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The Method of Exhaustion vs. Calculus

February 7, 2019

In Lecture 7 of the excellent DVD course Great Thinkers, Great Theorems (The Great Courses No. 1471) Professor Dunham of Muhlenberg College discusses Archimedes’ Measurement of a Circle. Archimedes’ first Proposition in effect gives us the formula for the area of a circle ─ though this is not quite how Archimedes puts it. He writes
“The area of any circle is equal to [that of] a right-angled triangin which one of the sides about the right angle is equal to the radius and the other to the circumference of the circle”
        This is ancient Greece, more particularly the ancient Greek ex-colony of Syracuse, and Archimedes had no algebra or modern notation ─ these innovations had to wait for Descartes to be born a millennium and a half later. Historians of mathematics never tire of telling us that the ‘Method of Exhaustion’, which Archimedes inherited from Eudoxus, is an unwieldy and clumsy tool compared to Calculus which up to a point it is. Nonetheless, Archimedes’ approach has several  advantages over the contemporary textbook one. For a start, Archimedes’ proof makes sense, is intuitively acceptable and thus far more accessible to the intelligent layman, even to the intelligent child ─ the older Victorian style textbooks invariably used simplified ‘exhaustion’ arguments and that’s how I learned my basic geometry. In comparison, the algebraic Calculus proof, though shorter, is tiresome in the extreme and doesn’t seem to have much to do with real circles and areas  ─ it makes one think of an immensely capable stage magician performing a series of  tricks.

To Archimedes then. First off, Archimedes does not turn the area of a circle into an abstract algebraic function: he compares the circle to something that we can see and draw, namely a right angled triangle with known area ½ base × height where the height is the radius of a circle (any circle) and the base is the same circle’s circumference straightened out. Area =  r × C) = ½ r C.
[Apologies but currently I have problems putting diagrams on this website, see later versions of this post.]
The proof is by double contradiction (as Professor Dunham puts it) and though these kind of proofs are often rather artificial, in this case the reasoning is both impeccable and even entertaining. Archimedes says in effect, “Let us suppose my claim is wrong, and that the Area of a circle is not equal to that of my triangle”. It might, for example, be less than the area of a circle.       We now ‘exhaust’ the circle in the inimitable Greek manner by fitting regular polygons (many sided figures with all sides and matching angles equal) inside the circle. We can start with an equilateral triangle which fits exactly into the circle (how to do this is demonstrated in Euclid Book I), we then double it to produce a hexagon (which, amazingly has six outer sides equal to the radius), double that to make a 12-sider, 24-sider, 48-sider &c. &c. All these doublings can actually be performed by following instructions given in Euclid Book I using only straight edge and compass.
Clearly ─ because we can actually see it happening ─ the gap between the total area of the inscribed polygon and the area of the circle gets less each time. Next, very reasonably, Archimedes asks us to accept that, in principle at least, by constructing polygons with an ever growing number of sides we can make the difference between the area of the polygon and the area of the circle as small as we please. (Difficulties with computer graphics  stop me giving too many diagrams at this point, see updated versions of this post.)
Now the polygon is made up of identical triangular sections where the ‘base’ is the outer side of each section and the ‘height’ is the length of the perpendicular from the centre to this base, technically known as the apothem. Note that the two other sides of the triangular section are radii and, important point, the radius is always greater than the apothem since the apothem is the shortest route to the base. Now, every triangular section of the (regular) polygon is the same,  so, if the polygon consists of 6 triangles, the total area will be 6 × ( ½ base × height) or, to put it another way, 6 bases × ½ h . The length ‘6 bases’ is the length of the perimeter of the polygon, the distance you would have to go if you walked all round it. We now increase the number of sides to 12, 24, 48, 96…. and so on. At each stage, what we end up  computing to give the area of the polygon is the perimeter of the inscribed polygon multiplied by half the height, or ½h × P where P is the  Perimeter.  [Apologies but currently I have problems putting diagrams on this website, see later versions of this post.]
What about the original right angled triangle? The area of Archimedes’ right angled triangle is½ base × height = ½ C × r   ─ where C is the circumference of the circle and r is the radius. Archimedes has suggested, for sake of argument, that the area of the right angled triangle is less than the area of the circle. Now, since the area of the inscribed polygon can be made as close to the area of the circle as we wish, it should be a better fit than the area of the right angled triangle. So, according to this supposition, we have the inequality
                   Area Triangle <  Area n-gon for large n < Area circle
        However, this can’t be right because the height of the triangle is r, the radius of the circle, and the height of any triangular section of any n-gon is the apothem which is always less than the radius. Also, the circumference of the circle, the base of the triangle, is clearly greater than the perimeter of any n-gon ─ because the length of a curve drawn between two points is always greater than that of a straight line.
We thus have a contradiction so we must reject the hypothesis that the area of the triangle < area of the circle.
Suppose, then, the area of the triangle is greater than the area of the circle. This time Archimedes’ constructs a series of regular polygons that fit neatly around the circle, i.e. the bases of each triangular section lie outside, not inside the circle. The number of sides is again increased without limit, making the difference between the area of the circumscribed polygon and the area of the circle as small as we wish. This time we find that the height of each triangular section is indeed the length of the radius but the perimeter is undeniably greater than the circumference of the circle. So this polygonal total area cannot be greater than the area of the right-angled triangle because the base of the triangle is equal to the circumference. Again a contradiction and the second hypothesis must be rejected.
Archimedes now delivers the coup de grace. Since the area of the right angled triangle is not smaller than the area of the circle and is not greater than the area of the circle, the only possibility left is that it is exactly equal to the area of the circle. But this is what he wanted to prove.
But what is the area of the right angled triangle in modern terms?
It is ½ radius × Circumference which in our terms (not his) comes to ½ r (2pr) = pr2 (since the ½ and 2 cancel). Archimedes does not write π which though a Greek letter was not given its modern mathematical meaning until the 18th century of our era (by Euler). But neither does he think π. For the Greeks curved and straight lines (along with certain pairs of straight lines) were incommensurable ─ could not strictly be compared because they lacked a basic common ‘measure’. Archimedes restricts himself to saying that the ratio of the diameter of a circle to the circumference (our p) lies between two limits: it was less than (3 + 1/7) and greater than (3 + 10/71). This is not the same thing as equating the circular ratio of two lengths to the irrational number we know today as π ─ since the Greeks did not use irrational numbers which for them were not true numbers. In effect, modern mathematics does (or at any rate appears to do) what the later Greeks knew to be impossible ─ it ‘squares the circle’, a figure that cannot be squared.
The only point in Archimedes’ excellent piece of reasoning that is somewhat questionable is the following. By making the circumference the base of his triangle, Archimedes is in effect straightening it out, something that is in practice impossible to do without falsifying its true length, if only by a little. Even when using a piece of string, the molecules are inevitably going to be more extended when measuring the circumference and more compressed when measuring a straight line. This incidentally is very much the sort of objection that already in ancient Greek times certain sophists and epicureans made against the new-fangled discipline of higher geometry so enthusiastically promoted by the rival school of Plato and his followers.

