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| {{DISPLAYTITLE:Liu Hui's {{pi}} algorithm}}
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| {{pi box}}
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| [[Image:Cutcircle2.svg|thumb|right|Liu Hui's method of calculating the area of a circle]]
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| '''Liu Hui's {{pi}} algorithm''' was invented by [[Liu Hui]] (fl. 3rd century), a mathematician of [[Wei Kingdom]]. Before his time, the ratio of the circumference of a circle to diameter was often taken experimentally as three in China, while [[Zhang Heng]] (78–139) rendered it as 3.1724 (from the proportion of the celestial circle to the diameter of the earth, {{math|92/29}}) or as <math>\pi \approx \sqrt{10} \approx 3.162</math>. Liu Hui was not satisfied with this value. He commented that it was too large and overshot the mark. Another mathematician [[Wan Fan]] (219–257) provided {{math|1=π ≈ 142/45 ≈ 3.156}}.<ref>Schepler, Herman C. (1950), “The Chronology of Pi”, Mathematics Magazine 23 (3): 165–170, ISSN 0025-570X.</ref> All these empirical {{pi}} values were accurate to two digits (i.e. one decimal place). Liu Hui was the first Chinese mathematician to provide a rigorous algorithm for calculation of {{pi}} to any accuracy. Liu Hui's own calculation with a 96-gon provided an accuracy of five digits: {{math|π ≈ 3.1416}}.
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| Liu Hui remarked in his commentary to the ''[[The Nine Chapters on the Mathematical Art]]'',<ref>Needham, Volume 3, 66.</ref> that the ratio of the circumference of an inscribed hexagon to the diameter of the circle was three, hence {{pi}} must be greater than three. He went on to provide a detailed step-by-step description of an iterative algorithm to calculate {{pi}} to any required accuracy based on bisecting polygons; he calculated {{pi}} to between 3.141024 and 3.142708 with a 96-gon; he suggested that 3.14 was a good enough approximation, and expressed {{pi}} as 157/50; he admitted that this number was a bit small. Later he invented an ingenious [[#Quick method|quick method]] to improve on it, and obtained {{math|π ≈ 3.1416}} with only a 96-gon, with an accuracy comparable to that from a 1536-gon. His most important contribution in this area was his simple iterative {{pi}} algorithm.
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| ==Area of a circle==
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| [[Image:Cutcircle.svg|thumb|right|The area within a circle is equal to the radius multiplied by half the circumference, or {{math|''A''}} = {{math|''r''}} x {{math|''C''}}/2 = {{math|''r''}} x {{math|''r''}} x {{pi}}.]] | |
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| Liu Hui argued:
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| :"''Multiply one side of a hexagon by the radius (of its circumcircle), then multiply this by three, to yield the area of a dodecagon; if we cut a hexagon into a dodecagon, multiply its side by its radius, then again multiply by six, we get the area of a 24-gon; the finer we cut, the smaller the loss with respect to the area of circle, thus with further cut after cut, the area of the resulting polygon will coincide and become one with the circle; there will be no loss''".
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| Apparently Liu Hui had already mastered the concept of the limit<ref>First noted by Japanese mathematician [[Yoshio Mikami]]</ref>
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| : <math>\lim_{N \to \infty}\text{area of }N\text{-gon} = \text{area of circle}. \, </math>
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| Further, Liu Hui proved that the area of a circle is half of its circumference multiplied by its radius. He said:
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| "''Between a polygon and a circle, there is excess radius. Multiply the excess radius by a side of the polygon. The resulting area exceeds the boundary of the circle''".
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| In the diagram {{math|''d''}} = excess radius. Multiplying {{math|''d''}} by one side results in oblong {{math|ABCD}} which exceeds the boundary of the circle. If a side of the polygon is small (i.e. there is a very large number of sides), then the excess radius will be small, hence excess area will be small.
