Archimedes’ Quadrature of the Parabola

15 Pages·2015·0.19 MB·English

Archimedes’ Quadrature of the Parabola
Notes for Math 1103
Mark Reeder
January 20, 2015.
1 Introduction
Archimedes to Dositheus greeting. 1
When I heard that Conon, who was my
friend in his lifetime, was dead, but that
you were aquainted with Conon and withal
versed in geometry, while I grieved for the
loss not only of a friend but of an ad-
mirable mathematician, I set myself the
task of communicating to you, as I had in-
tended to send to Conon, a certain geomet-
rical theorem which had not been investi-
gated before but has now been investigated
by me, and which I first discovered by means of mechanics and then exhibited by means of ge-
ometry.
Now some of the earlier geometers tried to prove it possible to find a rectilineal area equal to a
given circle and a given segment of a circle ... assuming lemmas not easily conceded 2, so that
it was recognized by most people that the problem was not solved.
But I am not aware that any one of my predecessors has attempted to square the segment bounded
by a straight line and a section of a right angled cone [a parabola], of which problem I have now
discovered the solution.
For it is here shown that every segment bounded by a straight line and a section of a right angled
cone [a parabola] is four-thirds of the triangle which has the same base and equal height of the
segment, and for the demonstration of this property the following lemma is assumed: that the
excess by which the greater of (two) unequal areas exceeds the less can, by being added to itself,
be made to exceed any given finite area.
1From the standard translation by T.L. Heath. Paragraph breaks and footnote added by me.
2A lemma is a preliminary result which in those days was often taken for granted. Archimedes is politely
reminding Dositheus that these lemmas are false.
1
The earlier geometers have also used this lemma; for it is by use of this same lemma that they
have shown that circles are to one another as the duplicate ratio of their diameters, and that
spheres are to one another in the triplicate ratio of their diameters ...
Archimedes says that if you take any segment of a parabola cut off by a line AB, and if C is
the point on the segment furthest from AB, then the area of the parabolic segment ABC is
four-thirds the area of the triangle ABC.
B
A
C
Why did Archimedes want to prove this? He observes that his predecessors were unable to find
the area of a circle (or equivalently, an ellipse). This was the most important unsolved problem
at that time (indeed, for the next 1500 years) so it may have seemed natural to Archimedes to
try another conic section. And perhaps he chose the parabola because it is a very special conic
section, lying on the border between ellipses and hyperbolas.
How did Archimedes figure out the ratio four-thirds, or even that the ratio is the same for
all parabolic segments? According to his preface, Archimedes used “mechanics”. This was a
method which he devised for comparing the area of two geometric figures by regarding area as
weight, and placing the figures on opposite sides of an imaginary see-saw. He frequently used
this method to make guesses which he then rigorously proved using geometry.
2 The idea of Archimedes’ proof
The point C furthest from AB is called the vertex of the parabolic segment. We will call the
triangle ABC the vertex triangle. Archimedes proves that the area of any parabolic segment
is four-thirds the area of its vertex triangle.
2
His main insight is that when we remove the vertex triangle ABC from the parabolic segment
we are left with two smaller parabolic segments which themselves have vertex triangles ACE
and BCD. Removing this layer of two triangles we get four even smaller parabolic segments,
whose four vertex triangles AFE,EGC,CHD,DIB form the next layer, and so on.
B
I
D
A
H
F
C
E
G
The first three layers of vertex triangles.
The idea of filling out a curved figure such as a parabolic segment with rectilineal figures such
as triangles is called the “Method of Exhaustion”. It was discovered prior to Archimedes, and
is the basis of what we now call Integral Calculus, but was in the old days called Quadrature.
3 Eudoxus and the Method of Exhaustion
If we had to name a single inventor of Integral Calculus, it would have to be Eudoxus of Cnidus
(approx. 390-340 BC). All manuscripts directly attributed to Eudoxus seem to have been lost,
3
but we know from commentaries by others that his work is preserved in Euclid’s Elements,
especially in books V and X. One of his results is the lemma quoted above in Archimedes’
preface, which we can rephrase more succinctly as follows:
Given a magnitude M and a smaller magnitude E, then some multiple of E exceeds M.
This is used in the Elements prove Proposition 1 of Book X:
Proposition X.1
Given a magnitude M and a smaller magnitude E, if we remove at least half of M, and then
remove at least half of the remainder, and from this remainder remove at least half, leaving a
smaller remainder, and so on, then eventually E will exceed the remainder.
