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| [[Image:Geometric series 14 ell.png|right|thumb|Archimedes' figure with ''a'' = 3/4]]
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| In [[mathematics]], the [[infinite series]] '''1/4 + 1/16 + 1/64 + 1/256 + · · ·''' is an example of one of the first infinite series to be summed in the [[history of mathematics]]; it was used by [[Archimedes]] circa 250–200 BC.<ref name = "Shawyer 3">Shawyer and Watson p. 3.</ref> As it is a [[geometric series]] with first term 1/4 and common ratio 1/4, its sum is
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| :<math>\frac {\frac 1 4} {1 - \frac 1 4}=\frac 1 3.</math>
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| ==Visual demonstrations==
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| [[Image:Geometric series 14 square2.svg|left|thumb|3''s'' = 1.]]
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| The series 1/4 + 1/16 + 1/64 + 1/256 + · · · lends itself to some particularly simple visual demonstrations because a [[Square (geometry)|square]] and a triangle both divide into four [[similarity (geometry)|similar]] pieces, each of which contains 1/4 the area of the original.
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| In the figure on the left,<ref name = "Nelsen 74">Nelsen and Alsina p. 74.</ref><ref name = "Ajose">Ajose and Nelson.</ref> if the large square is taken to have area 1, then the largest black square has area (1/2)(1/2) = 1/4. Likewise, the second largest black square has area 1/16, and the third largest black square has area 1/64. The area taken up by all of the black squares together is therefore 1/4 + 1/16 + 1/64 + · · ·, and this is also the area taken up by the gray squares and the white squares. Since these three areas cover the unit square, the figure demonstrates that
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| :<math>3\left(\frac14+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots\right) = 1.</math>
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| Archimedes' own illustration, adapted at top,<ref>Heath p.250</ref> was slightly different, being closer to the equation
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| [[Image:Geometric series triangle.svg|right|thumb|3''s'' = 1 again]]
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| :<math>\frac34+\frac{3}{4^2}+\frac{3}{4^3}+\frac{3}{4^4}+\cdots = 1.</math>
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| See below for details on Archimedes' interpretation.
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| The same geometric strategy also works for [[triangle]]s, as in the figure on the right:<ref name = "Nelsen 74" /><ref name = "Stein 46">Stein p. 46.</ref><ref name = "Mabry">Mabry.</ref> if the large triangle has area 1, then the largest black triangle has area 1/4, and so on. The figure as a whole has a [[self-similarity]] between the large triangle and its upper sub-triangle. A related construction making the figure similar to all three of its corner pieces produces the [[Sierpinski triangle]].<ref>Nelson and Alsina p.56</ref>
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| ==Archimedes==
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| [[Image:Quadrature.png|250px|thumb|This curve is a parabola. The dots on the [[secant line]] ''AE'' are equally spaced. Archimedes showed that the sum of the areas of triangles ''ABC'' and ''CDE'' is 1/4 of the area of triangle ''ACE''. He then constructs another layer of four triangles atop those, the sum of whose areas is 1/4 of the sum of the areas of ''ABC'' and ''CDE'', and then another layer of eight triangles atop that, having 1/4 of ''that'' area, and so on. He concluded that the area between the secant line and the curve is 4/3 the area of triangle ''ACE''.]]
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| Archimedes encounters the series in his work ''[[Quadrature of the Parabola]]''. He is finding the area inside a parabola by the method of exhaustion, and he gets a series of triangles; each stage of the construction adds an area 1/4 times the area of the previous stage. His desired result in that the total area is 4/3 the area of the first stage. To get there, he takes a break from parabolas to introduce an algebraic lemma:
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| '''Proposition 23.''' Given a series of areas ''A'', ''B'', ''C'', ''D'', … , ''Z'', of which ''A'' is the greatest, and each is equal to four times the next in order, then<ref>This is a quotation from Heath's English translation (p.249).</ref> | |
| :<math>A + B + C + D + \cdots + Z + \frac13 Z = \frac43 A.</math>
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| Archimedes proves the proposition by first calculating
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| :<math>\begin{array}{rcl}
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| \displaystyle B+C+\cdots+Z+\frac{B}{3}+\frac{C}{3}+\cdots+\frac{Z}{3} & = &\displaystyle \frac{4B}{3}+\frac{4C}{3}+\cdots+\frac{4Z}{3} \\[1em]
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| & = &\displaystyle \frac13(A+B+\cdots+Y).