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| As in the diagram, when {{math|''N'' → ∞}}, {{math|''d'' → 0}}, and {{math|ABCD → 0}}.
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| "''Multiply the side of a polygon by its radius, and the area doubles; hence multiply half the circumference by the radius to yield the area of circle''".
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| When {{math|''N'' → ∞}}, half the circumference of the {{math|''N''}}-gon approaches a semicircle, thus half a circumference of a circle multiplied by its radius equals the area of the circle. Liu Hui did not explain in detail this deduction. However it is self-evident by using Liu Hui's "in-out complement principle" which he provided elsewhere in ''The Nine Chapters on the Mathematical Art'': Cut up a geometric shape into parts, rearrange the parts to form another shape, the area of the two shapes will be identical.
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| Thus rearranging the six green triangles, three blue triangles and three red triangles into a rectangle with width = 3{{math|''L''}}, and height {{math|''R''}} shows that the area of the dodecagon = 3{{math|''RL''}}.
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| In general, multiplying half of the circumference of a {{math|''N''}}-gon by its radius yields the area of a 2{{math|''N''}}-gon. Liu Hui used this result repetitively in his {{pi}} algorithm.
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| ==Liu Hui's {{pi}} inequality==
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| [[Image:Liuhui Pi Inequality.svg|thumb|right|Liu Hui's {{pi}} inequality]]
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| Liu Hui proved an inequality involving {{pi}} by considering the area of inscribed polygons with {{math|''N''}} and 2{{math|''N''}} sides.
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| In the diagram, the yellow area represents the area of an {{math|''N''}}-gon, denoted by <math>A_N</math>, and the yellow area plus the green area represents the area of a 2{{math|''N''}}-gon, denoted by <math>A_{2N}</math>. Therefore the green area represents the difference between the areas of the 2{{math|''N''}}-gon and the ''N''-gon:
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| :<math>D_{2N}=A_{2N}-A_N.</math>
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| The red area is equal to the green area, and so is also <math>D_{2N}</math>. So
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| :Yellow area + green area + red area = <math>A_{2N} + D_{2N}.</math>
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| Let ''C'' represent the area of the circle. Then
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| :<math>A_{2N} < C < A_{2N} + D_{2N}.</math>
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| If the radius of the circle is taken to be 1, then we have Liu Hui's {{pi}} inequality:
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| :<math> A_{2N} < \pi < A_{2N} + D_{2N}.</math>
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| ==Iterative algorithm==
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| {{Original research|section|date=March 2009}}
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| [[Image:Liuhui geyuanshu.svg|thumb|right|Liu Hui's {{pi}} algorithm]]
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| Lui Hui began with an inscribed hexagon. Let {{math|M}} be the length of one side {{math|AB}} of hexagon, {{math|''r''}} is the radius of circle.
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| Bisect {{math|AB}} with line {{math|OPC}}, {{math|AC}} becomes one side of dodecagon, let its length be {{math|m}}.
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| {{math|AOP}}, {{math|APC}} are two right angle triangles. Liu Hui used [[Pythagorean theorem|Gou Gu theorem]] repetitively:
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| : <math>{} G^2 = r^2 - \left(\tfrac{M}{2}\right)^2</math>
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| : <math>{}G= \sqrt{r^2- \tfrac{M^2}{4}}</math>
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| : <math>{} j= r - G = r - \sqrt{r^2- \tfrac{M^2}{4}}</math>
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| : <math>{}m^2= \left(\tfrac{M}{2}\right)^2 + j^2</math>
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| : <math>{}m=\sqrt{m^2}.</math>
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| With {{math|''r''}} = 10 units, he obtained
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| : area of 48-gon <math>{}A_{96} = 313 {584 \over 625} </math>
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| : area of 96-gon <math>{}A_{192} = 314 {64 \over 625} </math>
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| : Difference of 96-gon and 48-gon:
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| :<math>{}D_{192} = 314\frac{64}{625} - 313\frac{584}{625} = \frac{105}{625}</math>
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| :from Liu Hui's {{pi}} inequality:
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| :<math> A_{2N} < C < A_{2N} + D_{2N}.</math>
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| :Since {{math|''r''}} = 10, {{math|''C''}} = <math> 100 \times \pi</math>
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| :therefore:
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| :<math>{}314\frac{64}{625}<100 \times \pi <314 \frac{64}{625} +\frac{105}{625}</math>
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| ::<math>{}314\frac{64}{625}<100 \times \pi <314 \frac{169}{625}</math>
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| :<math>{} 3.141024 < \pi < 3.142704.</math>
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| He never took {{pi}} as the average of the lower limit 3.141024 and upper limit 3.142704. Instead he suggested that 3.14 was a good enough approximation for {{pi}}, and expressed it as a fraction <math>\tfrac{157}{50}</math>; he pointed out this number is slightly less than the real thing.
