# Talk:The Quadrature of the Parabola

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## Moved from the end of the intro

In 1906 Heiberg suggested that Archimedes' proof was written as

4A/3 = A + A/4 + A/12[citation needed]

Archimedes' proof was also written as
1, 5/8, 14/27, 30/64, ....., Pn/n3, ..... tends at number 1/3, as n tends to infinity
where the numerator of the sequence terms is the nth square pyramidal number Pn.
see "Quadrature of the parabola with the square pyramidal number" in this talk page.--Ancora Luciano (talk) 18:50, 14 May 2013 (UTC)

## Proof by abstract mechanics

diagram for the mechanical proof

As Archimedes gave two proofs, can we have a section for the other one too, the proof by abstract mechanics? I'd do it, but don't really have the skills for a mathematical article.--Annielogue (talk) 15:31, 24 November 2012 (UTC)

## Quadrature of the parabola with the "square pyramidal number" (new proof)

This proof (possibly unpublished) of the Archimedes' theorem: "Quadrature of the parabolic segment" is obtained numerically, without the aid of Mathematical Analysis. Below we show a summary of the proof. The entire article is at the following web address:

Proposition: The area of ​​ parabolic segment is a third of the triangle ABC.

Divide AB and BC into 6 equal parts and use the green triangle as measurement unit of the areas.

The triangle ABC contains:

(1+3+5+7+9+11).6 = 62.6 = 63 green triangles.

The parabola circumscribed figure (in red) contains:

A(cir.) = 6.1 + 5.3 + 4.5 + 3.7 + 2.9 + 1.11 = 91 green triangles. (3)

The sum (3) can be written:

A(cir.) = 6 + 11 + 15 + 18 + 20 + 21 , that is:
6+
6+5+
6+5+4+
6+5+4+3+
6+5+4+3+2
6+5+4+3+2+1

or rather:

A(cir.) = sum of the squares of first 6 natural numbers !

Generally, for any number n of divisions of AB and BC, it is:

1. The triangle ABC contains n3 green triangles
2. An(cir.) = sum of the squares of first n natural numbers

So, the saw-tooth figure that circumscribes the parabolic segment can be expressed with the "square pyramidal number" of number theory! For the principle of mathematical induction, this circumstance (which was well hidden in (3)) we can reduce the proof to the simple check of the following statement:

the sequence of the areas ratio: 1, 5/8, 14/27, 30/64, ....., Pn/n3, ..... tends at number 1/3, as n tends to infinity (4a)

where the numerator of the sequence terms is the nth square pyramidal number Pn.

But (4a) state that: the area (measured in green triangles) of the circumscribed figure is one-third the area of ​​the triangle ABC, at the limit of n = infinity. End of proof

This proof is very beautiful! Notice its three essential steps:

1. Choice of equivalent triangles for measuring areas.
2. With this choice, the area of ​​triangle ABC measure n3 triangles.
3. Counting the number of triangles in the saw-tooth figure that encloses the parabolic segment and discovery that, for each number n of divisions, this number is the square pyramidal number !

The rest came by itself.--Ancora Luciano (talk) 18:49, 14 May 2013 (UTC)