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| \end{array}</math>
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| On the other hand,
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| :<math>\frac{B}{3}+\frac{C}{3}+\cdots+\frac{Y}{3} = \frac13(B+C+\cdots+Y).</math>
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| Subtracting this equation from the previous equation yields
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| :<math>B+C+\cdots+Z+\frac{Z}{3} = \frac13 A</math>
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| and adding ''A'' to both sides gives the desired result.<ref>This presentation is a shortened version of Heath p.250.</ref>
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| Today, a more standard phrasing of Archimedes' proposition is that the partial sums of the series {{nowrap|1 + 1/4 + 1/16 + · · ·}} are:
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| :<math>1+\frac{1}{4}+\frac{1}{4^2}+\cdots+\frac{1}{4^n}=\frac{1-\left(\frac14\right)^{n+1}}{1-\frac14}.</math>
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| This form can be proved by multiplying both sides by 1 − 1/4 and observing that all but the first and the last of the terms on the left-hand side of the equation cancel in pairs. The same strategy works for any finite [[geometric series]].
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| ==The limit==
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| Archimedes' Proposition 24 applies the finite (but indeterminate) sum in Proposition 23 to the area inside a parabola by a double ''[[reductio ad absurdum]]''. He does not ''quite''<ref>Modern authors differ on how appropriate it is to say that Archimedes summed the infinite series. For example, Shawyer and Watson (p.3) simply say he did; Swain and Dence say that "Archimedes applied an indirect limiting process"; and Stein (p.45) stops short with the finite sums.</ref> take the [[limit of a sequence|limit]] of the above partial sums, but in modern calculus this step is easy enough:
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| :<math>\lim_{n\to\infty} \frac{1-\left(\frac14\right)^{n+1}}{1-\frac14} = \frac{1}{1-\frac14} = \frac43.</math>
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| Since the sum of an infinite series is defined as the limit of its partial sums,
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| :<math>1+\frac14+\frac{1}{4^2}+\frac{1}{4^3}+\cdots = \frac43.</math>
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| ==Notes==
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| {{reflist}}
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| ==References==
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| <div class="references-small">
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| *{{cite journal |doi=10.2307/2690617 |title=Proof without Words: Geometric Series |author=Ajose, Sunday and Roger Nelsen |journal=Mathematics Magazine |volume=67 |issue=3 |pages=230 |jstor=2690617 |date=June 1994}}
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| *{{cite book |first=T. L. |last=Heath |title=The Works of Archimedes |year=1953 |origyear=1897 |publisher=Cambridge UP}} Page images at {{cite web |first=Bill |last=Casselman |title=Archimedes' quadrature of the parabola |url=http://www.math.ubc.ca/~cass/archimedes/parabola.html |accessdate=2007-03-22}} HTML with figures and commentary at {{cite web |first=Daniel E. |last=Otero |year=2002 |title=Archimedes of Syracuse |url=http://www.cs.xu.edu/math/math147/02f/archimedes/archpartext.html |accessdate=2007-03-22| archiveurl= http://web.archive.org/web/20070307141104/http://www.cs.xu.edu/math/math147/02f/archimedes/archpartext.html| archivedate= 07 March 2007 <!--Added by DASHBot-->}}
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| *{{cite journal |title=Proof without Words: <sup>1</sup>⁄<sub>4</sub> + (<sup>1</sup>⁄<sub>4</sub>)<sup>2</sup> + (<sup>1</sup>⁄<sub>4</sub>)<sup>3</sup> + · · · = <sup>1</sup>⁄<sub>3</sub> |first=Rick |last=Mabry |journal=Mathematics Magazine |volume=72 |issue=1 |pages=63 |jstor=2691318 |date=February 1999}}
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| *{{cite book |author=Nelsen, Roger B. and Claudi Alsina |title=Math Made Visual: Creating Images for Understanding Mathematics |year=2006 |publisher=MAA |isbn=0-88385-746-4}}
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| *{{cite book |author=Shawyer, Bruce and Bruce Watson |title=Borel's Methods of Summability: Theory and Applications |publisher=Oxford UP |year=1994 |isbn=0-19-853585-6}}
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| *{{cite book |first=Sherman K. |last=Stein |title=Archimedes: What Did He Do Besides Cry Eureka? |publisher=MAA |year=1999 |isbn=0-88385-718-9}}
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| *{{cite journal |doi=10.2307/2691014 |author=Swain, Gordon and Thomas Dence |title=Archimedes' Quadrature of the Parabola Revisited |journal=Mathematics Magazine |volume=71 |issue=2 |pages=123–30 |jstor=2691014 |date=April 1998}}
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| </div>
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| {{Series (mathematics)}}
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| {{DEFAULTSORT:1 4 + 1 16 + 1 64 + 1 256 +}}
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| [[Category:Geometric series]]
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| [[Category:Proof without words]]
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| [[zh:1/4+1/16+1/64+1/256+…]]
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Hi there, I am Alyson Boon although it is not the title on my birth certification. One of the issues she loves most is canoeing and she's been doing it for quite a whilst. Alaska is where he's usually been living. Office supervising is where her primary income comes from.
my web blog: live psychic reading