# Proofs Using a Quadrature Method of Archimedes

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Archimedes divided a segment of a parabola into increasingly many diminishing triangles. Having deduced the area of each triangle and observing that the areas formed a geometric sequence, he found the area by summing the geometric series [1].

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Contributed by: Jenda Vondra (January 2014)

Open content licensed under CC BY-NC-SA

## Snapshots

## Details

The variables , , , are positive integers and is real and non-negative.

This Demonstration plots the curve over the interval , . The vertical lines at intersect the curve and the line from to at red points; those points form nonoverlapping polygons. The regions between the curve and the secants joining the red points on the curve can be partially filled by polygons, too; this is the second layer. In general, the area of each polygon formed at layer at position on the interval is computed and approximated.

The area of the polygon with end vertices on the curve at and is precisely the sum of the areas of the yellow and the gray triangles with red dots at vertices shown in the diagram, so the expression for the area is

.

For example, .

The limit as gives the closed area above the curve, and is in principle the method of Riemann integration.

The area of the polygon with end vertices on the curve at and at layer at position on the interval is

, , with value shown in the diagram as "area of moving polygon".

Considering the polygons as a series of "layers" is key to the following algebra.

Re-indexing as used in the diagram plot rewrites the term of the infinite sequence of polygon areas in a layer as

, giving .

The sum of the first layer of polygon areas on the infinite curve is then and the sum of all polygon layers is

.

This can be simplified to

.

Rearranging the order of summation into tuples in order of size produces the single sum over , which is easier to compute:

.

Looking at the dissection diagram and its algebraic expression lets us see many polygon dissections for given closed areas.

Putting in the diagram plots the hyperbola and we observe a sum of polygon areas giving the identity

Putting we observe

Theorem 1

for .

Proof

The term is the area under the curve for .

The term is the total area of the right triangles above the curve at positive integer points, and by telescoping is constant for .

The term is the total area of all polygons on the infinite curve.

At the special value , gives the Euler–Mascheroni constant (see [2]).

The sum .

Rewriting shows the Vacca-type rational series for the Euler–Masceroni constant due to Sondow [3, 4].

for a positive integer greater than 1 [3].

Putting , we observe

Theorem 2

for .

Proof

The term alternates the areas under the curve. Writing the integrals gives the term as

,

with the factor excising the pole in the expression when and the term value is ; and are the ceiling and floor functions, respectively.

The term is the alternating sum of the right triangle areas above the curve at positive integer points.

The term alternates the summation of the areas of all the polygons on the curve.

At the special value , gives the constant (see [3, 4]); relates to a problem mentioned in [4].

References

[1] Wikipedia. "The Quadrature of the Parabola." (Oct 23, 2013) en.wikipedia.org/wiki/The_Quadrature_of _the _Parabola.

[2] Wikipedia. "Euler–Mascheroni Constant." (Jul 5, 2013) en.wikipedia.org/wiki/Euler-Mascheroni_constant.

[3] J. Sondow, "New Vacca-Type Rational Series for Euler's Constant and Its 'Alternating' Analog ."arxiv.org/abs/math/0508042v1.

[4] J. Sondow, "New Vacca-Type Rational Series for Euler's Constant and Its 'Alternating' Analog ," *Additive Number Theory* (D. Chudnovsky, ed.), New York, NY: Springer, 2010 pp. 331–340. link.springer.com/chapter/10.1007%2 F978-0-387-68361-4_ 23.

## Permanent Citation

"Proofs Using a Quadrature Method of Archimedes"

http://demonstrations.wolfram.com/ProofsUsingAQuadratureMethodOfArchimedes/

Wolfram Demonstrations Project

Published: January 7 2014