x4
8a3+
x6
16c5− 5x8
128a7+ · · · .
Here,Q = x2
a2 . With the first termA = a, we havethe second termB = 1
2AQ = 1
2· a · x2
a2 = x2
2a,
the third termC = −14
BQ = −14· x2
2a· x2
a2 = − x4
8a3 ,
the fourth termD = −36
CQ = −12
(
− x4
8a3
)
· x2
a2 = x6
16a5 ,. . . .
Here is a modern formulation of the binomial theorem: letm be arational number,
(1 + x)m = 1 +
(m
1
)
x +
(m
2
)
x2 + · · ·+(
m
k
)
xk + · · ·
where (m
k + 1
)
=m − k
k + 1
(m
k
)
,
(m
0
)
= 1.
320 Newton: The binomial theorem
It is easy to see that(
m
k
)
=m(m − 1) · · · (m − k + 1)
k!.
Exercise
Find the series expansion of 1√1−x2 up to5 terms.
Chapter 33
Newton’s method ofapproximate solution of anequation
To solve a polynomial equation
f(x) =
n∑
k−0
akxk = 0,
Newton started with an approximate solutionx∗ and sought a solutionof the formx = x∗ + p:
0 =n∑
k=0
akxk
=n∑
k=0
ak(x∗ + p)k
=n∑
k=0
ak(xk∗ + kxk−1
∗ p + · · · )
=
n∑
k=0
akxk∗ + p
n∑
k=0
kakxk−1∗ + higher order terms ofp
= f(x∗) + pf ′(x∗) + higher order terms ofp.
Ignoring the higher terms ofp, we havep ≈ − f(x∗)f ′(x∗)
. Therefore, ifx∗ is
an approximate solution, thenx∗ − f(x∗)f ′(x∗)
is a better one.
322 Newton’s method of approximate solution of an equation
Exercise
(1) Find the cube root of2 by iterating Newton’s method3 times.(2) Show that ify3 + y + xy − x3 − 2 = 0, then
y = 1 − 1
4x +
x2
64+ · · · .
Chapter 34
Newton’s reversion of series
Beginning with the area under the graphy = 11+x
:
z = x − 1
2x2 +
1
3x3 − 1
4x4 + − · · ·
Newton expressedx in terms ofz in the following way. Newton usedonly the first five terms of the series, and solved
1
5x5 − 1
4x4 − 1
3x3 +
1
2x2 + x − z = 0
for x (in terms ofz). Beginning with the first approximationz for x, hewrotex = z + p, substituted into the above equation, suppressing higherorder terms ofp and got
(
−1
2z2 +
1
3z3 − 1
4z4 +
1
5z5
)
+ p(1 − z + z2 − z3 + z4)
+ p2
(
−1
2+ z − 3z2
2+ 2z3
)
+ p3
(1
3− z + 2z2
)
+ p4
(
−1
4+ z
)
+1
5p5 = 0. (34.1)
Using only the first two terms, he obtained
p ≈ −−12z2 + 1
3z3 − 1
4z4 + 1
5z5
1 − z + z2 − z3 + z4=
1
2z2 + · · ·
324 Newton’s reversion of series
This gives the second approximationx ≈ z + 12z2. Then Newton put
p = 12z2 + q into (34.1) and obtain(
−1
6z3 +
1
8z4 − 1
20z5
)
+ q
(
1 − z +1
2z2
)
+ · · · = 0.
From this,
q = −−16z3 + 1
8z4 − 1
20z5
1 − z + 12z2)
=1
6z3 + · · · .
This gives the third approximation
x ≈ z +1
2z2 +
1
6z3 + · · · .
Continuing in the same way, Newton obtained
x = z +1
2z2 +
1
6z3 +
1
24z4 +
1
120z5 + · · · .
Now, sincez = log(1 + x) is equivalent toez − 1, this gives theexpansion
ez = 1 + z +1
2z2 +
1
6z3 +
1
24z4 +
1
120z5 + · · · .
Chapter 35
Newton: The series for sine andcosine
The series forarcsin x
The series ofarcsin x can be found from the quadrature of the circle.More precisely, the area under the circular arcAQ can be found in twoways:
1
2θ +
1
2x√
1 − x2 =
∫ x
0
√1 − x2dx.
x
y
O
θ
1 √1 − x2
P
Q
x
B
A
Note that∫ x
0
√1 − x2dx =
∫ x
0
(
1 − 1
2x2 − 1
8x4 − 1
16x6 − 5
128x8 + · · ·
)
dx
= x − 1
6x3 − 1
40x5 − 1
112x7 + · · ·
326 Newton: The series for sine and cosine
Therefore, withθ = arcsin x, we have
arcsin x = 2
∫ x
0
√1 − x2dx − x
√1 − x2
= 2
(
x − 1
6x3 − 1
40x5 − 1
112x7 + · · ·
)
− x
(
1 − 1
2x2 − 1
8x4 − 1
16x6 + · · ·
)
= x +1
6x3 +
3
40x5 +
5
112x7 + · · · .
Newton then obtain the series ofsin θ by reversing
θ = x +1
6x3 +
3
40x5 +
5
112x7 + · · · .
This is
sin θ = θ − 1
6θ3 +
1
120θ5 − + · · ·
From this, the series ofcos θ also follows.
Chapter 36
Wallis’ product
Wallis has obtained an infinite product expression forπ by interpola-tion. 1 Precisely,
4
π=
3 × 3 × 5 × 5 × 7 × 7 × 9 × 9 × 11 × 11 × · · ·2 × 4 × 4 × 6 × 6 × 8 × 8 × 10 × 10 × 12 × · · · .
Let In :=∫ π
20
sinn dx.(1) Establish the reduction formula
In =n − 1
n· In−2.
(2) From this we see that the subsequences(I2n) and (I2n+1) aremonotonic decreasing. Since everyIn is positive, these two subsequencesare convergent.
(3) We show that they indeed converge to the same limit. Sincesin2n+1 x ≤ sin2n x ≤ sin2n−1 x for x ∈
[0, π
2
], we haveI2n+1 ≤ I2n ≤
I2n−1, and
2n
2n + 1=
I2n+1
I2n−1≤ I2n
I2n−1≤ 1.
Therefore,limn→∞I2n
I2n−1= 1, and indeedlimn→∞ I2n = limn→∞ I2n+1.
(4) Now, from the recurrence relation and the initial valuesI0 = π2
1J. F. Scott,The Work of John Wallis, Chelsea, 1981; pp.47–60.
328 Wallis’ product
andI1 = 1, we have
I2n =1
2· 3
4· · · 2n − 1
2n· π
2=
(2n − 1)!!
(2n)!!· π
2,
I2n−1 =2
3· 4
5· · · 2n − 2
2n − 1=
(2n − 2)!!
(2n − 1)!!,
where, for convenience, we have put
k!! :=
{
k(k − 2) · · ·4 · 2, if k is even,
k(k − 2) · · ·3 · 1, if k is odd,
for a positive integerk. Now, since
I2n
I2n−1
=π
2× ((2n − 1)!!)2
(2n − 2)!!(2n)!!
=π
4× 3 × 3 × 5 × 5 × · · · × (2n − 1) × (2n − 1)
2 × 4 × 4 × 6 × · · · × (2n − 2) × (2n),
we conclude that
4
π= lim
n→∞
3 × 3 × 5 × 5 × · · · × (2n − 1) × (2n − 1)
2 × 4 × 4 × 6 × · · · × (2n − 2) × (2n).
Remark.By omitting the last fraction on the right hand side, this canalso be written as
4
π= lim
n→∞
3 × 3 × 5 × 5 × 7 × · · · × (2n − 3) × (2n − 1)
2 × 4 × 4 × 6 × 6 × · · · × (2n − 2) × (2n − 2).
Exercise
Suppose we construct a sequence of rectangles as follows. Webeginwith a square of area one. We then alternate adjoining a rectangle ofarea one alongside or on top of the previous rectangle. Find the limitingratio of length to height.
Chapter 37
Newton: Universal gravitation
Kepler’s three laws of planetary motion
From astronomical observations, Johannes Kepler formulated his fa-mous three laws of planetary motions.
Kepler’s first law : A planet revolves around the sun in an elliptical orbitwith the sun at one focus.
S
Kepler’s second law: Inside the ellipse, a focal radius sweeps off equalareas in equal times.
Kepler’s third law : The ratio of the cube of the major axis of an ellip-tical orbit and the square of the period of revolution is the same for allplanets in the solar system.
We set up a polar coordinate system with a pole atS. The area sweptout a focal radius is given bydA = 1
2ρ2dθ. The angular velocitydθ
dtis usually denoted byω. The quantitymass · ρ2ω is called theangu-lar momentum. Thus, dA
dt= 1
2ρ2ω is constant.Kepler’s second law is
equivalent to the principle of conservation of angular momentum.
330 Newton: Universal gravitation
Torque and angular momentum
Consider a force−→F acting on a particleP , of massm, moving at a
velocity−→v . With respect to an originO,(i) the torque is the vector product
−→OP ×−→
F ,
(ii) the angular momentum is−→OP × m−→v = m · −→OP × d
dt
(−→OP)
.
The torque is the rate of change of the angular momentum.
Proof.
d
dt
(
m · −→OP × d
dt
(−→OP))
=m · d
dt
(−→OP)
× d
dt
(−→OP)
+ m · −→OP × d2
dt2
(−→OP)
=−→OP ×
(
m · d2
dt2
(−→OP))
=−→OP ×−→
F .
From this we deduce thatthe planet experiences a central force field(which is along the lineOP ) if and only if the angular momentum isconstant.
Polar coordinate description of planetary orbit
Suppose the orbit of a planet is an ellipse of semi-major axisa and semi-minor axisb (so that it has Cartesian equation
x2
a2+
y2
b2= 1.
The foci are the pointsS = (−c, 0) andS ′ = (c, 0), wherec2 = a2− b2.We takeS as the origin and the positivex-axis as the pole and polar of apolar coordinate system. The polar equation of the ellipse is
ρ =b2
a(1 − ε cos θ)(37.1)
whereε := ca
is the eccentricity of the ellipse.
331
Radial and transverse accelerations
Every vector−→F can be resolved into the resultant of two orthogonal
vectors, one component−→F r along the radial direction, and another
−→F θ
in an orthogonal direction, called the transverse component. Let
−→e r =(cos θ, sin θ),−→e θ =(− sin θ, cos θ),
be theunit radial andunit transversevectors respectively.Let ρ := dρ
dtandρ := d2ρ
dt2.
Proposition 37.1.The acceleration vector has an orthogonal decompo-sition −→a = (ρ − ρω2)−→e r + (2ρω + ρω)−→e θ.
Proof. From−→OP = (ρ cos θ, ρ sin θ), we have the acceleration vector
−→a =d2
dt2−→OP =
(d2
dt2(ρ cos θ),
d2
dt2(ρ sin θ)
)
=d
dt
(d
dt(ρ cos θ),
d
dt(ρ sin θ)
)
=d
dt(ρ cos θ − ρω sin θ, ρ sin θ + ρω cos θ)
= (ρ cos θ − 2ρω sin θ − ρω sin θ − ρω2 cos θ,
ρ sin θ + 2ρω cos θ + ρω cos θ − ρω2 sin θ)
= (ρ − ρω2)(cos θ, sin θ) + (2ρω + ρω)(− sin θ, cos θ)
= (ρ − ρω2)−→e r + (2ρω + ρω)−→e θ.
Law of universal gravitation
Assume Kepler’s laws. From the second law,12ρ2ω = constant, we have
(1) 2ρρω + ρ2ω = 0. This means that−→a θ = 0 and the force is central:
−→a = (ρ − ρω2)−→e r.
