history of mathematics - National STEM Centre

Figure 4.6
48
The Greeks
together with the starting conditions from question 1, and the correct value of c0 , to
write an algorithm for your calculator to calculate and display values of Sn and Tn
until they differ by less than 10~ . You will need to put a pause statement into the
algorithm at each iteration.
5 What would be the modern way of establishing these recurrence relations?
Archimedes actually carried out four iterations starting from the regular hexagon in
Figure 4.4. Thus he reached the approximation 3.1403... < n < 3.1427 ....
The process which Archimedes used to find the value of n is an example of the
process of exhaustion. Archimedes showed that the only number which is greater
than the areas of all the inside regular polygons, and, at the same time, is less than
the areas of all the outside regular polygons is 3.141 59 ... .
A geometric argument shows that the circle has the property that its area, n, is
greater than the areas of all the inside regular polygons, and less than the areas of all
the outside regular polygons. The process of exhaustion is then used to show that
the area of the circle, n, is 3.141 59 .... All other possibilities have been exhausted.
Some of the ideas of modern integration can be traced back to the method which
Archimedes used to calculate his approximations to n.
Archimedes also showed that the area under a symmetrical parabolic arc is -| of the
area of the rectangle which surrounds it, as shown in Figure 4.6.
Archimedes's method, however, is not general. It relies on a geometric property of
the parabola, which it is helpful to prove as a first step.
Activity 4.7 The area under a parabolic arc
Figure 4.7a Figure 4.7b
In Figure 4. la, M is the mid-point of PR, and the length QM is the distance between
the parabola and the chord PR. In Figure 4.7b, V and K are the mid-points of PQ
and QR. The crucial step in Archimedes' s proof is that, for any parabola,
Suppose that the parabola has the equation f(x) = px 2 + qx + r , and that the points
A and B have ^-coordinates a and b respectively. Let b — a = h.