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| Liu Hui carried out his calculation with [[rod calculus]], and expressed his results with fractions. However, the iterative nature of Liu Hui's {{pi}} algorithm is quite clear:
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| : <math> 2-m^2 =\sqrt{2+(2-M^2)}\, ,</math>
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| in which {{math|''m''}} is the length of one side of next order polygon bisected from {{math|''M''}}, then repeat the same calculation, each step required only one addition, one square root extraction.
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| == Quick method ==
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| Calculation of square roots of irrational numbers was not an easy task in the third century with
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| [[counting rods]]. Liu Hui discovered a short cut by comparing the area differentials of polygons, and found that the proportion of the difference in area of successive order polygons was approximately 1/4.<ref>Yoshio Mikami: Ph.D. Dissertation 1932</ref>
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| Let {{math|D}}<sub>{{math|N}}</sub> denote the difference in areas of {{math|N}}-gon and ({{math|N}}/2)-gon
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| : <math>D_N = A_N - A_{N/2}\,</math> | |
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| He found:
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| :<math>D_{96} \approx \tfrac{1}{4} D_{48}</math>
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| :<math>D_{192} \approx \tfrac{1}{4} D_{96}</math> {{ref|1|1}}
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| Hence:
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| :<math>
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| \begin{align}
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| D_{384} & {} \approx \tfrac{1}{4} D_{192} \\
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| D_{768} & {} \approx \left(\tfrac{1}{4}\right)^2 D_{192} \\
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| D_{1536} & {} \approx \left(\tfrac{1}{4}\right)^3 D_{192} \\
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| D_{3072} & {} \approx \left(\tfrac{1}{4}\right)^4 D_{192} \\
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| & {} \ \ \vdots
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| \end{align}
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| </math>
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| Area of unit radius circle =
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| : <math>{}\pi = A_{192} + D_{384} + D_{768}+D_{1536}+D_{3072} + \cdots \approx A_{192} + F \cdot D_{192}. \, </math>
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| In which
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| : <math> F = \tfrac{1}{4} + \left(\tfrac{1}{4}\right)^2 + \left(\tfrac{1}{4}\right)^3 + \left(\tfrac{1}{4}\right)^4 + \cdots = \tfrac{1}{3}.</math> | |
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| That is all the subsequent excess areas add up amount to one third of the <math>D_{192}</math>
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| : area of unit circle<math> {}=\pi \approx A_{192} + \left(\tfrac{1}{3}\right)D_{192} \approx {3927 \over 1250} \approx 3.1416.\,</math>{{ref|2|2}}
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| Liu Hui was quite happy with this result because he had acquired the same result with the calculation for a 1536-gon, obtaining the area of a 3072-gon. This explains four questions:
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| # Why he stopped short at {{math|''A''}}<sub>192</sub> in his presentation of his algorithm. Because he discovered a quick method of improving the accuracy of {{pi}}, achieving same result of 1536-gon with only 96-gon. After all calculation of square roots was not a simple task with [[rod calculus]]. With the quick method, he only needed to perform one more subtraction, one more division (by 3) and one more addition, instead of four more square root extractions.