(2) πab =∫ T
0dA =
∫ T
012ρ2ωdt = 1
2ρ2ω · T . Therefore,ω = 2πab
T· 1
ρ2 ,and
332 Newton: Universal gravitation
(3) ρω2 = ρ(
2πabT
· 1ρ2
)2
= 4π2a2
T 2 · 1ρ2 · b2
ρ= 4π2a3
T 2 · 1ρ2 (1 − ε cos θ) from
(37.1). According to Kepler’s third law,4π2a3
T 2 is a constant. We rewrite(37.1) as
1
ρ=
a
b2(1 − ε cos θ),
and differentiate twice to obtain
−ρ
ρ2=
a
b2· ε sin θ · ω;
−ρ = ρ2 · a
b2· ε sin θ · ω
=2πab
T· a
b2· ε sin θ;
−ρ =2πab
T· a
b2· ε cos θ · ω
=2πab
T· a
b2· ρ2ω · ε cos θ
ρ2
=4π2a3
T 2· ε cos θ
ρ2.
Therefore the acceleration is
−→a = −→a r = (ρ − ρω2)−→e r
=
[
−4π2a3
T 2· ε cos θ
ρ2− 4π2a3
T 2· 1
ρ2(1 − ε cos θ)
]
−→e r
= −(
4π2a3
T 2
)1
ρ2· −→e r.
From this,−→F = m−→a = −4π2a3
T 2· m
ρ2· −→e r.
Let M be the mass of the sun. ThenG = 4πa3
T 2·M is a universal constant,and
F = −G · M · mρ2
.
This is Newton’s law of universal gravitation:a planet revolving aroundthe sun experiences acentripetal force which is proportional to the massof the planet, and inversely proportional to the square of its distancefrom the sun.
ccci
Supplement 25: Hilbert’s third problem: The equality of the vol-umes of two tetrahedra of equal bases and equal altitudes
In two letters to Gerling, Gauss1 expresses his regret that certain theo-rems of solid geometry depend upon the method of exhaustion,i.e., inmodem phraseology, upon the axiom of continuity (or upon theaxiom ofArchimedes). Gauss mentions in particular the theorem of Euclid, thattriangular pyramids of equal altitudes are to each other as their bases.Now the analogous problem in the plane has been solved: Gerling alsosucceeded in proving the equality of volume of symmetrical polyhedraby by dividing them into congruent parts. Nevertheless, it seems to meprobable that a general proof of this kind for the theorem of Euclid justmentioned is impossible, and it should be our task to give a rigorousproof of its impossibility. This would be obtained, as soon as we suc-ceeded inspecifying two tetrahedra of equal bases and equal altitudeswhich can in no way he split up into congruent tetrahedra, andwhichcannot be combined with congruent tetrahedra to form two polyhedrawhich themselves could be split up into congruent tetrahedra. 2
Combinatorial equivalence of polyhedra
We say that two polyhedra are combinatorially equivalent ifthey can bedivided into congruent parts. A polyhedron has a dihedral angle for eachtwo faces intersecting in an edge. We can therefore speak of its sum ofdihedral angles. If two polyhedraP with dihedral anglesθ1, . . . , θp andQ with dihedral anglesϕ1, . . . , ϕq are combinatorially equivalent, thenthere are positive integersk1, . . . , kp, kp+1 andm1, . . . , mq, mq+1 suchthat
k1θ1 + · · · kpθp + kp+1π = m1ϕ1 + · · ·+ m1ϕq + mq+1π.
Max Dehn’s proof of non-equivalence of the cube and regular tetra-hedron
Each dihedral angle of a cube is clearlyπ2. On the other hand, each
dihedral angle of a regular tetrahedron isθ = arccos 13. If the two are
1Werke, vol. 8, pp.241 and 244.2Since this was written Herr Dehn has succeeded in proving this impossibility. See his note[hUber
raumgleiche Polyheder in Nach. d. K. Geselsch. d Wiss. zu Gottingen, 1900, and a paper soon to appear inMath. Annalen [vol. 55, pp. 465–178].
cccii Newton: Universal gravitation
θ
combinatorially equivalent, then there are positive integersk andm suchthatkθ = m · π
2. This means thatcos kθ = 0 or±1. This is a contradic-
tion sincecos kθ cannot be an integer ifcos θ = 13.
Two tetrahedra which are not combinatorially equivalent 3
A
BC
D
A′
B′
C′
D′
P Q
The tetrahedronP hasAC = BC = DC and∠BCD = π2, and
AC perpendicular to the plane ofBCD. It has three dihedral anglesequal toπ
2and three equal toarccos 1√
3. This solid is not combinatorially
equivalent to a cube.The tetrahedronQ has dihedral angles equal toπ
2, π
2, π
6, π
4, arccos 1√
6
andarccos√
23. This is combinatorially equivalent to a trianglular prism
with baseB′C ′D′, which in turn is combinatorially equivalent to a cube.This shows that the solidsP andQ are not combinatorially equivalent.
3I learned this from my friend Professor M. K. Siu of the University of Hong Kong.
ccciii
Supplement 26: Archimedes’ calculation of the surface areaof asphere
Proposition 0.2 (SCI 33). The surface of any sphere is equal to fourtimes the greatest circle in it.
Lower bounds of surface of sphere
Proposition 0.3(SCI 21). A regular polygon of an even number of sidesbeing inscribed in a circle, asABC . . . A′ . . . C ′B′A, so thatAA′ is adiameter, if two angular points next but one to each other, asBB′, bejoined, and the other lines parallel toBB′ and joining pairs of angularpoints be drawn, asCC ′, DD′ . . . , then
(BB′ + CC ′ + · · · ) : AA′ = A′B : BA.
INLGKF
C'
C
A
E'B'
B E
D'
D
HA'
Proposition 0.4 (SCI 22). If a polygon be inscribed in a segment of acircle LAL′, so that all its sides excluding the base are equal and theirnumber is even, asLK . . .A . . .K ′L′, A being the middle point of thesegment, and if the linesBB′, CC ′ . . . , parallel to the baseLL′ andjoining pairs of angular points be drawn, then
(BB′ + CC ′ + · · ·+ LM) : AM = A′B : BA.
ccciv Newton: Universal gravitation
RHQGPF A'
KC
B
K'
B'
C'
A
M
L'
L
Proposition 0.5(SCI 23). The surface of the sphere is greater than thesurface described by the revolution of the polygon inscribed in the greatcircle about the diameter of the great circle.
Proposition 0.6 (SCI 24). If a regular polygonAB . . . A′ . . . B′A, thenumber of whose sides is a multiple of 4, be inscribed in a great circleof a sphere, and ifBB′ subtending two sides be joined, and all the otherlines parallel toBB′ ad joining pairs of angular points be drawn, thenthe surface of the figure inscribed in the sphere by the revolution of thepolygon about the diameterAA′ is equal to a circle the square of whoseradius is equal to the rectangle
BA(BB′ + CC ′ + · · · ).Proposition 0.7(SCI 25). The surface of the figure inscribed in a sphereas in the last propositions, consisting of portions of conical surfaces, isless than 4 times the greatest circle in the sphere.
Upper bounds of surface of sphere
Let a regular polygon, whose sides are a multiple of 4 in number, becircumscribed about a great circle of given sphere, asAB . . . A′ . . . B′A,and about the polygon, describe another circle, which will therefore havehe same centre as the great circle of the sphere.
Proposition 0.8(SCI 28). The surface of the figure circumscribed to thegiven sphere is greater than that of the sphere itself.
cccv
m'
m
a'a
B'
B
M'
M
A
OA'
Proposition 0.9(SCI 29). The surface of the figure circumscribed to thegiven sphere is equal to a circle the square on whose radius isequal to
AB(BB′ + CC ′ + · · · ).
Proposition 0.10(SCI 30). The surface of the figure circumscribed tothe given sphere is greater than 4 times the great circle of the sphere.
Proposition 0.11(SCI 32). If a regular polygon with4n sides be in-scribed in a great circle of a sphere, asab . . . a′ · · · b′a, and a similarpolygonAB . . .A′ . . . B′A be described about the great circle, and ifthe polygons revolve with the great circle about the diameters aa′, AA′
respectively, so that they describe the surfaces of solid figures inscribedin and circumscribed to the sphere respectively, then the surfaces of thecircumscribed and the inscribed figures are to one another inthe dupli-cate ratio of their sides.
Proof of SCI 33
Let C be a circle equal to four times the great circle.Then if C is not equal to the surface of the sphere, it must either be
less or greater.I. SupposeC less than the surface of the sphere.
It is then possible to find two linesβ > γ such that
β : γ < surface of sphere: C.
cccvi Newton: Universal gravitation
Let δ be a mean proportional betweenβ andγ.Suppose similar regular polygons with4n sides circumscribed about
and inscribed in a great circle such that the ratio of their sides is less thanthe ratioβ : δ. 4 Let the polygons with the circle revolve together about adiameter common to all, describing solids of revolution as before. Then,
surface of outer solid: surface of inner solid
=side of outer polygon2 : side of inner polygon2
<β2 : δ2
=β : γ
<surface of sphere: C.
But this is impossible, since the surface of the circumscribed solid isgreater than that of the sphere, while the surface of the inscribed solid isless thanC.
ThereforeC is not less than the surface of the sphere.II. SupposeC greater than the surface of the sphere. Take linesβ and
γ such thatβ : γ < C : surface of sphere.
Circumscribe and inscribe to the great circle similar regular polygons,as before, such that the ratio of their sides is less than the ratio β : δ (δbeing a mean proportional ofβ andγ), and suppose solids of revolutiongenerated in the usual manner. Then in this case,
surface of outer solid: surface of inner solid
=side of outer polygon2 : side of inner polygon2
<β2 : δ2
=β : γ
<C : surface of sphere.
But this is impossible, since the surface of the circumscribed solid isgreater thanC, while the surface of the inscribed solid is less than thatof the sphere.
4Proposition 3: Given two unequal magnitudes and a circle, itis possible to inscribe a polygon in thecircle and to describe another about it so that the side of thecircumscribed polygon may have to the side ofthe inscribed polygon a ratio less than that of the greater magnitude to the less.
1
ThusC is not greater than the surface of the sphere.Therefore, since it is neither greater nor less,C is equal to the surface
of the sphere.
Chapter 38
Logarithms
The logarithms were originally defined to facilitate multiplication ofnumbers. Leta > 1 be a fixed positive number Assuming the indexlaws
am · an = am+n,
am ÷ an = am−n,
(am)n = amn
to hold not only for positive integersm andn, but also arbitrary realnumbers, we first see the meaningam for negative and rational values ofm.
Given a positive numberx, we writex in the form ofay, and callythe logarithm ofx with basea. The index laws translate into
log(xy) = log x + log y,
logx
y= log x − log y,
log xy = y · log x.
The increasing property (and hence the continuity) of the logarithmsis clear: for positive numbersx andy,
log x < log y if and only if x < y.
The additive property of logarithms also translates into the state-ment that logarithm of the geometric mean of two positive numbers is
402 Logarithms
the arithmetic mean of their logarithms. This is how Euler computedlog10 5 = 0.698970003 · · · .
§106,Introductio
Let a = 10 be the base of the logarithm, which is usually the case inthe computed tables. We seek an approximate logarithm of thenumber5. Since5 lies between1 and10, whose logarithms are0 and1, in thefollowing manner we take successive square roots until we arrive at thenumber5 exactly.