 The Calculus derivation
The modern derivation of the formula for the area of a circle goes like this. Armed with a co-ordinate system (which Archimedes did not possess) we draw a circle centred at the origin and assess the ‘area under the curve’ ─ or rather the area of the first quadrant which is all we need (because we eventually multiply it by 4). By Pythagoras, r2 = x2 + y2 where r is the radius of the circle, or  y = √r2 – x2. We want the inner area between the right side of the x axis and the top part of the y axis, i.e. the limits of the integral are x = 0, x = r. To get out this definite integral you have to reverse the standard derivative of arcsin or sin−1, a formula that is impossible to remember. The integral turns out to be
[ ½ x √r2 – x2  + ½ r2 sin−1 x/r] taken between the limits x = r and x = 0.
This expression is guaranteed to make the non-mathematical reader give up in disgust and go to the pub instead, and, although I have on occasion taught Calculus to school age kids, I dislike the inverse trigonometrical functions and had to look up the answer in a book and differentiate the result to see if I got back to √r2 – x2. More to the point, this  hodgepodge seems to have nothing at all to do with circles or areas of anything at all though, if you do feed into the expression x = 0 and take this away from what you get with x = r you (eventually) end up with ¼ πr2  which, since this is a quadrant, makes the total area four times this or πr2.
But how do I know this Calculus result is right? Only by accepting various assumptions which underpin Calculus, assumptions which were, in the days of ‘infinitesimals’, extremely dubious as Bishop Berkeley pointed out at the time and have only been made rigorous by disconnecting  Calculus completely from physical reality. The Greeks (correctly) argued that curved lines and straight lines are different species and that areas bounded by curves can be approximated to any desired degree of precision but in general never the twain shall meet. The fact is that the circle cannot be squared and Calculus is being extremely helpful but disingenuous when it gives the area as exactly πr2. Like all irrationals, the ‘number’ π is not a real quantity but a numerical limit that real quantities can approach but never attain. There is no such thing as an object whose length is π, and you cannot perform any action a π number of times.
The ancient Method of Exhaustion rests on very different assumptions about the physical world than Calculus does and on much more sensible ones. Nevertheless, we needed Calculus in the bad old days because the labour of attaining a desired level of precision was horrendous (even after the invention of logarithms which reduced it considerably). Calculus, in its trendy post-Weierstrass version now obligatory in colleges of higher education, is admittedly an ingenious piece of pure mathematics which, by fancy footwork, effectively sidesteps issues about ‘infinitesimals’. But it is a piece of logic rather than mathematics. And anyway the writing is on the wall already for Calculus. Now that we have supercomputers, the trend is not to bother looking for analytic solutions but simply to slog it out numerically. Numerical Analysis, the art of progressive approximation, is rapidly becoming more important than Functional Analysis. In any case, the vast majority of differential equations (equations dealing with change and motion) are insoluble analytically so there is all too often no alternative to approximate methods. Practically speaking, we no longer need all this wretched philosophical lumber of ‘infinitesimals’ and infinite processes: the eminently sensible ancient Greek Method of Exhaustion, which successfully avoided actual infinity altogether, is, against all the odds, making a triumphant come-back . Traditional Calculus is a mighty cultural achievement and a necessary bridge to the present and what lies ahead,  but one day it will be probably put in the same class as those impressive looking species that eventually went extinct because they were unable to cope with real-life conditions. SH

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