Jigu suanjing (緝古算經,
Continuation of Ancient Mathematics) was the work of early
Tang dynasty calendarist and mathematician
Wang Xiaotong, written some time before the year 626, when he presented his work to the Emperor.
Jigu Suanjing was included as one of the requisite texts for Imperial examination; the amount of time required for the study of
Jigu Suanjing was three years, the same as for
The Nine Chapters on the Mathematical Art and
Haidao Suanjing.
The book began with presentations to the Emperor, followed by a pursuit problem similar to the one in Jiu Zhang Suan shu, followed by 13 three-dimensional geometry problems based mostly on engineering construction of astronomic observation tower, dike, barn, excavation of a canal bed etc., and 6 problems in right angled triangle plane geometry. Apart from the first problem which was solved by arithmetic, the problems deal with the solution of cubic equations, the first known Chinese work to deal with complete cubic equations, as such, it played important roles in the development for solution of high order polynomial equations in the history of Chinese mathematics. Before his time, The Nine Chapters on the Mathematical Art developed algorithm of solving simple cubic equation x^3=N numerically, often referred to as the "finding the root method".
Wang Xiaotong used an algebraic method to solve three-dimensional geometry problems, and his work is a major advance in Algebra in the history of Chinese mathematics.
Each problem in Jigu Suanjing follows the same format, the question part begins with "suppose we have such and such,... question:...how many are there?"; followed by "answer:", with concrete numbers; then followed by "The algorithm says:...", in which Wang Xiaotong detailed the reasoning and procedure for the construction of equations, with a terse description of the method of solution. The emphasis of the book is on how to solve engineering problems by construction of mathematical equations from geometric properties of the relevant problem.
In Jigu Suanjin, Wang established and solved 25 cubic equations, 23 of them from problem 2 to problem 18 have the form
x^3+px^2+qx=N, \,
The remaining two problems 19, and 20 each has a double quadratic equation:
:x^4+px^2+q=0
• Problem 3, two cubic equations:
:: x^3+\frac{3cd}{b-c}x^2+\frac{3(a+c)hd^2}{(H-h)(b-c)}x=\frac{6Vd^2}{(H-h)(b-c)}
:: x^3+5004x^2+1169953\frac{1}{3}x=41107188\frac{1}{3}；
• Problem 4 two cubic equations:
:: x^3+62x^2+696x=38448,\quad x=18;
:: x^3+594x^2=682803,\quad x=33;
• Problem 5
:: x^3+15x^2+66x-360,\quad x=3
• Problem 7:
:: x^3+(D+G)x^2+\left(DG +\frac{D^2}{3}\right)x=P-\frac{D^2G}{3}
:: X^+3\frac{hs}{D}x^2+3\left(\frac{hs}{D}\right)^2x=\frac{P'}{3}\frac{h^2}{D^2}
• Problem 8:
:: x^3+90x^2-839808,\quad x=72
• Problem 15:
:: x^3 +\frac{S}{2}x^2-\frac{P^2}{2S}=0。
• Problem 17:
::x^3 +\frac{5}{2}Dx^2+2D^2x = \frac{P^2}{2D} - \frac{D^2}{2}
• Problem 20:"Suppose the long side of a right angle triangle equals to sixteen and a half, the square of the product of the short side and the hypothenuse equals to one hundred sixty four and 14 parts of 25, question, what is the length of the short side ?"
Answer: "the length of the short side is eight and four fifth."
Algorithm:"Let the square of the square of product as 'shi' (the constant term), and let the square of the long side of right angle triangle be the 'fa' (the coefficient of the y term). Solve by 'finding the root method', then find the square root again."
The algorithm is about setting up a double quadratic equation:
:: x^4+\left(16\frac{1}{2}\right)^2x^2=\left(164\frac{14}{15}\right)^2。
where x is the short side.
Wang's work influence later Chinese mathematicians, like Jia Xian and Qin Jiushao of Song dynasty.
Read more...: Editions
Editions
During the Tang dynasty there were hand-copied Jigu Suanjing in circulation. During the Song dynasty there were 1084 government-printed edition copies. However, by the Ming dynasty the Tang dynasty hand-copied editions and Song dynasty printed editions were almost all lost; only a single copy of a Southern Song print survived. This copy was later obtained by early Qing dynasty publisher Mao Jin, who made an image hand copy (hand-copied character by character, following the printed form closely) of it. Mao Jin's image copy of Jigu Suanjing later became the source for a printed edition during the Qianlong era and was also incorporated into the Siku Quanshu. The Qianlong era printed edition disappeared, and only Mao Jin's image copy edition of Jigu Suanjing survived at the Forbidden City Museum. The copy in the Siku Quanshu still exists.