A = 1.000000; log A = 0.0000000; so that
B = 10.000000; log B = 1.0000000; C =√
AB
C = 3.162277; log C = 0.5000000; D =√
BC
D = 5.623413; log D = 0.7500000; E =√
CD
E = 4.216964; log E = 0.6250000; F =√
DE
F = 4.869674; log F = 0.6875000; G =√
DF
G = 5.232991; log G = 0.7187500; H =√
FG
H = 5.048065; log H = 0.7031250; I =√
FH
I = 4.958069; log I = 0.6953125; K =√
HI
K = 5.002865; log K = 0.6992187; L =√
KI
L = 4.980416; log L = 0.6972656; M =√
KL
M = 4.991627; log M = 0.6982421; N =√
KM
N = 4.997242; log N = 0.6987304; O =√
KN
O = 5.000052; log O = 0.6989745; P =√
NO
P = 4.998647; log P = 0.6988525; Q =√
OPQ = 4.999350; log Q = 0.6989135; R =
√OQ
R = 4.999701; log R = 0.6989440; S =√
OR
S = 4.999876; log S = 0.6989592; T =√
OS
T = 4.999963; log T = 0.6989668; V =√
OT
V = 5.000008; log V = 0.6989707; W =√
TV
W = 4.999984; log W = 0.6989687; X =√
WV
X = 4.999997; log X = 0.6989697; Y =√
V X
Y = 5.000003; log Y = 0.6989702; Z =√
XYZ = 5.000000; log Z = 0.6989700.
403
Here is a reorganization of the calculation by repeated bisections andextraction of square roots:
n x log x1 0
1 3.16227766016837933199889354443 0.53 4.2169650342858224856901335951 0.6254 4.8696752516586311493529748503 0.68757 4.9580682416846557488356043457 0.69531259 4.9804160612484109831271017907 0.69726562510 4.9916277163626864097472854701 0.698242187511 4.9972430050336106382451186179 0.6987304687513 4.9986478139349424228699400640 0.698852539062514 4.9993503664719326330925881066 0.6989135742187515 4.9997016797680793318287505450 0.69894409179687516 4.9998773456738244938535734160 0.698959350585937517 4.9999651809412098985610500405 0.698966979980468819 4.9999871399991559239181031834 0.698968887329101620 4.9999981195642941789135393844 0.69896984100341823 4.9999994920116317103993039017 0.698969960212707525 4.9999998351235249561542900839 0.698969990015029927 4.9999999209015019465236999950 0.698969997465610528 4.9999999637904909935480218118 0.698970001190900829 4.9999999852349856550200883161 0.698970003053546
30 4.9999999959572330202460981398 0.6989700039848685
26 5.0000000066794804084654257436 0.698970004916191124 5.0000001782354417470659257366 0.698970019817352322 5.0000008644593459643656083525 0.698970079421997121 5.0000036093559046401534626508 0.698970317840576218 5.0000090991535426660455335983 0.698970794677734412 5.0000530177516399948230149100 0.6989746093758 5.0028646105752330658522723684 0.699218756 5.0480657166674707699374825934 0.7031255 5.2329911468149468809767055026 0.718752 5.6234132519034908039495103978 0.75
10 1
Change of basis of logarithms
How are the logarithms of a number to different bases related? Begin-ning withx = blogb x, if we take logarithms to basea, then
loga x = loga blogb x = (logb x)(loga b).
Therefore,
logb x =loga x
loga b.
404 Logarithms
Chapter 39
Euler’s introduction to e and theexponential functions
Chapter 7 ofIntroduction to the Analysis of the Infinite, I, Euler beginswith the general question of calculating the various powersof a numbera > 1.
§114,Introductio
Sincea0 = 1, when the exponent ona increases, the power itself in-creases, provided thata > 1. It follows that if the exponent is infinitelysmall and positive, then the power also exceeds1 by an infinitely smallnumber. Letaε = 1+δ, whereδ is also an infinitely small number. Eulerassumesδ in the formkε. Thus,aε = 1 + kε. In this case,
ε = loga(1 + kε).
For a = 10, from the table of common logarithms, we look for anumber which exceeds1 by the smallest possible amount, for instance,1 + 1
1000000, so thatkε = 1
1000000. Then,
log10
(
1 +1
1000000
)
= log10
1000001
1000000= 0.0000004329 = ε.
Sincekε = 0.000001, it follows that 1k
= 43429100000
, andk = 10000043429
=2.30258. We see thatk is a finite number which depends on the value ofthe basea. If a different base had been chosen, then the logarithms ofthe same number1 + kε will differ from the logarithm already given. Itfollows that a different value ofk will result.
406 Euler’s introduction to e and the exponential functions
Remarks.(1) k = limε→0aε−1
ε; it is the derivative ofax atx = 0.
(2) limε→0loga(1+kε)
ε= 1.
§115
Sinceaε = 1 + kε, we haveajε = (1 + kε)j whatever value we assignto j. It follows that
ajε = 1 +j
1kε +
j(j − 1)
1 · 2 k2ε2 +j(j − 1)(j − 2)
1 · 2 · 3 k3ε + · · ·
If we let j = zε, wherez is any finite number, sinceε is infinitely small,
thenj is infinitely large. Then we haveε = zj, whereε is represented
by a fraction with infinite denominator, so thatε is infinitely small, as itshould be. When we substitutez
jfor ε then
az =
(
1 +kz
j
)j
= 1 +1
1kz +
1(j − 1)
1 · 2j k2z2 +1(j − 1)(j − 2)
1 · 2j · 3j k3z3
+1(j − 1)(j − 2)(j − 3)
1 · 2j · 3j · 4j k4z4 + · · ·
This equation is true provided an infinitely large number is substitutedfor j, but thenk is a finite number depending ona, as we have just seen.
§116
Sincej is infinitely large,j−1j
= 1, and the larger the number we substi-
tute forj, the closer the value of the fractionj−1j
comes to1. Therefore,
if j is a number larger than any assignable number, thenj−1j
is equal to
1. For the same reason,j−2j
= 1, j−3j
= 1, and so forth. It follows thatj−12j
= 12, j−2
3j= 1
3, j−3
4j= 1
4and so forth. When we substitute these
values, we obtain
az = 1 +kz
1+
k2z2
1 · 2 +k3z3
1 · 2 · 3 +k4z4
1 · 2 · 3 · 4 + · · · . (39.1)
This equation represents a relationship between the numbers a andk,since when we letz = 1, we have
a = 1 +k
1+
k2
1 · 2 +k3
1 · 2 · 3 +k4
1 · 2 · 3 · 4 + · · · . (39.2)
407
Whena = 10, thenk is approximately equal to2.30258 as we havealready seen.
§122
Since we are free to choose the basea for the system of logarithms, wenow choosea in such a way thatk = 1. Suppose thatk = 1, then theseries found in§116,
1 +1
1+
1
1 · 2 +1
1 · 2 · 3 +1
1 · 2 · 3 · 4 + · · ·
is equal toa. If the terms are represented as decimal fractions andsummed, we obtain the value for
a = 2.71828182845904523536028 · · · .
When this base is chosen the logarithms are called natural orhyper-bolic. The latter name is used since the quadrature of a hyperbolic canbe expressed through these logarithms. For the sake of brevity for thisnumber2.718281828459 · · · we will use the symbole, which will de-note the base for natural or hyperbolic logarithms, which corresponds tothe valuek = 1, and
e = 1 +1
1+
1
1 · 2 +1
1 · 2 · 3 +1
1 · 2 · 3 · 4 + · · · .
Remarks.(1) With a = e andk = 1, (39.1) now reads, as Euler gives in§123,
ez = 1 +z
1+
z2
1 · 2 +z3
1 · 2 · 3 +z4
1 · 2 · 3 · 4 + · · · .
Now, with z = k, this isa according to (39.2). It follows thata = ek
andk = log a.(2) limε→0
aε−1ε
= log a. In particular,limε→0eε−1
ε= 1.
Theorem 39.1. ddx
(ex) = ex.
Proof.
d
dx(ex) = lim
h→0
ex+h − ex
h= lim
h→0
ex(eh − 1)
h= ex·lim
h→0
eh − 1
h= ex·1 = ex.
Corollary 39.2. ddx
(ax) = (log a) · ax.
408 Euler’s introduction to e and the exponential functions
Chapter 40
Euler: Natural logarithms
§118,Introductio
Sinceeε = 1 + ε, whereε is an infinitely small fraction and
e = 1 +1
1+
1
1 · 2 +1
1 · 2 · 3 +1
1 · 2 · 3 · 4 + · · · ,
if e is taken as the base of the logarithm, then
ε = log(1 + ε)
andjε = log(1 + ε)j . It is clear that the larger the number chosen forj the more(1 + ε)j will exceed1. If we let j be an infinite number, thevalue of the power(1 + ε)j becomes greater than any number greaterthan1. Now if we let(1+ ε)j = 1+x, thenlog(1+x) = jε. Sincejε isa finite number, namely the logarithm of1 + x, it is clear thatj must bean infinitely large number; otherwise,jε could not have a finite value.
§119
Since we have let(1 + ε)j = 1 + x, we have1 + ε = (1 + x)1j and
ε = (1 + x)1j − 1 so that
log(1 + x) = j(1 + x)1j − j
410 Euler: Natural logarithms
wherej is a number infinitely large. But we have1
(1 + x)1j = 1 +
1
jx − 1(j − 1)
j · 2j x2 +1(j − 1)(2j − 1)
j · 2j · 3j x3
− 1(j − 1)(2j − 1)(3j − 1)
j · 2j · 3j · 4j x4 + · · ·
Sincej is an infinite number,j−12
= 12, 2j−1
3j= 2
3, 3j−1
4j= 3
4, etc. Now it
follows that
j(1 + x)1j = j +
x
1− x2
2+
x3
3− x4
4+ · · ·
As a result we have
log(1 + x) =x
1− x2
2+
x3
3− x4
4+ · · · .
Theorem 40.1. ddx
(log x) = 1x.
Proof.
d
dx(log x) = lim
h→0
log(x + h) − log x
h= lim
h→0
log(1 + h
x
)
h
= limh→0
1
x· log
(1 + h
x
)
hx
=1
x· lim
h→0
log(1 + h
x
)
hx
=1
x.
Corollary 40.2. ddx
(loga x) = 1x log a
.
Theorem 40.3.log a = limn→∞ n( n√
a − 1).
In §123, Euler also gives the series
log1 + x
1 − x=
2x
1+
2x3
3+
2x5
5+
2x7
7+
2x9
9+ · · ·
and remarks that this series is strongly convergent if we substitute anextremely small fraction forx. For instance, ifx = 1
5, then
log3
2=
2
1 · 5 +2
3 · 53+
2
5 · 55+
2
7 · 57+ · · ·
1In §72 Euler has explained the binomial theorem for a rational exponent.
Chapter 41
Euler’s formulaeiv = cos v + i sin v
Chapter VIII ofIntroductio, I: On transcendental quantities which arisefrom the circle
§126
After having considered logarithms and exponentials, we must now turnto circular arcs with their sines and cosines. This is not only becausethese are further genera of transcendental quantities, butalso since theyarise from logarithms and exponentials when complex valuesare used.
Then Euler gives the value ofπ to 127 places after the decimal point:1
π = 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 3751058209 74944 59230 78164 06286 20899 86280 34825 34211 7067982148 08651 32823 06647 09384 46 · · ·
§130
Recall addition formulas
sin(y ± z) = sin y cos z ± cos y sin z,
cos(y ± z) = cos y cos z ∓ sin y sin z.
1Both the Lausanne 1748 edition and the English translation (Springer) print7 (instead of8) for the 113thdigit after the decimal point.
412 Euler’s formula eiv = cos v + i sin v
Convert products into sums:
sin y cos z =1
2[sin(y + z) + sin(y − z)]
cos y sin z =1
2[sin(y + z) − sin(y − z)]
cos y cos z =1
2[cos(y + z) + cos(y − z)]
sin y sin z =−1
2[cos(y + z) − cos(y − z)].
Also establish the half angle formulas:
cosv
2=
√
1 + cos v
2, sin
v
2=
√
1 − cos v
2.
