Talk:Quadrature of the Parabola

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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, ....., P_n/n³, ..... tends at number 1/3, as n tends to infinity

where the numerator of the sequence terms is the n^th 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)[reply]

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)[reply]

This proof (possibly unpublished) of the Archimedes' theorem: "Quadrature of the parabolic segment" is obtained numerically, without the aid of Mathematical Analysis.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:

https://sites.google.com/site/leggendoarchimede

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 = 6^2.6 = 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. A_n(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, ....., P_n/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 P_n.

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)[reply]

In this section, there are some diagrams, and this statement:

"Archimedes proves that the area of each green triangle is one eighth of the area of the blue triangle. From a modern point of view, this is because the green triangle has half the width and a fourth of the height:[1]"

It is untrue that the green triangle has a fourth of the height of the blue triangle. The "height" being referred to in Archimedes proof is the height from the vertex to the midpoint of each green triangle base. It is most definitely not the height of the green triangle by any stretch. It shouldn't be called "height" but rather vertical distance from apex to base. 2603:7080:AFF0:60:289F:139F:D4E7:51A4 (talk) 16:40, 6 August 2022 (UTC)[reply]

Note [1] says: "The green triangle has half of the width of blue triangle by construction. The statement about the height follows from the geometric properties of a parabola, and is easy to prove using modern analytic geometry."

THIS PRESENT ARTICLE WOULD BE MADE HUGELY MORE USEFUL IF IT WERE TO INCLUDE _HERE_ THE ("EASY") PROOF OF THE GREEN TRIANGLE HAVING A HEIGHT THAT IS ONE-FOURTH OF THE BLUE TRIANGLE.

I'm too dumb to do this - I came to this article to learn something, and I'm not able to understand it fully because of this omission. The point of an encyclopedia is to provide information to people who don't already have that information. In this respect, this article fails at this point.

Please could someone smart fill in this bit?

115.64.142.162 (talk) 09:30, 18 November 2015 (UTC)[reply]

Bump. Yes, if it's so easy, can't someone smart put the proof here in the article? I came to this article to find out how it works, and I have been disappointed.

The summing of the series (which, for me, is fairly easy) is included, but the basis on which it is built - subsequent triangles being 1/8th the size of previous ones, needs to be spelled out too.

The comment(s) below were originally left at Talk:Quadrature of the Parabola/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Last edited at 19:24, 29 January 2008 (UTC). Substituted at 02:38, 5 May 2016 (UTC)

A better suggestion is: The limit (not sum!) of 1/4 + 1/16 + 1/64 + ... is not greater than 1/3 and not less than 1/3, therefore it must be a third. 2603:7080:AFF0:60:289F:139F:D4E7:51A4 (talk) 18:05, 6 August 2022 (UTC)[reply]

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