1 Archimedes: the First Mathematical Physicist

Home , Central angle, Measurement of a Circle

1 Archimedes: The ﬁrst mathematical physicist

1.1 Quadrature of a parabola Problem Compute the area of a section of a parabola, delimited by a parabola and an arbitrary secant. Tools and Methods Some results from Euclid and method of triangulation, along with geometric progression. Notation

_ AB is the base and P , the vertex (the point farthest from AB of the parabolic section AP B .

M is the midpoint of AB and M1 is the midpoint of AM. Deﬁne P2 and M2 in a similar fashion. Step 1: 1 1 Now, a(4AP M1) = 2 a(4AP M) (easy) and a(4AP1P ) = 2 a(4AP M1) (based on Euclid’s results). 1 Thus, a(4AP1P ) + a(4PP2B) = 4 a(4AP B). Write α = a(4AP B). 1 If P1 is the polygon AP1PP2B, then a(P1) = α + 4 α. Step 2:

Repeat the argument with the parabolic section AP1P on base AP , as with PP2B on base PB. 1 1 If P2 is the resulting nine-sided polygon, then a(P2) = α + 4 α + 42 α. Pn α In general, if Pn is the polygon resulting after n steps,, then a(Pn) = 0 4k .

Pn 1 1 1 4 Identity: 0 4k + 3 . 4n = 3 . Observe that: n−1 n n−1 rn 1 if r < 1, then (1 + r + ... + r )(1 − r) = 1 − r and hence (1 + r + ... + r ) + 1−r = 1−r .

1 1 _ Final Step: Given ε > 0, choose n such that 3 . 4n < ε and thus area of the parabolic segment AP B 4 4 _ is at least [ 3 −ε]α and hence at least 3 α. Intuitively, area of the region AP B \Pn is inﬁnitesimally small (too small to measure). Thus, _ 4 a( AP B ) = a(4AP B) 3

1 1.2 Measurement of a Circle Suppose A denotes area of a circle of radius r and C its circumference. It was known that. 2 A = π1r and C = 2π2r,implying that the length of an arc with central angle θ is rθ. Archimedes showed that 1 A = rC ⇒ π = π (= π) 2 1 2 Basic Geometry Suppose Pn and Qn are regular n-gons, inscribed in and circumscribing the circle. Each of Pn and Qn consists of n congruent isosceles triangles with central angle 2π/n. Suppose base and height of Pn are denoted by sn and rn and perimeter by pn and base, height and perimeter of Qn by 2tn, r and qn. π π π π Clearly, sn = 2 sin( n ), tn = tan( n ) and pn = nsn = 2n sin( n ) and qn = 2ntn = 2n tan( n ).

Lemma 1.2.1. : Clearly, pn < C < qn Proof. Recall that if θ ∈ (0, π/2), then sinθ < θ < tanθ

Corollary 1.2.2. Thus, 1 1 (i)a(P ) < rC (ii)a(Q ) > rC n 2 n 2 Lemma 1.2.3. Given ε > 0, one can inscribe and circumscribe regular n-gons Pn and Qn such that A − a(Pn) < ε and a(Qn) − A < ε. Proof is deferred while we show how this proves Archimedes’s result. 1 Corollary 1.2.4. A = 2 rC Proof. To prove his result, Archimedes uses an argument, known as reductio ad absurdum. 1 1 Case 1: Assume A > 2 rC and let A − 2 rC = ε. By Lemma 1.2.3, we choose n so that A − a(Pn) < ε. It follows that

a(Pn) ≥ A − ε 1 = rC 2 1 1 Case 2: Assume A < 2 rC and let 2 rC − A = ε.

a(Qn) ≤ A + ε 1 = rC 2 In each case, we have a contradiction. By trichotomy, 1 A = rC 2

2 This refers to Lemma 1.2.2. Start with a square P0 and extend to an octagon

Observe that a(4AGH) < a(4T AG) since

P1. Write E0 = A−a(P0) and E1 = A−a(P1). _ Now E0 = 4a( EKF ). T G > GH(∠T AG is a right angle). E (= area between the square and the circle) = 0 _ 4a( T EAH ). E0 − E1 = a(P1) − a(P0) If E1 = area between the square and the octagon, = 4a(4EFK) then E0 − E1 = 4a(4TFG). Now, = 2a( EFF 0E0) _ _ _ a T AH = a(4T AG) + a AGH > 2a EKF 1 < a(4T AG) + a(4AGH) > E 2 0 < 2a(4T AG) Thus E > 2E and continuing the pattern, we 0 1 i.e. E0/4 < (E0 − E1)/2. Thus, inscribe Pn, write En = A − a(Pn). Now, E 1 E < 0 E < E 1 2 n+1 2n 0 As before we conclude that if E = a(Q ) − A, One is not permitted to take limits, so we rely on n n then the principle of Eudoxus, which we illustrate. 1 E < E Suppose ε > 0 is given. Choose N such that n+1 2n 0 (N + 1)ε > E0. We illustrate the process. By the same argument as before, given ε, we Suppose 5ε > E0. It follows that 3ε > E1 and may choose n such that a(Qn) − A < ε. 2ε > E2, resulting in ε > E3. In general, N steps sufﬁce and we get, ε > En. In other words, A − a(Pn) < ε.

3 2 Geometric Series and Extensions

Oresme (1350) found the following formula for an integer k > 1: a a 1 a 1 a 1 + (1 − ) + (1 − )2 + ... + (1 − )n + ... = a k k k k k k k It is easy to see that this encompasses Archimedes’s use of: 1 1 4 1 + + ( )2 + ... = 4 4 3

Swineshead posed and solved a problem which when posed as describing motion of a particle reads as follows: If a point moves throughout the first half of a certain time interval with a constant velocity, throughout the next quarter of the interval at double the initial velocity, throughout the following eighth at triple the initial velocity, and so on ad infinitum; then the average velocity during the whole time interval will be double the initial velocity. Taking both the time interval and the initial velocity as unity, this is equivalent to the following: 1 2 3 n + + + ... + + ... = 2 2 4 8 2n Swineshead gave a long and tedious proof verbal proof. Again, Oresme came to the rescue and gave a geometric method of summing the series. In his ”Treatise on Configurations”, Orseme used n two dissections of the graph of the graph of the Swineshead motion. It is clear that a(An) = 2n

1 and a(Bn) = n . 2 P P P Exercise: Show that a(An) = a(Bn) to conclude that a(An) = 2.

Historical Note: Nicole Orseme taught at University of Paris and in a commentary he wrote on Euclid’s ”Elements”, he made a more detailed analysis of motion.