§131
Conversion of sums into products:
sin a + sin b = 2 sina + b
2cos
a − b
2,
sin a − sin b = 2 sina − b
2cos
a + b
2,
cos a + cos b = 2 cosa + b
2cos
a − b
2,
cos a − cos b = −2 sina + b
2sin
a − b
2.
§132
Sincesin2 z + cos2 z = 1, we have the factors(cos z + i sin z)(cos z −i sin z) = 1. Although these factors are complex, still they are quiteuseful in combining and multiplying arcs.
413
Then Euler established the formula
(cos x±i sin x)(cos y±i sin y)(cos z±i sin z) = cos(x+y+z)±i sin(x+y+z).
2
(cos z ± i sin z)n = cos nz ± i sin nz.
§133 (continued)
It follows that
cos nz =1
2[(cos z + i sin z)n + (cos z − i sin z)n],
sin nz =1
2[(cos z + i sin z)n − (cos z − i sin z)n].
Expanding the binomials we obtain the following series
cos nz = (cos z)n − n(n − 1)
1 · 2 (cos z)n−2(sin z)2
+n(n − 1)(n − 2)(n − 3)
1 · 2 · 3 · 4 (cos z)n−4(sin z)4
− n(n − 1)(n − 2)(n − 3)(n − 4)(n − 5)
1 · 2 · 3 · 4 · 5 · 6 (cos z)n−6(sin z)6 + · · · ;
sin nz =n
1(cos z)n−1 sin z − n(n − 1)(n − 2)
1 · 2 · 3 (cos z)n−3(sin z)3
+n(n − 1)(n − 2)(n − 3)(n − 4)
1 · 2 · 3 · 4 · 5 (cos z)n−5(sin z)5 − · · ·
§134
Let the arcz be infinitely small, thensin z = z andcos z = 1. If n is aninfinitely large number, so thatnz is a finite number, saynz = v, then,sincesin z = z = v
n, we have
2[sic] The correct formula should read
(cos x ± i sinx)(cos y ± i sin y)(cos z ± i sin z) = cos(x ± y ± z) + i sin(x ± y ± z).
and deduce (in§133) de Moivre’s theorem
414 Euler’s formula eiv = cos v + i sin v
cos v = 1 − v2
2!+
v4
4!− v6
6!+ · · · ,
sin v = v − v3
3!+
v5
5!− v7
7!+ · · ·
It follows that if v is a given arc, by means of these series, the sine andcosine can be found. . . .
§135
Once sines and cosines have been computed, tangents and cotangentscan be found in the ordinary way. However,, since the multiplication anddivision of such gigantic numbers is so inconvenient, a different methodof expressing these functions is desirable. . . .
§138
Once again we use the formulas in§133, where we letz be an infinitelysmall arc and letn be an infinitely large numberj, so thatjz has a finitevalue v. Now we havenz = v and z = v
j, so thatsin z = v
jand
cos z = 1. With these substitutions,
cos v =(1 + iv
j)j + (1 − iv
j)j
2
and
sin v =(1 + iv
j)j − (1 − iv
j)j
2i.
In the preceding chapter we saw that(1+ zj)j = ez wheree is the base of
the natural logarithm. When we letz = iv and thenz = −iv we obtain
cos v =eiv + e−iv
2and sin v =
e−v − e−iv
2i.
From these equations we understand how complex exponentials can beexpressed by real sines and cosines, since
eiv = cos v + i sin v and e−iv = cos v − i sin v.
415
§140
Since sin zcos z
= tan z, the arcz can now be expressed through the tangentas follows:z = 1
2ilog(
1+i tan z1−i tan z
). We have seen in§123 that
log
(1 + z
1 − z
)
=2z
1+
2z3
3+
2z5
5+
2z7
7+ · · · .
When we substitutei tan z for z we obtain
z =tan z
1− tan3 z
3+
tan5 z
5− tan7 z
7+ · · · .
If we let t = tan z so thatz is the arc whose tangent ist, which we willindicate byarctan t, thenz = arctan t. When we know the tangent oft,the corresponding arcz is given by
z =t
1− t3
3+
t5
5− t7
7+
t9
7− · · · .
If the tangent is equal to the unit radius, then the arcz is equal to45degrees orz = π
4and
π
4= 1 − 1
3+
1
5− 1
7+ · · ·
This series, which was first discovered by Leibniz, can be used to findthe value of the circumference of the circle.
416 Euler’s formula eiv = cos v + i sin v
Chapter 42
Summation of powers ofintegers
Let Sk(n) := 1k + 2k + · · · + nk. We make use of the series expansionof ez:
ez = 1 + z +z2
2!+ · · ·+ zk
k!+ · · ·
and analogous series fore2z, e3z, . . . ,enz:
ez = 1 + z +z2
2!+ · · ·+ zk
k!+ · · ·
e2z = 1 + 2z +22z2
2!+ · · ·+ 2kzk
k!+ · · ·
...
e(n−1)z = 1 + (n − 1)z +(n − 1)2z2
2!+ · · · + (n − 1)kzk
k!+ · · · .
Combining these (with1 = 1), we have
n + S1(n − 1)z +S2(n − 1)z2
2!+ · · ·+ Sk(n − 1)zk
k!+ · · ·
for the series expansion of
1 + ez + e2z + · · ·+ e(n−1)z =enz − 1
ez − 1.
418 Summation of powers of integers
The series expansion of the function on the right hand side can be foundfrom
enz − 1
ez − 1=
enz − 1
z· z
ez − 1.
(1) The first factor clearly has series expansion
enz − 1
z= n +
n2
2!z + · · ·+ nk+1
(k + 1)!zk + · · ·
(2) Suppose we write
z
ez − 1= B0 +
B1
1!z +
B2
2!z2 + · · ·+ Bk
k!zk + · · ·
These coefficientsB0, B1, . . . , Bk, . . . are called theBernoulli num-bers.
The product of these two series has beginning termn, and subse-quently, for eachk = 1, 2, . . . , the coefficient ofzk equal to
∑
i+j=k
Bi
i!· nj+1
(j + 1)!=
1
(k + 1)!
k∑
i=0
(k + 1
i
)
Bink+1−i
Now comparison gives
Sk(n − 1) =1
k + 1
k∑
i=0
(k + 1
i
)
Bink+1−i.
Remarks.(1) If we agree, in the binomial expansion of(B + 1)k+1, toreplace everyBi by the numberBi, then the above formula can be simplywritten as
Sk(n − 1) =(B + n)k+1 − Bk+1
k + 1. (42.1)
(2) This symbolic device can further help computingSk(n−1) easily,without actually calculating the Bernoulli numbers. Comparing (42.1)with
k · Sk−1(n − 1) = (B + n)k − Bk,
we note that “formally”
Sk(n − 1) =
∫
kSk−1(n − 1)dn + cn
419
where the constantc so chosen that the sum of the coefficients is equalto 1.
(3) An analogous expression holds forSk(n):
Sk(n) =
∫
kSk−1(n)dn + cn
where the constantc so chosen that the sum of the coefficients is equalto 1.
Example 42.1.SinceS1(n) = 12n2 + 1
2n, we have
S2(n) =
∫
(n2 + n)dn + c2n (c2 appropriately chosen)
=1
3n3 +
1
2n2 +
1
6n;
S3(n) =
∫
(n3 +3
2n2 +
1
2n)dn + c3n (c3 appropriately chosen)
=1
4n4 +
1
2n3 +
1
4n2; (c3 = 0)
S4(n) =
∫
(n4 + 2n3 + n2)dn + c4n (c4 appropriately chosen)
=1
5n5 +
1
2n4 +
1
3n3 − 1
30n;
S5(n) =
∫ (
n5 +5
2n4 +
5
3n3 − 1
6n
)
dn + c5n (c5 appropriately chosen)
=1
6n6 +
1
2n5 +
5
12n4 − 1
12n2; (c5 = 0)
...
We summarize these formulas, with a few more, in factored form:
420 Summation of powers of integers
S1(n) =1
2n(n + 1),
S2(n) =1
6n(n + 1)(2n + 1),
S3(n) =1
4n
2(n + 1)2,
S4(n) =1
30n(n + 1)(2n + 1)(3n
2 + 3n − 1),
S5(n) =1
12n
2(n + 1)2(2n2 + 2n − 1),
S6(n) =1
42n(n + 1)(2n + 1)(3n
4 + 6n3− 3n + 1),
S7(n) =1
24n
2(n + 1)2(3n4 + 6n
3− n
2− 4n + 2),
S8(n) =1
90n(n + 1)(2n + 1)(5n
6 + 15n5 + 5n
4− 15n
3− n
2 + 9n − 3),
S9(n) =1
20n
2(n + 1)2(n2 + n − 1)(2n4 + 4n
3− n
2− 3n + 3),
S10(n) =1
66n(n + 1)(2n + 1)(n2 + n − 1)(3n
6 + 9n5 + 2n
4− 11n
3 + 3n2 + 10n − 5),
...
Chapter 43
Series expansions for thetangent and secant functions
From the series expansion ofzez−1
involving the Bernoulli numbers,namely,
z
2· ez + 1
ez − 1=
∞∑
k=0
B2k
(2k)!z2k,
if we replacez by iz, this becomes
∞∑
k=0
(−1)kB2k
(2k)!z2k =
iz
2· eiz + 1
eiz − 1
=iz
2· e
iz2 + e−
iz2
eiz2 − e−
iz2
=iz
2· 2 cos iz
2
2i sin iz2
=z
2· cot
z
2.
Therefore, we also have
z cot z =∞∑
k=0
(−1)k22kB2k
(2k)!z2k,
2z cot 2z =
∞∑
k=0
(−1)k24kB2k
(2k)!z2k.
422 Series expansions for the tangent and secant functions
Since these two series have equal constant terms, by taking their differ-ence and canceling a common factorz, we obtain
cot z − 2 cot 2z =∞∑
k=1
(−1)k(22k − 24k)B2k
(2k)!z2k−1.
This serendipitously leads to the series expansion oftan z, sincecot z−2 cot 2z = 1
tan z− 1−tan2 z
tan z= tan z. Therefore,
tan z =
∞∑
k=1
(−1)k−122k(22k − 1)B2k
(2k)!z2k−1.
The beginning terms are
tan z = z +1
3z3 +
5
12z5 +
17
315z7 + · · · .
Exercise
Verify the formula
cot z + tanz
2=
1
sin zand make use of it to find the series expansion ofz
sin z.
sec x
Since the secant functionsec z is an even function, we try to find theseries expansion ofsec x in the form
sec x = E0 −E2
2!z2 +
E4
4!z4 − E6
6!z6 + · · · + (−1)kE2k
(2k)!z2k + · · ·
Sincecos x sec x = 1, we require(
1 − z2
2!+
z4
4!+ · · ·
)(
E0 −E2
2!z2 +
E4
4!z4 + · · ·
)
≡ 1.
For each positive integern, the coefficient ofz2n of this product mustbe zero. It follows that
E2n +
(2n
2
)
E2n−2 +
(2n
4
)
E2n−4 + · · · + E0 = 0.
423
If we agree to replace in the binomial expansion of(E ± 1)2n everyEj
by Ej , this relation can be rewritten succinctly as
(E + 1)2n + (E − 1)2n = 0.
The numbersE2k are called theEuler numbers. The beginning onesare
E0 = 1, E2 = −1, E4 = 5, E6 = −61, E8 = 1385, . . .
424 Series expansions for the tangent and secant functions
Chapter 44
Euler’s first calculation of1 +
1
22k +1
32k +1
42k +1
52k + · · ·
Euler’s papers on these series:
• Volume 14: 41(1734/5), 61(1743), 63(1743), 130(1740).
• Volume 15: 393(1769), 597(1785).
• Volume 16 (part 1): 664(1790),
• Volume 16 (part 2): 736(1809), 746(1812).
