Editing Archimedes (section)

=== Method of exhaustion ===
{{main|Method of exhaustion}}
[[File:PiArchimede4.svg|thumb|Archimedes calculates the side of the 12-gon from that of the [[hexagon]] and for each subsequent doubling of the sides of the regular polygon]]
In ''[[Quadrature of the Parabola]]'', Archimedes states that a certain proposition in [[Euclid's Elements]] demonstrating that the area of a circle is proportional to its diameter was proven using a lemma now known as the [[Archimedean property]], that “the excess by which the greater of two unequal regions exceed the lesser, if added to itself, can exceed any given bounded region.” Prior to Archimedes, [[Eudoxus of Cnidus]] and other earlier mathematicians{{efn|While Eudoxus is often credited for the whole of Euclid's Book XII, Archimedes explicitly credits him only with the proofs of XII.7 and XII.10, that the volume of a pyramid and a cone are, respectively, one-third of the volume of the rectangular prism and cone with the same base and height.{{harvnb|Acerbi|2018|p=279}}}} applied this lemma, a technique now referred to as the "method of exhaustion," to find the volume of a [[tetrahedron]], [[cylinder]], [[cone]], and [[sphere]], for which proofs are given in book XII of [[Euclid's Elements]].{{sfn|Acerbi|2018|p=279}}

In ''[[Measurement of a Circle]]'', Archimedes employed this method to show that the area of a circle is the same as a right triangle whose base and height are equal to its radius and circumference.{{sfn|Acerbi|2018|p=280}} He then approximated the ratio between the radius and the circumference, the value of [[Pi|{{pi}}]], by drawing a larger [[regular hexagon]] outside a [[circle]] then a smaller regular hexagon inside the circle, and progressively doubling the number of sides of each [[regular polygon]], calculating the length of a side of each polygon at each step. As the number of sides increases, it becomes a more accurate approximation of a circle. After four such steps, when the polygons had 96 sides each, he was able to determine that the value of {{pi}} lay between 3{{sfrac|1|7}} (approx. 3.1429) and 3{{sfrac|10|71}} (approx. 3.1408), consistent with its actual value of approximately 3.1416.<ref>{{cite web |title=The Computation of Pi by Archimedes |author=McKeeman, Bill |author-link=William M. McKeeman |website=Matlab Central |url=http://www.mathworks.com/matlabcentral/fileexchange/29504-the-computation-of-pi-by-archimedes/content/html/ComputationOfPiByArchimedes.html#37 |access-date=30 October 2012}}</ref> In the same treatise, he also asserts that the value of the [[square root]] of 3 as lying between {{sfrac|265|153}} (approximately 1.7320261) and {{sfrac|1351|780}} (approximately 1.7320512), which he may have derived from a similar method.<ref>{{Cite web |title=Of Calculations Past and Present: The Archimedean Algorithm |url=https://www.maa.org/programs/maa-awards/writing-awards/of-calculations-past-and-present-the-archimedean-algorithm |access-date=14 April 2021 |website=maa.org |publisher=Mathematical Association of America}}</ref>

[[File:Parabolic segment and inscribed triangle.svg|thumb|upright=.8|A proof that the area of the [[parabola|parabolic]] segment in the upper figure is equal to 4/3 that of the inscribed triangle in the lower figure from ''[[Quadrature of the Parabola]]'']]
In ''[[Quadrature of the Parabola]]'', Archimedes used this technique to prove that the area enclosed by a [[parabola]] and a straight line is {{sfrac|4|3}} times the area of a corresponding inscribed [[triangle]] as shown in the figure at right, expressing the solution to the problem as an [[Series (mathematics)#History of the theory of infinite series|infinite]] [[geometric series]] with the [[Geometric series#Common ratio r|common ratio]] {{sfrac|1|4}}:{{sfn|Netz|2022|p=139}}

:<math>\sum_{n=0}^\infty 4^{-n} = 1 + 4^{-1} + 4^{-2} + 4^{-3} + \cdots = {4\over 3}. \;</math>

If the first term in this series is the area of the triangle, then the second is the sum of the areas of two triangles whose bases are the two smaller [[secant line]]s, and whose third vertex is where the line that is parallel to the parabola's axis and that passes through the midpoint of the base intersects the parabola, and so on. This proof uses a variation of the series {{nowrap|[[1/4 + 1/16 + 1/64 + 1/256 + · · ·]]}} which sums to&nbsp;{{sfrac|1|3}}.

He also used this technique in order to measure the surface areas of a sphere and cone,<ref>''[[On the Sphere and Cylinder]]'' 13-14, 33-34, 42, 44</ref> to calculate the area of an ellipse,<ref>''[[On Conoids and Spheroids]]'' 4</ref> and to find the area contained within an [[Archimedean spiral]].<ref>''[[On Spirals]]'', 24-25</ref>{{sfn|Acerbi|2018|p=280}}