That is, from M we remove M ≥ 1M, leaving remainder R = M−M , and from R we remove
1 2 1 1 1
M ≥ 1R , leaving remainder R = R − M , and from R we remove M ≥ 1R , and so on.
2 2 1 2 1 1 2 3 2 2
Prop. X.1 says that eventually R < E.
n
This is more or less how Prop. X.1 is stated in the Elements. Today we would rephrase Prop.
X.1 as follows.
Proposition X.1, version 2
If M ,M ,M ,... are obtained as above then for any number E > 0, we shall have
1 2 3
M −(M +M +···+M ) < E for large enough n.
1 2 n
In modern notation we write this as
lim(M +M +···+M ) = M. (1)
1 2 n
n→∞
Despite the presence of “infinity” 3 there is nothing mystical here. Equation (1) means precisely
what is stated in Prop. X.1 version 2.
The letter E stands for “error”. Think of the error as an requirement for accuracy that you
don’t know in advance, so you must be able to respond to any requirement for accuracy by
3Theinfinitysymbol∞wasfirstusedin1655byJohnWallis, apredecessorofNewton, fromwhomwe’llhear
much more later.
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taking enough terms. A smaller E will require more terms. Thus (1) means that the finite sums
M +M +···+M give an arbitrarily good approximation to M.
1 2 n
This is one of the key ideas in our course. Typically M ,M ,... will be simple quantities such as
1 2
integers, orareasoftrianglesandM willbeadifficultquantitysuchasπ ortheareaofaparabolic
segment. If we can find simple quantities M , M ,..., each more than half of the remainder,
1 2
then by Prop. X.1 we will know M in the sense that we have arbitrarily good approximations
to M.
Sometimes we know M in advance, and it is the approximation itself that is useful.
Example 1: Suppose M is a line segment of length one. Remove exactly half, leaving a
remainder of one-half. Remove half of this remainder, leaving one-fourth. Remove half of this
new remainder, leaving one eighth, and so on. So M = 1/2, M = 1/4, M = 1/8 and
1 2 3
M = 1/2n, so we have
n
1 1 1
M +M +···+M = + +···+ ,
1 2 n 2 4 2n
and the remainder at the nth stage is
1
R = M −(M +M +···+M ) = .
n 1 2 n 2n
Prop. X.1 guarantees that for any E > 0 we will have R < E if n is large enough. Indeed this
n
will be the case if 2n > 1/E. For example if E = 0.1 then n ≥ 4 works. If E = .01 then n ≥ 7
works, and so on. Thus Prop. X.1 is confirmed in this example. In our modern notation we
express this as
(cid:18) (cid:19)
1 1 1 1
lim + + +···+ = 1.
n→∞ 2 4 8 2n
Example 2: Suppose M is a square of side-length one. Remove all but the lower right quarter-
square. From this remainder remove all but its lower-right quarter-square, and so on. At each
stage we’re removing three-fourths of a square, leaving a quarter-square, so M = 3, M =
1 4 2
3 · 1 = 3 , M = 3 · 1 = 3 and M = 3 , so we have
4 4 42 3 4 42 43 n 4n
3 3 3
M +M +···+M = + +···+ ,
1 2 n 4 42 4n
and the remainder at the nth stage is
1
R = M −(M +M +···+M ) = .
n 1 2 n 4n
Prop. X.1 guarantees that for any E > 0 we’ll have R < E if n is large enough. Indeed this
n
will be the case if 4n > 1/E. For example if E = 0.1 then n ≥ 2 works. If E = .01 then n ≥ 4
5
works, and so on. Thus Prop. X.1 is confirmed again. In our modern notation we express this
as
(cid:18) (cid:19)
3 3 3 3
lim + + +··· = 1.
n→∞ 4 42 43 4n
If we divide everything by 3, we get
(cid:18) (cid:19)
1 1 1 1 1
lim + + +··· = .
n→∞ 4 42 43 4n 3
This example 2 is used by Archimedes, as one of two applications that he makes of Prop. X.1.
4 Properties of the Parabola
We now begin Archimedes’ proof of his area formula for the parabola. He starts with some
properties of the parabola that were well-known in ancient times, but may be unfamilar to us.
Tangent Property: The tangent line to the segment at the vertex is parallel to the secant.
That is, PQ is parallel to AB.
B
Q
A
C
P
Bisecting Property: The line through C parallel to the axis bisects the secant.
That is, M is the midpoint of AB.
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axis B
M
A
C
Equation of the Parabola: If Y on CM and X on the parabola are such that XY is parallel
to the secant AB then CY is to CM as the square on XY is to the square on AM.
