Friday, March 13, 2009

Quadrature of the Parabola

Archimedes was the first person in recorded history to use the summation of an infinite series. He used it to determine the area of a parabola. The purpose of this blog entry is to study a little math history, aided by Clojure.

For simplicity, consider the area enclosed by the line y = 1 and the parabola y = x ^ 2. Archimedes concluded that the bounded area is 4/3 times the area of an inscribed triangle (imagine a triangle from 0,0 to 1,1 to -1,1). In his proof, Archimedes showed that the infinite series 1/1 + 1/4 + 1/16 + 1/64... converged to 4/3. We can express this as the summation from 1 to infinity of the function 1 / (4 ^ (n - 1)).

To express this in Clojure, we first need to define a power function. We could use Math/pow from Java but Greeks didn't have a double. So we will define a simple power function that operates with rationals, with the power argument restricted to being a whole number.

(defn power [i n] (apply * (replicate n i)))

Note that (power 4 0) does the right thing because (apply * nil) returns 1.

Next, we define a function to express one term in the series:

(defn one-term [n] (/ 1 (power 4 (dec n))))

We also need a general summation function:

(defn summation [from to f] (apply + (map f (range from (inc to)))))

Test it with:

(summation 1 10 (fn [x] x)) => 55.

One of the benefits of a lambda calculus based language, such as Clojure, is that we can pass an unapplied function (e.g., one-term) as data in the third argument to summation.

Finally, we define a function called area-parabola:

(defn area-parabola [base height] (* (/ (* base height) 2) (summation 1 100 one-term)))

I use 100 as the upper range because it is good enough and my computer doesn't do infinity well. The answer comes back as a rational number with a very large numerator and denominator due to Clojure repeatedly finding the least common denominator during each iteration. Cast the result as a double in order to see it in decimal notation. Our original problem defined by the line y = 1 and the parabola y = x ^ 2, inscribes a triangle with a base of 2 and a height of 1, (i.e., an area of 1). Therefore, the answer is:

(double (area-parabola 2 1)) => 1.3333333333, which is approximately 4/3.

What was the mechanism that Archimedes used to conclude that the answer was 4/3? He did not divide some large numerator by a large denominator, nor did he add up smaller and smaller decimals. He proved it geometrically. I tried to go through an 1897 translation of his proof but I didn't make it. I cheated and took a look at Proposition 24 at the end where he concludes that the answer can't be greater than 4/3 or less than 4/3.

In 1906, a hidden document authored by Archimedes was found in a monastery in Constantinople. Entitled "The Method", some argue that it describes the essence of what would be called integral calculus, over 1800 years later.

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