During the Qing dynasty, study of Jigu Suangjing was in vogue; half a dozen books devoted to the study of Jigu Suanjing by mathematicians were published, some of which concentrated on filling the gaps left by many missing characters due to age, and some devoted to the detail elaboration of algorithm either from geometry point of view (Li Huang) or from Tian yuan shu (Zhang Dunren).
In 1963, Chinese mathematics historian Qian Baocong published his annotated The Ten Computational Canons, which included Jigu Suanjing.
Jigu Suanjing was introduced to the English speaking world by Alexander Wylie in his book Notes on Chinese Literature.
The text above has been excerpted automatically from Wikipedia - please correct any errors in the
original article.
《
緝古算經》，原名《緝古算術》，初唐數學家
王孝通著于武德九年〔626年〕前所著。後被列入算經十書，改名為《緝古算經》。
《緝古算經》一書在中國數學史上有重要影響，王孝通在書中將幾何問題代數化，在世界上首次系統地創立三次多項式方程，對代數學的發展，有重要意義。王孝通在此書中建立 25個三次方程，其中自第二問至第十八問中的23個三次方程有如下形式：
x^3+px^2+q=0
剩下第十九問、二十問各有一個雙二次方程：
x^4+px^2+q=0。
Read more...: 內容 第一問 第二問 第三問 第四問 第五問 第六問 第七問 第八問 第九問 第十問 第十一問 第十二問 第十三問 第十四問 第十五問 第十六問 第十七問 第十八問 第十九問 第二十問 版本
內容
《緝古算經》全書共二十問，書首為《上緝古算術表》。各問題的形式大致相同，每問以「假令」開頭，以「問：……各幾何？」或「問：……個多少？」結尾；隨後是答案：「答曰……」；最後一段是「術曰」，詳細敘述建立方程的理論依據和具體程序。每題都有答案，但關于解題方法，王孝通則言簡意駭。
第一問
「假令天正十一月朔夜半，日在鬥十度七百分度之四百八十。以章歲為母，朔月行定分九千，朔日定小餘一萬，日法二萬，章歲七百，亦名行分法。今不取加時日度。問：天正朔夜半之時月在何處？」。這是一道天文題，求半夜時月亮的赤道經度，王孝通用算術解題。
第二問
假令太史造仰觀臺，上廣袤少，下廣袤多。上下廣差二丈，上下袤差四丈，上廣袤差三丈，高多上廣一十一丈，甲縣差一千四百一十八人，乙縣差三千二百二十二人，夏程人功常積七十五尺，限五日役臺畢。羨道從臺南面起，上廣多下廣一丈二尺，少袤一百四尺，高多袤四丈。甲縣一十三鄉，乙縣四十三鄉，每鄉別均賦常積六千三百尺，限一日役羨道畢。二縣差到人共造仰觀臺，二縣鄉人共造羨道，皆從先給甲縣，以次與乙縣。臺自下基給高，道自初登給袤。問：臺道廣、高、袤及縣別給高、廣、袤各幾何？」。
對于這個建造觀象台和台道的廣度、高度、深度的計算，王孝通列出三個
x^3+px^2+qx=n 形式的三次方程式，和一個x^3+px^2=n形式的三次方程式 。
第三問
