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== Quadrature of the Parabola with the "square pyramidal number" (new proof) ==
[[File:Figuraxy.jpg|thumb|Quadrature of the Parabola]]
I found that the "square pyramidal number" can be used to prove the Archimedes' theorem on the area of parabolic segment. The entire proof, carried out without the use of "mathematical analysis", one can readis at the following web adress:
https://sitesdrive.google.com/sitefile/leggendoarchimeded/0B4iaQ-gBYTaJMDJFd2FFbkU2TU0/view?usp=sharing
Below we show a summary of the proof.
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)<sup>.</sup>6 = 6<sup>2</sup><sup>.</sup>6 = 6<sup>3</sup> green triangles.
The parabola circumscribed figure (in red) contains:
:A(cir.) = 6<sup><small>.</small></sup>1 + 5<sup><small>.</small></sup>3 + 4<sup><small>.</small></sup>5 + 3<sup><small>.</small></sup>7 + 2<sup><small>.</small></sup>9 + 1<sup><small>.</small></sup>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:
# The triangle ABC contains n<sup><small>3</small></sup> green triangles
# A<sub>n</sub>(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<sub>n</sub>/n<sup>3</sup>, ..... 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<sub>n</sub>.
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:
# Choice of ''equivalent'' triangles for measuring areas.
# With this choice, the area of triangle ABC measure n<sup>3</sup> triangles.
# 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.--[[User:Ancora Luciano|Ancora Luciano]] ([[User talk:Ancora Luciano|talk]]) 18:47, 14 May 2013 (UTC)
:That does indeed seem to be a valid and nice proof (or at least something that could be made into a proof with a little more care about why the limit of the saw-tooth areas converges to the parabola area (in contrast to situations like [http://mathworld.wolfram.com/DiagonalParadox.html this one] for which the convergence argument doesn't work). But either it's not new or it doesn't (yet) belong here; see [[WP:NOR]]. So if you think it's new, the appropriate thing to do would be to get it properly published elsewhere first, so that the publication can be used as a [[WP:RS|reliable source]] here. Personally maintained web sites are not adequate for this purpose. —[[User:David Eppstein|David Eppstein]] ([[User talk:David Eppstein|talk]]) 22:48, 14 May 2013 (UTC)
:::I revised this demonstration, according to the preceding note of prof. David Eppstein. Its final edition was published. Overview and references are in the MATHDI database at no. ME 06644800.
== Sum of the first ''n'' squares (geometrical proof) ==
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