Paper 41:De Summis Serierum Reciprocarum (1734)
In this paper, Euler usedp to denoteπ, andq for π2. 1 2
§§3,4 begin with the series for sine and cosine.
1First occurrence of the signπ. According to Cajori,History of Mathematical Notations, Art. 396,“[t]he modern notation for3.14159 · · · was introduced in 1706. It was in that year that William Jonesmadehimself noted, without being aware that he was doing anything noteworthy, through his designation of theratio of the length of the circle to its diameter by the letterπ. He took this step with ostentation”.
2Euler’s use ofπ. Ibid., Art. 397: In 1734 Euler employedp instead ofπ andg instead of π2
. Ina letter of April 16, 1738, from Stirling to Euler, as well as in Euler’s reply, the letterp is used. But in1736 he designated the ratio by the sign1 : π and thus either consciously adopted the notation of Jones orindependently fell upon it. . . . But the letter is not restricted to this use in hismechanica, and the definitionof π is repeated when it is taken for 3.14159. . . . He represented 3.1415. . . again byπ in 1737 (in a paperprinted in 1744), in 1743, in 1746, and in 1748. Euler and Goldbach usedπ = 3.1415 . . . repeatedly intheir correspondence in 1739. Johann Bernoulli used in 1739, in his correspondence with Euler, the letterc,(circumferentia), but in a letter of 1740 he began to useπ. Likewise, Nikolaus Bernoulli employedπ in hisletters to euler of 1742. Particularly favorable for wider adoption was the appearance ofπ for 3.1415 . . . inEuler’s Introductio in analysin infinitorum(1748). In most of his later publications, Euler clung toπ as hisdesignation of 3.1415. . . .
426 Euler’s first calculation of 1 +1
22k +1
32k +1
42k +1
52k + · · ·
y = s − s3
3!+
s5
5!− s7
7!+ · · · ,
x = 1 − s2
2!+
s4
4!− s6
6!+ · · · .
§5. If, for a fixedy, the roots of the equation
0 = 1 − s
y+
s3
3!y− s5
5!y+ · · ·
areA, B, C, D, E etc., then the “infinite polynomial” factors into theproduct of
1 − s
A, 1 − s
B, 1 − s
C, 1 − s
D, . . .
By comparing coefficients,
1
y=
1
A+
1
B+
1
C+
1
D+ · · ·
§7. If A is an acute angle (smallest arc) withsin A = y, then all theangles with sine equal toy are
A, ±π − A, ±2π + A, ±3π − A, ±4π + A, . . .
The sums of these reciprocals is1y. Sum of these reciprocals taking two
at a time is 0, taking three at a time is−13!y
. etc.§8. In general, if
a + b + c + d + e + f + · · · = α,
ab + ac + bc + · · · = β,
abc + abd + bcd + · · · = γ,
thena2 + b2 + c2 + · · · = α2 − 2β,
a3 + b3 + c3 + · · · = α3 − 3αβ + 3γ,
anda4 + b4 + c4 + · · · = α4 − 4α2β + 4αγ + 2β2 − 4δ.
427
Euler denotes byQ the sum of the squares,R the sum of cubes,S thesum of fourth powers etc. and wrote
P = α,
Q = Pα − 2β,
R = Qα − Pβ + 3γ,
S = Rα − Qβ + Pγ − 4δ,
T = Sα − Rβ + Qγ − Pδ + 5ǫ,
...
Now he applies these to
α =1
y, β = 0, γ =
−1
3!y, δ = 0, ǫ =
1
5!y, . . .
and obtains
P =1
y, Q =
1
y2, R =
Q
y− 1
2!y, S =
R
y− P
3!y,
and
T =S
y− Q
3!y+
1
4!y,
V =T
y− R
3!y+
P
5!y,
W =V
y− S
3!y+
Q
5!y− 1
6!y.
§10. Now Euler takesy = 1. All the angles with sines equal to 1 are
π
2,
π
2,−3π
2,−3π
2,
5π
2,
5π
2,−7π
2,−7π
2,
9π
2,9π
2, . . . .
From these,
4
π
(
1 − 1
3+
1
5− 1
7+
1
9− 1
11+ · · ·
)
= 1.
This gives Leibniz’s famous series
1 − 1
3+
1
5− 1
7+
1
9− 1
11+ · · · =
π
4.
428 Euler’s first calculation of 1 +1
22k +1
32k +1
42k +1
52k + · · ·
§11. Noting thatP = α = 1, β = 0, so thatQ = P = 1, Eulerproceeded to obtain
1 +1
32+
1
52+
1
72+ · · · = π2
8.
From this, he deduced that
1 +1
22+
1
32+
1
42+
1
52+
1
62+
1
72+ · · · =
π
6.
Continuing with
R =1
2, S =
1
3, T =
5
24, V =
2
15, W =
6
720, X =
17
315, . . .
Euler obtained
1 − 1
33+
1
53− 1
73+
1
93− · · · =
π3
32,
1 +1
34+
1
54+
1
74+
1
94+ · · · =
π4
96,
1 +1
24+
1
34+
1
44+
1
54+ · · · =
π4
90,
1 − 1
35+
1
55− 1
75+
1
95− · · · =
5π5
1536,
1 +1
36+
1
56+
1
76+
1
96+ · · · =
π6
960,
1 +1
26+
1
36+
1
46+
1
56+ · · · =
π6
945,
1 − 1
37+
1
57− 1
77+
1
97− · · · =
61π7
184320,
1 +1
38+
1
58+
1
78+
1
98+ · · · =
17π8
161280,
1 +1
28+
1
38+
1
48+
1
58+ · · · =
π8
9450.
Then Euler remarked that the general values of
1 +1
2n+
1
3n+
1
4n+ · · ·
429
can be determined from the sequence
1, 1,1
2,
1
3,
5
24,
2
15,
61
720,
17
315, . . .
In the remaining articles,§16 – 18, Euler gave an alternative methodof determining these sums. This time, he consideredy = 0 and madeuse of
0 = s − s3
3!+
s5
5!− s7
7!+ · · · .
Removing the obvious factors, he noted that the roots of
0 = 1 − s2
3!+
s4
5!− s6
7!+ · · ·
are±π, ±2π, ±3π, . . .
so that
1−s2
3!+
s4
5!−s6
7!+· · · =
(
1 − s2
π2
)(
1 − s2
4π2
)(
1 − s2
9π2
)(
1 − s2
16π2
)
· · ·
§17. With
α =1
3!, β =
1
5!, γ =
1
7!, . . .
and
P =1
π2+
1
4π2+
1
9π2+
1
16π2+ · · · ,
Q the sum of the squares of these terms,R the sum of the cubes, etc.,Euler now obtained
P = α =1
3!=
1
6,
Q = Pα − 2β =1
90,
R = Qα − Pβ + 3γ =1
945,
S = Rα − Qβ + Pγ − 4δ =1
9450,
T = Sα − Rβ + Qγ − Pδ + 5ǫ =1
93555,
...
V = Tα − Sβ + Rγ − Qδ + Pǫ − 6ζ =691
6825 · 93555.
430 Euler’s first calculation of 1 +1
22k +1
32k +1
42k +1
52k + · · ·
In §18, he summarized the formulas which made him famous through-out Europe (for the first time):
1 +1
22+
1
32+
1
42+
1
52+ · · · =
π2
6,
1 +1
24+
1
34+
1
44+
1
54+ · · · =
π4
90,
1 +1
26+
1
36+
1
46+
1
56+ · · · =
π6
945,
1 +1
28+
1
38+
1
48+
1
58+ · · · =
π8
9450,
1 +1
210+
1
310+
1
410+
1
510+ · · · =
π10
93555,
1 +1
212+
1
312+
1
412+
1
512+ · · · =
691π12
6825 · 93555,
...
§17. With
α =1
3!, β =
1
5!, γ =
1
7!, . . .
and
P =1
π2+
1
4π2+
1
9π2+
1
16π2+ · · · ,
Q the sum of the squares of these terms,R the sum of the cubes, etc.,Euler now obtained
P = α =1
3!=
1
6,
Q = Pα − 2β =1
90,
R = Qα − Pβ + 3γ =1
945,
S = Rα − Qβ + Pγ − 4δ =1
9450,
T = Sα − Rβ + Qγ − Pδ + 5ǫ =1
93555,
...
V = Tα − Sβ + Rγ − Qδ + Pǫ − 6ζ =691
6825 · 93555.
431
In §18, he summarized the formulas which made him famous through-out Europe (for the first time):
1 +1
22+
1
32+
1
42+
1
52+ · · · =
π2
6,
1 +1
24+
1
34+
1
44+
1
54+ · · · =
π4
90,
1 +1
26+
1
36+
1
46+
1
56+ · · · =
π6
945,
1 +1
28+
1
38+
1
48+
1
58+ · · · =
π8
9450,
1 +1
210+
1
310+
1
410+
1
510+ · · · =
π10
93555,
1 +1
212+
1
312+
1
412+
1
512+ · · · =
691π12
6825 · 93555,
...
432 Euler’s first calculation of 1 +1
22k +1
32k +1
42k +1
52k + · · ·
Chapter 45
Euler: Triangulation of convexpolygon
Euler’s problem: To determine the numberan of ways of partitioning aconvex(n + 3)-gon into triangles by joining nonintersecting diagonals.
Denote byP0, P1, P2, . . . , Pn+1, Pn+2 the vertices of a convex(n+3)-gon. There is exactly one triangle containingP0Pn+2 as a side, the thirdvertex being eitherP1, Pn+1 or Pk+2 for somek = 0, 1, 2, . . . , n − 2.
In the last case, the triangle separates the(n + 3)-gon into a(k + 3)-gon (with verticesP0, P1, . . . ,Pk+2) and an(n+k−1)-gon (with verticesPk+2, . . . ,Pn+1, Pn+2). It follows that forn ≥ 2,
an = 2an−1 +
n−2∑
k=0
akan−2−k. (45.1)
Denote byF (x) := a0 +a1x+a2x2 + · · ·+anxn + · · · . Note thata0 = 1
434 Euler: Triangulation of convex polygon
anda1 = 2. Also, in (3), the term∑n−2
k=0 akan−2−k is the coefficient ofxn−2 in the product seriesF (x) · F (x) = F (x)2. It follows that
F (x) − 1 − 2x =∞∑
n=2
anxn
= 2x∞∑
n=2
an−1xn−1 + x2
∞∑
n=2
( ∞∑
k=0
akan−2−k
)
xn−2
= 2x(F (x) − 1) + x2F (x)2.
From this, we have
x2F (x)2 − (1 − 2x)F (x) + 1 = 0.
This is a quadratic equation inF (x) with solution1
F (x) =1 − 2x −
√1 − 4x
2x2.
By the binomial theorem,
F (x) = − 1
2x2
∞∑
n=2
( 12
n
)
(−4x)n
=
∞∑
n=2
1
n
(2n − 2
n − 2
)
xn−2
=
∞∑
n=0
1
n + 2
(2n + 2
n + 1
)
xn.
Consequently,
an =1
n + 2
(2n + 2
n + 1
)
.
an is called then-th Catalan number.
n 0 1 2 3 4 5 6 7 8 9 10an 1 2 5 14 42 132 429 1430 4862 16796 58786
1F (x) would not be a power series if the sign of the square root is positive.
cdi
Supplement 38: Hurwitz’s calculation of logarithms
Let a be a positive number. Define two sequences(bn) and(cn) by
bn =2n(
a1
2n − 1)
,
cn =2n
(
1 − 1
a1
2n
)
.
Theorem 37.1.The sequences(bn) and(cn) converge to the same limitlog a.
Implementation: Givena > 0, let a0 = a andb0 = a − 1. Define
an+1 =√
an,
bn+1 =2bn
an+1 + 1.