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| # Why he preferred to calculate {{pi}} through calculation of areas instead of circumferences of successive polygons, because the quick method required information about the difference in '''areas''' of successive polygons.
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| # Who was the true author of the paragraph containing calculation of <math>\pi = {3927 \over 1250}.</math>
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| # That famous paragraph began with "A Han dynasty bronze container in the military warehouse of [[Jin Dynasty (265–420)|Jin dynasty]]....". Many scholars, among them [[Yoshio Mikami]] and [[Joseph Needham]], believed that the "Han dynasty bronze container" paragraph was the work of Liu Hui and not Zu Chongzhi as other believed, because of the strong correlation of the two methods through area calculation, and because there was not a single word mentioning Zu's 3.1415926 < {{pi}} < 3.1415927 result obtained through 12288-gon.
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| ==Later developments==
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| Liu Hui established a solid algorithm for calculation of {{pi}} to any accuracy.
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| *[[Zu Chongzhi]] was familiar with Liu Hui's work, and obtained greater accuracy by applying his algorithm to a 12288-gon.
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| :From Liu Hui's formula for 2{{math|N}}-gon:
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| :<math> A_{2N} = m_{N} \times r</math>
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| :For 12288-gon inscribed in a unit radius circle:
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| :<math> A_{24576}=3.14159261864 < \pi </math>.
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| :From Liu Hui's {{pi}} inequality:
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| :<math> A_{24576} < \pi < A_{24576} +D_{24576}</math>
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| :In which <math> D_{24576}=A_{24576}- A_{12288}=0.0000001021</math>
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| :<math> A_{24576}=3.14159261864 < \pi <3.14159261864 +0.0000001021</math>.
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| :Therefore
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| :<math>3.14159261864 < \pi <3.141592706934</math>
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| Truncated to eight significant digits:
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| :<math> 3.1415926 < \pi <3.1415927</math>.
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| That was the famous Zu Chongzhi {{pi}} inequality.
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| Zu Chongzhi then used the interpolation formula by [[He Chengtian]] (何承天, 370-447) and obtained an approximating fraction: <math>\pi \approx {355 \over 113}</math>.
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| *[[Yuan dynasty]] mathematician [[Zhao Yu Xin]] worked on a variation of Liu Hui's {{pi}} algorithm, by bisecting an inscribed square instead of a hexagon.
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| ==Significance of Liu Hui's algorithm==
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| Liu Hui's {{pi}} algorithm was one of his most important contributions to ancient Chinese mathematics. It was based on calculation of {{math|N}}-gon area, in contrast to the Archimedean algorithm based on polygon circumference. Archimedes used a circumscribed 96-gon to obtain an upper limit <math>\pi < 22/7=3.142857</math>, and an inscribed 96-gon to obtain the lower limit <math>223/71=3.140845</math>. Liu Hui was able to obtain both his upper limit 3.142704 and lower limit 3.141024 with only an inscribed 96-gon. Furthermore, both the Liu Hui limits were tighter than Archimedes's: 3.140845 < 3.141024 < {{pi}} < 3.142704 < 3.142857. With his method Zu Chongzhi obtained the result: 3.1415926 < {{pi}} < 3.1415927, which held the world record for the most accurate value of {{pi}} for 1200 years, even by 1600 in Europe, mathematician Adriaen Anthoniszoom and his son obtained {{pi}} value of 3.1415929, accurate only to 7 digit, still 3 digits short of Zu's result<ref>Robert Temple, The Genius of China, a refined value of pi, p144-145, ISBN 1-85375-292-4</ref>
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| ==Notes==
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| {{Original research|section|date=March 2009}}
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| :{{note|1|1}} Correct value: 0.2502009052
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| :{{note|2|2}} Correct values:
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| :<math>A_{192}= 3.1410319509</math>
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| :<math>D_{192}=0.0016817478</math>
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| :<math>\pi \approx A_{192}+ \frac{1}{3} D_{192}\approxeq 3.1410319509 +0.0016817478/3</math>
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| :<math>\pi \approx 3.1410319509 +0.0005605826</math>
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| :<math>\pi \approx 3.1415925335.</math>
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| <br>
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| Liu Hui's quick method was potentially able to deliver almost the same result of 12288-gon (3.141592516588) with only 96-gon.