「假令築堤，西頭上、下廣差六丈八尺二寸，東頭上、下廣差六尺二寸。東頭高少於西頭高三丈一尺，上廣多東頭高四尺九寸，正袤多於東頭高四百七十六尺九寸。甲縣六千七百二十四人，乙縣一萬六千六百七十七人，丙縣一萬九千四百四十八人，丁縣一萬二千七百八十一人。四縣每人一日穿土九石九鬥二升。每人一日築常積一十一尺四寸十三分寸之六。穿方一尺得土八鬥。古人負土二鬥四升八合，平道行一百九十二步，一日六十二到。今隔山渡水取土，其平道只有一十一步，山斜高三十步，水寬一十二步，上山三當四，下山六當五，水行一當二，平道踟躕十加一，載輸一十四步。減計一人作功為均積。四縣共造，一日役華。今從東頭與甲，其次與乙、丙、丁。問：給斜、正袤與高，及下廣，並每人一日自穿、運、築程功，及堤上、下高、廣各幾何？
王孝通解此題，建立了一個二次方程，兩個三次方程：
第一個三次方程：
• x^3+\frac{3cd}{b-c}x^2+\frac{3(a+c)hd^2}{(H-h)(b-c)}x=\frac{6Vd^2}{(H-h)(b-c)}
第二個三次方程：
• x^3+5004x^2+1169953\frac{1}{3}x=41107188\frac{1}{3}；
王孝通求得其解：31
第四問
「假令築龍尾堤，其堤從頭高、上闊以次低狹至尾。上廣多，下廣少，堤頭上下廣差六尺，下廣少高一丈二尺，少袤四丈八尺。甲縣二千三百七十五人，乙縣二千三百七十八人，丙縣五千二百四十七人。各人程功常積一尺九寸八分，一日役畢，三縣共築。今從堤尾與甲縣，以次與乙、丙。問：龍尾堤從頭至尾高、袤、廣及各縣別給高、袤、廣各多少。」
王孝通 兩個三次方程。
• x^3+62x^2+696x=38448；解之得 x=18尺；
• x^3+594x^2=682803 解之得 x=33尺；
第五問
「假令穿河，袤一裏二百七十六步，下廣六步一尺二寸；北頭深一丈八尺六寸，上廣十二步二尺四寸；南頭深二百四十一尺八寸；上廣八十六步四尺八寸。運土於河西岸造漘，北頭高二百二十三尺二寸，南頭無高，下廣四百六尺七寸五厘，袤與河同。甲郡二萬二千三百二十人，乙郡六萬八千七十六人，丙郡五萬九千九百八十五人，丁郡三萬七千九百四十四人。自穿、負、築，各人程功常積三尺七寸二分。限九十六日役，河漘俱了。四郡分共造漘，其河自北頭先給甲郡，以次與乙，合均賦積尺。問：逐郡各給斜、正袤，上廣及深，並漘上廣各多少？」
解二次方程，三次方程個一。
三次方程：x^3+15x^2+66x-360 得 方倉上底邊 x=3尺，下底邊=9尺，高=12尺。
第六問
「假令四郡輸粟，斛法二尺五寸，一人作功為均。自上給甲，以次與乙。其甲郡輸粟三萬八千七百四十五石六鬥，乙郡輸粟三萬四千九百五石六鬥，丙郡輸粟，二萬六千二百七十石四鬥，丁郡輸粟一萬四千七十八石四鬥。四郡共穿窖，上袤多於上廣一丈，少於下袤三丈，多於深六丈，少於下廣一丈。各計粟多少，均出丁夫。自穿、負、築，冬程人功常積一十二尺，一日役。問：窖上下廣、袤、深，郡別出人及窖深、廣各多少？」
解兩個三次方程。
第七問
「假令亭倉上小下大，上下方差六尺，高多上方九尺，容粟一百八十七石二鬥。今已運出五十石四斗。問：倉上下方、高及余粟深、上方各多少？」
求穀倉上邊長，王孝通所述方法，相當于解一個三次方程：
x^3+(D+G)x^2+(DG +\frac{D^2}{3})x=P-\frac{D^2G}{3}
為求余粟深度，王孝通的辦法是建立又一個三次方程：
X^+3\frac{hs}{D}x^2+3(\frac{hs}{D})^2x=\frac{P'}{3}\frac{h^2}{D^2}
第八問
「假令芻甍上袤三丈，下袤九丈，廣六丈，高一十二丈。有甲縣六百三十二人，乙縣二百四十三人。夏程人功當積三十六尺，限八日役。自穿築，二縣共造。今甲縣先到。問：自下給高、廣、袤、各多少？」是關于建築觀象台、河堤、糧窖等工程中的土方問題。
解一個三次方程：
x^3+90*x^2-839808
得 乙縣工程 高 x=72尺; 甲縣工程高=120-72=48尺、上廣=36尺 、袤=66尺。
第九問
「假令圓囤上小下大，斛法二尺五寸，以率徑一周三。上下周差一丈二尺，高多上周一丈八尺，容粟七百五斛六鬥。今已運出二百六十六石四鬥。問：殘粟去口、上下周、高各多少？」