That is,
CY XY2
= .
CM AM2
B
M
A
Y
C
X
These properties were known to Archimedes from Euclid’s Conics, which is work of Euclid that
has been completely lost. They are also in Apollonius’ Conics, which has survived and is the
main existing reference for the ancient Greek approach to conic sections. The ancient proofs of
the above properties are purely geometrical and quite involved. We can prove them more easily
using coordinates and algebra, as follows.
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Let us put cartesian coordinates s,t on the plane so that the axis of the parabola is the s-axis
which intersects the parabola at (0,0), and the t-axis is perpendicular to the s-axis. Then the
equation of the parabola is s = kt2 for some constant k.
s B
M
A
Y
X
C
t
Let A = (a,ka2), B = (b,kb2), so that AB has slope
kb2 −ka2
m = = k(a+b).
b−a
Then the equation of AB is s = m(t−a)+ka2 = mt−kab.
Fix any point C = (c,kc2) on the parabola with a ≤ c ≤ b, and let f(t) be the distance from C
to a point (t,mt−kab) on AB varying with t. Minimizing f(t) for a ≤ t ≤ b, we find that the
distance from C to AB is given by the function
(mc−kc2 −kab)2
g(c) = .
1+m2
The function g(c) (of the variable c) attains its maximum when mc − kc2 − kab does, which
means
m a+b
c = = .
2k 2
Thus the vertex of the parabolic segment is C = (c,kc2), where c = m/2k. Now we can prove
the three properties of the parabola that we need.
The tangent line to the parabola s = kt2 at C has slope 2kc = m, which is also the slope of AB,
hence the tangent property holds.
The line through C parallel to the s-axis is the line t = c, which meets AB at the point
(cid:0) (cid:1)
M = (c,mc−kab) = 1(a+b), 1(ka2 +kb2)
2 2
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whose coordinates are the averages of those of A = (a,ka2) and B = (b,kb2). Hence M is the
midpoint of AB, and the bisecting property holds.
For the equation, we first check that
k(b−a)2 (b−a)2
CM = and AM2 = ·(1+m2).
4 4
Next, Y = (c,y) for some number y, so that
CY = y −kc2.
Finally, X = (x,kx2) for some number x. Since XY has slope m and passes through (x,kx2), it
follows that y and x satisfy the equation
y −kx2 = m(c−x),
which can also be written as
y −kc2 = k(c−x)2.
Now we have
XY2 = (c−x)2 +(y −kx2) = (c−x)2 +m2(c−x)2 = (c−x)2(1+m2).
It now follows that
XY2 4(c−x)2 4(y −kc2) CY
= = = ,
AM2 (b−a)2 k(b−a)2 CM
as claimed.
5 Exhausting the Parabola
Remove the vertex triangle ABC, so M = area(ABC). From the remaining two segments
1
remove their vertex triangles AEC and BCD, so that M = area(AEC) + area(BCD). From
2
the remaining four segments remove their vertex triangles AFE, EGC, CHD, DIB, whose
sum is M , and so on. In order to invoke Prop. X.1 we must show that the area of a vertex
3
triangle is at least one-half of its parabolic segment. Since our original parabolic segment is
completely general, it suffices to consider the vertex triangle ABC.
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B
M
Q
A
C
P
Draw parallels to the axis through A,B and C, and let PQ be the tangent line to the parabola
through C. By the tangent property, ABQP is a parallelogram. Since a diagonal divides the
area of a parallelogram in half, it follows that the vertex triangle ABC is exactly half of ABQP.
But ABQP contains the parabolic segment, so ABC is at least half of the parabolic segment,
as claimed.
It now follows from Prop. X.1 that the area S of the segment is given (in our modern notation)
by the limit
S = lim(M +M +···+M ),
1 2 n
n→∞
where M is the area of the nth layer of vertex triangles.
n
In the next section we will show that each layer is one-fourth of the layer immediately above it.
That is, M = 1M , M = 1M = 1 M , and so on. Then by the result in example 2 of section
2 4 1 3 4 2 42 1
3, we have
S = lim(M +M +M +···+M )
1 2 3 n
n→∞
(cid:18) (cid:19)
1 1 1
= lim M + M + M +···+ M
n→∞ 1 4 1 42 1 4n 1
(cid:18) (cid:19)
1 1 1
= M +M · lim + +···+
1 1 n→∞ 4 42 4n
1
= M +M ·
1 1
3
4
= ·M .
1
3
10