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| == See also ==
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| * [[Method of exhaustion]]
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| == References ==
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| {{reflist|2}}
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| == Further reading ==
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| *Needham, Joseph (1986). ''Science and Civilization in China'': Volume 3, Mathematics and the Sciences of the Heavens and the Earth. Taipei: Caves Books, Ltd.
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| * Wu Wenjun ed, ''History of Chinese Mathematics'' Vol III (in Chinese) ISBN 7-303-04557-0
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| | |
| {{DEFAULTSORT:Liu Hui's Pi Algorithm}}
| |
| [[Category:Pi algorithms]]
| |
| [[Category:Chinese mathematics]]
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| [[Category:Cao Wei]]
| |
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For a tasty, heаlthful snack, consider creating a puree of berries, pears, or peaches. Ƭhis will make a distribute which iѕ sweet which you cаn eat being a dip for pіta potato chips or pretzels. Explore all of thе innovatіve methods to use frеsh fruits inside your meals and snacks, to helр keep iѕsues intriguing and flavorful.
Clean beets are a good wholesome food. Refreshing beetѕ aгe full օf dietary fiber and essential nutrients, but canned involve a lot of sea salt. Steam some beets and will include beetroot to the salad.
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Use wholegrain or ѕprοuteɗ loaves of bread rather than white coloreɗ a lߋaf of bread. The glycemic crawl with this breads is low, which reduces hungеr ρangs, safeguards in oppositіon tο heart problems, and assistѕ weight management. These breads will also ƅe full of essentіаl essential fatty асids and fiber that permit your digеstive tract to fսnction morе effectively.
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It might Ьe helpful so that you can embrace somebody in your new diet plan. Search for a person who has both achieѵed their goals or peгhaps is starting their quest such as you. What is important iѕ the individual caггies a joint curiosity and will be your ally.
Natural veggies are still the fіnest snack food. They гemove cravіngs for food therefore makіng you truly feel total whilst providing you with a lot of minerals and vitamins. Theʏ are no more difficult to гetail store and tгy to eat than processed food. They are also less of a trouble to get reɑdy than commߋn snack food items. Uncooked vеgetableѕ are a fantastic betաeen dish goody.
Enable your fɑults go. If one makes on your own truly feel worse aboսt thiѕ, your old habits will attempt to come back. Just think of it as а cheat day time and selеct back up that you remaining οff. Overcoming yourself up is totally useless and ѡon't assist.
Rеmove things wҺich are unhealthy for you like sugary or salty snack foods. Instead of eating these junk foods, consider consuming a lot more muсh healthier foods, like fresɦ and freezing vegetables or beans.
Consume a Mediterranean influenced dіet program. Mediterranean food items are ideal for your life еxpectancy and heart. Try using pita loaves of bread to үour sandwiches as opposеd to the normal white coloreɗ breadѕ, and check out out wɦolegrain pasta too. Seed proԀucts and nut prоducts are loaded with essеntial nutrients and useful boԁy fat, so add more those to your diet plan. Change butter out for extra virgin olive oil, and try to еat as numerous vegatables and fruits thаt you can. Concentrate on consuming more sea food than red meat also.
Your diеt might bе enhanced in many diffeгent tеchniques. Set these idеas to work and you can create a ɦealthy eating plan to your loved ones. You will end up blessed with more healthy, a lot more full of energy bodies, in addition to a lifestyle it is possіble to bеcome older with. Recall that you will be achievіng thiѕ for yourself!