Thenlimn→∞ an = 1 andlimn→∞ bn = log a.Calculations oflog 2 andlog 3:
n an bn
1 1.414213562 0.82842712472 1.189207115 0.756828463 1.090507733 0.72406186134 1.044273782 0.70838051885 1.021897149 0.70070875696 1.010889286 0.69691430737 1.005429901 0.69502734248 1.002711275 0.69408641299 1.00135472 0.6936165848
10 1.000677131 0.6933818297
n an bn
1 1.732050808 1.4641016152 1.316074013 1.2642960523 1.14720269 1.1776215244 1.071075483 1.1372077295 1.034927767 1.1176885476 1.017313996 1.1080957647 1.008619847 1.103340458 1.004300676 1.1009729879 1.002148031 1.099791793
10 1.001073439 1.09920183
Exercise
(1) Use Newton’s method to find an approximation of the solution ofx ln x − 1 = 0.
(2) Consider the graph ofy = 1x. Let a > b > 0.
(a) Calculate the equation of the tangent at the point(c, 1c), where
c = a+b2
.(b) Show that the area under the tangent atc, above thex-axis, and
between the vertical linesx = a andx = b, is 2(a−b)a+b
.(c) Make use of (b) to show that
a + b
2>
a − b
log a − log b>
√ab.
cdii Euler: Triangulation of convex polygon
Supplement 40: The function log xx
How doab andba compare, for two given numbersa > b > 1? This isequivalent to comparinga
1a andb
1b . We examine the functionx
1x for its
increasing and decreasing properties. It is more convenient to consider,taking logarithms, the functionF (x) := log x
xfor x > 1.
e ?
?
1 2 3 4 5 6 7 8 9 10
Now, since
F ′(x) =1 − log x
x2,
F ′′(x) =−3 + 2 log x
x3,
F (x) has a maximum atx = e. It is increasing on[1, e] and decreasingon [e,∞]. Therefore, and an inflection atx = e
32 . It follows that
(i) ab > ba if e > a > b, and(ii) ab < ba if a > b ≥ e.
For example,πe < eπ.
Exercise
(1) Prove that ifn is a natural number,(i) n+1
√n < n
√n + 1 for n < 7,
(ii) n+1√
n > n√
n + 1 for n ≥ 7.(2) Show that then-th derivative oflog x
xis
(−1)n+1n!
xn+1
(
log x −(
1 +1
2+ · · · + 1
n
))
.
cdiii
Example 37.1.The only positive integer solution of the equationab =ba, a < b is (a, b) = (2, 4).
Proof. This is equivalent toa√
a = b√
b. Since the functionxx is strictlydecreasing forx > e, with limx→∞ x
1x = 1,
3√
3 > 4√
4 > 5√
5 > 6√
6 > · · · > n√
n > · · · > 1√
1,||2√
2
the only solution is(a, b) = (2, 4). 2
Exercise
(1) For a givena > 0, find the equation of the tangent line from theorigin to the graph ofy = ax.
The slope ofy = ax atx = p is (log a)ap. The equation of the tangentline is y − ap = (log a)ap(x − p). This line passes through the originif and only if −ap = −p(log a)ap, p = 1
log a= loga e. The point of
tangency is(loga e, e) and the slope of the tangent ise log a.
2Sierpinski,Elementary Theory of Numbers, Warsaw, 1964; pp.106–107.
cdiv Euler: Triangulation of convex polygon
Supplement 42A: The odd Bernoulli numbers are zero exceptB1 =−1
2
The Bernoulli numbersBn, n = 0, 1, . . . , are, by definition, the coeffi-cients in the series expansion ofz
ez−1:
z
ez − 1= B0 + B1z +
B2
2!z2 +
B3
3!z3 + · · ·+ Bk
k!zk + · · · .
Clearly,B0 = 1. It is quite easy to check thatB1 = −12
(exercise).One important observation is that all subsequent odd-indexed Bernoullinumbers are zero. This interesting fact follows from the fact that thefunction
g(z) :=z
ez − 1+
z
2
is an odd function,i.e., g(−z) = −g(z). For this, we only need to verifythat z
ez − 1+
z
e−z − 1+ z = 0.
This follows from
1
ez − 1+
1
e−z − 1+1 =
1
ez − 1+
ez
1 − ez+1 =
1 − ez + (ez − 1)
ez − 1= 0.
Now, sinceg(z) is an odd function, we conclude thatB1 = −12
andB2k+1 = 0 for all k ≥ 1. In terms of the even-indexed Bernoulli num-bers, we have
z
ez − 1= 1 − z
2+
∞∑
k=1
B2k
(2k)!z2k.
Here are the beginning even-indexed Bernoulli numbers:
B2 =1
6, B4 =
−1
30, B6 =
1
42, B8 =
−1
30, B10 =
5
66, B12 =
−691
2730.
cdv
Supplement 42B: Jacob Bernoulli’s summation of the powers of nat-ural numbers
Jakob Bernoulli3 arranged the binomial coefficients in the table belowand made use of them to sum the powers of natural numbers.
1 0 0 0 0 0 01 1 0 0 0 0 01 2 1 0 0 0 01 3 3 1 0 0 01 4 6 4 1 0 01 5 10 10 5 1 01 6 15 20 15 6 11 7 21 35 35 21 7
Let the series of natural numbers 1, 2, 3, 4, 5, etc. up ton be given,and let it be required to find their sum, the sum of the squares,cubes,etc. [Bernoulli then gave a simple derivation of the formula
∫
n =1
2n2 + n,
for the sum of the firstn natural numbers. He then continued]. A termin the third column is generally taken to be
(n − 1)(n − 2)
1 · 2 =n2 − 3n + 2
2,
and the sum of all terms (that is, of alln2−3n+22
) is
n(n − 1)(n − 2)
1 · 2 · 3 =n3 − 3n2 + 2n
6,
and ∫1
2n2 =
n3 − 3n2 + 2n
6+
∫3
2n −
∫
1;
but∫
32n = 3
2
∫n = 3
4n2 + 3
4n and
∫1 = n. Substituting, we have
1
2n2 =
n3 − 3n2 + 2n
6=
3n2 + 3n
4− n =
1
6n3 +
1
4n2 +
1
12n,
3Ars Conjectandi, 1713.
cdvi Euler: Triangulation of convex polygon
of which the double∫
n2 (the sum of the squares of alln) 4
=1
3n3 +
1
2n2 +
1
6n.
A term of the fourth column is generally
(n − 1)(n − 2)(n − 3)
1 · 2 · 3 =n3 − 6n2 + 11n − 6
6,
and the sum of all terms is
n(n − 1)(n − 2)(n − 3)
1 · 2 · 3 · 4 =n4 − 6n3 + 11n2 − 6n
24.
it must certainly be that∫
1
6n3 −
∫
n2 +
∫11
6n −
∫
1 =n4 − 6n2 + 11n2 − 6n
24.
Hence,∫
1
6n3 =
n4 − 6n3 + 11n2 − 6n
24+
∫
n2 −∫
11
6n +
∫
1.
And before it was found that . . . When all substitutions are made, thefollowing results:
∫
n3 =1
4n4 +
1
2n3 +
1
24n2.
Thus, we can step by step reach higher and higher powers and withslight effort form the following table:
∫n1 = 1
2n2 + 1
2n,
∫n2 = 1
3n3 + 1
2n2 + 1
6n,
∫n3 = 1
4n4 + 1
2n3 + 1
4n2
∫n4 = 1
5n5 + 1
2n4 + 1
3n3
−130
n∫
n5 = 16n6 + 1
2n5 + 5
12n4
−112
n2
∫n6 = 1
7n7 + 1
2n6 + 1
2n5
− 16n3 + 1
42n
∫n7 = 1
8n8 + 1
2n7 + 7
12n6
−724
n4 + 112
n2
∫n8 = 1
9n9 + 1
2n8 + 2
3n7
−715
n5 + 29n3
−130
n∫
n9 = 110
n10 + 12n9 + 3
4n8
− 710
n6 + 12n4
− 320
n2∫
n10 = 111
n11 + 12n10 + 5
6n9
−n7 +n5−
12n3 + 5
66n.
4More generally,(kk
)+(k+1
k
)+ · · · +
(nk
)=(n+1k+1
). This identity can be established by considering
the number of(k + 1)−element subsets of{1, 2, 3, . . . , n + 1}, noting that the greatest elementm of eachsubset must be one ofk + 1, k + 2, . . . , n + 1, and that there are exactly
(m−1
k
)subsets with greatest
elementm.
cdvii
Whoever will examine the series as to their regularity may beable tocontinue the table. Takingc to be the power of any exponent, the sum ofall nc or∫
nc =1
c + 1nc+1 +
1
2nc +
c
2Anc−1 +
c(c − 1)(c − 2)
2 · 3 · 4 Bnc−3
+c(c − 1)(c − 2)(c − 3)(c − 4)
2 · 3 · 4 · 5 · 6 Cnc−5
+c(c − 1)(c − 2)(c − 3)(c − 4)(c − 5)(c − 6)
2 · 3 · 4 · 5 · 6 · 7 · 8 Dnc−7 + · · · ,
the exponents ofn continually decreasing by 2 untiln or n2 is reached.The capital lettersA, B, C, D denote in order the coefficients of the lastterms in the expressions for
∫n2,∫
n4,∫
n6,∫
n8 etc., namely,
A =1
6, B = − 1
30, C =
1
42, D = − 1
30.
The coefficients are such that each one completes the others in the sameexpression to unity. Thus,D must have the value− 1
30, because
1
9+
1
2+
2
3− 7
15+
2
9− 1
30= 1.
cdviii Euler: Triangulation of convex polygon
Supplement 45: The Bell numbers
What is the sum of the series
Mn :=∞∑
k=0
kn
k!?
M0 = e, M1 = e,
Mn =∞∑
k=0
kn
k!=
∞∑
k=1
kn
k!=
∞∑
k=1
kn−1
(k − 1)!
=∞∑
k=0
(k + 1)n−1
k!=
∞∑
k=0
n−1∑
j=0
(n − 1
j
)kj
k!=
n−1∑
j=0
(n − 1
j
)
Mj .
M2 = 2e, M3 = 5e, . . .What isMn in general?If we write Mn = mne, the beginning terms of(mn) are
1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, . . .
This is sequenceA000110, called the sequence of Bell or exponen-tial numbers.mn is the number of ways of placingn labelled balls inton indistinguishable boxes.
Exercise
(1) Let r > 1 be a fixed real number. The sumsSn :=∑∞
k=0kn
rk satisfy
Sn =1
r − 1
n−1∑
j=0
(n
j
)
Sj,
with initial valueS0 =∑∞
k=01rk = 1
1− 1r
= rr−1
.
(2) Let Pn,k be the number of permutations ofn (distinct) objectstakenk at a time. Show that
n∑
k=0
Pn,k = ⌊n!e⌋.
Chapter 46
Infinite Series with positiveterms
Comparison test
The geometric series
a + ar + ar2 + · · ·+ arn−1 + · · ·is convergent to a
1−rif and only if the common ratior satisfies−1 <
r < 1.Given two series withpositive terms
∞∑
n=1
an = a1 + a2 + · · ·+ an + · · ·∞∑
n=1
bn = b1 + b2 + · · ·+ bn + · · ·
satisfyingan ≤ bn for n ≥ N.
(1) If∑
bn converges, then so does∑
an.(2) If
∑an diverges, then so does
∑bn.
Example 46.1.The harmonic series∞∑
n=1
1
n= 1 +
1
2+
1
3+ · · · + 1
n+ · · ·
is divergent.