解兩個三次方程。
第十問
「假令有粟二萬三千一百二十斛七鬥三升，欲作方倉一，圓窖一，盛各滿中而粟適盡。令高、深等，使方面少於圓徑九寸，多於高二丈九尺八寸，率徑七，周二十二。問：方、徑、深多少？」
解一個三次方程。
第十一問
「假令有粟一萬六千三百四十八石八鬥，欲作方倉四、圓窖三，令高、深等，方面少於圓徑一丈，多於高五尺，斛法二尺五寸，率徑七，周二十二。問：方、高、徑多少？」
解一個三次方程。
第十二問
「假令有粟三千七十二石，欲作方倉一、圓窖一，令徑與方等，方於窖深二尺，少於倉高三尺，盛各滿中而粟適盡（圓率、斛法並與前同）。問：方、徑、高、深各多少？」
解一個三次方程。
第十三問
「假令有粟五千一百四十石，欲作方窖、圓窖各一，令口小底大，方面於圓徑等，兩深亦同，其深少於下方七尺，多於上方一丈四尺，盛各滿中而粟適盡（圓率、斛法並與前同）。問：方、徑、深各多少？」
解一個三次方程。
第十四問
「假令有粟二萬六千三百四十二石四鬥，欲作方窖六、圓窖四，令口小底大，方面與圓徑等，其深亦同，令深少於下方七尺，多於上方一丈四尺，盛各滿中而粟適盡（圓率、斛法並與前同）。問上下方、深數各多少？」
解一個三次方程。
第十五問
「假令有句股相乘冪七百六十五分之一，弦多於句三十六十分之九。問：三事各多少？」
解一個三次方程：
x^3 +\frac{S}{2}x^2-\frac{P^2}{2S}=0。
第十六問
「假令有股弦相乘冪四千七百三十九五分之三，句少於弦五十四五分之二。問：股多少？」
解一個三次方程。
第十七問
「假令有句弦相乘冪一千三百三十七二十分之一，弦多股一、十分之一。問：股多少？」
解一個三次方程。「答曰：九十二五分之二。」
「術曰：冪自乘，倍多而一，為立冪。又多再自乘，半之，減立冪，余為實。又多數自乘，倍之，為方法。又置多數，五之，二而一，為廉法，從。開立方除之，即股（句弦相乘冪自乘，即句冪乘弦冪之積。故以倍股弦差而一，得一股與半差為方，令多再自乘半之為隅，橫虛二立廉……倍之為從隅……多為上廣即二多……法故五之二而一）。」
王孝通所述，相當于建立一個三次方程：
::x^3 +\frac{5}{2}Dx^2+2D^2x=\frac{P^2}{2D}-\frac{D^2}{2}
第十八問
「假令有股弦相乘冪四千七百三十九五分之三，句少於弦五十四五分之二。問：股多少？」
解一個三次方程。
第十九問
「假令有股弦相乘冪七百二十六，句七、十分之七。問：股多少？」
解一個雙二次方程。
第二十問
「假令有股十六二分之一，句弦相乘冪一百六十四二十五分之十四。問：句多少？」
解一個雙二次方程：
x^4+(16\frac{1}{2})^2x^2=(164\frac{14}{15})^2。
版本
《緝古算經》在唐代就有抄本，宋元豐七年（1084年）有秘書監趙彥若等校定刊本，但到明代，刊本幾乎遺失，僅存章丘李開先所藏一部南宋刊本。清代毛晉獲得《緝古算經》，影抄傳世。《緝古算經》影抄本後歸常熟毛扆汲古閣收藏；清乾隆年間孔繼涵得毛扆汲古閣所藏宋元豐七年《緝古算經》影抄本和其他算書六種，連同戴東原從永樂大典中編輯出的《海島算經》等書合為十部，一同刻印刊行；孔繼涵所刻《緝古算經》，世稱為微波謝本。同時《四庫全書》又收入吏部侍郎王傑所藏《緝古算經》的毛晉影抄本。微波謝本後佚，影抄本現存北京故宮博物院。
清代中期，研究《緝古算經》之風盛行，先後有李潢《緝古算經考注》二卷，張敦仁《緝古算經細草》一卷，陳杰《緝古算經細草》一卷，《緝古算經注》二卷，《緝古算經音義》一卷，及按微波謝本抄錄的《緝古算經經文》一卷；揭廷鏘《緝古算經考注圖草》一卷。
1963年中華書局出版錢寶琮校點多《算經十書》，其中包括《緝古算經》
1998你 郭書春 校點 《緝古算經》 《算經十書》 卷2 遼寧教育出版社。
The text above has been excerpted automatically from Wikipedia - please correct any errors in the
original article.