502 Infinite Series with positive terms
Example 46.2.The numbere The series
1 +1
1!+
1
2!+
1
3!+ · · ·+ 1
n!+ · · ·
is convergent by comparison with the geometric series
1 +1
2+
1
22+
1
23+ · · ·+ 1
2n+ · · ·
sincen! > 2n whenn ≥ 4. Denote bye the sum of the first series.
e −n∑
k=0
1
k!=
∞∑
k=n+1
1
k!=
1
(n + 1)!
(
1 +∞∑
k=1
1
(n + 2) · · · (n + k + 1)
)
<1
(n + 1)!
(
1 +∞∑
k=1
1
(n + 1)k
)
=1
(n + 1)!· 1
1 − 1n+1
=1
n · n!
Exercise
1. The numbere is irrational.2. Show that for each positive integerk,
∑∞n=1
nk
n!is a positive integer
multiple of e. Specifically,
∞∑
n=1
n
n!= e,
∞∑
n=1
n2
n!= 2e,
∞∑
n=1
n3
n!= 5e.
Example 46.3.. Thep−series
∞∑
n=1
1
np= 1 +
1
2p+
1
3p+ · · · + 1
np+ · · ·
is convergent if and only ifp > 1.
The ratio test
If limn→∞an+1
an= ℓ < 1, then the series converges.
503
Proof. Let r be a number betweenℓ and 1. Eventually, (say, forn > N),all ratios an+1
anare smaller thanr. Compare the series
∞∑
n=N
an
with the geometric seriesaN (1 + r + r2 + · · ·+ rn−N + · · · ):
an =an
an−1
· an−1
an−2
· · · aN+1
aN
· aN < aN · rn−N .
By the comparison test, the series∑∞
n=N an converges, since the latterone does.
Exercise
1. Test for convergence:
∑ n!
3n,
∑ (n!)2
2n2 ,∑ 2nn!
nn,
∑ 3nn!
nn.
2. Give examples to show that the comparison test fails whenℓ = 1.
3. Show that the series
1 +1
22+
1
33+
1
42+
1
53+ · · · + 1
(2n)2+
1
(2n + 1)3+ · · ·
converges but that the ratioan+1
anoscillates between 0 and+∞.
What if the ratio test fails?
Gauss’ test: Suppose
an+1
an= 1 − a
n+ O(
1
n2) as n → ∞.
The series∑
an converges ifa > 1 and diverges ifa ≤ 1.
Example 46.4.Consider the series
1 +∞∑
n=1
β(β + 1) · · · (β + n − 1)
n!,
504 Infinite Series with positive terms
whereβ is a positive constant. This series is clearly divergent ifβ ≥ 1.Assume0 < β < 1. Here,
an+1
an=
β + n
1 + n→ 1 as n → ∞.
Here, the ratio test fails. However,
an+1
an
= (1 +β
n)(1 − 1
n+
1
n2− + · · · ) = 1 − 1 − β
n+ O(
1
n2).
It follows that this series diverges by Gauss’ test. This example showsthat the binomial series for(1 + x)−β does not converge atx = −1.1 This is a special case of the following example due to Gauss. Gaussdiscovered his test in this context.
Example 46.5.The hypergeometric series
1 +
∞∑
n=1
a(a + 1) · · · (a + n − 1)b(b + 1) · · · (b + n − 1)
n!c(c + 1) · · · (c + n − 1)
converges ifc > a + b and diverges ifc ≤ a + b.
Root Test
If limn→∞ n√
an = ℓ < 1, then the series converges.
Exercise
1. Apply the root test to each of the series above.2. Give examples to show that the root test fails whenℓ = 1.
The integral test
Letf(x) be a decreasing, positive function on[N,∞] satisfyinglimx→∞ f(x) =0. Then the series
∑f(n) and the improper integral
∫ ∞
N
f(x)dx
converge or diverge simultaneously.1See§5.5.4 below.
505
Exercise
1. Test for convergence:
∞∑
n=2
1
n · log n,
∞∑
n=2
1
n(log n)2.
2. Show that
2√
n − 2 < 1 +1√2
+1√3
+ · · ·+ 1√n
< 2√
n − 1.
Also,π
2<
1
2√
1+
1
3√
2+
1
4√
3+ · · · <
1
2(π + 1).
3. LetHn := 1 + 12
+ 13
+ · · · + 1n. Test for convergence:
1 +1
2H2+
1
3H3+ · · ·+ 1
nHn+ · · ·
4. Show that the two sequences(xn) and(yn) defined by
xn := Hn − log n, yn := Hn−1 − log n
have a common limitγ. 2
5. If the series of real numbers∑∞
n=1 an converges, does∑∞
n=1 a3n
converge?
2This is the Euler constantγ ≈ 0.5772156649 · · · . It is not known ifγ is a rational number.
506 Infinite Series with positive terms
Chapter 47
The harmonic series
The harmonic series
1 +1
2+
1
3+ · · · + 1
n+ · · ·
is divergent. This means that the sequence(Hn) given by
Hn = 1 +1
2+
1
3+ · · · + 1
n
is divergent.
Exercise
(1) Is the series∑
1Hn
convergent or divergent?
(2) For a positive integerk, find the sum of the series
∞∑
n=1
1
n(n + k).
508 The harmonic series
Solution. ForN > k, theN-th partial sum is
SN :=
N∑
n=1
1
n(n + k)=
1
k
N∑
n=1
(1
n− 1
n + k
)
=1
k
N∑
n=1
1
n− 1
k
N∑
n=1
1
n + k
=1
k
N∑
n=1
1
n− 1
k
N+k∑
n=k+1
1
n
=1
k
k∑
n=1
1
n− 1
k
N+k∑
n=N+1
1
n
>Hk
k− 1
N.
Therefore,Hk
k− 1
N< SN < Hk
k. Since the two ends have the same limit
Hk
kasN → ∞, the series converges to the same limit.
Euler’s constantγ
Theorem 47.1.The sequences(xn) and(yn) defined by
xn = Hn − log(n + 1),
yn = Hn − log n,
have a common limit.
Proof. Clearly,xn < yn. We show that[xn, yn] is a sequence of nestedintervals withlimn→∞(yn − xn) = 0.
(1) The sequence(xn) is increasing:
xn−xn−1 =1
n−log
n + 1
n=
1
n−log
(
1 +1
n
)
=1
n
(
1 − log
(
1 +1
n
)n)
> 0,
since(1 + 1
n
)n< e.
(2) Similarly, the sequence(yn) is decreasing:
yn − yn−1 =1
n− log
n
n − 1=
1
n
(
1 − log
(
1 +1
n − 1
)n)
< 0
since(1 + 1
n−1
)n> e.
509
(3) yn − xn = log(n + 1) − log n = log(1 + 1
n
). It follows that
limn→∞(yn − xn) = 0.Therefore,(xn) and(yn) converge to a common limit.
The common limit is calledEuler’s constant, usually denoted byγ. 1
γ ≈ 0.5772156649015328606 · · · .
Remark.It is not known ifγ is rational or irrational.
1Euler usedC. The use ofγ was initiated by Lorenzo Mascheroni (1750–1800).
510 The harmonic series
Chapter 48
The alternating harmonic series
Theorem 48.1.The alternating harmonic series
1 − 1
2+
1
3− 1
4+ − · · ·+ (−1)n−1
n+ · · ·
converges tolog 2.
It is enough to calculate the limit of the even partial sums.
s2n = 1 − 1
2+
1
3− 1
4+ − · · ·+ 1
2n − 1− 1
2n
= H2n − 2
(1
2+
1
4+ · · · + 1
2n
)
= H2n − Hn.
First proof: Euler’s constant
We writeHn = log n + γn, with limn→∞ γn = γ, Euler’s constant. Withthis,
s2n = H2n − Hn = log(2n) − γ2n − log n + γn = log 2 − (γ2n − γn).
Sincelimn→∞(γ2n − γn) = γ − γ = 0, we have
limn→∞
s2n = log 2 − limn→∞
(γ2n − γn) = log 2.
Therefore the alternating harmonic series converges tolog 2.
512 The alternating harmonic series
Second proof: Riemann integral
Note that
H2n − Hn =1
n + 1+
1
n + 2+ · · ·+ 1
n + n
=1
n
(1
1 + 1n
+1
1 + 2n
+ · · ·+ 1
1 + nn
)
.
This is the (lower) Riemann sum for the functionf(x) = 11+x
on theinterval[0, 1] with n equal subdivisions. Therefore,
limn→∞
(H2n−Hn) =
∫ 1
0
1
1 + xdx = [log(1 + x)]10 = log 2−log 1 = log 2.
Historical notes
The alternating harmonic series is also called the Mercatorseries. Nico-las Mercator (1620–1687) found the series
log(1 + x) = x − x2
2+
x3
3− + · · ·+ (−1)n−1xn
n+ · · ·
by considering the area under the graph ofy = 11+x
from 0 to x.Wallis pointed out that for the series to converge, it is necessary that
x < 1. The Mercator seriesAbel’s limit theoremTo compute the area under the graph ofy = 1
1+x, Mercator did not
directly the series expansion
1
1 + x= 1 − x + x2 − · · ·+ (−1)nxn + · · ·
and integrate term by term. This Leibniz did for his famous series
π
4= 1 − 1
3+
1
5− 1
7+ − · · · + (−1)n−1
2n − 1+ · · ·
Leibniz did not explicitly used a series expansion forarctan x (alsoknown as the Gregory series).1 Rather, by considering the area of a
1For a derivation of the Gregory series, see Euler,Introductio, I §140; also Chapter 41.
513
quadrant of a unit circle, he arrived at
π
4= · · ·
= 1 −∫ 1
0
z2
1 + z2dz
= 1 −∫ 1
0
z2(1 − z2 + z4 − · · · )dz geometric series
= 1 −[z3
3− z5
5+
z7
7− + · · ·
]1
0
integrated term by term
= 1 − 1
3+
1
5− 1
7+ − · · · .
The convergence question forz = 1 was ignored.
514 The alternating harmonic series
Chapter 49
Conditionally convergent series
A series∑
an is absolutely convergent if∑
|an| is convergent. An ab-solutely convergent series is convergent. A series
∑an is conditionally
convergent if∑
an converges but∑
|an| does not.If a series with positive terms is convergent, then it is absolutely con-
vergent.If k ≥ 2 is an integer, the series
∑ (−1)n−1
nk is absolutely convergentsince
∑1
nk is convergent.
Euler,Introductio, I, §170, has computed the sumσk :=∑∞
n=1(−1)n−1
nk
in terms ofζ(k) :=∑∞
n=11
nk :
σk =2k−1 − 1
2k−1· ζ(k). (49.1)
Euler proved this by a rearrangement of the seriesζ(k) =∑
1nk . Note
thatζ(k)
2k=
1
2k+
1
4k+ · · ·+ 1
(2n)k+ · · · .
Subtracting twice of this fromζ(k), we obtain (49.1) above:
ζ(k)− ζ(k)
2k−1=
2k−1 − 1
2k−1·ζ(k) = 1− 1
2k+
1
3k− 1
4k+· · ·+(−1)n−1
nk+· · · .
In particular, fork = 2,
1 − 1
22+
1
32− 1
42+ − · · ·+ (−1)n−1
n2+ · · · =
1
2· π2
6=
π2
12.
516 Conditionally convergent series
Theorem 49.1.If∑∞
n=1 an converges absolutely tos, then for every per-mutationπ of the indices1, 2, . . . , the series
∑∞n=1 aπ(n) also converges
to s.
The alternating harmonic series is conditionally convergent since theharmonic series is divergent.
If we rearrange the alternating harmonic series by choosing, alter-nately and in their natural order,p positive terms followed byq negativeterms, the resulting series converges to
σp,q = log 2 +1
2log
p
q=
1
2log
4p
q.
For example,
1 +1
3+
1
5−
1
2−
1
4+
1
7+
1
9+
1
11−
1
6−
1
8+ + + −− · · · = σ3,2 =
1
2log 6.
Theorem 49.2(Riemann). A conditionally convergent series can be re-arranged to converge to any prescribed real number (including±∞).
Proof. Each of the subseries of positive terms and of negative termsmustbe divergent. Proceed termwise in the positive subseries until the partialsum first exceedsr. Then proceed termwise in the negative subseriesuntil the partial sum first falls belowr. Repeat this alternation of stepsindefinitely, leading to a rearrangement of the series. The difference atthe end of each step betweenr and the partial sum of the rearrangementsubseries at that point is less in magnitude than the magnitude of the lastselected term. Sincexn → 0 asn → 0, the partial sums converge tor.
Exercise
Let∑
xn be a conditionally convergent series of real numbers. For anarbitrary real numberr, show that there is a conditionally convergentsubseries (obtained by deletion of terms without rearrangement) whichconverges tor.
Proceed termwise deleting all negative terms until the sum of the re-tained positive terms first exceedsr. Now, delete positive terms until thesum of all retained terms first falls belowr. Repeat this alternation ofsteps indefinitely. The difference at the end of each step betweenr andthe partial sum of the subseries at that point is less in magnitude than the
517
magnitude of the last retained term. Sincexn → 0 asn → 0, the partialsums converge tor. [?].
Almost alternating harmonic series
Example 49.1.Consider the series
1 − 1
2− 1
3+
1
4+
1
5+
1
6−−−− + + + + + · · · (49.2)
resulting from modifying the signs of the terms of the harmonic series,alternately taking one positive, two negative, three positive, four negativeterms etc. We claim that this series is convergent.1
Let an be the sum of the terms in then-th group containing terms ofthe same sign(−1)n−1: if m = 1
2n(n − 1), then
an =1
m + 1+
1
m + 2+ · · · + 1
m + n.
Note that(i) 0 < an < n
m+n= 2
n+1, andlimn→∞ an = 0,
(ii) the sequence(an) is decreasing:
an − an+1 =
n∑
k=1
(1
m + k− 1
m + n + k
)
− 1
m + 2n + 1
=n∑
k=1
n
(m + k)(m + n + k)− 1
m + 2n + 1
>n∑
k=1
n
(m + n)(m + 2n)− 1
m + 2n + 1
=n2
(m + n)(m + 2n)− 1
m + 2n + 1
=2
(n + 2)(n + 3)> 0.
Therefore, the alternating series test,∑
n→∞(−1)n−1an converges.From this we infer that the series (49.2) converges. Letsm be the
m-th partial sum. If12n(n − 1) ≤ m < 1
2n(n + 1), then
s′n−1 − an < sm < s′n−1 + an,
1Problem E824,American Math. Monthly, , 55 (1948) 365; solution, 56 (1949) 263–265.
518 Conditionally convergent series
wheres′n−1 is the(n−1)-st partial sum of the alternating series∑
(−1)(n−1)an.Sincelimm→∞ an = 0 we conclude that(sm) converges to the same limitof (s′n−1), which is the sum of the sum of
∑
n→∞(−1)n−1an.
Example 49.2.On the other hand, the series
1 − 1
2− 1
3+
1
4+
1
5+
1
6+
1
7−−−−−−−− · · · (49.3)
resulting from modifying the signs of the terms of the harmonic series,alternately taking one positive, two negative, four positive, eight negativeterms etc, is divergent. Again, letbn be the sum of the terms in then-thgroup containing terms of the same sign(−1)n−1. Here,
bn =1
2n−1+
1
2n−1 + 2+ · · ·+ 1
2n − 1> 2n−1 · 1
2n=
1
2.
Since the sequence(bn) does not converge to0, the series∑
bn diverges;so does the series (49.3).
Exercise
Is the series∞∑
n=1
(−1)⌊√
n−1⌋
n= 1−1
2−1
3−1
4+
1
5+
1
6+
1
7+
1
8+
1
9−−−−−−−+ · · ·
convergent or divergent?
di
Supplement : Series with Arbitrary Terms
Alternating series
Let a0 > a1 > a2 > · · · be a sequence of positive numbers descendingto 0. The alternating series
∞∑
n=0
(−1)nan = a0 − a1 + a2 − a3 + − · · ·
converges.
Associativity
The sum of a convergent series (with arbitrary terms)
a1 + a2 + · · ·+ an + · · ·is unaffected by bracketing the terms of the series. Thus, pairing the
terms of the alternating harmonic series
1 − 1
2+
1
3− 1
4+ · · ·+ (−1)n−1
n· · ·
gives(
1 − 1
2
)
+
(1
3− 1
4
)
+ · · ·+(
1
2n − 1− 1
2n
)
+ · · ·
=1
1 · 2 +1
3 · 4 + · · ·+ 1
(2n − 1)(2n)+ · · · .
Both series converge tolog 2. 2
Rearrangement
Rearrange the terms of the alternating harmonic series to form the newseries
1−1
2−1
4+
1
3−1
6−1
8+
1
5− 1
10− 1
12+−−· · ·+ 1
2n + 1− 1
4n + 2− 1
4n + 4+−−· · ·
It is known that this series converges to12log 2, which is different from
the sum of the alternating harmonic series. The invariance of the sum ofa series under rearrangement is a characteristic ofabsolutelyconvergentseries.
2See§5.5.3 below.
dii Conditionally convergent series
Theorem
Let∑
an be anabsolutelyconvergent series,i.e., the series∑
|an| isconvergent. Every rearrangement of the series
∑an converges to the
same sum.
Theorem (Riemann)
Let∑
an be aconditionallyconvergent series, andσ an arbitrary realnumber. There is a rearrangement of the series which converges toσ.There are also rearrangements that are convergent.
diii
Supplement : Power Series
Radius of convergence
Consider a power series
a0 + a1x + a2x2 + · · · + anxn + · · · .
There is a numberr ≥ 0 such that the series(i) converges absolutely in the open interval(−r, r), and(ii) diverges outside the closed interval[−r, r].
Example 45.3.The power series
∞∑
n=0
xn = 1 + x + x2 + · · ·+ xn + · · ·
converges to 11−x
in the open interval(−1, 1).
Calculation of radius of convergence
The radius of convergence is usually determined by one of thefollowingmethods.
(1) If limn // ∞
∣∣∣∣an+1
an
∣∣∣∣= ℓ, then the radius of convergence isr = 1
ℓ.
(2) If limn // ∞n√
|an| = ℓ, then the radius of convergence isr = 1ℓ.
Exercise
Find the interval of convergence of each of the power series
∑
xn,∑ 1
nxn,
∑ 1
n2xn,
∑ (−1)n
nxn,
∑ n!
nnxn.
Theorem 45.3(Cauchy). Given a power series∑
anxn, let
µ = lim n√
|an|.
Then, the power series has radius of convergence1µ.
div Conditionally convergent series
Functions defined by Power Series
Suppose the power series
a0 + a1x + a2x2 + · · · + anxn + · · ·
has radius of convergencer > 0, and defines a functionf(x) on theinterval(−r, r). Then the power series∞∑
n=0
(n+1)an+1xn = a1 +2a2x+ · · ·+nanxn−1 +(n+1)an+1x
n + · · ·
and ∞∑
n=1
an−1
nxn = a0x +
a1
2x2 + · · ·+ an−1
nxn + · · ·
have the same radius of convergence, and defines respectively the deriva-tive f ′(x) and the integral
∫ x
0f(t)dt in the interval(−r, r).
45.0.1 Abel’s Theorem3
Let r be the radius of convergence of a power series defining a functionf(x) on (−r, r). If the series converges atx = r, then it converges tolimx // r− f(x).
Example 45.4.The exponential function The exponential functionex isthe unique solution of the differential equation
dy
dx= y, y(0) = 1.
Write y = a0 +∑∞
n=1an
n!xn. Clearly,a0 = 1 sincey(0) = 1. Differen-
tiation gives
y′ =
∞∑
n=1
an
(n − 1)!xn−1 =
∞∑
n=0
an+1
n!xn.
By comparison, we have
an+1 = an for n = 0, 1, 2, . . . .
It follows thatan = 1 for eachn, and
ex = 1 +∞∑
n=1
1
n!xn.
This has infinite radius of convergence, and is convergent everywhere.3See K. Knopp, Theory and Application of Infinite Series, 1950, Dover reprint, pp.177–178.
dv
Power series expansions of elementary functions
sin x =
∞∑
n=0
(−1)n
(2n + 1)!x2n+1, −∞ < x < ∞;
cos x =
∞∑
n=0
(−1)n
(2n)!x2n, −∞ < x < ∞;
log(1 + x) =
∞∑
n=1
(−1)n−1
nxn, −1 < x ≤ 1.
Since the series forlog(1 + x) converges atx = 1, it converges tolog 2 at this point:
log 2 = 1 − 1
2+
1
3− 1
4+ − · · ·
log1 + x
1 − x= 2
∞∑
n=0
1
2n + 1x2n+1.
This has interval of convergence(−1, 1). With x = 12, we obtain
log 3 = 1 +
∞∑
n=1
1
(2n + 1)22n= 1 +
1
3 · 22+
1
5 · 24+
1
7 · 26+ · · ·
Exercise
1. How would you make use of this series to computelog 2?2. Find the sum of the series
1
1 · 2 +1
3 · 4 +1
5 · 6 + · · · .
The binomial series
If k is a rational number other than a positive integer,
(1 + x)k = 1 +∞∑
n=1
(k
n
)
xn
dvi Conditionally convergent series
with interval of convergence(−1, 1).Here,
(kn
)is the binomial coefficient:
(k
n
)
=k(k − 1) · · · (k − n + 1)
n!.
Exercise
Show that(
1/2
n
)
=(−1)n−1
22n−1n
(2n − 2
n − 1
)
,
(−1/2
n
)
=(−1)n
22n
(2n
n
)
.
Example 45.5.The arctangent series
arctan x = x − 1
3x3 +
1
5x5 − 1
7x7 + − · · ·
In particular,π
4= 1 − 1
3+
1
5− 1
7+ − · · ·
Example 45.6.The arcsine series Sinceddx
(arcsin x) = (1 − x2)−1/2,we integrate the series
(1 − x2)−1/2 = 1 +∑
n=0
(−1/2
n
)
xn
to obtain
arcsin x = x +1
2· x3
3+
1 · 32 · 4 · x5
5+
1 · 3 · 52 · 4 · 6 · x7
7+ · · ·
Exercise
1. Find the sum of the series
1
1 · 2 · 3 +1
2 · 3 · 4 +1
3 · 4 · 5 + · · ·+ 1
n(n + 1)(n + 2)+ · · ·
dvii
and show that
1
1 · 2 · 3+1
3 · 4 · 5+1
5 · 6 · 7+· · ·+ 1
(2n − 1)(2n)(2n + 1)+· · · = log 2−1
2.
2. Show that
1
2(log(1 − x))2 =
1
2x2 +
1
3(1 +
1
2)x3 +
1
4(1 +
1
2+
1
3)x4 + · · ·
Does the series converge atx = −1?3. (Gauss)
1
a− 1
a + b+
1
a + 2b− 1
a + 3b+ · · · =
∫ 1
0
ta−1
1 + tbdt, a, b > 0.
Thus, the series can be evaluated in finite terms ifba
is rational.Deduce that
1 − 1
4+
1
7− 1
10+ − · · · =
1
3(
π√3
+ log 2),
1
2− 1
5+
1
8− 1
11+ − · · · =
1
3(
π√3− log 2).
Theorem (The Binomial Theorem)
If k is a rational number,
(1 + x)k = 1 +∞∑
n=1
(k
n
)
xn
with interval of convergence[−1, 1].Here,
(kn
)is the binomial coefficient:
(k
n
)
=k(k − 1) · · · (k − n + 1)
n!.
